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\ihead{Errata to ``... Double Partially Ordered Sets''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{A Self-Dual Hopf Algebra on Double Partially Ordered Sets}

\textit{Claudia Malvenuto and Christophe Reutenauer}

\texttt{\href{https://arxiv.org/abs/0905.3508v1}{\texttt{0905.3508v1.pdf}}}

version of 21 May 2009 (\href{https://arxiv.org/abs/0905.3508v1}{arXiv
preprint arXiv:0905.3508v1})

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the article \textquotedblleft A
Self-Dual Hopf Algebra on Double Partially Ordered Sets\textquotedblright\ by
the numbers under which they appear in this article (specifically, in its
version of 21 May 2009, posted on the arXiv as arXiv preprint arXiv:0905.3508v1).

\setcounter{section}{4}

\section{Errata}

\begin{itemize}
\item \textbf{Various places (including Theorem 2.1):} You claim that
$\mathbb{Z}\mathbf{D}$ is a self-dual Hopf algebra. The notion of self-duality
that you are using, however, differs from the notions of self-duality commonly
used in literature. In particular, a standard definition of a self-dual Hopf
algebra $H$ requires it to have a bilinear form which provides a Hopf algebra
isomorphism $H\rightarrow H^{\ast}$. This is not satisfied for $\mathbb{Z}%
\mathbf{D}$: In fact, the bilinear form $\left\langle \cdot,\cdot\right\rangle
$ on $\mathbb{Z}\mathbf{D}$ gives rise to a Hopf algebra homomorphism
$\mathbb{Z}\mathbf{D}\rightarrow\left(  \mathbb{Z}\mathbf{D}\right)  ^{\ast}$
which is not bijective. (It is in fact injective, but you do not prove this;
instead, this is a consequence of \cite[Theorem 36 1.]{Foissy11}.)

\item \textbf{Page 2, \S 2:} Replace \textquotedblleft if $x\in J$ and
$x<y$\textquotedblright\ by \textquotedblleft if $x\in S$ and $x<y$%
\textquotedblright.

\item \textbf{Page 3:} Replace \textquotedblleft We define a pairing
$\left\langle ,\right\rangle :\mathbb{Z}\mathbf{D}\times\mathbb{Z}%
\mathbf{D}\rightarrow\mathbb{Z}$ for any double posets $E,F$ by:%
\[
\left\langle E,F\right\rangle :=\left\vert \left\{  \alpha:E\rightarrow
F,\ \alpha\text{ is a picture}\right\}  \right\vert
\]
\textquotedblright\ by \textquotedblleft We define a pairing $\left\langle
,\right\rangle :\mathbb{Z}\mathbf{D}\times\mathbb{Z}\mathbf{D}\rightarrow
\mathbb{Z}$ by:%
\[
\left\langle E,F\right\rangle :=\left\vert \left\{  \alpha:E\rightarrow
F,\ \alpha\text{ is a picture}\right\}  \right\vert
\ \ \ \ \ \ \ \ \ \ \text{for any double posets }E,F.
\]
\textquotedblright.

\item \textbf{Page 3:} Replace \textquotedblleft$\left\langle E\otimes
G,\delta G\right\rangle $\textquotedblright\ by \textquotedblleft$\left\langle
E\otimes F,\delta G\right\rangle $\textquotedblright.

\item \textbf{Page 3, proof of Theorem 2.1:} Replace \textquotedblleft$I\cap
F$ is an inferior ideal of $\left(  E,<_{1}\right)  $\textquotedblright\ by
\textquotedblleft$I\cap F$ is an inferior ideal of $\left(  F,<_{1}\right)
$\textquotedblright.

\item \textbf{Page 4, proof of Theorem 2.1:} In \textquotedblleft$I$ is an
inferior ideal of G\textquotedblright, replace the \textquotedblleft
G\textquotedblright\ by a \textquotedblleft$G$\textquotedblright\ (in mathmode).

\item \textbf{Page 4, proof of Theorem 2.1:} In \textquotedblleft take
$g,g^{\prime}$ in G\textquotedblright, replace the \textquotedblleft
G\textquotedblright\ by a \textquotedblleft$G$\textquotedblright\ (in mathmode).

\item \textbf{Page 4, proof of Theorem 2.1:} Replace \textquotedblleft
ordre\textquotedblright\ by \textquotedblleft order\textquotedblright.

\item \textbf{Page 4, proof of Theorem 2.1:} Replace \textquotedblleft a
bijection $\alpha:E\rightarrow G$\textquotedblright\ by \textquotedblleft a
bijection $\alpha:EF\rightarrow G$\textquotedblright.

\item \textbf{Page 4, proof of Theorem 2.1:} Replace \textquotedblleft doubles
posets\textquotedblright\ by \textquotedblleft double posets\textquotedblright.

\item \textbf{Page 4, \S 2.2:} Replace \textquotedblleft over all $\pi
$-partitions\textquotedblright\ by \textquotedblleft over all $\pi$-partitions
$x$ into $X$\textquotedblright.

\item \textbf{Page 4, \S 2.2:} Replace \textquotedblleft algebra of
quasi-symmetric function\textquotedblright\ by \textquotedblleft algebra of
quasi-symmetric functions\textquotedblright.

\item \textbf{Page 4, \S 2.2:} Replace \textquotedblleft commuting variables
Y\textquotedblright\ by \textquotedblleft commuting variables $Y$%
\textquotedblright\ (the \textquotedblleft$Y$\textquotedblright\ should be in mathmode).

\item \textbf{Page 5, \S 2.3:} Replace \textquotedblleft Let $\phi:\left(
E,<_{1}\right)  \rightarrow\left(  F,<_{2}\right)  $ be a
bijection\textquotedblright\ by \textquotedblleft Let $\phi:E\rightarrow F$ be
a bijection\textquotedblright. (You really mean a bijection between sets here,
and poset structures are not relevant to it.)

\item \textbf{Page 5, \S 2.3:} The internal product is non-unital (i.e., there
exists no element $e$ of $\mathbb{Z}\mathbf{D}$ such that every $b\in
\mathbb{Z}\mathbf{D}$ satisfies $e\circ b=b\circ e=e$). When I hear
\textquotedblleft product\textquotedblright, I normally tend to expect that it
has a unit, which is why I think this should be pointed out explicitly.

On the other hand, what is true is that the internal product is associative.
See Proposition \ref{prop.assoc}\ below for a proof.

\item \textbf{Page 5, Lemma 2.1:} Replace \textquotedblleft and
and\textquotedblright\ by \textquotedblleft and\textquotedblright.

\item \textbf{Page 5, Lemma 2.1:} Replace \textquotedblleft and $\beta$ a
picture\textquotedblright\ by \textquotedblleft and $\beta$ is a
picture\textquotedblright.

\item \textbf{Page 5, proof of Lemma 2.1:} Replace \textquotedblleft the
projection $F\times G\rightarrow F$\textquotedblright\ by \textquotedblleft
the projection $F\times_{\psi}G\rightarrow F$\textquotedblright.

\item \textbf{Page 5, proof of Lemma 2.1:} Replace \textquotedblleft the
projection $E\times F\rightarrow F$\textquotedblright\ by \textquotedblleft
the projection $E\times_{\psi}F\rightarrow F$\textquotedblright.

\item \textbf{Page 5, proof of Lemma 2.1:} In \textquotedblleft Now $\beta$
maps bijectively $E$ into $F\times_{\phi}G$\textquotedblright, replace
\textquotedblleft$\phi$\textquotedblright\ by \textquotedblleft$\psi
$\textquotedblright.

\item \textbf{Page 5, proof of Lemma 2.1:} Replace \textquotedblleft%
$\alpha^{-1}(\psi\left(  f\right)  =\left(  \phi^{-1}\left(  f\right)
,f\right)  $\textquotedblright\ by \textquotedblleft$\alpha^{-1}\left(
\psi\left(  f\right)  \right)  =\left(  \phi^{-1}\left(  f\right)  ,f\right)
$\textquotedblright.

\item \textbf{Page 5, proof of Lemma 2.1:} Replace \textquotedblleft%
$\beta\left(  e\right)  =\left(  \phi\left(  e\right)  ,\psi(\phi\left(
e\right)  \right)  $\textquotedblright\ by \textquotedblleft$\beta\left(
e\right)  =\left(  \phi\left(  e\right)  ,\psi\left(  \phi\left(  e\right)
\right)  \right)  $\textquotedblright.

\item \textbf{Page 6, proof of Lemma 2.1:} Replace \textquotedblleft%
$\beta^{\prime}=\beta^{\prime}$\textquotedblright\ by \textquotedblleft%
$\beta^{\prime}=\beta$\textquotedblright.

\item \textbf{Pages 6-7, \S 3.1:} Here you define a bialgebra structure on
$\mathbb{Z}S$ and claim that it is the one constructed in [MR]. It would be
helpful to make this somewhat more precise: In [MR] you have defined two
different (albeit isomorphic) bialgebra structures on $\mathbb{Z}S$, and the
one that you are talking about is the one that you have called $\left(
\mathbb{Z}S,\ast^{\prime},\Delta^{\prime}\right)  $ in [MR] (rather than the
one you have called $\left(  \mathbb{Z}S,\ast,\Delta\right)  $ or simply
$\mathbb{Z}S$ in [MR]).

\item \textbf{Page 7:} Replace \textquotedblleft each special double poset on
the sum\textquotedblright\ to \textquotedblleft each special double poset to
the sum\textquotedblright.

\item \textbf{Page 7:} Replace \textquotedblleft diagramm\textquotedblright%
\ by \textquotedblleft diagram\textquotedblright.

\item \textbf{Page 7, proof of Theorem 3.1:} \textquotedblleft a
lower\textquotedblright\ should be \textquotedblleft an
inferior\textquotedblright.

\item \textbf{Page 8, proof of Theorem 3.1:} Replace \textquotedblleft%
$st\left(  v\right)  =\sigma$\textquotedblright\ by \textquotedblleft%
$st\left(  v\right)  =\beta$\textquotedblright.

\item \textbf{Page 8, proof of Lemma 3.1:} The proof should start with the
sentence \textquotedblleft Write the special double posets $\pi$ and
$\pi^{\prime}$ in the forms $\pi=\left(  P,<_{1},<_{2}\right)  $ and
$\pi^{\prime}=\left(  P^{\prime},<_{1},<_{2}\right)  $,
respectively\textquotedblright, so that $P$ and $P^{\prime}$ are defined.

\item \textbf{Page 8, proof of Theorem 3.2:} Replace \textquotedblleft let
$\phi$ be increasing $\left(  E,<_{1}\right)  \rightarrow\left(
F,<_{2}\right)  =\left\{  1,\ldots,n\right\}  $\textquotedblright\ by
\textquotedblleft let $\phi$ be an increasing bijection $\left(
E,<_{1}\right)  \rightarrow\left(  F,<_{2}\right)  =\left\{  1,\ldots
,n\right\}  $\textquotedblright.

\item \textbf{Page 8, proof of Theorem 3.2:} Replace \textquotedblleft as in
Section 2.2\textquotedblright\ by \textquotedblleft as in Section
2.3\textquotedblright.

\item \textbf{Page 8, proof of Theorem 3.2:} Replace \textquotedblleft%
$\sigma-1\circ\phi^{-1}$\textquotedblright\ by \textquotedblleft$\sigma
^{-1}\circ\phi^{-1}$\textquotedblright.

\item \textbf{Page 8, proof of Theorem 3.2:} Replace \textquotedblleft%
$L\left(  \pi\right)  L\left(  \pi^{\prime}\right)  $\textquotedblright\ by
\textquotedblleft$L\left(  \pi\right)  \circ L\left(  \pi^{\prime}\right)
$\textquotedblright.

\item \textbf{Page 9, \S 3.1:} Replace \textquotedblleft$c_{1}...,c_{k}%
$\textquotedblright\ by \textquotedblleft$c_{1},...,c_{k}$\textquotedblright.

\item \textbf{Page 9, \S 3.1:} Replace \textquotedblleft Section
3\textquotedblright\ in \textquotedblleft Section 2.2\textquotedblright\ in
\textquotedblleft Recall that the bialgebra homomorphism $\Gamma
:\mathbb{Z}\mathbf{D}\rightarrow\mathbf{QSym}$ has been defined in Section
3\textquotedblright.

\item \textbf{Page 9, \S 3.2:} The formulation \textquotedblleft the word
obtained from $w$ by exchanging $1$ and $k$ in $w$ , then $2$ and $k-1$, and
so on\textquotedblright\ is ambiguous (the \textquotedblleft and so
on\textquotedblright\ could be misunderstood to end at \textquotedblleft$k$
and $1$\textquotedblright, which would entail that the complement of $w$ is
$w$ itself). I think a clearer definition of the complement would be: If
$w=w_{1}w_{2}\ldots w_{n}$ is a word (with $w_{1},w_{2},\ldots,w_{n}$ being
its letters) whose letters belong to $\left\{  1,2,3,\ldots\right\}  $, and
$k$ is the highest letter appearing in $w$, then the \textit{complement} of
$w$ is defined as the word $\left(  k+1-w_{1}\right)  \left(  k+1-w_{2}%
\right)  \ldots\left(  k+1-w_{n}\right)  $.

\item \textbf{Page 9, \S 3.2:} You define the weight of a word as follows:
\textquotedblleft The \textit{weight} of a word $w$ is the partition
$\nu=1^{n_{1}}2^{n_{2}}...$, where $n_{i}$ is the number of $i$'s in $w$. For
the word above, it is the partition $1^{4}2^{3}3^{1}$.\textquotedblright\ This
notion of a weight (while clearly well-defined) is very unlikely to be the one
that you want: indeed, it makes some of your statements below
(\textquotedblleft there is a well-known bijection between standard Young
tableaux of shape $\nu$ and lattice permutations of weight $\nu$%
\textquotedblright, and Theorem 3.4) false. Instead, I believe that you want
to define a different notion of weight, namely the following one:
\textquotedblleft The \textit{weight} of a word $w$ (on the symbols
$1,2,3,\ldots$) is defined as the sequence $\left(  n_{1},n_{2},n_{3}%
,\ldots\right)  $, where $n_{i}$ is the number of $i$'s in $w$. When $w$ is a
lattice permutation (but also in some other cases), this sequence is a
partition. For example, the weight of the word $11122132$ is $\left(
4,3,1,0,0,0,\ldots\right)  =\left(  4,3,1\right)  $.\textquotedblright

\item \textbf{Page 9, \S 3.2:} Replace \textquotedblleft$w$\textit{ fits into
}$\pi$\textquotedblright\ by \textquotedblleft$a_{1}a_{2}\ldots a_{n}$
\textit{fits into }$\pi$\textquotedblright.

\item \textbf{Page 9, \S 3.2:} Replace \textquotedblleft$f\left(
\omega\left(  1\right)  \right)  ...f\left(  \omega\left(  n\right)  \right)
$\textquotedblright\ by \textquotedblleft$f\left(  \omega^{-1}\left(
1\right)  \right)  ...f\left(  \omega^{-1}\left(  n\right)  \right)
$\textquotedblright.

\item \textbf{Page 9, \S 3.2:} Replace \textquotedblleft we have that $\tau$
fits into $\pi$\textquotedblright\ by \textquotedblleft we say that $\tau$
fits into $\pi$\textquotedblright.

\item \textbf{Page 9, \S 3.2:} Replace \textquotedblleft and where $<_{2}$ is
given on the elements of $E_{\nu}$ by $\left(  x,y\right)  <_{2}\left(
x^{\prime},y^{\prime}\right)  $ if and only if either $y>y^{\prime}$, or
$y=y^{\prime}$ and $x<x^{\prime}$\textquotedblright\ by \textquotedblleft and
where $<_{2}$ is given on the elements of $E_{\nu}$ by $\left(  x,y\right)
<_{2}\left(  x^{\prime},y^{\prime}\right)  $ if and only if either
$x>x^{\prime}$, or $x=x^{\prime}$ and $y<y^{\prime}$ (in other words, $<_{2}$
compares two cells of $E_{\nu}$ by decreasing row number, or, when the rows
are equal, by increasing column number)\textquotedblright. This definition is
needed to ensure that the reading word in the example on page 10 is the right one.

\item \textbf{Page 9, \S 3.2:} Let's be self-contained: After
\textquotedblleft Recall that there is a well-known bijection between standard
tableaux of shape $\nu$ and lattice permutations of weight $\nu$, see [S1]
Prop.7.10.3 (d).\textquotedblright, add: \textquotedblleft Explicitly, this
bijection sends a standard tableau $T$ with $n$ entries to the lattice
permutation $r_{1}r_{2}\ldots r_{n}$, where $r_{k}$ is the number of the row
in which the entry $k$ lies in $T$.\textquotedblright.

\item \textbf{Page 10, Theorem 3.4:} Replace \textquotedblleft$\left(  \pi
,\pi_{\nu}\right)  $\textquotedblright\ by \textquotedblleft$\left\langle
\pi,\pi_{\nu}\right\rangle $\textquotedblright.

\item \textbf{Page 10:} Replace \textquotedblleft\lbrack Ma]
(9.2)\textquotedblright\ by \textquotedblleft\lbrack Ma] (Chapter 1,
(9.2))\textquotedblright. In the same sentence, replace \textquotedblleft%
\lbrack S1] Th.A.1.3.3\textquotedblright\ by \textquotedblleft\lbrack S1]
Theorem A1.5.3\textquotedblright.

Another reference for the same result is \cite[Theorem 1.2]{Gashar98} for
$\theta=\varnothing$.

\item \textbf{Page 10:} You claim that part (ii) of Theorem 3.4
\textquotedblleft is the classical formulation of the Littlewood-Richardson
rule\textquotedblright. I find this misleading, since Theorem 3.4 comes
nowhere close to \textbf{proving} the Littlewood-Richardson rule (about the
Hall inner product of a skew Schur functions with a Schur function). What is
true is that Theorem 3.4 \textbf{looks like} the Littlewood-Richardson rule,
but it differs from it in that it gives an expression for a scalar product
$\left\langle \pi,\pi_{\nu}\right\rangle $ on $\mathbb{Z}\mathbf{D}$, while
the Littlewood-Richardson rule gives an expression for a Hall inner product
$\left\langle s_{\lambda/\mu},s_{\nu}\right\rangle $ on the ring of symmetric
functions. That the expressions are equal (when $\pi$ is a skew Ferrers
diagram $\pi_{\lambda/\mu}$) makes the two results similar, but I do not think
it allows to derive one from the other. Or am I missing something simple?

\item \textbf{Page 10, proof of Lemma 3.2:} Replace \textquotedblleft$\tau$
fits into $\pi$\textquotedblright\ by \textquotedblleft a permutation $\tau$
fits into $\pi$\textquotedblright.

\item \textbf{Page 10:} You write that Proposition 3.1 \textquotedblleft is
equivalent to a result of Stanley, see [G] Th.1 or [S1]
Th.7.19.14\textquotedblright. There is no Theorem 7.19.14 in [S1]; I suspect
that you mean Theorem 7.19.4 instead. (Besides, the equivalence of your
Proposition 3.1 to the result of Stanley that you mention is not completely
obvious. See below for a different proof of Proposition 3.1, which in my
opinion is simpler than deriving it from Stanley's result.)

\item \textbf{Page 10, example below Proposition 3.2:} \textquotedblleft whose
inverse is $5\ 3\ 9\ 2\ 6\ 10\ 11\ 1\ 4\ 6\ 7$\textquotedblright\ should be
\textquotedblleft whose inverse is $5\ 3\ 9\ 2\ 6\ 10\ 11\ 1\ 4\ 7\ 8$%
\textquotedblright\ (the last two letters were wrong).

\item \textbf{Page 11, Lemma 3.3:} Replace \textquotedblleft$\nu=\left(
\nu_{1}>...>\nu_{k}>0\right)  $\textquotedblright\ by \textquotedblleft%
$\nu=\left(  \nu_{1}\geq...\geq\nu_{k}>0\right)  $\textquotedblright.

\item \textbf{Page 11, proof of proposition:} \textquotedblleft Proof of
proposition\textquotedblright\ should be \textquotedblleft Proof of
proposition 3.2\textquotedblright\ (it is not the only proposition around).

\item \textbf{Page 11, proof of proposition:} Replace \textquotedblleft Lemma
5.2\textquotedblright\ by \textquotedblleft Lemma 3.3\textquotedblright\ twice.

\item \textbf{Page 11, proof of proposition:} Replace \textquotedblleft done
in(i)\textquotedblright\ by \textquotedblleft done in (i)\textquotedblright.

\item \textbf{Page 12, proof of theorem:} \textquotedblleft complements of
lattice permutation\textquotedblright\ $\rightarrow$ \textquotedblleft
complements of lattice permutations\textquotedblright.
\end{itemize}

\section{Additional details and proofs}

\begin{statement}
\textbf{Note:} The below addenda were written in approx. 2014, as an exercise
in understanding and formalizing some arguments in this paper and in the
theory of combinatorial Hopf algebras in general. I was neither trying to be
concise nor good at mathematical writing; instead I was attempting to prepare
the arguments for formal verification (a project that has yet to materialize).
--DG, 2024
\end{statement}

\subsection{Page 5, \S 2.3: associativity of the internal product}

Let me add an additional fact: The internal product $\circ$ is associative. In
other words:

\begin{proposition}
\label{prop.assoc}Any three elements $E$, $F$ and $G$ of $\mathbb{Z}%
\mathbf{D}$ satisfy $\left(  E\circ F\right)  \circ G=E\circ\left(  F\circ
G\right)  $.
\end{proposition}

Before we prove this, let us first establish some notation.

\begin{definition}
\label{def.newbin}Let $A$ and $B$ be two sets. Let $<_{B}$ denote the smaller
relation of a partial order on $B$. (We write this relation $<_{B}$ in infix
notation; this means that, if $b$ and $b^{\prime}$ are two elements of $B$,
then we write $b<_{B}b^{\prime}$ for $\left(  b,b^{\prime}\right)  \in\left(
<_{B}\right)  $.) Let $f:A\rightarrow B$ be a map.

We define a new binary relation $\left(  <_{B}\right)  ^{f}$ on the set $A$ as
follows: For any two elements $a$ and $a^{\prime}$ of $A$, we set $a\left(
<_{B}\right)  ^{f}a^{\prime}$ if and only if $f\left(  a\right)  <_{B}f\left(
a^{\prime}\right)  $. (Here, we again write the relation $\left(
<_{B}\right)  ^{f}$ in infix notation.)
\end{definition}

\begin{proposition}
\label{prop.assoc.AB}Let $A$ and $B$ be two sets. Let $<_{B}$ denote the
smaller relation of a partial order on $B$. Let $f:A\rightarrow B$ be a map.
Then, the binary relation $\left(  <_{B}\right)  ^{f}$ is the smaller relation
of a partial order on $A$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.assoc.AB}.]Let us write both relations $<_{B}$
and $\left(  <_{B}\right)  ^{f}$ in infix notation.

We know that the relation $<_{B}$ is the smaller relation of a partial order.
In other words, the relation $<_{B}$ is irreflexive, transitive and asymmetric.

We now notice that the relation $\left(  <_{B}\right)  ^{f}$ is
irreflexive\footnote{\textit{Proof.} Let $a\in A$ be such that $a\left(
<_{B}\right)  ^{f}a$. We recall that $a\left(  <_{B}\right)  ^{f}a$ holds if
and only if $f\left(  a\right)  <_{B}f\left(  a\right)  $ (according to the
definition of \textquotedblleft$a\left(  <_{B}\right)  ^{f}a$%
\textquotedblright). Hence, $f\left(  a\right)  <_{B}f\left(  a\right)  $ must
hold (because $a\left(  <_{B}\right)  ^{f}a$ holds). But this contradicts the
fact that the relation $<_{B}$ is irreflexive. We thus have found a
contradiction.
\par
Now, let us forget that we fixed $a$. We thus have found a contradiction for
each $a\in A$ satisfying $a\left(  <_{B}\right)  ^{f}a$. Therefore, there
exists no $a\in A$ satisfying $a\left(  <_{B}\right)  ^{f}a$. In other words,
the relation $\left(  <_{B}\right)  ^{f}$ is irreflexive. Qed.},
transitive\footnote{\textit{Proof.} Let $a$, $a^{\prime}$ and $a^{\prime
\prime}$ be three elements of $A$ such that $a\left(  <_{B}\right)
^{f}a^{\prime}$ and $a^{\prime}\left(  <_{B}\right)  ^{f}a^{\prime\prime}$. We
shall prove that $a\left(  <_{B}\right)  ^{f}a^{\prime\prime}$.
\par
We recall that $a\left(  <_{B}\right)  ^{f}a^{\prime}$ holds if and only if
$f\left(  a\right)  <_{B}f\left(  a^{\prime}\right)  $ (according to the
definition of \textquotedblleft$a\left(  <_{B}\right)  ^{f}a^{\prime}%
$\textquotedblright). Thus, $f\left(  a\right)  <_{B}f\left(  a^{\prime
}\right)  $ must hold (since $a\left(  <_{B}\right)  ^{f}a^{\prime}$ holds).
\par
We recall that $a^{\prime}\left(  <_{B}\right)  ^{f}a^{\prime\prime}$ holds if
and only if $f\left(  a^{\prime}\right)  <_{B}f\left(  a^{\prime\prime
}\right)  $ (according to the definition of \textquotedblleft$a^{\prime
}\left(  <_{B}\right)  ^{f}a^{\prime\prime}$\textquotedblright). Thus,
$f\left(  a^{\prime}\right)  <_{B}f\left(  a^{\prime\prime}\right)  $ must
hold (since $a^{\prime}\left(  <_{B}\right)  ^{f}a^{\prime\prime}$ holds).
\par
From $f\left(  a\right)  <_{B}f\left(  a^{\prime}\right)  $ and $f\left(
a^{\prime}\right)  <_{B}f\left(  a^{\prime\prime}\right)  $, we obtain
$f\left(  a\right)  <_{B}f\left(  a^{\prime\prime}\right)  $ (since the
relation $<_{B}$ is transitive).
\par
We recall that $a\left(  <_{B}\right)  ^{f}a^{\prime\prime}$ holds if and only
if $f\left(  a\right)  <_{B}f\left(  a^{\prime\prime}\right)  $ (according to
the definition of \textquotedblleft$a\left(  <_{B}\right)  ^{f}a^{\prime
\prime}$\textquotedblright). Thus, $a\left(  <_{B}\right)  ^{f}a^{\prime
\prime}$ must hold (since $f\left(  a\right)  <_{B}f\left(  a^{\prime\prime
}\right)  $ holds).
\par
Now, let us forget that we fixed $a$, $a^{\prime}$ and $a^{\prime\prime}$. We
thus have shown that if $a$, $a^{\prime}$ and $a^{\prime\prime}$ are three
elements of $A$ such that $a\left(  <_{B}\right)  ^{f}a^{\prime}$ and
$a^{\prime}\left(  <_{B}\right)  ^{f}a^{\prime\prime}$, then $a\left(
<_{B}\right)  ^{f}a^{\prime\prime}$. In other words, the relation $\left(
<_{B}\right)  ^{f}$ is transitive. Qed.} and
asymmetric\footnote{\textit{Proof.} It is known that every irreflexive and
transitive relation is asymmetric. Applying this to the relation $\left(
<_{B}\right)  ^{f}$, we conclude that the relation $\left(  <_{B}\right)
^{f}$ is asymmetric (since $\left(  <_{B}\right)  ^{f}$ is irreflexive and
transitive).}. In other words, the relation $\left(  <_{B}\right)  ^{f}$ is
the smaller relation of a partial order on $A$ (because the smaller relations
of partial orders are characterized by being irreflexive, transitive and
asymmetric). This proves Proposition \ref{prop.assoc.AB}.
\end{proof}

The following proposition, while not needed in the proof of Proposition
\ref{prop.assoc}, will be useful much later:

\begin{proposition}
\label{prop.assoc.AB.total}Let $A$ and $B$ be two sets. Let $<_{B}$ denote the
smaller relation of a total order on $B$. Let $f:A\rightarrow B$ be an
injective map. Then, the binary relation $\left(  <_{B}\right)  ^{f}$ is the
smaller relation of a total order on $A$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.assoc.AB.total}.]Let us write both relations
$<_{B}$ and $\left(  <_{B}\right)  ^{f}$ in infix notation.

The relation $<_{B}$ is the smaller relation of a total order on $B$, and
therefore also the smaller relation of a partial order on $B$. Hence,
Proposition \ref{prop.assoc.AB} shows that the binary relation $\left(
<_{B}\right)  ^{f}$ is the smaller relation of a partial order on $A$. It now
remains to prove that this partial order is total.

Every two distinct elements $a$ and $a^{\prime}$ of $A$ satisfy either
$a\left(  <_{B}\right)  ^{f}a^{\prime}$ or $a^{\prime}\left(  <_{B}\right)
^{f}a$\ \ \ \ \footnote{\textit{Proof.} Let $a$ and $a^{\prime}$ be two
distinct elements of $A$. Then, $a\neq a^{\prime}$ (since $a$ and $a^{\prime}$
are distinct). If we had $f\left(  a\right)  =f\left(  a^{\prime}\right)  $,
then we would have $a=a^{\prime}$ (since the map $f$ is injective), which
would contradict $a\neq a^{\prime}$. Hence, we cannot have $f\left(  a\right)
=f\left(  a^{\prime}\right)  $. Thus, we have $f\left(  a\right)  \neq
f\left(  a^{\prime}\right)  $. In other words, the elements $f\left(
a\right)  $ and $f\left(  a^{\prime}\right)  $ of $B$ are distinct.
\par
But we know that $<_{B}$ is the smaller relation of a total order on $B$.
Thus, if $b$ and $b^{\prime}$ are two distinct elements of $B$, then either
$b<_{B}b^{\prime}$ or $b^{\prime}<_{B}b$. Applying this to $b=f\left(
a\right)  $ and $b^{\prime}=f\left(  a^{\prime}\right)  $, we conclude that
either $f\left(  a\right)  <_{B}f\left(  a^{\prime}\right)  $ or $f\left(
a^{\prime}\right)  <_{B}f\left(  a\right)  $. We thus are in one of the
following two cases:
\par
\textit{Case 1:} We have $f\left(  a\right)  <_{B}f\left(  a^{\prime}\right)
$.
\par
\textit{Case 2:} We have $f\left(  a^{\prime}\right)  <_{B}f\left(  a\right)
$.
\par
Let us first consider Case 1. In this case, we have $f\left(  a\right)
<_{B}f\left(  a^{\prime}\right)  $. Thus, $a\left(  <_{B}\right)
^{f}a^{\prime}$ (because we have $a\left(  <_{B}\right)  ^{f}a^{\prime}$ if
and only if $f\left(  a\right)  <_{B}f\left(  a^{\prime}\right)  $ (according
to the definition of \textquotedblleft$a\left(  <_{B}\right)  ^{f}a^{\prime}%
$\textquotedblright)). Hence, either $a\left(  <_{B}\right)  ^{f}a^{\prime}$
or $a^{\prime}\left(  <_{B}\right)  ^{f}a$. We thus have shown that $\left(
\text{either }a\left(  <_{B}\right)  ^{f}a^{\prime}\text{ or }a^{\prime
}\left(  <_{B}\right)  ^{f}a\right)  $ in Case 1.
\par
Let us now consider Case 2. In this case, we have $f\left(  a^{\prime}\right)
<_{B}f\left(  a\right)  $. Thus, $a^{\prime}\left(  <_{B}\right)  ^{f}a$
(because we have $a^{\prime}\left(  <_{B}\right)  ^{f}a$ if and only if
$f\left(  a^{\prime}\right)  <_{B}f\left(  a\right)  $ (according to the
definition of \textquotedblleft$a^{\prime}\left(  <_{B}\right)  ^{f}%
a$\textquotedblright)). Hence, either $a\left(  <_{B}\right)  ^{f}a^{\prime}$
or $a^{\prime}\left(  <_{B}\right)  ^{f}a$. We thus have shown that $\left(
\text{either }a\left(  <_{B}\right)  ^{f}a^{\prime}\text{ or }a^{\prime
}\left(  <_{B}\right)  ^{f}a\right)  $ in Case 2.
\par
Now, we have proven $\left(  \text{either }a\left(  <_{B}\right)
^{f}a^{\prime}\text{ or }a^{\prime}\left(  <_{B}\right)  ^{f}a\right)  $ in
each of the two Cases 1 and 2. Since these two Cases cover all possibilities,
this yields that $\left(  \text{either }a\left(  <_{B}\right)  ^{f}a^{\prime
}\text{ or }a^{\prime}\left(  <_{B}\right)  ^{f}a\right)  $ always holds.
Qed.}. Thus, the binary relation $\left(  <_{B}\right)  ^{f}$ is the smaller
relation of a total order on $A$ (since we already know that the binary
relation $\left(  <_{B}\right)  ^{f}$ is the smaller relation of a partial
order on $A$). This proves Proposition \ref{prop.assoc.AB.total}.
\end{proof}

We now show the main tool with which we will work in our proof of Proposition
\ref{prop.assoc}:

\begin{lemma}
\label{lem.assoc.fiber}Let $\left(  E,<_{E1},<_{E2}\right)  $ and $\left(
F,<_{F1},<_{F2}\right)  $ be two double posets.

\textbf{(a)} For every bijection $\phi:E\rightarrow F$, we have%
\[
\left(  E,<_{E1},<_{E2}\right)  \times_{\phi}\left(  F,<_{F1},<_{F2}\right)
\cong\left(  F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)
\ \ \ \ \ \ \ \ \ \ \text{as double posets.}%
\]


\textbf{(b)} We have%
\[
\left(  E,<_{E1},<_{E2}\right)  \circ\left(  F,<_{F1},<_{F2}\right)
=\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\left(  F,<_{F1}%
,\left(  <_{E2}\right)  ^{\phi^{-1}}\right)
\]
in $\mathbb{Z}\mathbf{D}$.

\textbf{(c)} For every bijection $\phi:E\rightarrow F$, we have%
\[
\left(  E,<_{E1},<_{E2}\right)  \times_{\phi}\left(  F,<_{F1},<_{F2}\right)
\cong\left(  E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)
\ \ \ \ \ \ \ \ \ \ \text{as double posets.}%
\]


\textbf{(d)} We have%
\[
\left(  E,<_{E1},<_{E2}\right)  \circ\left(  F,<_{F1},<_{F2}\right)
=\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\left(  E,\left(
<_{F1}\right)  ^{\phi},<_{E2}\right)
\]
in $\mathbb{Z}\mathbf{D}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.assoc.fiber}.]We shall use the abbreviation $E$ for
$\left(  E,<_{E1},<_{E2}\right)  $, and we shall use the abbreviation $F$ for
$\left(  F,<_{F1},<_{F2}\right)  $.

The definition of the set $E\times_{\phi}F$ yields%
\begin{align}
E\times_{\phi}F  &  =\left\{  \left(  e,f\right)  \in E\times F\ \mid
\ \phi\left(  e\right)  =f\right\} \nonumber\\
&  =\left\{  \left(  u,v\right)  \in E\times F\ \mid\ \phi\left(  u\right)
=v\right\}  \label{pf.lem.assoc.fiber.set}%
\end{align}
(here, we renamed the indices $e$ and $f$ as $u$ and $v$). Clearly, this shows
that $E\times_{\phi}F\subseteq E\times F$. The definition of the double poset
$E\times_{\phi}F$ shows that $E\times_{\phi}F=\left(  E\times_{\phi}%
F,<_{1},<_{2}\right)  $, where any two elements $\left(  e,f\right)  \in
E\times_{\phi}F$ and $\left(  e^{\prime},f^{\prime}\right)  \in E\times_{\phi
}F$ satisfy%
\begin{equation}
\left(  \left(  e,f\right)  <_{1}\left(  e^{\prime},f^{\prime}\right)  \text{
if and only if }f<_{F1}f^{\prime}\right)  \label{pf.lem.assoc.fiber.less1}%
\end{equation}
and%
\begin{equation}
\left(  \left(  e,f\right)  <_{2}\left(  e^{\prime},f^{\prime}\right)  \text{
if and only if }e<_{E2}e^{\prime}\right)  . \label{pf.lem.assoc.fiber.less2}%
\end{equation}


\textbf{(a)} Let $\phi$ be a bijection $E\rightarrow F$. Hence, $\phi^{-1}$ is
a bijection $F\rightarrow E$. Thus, Proposition \ref{prop.assoc.AB} (applied
to $F$, $E$, $<_{E2}$ and $\phi^{-1}$ instead of $A$, $B$, $<_{B}$ and $f$)
shows that the binary relation $\left(  <_{E2}\right)  ^{\phi^{-1}}$ is the
smaller relation of a partial order on $F$. Hence, $\left(  F,<_{F1},\left(
<_{E2}\right)  ^{\phi^{-1}}\right)  $ is a well-defined double poset.

Define a map $p_{2}:E\times_{\phi}F\rightarrow F$ by $\left(  p_{2}\left(
e,f\right)  =f\text{ for every }\left(  e,f\right)  \in E\times_{\phi
}F\right)  $. Thus, $p_{2}$ is the projection on the second component. On the
other hand, consider the map $\left(  \phi^{-1},\operatorname*{id}\right)
:F\rightarrow E\times_{\phi}F$ which sends every $f\in F$ to $\left(
\phi^{-1}\left(  f\right)  ,\underbrace{\operatorname*{id}\left(  f\right)
}_{=f}\right)  =\left(  \phi^{-1}\left(  f\right)  ,f\right)  \in
E\times_{\phi}F$\ \ \ $\ $\footnote{Let us check that this map $\left(
\phi^{-1},\operatorname*{id}\right)  $ is well-defined:
\par
Fix $f\in F$. Then, $\left(  \phi^{-1}\left(  f\right)  ,f\right)  $ is an
element of $E\times F$ satisfying $\phi\left(  \phi^{-1}\left(  f\right)
\right)  =f$. In other words, $\left(  \phi^{-1}\left(  f\right)  ,f\right)  $
is an element $\left(  u,v\right)  $ of $E\times F$ satisfying $\phi\left(
u\right)  =v$. In other words,%
\[
\left(  \phi^{-1}\left(  f\right)  ,f\right)  \in\left\{  \left(  u,v\right)
\in E\times F\ \mid\ \phi\left(  u\right)  =v\right\}  =E\times_{\phi
}F\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.assoc.fiber.set})}\right)
.
\]
\par
Now, let us forget that we fixed $f$. We thus have shown that $\left(
\phi^{-1}\left(  f\right)  ,f\right)  \in E\times_{\phi}F$ for every $f\in F$.
Thus, the map $\left(  \phi^{-1},\operatorname*{id}\right)  $ is well-defined,
qed.}. The maps $p_{2}$ and $\left(  \phi^{-1},\operatorname*{id}\right)  $
are mutually inverse\footnote{\textit{Proof.} Every $f\in F$ satisfies%
\begin{align*}
\left(  p_{2}\circ\left(  \phi^{-1},\operatorname*{id}\right)  \right)
\left(  f\right)   &  =p_{2}\left(  \underbrace{\left(  \phi^{-1}%
,\operatorname*{id}\right)  \left(  f\right)  }_{\substack{=\left(  \phi
^{-1}\left(  f\right)  ,\operatorname*{id}\left(  f\right)  \right)
\\\text{(by the definition of }\left(  \phi^{-1},\operatorname*{id}\right)
\text{)}}}\right)  =p_{2}\left(  \phi^{-1}\left(  f\right)
,\operatorname*{id}\left(  f\right)  \right)  =\operatorname*{id}\left(
f\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }p_{2}\right)  .
\end{align*}
In other words, $p_{2}\circ\left(  \phi^{-1},\operatorname*{id}\right)
=\operatorname*{id}$.
\par
On the other hand, let $\left(  e,f\right)  \in E\times_{\phi}F$. Then,
$\left(  e,f\right)  \in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in
E\times F\ \mid\ \phi\left(  u\right)  =v\right\}  $ (according to
(\ref{pf.lem.assoc.fiber.set})). In other words, $\left(  e,f\right)  $ is an
element $\left(  u,v\right)  $ of $E\times F$ satisfying $\phi\left(
u\right)  =v$. In other words, $\left(  e,f\right)  $ is an element of
$E\times F$ and satisfies $\phi\left(  e\right)  =f$. Now, $p_{2}\left(
e,f\right)  =f$ (by the definition of $p_{2}$) and $\phi^{-1}\left(  f\right)
=e$ (since $\phi\left(  e\right)  =f$). Now,%
\begin{align*}
\left(  \left(  \phi^{-1},\operatorname*{id}\right)  \circ p_{2}\right)
\left(  e,f\right)   &  =\left(  \phi^{-1},\operatorname*{id}\right)  \left(
\underbrace{p_{2}\left(  e,f\right)  }_{=f}\right)  =\left(  \phi
^{-1},\operatorname*{id}\right)  \left(  f\right) \\
&  =\left(  \underbrace{\phi^{-1}\left(  f\right)  }_{=e}%
,\underbrace{\operatorname*{id}\left(  f\right)  }_{=f}\right)  =\left(
e,f\right)  =\operatorname*{id}\left(  e,f\right)  .
\end{align*}
\par
Let us now forget that we fixed $\left(  e,f\right)  $. We thus have shown
that $\left(  \left(  \phi^{-1},\operatorname*{id}\right)  \circ p_{2}\right)
\left(  e,f\right)  =\operatorname*{id}\left(  e,f\right)  $ for every
$\left(  e,f\right)  \in E\times_{\phi}F$. In other words, $\left(  \phi
^{-1},\operatorname*{id}\right)  \circ p_{2}=\operatorname*{id}$. Combined
with $p_{2}\circ\left(  \phi^{-1},\operatorname*{id}\right)
=\operatorname*{id}$, this yields that the maps $p_{2}$ and $\left(  \phi
^{-1},\operatorname*{id}\right)  $ are mutually inverse. Qed.}. Hence, these
two maps $p_{2}$ and $\left(  \phi^{-1},\operatorname*{id}\right)  $ are
invertible and satisfy $p_{2}^{-1}=\left(  \phi^{-1},\operatorname*{id}%
\right)  $.

The map $p_{2}$ is a poset homomorphism $\left(  E\times_{\phi}F,<_{1}\right)
\rightarrow\left(  F,<_{F1}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $g$
and $g^{\prime}$ be two elements of $E\times_{\phi}F$ satisfying
$g<_{1}g^{\prime}$. We shall show that $p_{2}\left(  g\right)  <_{F1}%
p_{2}\left(  g^{\prime}\right)  $.
\par
Indeed, $g\in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in E\times
F\ \mid\ \phi\left(  u\right)  =v\right\}  $. In other words, $g$ has the form
$\left(  u,v\right)  $ for some $\left(  u,v\right)  \in E\times F$ satisfying
$\phi\left(  u\right)  =v$. Let us denote this $\left(  u,v\right)  $ by
$\left(  e,f\right)  $. Thus, $\left(  e,f\right)  $ is an element of $E\times
F$ and satisfies $\phi\left(  e\right)  =f$ and $g=\left(  e,f\right)  $.
Applying the map $p_{2}$ to both sides of the equality $g=\left(  e,f\right)
$, we obtain $p_{2}\left(  g\right)  =p_{2}\left(  e,f\right)  =f$ (by the
definition of $p_{2}$).
\par
Also, $g^{\prime}\in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in E\times
F\ \mid\ \phi\left(  u\right)  =v\right\}  $. In other words, $g^{\prime}$ has
the form $\left(  u,v\right)  $ for some $\left(  u,v\right)  \in E\times F$
satisfying $\phi\left(  u\right)  =v$. Let us denote this $\left(  u,v\right)
$ by $\left(  e^{\prime},f^{\prime}\right)  $. Thus, $\left(  e^{\prime
},f^{\prime}\right)  $ is an element of $E\times F$ and satisfies $\phi\left(
e^{\prime}\right)  =f^{\prime}$ and $g^{\prime}=\left(  e^{\prime},f^{\prime
}\right)  $. Applying the map $p_{2}$ to both sides of the equality
$g^{\prime}=\left(  e^{\prime},f^{\prime}\right)  $, we obtain $p_{2}\left(
g^{\prime}\right)  =p_{2}\left(  e^{\prime},f^{\prime}\right)  =f^{\prime}$
(by the definition of $p_{2}$).
\par
We have $\left(  e,f\right)  =g<_{1}g^{\prime}=\left(  e^{\prime},f^{\prime
}\right)  $. Due to (\ref{pf.lem.assoc.fiber.less1}), this yields
$f<_{F1}f^{\prime}$. In other words, $p_{2}\left(  g\right)  <_{F1}%
p_{2}\left(  g^{\prime}\right)  $ (since $p_{2}\left(  g\right)  =f$ and
$p_{2}\left(  g^{\prime}\right)  =f^{\prime}$).
\par
Now, let us forget that we fixed $g$ and $g^{\prime}$. We thus have proven
that if $g$ and $g^{\prime}$ are two elements of $E\times_{\phi}F$ satisfying
$g<_{1}g^{\prime}$, then $p_{2}\left(  g\right)  <_{F1}p_{2}\left(  g^{\prime
}\right)  $. In other words, the map $p_{2}$ is a strictly order-preserving
map $\left(  E\times_{\phi}F,<_{1}\right)  \rightarrow\left(  F,<_{F1}\right)
$. As a consequence, this map $p_{2}$ is a poset homomorphism $\left(
E\times_{\phi}F,<_{1}\right)  \rightarrow\left(  F,<_{F1}\right)  $ (since any
strictly order-preserving map between two posets is a poset homomorphism).
Qed.} and a poset homomorphism $\left(  E\times_{\phi}F,<_{2}\right)
\rightarrow\left(  F,\left(  <_{E2}\right)  ^{\phi^{-1}}\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $g$ and $g^{\prime}$ be two elements of
$E\times_{\phi}F$ satisfying $g<_{2}g^{\prime}$. We shall show that
$p_{2}\left(  g\right)  \left(  <_{E2}\right)  ^{\phi^{-1}}p_{2}\left(
g^{\prime}\right)  $.
\par
Indeed, $g\in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in E\times
F\ \mid\ \phi\left(  u\right)  =v\right\}  $. In other words, $g$ has the form
$\left(  u,v\right)  $ for some $\left(  u,v\right)  \in E\times F$ satisfying
$\phi\left(  u\right)  =v$. Let us denote this $\left(  u,v\right)  $ by
$\left(  e,f\right)  $. Thus, $\left(  e,f\right)  $ is an element of $E\times
F$ and satisfies $\phi\left(  e\right)  =f$ and $g=\left(  e,f\right)  $.
Applying the map $p_{2}$ to both sides of the equality $g=\left(  e,f\right)
$, we obtain $p_{2}\left(  g\right)  =p_{2}\left(  e,f\right)  =f$ (by the
definition of $p_{2}$).
\par
Also, $g^{\prime}\in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in E\times
F\ \mid\ \phi\left(  u\right)  =v\right\}  $. In other words, $g^{\prime}$ has
the form $\left(  u,v\right)  $ for some $\left(  u,v\right)  \in E\times F$
satisfying $\phi\left(  u\right)  =v$. Let us denote this $\left(  u,v\right)
$ by $\left(  e^{\prime},f^{\prime}\right)  $. Thus, $\left(  e^{\prime
},f^{\prime}\right)  $ is an element of $E\times F$ and satisfies $\phi\left(
e^{\prime}\right)  =f^{\prime}$ and $g^{\prime}=\left(  e^{\prime},f^{\prime
}\right)  $. Applying the map $p_{2}$ to both sides of the equality
$g^{\prime}=\left(  e^{\prime},f^{\prime}\right)  $, we obtain $p_{2}\left(
g^{\prime}\right)  =p_{2}\left(  e^{\prime},f^{\prime}\right)  =f^{\prime}$
(by the definition of $p_{2}$).
\par
We have $\left(  e,f\right)  =g<_{2}g^{\prime}=\left(  e^{\prime},f^{\prime
}\right)  $. Due to (\ref{pf.lem.assoc.fiber.less2}), this yields
$e<_{E2}e^{\prime}$. Since $\phi\left(  e\right)  =f$, we have $e=\phi
^{-1}\left(  f\right)  $, so that $\phi^{-1}\left(  f\right)  =e<_{E2}%
e^{\prime}=\phi^{-1}\left(  f^{\prime}\right)  $ (since $\phi\left(
e^{\prime}\right)  =f^{\prime}$).
\par
But $f\left(  <_{E2}\right)  ^{\phi^{-1}}f^{\prime}$ holds if and only if
$\phi^{-1}\left(  f\right)  <_{E2}\phi^{-1}\left(  f^{\prime}\right)  $ (due
to the definition of \textquotedblleft$f\left(  <_{E2}\right)  ^{\phi^{-1}%
}f^{\prime}$\textquotedblright). Thus, $f\left(  <_{E2}\right)  ^{\phi^{-1}%
}f^{\prime}$ holds (since $\phi^{-1}\left(  f\right)  <_{E2}\phi^{-1}\left(
f^{\prime}\right)  $ holds). In other words, $p_{2}\left(  g\right)  \left(
<_{E2}\right)  ^{\phi^{-1}}p_{2}\left(  g^{\prime}\right)  $ holds (since
$p_{2}\left(  g\right)  =f$ and $p_{2}\left(  g^{\prime}\right)  =f^{\prime}%
$).
\par
Now, let us forget that we fixed $g$ and $g^{\prime}$. We thus have proven
that if $g$ and $g^{\prime}$ are two elements of $E\times_{\phi}F$ satisfying
$g<_{2}g^{\prime}$, then $p_{2}\left(  g\right)  \left(  <_{E2}\right)
^{\phi^{-1}}p_{2}\left(  g^{\prime}\right)  $. In other words, the map $p_{2}$
is a strictly order-preserving map $\left(  E\times_{\phi}F,<_{2}\right)
\rightarrow\left(  F,\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  $. As a
consequence, this map $p_{2}$ is a poset homomorphism $\left(  E\times_{\phi
}F,<_{2}\right)  \rightarrow\left(  F,\left(  <_{E2}\right)  ^{\phi^{-1}%
}\right)  $ (since any strictly order-preserving map between two posets is a
poset homomorphism). Qed.}. Hence, the map $p_{2}$ is a homomorphism of double
posets $\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \rightarrow\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  $.

On the other hand, the map $p_{2}^{-1}$ is a poset homomorphism $\left(
F,<_{F1}\right)  \rightarrow\left(  E\times_{\phi}F,<_{1}\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $f$ and $f^{\prime}$ be two elements of
$F$ satisfying $f<_{F1}f^{\prime}$. We shall prove that $p_{2}^{-1}\left(
f\right)  <_{1}p_{2}^{-1}\left(  f^{\prime}\right)  $.
\par
We have $\underbrace{p_{2}^{-1}}_{=\left(  \phi^{-1},\operatorname*{id}%
\right)  }\left(  f\right)  =\left(  \phi^{-1},\operatorname*{id}\right)
\left(  f\right)  =\left(  \phi^{-1}\left(  f\right)
,\underbrace{\operatorname*{id}\left(  f\right)  }_{=f}\right)  =\left(
\phi^{-1}\left(  f\right)  ,f\right)  $, so that $\left(  \phi^{-1}\left(
f\right)  ,f\right)  =p_{2}^{-1}\left(  f\right)  \in E\times_{\phi}F$.
\par
Also, $\underbrace{p_{2}^{-1}}_{=\left(  \phi^{-1},\operatorname*{id}\right)
}\left(  f^{\prime}\right)  =\left(  \phi^{-1},\operatorname*{id}\right)
\left(  f^{\prime}\right)  =\left(  \phi^{-1}\left(  f^{\prime}\right)
,\underbrace{\operatorname*{id}\left(  f^{\prime}\right)  }_{=f^{\prime}%
}\right)  =\left(  \phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime}\right)  $,
so that $\left(  \phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime}\right)
=p_{2}^{-1}\left(  f^{\prime}\right)  \in E\times_{\phi}F$.
\par
Now, (\ref{pf.lem.assoc.fiber.less1}) (applied to $\phi^{-1}\left(  f\right)
$ and $\phi^{-1}\left(  f^{\prime}\right)  $ instead of $e$ and $e^{\prime}$)
yields $\left(  \left(  \phi^{-1}\left(  f\right)  ,f\right)  <_{1}\left(
\phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime}\right)  \text{ if and only if
}f<_{F1}f^{\prime}\right)  $ (since $\left(  \phi^{-1}\left(  f\right)
,f\right)  \in E\times_{\phi}F$ and $\left(  \phi^{-1}\left(  f^{\prime
}\right)  ,f^{\prime}\right)  \in E\times_{\phi}F$). Thus, we have $\left(
\phi^{-1}\left(  f\right)  ,f\right)  <_{1}\left(  \phi^{-1}\left(  f^{\prime
}\right)  ,f^{\prime}\right)  $ (since $f<_{F1}f^{\prime}$). In other words,
$p_{2}^{-1}\left(  f\right)  <_{1}p_{2}^{-1}\left(  f^{\prime}\right)  $
(since $p_{2}^{-1}\left(  f\right)  =\left(  \phi^{-1}\left(  f\right)
,f\right)  $ and $p_{2}^{-1}\left(  f^{\prime}\right)  =\left(  \phi
^{-1}\left(  f^{\prime}\right)  ,f^{\prime}\right)  $).
\par
Now, let us forget that we fixed $f$ and $f^{\prime}$. We thus have shown that
if $f$ and $f^{\prime}$ are two elements of $F$ satisfying $f<_{F1}f^{\prime}%
$, then $p_{2}^{-1}\left(  f\right)  <_{1}p_{2}^{-1}\left(  f^{\prime}\right)
$. In other words, the map $p_{2}^{-1}$ is a strictly order-preserving map
$\left(  F,<_{F1}\right)  \rightarrow\left(  E\times_{\phi}F,<_{1}\right)  $.
As a consequence, this map $p_{2}^{-1}$ is a poset homomorphism $\left(
F,<_{F1}\right)  \rightarrow\left(  E\times_{\phi}F,<_{1}\right)  $ (since any
strictly order-preserving map between two posets is a poset homomorphism).
Qed.} and a poset homomorphism $\left(  F,\left(  <_{E2}\right)  ^{\phi^{-1}%
}\right)  \rightarrow\left(  E\times_{\phi}F,<_{2}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $f$ and $f^{\prime}$ be two elements of
$F$ satisfying $f\left(  <_{E2}\right)  ^{\phi^{-1}}f^{\prime}$. We shall
prove that $p_{2}^{-1}\left(  f\right)  <_{2}p_{2}^{-1}\left(  f^{\prime
}\right)  $.
\par
We have $f\left(  <_{E2}\right)  ^{\phi^{-1}}f^{\prime}$ if and only if
$\phi^{-1}\left(  f\right)  <_{E2}\phi^{-1}\left(  f^{\prime}\right)  $ (due
to the definition of \textquotedblleft$f\left(  <_{E2}\right)  ^{\phi^{-1}%
}f^{\prime}$\textquotedblright). Thus, we have $\phi^{-1}\left(  f\right)
<_{E2}\phi^{-1}\left(  f^{\prime}\right)  $ (since we have $f\left(
<_{E2}\right)  ^{\phi^{-1}}f^{\prime}$).
\par
We have $\underbrace{p_{2}^{-1}}_{=\left(  \phi^{-1},\operatorname*{id}%
\right)  }\left(  f\right)  =\left(  \phi^{-1},\operatorname*{id}\right)
\left(  f\right)  =\left(  \phi^{-1}\left(  f\right)
,\underbrace{\operatorname*{id}\left(  f\right)  }_{=f}\right)  =\left(
\phi^{-1}\left(  f\right)  ,f\right)  $, so that $\left(  \phi^{-1}\left(
f\right)  ,f\right)  =p_{2}^{-1}\left(  f\right)  \in E\times_{\phi}F$.
\par
Also, $\underbrace{p_{2}^{-1}}_{=\left(  \phi^{-1},\operatorname*{id}\right)
}\left(  f^{\prime}\right)  =\left(  \phi^{-1},\operatorname*{id}\right)
\left(  f^{\prime}\right)  =\left(  \phi^{-1}\left(  f^{\prime}\right)
,\underbrace{\operatorname*{id}\left(  f^{\prime}\right)  }_{=f^{\prime}%
}\right)  =\left(  \phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime}\right)  $,
so that $\left(  \phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime}\right)
=p_{2}^{-1}\left(  f^{\prime}\right)  \in E\times_{\phi}F$.
\par
Now, (\ref{pf.lem.assoc.fiber.less2}) (applied to $\phi^{-1}\left(  f\right)
$ and $\phi^{-1}\left(  f^{\prime}\right)  $ instead of $e$ and $e^{\prime}$)
yields $\left(  \left(  \phi^{-1}\left(  f\right)  ,f\right)  <_{2}\left(
\phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime}\right)  \text{ if and only if
}\phi^{-1}\left(  f\right)  <_{E2}\phi^{-1}\left(  f^{\prime}\right)  \right)
$ (since $\left(  \phi^{-1}\left(  f\right)  ,f\right)  \in E\times_{\phi}F$
and $\left(  \phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime}\right)  \in
E\times_{\phi}F$). Thus, we have $\left(  \phi^{-1}\left(  f\right)
,f\right)  <_{2}\left(  \phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime
}\right)  $ (since $\phi^{-1}\left(  f\right)  <_{E2}\phi^{-1}\left(
f^{\prime}\right)  $). In other words, $p_{2}^{-1}\left(  f\right)  <_{2}%
p_{2}^{-1}\left(  f^{\prime}\right)  $ (since $p_{2}^{-1}\left(  f\right)
=\left(  \phi^{-1}\left(  f\right)  ,f\right)  $ and $p_{2}^{-1}\left(
f^{\prime}\right)  =\left(  \phi^{-1}\left(  f^{\prime}\right)  ,f^{\prime
}\right)  $).
\par
Now, let us forget that we fixed $f$ and $f^{\prime}$. We thus have shown that
if $f$ and $f^{\prime}$ are two elements of $F$ satisfying $f\left(
<_{E2}\right)  ^{\phi^{-1}}f^{\prime}$, then $p_{2}^{-1}\left(  f\right)
<_{2}p_{2}^{-1}\left(  f^{\prime}\right)  $. In other words, the map
$p_{2}^{-1}$ is a strictly order-preserving map $\left(  F,\left(
<_{E2}\right)  ^{\phi^{-1}}\right)  \rightarrow\left(  E\times_{\phi}%
F,<_{2}\right)  $. As a consequence, this map $p_{2}^{-1}$ is a poset
homomorphism $\left(  F,\left(  <_{E2}\right)  ^{\phi^{-1}}\right)
\rightarrow\left(  E\times_{\phi}F,<_{2}\right)  $ (since any strictly
order-preserving map between two posets is a poset homomorphism). Qed.}.
Hence, the map $p_{2}^{-1}$ is a homomorphism of double posets $\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  \rightarrow\left(
E\times_{\phi}F,<_{1},<_{2}\right)  $.

Now, we know that the map $p_{2}$ is a homomorphism of double posets
\newline$\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \rightarrow\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  $, while its inverse
$p_{2}^{-1}$ is a homomorphism of double posets $\left(  F,<_{F1},\left(
<_{E2}\right)  ^{\phi^{-1}}\right)  \rightarrow\left(  E\times_{\phi}%
F,<_{1},<_{2}\right)  $. In other words, $p_{2}$ is an isomorphism of double
posets $\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \rightarrow\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  $. Thus, the double
posets $\left(  E\times_{\phi}F,<_{1},<_{2}\right)  $ and $\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  $ are isomorphic. That
is, we have $\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \cong\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  $ as double posets.
Hence,%
\begin{equation}
E\times_{\phi}F=\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \cong\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)
\label{pf.lem.assoc.fiber.a.almost}%
\end{equation}
as double posets. Thus,%
\[
\underbrace{\left(  E,<_{E1},<_{E2}\right)  }_{=E}\times_{\phi}%
\underbrace{\left(  F,<_{F1},<_{F2}\right)  }_{=F}=E\times_{\phi}F\cong\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)
\]
as double posets. This proves Lemma \ref{lem.assoc.fiber} \textbf{(a)}.

\textbf{(b)} The definition of $E\circ F$ shows that $E\circ F$ is the sum of
$E\times_{\phi}F$ for all increasing bijections $\phi:\left(  E,<_{E1}\right)
\rightarrow\left(  F,<_{F2}\right)  $. In other words,%
\begin{align*}
E\circ F  &  =\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection
}\left(  E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }%
}\underbrace{E\times_{\phi}F}_{\substack{=\left(  F,<_{F1},\left(
<_{E2}\right)  ^{\phi^{-1}}\right)  \\\text{(according to
(\ref{pf.lem.assoc.fiber.a.almost}))}}}\\
&  =\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\left(  F,<_{F1}%
,\left(  <_{E2}\right)  ^{\phi^{-1}}\right)
\end{align*}
in $\mathbb{Z}\mathbf{D}$. Thus,%
\[
\underbrace{\left(  E,<_{E1},<_{E2}\right)  }_{=E}\circ\underbrace{\left(
F,<_{F1},<_{F2}\right)  }_{=F}=E\circ F=\sum_{\substack{\phi\text{ is an
increasing}\\\text{bijection }\left(  E,<_{E1}\right)  \rightarrow\left(
F,<_{F2}\right)  }}\left(  F,<_{F1},\left(  <_{E2}\right)  ^{\phi^{-1}%
}\right)
\]
in $\mathbb{Z}\mathbf{D}$. This proves Lemma \ref{lem.assoc.fiber}
\textbf{(b)}.

\textbf{(c)} Let $\phi$ be a bijection $E\rightarrow F$. Proposition
\ref{prop.assoc.AB} (applied to $E$, $F$, $<_{F1}$ and $\phi$ instead of $A$,
$B$, $<_{B}$ and $f$) shows that the binary relation $\left(  <_{F1}\right)
^{\phi}$ is the smaller relation of a partial order on $E$. Hence, $\left(
E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)  $ is a well-defined double poset.

Define a map $p_{1}:E\times_{\phi}F\rightarrow E$ by $\left(  p_{1}\left(
e,f\right)  =e\text{ for every }\left(  e,f\right)  \in E\times_{\phi
}F\right)  $. Thus, $p_{1}$ is the projection on the first component. On the
other hand, consider the map $\left(  \operatorname*{id},\phi\right)
:E\rightarrow E\times_{\phi}F$ which sends every $e\in E$ to $\left(
\underbrace{\operatorname*{id}\left(  e\right)  }_{=e},\phi\left(  e\right)
\right)  =\left(  e,\phi\left(  e\right)  \right)  \in E\times_{\phi}%
F$\ \ \ $\ $\footnote{Let us check that this map $\left(  \operatorname*{id}%
,\phi\right)  $ is well-defined:
\par
Fix $e\in E$. Then, $\left(  e,\phi\left(  e\right)  \right)  $ is an element
of $E\times F$ satisfying $\phi\left(  e\right)  =\phi\left(  e\right)  $. In
other words, $\left(  e,\phi\left(  e\right)  \right)  $ is an element
$\left(  u,v\right)  $ of $E\times F$ satisfying $\phi\left(  u\right)  =v$.
In other words,%
\[
\left(  e,\phi\left(  e\right)  \right)  \in\left\{  \left(  u,v\right)  \in
E\times F\ \mid\ \phi\left(  u\right)  =v\right\}  =E\times_{\phi
}F\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.assoc.fiber.set})}\right)
.
\]
\par
Now, let us forget that we fixed $e$. We thus have shown that $\left(
e,\phi\left(  e\right)  \right)  \in E\times_{\phi}F$ for every $e\in E$.
Thus, the map $\left(  \operatorname*{id},\phi\right)  $ is well-defined,
qed.}. The maps $p_{1}$ and $\left(  \operatorname*{id},\phi\right)  $ are
mutually inverse\footnote{\textit{Proof.} Every $e\in E$ satisfies%
\begin{align*}
\left(  p_{1}\circ\left(  \operatorname*{id},\phi\right)  \right)  \left(
e\right)   &  =p_{1}\left(  \underbrace{\left(  \operatorname*{id}%
,\phi\right)  \left(  e\right)  }_{\substack{=\left(  \operatorname*{id}%
\left(  e\right)  ,\phi\left(  e\right)  \right)  \\\text{(by the definition
of }\left(  \operatorname*{id},\phi\right)  \text{)}}}\right)  =p_{1}\left(
\operatorname*{id}\left(  e\right)  ,\phi\left(  e\right)  \right)
=\operatorname*{id}\left(  e\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }p_{1}\right)  .
\end{align*}
In other words, $p_{1}\circ\left(  \operatorname*{id},\phi\right)
=\operatorname*{id}$.
\par
On the other hand, let $\left(  e,f\right)  \in E\times_{\phi}F$. Then,
$\left(  e,f\right)  \in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in
E\times F\ \mid\ \phi\left(  u\right)  =v\right\}  $ (according to
(\ref{pf.lem.assoc.fiber.set})). In other words, $\left(  e,f\right)  $ is an
element $\left(  u,v\right)  $ of $E\times F$ satisfying $\phi\left(
u\right)  =v$. In other words, $\left(  e,f\right)  $ is an element of
$E\times F$ and satisfies $\phi\left(  e\right)  =f$. Now, $p_{1}\left(
e,f\right)  =e$ (by the definition of $p_{1}$). Now,%
\[
\left(  \left(  \operatorname*{id},\phi\right)  \circ p_{1}\right)  \left(
e,f\right)  =\left(  \operatorname*{id},\phi\right)  \left(  \underbrace{p_{1}%
\left(  e,f\right)  }_{=e}\right)  =\left(  \operatorname*{id},\phi\right)
\left(  e\right)  =\left(  \underbrace{\operatorname*{id}\left(  e\right)
}_{=e},\underbrace{\phi\left(  e\right)  }_{=f}\right)  =\left(  e,f\right)
=\operatorname*{id}\left(  e,f\right)  .
\]
\par
Let us now forget that we fixed $\left(  e,f\right)  $. We thus have shown
that $\left(  \left(  \operatorname*{id},\phi\right)  \circ p_{1}\right)
\left(  e,f\right)  =\operatorname*{id}\left(  e,f\right)  $ for every
$\left(  e,f\right)  \in E\times_{\phi}F$. In other words, $\left(
\operatorname*{id},\phi\right)  \circ p_{1}=\operatorname*{id}$. Combined with
$p_{1}\circ\left(  \operatorname*{id},\phi\right)  =\operatorname*{id}$, this
yields that the maps $p_{1}$ and $\left(  \operatorname*{id},\phi\right)  $
are mutually inverse. Qed.}. Hence, these two maps $p_{1}$ and $\left(
\operatorname*{id},\phi\right)  $ are invertible and satisfy $p_{1}%
^{-1}=\left(  \operatorname*{id},\phi\right)  $.

The map $p_{1}$ is a poset homomorphism $\left(  E\times_{\phi}F,<_{1}\right)
\rightarrow\left(  E,\left(  <_{F1}\right)  ^{\phi}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $g$ and $g^{\prime}$ be two elements of
$E\times_{\phi}F$ satisfying $g<_{1}g^{\prime}$. We shall show that
$p_{1}\left(  g\right)  \left(  <_{F1}\right)  ^{\phi}p_{1}\left(  g^{\prime
}\right)  $.
\par
Indeed, $g\in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in E\times
F\ \mid\ \phi\left(  u\right)  =v\right\}  $. In other words, $g$ has the form
$\left(  u,v\right)  $ for some $\left(  u,v\right)  \in E\times F$ satisfying
$\phi\left(  u\right)  =v$. Let us denote this $\left(  u,v\right)  $ by
$\left(  e,f\right)  $. Thus, $\left(  e,f\right)  $ is an element of $E\times
F$ and satisfies $\phi\left(  e\right)  =f$ and $g=\left(  e,f\right)  $.
Applying the map $p_{1}$ to both sides of the equality $g=\left(  e,f\right)
$, we obtain $p_{1}\left(  g\right)  =p_{1}\left(  e,f\right)  =e$ (by the
definition of $p_{1}$).
\par
Also, $g^{\prime}\in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in E\times
F\ \mid\ \phi\left(  u\right)  =v\right\}  $. In other words, $g^{\prime}$ has
the form $\left(  u,v\right)  $ for some $\left(  u,v\right)  \in E\times F$
satisfying $\phi\left(  u\right)  =v$. Let us denote this $\left(  u,v\right)
$ by $\left(  e^{\prime},f^{\prime}\right)  $. Thus, $\left(  e^{\prime
},f^{\prime}\right)  $ is an element of $E\times F$ and satisfies $\phi\left(
e^{\prime}\right)  =f^{\prime}$ and $g^{\prime}=\left(  e^{\prime},f^{\prime
}\right)  $. Applying the map $p_{1}$ to both sides of the equality
$g^{\prime}=\left(  e^{\prime},f^{\prime}\right)  $, we obtain $p_{1}\left(
g^{\prime}\right)  =p_{1}\left(  e^{\prime},f^{\prime}\right)  =e^{\prime}$
(by the definition of $p_{1}$).
\par
We have $\left(  e,f\right)  =g<_{1}g^{\prime}=\left(  e^{\prime},f^{\prime
}\right)  $. Due to (\ref{pf.lem.assoc.fiber.less1}), this yields
$f<_{F1}f^{\prime}$. Thus, $\phi\left(  e\right)  =f<_{F1}f^{\prime}%
=\phi\left(  e^{\prime}\right)  $ (since $\phi\left(  e^{\prime}\right)
=f^{\prime}$).
\par
But $e\left(  <_{F1}\right)  ^{\phi}e^{\prime}$ holds if and only if
$\phi\left(  e\right)  <_{F1}\phi\left(  e^{\prime}\right)  $ (due to the
definition of \textquotedblleft$e\left(  <_{F1}\right)  ^{\phi}e^{\prime}%
$\textquotedblright). Thus, $e\left(  <_{F1}\right)  ^{\phi}e^{\prime}$ holds
(since $\phi\left(  e\right)  <_{F1}\phi\left(  e^{\prime}\right)  $ holds).
In other words, $p_{1}\left(  g\right)  \left(  <_{F1}\right)  ^{\phi}%
p_{1}\left(  g^{\prime}\right)  $ holds (since $p_{1}\left(  g\right)  =e$ and
$p_{1}\left(  g^{\prime}\right)  =e^{\prime}$).
\par
Now, let us forget that we fixed $g$ and $g^{\prime}$. We thus have proven
that if $g$ and $g^{\prime}$ are two elements of $E\times_{\phi}F$ satisfying
$g<_{1}g^{\prime}$, then $p_{1}\left(  g\right)  \left(  <_{F1}\right)
^{\phi}p_{1}\left(  g^{\prime}\right)  $. In other words, the map $p_{1}$ is a
strictly order-preserving map $\left(  E\times_{\phi}F,<_{1}\right)
\rightarrow\left(  E,\left(  <_{F1}\right)  ^{\phi}\right)  $. As a
consequence, this map $p_{1}$ is a poset homomorphism $\left(  E\times_{\phi
}F,<_{1}\right)  \rightarrow\left(  E,\left(  <_{F1}\right)  ^{\phi}\right)  $
(since any strictly order-preserving map between two posets is a poset
homomorphism). Qed.} and a poset homomorphism $\left(  E\times_{\phi}%
F,<_{2}\right)  \rightarrow\left(  E,<_{E2}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $g$ and $g^{\prime}$ be two elements of
$E\times_{\phi}F$ satisfying $g<_{2}g^{\prime}$. We shall show that
$p_{1}\left(  g\right)  <_{E2}p_{1}\left(  g^{\prime}\right)  $.
\par
Indeed, $g\in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in E\times
F\ \mid\ \phi\left(  u\right)  =v\right\}  $. In other words, $g$ has the form
$\left(  u,v\right)  $ for some $\left(  u,v\right)  \in E\times F$ satisfying
$\phi\left(  u\right)  =v$. Let us denote this $\left(  u,v\right)  $ by
$\left(  e,f\right)  $. Thus, $\left(  e,f\right)  $ is an element of $E\times
F$ and satisfies $\phi\left(  e\right)  =f$ and $g=\left(  e,f\right)  $.
Applying the map $p_{1}$ to both sides of the equality $g=\left(  e,f\right)
$, we obtain $p_{1}\left(  g\right)  =p_{1}\left(  e,f\right)  =e$ (by the
definition of $p_{1}$).
\par
Also, $g^{\prime}\in E\times_{\phi}F=\left\{  \left(  u,v\right)  \in E\times
F\ \mid\ \phi\left(  u\right)  =v\right\}  $. In other words, $g^{\prime}$ has
the form $\left(  u,v\right)  $ for some $\left(  u,v\right)  \in E\times F$
satisfying $\phi\left(  u\right)  =v$. Let us denote this $\left(  u,v\right)
$ by $\left(  e^{\prime},f^{\prime}\right)  $. Thus, $\left(  e^{\prime
},f^{\prime}\right)  $ is an element of $E\times F$ and satisfies $\phi\left(
e^{\prime}\right)  =f^{\prime}$ and $g^{\prime}=\left(  e^{\prime},f^{\prime
}\right)  $. Applying the map $p_{1}$ to both sides of the equality
$g^{\prime}=\left(  e^{\prime},f^{\prime}\right)  $, we obtain $p_{1}\left(
g^{\prime}\right)  =p_{1}\left(  e^{\prime},f^{\prime}\right)  =e^{\prime}$
(by the definition of $p_{1}$).
\par
We have $\left(  e,f\right)  =g<_{2}g^{\prime}=\left(  e^{\prime},f^{\prime
}\right)  $. Due to (\ref{pf.lem.assoc.fiber.less2}), this yields
$e<_{E2}e^{\prime}$. In other words, $p_{1}\left(  g\right)  <_{E2}%
p_{1}\left(  g^{\prime}\right)  $ (since $p_{1}\left(  g\right)  =e$ and
$p_{1}\left(  g^{\prime}\right)  =e^{\prime}$).
\par
Now, let us forget that we fixed $g$ and $g^{\prime}$. We thus have proven
that if $g$ and $g^{\prime}$ are two elements of $E\times_{\phi}F$ satisfying
$g<_{2}g^{\prime}$, then $p_{1}\left(  g\right)  <_{E2}p_{1}\left(  g^{\prime
}\right)  $. In other words, the map $p_{1}$ is a strictly order-preserving
map $\left(  E\times_{\phi}F,<_{2}\right)  \rightarrow\left(  E,<_{E2}\right)
$. As a consequence, this map $p_{1}$ is a poset homomorphism $\left(
E\times_{\phi}F,<_{2}\right)  \rightarrow\left(  E,<_{E2}\right)  $ (since any
strictly order-preserving map between two posets is a poset homomorphism).
Qed.}. Hence, the map $p_{1}$ is a homomorphism of double posets $\left(
E\times_{\phi}F,<_{1},<_{2}\right)  \rightarrow\left(  E,\left(
<_{F1}\right)  ^{\phi},<_{E2}\right)  $.

On the other hand, the map $p_{1}^{-1}$ is a poset homomorphism $\left(
E,\left(  <_{F1}\right)  ^{\phi}\right)  \rightarrow\left(  E\times_{\phi
}F,<_{1}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $e$ and $e^{\prime}$
be two elements of $E$ satisfying $e\left(  <_{F1}\right)  ^{\phi}e^{\prime}$.
We shall prove that $p_{1}^{-1}\left(  e\right)  <_{1}p_{1}^{-1}\left(
e^{\prime}\right)  $.
\par
We have $e\left(  <_{F1}\right)  ^{\phi}e^{\prime}$ if and only if
$\phi\left(  e\right)  <_{F1}\phi\left(  e^{\prime}\right)  $ (due to the
definition of \textquotedblleft$e\left(  <_{F1}\right)  ^{\phi}e^{\prime}%
$\textquotedblright). Thus, we have $\phi\left(  e\right)  <_{F1}\phi\left(
e^{\prime}\right)  $ (since we have $e\left(  <_{F1}\right)  ^{\phi}e^{\prime
}$).
\par
We have $\underbrace{p_{1}^{-1}}_{=\left(  \operatorname*{id},\phi\right)
}\left(  e\right)  =\left(  \operatorname*{id},\phi\right)  \left(  e\right)
=\left(  \underbrace{\operatorname*{id}\left(  e\right)  }_{=e},\phi\left(
e\right)  \right)  =\left(  e,\phi\left(  e\right)  \right)  $, so that
$\left(  e,\phi\left(  e\right)  \right)  =p_{1}^{-1}\left(  e\right)  \in
E\times_{\phi}F$.
\par
Also, $\underbrace{p_{1}^{-1}}_{=\left(  \operatorname*{id},\phi\right)
}\left(  e^{\prime}\right)  =\left(  \operatorname*{id},\phi\right)  \left(
e^{\prime}\right)  =\left(  \underbrace{\operatorname*{id}\left(  e^{\prime
}\right)  }_{=e^{\prime}},\phi\left(  e^{\prime}\right)  \right)  =\left(
e^{\prime},\phi\left(  e^{\prime}\right)  \right)  $, so that $\left(
e^{\prime},\phi\left(  e^{\prime}\right)  \right)  =p_{1}^{-1}\left(
e^{\prime}\right)  \in E\times_{\phi}F$.
\par
Now, (\ref{pf.lem.assoc.fiber.less1}) (applied to $\phi\left(  e\right)  $ and
$\phi\left(  e^{\prime}\right)  $ instead of $f$ and $f^{\prime}$) yields
$\left(  \left(  e,\phi\left(  e\right)  \right)  <_{1}\left(  e^{\prime}%
,\phi\left(  e^{\prime}\right)  \right)  \text{ if and only if }\phi\left(
e\right)  <_{F1}\phi\left(  e^{\prime}\right)  \right)  $ (since $\left(
e,\phi\left(  e\right)  \right)  \in E\times_{\phi}F$ and $\left(  e^{\prime
},\phi\left(  e^{\prime}\right)  \right)  \in E\times_{\phi}F$). Thus, we have
$\left(  e,\phi\left(  e\right)  \right)  <_{1}\left(  e^{\prime},\phi\left(
e^{\prime}\right)  \right)  $ (since $\phi\left(  e\right)  <_{F1}\phi\left(
e^{\prime}\right)  $). In other words, $p_{1}^{-1}\left(  e\right)  <_{1}%
p_{1}^{-1}\left(  e^{\prime}\right)  $ (since $p_{1}^{-1}\left(  e\right)
=\left(  e,\phi\left(  e\right)  \right)  $ and $p_{1}^{-1}\left(  e^{\prime
}\right)  =\left(  e^{\prime},\phi\left(  e^{\prime}\right)  \right)  $).
\par
Now, let us forget that we fixed $e$ and $e^{\prime}$. We thus have shown that
if $e$ and $e^{\prime}$ are two elements of $E$ satisfying $e\left(
<_{F1}\right)  ^{\phi}e^{\prime}$, then $p_{1}^{-1}\left(  e\right)
<_{1}p_{1}^{-1}\left(  e^{\prime}\right)  $. In other words, the map
$p_{1}^{-1}$ is a strictly order-preserving map $\left(  E,\left(
<_{F1}\right)  ^{\phi}\right)  \rightarrow\left(  E\times_{\phi}%
F,<_{1}\right)  $. As a consequence, this map $p_{1}^{-1}$ is a poset
homomorphism $\left(  E,\left(  <_{F1}\right)  ^{\phi}\right)  \rightarrow
\left(  E\times_{\phi}F,<_{1}\right)  $ (since any strictly order-preserving
map between two posets is a poset homomorphism). Qed.} and a poset
homomorphism $\left(  E,<_{E2}\right)  \rightarrow\left(  E\times_{\phi
}F,<_{2}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $e$ and $e^{\prime}$
be two elements of $E$ satisfying $e<_{E2}e^{\prime}$. We shall prove that
$p_{1}^{-1}\left(  e\right)  <_{2}p_{1}^{-1}\left(  e^{\prime}\right)  $.
\par
We have $\underbrace{p_{1}^{-1}}_{=\left(  \operatorname*{id},\phi\right)
}\left(  e\right)  =\left(  \operatorname*{id},\phi\right)  \left(  e\right)
=\left(  \underbrace{\operatorname*{id}\left(  e\right)  }_{=e},\phi\left(
e\right)  \right)  =\left(  e,\phi\left(  e\right)  \right)  $, so that
$\left(  e,\phi\left(  e\right)  \right)  =p_{1}^{-1}\left(  e\right)  \in
E\times_{\phi}F$.
\par
Also, $\underbrace{p_{1}^{-1}}_{=\left(  \operatorname*{id},\phi\right)
}\left(  e^{\prime}\right)  =\left(  \operatorname*{id},\phi\right)  \left(
e^{\prime}\right)  =\left(  \underbrace{\operatorname*{id}\left(  e^{\prime
}\right)  }_{=e^{\prime}},\phi\left(  e^{\prime}\right)  \right)  =\left(
e^{\prime},\phi\left(  e^{\prime}\right)  \right)  $, so that $\left(
e^{\prime},\phi\left(  e^{\prime}\right)  \right)  =p_{1}^{-1}\left(
e^{\prime}\right)  \in E\times_{\phi}F$.
\par
Now, (\ref{pf.lem.assoc.fiber.less2}) (applied to $\phi\left(  e\right)  $ and
$\phi\left(  e^{\prime}\right)  $ instead of $f$ and $f^{\prime}$) yields
$\left(  \left(  e,\phi\left(  e\right)  \right)  <_{2}\left(  e^{\prime}%
,\phi\left(  e^{\prime}\right)  \right)  \text{ if and only if }%
e<_{E2}e^{\prime}\right)  $ (since $\left(  e,\phi\left(  e\right)  \right)
\in E\times_{\phi}F$ and $\left(  e^{\prime},\phi\left(  e^{\prime}\right)
\right)  \in E\times_{\phi}F$). Thus, we have $\left(  e,\phi\left(  e\right)
\right)  <_{2}\left(  e^{\prime},\phi\left(  e^{\prime}\right)  \right)  $
(since $e<_{E2}e^{\prime}$). In other words, $p_{1}^{-1}\left(  e\right)
<_{2}p_{1}^{-1}\left(  e^{\prime}\right)  $ (since $p_{1}^{-1}\left(
e\right)  =\left(  e,\phi\left(  e\right)  \right)  $ and $p_{1}^{-1}\left(
e^{\prime}\right)  =\left(  e^{\prime},\phi\left(  e^{\prime}\right)  \right)
$).
\par
Now, let us forget that we fixed $e$ and $e^{\prime}$. We thus have shown that
if $e$ and $e^{\prime}$ are two elements of $E$ satisfying $e<_{E2}e^{\prime}%
$, then $p_{1}^{-1}\left(  e\right)  <_{2}p_{1}^{-1}\left(  e^{\prime}\right)
$. In other words, the map $p_{1}^{-1}$ is a strictly order-preserving map
$\left(  E,<_{E2}\right)  \rightarrow\left(  E\times_{\phi}F,<_{2}\right)  $.
As a consequence, this map $p_{1}^{-1}$ is a poset homomorphism $\left(
E,<_{E2}\right)  \rightarrow\left(  E\times_{\phi}F,<_{2}\right)  $ (since any
strictly order-preserving map between two posets is a poset homomorphism).
Qed.}. Hence, the map $p_{1}^{-1}$ is a homomorphism of double posets $\left(
E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)  \rightarrow\left(
E\times_{\phi}F,<_{1},<_{2}\right)  $.

Now, we know that the map $p_{1}$ is a homomorphism of double posets
\newline$\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \rightarrow\left(
E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)  $, while its inverse
$p_{1}^{-1}$ is a homomorphism of double posets $\left(  E,\left(
<_{F1}\right)  ^{\phi},<_{E2}\right)  \rightarrow\left(  E\times_{\phi}%
F,<_{1},<_{2}\right)  $. In other words, $p_{1}$ is an isomorphism of double
posets $\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \rightarrow\left(
E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)  $. Thus, the double posets
$\left(  E\times_{\phi}F,<_{1},<_{2}\right)  $ and $\left(  E,\left(
<_{F1}\right)  ^{\phi},<_{E2}\right)  $ are isomorphic. That is, we have
$\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \cong\left(  E,\left(
<_{F1}\right)  ^{\phi},<_{E2}\right)  $ as double posets. Hence,%
\begin{equation}
E\times_{\phi}F=\left(  E\times_{\phi}F,<_{1},<_{2}\right)  \cong\left(
E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)
\label{pf.lem.assoc.fiber.b.almost}%
\end{equation}
as double posets. Thus,%
\[
\underbrace{\left(  E,<_{E1},<_{E2}\right)  }_{=E}\times_{\phi}%
\underbrace{\left(  F,<_{F1},<_{F2}\right)  }_{=F}=E\times_{\phi}F\cong\left(
E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)
\]
as double posets. This proves Lemma \ref{lem.assoc.fiber} \textbf{(c)}.

\textbf{(d)} The definition of $E\circ F$ shows that $E\circ F$ is the sum of
$E\times_{\phi}F$ for all increasing bijections $\phi:\left(  E,<_{E1}\right)
\rightarrow\left(  F,<_{F2}\right)  $. In other words,%
\begin{align*}
E\circ F &  =\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection
}\left(  E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }%
}\ \ \underbrace{E\times_{\phi}F}_{\substack{=\left(  E,\left(  <_{F1}\right)
^{\phi},<_{E2}\right)  \\\text{(according to
(\ref{pf.lem.assoc.fiber.b.almost}))}}}\\
&  =\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\left(  E,\left(
<_{F1}\right)  ^{\phi},<_{E2}\right)
\end{align*}
in $\mathbb{Z}\mathbf{D}$. Thus,%
\[
\underbrace{\left(  E,<_{E1},<_{E2}\right)  }_{=E}\circ\underbrace{\left(
F,<_{F1},<_{F2}\right)  }_{=F}=E\circ F=\sum_{\substack{\phi\text{ is an
increasing}\\\text{bijection }\left(  E,<_{E1}\right)  \rightarrow\left(
F,<_{F2}\right)  }}\left(  E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)
\]
in $\mathbb{Z}\mathbf{D}$. This proves Lemma \ref{lem.assoc.fiber}
\textbf{(d)}.
\end{proof}

Now, we can finally prove Proposition \ref{prop.assoc}:

\begin{proof}
[Proof of Proposition \ref{prop.assoc}.]Let $E$, $F$ and $G$ be three elements
of $\mathbb{Z}\mathbf{D}$. We have to prove the equality $\left(  E\circ
F\right)  \circ G=E\circ\left(  F\circ G\right)  $. Since this equality is
$\mathbb{Z}$-linear in each of $E$, $F$ and $G$, we can WLOG assume that $E$,
$F$ and $G$ are (isomorphism classes of) double posets (since (isomorphism
classes of) double posets span the $\mathbb{Z}$-module $\mathbb{Z}\mathbf{D}%
$). Assume this. Let us write the double poset $E$ in the form $E=\left(
E,<_{E1},<_{E2}\right)  $. Let us write the double poset $F$ in the form
$F=\left(  F,<_{F1},<_{F2}\right)  $. Let us write the double poset $G$ in the
form $\left(  G,<_{G1},<_{G2}\right)  $.

We have%
\begin{align*}
&  \underbrace{E}_{=\left(  E,<_{E1},<_{E2}\right)  }\circ\underbrace{F}%
_{=\left(  F,<_{F1},<_{F2}\right)  }\\
&  =\left(  E,<_{E1},<_{E2}\right)  \circ\left(  F,<_{F1},<_{F2}\right)
=\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\left(  F,<_{F1}%
,\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.assoc.fiber} \textbf{(b)}}\right)  \\
&  =\sum_{\substack{\alpha\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\left(  F,<_{F1}%
,\left(  <_{E2}\right)  ^{\alpha^{-1}}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }\phi\text{ as }\alpha\right)  .
\end{align*}
Thus,%
\begin{align}
&  \underbrace{\left(  E\circ F\right)  }_{=\sum_{\substack{\alpha\text{ is an
increasing}\\\text{bijection }\left(  E,<_{E1}\right)  \rightarrow\left(
F,<_{F2}\right)  }}\left(  F,<_{F1},\left(  <_{E2}\right)  ^{\alpha^{-1}%
}\right)  }\circ\underbrace{G}_{=\left(  G,<_{G1},<_{G2}\right)  }\nonumber\\
&  =\left(  \sum_{\substack{\alpha\text{ is an increasing}\\\text{bijection
}\left(  E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\alpha^{-1}}\right)  \right)  \circ\left(
G,<_{G1},<_{G2}\right)  \nonumber\\
&  =\sum_{\substack{\alpha\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\underbrace{\left(
F,<_{F1},\left(  <_{E2}\right)  ^{\alpha^{-1}}\right)  \circ\left(
G,<_{G1},<_{G2}\right)  }_{\substack{=\sum_{\substack{\phi\text{ is an
increasing}\\\text{bijection }\left(  F,<_{F1}\right)  \rightarrow\left(
G,<_{G2}\right)  }}\left(  F,\left(  <_{G1}\right)  ^{\phi},\left(
<_{E2}\right)  ^{\alpha^{-1}}\right)  \\\text{(by Lemma \ref{lem.assoc.fiber}
\textbf{(d)}, applied to }F\text{, }<_{F1}\text{, }\left(  <_{E2}\right)
^{\alpha^{-1}}\text{, }G\text{, }<_{G1}\text{ and }<_{G2}\\\text{instead of
}E\text{, }<_{E1}\text{, }<_{E2}\text{, }F\text{, }<_{F1}\text{ and }%
<_{F2}\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since the operation
}\circ\text{ is }\mathbb{Z}\text{-bilinear}\right)  \nonumber\\
&  =\sum_{\substack{\alpha\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\ \ \sum
_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}\left(  F,\left(
<_{G1}\right)  ^{\phi},\left(  <_{E2}\right)  ^{\alpha^{-1}}\right)
\nonumber\\
&  =\sum_{\substack{\alpha\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\ \ \sum
_{\substack{\beta\text{ is an increasing}\\\text{bijection }\left(
F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}\left(  F,\left(
<_{G1}\right)  ^{\beta},\left(  <_{E2}\right)  ^{\alpha^{-1}}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }\phi\text{ as }\beta\text{ in the inner sum}\right)
\nonumber\\
&  =\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\ \ \sum
_{\substack{\beta\text{ is an increasing}\\\text{bijection }\left(
F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}\left(  F,\left(
<_{G1}\right)  ^{\beta},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)
\label{pf.prop.assoc.(EF)G}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }\alpha\text{ as }\phi\text{ in the outer sum}\right)
.\nonumber
\end{align}
On the other hand,%
\begin{align*}
&  \underbrace{F}_{=\left(  F,<_{F1},<_{F2}\right)  }\circ\underbrace{G}%
_{=\left(  G,<_{G1},<_{G2}\right)  }\\
&  =\left(  F,<_{F1},<_{F2}\right)  \circ\left(  G,<_{G1},<_{G2}\right)
=\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}\left(  F,\left(
<_{G1}\right)  ^{\phi},<_{F2}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.assoc.fiber} \textbf{(d)}, applied to }\\
F\text{, }<_{F1}\text{, }<_{F2}\text{, }G\text{, }<_{G1}\text{ and }<_{G2}\\
\text{instead of }E\text{, }<_{E1}\text{, }<_{E2}\text{, }F\text{, }%
<_{F1}\text{ and }<_{F2}%
\end{array}
\right)  \\
&  =\sum_{\substack{\beta\text{ is an increasing}\\\text{bijection }\left(
F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}\left(  F,\left(
<_{G1}\right)  ^{\beta},<_{F2}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }\phi\text{ as }\beta\right)  .
\end{align*}
Thus,%
\begin{align*}
&  \underbrace{E}_{=\left(  E,<_{E1},<_{E2}\right)  }\circ\underbrace{\left(
F\circ G\right)  }_{=\sum_{\substack{\beta\text{ is an increasing}%
\\\text{bijection }\left(  F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}%
\right)  }}\left(  F,\left(  <_{G1}\right)  ^{\beta},<_{F2}\right)  }\\
&  =\left(  E,<_{E1},<_{E2}\right)  \circ\left(  \sum_{\substack{\beta\text{
is an increasing}\\\text{bijection }\left(  F,<_{F1}\right)  \rightarrow
\left(  G,<_{G2}\right)  }}\left(  F,\left(  <_{G1}\right)  ^{\beta}%
,<_{F2}\right)  \right)  \\
&  =\sum_{\substack{\beta\text{ is an increasing}\\\text{bijection }\left(
F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}\underbrace{\left(
E,<_{E1},<_{E2}\right)  \circ\left(  F,\left(  <_{G1}\right)  ^{\beta}%
,<_{F2}\right)  }_{\substack{=\sum_{\substack{\phi\text{ is an increasing}%
\\\text{bijection }\left(  E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}%
\right)  }}\left(  F,\left(  <_{G1}\right)  ^{\beta},\left(  <_{E2}\right)
^{\phi^{-1}}\right)  \\\text{(by Lemma \ref{lem.assoc.fiber} \textbf{(b)},
applied to }\left(  <_{G1}\right)  ^{\beta}\\\text{instead of }<_{F1}\text{)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since the operation
}\circ\text{ is }\mathbb{Z}\text{-bilinear}\right)  \\
&  =\underbrace{\sum_{\substack{\beta\text{ is an increasing}\\\text{bijection
}\left(  F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}\ \ \sum
_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}}_{=\sum
_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\ \ \sum
_{\substack{\beta\text{ is an increasing}\\\text{bijection }\left(
F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}}\left(  F,\left(
<_{G1}\right)  ^{\beta},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  \\
&  =\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection }\left(
E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\ \ \sum
_{\substack{\beta\text{ is an increasing}\\\text{bijection }\left(
F,<_{F1}\right)  \rightarrow\left(  G,<_{G2}\right)  }}\left(  F,\left(
<_{G1}\right)  ^{\beta},\left(  <_{E2}\right)  ^{\phi^{-1}}\right)  .
\end{align*}
Compared with (\ref{pf.prop.assoc.(EF)G}), this yields $\left(  E\circ
F\right)  \circ G=E\circ\left(  F\circ G\right)  $. This proves Proposition
\ref{prop.assoc}.
\end{proof}

\subsection{Page 5, proof of Lemma 2.1: an alternative argument}

In the proof of Lemma 2.1, you take several lines (starting with
\textquotedblleft Moreover, let $\left(  f,g\right)  ,\left(  f^{\prime
},g^{\prime}\right)  \in F\times_{\psi}G$\textquotedblright\ and ending with
\textquotedblleft Finally $e<_{2}e^{\prime}$\textquotedblright) in order to
prove that the map $\beta^{-1}:\left(  F\times_{\psi}G,<_{1}\right)
\rightarrow\left(  E,<_{2}\right)  $ is increasing. Here is a different proof
of this: We have $\beta=\left(  \operatorname*{id},\psi\right)  \circ\phi$.
Thus, $\beta$ is a composition of two bijections (since $\left(
\operatorname*{id},\psi\right)  $ and $\phi$ are bijections), thus itself a
bijection. In other words, $\beta$ is invertible. For every $e\in E$, we have%
\begin{align*}
\underbrace{\beta}_{=\left(  \operatorname*{id},\psi\right)  \circ\phi}\left(
e\right)   &  =\left(  \left(  \operatorname*{id},\psi\right)  \circ
\phi\right)  \left(  e\right)  =\left(  \operatorname*{id},\psi\right)
\left(  \phi\left(  e\right)  \right) \\
&  =\left(  \underbrace{\operatorname*{id}\left(  \phi\left(  e\right)
\right)  }_{=\phi\left(  e\right)  },\psi\left(  \phi\left(  e\right)
\right)  \right)  =\left(  \phi\left(  e\right)  ,\psi\left(  \phi\left(
e\right)  \right)  \right)  .
\end{align*}
Thus, for every $e\in E$, we have%
\begin{align*}
&  \left(  \left(  p_{1}\circ\alpha^{-1}\circ p_{2}\right)  \circ\beta\right)
\left(  e\right) \\
&  =\left(  p_{1}\circ\alpha^{-1}\circ p_{2}\right)  \left(  \underbrace{\beta
\left(  e\right)  }_{=\left(  \phi\left(  e\right)  ,\psi\left(  \phi\left(
e\right)  \right)  \right)  }\right)  =\left(  p_{1}\circ\alpha^{-1}\circ
p_{2}\right)  \left(  \phi\left(  e\right)  ,\psi\left(  \phi\left(  e\right)
\right)  \right) \\
&  =p_{1}\left(  \alpha^{-1}\left(  \underbrace{p_{2}\left(  \phi\left(
e\right)  ,\psi\left(  \phi\left(  e\right)  \right)  \right)  }%
_{\substack{=\psi\left(  \phi\left(  e\right)  \right)  \\\text{(by the
definition of }p_{2}\text{)}}}\right)  \right)  =p_{1}\left(
\underbrace{\alpha^{-1}\left(  \psi\left(  \phi\left(  e\right)  \right)
\right)  }_{=\left(  \alpha^{-1}\circ\psi\right)  \left(  \phi\left(
e\right)  \right)  }\right) \\
&  =p_{1}\left(  \left(  \alpha^{-1}\circ\psi\right)  \left(  \phi\left(
e\right)  \right)  \right)  =p_{1}\left(  \underbrace{\left(  \phi
^{-1},\operatorname*{id}\right)  \left(  \phi\left(  e\right)  \right)
}_{=\left(  \phi^{-1}\left(  \phi\left(  e\right)  \right)
,\operatorname*{id}\left(  \phi\left(  e\right)  \right)  \right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\alpha^{-1}\circ\underbrace{\psi
}_{=\alpha\circ\left(  \phi^{-1},\operatorname*{id}\right)  }=\alpha^{-1}%
\circ\alpha\circ\left(  \phi^{-1},\operatorname*{id}\right)  =\left(
\phi^{-1},\operatorname*{id}\right)  \right) \\
&  =p_{1}\left(  \phi^{-1}\left(  \phi\left(  e\right)  \right)
,\operatorname*{id}\left(  \phi\left(  e\right)  \right)  \right)  =\phi
^{-1}\left(  \phi\left(  e\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }p_{1}\right) \\
&  =e=\operatorname*{id}\left(  e\right)  .
\end{align*}
In other words, $\left(  p_{1}\circ\alpha^{-1}\circ p_{2}\right)  \circ
\beta=\operatorname*{id}$. Since $\beta$ is invertible, this yields that%
\begin{equation}
\beta^{-1}=p_{1}\circ\alpha^{-1}\circ p_{2}. \label{add.p5.beta-1}%
\end{equation}
But $\alpha$ is a picture, and thus $\alpha^{-1}$ is increasing as a map from
$\left(  G,<_{1}\right)  $ to $\left(  E\times_{\phi}F,<_{2}\right)  $. Now,
we know that:

\begin{itemize}
\item the map $p_{2}$ is increasing as a map from $\left(  F\times_{\psi
}G,<_{1}\right)  $ to $\left(  G,<_{1}\right)  $;

\item the map $\alpha^{-1}$ is increasing as a map from $\left(
G,<_{1}\right)  $ to $\left(  E\times_{\phi}F,<_{2}\right)  $;

\item the map $p_{1}$ is increasing as a map from $\left(  E\times_{\phi
}F,<_{2}\right)  $ to $\left(  E,<_{2}\right)  $.
\end{itemize}

Combining these, we conclude that the composition $p_{1}\circ\alpha^{-1}\circ
p_{2}$ is increasing as a map from $\left(  F\times_{\psi}G,<_{1}\right)  $ to
$\left(  E,<_{2}\right)  $. In other words, $\beta^{-1}$ is increasing as a
map from $\left(  F\times_{\psi}G,<_{1}\right)  $ to $\left(  E,<_{2}\right)
$ (since $\beta^{-1}=p_{1}\circ\alpha^{-1}\circ p_{2}$).

\subsection{Page 6, proof of Lemma 2.1: an alternative argument}

In the proof of Lemma 2.1, you spend several lines to prove that the map
$\alpha^{-1}:\left(  G,<_{1}\right)  \rightarrow\left(  E\times_{\phi}%
F,<_{2}\right)  $ is increasing. Here is a different proof of this:

We have $\psi=\alpha\circ\left(  \phi^{-1},\operatorname*{id}\right)
$\ \ \ \ \footnote{\textit{Proof.} We know that the inverse $p_{2}^{-1}$ of
the bijection $p_{2}:E\times_{\psi}F\rightarrow F$ is given by $p_{2}%
^{-1}=\left(  \phi^{-1},\operatorname*{id}\right)  $. Thus,
$\underbrace{\alpha}_{=\psi\circ p_{2}}\circ\underbrace{\left(  \phi
^{-1},\operatorname*{id}\right)  }_{=p_{2}^{-1}}=\psi\circ\underbrace{p_{2}%
\circ p_{2}^{-1}}_{=\operatorname*{id}}=\psi$, qed.} and $\beta=\left(
\operatorname*{id},\psi\right)  \circ\phi\ \ \ \ $\footnote{\textit{Proof.} We
know that the inverse $p_{1}^{-1}$ of the bijection $p_{1}:F\times_{\psi
}G\rightarrow G$ is given by $p_{1}^{-1}=\left(  \operatorname*{id}%
,\psi\right)  $. Thus, $\underbrace{\left(  \operatorname*{id},\psi\right)
}_{=p_{1}^{-1}}\circ\underbrace{\phi}_{=p_{1}\circ\beta}=\underbrace{p_{1}%
^{-1}\circ p_{1}}_{=\operatorname*{id}}\circ\beta=\beta$, qed$.$}. Thus, we
can prove that (\ref{add.p5.beta-1}) holds (similarly to how we did it above).
In other words, $\beta^{-1}=p_{1}\circ\alpha^{-1}\circ p_{2}$. Thus,
$p_{1}^{-1}\circ\underbrace{\beta^{-1}}_{=p_{1}\circ\alpha^{-1}\circ p_{2}%
}\circ p_{2}^{-1}=\underbrace{p_{1}^{-1}\circ p_{1}}_{=\operatorname*{id}%
}\circ\alpha^{-1}\circ\underbrace{p_{2}\circ p_{2}^{-1}}_{=\operatorname*{id}%
}=\alpha^{-1}$, so that $\alpha^{-1}=p_{1}^{-1}\circ\beta^{-1}\circ p_{2}%
^{-1}$.

But $\beta$ is a picture, and thus $\beta^{-1}$ is increasing as a map from
$\left(  F\times_{\psi}G,<_{1}\right)  $ to $\left(  E,<_{2}\right)  $. Now,
we know that:

\begin{itemize}
\item the map $p_{2}^{-1}$ is increasing as a map from $\left(  G,<_{1}%
\right)  $ to $\left(  F\times_{\psi}G,<_{1}\right)  $;

\item the map $\beta^{-1}$ is increasing as a map from $\left(  F\times_{\psi
}G,<_{1}\right)  $ to $\left(  E,<_{2}\right)  $;

\item the map $p_{1}^{-1}$ is increasing as a map from $\left(  E,<_{2}%
\right)  $ to $\left(  E\times_{\phi}F,<_{2}\right)  $.
\end{itemize}

Combining these, we conclude that the composition $p_{1}^{-1}\circ\beta
^{-1}\circ p_{2}^{-1}$ is increasing as a map from $\left(  G,<_{1}\right)  $
to $\left(  E\times_{\phi}F,<_{2}\right)  $. In other words, $\alpha^{-1}$ is
increasing as a map from $\left(  G,<_{1}\right)  $ to $\left(  E\times_{\phi
}F,<_{2}\right)  $ (since $\alpha^{-1}=p_{1}^{-1}\circ\beta^{-1}\circ
p_{2}^{-1}$).

\subsection{Page 9, \S 3.1: descent compositions vs. descent sets}

On page 9, you define the descent composition of a permutation $\sigma\in
S_{n}$ in terms of its ascending runs. Let me give an alternative description
of descent compositions:

\begin{proposition}
\label{prop.C(sigma)}For every composition $\alpha=\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  $, we denote the number $\alpha
_{1}+\alpha_{2}+\cdots+\alpha_{\ell}$ by $\left\vert \alpha\right\vert $ and
call it the \textit{size} of the composition $\alpha$, and we say that
$\alpha$ is a \textit{composition of }$\left\vert \alpha\right\vert $.

For every composition $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right)  $, define a set $D\left(  \alpha\right)  $ by%
\begin{align}
D\left(  \alpha\right)   &  =\left\{  \alpha_{1},\alpha_{1}+\alpha_{2}%
,\alpha_{1}+\alpha_{2}+\alpha_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots
+\alpha_{\ell-1}\right\} \label{eq.prop.C(sigma).Dalpha0}\\
&  =\left\{  \alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}\ \mid\ p\in\left\{
1,2,\ldots,\ell-1\right\}  \right\}  . \label{eq.prop.C(sigma).Dalpha}%
\end{align}
(Here, the set $\left\{  1,2,\ldots,q\right\}  $ is to be understood as the
empty set whenever $q<1$. In particular, $\left\{  1,2,\ldots,0-1\right\}
=\varnothing$. Thus, if $\alpha$ is the empty composition $\left(  {}\right)
$, then $D\left(  \alpha\right)  =\varnothing$.)

\textbf{(a)} For every composition $\alpha=\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  $, we have%
\begin{equation}
D\left(  \alpha\right)  \subseteq\left\{  1,2,\ldots,\left\vert \alpha
\right\vert -1\right\}  , \label{eq.prop.C(sigma).a.0}%
\end{equation}%
\begin{equation}
\left\vert D\left(  \alpha\right)  \right\vert =\max\left\{  \ell-1,0\right\}
\label{eq.prop.C(sigma).a.1}%
\end{equation}
and%
\begin{align}
&  \left(  \alpha_{1},\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha_{2}+\alpha
_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}\right) \nonumber\\
&  =\left(  \text{the increasing list of }D\left(  \alpha\right)  \right)  .
\label{eq.prop.C(sigma).a.2}%
\end{align}
(Here, if $S$ is any finite set of integers, then the \textit{increasing list}
of $S$ is defined to be the list of all elements of $S$ in increasing order
(with each element of $S$ appearing precisely once). This list is uniquely determined.)

\textbf{(b)} For every $n\in\mathbb{N}$ and every subset $I$ of $\left\{
1,2,\ldots,n-1\right\}  $, there exists a unique composition $\alpha$ of $n$
such that $D\left(  \alpha\right)  =I$.

\textbf{(c)} Let $n\in\mathbb{N}$. Let $\sigma\in S_{n}$ be a permutation. Let
$\operatorname*{Des}\sigma$ denote the subset%
\[
\left\{  i\in\left\{  1,2,\ldots,n-1\right\}  \ \mid\ \sigma\left(  i\right)
>\sigma\left(  i+1\right)  \right\}
\]
of $\left\{  1,2,\ldots,n-1\right\}  $. Proposition \ref{prop.C(sigma)}
\textbf{(b)} (applied to $I=\operatorname*{Des}\sigma$) yields that there
exists a unique composition $\alpha$ of $n$ such that $D\left(  \alpha\right)
=\operatorname*{Des}\sigma$. Consider this $\alpha$. Then, $\alpha=C\left(
\sigma\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.C(sigma)}.]\textbf{(a)} Let $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $. Then, $\alpha_{1}$,
$\alpha_{2}$, $\ldots$, $\alpha_{\ell}$ are positive integers (since $\alpha$
is a composition). Now, every $p\in\left\{  1,2,\ldots,\ell-1\right\}  $
satisfies $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}\in\left\{  1,2,\ldots
,\left\vert \alpha\right\vert -1\right\}  $\ \ \ \ \footnote{\textit{Proof.}
Let $p\in\left\{  1,2,\ldots,\ell-1\right\}  $. Then, $p\geq1$ and $p\leq
\ell-1$, so that $p\geq1>0$ and $\ell-\underbrace{p}_{\leq\ell-1<\ell}%
>\ell-\ell=0$. Now, $\alpha_{1}$, $\alpha_{2}$, $\ldots$, $\alpha_{p}$ are
positive integers (since $\alpha_{1}$, $\alpha_{2}$, $\ldots$, $\alpha_{\ell}$
are positive integers). Thus, $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}$ is a
nonempty sum of positive integers (in fact, it is nonempty because the number
of its addends is $p>0$), and thus a positive integer. Also, $\alpha_{p+1}$,
$\alpha_{p+2}$, $\ldots$, $\alpha_{\ell}$ are positive integers (since
$\alpha_{1}$, $\alpha_{2}$, $\ldots$, $\alpha_{\ell}$ are positive integers).
Thus, $\alpha_{p+1}+\alpha_{p+2}+\cdots+\alpha_{\ell}$ is a nonempty sum of
positive integers (in fact, it is nonempty because the number of its addends
is $\ell-p>0$), and thus a positive integer. Hence, $\alpha_{p+1}+\alpha
_{p+2}+\cdots+\alpha_{\ell}>0$.
\par
Now, the definition of $\left\vert \alpha\right\vert $ yields%
\[
\left\vert \alpha\right\vert =\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
}=\left(  \alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}\right)  +\underbrace{\left(
\alpha_{p+1}+\alpha_{p+2}+\cdots+\alpha_{\ell}\right)  }_{>0}>\alpha
_{1}+\alpha_{2}+\cdots+\alpha_{p}.
\]
Hence, $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}<\left\vert \alpha\right\vert
$. Since $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}$ is a positive integer, this
shows that $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}$ is a positive integer
smaller than $\left\vert \alpha\right\vert $. Thus, $\alpha_{1}+\alpha
_{2}+\cdots+\alpha_{p}\in\left\{  1,2,\ldots,\left\vert \alpha\right\vert
-1\right\}  $, qed.}. Hence,%
\[
\left\{  \alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}\ \mid\ p\in\left\{
1,2,\ldots,\ell-1\right\}  \right\}  \subseteq\left\{  1,2,\ldots,\left\vert
\alpha\right\vert -1\right\}  .
\]
Thus, (\ref{eq.prop.C(sigma).Dalpha}) becomes%
\[
D\left(  \alpha\right)  =\left\{  \alpha_{1}+\alpha_{2}+\cdots+\alpha
_{p}\ \mid\ p\in\left\{  1,2,\ldots,\ell-1\right\}  \right\}  \subseteq
\left\{  1,2,\ldots,\left\vert \alpha\right\vert -1\right\}  .
\]
This proves (\ref{eq.prop.C(sigma).a.0}).

Next, let us prove (\ref{eq.prop.C(sigma).a.2}). Indeed, every $i\in\left\{
1,2,\ldots,\ell-2\right\}  $ satisfies $\alpha_{1}+\alpha_{2}+\cdots
+\alpha_{i}<\alpha_{1}+\alpha_{2}+\cdots+\alpha_{i+1}$%
\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,\ell-2\right\}
$. Then, $\alpha_{i+1}$ is a positive integer (since $\alpha_{1}$, $\alpha
_{2}$, $\ldots$, $\alpha_{\ell}$ are positive integers), so that $\alpha
_{i+1}>0$. Now, $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{i+1}=\left(  \alpha
_{1}+\alpha_{2}+\cdots+\alpha_{i}\right)  +\underbrace{\alpha_{i+1}}%
_{>0}>\alpha_{1}+\alpha_{2}+\cdots+\alpha_{i}$. In other words, $\alpha
_{1}+\alpha_{2}+\cdots+\alpha_{i}<\alpha_{1}+\alpha_{2}+\cdots+\alpha_{i+1}$,
qed.}. In other words,%
\[
\alpha_{1}<\alpha_{1}+\alpha_{2}<\alpha_{1}+\alpha_{2}+\alpha_{3}%
<\cdots<\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}.
\]
Thus, the list $\left(  \alpha_{1},\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha
_{2}+\alpha_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}\right)  $
is strictly increasing. Hence, all entries of this list are pairwise distinct.

But $D\left(  \alpha\right)  =\left\{  \alpha_{1},\alpha_{1}+\alpha_{2}%
,\alpha_{1}+\alpha_{2}+\alpha_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots
+\alpha_{\ell-1}\right\}  $. Therefore, the list $\left(  \alpha_{1}%
,\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha_{2}+\alpha_{3},\ldots,\alpha
_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}\right)  $ is a list of all elements of
$D\left(  \alpha\right)  $. Thus, this list is a list of all elements of
$D\left(  \alpha\right)  $ in increasing order (since it is strictly
increasing). Moreover, each element of $D\left(  \alpha\right)  $ appears
precisely once in this list (since each element of $D\left(  \alpha\right)  $
must appear at least once in this list (because this list is a list of all
elements of $D\left(  \alpha\right)  $) and must appear at most once in this
list (since all entries of this list are pairwise distinct)). Hence, the list
$\left(  \alpha_{1},\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha_{2}+\alpha
_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}\right)  $ is the list
of all elements of $D\left(  \alpha\right)  $ in increasing order (with each
element of $D\left(  \alpha\right)  $ appearing precisely once). In other
words, the list $\left(  \alpha_{1},\alpha_{1}+\alpha_{2},\alpha_{1}%
+\alpha_{2}+\alpha_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1}\right)  $ is the increasing list of $D\left(  \alpha\right)  $ (since the
increasing list of $D\left(  \alpha\right)  $ is the list of all elements of
$D\left(  \alpha\right)  $ in increasing order (with each element of $D\left(
\alpha\right)  $ appearing precisely once) (due to the definition of
\textquotedblleft increasing list\textquotedblright)). This proves
(\ref{eq.prop.C(sigma).a.2}).

It is known that every finite subset $S$ of $\mathbb{Z}$ satisfies%
\[
\left\vert S\right\vert =\left(  \text{the length of the increasing list of
}S\right)  .
\]
Applying this to $S=D\left(  \alpha\right)  $, we obtain%
\begin{align*}
\left\vert D\left(  \alpha\right)  \right\vert  &  =\left(  \text{the length
of }\underbrace{\text{the increasing list of }D\left(  \alpha\right)
}_{\substack{=\left(  \alpha_{1},\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha
_{2}+\alpha_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}\right)
\\\text{(by (\ref{eq.prop.C(sigma).a.2}))}}}\right) \\
&  =\left(  \text{the length of the list }\left(  \alpha_{1},\alpha_{1}%
+\alpha_{2},\alpha_{1}+\alpha_{2}+\alpha_{3},\ldots,\alpha_{1}+\alpha
_{2}+\cdots+\alpha_{\ell-1}\right)  \right) \\
&  =\left\vert \left\{  1,2,\ldots,\ell-1\right\}  \right\vert =\max\left\{
\ell-1,0\right\}  .
\end{align*}
This proves (\ref{eq.prop.C(sigma).a.1}). Thus, Proposition
\ref{prop.C(sigma)} \textbf{(a)} is proven.

\textbf{(b)} Let $n\in\mathbb{N}$. Let $I$ be a subset of $\left\{
1,2,\ldots,n-1\right\}  $.

If $n=0$, then Proposition \ref{prop.C(sigma)} \textbf{(b)} is easy to
prove\footnote{\textit{Proof.} Assume that $n=0$. Then, $I\subseteq\left\{
1,2,\ldots,n-1\right\}  =\varnothing$, so that $I=\varnothing$. Now, there
exists a composition $\alpha$ of $0$ such that $D\left(  \alpha\right)
=\varnothing$ (namely, $\alpha=\left(  {}\right)  $). Also, there exists at
most one composition $\alpha$ of $0$ such that $D\left(  \alpha\right)
=\varnothing$ (in fact, there exists at most one composition $\alpha$ of $0$
(namely, $\left(  {}\right)  $)). Hence, there exists a unique composition
$\alpha$ of $0$ such that $D\left(  \alpha\right)  =\varnothing$ (since there
exists a composition $\alpha$ of $0$ such that $D\left(  \alpha\right)
=\varnothing$). In other words, there exists a unique composition $\alpha$ of
$n$ such that $D\left(  \alpha\right)  =I$ (since $n=0$ and $I=\varnothing$).
In other words, Proposition \ref{prop.C(sigma)} \textbf{(b)} is proven.}.
Hence, for the rest of this proof, we can WLOG assume that we don't have
$n=0$. Assume this.

We have $n\neq0$ (since we don't have $n=0$) and thus $n>0$ (since
$n\in\mathbb{N}$).

Let $\left(  i_{1},i_{2},\ldots,i_{k}\right)  $ be the increasing list of the
set $I$. Then, $\left(  i_{1},i_{2},\ldots,i_{k}\right)  $ is the list of all
elements of $I$ in increasing order (with each element of $I$ appearing
precisely once) (according to the definition of the \textquotedblleft
increasing list\textquotedblright). As a consequence, we have $I=\left\{
i_{1},i_{2},\ldots,i_{k}\right\}  $ and $i_{1}<i_{2}<\cdots<i_{k}$. Let us
extend this list $\left(  i_{1},i_{2},\ldots,i_{k}\right)  $ to a list
$\left(  i_{0},i_{1},i_{2},\ldots,i_{k},i_{k+1}\right)  $ by defining two
additional elements $i_{0}$ and $i_{k+1}$ by $i_{0}=0$ and $i_{k+1}=n$. Notice
that $i_{0}<i_{1}<i_{2}<\cdots<i_{k}<i_{k+1}$\ \ \ \ \footnote{\textit{Proof.}
Let us first assume that $k\neq0$. Then, $k\geq1$ (since $k\in\mathbb{N}$),
and thus the integers $i_{1}$ and $i_{k}$ are well-defined. Now, $i_{1}%
\in\left\{  i_{1},i_{2},\ldots,i_{k}\right\}  =I\subseteq\left\{
1,2,\ldots,n-1\right\}  $, so that $i_{1}\geq1$ and thus $i_{1}>0$. Hence,
$i_{0}=0<i_{1}$. Furthermore, $k\geq1$, so that $i_{k}\in\left\{  i_{1}%
,i_{2},\ldots,i_{k}\right\}  =I\subseteq\left\{  1,2,\ldots,n-1\right\}  $,
thus $i_{k}\leq n-1<n=i_{k+1}$.
\par
Combining $i_{0}<i_{1}$ with $i_{1}<i_{2}<\cdots<i_{k}$, we obtain
$i_{0}<i_{1}<i_{2}<\cdots<i_{k}$. Combining this with $i_{k}<i_{k+1}$, we
obtain $i_{0}<i_{1}<i_{2}<\cdots<i_{k}<i_{k+1}$.
\par
Let us now forget that we assumed that $k\neq0$. We thus have proven
$i_{0}<i_{1}<i_{2}<\cdots<i_{k}<i_{k+1}$ under the assumption that $k\neq0$.
Thus, for the rest of this proof, we can WLOG assume that we don't have
$k\neq0$.
\par
We have $k=0$ (since we don't have $k\neq0$). Thus, $i_{k+1}=i_{0+1}=i_{1}$,
so that $i_{1}=i_{k+1}=n>0=i_{0}$, so that $i_{0}<i_{1}$. This rewrites as
$i_{0}<i_{1}<i_{2}<\cdots<i_{k}<i_{k+1}$ (since $k=0$). Thus, $i_{0}%
<i_{1}<i_{2}<\cdots<i_{k}<i_{k+1}$ is proven.}. Thus, $i_{p-1}<i_{p}$ for
every $p\in\left\{  1,2,\ldots,k+1\right\}  $. In other words, $i_{p}%
-i_{p-1}>0$ for every $p\in\left\{  1,2,\ldots,k+1\right\}  $. Hence,
$i_{p}-i_{p-1}$ is a positive integer for every $p\in\left\{  1,2,\ldots
,k+1\right\}  $. In other words, $i_{1}-i_{0}$, $i_{2}-i_{1}$, $\ldots$,
$i_{k+1}-i_{k}$ are positive integers. Hence, $\left(  i_{1}-i_{0},i_{2}%
-i_{1},\ldots,i_{k+1}-i_{k}\right)  $ is a composition. Let us denote this
composition by $\tau$. We then have $\tau=\left(  i_{1}-i_{0},i_{2}%
-i_{1},\ldots,i_{k+1}-i_{k}\right)  $, so that $\tau$ is a composition of
$n$\ \ \ \ \footnote{\textit{Proof.} From $\tau=\left(  i_{1}-i_{0}%
,i_{2}-i_{1},\ldots,i_{k+1}-i_{k}\right)  $, we obtain%
\begin{align*}
\left\vert \tau\right\vert  &  =\left\vert \left(  i_{1}-i_{0},i_{2}%
-i_{1},\ldots,i_{k+1}-i_{k}\right)  \right\vert =\left(  i_{1}-i_{0}\right)
+\left(  i_{2}-i_{1}\right)  +\cdots+\left(  i_{k+1}-i_{k}\right)  =\sum
_{p=1}^{k+1}\left(  i_{p}-i_{p-1}\right) \\
&  =\sum_{p=1}^{k+1}i_{p}-\sum_{p=1}^{k+1}i_{p-1}=\underbrace{\sum_{p=1}%
^{k+1}i_{p}}_{=\sum_{p=1}^{k}i_{p}+i_{k+1}}-\underbrace{\sum_{p=0}^{k}i_{p}%
}_{=i_{0}+\sum_{p=1}^{k}i_{p}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }p\text{ for
}p-1\text{ in the second sum}\right) \\
&  =\left(  \sum_{p=1}^{k}i_{p}+i_{k+1}\right)  -\left(  i_{0}+\sum_{p=1}%
^{k}i_{p}\right)  =i_{k+1}-\underbrace{i_{0}}_{=0}=i_{k+1}=n.
\end{align*}
Thus, $\tau$ is a composition of $n$, qed.}. Moreover, $D\left(  \tau\right)
=I$\ \ \ \ \footnote{\textit{Proof.} Every $p\in\left\{  1,2,\ldots,k\right\}
$ satisfies%
\begin{align}
&  \left(  i_{1}-i_{0}\right)  +\left(  i_{2}-i_{1}\right)  +\cdots+\left(
i_{p}-i_{p-1}\right) \nonumber\\
&  =\sum_{q=1}^{p}\left(  i_{q}-i_{q-1}\right)  =\sum_{q=1}^{p}i_{q}%
-\sum_{q=1}^{p}i_{q-1}=\underbrace{\sum_{q=1}^{p}i_{q}}_{=\sum_{q=1}%
^{p-1}i_{q}+i_{p}}-\underbrace{\sum_{q=0}^{p-1}i_{q}}_{=i_{0}+\sum_{q=1}%
^{p-1}i_{q}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }q\text{ for
}q-1\text{ in the second sum}\right) \nonumber\\
&  =\left(  \sum_{q=1}^{p-1}i_{q}+i_{p}\right)  +\left(  i_{0}+\sum
_{q=1}^{p-1}i_{q}\right)  =i_{p}-\underbrace{i_{0}}_{=0}=i_{p}.
\label{pf.prop.C(sigma).b.3}%
\end{align}
\par
Now, from (\ref{eq.prop.C(sigma).Dalpha}) (applied to $\tau$, $k+1$ and
$\left(  i_{1}-i_{0},i_{2}-i_{1},\ldots,i_{k+1}-i_{k}\right)  $ instead of
$\alpha$, $\ell$ and $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  $), we obtain%
\begin{align*}
D\left(  \tau\right)   &  =\left\{  \left(  i_{1}-i_{0}\right)  +\left(
i_{2}-i_{1}\right)  +\cdots+\left(  i_{p}-i_{p-1}\right)  \ \mid\ p\in\left\{
1,2,\ldots,\underbrace{\left(  k+1\right)  -1}_{=k}\right\}  \right\} \\
&  =\left\{  \underbrace{\left(  i_{1}-i_{0}\right)  +\left(  i_{2}%
-i_{1}\right)  +\cdots+\left(  i_{p}-i_{p-1}\right)  }_{\substack{=i_{p}%
\\\text{(by (\ref{pf.prop.C(sigma).b.3}))}}}\ \mid\ p\in\left\{
1,2,\ldots,k\right\}  \right\} \\
&  =\left\{  i_{p}\ \mid\ p\in\left\{  1,2,\ldots,k\right\}  \right\}
=\left\{  i_{1},i_{2},\ldots,i_{k}\right\}  =I,
\end{align*}
qed.}. Hence, $\tau$ is a composition of $n$ satisfying $D\left(  \tau\right)
=I$. In other words, $\tau$ is a composition $\alpha$ of $n$ such that
$D\left(  \alpha\right)  =I$.

On the other hand, there exists at most one composition $\alpha$ of $n$ such
that $D\left(  \alpha\right)  =I$.\ \ \ \ \footnote{\textit{Proof.} Let
$\beta$ and $\gamma$ be two compositions $\alpha$ of $n$ such that $D\left(
\alpha\right)  =I$. We shall show that $\beta=\gamma$.
\par
We know that $\beta$ is a composition $\alpha$ of $n$ such that $D\left(
\alpha\right)  =I$. In other words, $\beta$ is a composition of $n$ and
satisfies $D\left(  \beta\right)  =I$. Let us write the composition $\beta$ in
the form $\left(  \beta_{1},\beta_{2},\ldots,\beta_{\ell}\right)  $. Then,
$\left\vert \beta\right\vert =n$ (since $\beta$ is a composition of $n$).
Since $\left\vert \beta\right\vert =\beta_{1}+\beta_{2}+\cdots+\beta_{\ell}$
(by the definition of $\left\vert \beta\right\vert $), this rewrites as
$\beta_{1}+\beta_{2}+\cdots+\beta_{\ell}=n$. If $\ell=0$, then $\beta
_{1}+\beta_{2}+\cdots+\beta_{\ell}=\left(  \text{empty sum}\right)  =0$, which
contradicts $\beta_{1}+\beta_{2}+\cdots+\beta_{\ell}=n\neq0$. Hence, we cannot
have $\ell=0$. Thus, $\ell>0$ (since $\ell\in\mathbb{N}$).
\par
The equality (\ref{eq.prop.C(sigma).a.2}) (applied to $\beta$ and $\left(
\beta_{1},\beta_{2},\ldots,\beta_{\ell}\right)  $ instead of $\alpha$ and
$\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $) yields%
\begin{align*}
&  \left(  \beta_{1},\beta_{1}+\beta_{2},\beta_{1}+\beta_{2}+\beta_{3}%
,\ldots,\beta_{1}+\beta_{2}+\cdots+\beta_{\ell-1}\right) \\
&  =\left(  \text{the increasing list of }\underbrace{D\left(  \beta\right)
}_{=I}\right)  =\left(  \text{the increasing list of }I\right) \\
&  =\left(  i_{1},i_{2},\ldots,i_{k}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  i_{1},i_{2},\ldots,i_{k}\right)  \text{ is the increasing
list of }I\right)  .
\end{align*}
Hence,%
\begin{align}
&  \left(  \underbrace{0}_{=i_{0}},\underbrace{\beta_{1},\beta_{1}+\beta
_{2},\beta_{1}+\beta_{2}+\beta_{3},\ldots,\beta_{1}+\beta_{2}+\cdots
+\beta_{\ell-1}}_{=\left(  i_{1},i_{2},\ldots,i_{k}\right)  }%
,\underbrace{\beta_{1}+\beta_{2}+\cdots+\beta_{\ell}}_{=n=i_{k+1}}\right)
\nonumber\\
&  =\left(  i_{0},i_{1},i_{2},\ldots,i_{k},i_{k+1}\right)  .
\label{pf.prop.C(sigma).i0ik}%
\end{align}
Therefore,%
\begin{align*}
&  \left(  \text{the length of the list }\underbrace{\left(  0,\beta_{1}%
,\beta_{1}+\beta_{2},\beta_{1}+\beta_{2}+\beta_{3},\ldots,\beta_{1}+\beta
_{2}+\cdots+\beta_{\ell-1},\beta_{1}+\beta_{2}+\cdots+\beta_{\ell}\right)
}_{=\left(  i_{0},i_{1},i_{2},\ldots,i_{k},i_{k+1}\right)  }\right) \\
&  =\left(  \text{the length of the list }\left(  i_{0},i_{1},i_{2}%
,\ldots,i_{k},i_{k+1}\right)  \right)  =k+2,
\end{align*}
so that%
\begin{align*}
&  k+2\\
&  =\left(  \text{the length of the list }\left(  0,\beta_{1},\beta_{1}%
+\beta_{2},\beta_{1}+\beta_{2}+\beta_{3},\ldots,\beta_{1}+\beta_{2}%
+\cdots+\beta_{\ell-1},\beta_{1}+\beta_{2}+\cdots+\beta_{\ell}\right)  \right)
\\
&  =\ell+1
\end{align*}
and therefore $k=\ell-1$. Now, (\ref{pf.prop.C(sigma).i0ik}) yields%
\begin{equation}
\beta_{1}+\beta_{2}+\cdots+\beta_{p}=i_{p}\ \ \ \ \ \ \ \ \ \ \text{for every
}p\in\left\{  0,1,\ldots,\ell\right\}  . \label{pf.prop.C(sigma).i0ik2}%
\end{equation}
\par
Now, let us fix $p\in\left\{  1,2,\ldots,\ell\right\}  $. Then,
(\ref{pf.prop.C(sigma).i0ik2}) yields $\beta_{1}+\beta_{2}+\cdots+\beta
_{p}=i_{p}$. Meanwhile, (\ref{pf.prop.C(sigma).i0ik2}) (applied to $p-1$
instead of $p$) yields $\beta_{1}+\beta_{2}+\cdots+\beta_{p-1}=i_{p-1}$. Now,
$\underbrace{\left(  \beta_{1}+\beta_{2}+\cdots+\beta_{p}\right)  }_{=i_{p}%
}-\underbrace{\left(  \beta_{1}+\beta_{2}+\cdots+\beta_{p-1}\right)
}_{=i_{p-1}}=i_{p}-i_{p-1}$, so that%
\[
i_{p}-i_{p-1}=\left(  \beta_{1}+\beta_{2}+\cdots+\beta_{p}\right)  -\left(
\beta_{1}+\beta_{2}+\cdots+\beta_{p-1}\right)  =\beta_{p}.
\]
Thus, $\beta_{p}=i_{p}-i_{p-1}$.
\par
Let us now forget that we fixed $p$. We thus have shown that $\beta_{p}%
=i_{p}-i_{p-1}$ for every $p\in\left\{  1,2,\ldots,\ell\right\}  $. Thus,
\begin{align*}
\left(  \beta_{1},\beta_{2},\ldots,\beta_{\ell}\right)   &  =\left(
i_{1}-i_{0},i_{2}-i_{1},\ldots,i_{\ell}-i_{\ell-1}\right)  =\left(
i_{1}-i_{0},i_{2}-i_{1},\ldots,i_{k+1}-\underbrace{i_{\left(  k+1\right)  -1}%
}_{=i_{k}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell=k+1\text{ (since }%
k=\ell-1\text{)}\right) \\
&  =\left(  i_{1}-i_{0},i_{2}-i_{1},\ldots,i_{k+1}-i_{k}\right)  =\tau,
\end{align*}
so that $\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{\ell}\right)  =\tau$.
\par
We thus have shown that $\beta=\tau$. The same argument (but applied to
$\gamma$ instead of $\beta$) shows that $\gamma=\tau$. Thus, $\beta=\gamma$.
\par
Let us now forget that we fixed $\beta$ and $\gamma$. We thus have shown that
if $\beta$ and $\gamma$ are two compositions $\alpha$ of $n$ such that
$D\left(  \alpha\right)  =I$, then $\beta=\gamma$. In other words, there
exists at most one composition $\alpha$ of $n$ such that $D\left(
\alpha\right)  =I$. This completes the proof.} Since we know that such an
$\alpha$ exists (namely, $\tau$), we can thus conclude that there exists a
unique composition $\alpha$ of $n$ such that $D\left(  \alpha\right)  =I$.
This proves Proposition \ref{prop.C(sigma)} \textbf{(b)}.

\textbf{(c)} We know that $\alpha$ is the unique composition of $n$ such that
$D\left(  \alpha\right)  =\operatorname*{Des}\sigma$. Thus, $\alpha$ is a
composition of $n$ and satisfies $D\left(  \alpha\right)  =\operatorname*{Des}%
\sigma$.

Let us write the composition $\alpha$ in the form $\alpha=\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $. Then, $\left\vert
\alpha\right\vert =\left\vert \left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right)  \right\vert =\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
}$, so that $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}=\left\vert
\alpha\right\vert =n$ (since $\alpha$ is a composition of $n$).

From (\ref{eq.prop.C(sigma).a.2}), we obtain%
\begin{align*}
&  \left(  \alpha_{1},\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha_{2}+\alpha
_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}\right) \\
&  =\left(  \text{the increasing list of }\underbrace{D\left(  \alpha\right)
}_{=\operatorname*{Des}\sigma}\right)  =\left(  \text{the increasing list of
}\operatorname*{Des}\sigma\right)  .
\end{align*}


Let us first prove that the word $\sigma=\left(  \sigma\left(  1\right)
,\sigma\left(  2\right)  ,\ldots,\sigma\left(  n\right)  \right)  $ can be
split into $\ell$ consecutive ascending runs of lengths $\alpha_{1}$,
$\alpha_{2}$, $\ldots$, $\alpha_{\ell}$.

For every $p\in\left\{  0,1,\ldots,\ell\right\}  $, let $i_{p}$ denote the
nonnegative integer $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}$. Then, the
definition of $i_{0}$ yields $i_{0}=\alpha_{1}+\alpha_{2}+\cdots+\alpha
_{0}=\left(  \text{empty sum}\right)  =0$. Also, the definition of $i_{\ell}$
yields $i_{\ell}=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}=n$. Every
$p\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies%
\begin{equation}
i_{p}-i_{p-1}=\alpha_{p}. \label{pf.prop.C(sigma).c.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.C(sigma).c.1}):} Let $p\in\left\{
1,2,\ldots,\ell\right\}  $. Then, the definition of $i_{p}$ yields
$i_{p}=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}$. But the definition of
$i_{p-1}$ yields $i_{p-1}=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p-1}$. Hence,%
\[
i_{p}-i_{p-1}=\left(  \alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}\right)  -\left(
\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p-1}\right)  =\alpha_{p}.
\]
This proves (\ref{pf.prop.C(sigma).c.1}).} Moreover, $i_{0}<i_{1}%
<\cdots<i_{\ell}$\ \ \ \ \footnote{\textit{Proof.} Let $p\in\left\{
1,2,\ldots,\ell\right\}  $. Then, $\alpha_{p}$ is a positive integer (since
$\alpha_{1}$, $\alpha_{2}$, $\ldots$, $\alpha_{\ell}$ are positive integers
(since $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ is a
composition)). Now, (\ref{pf.prop.C(sigma).c.1}) yields $i_{p}-i_{p-1}%
=\alpha_{p}>0$ (since $\alpha_{p}$ is a positive integer). In other words,
$i_{p-1}<i_{p}$.
\par
Let us now forget that we fixed $p$. We thus have proven that $i_{p-1}<i_{p}$
for every $p\in\left\{  1,2,\ldots,\ell\right\}  $. In other words,
$i_{0}<i_{1}<\cdots<i_{\ell}$, qed.}.

But for every $q\in\left\{  1,2,\ldots,\ell\right\}  $, the list $\left(
\sigma\left(  i_{q-1}+1\right)  ,\sigma\left(  i_{q-1}+2\right)
,\ldots,\sigma\left(  i_{q}\right)  \right)  $ is strictly
increasing\footnote{\textit{Proof.} Let $q\in\left\{  1,2,\ldots,\ell\right\}
$. Since $q\leq\ell$ and $i_{0}<i_{1}<\cdots<i_{\ell}$, we have $i_{q}\leq
i_{\ell}=n$. Because of this, and because of $\underbrace{i_{q-1}}_{\geq
0}+1\geq1$, it is clear that each of the integers $i_{q-1}+1$, $i_{q-1}+2$,
$\ldots$, $i_{q}$ belongs to $\left\{  1,2,\ldots,n\right\}  $. Hence, the
list $\left(  \sigma\left(  i_{q-1}+1\right)  ,\sigma\left(  i_{q-1}+2\right)
,\ldots,\sigma\left(  i_{q}\right)  \right)  $ is well-defined.
\par
Now, let $g\in\left\{  1,2,\ldots,i_{q}-i_{q-1}-1\right\}  $. We shall now
show that $\sigma\left(  i_{q-1}+g\right)  <\sigma\left(  i_{q-1}+\left(
g+1\right)  \right)  $.
\par
Indeed, assume the contrary. Then, $\sigma\left(  i_{q-1}+g\right)  \geq
\sigma\left(  i_{q-1}+\left(  g+1\right)  \right)  $. Let $h=i_{q-1}+g$. Then,%
\[
\sigma\left(  \underbrace{h}_{=i_{q-1}+g}\right)  =\sigma\left(
i_{q-1}+g\right)  \geq\sigma\left(  i_{q-1}+\left(  g+1\right)  \right)
=\sigma\left(  \underbrace{i_{q-1}+g}_{=h}+1\right)  =\sigma\left(
h+1\right)  .
\]
Moreover, recall that $g\in\left\{  1,2,\ldots,i_{q}-i_{q-1}-1\right\}  $.
Thus, $g\geq1$ and $g\leq i_{q}-i_{q-1}-1$. Combining $h=i_{q-1}%
+\underbrace{g}_{\geq1}\geq\underbrace{i_{q-1}}_{\geq0}+1\geq1$ with
$h=i_{q-1}+\underbrace{g}_{\leq i_{q}-i_{q-1}-1}\leq i_{q-1}+\left(
i_{q}-i_{q-1}-1\right)  =\underbrace{i_{q}}_{\leq n}-1\leq n-1$, we obtain
$1\leq h\leq n-1$, so that $h\in\left\{  1,2,\ldots,n-1\right\}  $.
\par
The map $\sigma$ is a permutation, thus bijective, thus injective. Hence,
$\sigma\left(  u\right)  \neq\sigma\left(  v\right)  $ for any two distinct
elements $u$ and $v$ of $\left\{  1,2,\ldots,n\right\}  $. Applying this to
$u=h$ and $v=h+1$, we obtain $\sigma\left(  h\right)  \neq\sigma\left(
h+1\right)  $. Combined with $\sigma\left(  h\right)  \geq\sigma\left(
h+1\right)  $, we obtain $\sigma\left(  h\right)  >\sigma\left(  h+1\right)
$. Now, we know that $h$ is an element of $\left\{  1,2,\ldots,n-1\right\}  $
satisfying $\sigma\left(  h\right)  >\sigma\left(  h+1\right)  $. In other
words,%
\begin{align*}
h  &  \in\left\{  i\in\left\{  1,2,\ldots,n-1\right\}  \ \mid\ \sigma\left(
i\right)  >\sigma\left(  i+1\right)  \right\}  =\operatorname*{Des}\sigma\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\operatorname*{Des}\sigma=\left\{  i\in\left\{  1,2,\ldots
,n-1\right\}  \ \mid\ \sigma\left(  i\right)  >\sigma\left(  i+1\right)
\right\} \\
\text{(by the definition of }\operatorname*{Des}\sigma\text{)}%
\end{array}
\right) \\
&  =D\left(  \alpha\right)  =\left\{  \alpha_{1}+\alpha_{2}+\cdots+\alpha
_{p}\ \mid\ p\in\left\{  1,2,\ldots,\ell-1\right\}  \right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.prop.C(sigma).Dalpha})}\right)
.
\end{align*}
In other words, there exists some $p\in\left\{  1,2,\ldots,\ell-1\right\}  $
such that $h=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}$. Let us consider this
$p$.
\par
The definition of $i_{p}$ yields $i_{p}=\alpha_{1}+\alpha_{2}+\cdots
+\alpha_{p}=h$ (since $h=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{p}$).
\par
Now, let us assume (for the sake of contradiction) that $p\leq q-1$. Then,
$i_{p}\leq i_{q-1}$ (since $i_{0}<i_{1}<\cdots<i_{\ell}$); but this
contradicts $i_{p}=h\geq i_{q-1}+1>i_{q-1}$. This contradiction shows that our
assumption (that $p\leq q-1$) was wrong. Hence, we cannot have $p\leq q-1$. We
thus must have $p>q-1$. Since $p$ and $q-1$ are integers, this yields
$p\geq\left(  q-1\right)  +1=q$. Thus, $q\leq p$, so that $i_{q}\leq i_{p}$
(since $i_{0}<i_{1}<\cdots<i_{\ell}$), so that $i_{q}\leq i_{p}=h\leq
i_{q}-1<i_{q}$. But this is absurd. This contradiction shows that our
assumption was wrong. Hence, $\sigma\left(  i_{q-1}+g\right)  <\sigma\left(
i_{q-1}+\left(  g+1\right)  \right)  $ is proven.
\par
Let us now forget that we fixed $g$. Thus we have proven that $\sigma\left(
i_{q-1}+g\right)  <\sigma\left(  i_{q-1}+\left(  g+1\right)  \right)  $ for
every $g\in\left\{  1,2,\ldots,i_{q}-i_{q-1}-1\right\}  $. In other words, the
list $\left(  \sigma\left(  i_{q-1}+1\right)  ,\sigma\left(  i_{q-1}+2\right)
,\ldots,\sigma\left(  i_{q}\right)  \right)  $ is strictly increasing. Qed.}.
Thus, for every $q\in\left\{  1,2,\ldots,\ell\right\}  $, the list
\newline$\left(  \sigma\left(  i_{q-1}+1\right)  ,\sigma\left(  i_{q-1}%
+2\right)  ,\ldots,\sigma\left(  i_{q}\right)  \right)  $ is an ascending run
of the word $\sigma$. This ascending run has length $\alpha_{q}$%
\ \ \ \ \footnote{\textit{Proof.} Fix $q\in\left\{  1,2,\ldots,\ell\right\}
$. We need to show that the list $\left(  \sigma\left(  i_{q-1}+1\right)
,\sigma\left(  i_{q-1}+2\right)  ,\ldots,\sigma\left(  i_{q}\right)  \right)
$ has length $\alpha_{q}$.
\par
But we have $i_{q-1}<i_{q}$ (since $i_{0}<i_{1}<\cdots<i_{\ell}$). Thus, the
list $\left(  \sigma\left(  i_{q-1}+1\right)  ,\sigma\left(  i_{q-1}+2\right)
,\ldots,\sigma\left(  i_{q}\right)  \right)  $ has length $i_{q}%
-i_{q-1}=\alpha_{q}$ (by (\ref{pf.prop.C(sigma).c.1}), applied to $p=q$). This
completes our proof.}. Hence, for every $q\in\left\{  1,2,\ldots,\ell\right\}
$, the list $\left(  \sigma\left(  i_{q-1}+1\right)  ,\sigma\left(
i_{q-1}+2\right)  ,\ldots,\sigma\left(  i_{q}\right)  \right)  $ is an
ascending run of the word $\sigma$ having length $\alpha_{q}$. Thus, for $q$
ranging over the set $\left\{  1,2,\ldots,\ell\right\}  $, these lists
\newline$\left(  \sigma\left(  i_{q-1}+1\right)  ,\sigma\left(  i_{q-1}%
+2\right)  ,\ldots,\sigma\left(  i_{q}\right)  \right)  $ are $\ell$ ascending
runs of the word $\sigma$ having lengths $\alpha_{1}$, $\alpha_{2}$, $\ldots$,
$\alpha_{\ell}$. These $\ell$ ascending runs are clearly consecutive, and
cover the whole word $\sigma$ (since $i_{0}=0$ and $i_{\ell}=n$). Hence, the
word $\sigma$ can be split into $\ell$ consecutive ascending runs of lengths
$\alpha_{1}$, $\alpha_{2}$, $\ldots$, $\alpha_{\ell}$ (namely, into the
ascending runs $\left(  \sigma\left(  i_{q-1}+1\right)  ,\sigma\left(
i_{q-1}+2\right)  ,\ldots,\sigma\left(  i_{q}\right)  \right)  $ for
$q\in\left\{  1,2,\ldots,\ell\right\}  $). In particular, the word $\sigma$
has $\ell$ consecutive ascending runs of lengths $\alpha_{1}$, $\alpha_{2}$,
$\ldots$, $\alpha_{\ell}$.

Let now $\left(  c_{1},c_{2},\ldots,c_{k}\right)  $ be any composition of $n$
such that the word $\sigma$ has $k$ consecutive ascending runs of lengths
$c_{1}$, $c_{2}$, $\ldots$, $c_{k}$, and such that $k\leq\ell$. We shall prove
that $\left(  c_{1},c_{2},\ldots,c_{k}\right)  =\alpha$.

This is obvious if $n=0$\ \ \ \ \footnote{\textit{Proof.} Assume that $n=0$.
Then, $\left(  c_{1},c_{2},\ldots,c_{k}\right)  $ is a composition of $n$. In
other words, $\left(  c_{1},c_{2},\ldots,c_{k}\right)  $ is a composition of
$0$ (since $n=0$). Hence, $\left(  c_{1},c_{2},\ldots,c_{k}\right)
=\varnothing$ (since $\varnothing$ is the only composition of $0$). On the
other hand, $\alpha$ is a composition of $n$. In other words, $\alpha$ is a
composition of $0$ (since $n=0$). Hence, $\alpha=\varnothing$ (since
$\varnothing$ is the only composition of $0$). Thus, $\left(  c_{1}%
,c_{2},\ldots,c_{k}\right)  =\varnothing=\alpha$, qed.}. Hence, for the rest
of the proof of $\left(  c_{1},c_{2},\ldots,c_{k}\right)  =\alpha$, we can
WLOG assume that we don't have $n=0$. Assume this. We have $n\geq1$ (since
$n\in\mathbb{N}$ and since we don't have $n=0$). Hence, $\ell\geq
1$\ \ \ \ \footnote{\textit{Proof.} We have $i_{\ell}=n\geq1>0=i_{0}$, so that
$i_{\ell}\neq i_{0}$ and thus $\ell\neq0$. Since $\ell\in\mathbb{N}$, this
yields $\ell\geq1$, qed.} and $k\geq1$\ \ \ \ \footnote{\textit{Proof.} Assume
the contrary. Then, $k<1$. Since $k\in\mathbb{N}$, this shows that $k=0$.
Hence, $\left\vert \left(  c_{1},c_{2},\ldots,c_{k}\right)  \right\vert
=\left\vert \underbrace{\left(  c_{1},c_{2},\ldots,c_{0}\right)  }_{=\left(
{}\right)  =\varnothing}\right\vert =\left\vert \varnothing\right\vert =0$.
Compared with $\left\vert \left(  c_{1},c_{2},\ldots,c_{k}\right)  \right\vert
=n$ (since $\left(  c_{1},c_{2},\ldots,c_{k}\right)  $ is a composition of
$n$), this yields $n=0$, so that $0=n\geq1$. But this is absurd. This
contradiction shows that our assumption was wrong, qed.}. Thus, $\ell-1\geq0$
(since $\ell\geq1$) and $k-1\geq0$ (and $k\geq1$).

We know that $\left(  c_{1},c_{2},\ldots,c_{k}\right)  $ is a composition of
$n$. Thus, $\left\vert \left(  c_{1},c_{2},\ldots,c_{k}\right)  \right\vert
=n$, so that $n=\left\vert \left(  c_{1},c_{2},\ldots,c_{k}\right)
\right\vert =c_{1}+c_{2}+\cdots+c_{k}$.

The word $\sigma$ has $k$ consecutive ascending runs of lengths $c_{1}$,
$c_{2}$, $\ldots$, $c_{k}$. In other words, there exist $k+1$ elements $j_{0}%
$, $j_{1}$, $\ldots$, $j_{k}$ of $\left\{  0,1,\ldots,n\right\}  $ satisfying
the following properties:

\begin{itemize}
\item We have $j_{0}\leq j_{1}\leq\cdots\leq j_{k}$.

\item For every $q\in\left\{  1,2,\ldots,k\right\}  $, the list $\left(
\sigma\left(  j_{q-1}+1\right)  ,\sigma\left(  j_{q-1}+2\right)
,\ldots,\sigma\left(  j_{q}\right)  \right)  $ is an ascending run of $\sigma$
having length $c_{q}$.
\end{itemize}

Consider these elements $j_{0}$, $j_{1}$, $\ldots$, $j_{k}$%
.\ \ \ \ \footnote{Visually speaking, these elements $j_{0}$, $j_{1}$,
$\ldots$, $j_{k}$ are the borders of the $k$ ascending runs.} It is easy to
see that
\begin{equation}
c_{q}=j_{q}-j_{q-1}\ \ \ \ \ \ \ \ \ \ \text{for every }q\in\left\{
1,2,\ldots,k\right\}  . \label{pf.prop.C(sigma).c.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.C(sigma).c.6}):} Let $q\in\left\{
1,2,\ldots,k\right\}  $. Then, one of the two above properties of the elements
$j_{0}$, $j_{1}$, $\ldots$, $j_{k}$ shows that the list $\left(  \sigma\left(
j_{q-1}+1\right)  ,\sigma\left(  j_{q-1}+2\right)  ,\ldots,\sigma\left(
j_{q}\right)  \right)  $ is an ascending run of $\sigma$ having length $c_{q}%
$. Hence, this list $\left(  \sigma\left(  j_{q-1}+1\right)  ,\sigma\left(
j_{q-1}+2\right)  ,\ldots,\sigma\left(  j_{q}\right)  \right)  $ has length
$c_{q}$. In other words,%
\begin{equation}
\left(  \text{the length of the list }\left(  \sigma\left(  j_{q-1}+1\right)
,\sigma\left(  j_{q-1}+2\right)  ,\ldots,\sigma\left(  j_{q}\right)  \right)
\right)  =c_{q}. \label{pf.prop.C(sigma).c.6.pf.1}%
\end{equation}
\par
But $j_{q-1}\leq j_{q}$ (since $j_{0}\leq j_{1}\leq\cdots\leq j_{k}$). Hence,
the list $\left(  \sigma\left(  j_{q-1}+1\right)  ,\sigma\left(
j_{q-1}+2\right)  ,\ldots,\sigma\left(  j_{q}\right)  \right)  $ has length
$j_{q}-j_{q-1}$. In other words,%
\[
\left(  \text{the length of the list }\left(  \sigma\left(  j_{q-1}+1\right)
,\sigma\left(  j_{q-1}+2\right)  ,\ldots,\sigma\left(  j_{q}\right)  \right)
\right)  =j_{q}-j_{q-1}.
\]
Compared with (\ref{pf.prop.C(sigma).c.6.pf.1}), this yields $c_{q}%
=j_{q}-j_{q-1}$. This proves (\ref{pf.prop.C(sigma).c.6}).} It is now easy to
see that $\left(  j_{0}=0\text{ and }j_{k}=n\right)  $%
\ \ \ \ \footnote{\textit{Proof.} We know that $j_{0}$, $j_{1}$, $\ldots$,
$j_{k}$ are elements of $\left\{  0,1,\ldots,n\right\}  $. In particular,
$j_{0}$ and $j_{k}$ are elements of $\left\{  0,1,\ldots,n\right\}  $. Since
$j_{k}$ is an element of $\left\{  0,1,\ldots,n\right\}  $, we have $j_{k}\leq
n$, so that $n\geq j_{k}$. Combined with%
\begin{align*}
n  &  =c_{1}+c_{2}+\cdots+c_{k}=\sum_{q=1}^{k}\underbrace{c_{q}}%
_{\substack{=j_{q}-j_{q-1}\\\text{(by (\ref{pf.prop.C(sigma).c.6}))}}%
}=\sum_{q=1}^{k}\left(  j_{q}-j_{q-1}\right)  =\sum_{q=1}^{k}j_{q}-\sum
_{q=1}^{k}j_{q-1}\\
&  =\underbrace{\sum_{q=1}^{k}j_{q}}_{=\sum_{q=1}^{k-1}j_{q}+j_{k}%
}-\underbrace{\sum_{q=0}^{k-1}j_{q}}_{=j_{0}+\sum_{q=1}^{k-1}j_{q}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }q\text{ for
}q-1\text{ in the second sum}\right) \\
&  =\left(  \sum_{q=1}^{k-1}j_{q}+j_{k}\right)  -\left(  j_{0}+\sum
_{q=1}^{k-1}j_{q}\right)  =j_{k}-\underbrace{j_{0}}_{\substack{\geq
0\\\text{(since }j_{0}\in\left\{  0,1,\ldots,n\right\}  \text{)}}}\leq j_{k},
\end{align*}
this yields $n=j_{k}$. Thus, $j_{k}=n$. Now, $n=\underbrace{j_{k}}_{=n}%
-j_{0}=n-j_{0}$, so that $0=-j_{0}$ and thus $j_{0}=0$. Hence, $\left(
j_{0}=0\text{ and }j_{k}=n\right)  $, qed.}. Furthermore,%
\begin{equation}
\operatorname*{Des}\sigma\subseteq\left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}
\label{pf.prop.C(sigma).c.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.C(sigma).c.7}):} Let $h\in
\operatorname*{Des}\sigma$. Then,%
\[
h\in\operatorname*{Des}\sigma=\left\{  i\in\left\{  1,2,\ldots,n-1\right\}
\ \mid\ \sigma\left(  i\right)  >\sigma\left(  i+1\right)  \right\}
\]
(by the definition of $\operatorname*{Des}\sigma$). In other words, $h$ is an
element of $\left\{  1,2,\ldots,n-1\right\}  $ and satisfies $\sigma\left(
h\right)  >\sigma\left(  h+1\right)  $.
\par
We have $h\in\left\{  1,2,\ldots,n-1\right\}  $, so that $1\leq h\leq n-1$.
Now, $h\leq n-1<n=j_{k}$ (since $j_{k}=n$). Thus, there exists at least one
$q\in\left\{  0,1,\ldots,k\right\}  $ satisfying $h\leq j_{q}$ (namely,
$q=k$). Hence, there exists the \textbf{smallest} $q\in\left\{  0,1,\ldots
,k\right\}  $ satisfying $h\leq j_{q}$. Let us denote this smallest $q$ by
$r$. Thus, $r$ is an element $q\in\left\{  0,1,\ldots,k\right\}  $ satisfying
$h\leq j_{q}$. In other words, $r$ is an element of $\left\{  0,1,\ldots
,k\right\}  $ satisfying $h\leq j_{r}$. We have $j_{0}=0<1\leq h\leq j_{r}$,
so that $j_{0}\neq j_{r}$ and thus $0\neq r$. Hence, $r\neq0$. Since
$r\in\left\{  0,1,\ldots,k\right\}  $ and $r\neq0$, we have $r\in\left\{
0,1,\ldots,k\right\}  \setminus\left\{  0\right\}  =\left\{  1,2,\ldots
,k\right\}  $. Thus, $r-1\in\left\{  0,1,\ldots,k-1\right\}  \subseteq\left\{
0,1,\ldots,k\right\}  $.
\par
But%
\begin{equation}
\text{every }q\in\left\{  0,1,\ldots,k\right\}  \text{ satisfying }h\leq
j_{q}\text{ must satisfy }q\geq r \label{pf.prop.C(sigma).c.7.pf.1}%
\end{equation}
(because $r$ is the \textbf{smallest} $q\in\left\{  0,1,\ldots,k\right\}  $
satisfying $h\leq j_{q}$).
\par
Let us assume (for the sake of contradiction) that $h\leq j_{r-1}$. Then,
(\ref{pf.prop.C(sigma).c.7.pf.1}) (applied to $q=r-1$) yields $r-1\geq r$
(since $r-1\in\left\{  0,1,\ldots, k\right\}  $). But this is absurd. This
contradiction shows that our assumption (that $h\leq j_{r-1}$) is false. Thus,
we do not have $h\leq j_{r-1}$. Hence, we have $h>j_{r-1}$. Thus, $h\geq
j_{r-1}+1$ (since $h$ and $j_{r-1}$ are integers), so that $h-j_{r-1}\geq1$.
\par
Let us assume (for the sake of contradiction) that $h\neq j_{r}$. Combined
with $h\leq j_{r}$, this yields $h<j_{r}$. Thus, $h\leq j_{r}-1$ (since $h$
and $j_{r}$ are integers), so that $\underbrace{h}_{\leq j_{r}-1}%
-\,j_{r-1}\leq j_{r}-1-j_{r-1}=j_{r}-j_{r-1}-1$. Combined with $h-j_{r-1}%
\geq1$, this yields $1\leq h-j_{r-1}\leq j_{r}-j_{r-1}-1$, so that
$h-j_{r-1}\in\left\{  1,2,\ldots,j_{r}-j_{r-1}-1\right\}  $.
\par
But recall that, for every $q\in\left\{  1,2,\ldots,k\right\}  $, the list
$\left(  \sigma\left(  j_{q-1}+1\right)  ,\sigma\left(  j_{q-1}+2\right)
,\ldots,\sigma\left(  j_{q}\right)  \right)  $ is an ascending run of $\sigma$
having length $c_{q}$. Applying this to $q=r$, we conclude that the list
$\left(  \sigma\left(  j_{r-1}+1\right)  ,\sigma\left(  j_{r-1}+2\right)
,\ldots,\sigma\left(  j_{r}\right)  \right)  $ is an ascending run of $\sigma$
having length $c_{r}$. In particular, this list is an ascending run, and
therefore increasing. In other words, every $g\in\left\{  1,2,\ldots
,j_{r}-j_{r-1}-1\right\}  $ satisfies $\sigma\left(  j_{r-1}+g\right)
\leq\sigma\left(  j_{r-1}+\left(  g+1\right)  \right)  $. Applying this to
$g=h-j_{r-1}$, we obtain%
\[
\sigma\left(  j_{r-1}+\left(  h-j_{r-1}\right)  \right)  \leq\sigma\left(
j_{r-1}+\left(  \left(  h-j_{r-1}\right)  +1\right)  \right)
\]
(since $h-j_{r-1}\in\left\{  1,2,\ldots,j_{r}-j_{r-1}-1\right\}  $). Since
$j_{r-1}+\left(  h-j_{r-1}\right)  =h$ and $j_{r-1}+\left(  \left(
h-j_{r-1}\right)  +1\right)  =h+1$, this rewrites as $\sigma\left(  h\right)
\leq\sigma\left(  h+1\right)  $. This contradicts $\sigma\left(  h\right)
>\sigma\left(  h+1\right)  $.
\par
This contradiction shows that our assumption (that $h\neq j_{r}$) was wrong.
Hence, we must have $h=j_{r}$. Thus, $j_{r}=h<j_{k}$, so that $j_{r}\neq
j_{k}$ and thus $r\neq k$. Combining this with $r\in\left\{  1,2,\ldots
,k\right\}  $, we obtain $r\in\left\{  1,2,\ldots,k\right\}  \setminus\left\{
k\right\}  =\left\{  1,2,\ldots,k-1\right\}  $. Now, $h=j_{r}\in\left\{
j_{1},j_{2},\ldots,j_{k-1}\right\}  $ (since $r\in\left\{  1,2,\ldots
,k-1\right\}  $).
\par
Now, let us forget that we fixed $h$. We thus have shown that every
$h\in\operatorname*{Des}\sigma$ satisfies $h\in\left\{  j_{1},j_{2}%
,\ldots,j_{k-1}\right\}  $. In other words, $\operatorname*{Des}%
\sigma\subseteq\left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}  $. This proves
(\ref{pf.prop.C(sigma).c.7}).}. But $\left\vert
\underbrace{\operatorname*{Des}\sigma}_{=D\left(  \alpha\right)  }\right\vert
=\left\vert D\left(  \alpha\right)  \right\vert =\max\left\{  \ell
-1,0\right\}  $ (by (\ref{eq.prop.C(sigma).a.1})). But $\ell-1\geq0$ and thus
$\max\left\{  \ell-1,0\right\}  =\ell-1$. Hence, $\left\vert
\operatorname*{Des}\sigma\right\vert =\max\left\{  \ell-1,0\right\}  =\ell-1$.
But $\left\vert \left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}  \right\vert \leq
k-1$ (since $j_{1}$, $j_{2}$, $\ldots$, $j_{k-1}$ are $k-1$ integers (since
$k-1\geq0$)). Now,%
\[
\left\vert \operatorname*{Des}\sigma\right\vert =\underbrace{\ell
}_{\substack{\geq k\\\text{(since }k\leq\ell\text{)}}}-\,1\geq k-1\geq
\left\vert \left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}  \right\vert
\]
(since $\left\vert \left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}  \right\vert
\leq k-1$). But it is clear that if $S$ is a finite set, and if $T$ is a
subset of $S$ satisfying $\left\vert T\right\vert \geq\left\vert S\right\vert
$, then we must have $T=S$. Applying this to $S=\left\{  j_{1},j_{2}%
,\ldots,j_{k-1}\right\}  $ and $T=\operatorname*{Des}\sigma$, we obtain%
\begin{equation}
\operatorname*{Des}\sigma=\left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}
\label{pf.prop.C(sigma).c.9}%
\end{equation}
(since $\operatorname*{Des}\sigma\subseteq\left\{  j_{1},j_{2},\ldots
,j_{k-1}\right\}  $ and $\left\vert \operatorname*{Des}\sigma\right\vert
\geq\left\vert \left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}  \right\vert $).

On the other hand, $D\left(  \left(  c_{1},c_{2},\ldots,c_{k}\right)  \right)
=\left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} For every $p\in\left\{  1,2,\ldots
,k-1\right\}  $, we have%
\begin{align}
c_{1}+c_{2}+\cdots+c_{p}  &  =\sum_{q=1}^{p}\underbrace{c_{q}}%
_{\substack{=j_{q}-j_{q-1}\\\text{(by (\ref{pf.prop.C(sigma).c.6}))}}%
}=\sum_{q=1}^{p}\left(  j_{q}-j_{q-1}\right)  =\sum_{q=1}^{p}j_{q}-\sum
_{q=1}^{p}j_{q-1}\nonumber\\
&  =\underbrace{\sum_{q=1}^{p}j_{q}}_{=\sum_{q=1}^{p-1}j_{q}+j_{p}%
}-\underbrace{\sum_{q=0}^{p-1}j_{q}}_{=j_{0}+\sum_{q=1}^{p-1}j_{q}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }q\text{ for
}q-1\text{ in the second sum}\right) \nonumber\\
&  =\left(  \sum_{q=1}^{p-1}j_{q}+j_{p}\right)  -\left(  j_{0}+\sum
_{q=1}^{p-1}j_{q}\right)  =j_{p}-\underbrace{j_{0}}_{=0}=j_{p}.
\label{pf.prop.C(sigma).c.9.pf.1}%
\end{align}
\par
Now, applying (\ref{eq.prop.C(sigma).Dalpha}) to $\left(  c_{1},c_{2}%
,\ldots,c_{k}\right)  $, $k$ and $c_{i}$ instead of $\alpha$, $\ell$ and
$\alpha_{i}$, we obtain%
\begin{align*}
D\left(  \left(  c_{1},c_{2},\ldots,c_{k}\right)  \right)   &  =\left\{
\underbrace{c_{1}+c_{2}+\cdots+c_{p}}_{\substack{=j_{p}\\\text{(by
(\ref{pf.prop.C(sigma).c.9.pf.1}))}}}\ \mid\ p\in\left\{  1,2,\ldots
,k-1\right\}  \right\} \\
&  =\left\{  j_{p}\ \mid\ p\in\left\{  1,2,\ldots,k-1\right\}  \right\}
=\left\{  j_{1},j_{2},\ldots,j_{k-1}\right\}  .
\end{align*}
This proves (\ref{pf.prop.C(sigma).c.9}).}. Compared with
(\ref{pf.prop.C(sigma).c.9}), this yields $D\left(  \left(  c_{1},c_{2}%
,\ldots,c_{k}\right)  \right)  =\operatorname*{Des}\sigma$.

Now, recall that $\alpha$ was defined as the unique composition of $n$ such
that $D\left(  \alpha\right)  =\operatorname*{Des}\sigma$. In other words,
$\alpha$ is the unique composition $\beta$ of $n$ such that $D\left(
\beta\right)  =\operatorname*{Des}\sigma$. The \textquotedblleft
uniqueness\textquotedblright\ part of this statement shows that every
composition $\beta$ of $n$ such that $D\left(  \beta\right)
=\operatorname*{Des}\sigma$ must satisfy $\beta=\alpha$. Applying this to
$\beta=\left(  c_{1},c_{2},\ldots,c_{k}\right)  $, we obtain $\left(
c_{1},c_{2},\ldots,c_{k}\right)  =\alpha$ (since $\left(  c_{1},c_{2}%
,\ldots,c_{k}\right)  $ is a composition of $n$ and satisfies $D\left(
\left(  c_{1},c_{2},\ldots,c_{k}\right)  \right)  =\operatorname*{Des}\sigma
$). Thus, $\left(  c_{1},c_{2},\ldots,c_{k}\right)  =\alpha$ is proven.

Now, let us forget that we fixed $\left(  c_{1},c_{2},\ldots,c_{k}\right)  $.
We thus have shown the following fact:

\begin{statement}
\textit{Fact A:} If $\left(  c_{1},c_{2},\ldots,c_{k}\right)  $ is any
composition of $n$ such that the word $\sigma$ has $k$ consecutive ascending
runs of lengths $c_{1}$, $c_{2}$, $\ldots$, $c_{k}$, and such that $k\leq\ell
$, then $\left(  c_{1},c_{2},\ldots,c_{k}\right)  =\alpha$.
\end{statement}

We also know that $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  $ is a composition of $n$ with the property that the word $\sigma$
has $\ell$ consecutive ascending runs of lengths $\alpha_{1}$, $\alpha_{2}$,
$\ldots$, $\alpha_{\ell}$. In other words, $\alpha$ is a composition $\left(
c_{1},c_{2},\ldots,c_{k}\right)  $ of $n$ such that the word $\sigma$ has $k$
consecutive ascending runs of lengths $c_{1}$, $c_{2}$, $\ldots$, $c_{k}$.
Moreover, among all such compositions $\left(  c_{1},c_{2},\ldots
,c_{k}\right)  $, the composition $\alpha$ is the one with the minimum $k$
(because of Fact A, and since $\alpha$ has $k=\ell$). In other words, we have
proven the following fact:

\begin{statement}
\textit{Fact B:} Among all compositions $\left(  c_{1},c_{2},\ldots
,c_{k}\right)  $ of $n$ such that the word $\sigma$ has $k$ consecutive
ascending runs of lengths $c_{1}$, $c_{2}$, $\ldots$, $c_{k}$, the composition
$\alpha$ is the one with the minimum $k$.
\end{statement}

Recall the definition of $C\left(  \sigma\right)  $. This definition says
that, among all compositions $\left(  c_{1},c_{2},\ldots,c_{k}\right)  $ of
$n$ such that the word $\sigma$ has $k$ consecutive ascending runs of lengths
$c_{1}$, $c_{2}$, $\ldots$, $c_{k}$, the composition $C\left(  \sigma\right)
$ is the one with the minimum $k$. Comparing this description of $C\left(
\sigma\right)  $ with the description of $\alpha$ given in Fact B, we obtain
that $\alpha=C\left(  \sigma\right)  $. This proves Proposition
\ref{prop.C(sigma)} \textbf{(c)}.
\end{proof}

\subsection{Page 9, Corollary 3.3: a proof}

For the sake of completeness, let me give a proof of Corollary 3.3 using
\cite[Theorem 5.2.11]{GriRei15}. But first, let me recall your definition of
the notion of a $\pi$-partition:

\begin{definition}
\label{def.pi-par}Let $\pi=\left(  E,<_{1},<_{2}\right)  $ be a double poset.
Let $X$ be a totally ordered set. A $\pi$\textit{-partition} into $X$ means a
map $x:E\rightarrow X$ such that any two elements $e$ and $e^{\prime}$ of $E$
satisfy%
\[
\left(  e<_{1}e^{\prime}\text{ implies }x\left(  e\right)  \leq x\left(
e^{\prime}\right)  \right)
\]
and%
\[
\left(  e<_{1}e^{\prime}\text{ and }e\geq_{2}e^{\prime}\text{ implies
}x\left(  e\right)  <x\left(  e^{\prime}\right)  \right)  .
\]

\end{definition}

\textit{Proof of Corollary 3.3.} Let $\pi$ be a special double poset. We shall
show that $\Gamma\left(  \pi\right)  =\left(  F\circ L\right)  \left(
\pi\right)  $.

We use the definition of $\mathbf{QSym}$ given in \cite[Definition
5.1.5]{GriRei15} (where it is denoted by $\operatorname*{QSym}$). The
definition of $\Gamma\left(  \pi\right)  $ then rewrites as follows:%
\begin{equation}
\Gamma\left(  \pi\right)  =\sum_{\substack{f\text{ is a }\pi\text{-partition}%
\\\text{into }\left\{  1,2,3,\ldots\right\}  }}\ \ \prod_{e\in E}x_{f\left(
e\right)  }. \label{add.p9.FGamma}%
\end{equation}


Write $\pi$ in the form $\pi=\left(  E,<_{1},<_{2}\right)  $. Then, $<_{2}$ is
a total order (since $\pi$ is special). Hence, we can WLOG assume that
$E=\left\{  1,2,\ldots,n\right\}  $ for some $n\in\mathbb{N}$, and that the
second order $<_{2}$ of $\pi$ is the natural order on $\left\{  1,2,\ldots
,n\right\}  $ (since we can achieve this by relabelling the elements of $E$).
Assume this. Then, $\left(  E,<_{1}\right)  $ becomes a labelled poset in the
sense of \cite[Definition 5.2.1]{GriRei15} (since $E\subseteq\mathbb{Z}$).
Also, $<_{2}$ is the natural order on $\left\{  1,2,\ldots,n\right\}  $. In
other words, $<_{2}$ is the restriction of $<_{\mathbb{Z}}$ to $E$ (where
$<_{\mathbb{Z}}$ is the natural order on $\mathbb{Z}$). Thus, $\geq_{2}$ is
the restriction of $\geq_{\mathbb{Z}}$ to $E$.

We recall the definition of a $P$-partition (for $P$ being a labelled poset)
given in \cite[Definition 5.2.1]{GriRei15}. It is easy to see that the
$\left(  E,<_{1}\right)  $-partitions are the same as the $\pi$-partitions
into $\left\{  1,2,3,\ldots\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let
$x$ be a $\pi$-partition into $\left\{  1,2,3,\ldots\right\}  $. By the
definition of a $\pi$-partition, we then know that $x$ is a map $E\rightarrow
\left\{  1,2,3,\ldots\right\}  $ such that any two elements $e$ and
$e^{\prime}$ of $E$ satisfy%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ implies }x\left(  e\right)  \leq x\left(
e^{\prime}\right)  \right)  \label{add.p9.c3.3.pf.1}%
\end{equation}
and%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ and }e\geq_{2}e^{\prime}\text{ implies
}x\left(  e\right)  <x\left(  e^{\prime}\right)  \right)
\label{add.p9.c3.3.pf.2}%
\end{equation}
(since $x$ is a $\pi$-partition). Thus, any two elements $e$ and $e^{\prime}$
of $E$ satisfy%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ and }e\geq_{\mathbb{Z}}e^{\prime}\text{ implies
}x\left(  e\right)  <x\left(  e^{\prime}\right)  \right)  .
\label{add.p9.c3.3.pf.2a}%
\end{equation}
\par
(\textit{Proof of (\ref{add.p9.c3.3.pf.2a}):} Let $e$ and $e^{\prime}$ be two
elements of $E$. Assume that $e<_{1}e^{\prime}$ and $e\geq_{\mathbb{Z}%
}e^{\prime}$. We have $e\geq_{\mathbb{Z}}e^{\prime}$. In other words,
$e\geq_{2}e^{\prime}$ (since $\geq_{2}$ is the restriction of $\geq
_{\mathbb{Z}}$ to $E$). Thus, (\ref{add.p9.c3.3.pf.2}) shows that $x\left(
e\right)  <x\left(  e^{\prime}\right)  $. This proves (\ref{add.p9.c3.3.pf.2a}%
).)
\par
Now, if $i$ and $j$ are two elements of $E$, then%
\begin{equation}
\left(  i<_{1}j\text{ and }i<_{\mathbb{Z}}j\text{ implies }x\left(  i\right)
\leq x\left(  j\right)  \right)  \label{add.p9.c3.3.pf.3}%
\end{equation}
(according to (\ref{add.p9.c3.3.pf.1}), applied to $e=i$ and $e^{\prime}=j$)
and%
\begin{equation}
\left(  i<_{1}j\text{ and }i>_{\mathbb{Z}}j\text{ implies }x\left(  i\right)
<x\left(  j\right)  \right)  \label{add.p9.c3.3.pf.4}%
\end{equation}
(according to (\ref{add.p9.c3.3.pf.2a}), applied to $e=i$ and $e^{\prime}=j$).
In other words, $x$ is an $\left(  E,<_{1}\right)  $-partition (due to the
definition of an $\left(  E,<_{1}\right)  $-partition).
\par
Now, let us forget that we fixed $x$. We thus have shown that if $x$ is a
$\pi$-partition into $\left\{  1,2,3,\ldots\right\}  $, then $x$ is an
$\left(  E,<_{1}\right)  $-partition. In other words,%
\begin{equation}
\text{every }\pi\text{-partition into }\left\{  1,2,3,\ldots\right\}  \text{
is an }\left(  E,<_{1}\right)  \text{-partition.} \label{add.p9.c3.3.pf.5}%
\end{equation}
\par
Let now $y$ be an $\left(  E,<_{1}\right)  $-partition. According to the
definition of an $\left(  E,<_{1}\right)  $-partition, we thus know that $y$
is a map $E\rightarrow\left\{  1,2,3,\ldots\right\}  $ such that any two
elements $i$ and $j$ of $E$ satisfy%
\begin{equation}
\left(  i<_{1}j\text{ and }i<_{\mathbb{Z}}j\text{ implies }y\left(  i\right)
\leq y\left(  j\right)  \right)  \label{add.p9.c3.3.pf.11}%
\end{equation}
and%
\begin{equation}
\left(  i<_{1}j\text{ and }i>_{\mathbb{Z}}j\text{ implies }y\left(  i\right)
<y\left(  j\right)  \right)  \label{add.p9.c3.3.pf.12}%
\end{equation}
(since $y$ is an $\left(  E,<_{1}\right)  $-partition).
\par
Now, let $e$ and $e^{\prime}$ be two elements of $E$.
\par
Assume that $e<_{1}e^{\prime}$. We shall now prove that $y\left(  e\right)
\leq y\left(  e^{\prime}\right)  $. If $e<_{\mathbb{Z}}e^{\prime}$, then this
follows immediately from (\ref{add.p9.c3.3.pf.11}) (applied to $e=i$ and
$e^{\prime}=j$). Hence, for the rest of this proof, we can WLOG assume that we
don't have $e<_{\mathbb{Z}}e^{\prime}$. Assume this. We have $e\geq
_{\mathbb{Z}}e^{\prime}$ (since we don't have $e<_{\mathbb{Z}}e^{\prime}$).
But $e\neq e^{\prime}$ (since $e<_{1}e^{\prime}$, and since $<_{1}$ is a
partial order). Combined with $e\geq_{\mathbb{Z}}e^{\prime}$, this yields
$e>_{\mathbb{Z}}e^{\prime}$. Thus, $y\left(  e\right)  <y\left(  e^{\prime
}\right)  $ (by (\ref{add.p9.c3.3.pf.12})), so that $y\left(  e\right)  \leq
y\left(  e^{\prime}\right)  $. Thus, $y\left(  e\right)  \leq y\left(
e^{\prime}\right)  $ is proven.
\par
Let us now forget that we assumed that $e<_{1}e^{\prime}$. We thus have shown
that%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ implies }y\left(  e\right)  \leq y\left(
e^{\prime}\right)  \right)  . \label{add.p9.c3.3.pf.13}%
\end{equation}
\par
Now, assume that $e<_{1}e^{\prime}$ and $e\geq_{2}e^{\prime}$. We have
$e\geq_{2}e^{\prime}$. In other words, $e\geq_{\mathbb{Z}}e^{\prime}$ (since
$\geq_{2}$ is the restriction of $\geq_{\mathbb{Z}}$ to $E$). But
$e<_{1}e^{\prime}$, so that $e\neq e^{\prime}$ (since $<_{1}$ is a partial
order). Combined with $e\geq_{\mathbb{Z}}e^{\prime}$, this yields
$e>_{\mathbb{Z}}e^{\prime}$. Thus, (\ref{add.p9.c3.3.pf.12}) (applied to $i=e$
and $j=e^{\prime}$) yields $y\left(  e\right)  <y\left(  e^{\prime}\right)  $.
\par
Let us now forget that we assumed that $e<_{1}e^{\prime}$ and $e\geq
_{2}e^{\prime}$. We thus have shown that%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ and }e\geq_{2}e^{\prime}\text{ implies
}y\left(  e\right)  <y\left(  e^{\prime}\right)  \right)  .
\label{add.p9.c3.3.pf.14}%
\end{equation}
\par
Now, let us forget that we fixed $e$ and $e^{\prime}$. We thus have proven
that any two elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{add.p9.c3.3.pf.13}) and (\ref{add.p9.c3.3.pf.14}). In other words, $y$
is a $\pi$-partition into $\left\{  1,2,3,\ldots\right\}  $ (according to the
definition of a $\pi$-partition).
\par
Let us now forget that we fixed $y$. We thus have proven that if $y$ is an
$\left(  E,<_{1}\right)  $-partition, then $y$ is a $\pi$-partition into
$\left\{  1,2,3,\ldots\right\}  $. In other words,%
\[
\text{every }\left(  E,<_{1}\right)  \text{-partition is a }\pi
\text{-partition into }\left\{  1,2,3,\ldots\right\}  \text{.}%
\]
Combining this with (\ref{add.p9.c3.3.pf.5}), we conclude that the $\left(
E,<_{1}\right)  $-partitions are the same as the $\pi$-partitions into
$\left\{  1,2,3,\ldots\right\}  $. Qed.}. But now recall the definition of
$F_{P}$ (for $P$ being a labelled poset) given in \cite[Definition
5.2.1]{GriRei15}. This yields%
\[
F_{\left(  E,<_{1}\right)  }=\sum_{f\in\mathcal{A}\left(  \left(
E,<_{1}\right)  \right)  }\mathbf{x}_{f},
\]
where $\mathcal{A}\left(  \left(  E,<_{1}\right)  \right)  $ denotes the set
of all $\left(  E,<_{1}\right)  $-partitions and where $\mathbf{x}_{f}$ is
defined as $\prod_{i\in E}x_{f\left(  i\right)  }$. Thus,%
\begin{align}
F_{\left(  E,<_{1}\right)  }  &  =\underbrace{\sum_{f\in\mathcal{A}\left(
\left(  E,<_{1}\right)  \right)  }}_{\substack{=\sum_{f\text{ is a }\left(
E,<_{1}\right)  \text{-partition}}\\\text{(since }\mathcal{A}\left(  \left(
E,<_{1}\right)  \right)  \text{ is the set}\\\text{of all }\left(
E,<_{1}\right)  \text{-partitions)}}}\underbrace{\mathbf{x}_{f}}_{=\prod_{i\in
E}x_{f\left(  i\right)  }}=\underbrace{\sum_{f\text{ is a }\left(
E,<_{1}\right)  \text{-partition}}}_{\substack{=\sum_{\substack{f\text{ is a
}\pi\text{-partition}\\\text{into }\left\{  1,2,3,\ldots\right\}
}}\\\text{(since the }\left(  E,<_{1}\right)  \text{-partitions are}%
\\\text{the same as the }\pi\text{-partitions}\\\text{into }\left\{
1,2,3,\ldots\right\}  \text{)}}}\underbrace{\prod_{i\in E}x_{f\left(
i\right)  }}_{\substack{=\prod_{e\in E}x_{f\left(  e\right)  }\\\text{(here,
we renamed}\\\text{the index }i\text{ as }e\text{)}}}\nonumber\\
&  =\sum_{\substack{f\text{ is a }\pi\text{-partition}\\\text{into }\left\{
1,2,3,\ldots\right\}  }}\ \ \prod_{e\in E}x_{f\left(  e\right)  }%
=\Gamma\left(  \pi\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{add.p9.FGamma})}\right)  . \label{add.p9.FGamma.pf.F=Gamma}%
\end{align}


Let us now compute $F\left(  L\left(  \pi\right)  \right)  $.

Recall that you have defined the notion of a \textquotedblleft linear
extension\textquotedblright\ of a special double poset; in particular, this
yields the notion of a \textquotedblleft linear extension\textquotedblright%
\ of $\pi$. On the other hand, in \cite[\S 5.2]{GriRei15}, the notion of a
\textquotedblleft linear extension\textquotedblright\ of a labelled poset is
defined; in particular, this yields the notion of a \textquotedblleft linear
extension\textquotedblright\ of $\left(  E,<_{1}\right)  $. Now, we shall see
that these two notions agree: Namely,%
\begin{equation}
\text{the linear extensions of }\left(  E,<_{1}\right)  \text{ are precisely
the linear extensions of }\pi. \label{add.p9.FGamma.pf.linex}%
\end{equation}
\footnote{\textit{Proof of (\ref{add.p9.FGamma.pf.linex}):} A linear extension
of $\pi$ is the same as a total order on the set $E$ which extends the first
order $<_{1}$ of $\pi$ (according to the definition of a \textquotedblleft
linear extension\textquotedblright\ of $\pi$). In other words, a linear
extension of $\pi$ is the same as a total order on the set $E$ which extends
the partial order $<_{1}$. In other words, a linear extension of $\pi$ is the
same as a total order on the set $E$ which extends $\left(  E,<_{1}\right)  $
as a poset.
\par
But a linear extension of $\left(  E,<_{1}\right)  $ is the same as a linear
order on the set $E$ which extends $\left(  E,<_{1}\right)  $ as a poset
(according to the definition of \textquotedblleft linear
extension\textquotedblright\ given in \cite[\S 5.2]{GriRei15}). Since
\textquotedblleft linear order\textquotedblright\ is synonymous for
\textquotedblleft total order\textquotedblright, this rewrites as follows: A
linear extension of $\left(  E,<_{1}\right)  $ is the same as a total order on
the set $E$ which extends $\left(  E,<_{1}\right)  $ as a poset. In other
words, a linear extension of $\left(  E,<_{1}\right)  $ is the same as a
linear extension of $\pi$ (since a linear extension of $\pi$ is the same as a
total order on the set $E$ which extends $\left(  E,<_{1}\right)  $ as a
poset). In other words, the linear extensions of $\left(  E,<_{1}\right)  $
are precisely the linear extensions of $\pi$. This proves
(\ref{add.p9.FGamma.pf.linex}).}

You identify linear extensions of $\pi$ with a certain kind of words, and also
with permutations in $S_{n}$. Similarly, in \cite[\S 5.2]{GriRei15}, linear
extensions of $\left(  E,<_{1}\right)  $ are identified with a certain kind of
words. It is clear that these identifications are the same -- i.e., if
$\sigma$ is a linear extension of $\pi$, then the word with which the linear
extension $\sigma$ of $\pi$ is identified is identical with the word with
which the linear extension $\sigma$ of $\left(  E,<_{1}\right)  $ is
identified.\footnote{This is because both of these words are defined in the
same way (namely, as the list of all elements of $E$ in $\sigma$-increasing
order).} Following \cite[\S 5.2]{GriRei15}, we use the notation $\mathcal{L}%
\left(  P\right)  $ for the set of all linear extensions of a labelled poset
$P$. Then, \cite[Theorem 5.2.11]{GriRei15} says that $F_{P}\left(
\mathbf{x}\right)  =\sum_{w\in\mathcal{L}\left(  P\right)  }F_{w}\left(
\mathbf{x}\right)  $ for every labelled poset $P$. Applying this to $P=\left(
E,<_{1}\right)  $, we obtain%
\begin{align*}
F_{\left(  E,<_{1}\right)  }\left(  \mathbf{x}\right)   &  =\underbrace{\sum
_{w\in\mathcal{L}\left(  \left(  E,<_{1}\right)  \right)  }}_{\substack{=\sum
_{\substack{w\text{ is a linear}\\\text{extension of }\left(  E,<_{1}\right)
}}\\\text{(since }\mathcal{L}\left(  \left(  E,<_{1}\right)  \right)  \text{
is the set of all}\\\text{linear extensions of }\left(  E,<_{1}\right)
\text{)}}}F_{w}\left(  \mathbf{x}\right)  =\underbrace{\sum_{\substack{w\text{
is a linear}\\\text{extension of }\left(  E,<_{1}\right)  }}}_{\substack{=\sum
_{\substack{w\text{ is a linear}\\\text{extension of }\pi}}\\\text{(since the
linear extensions}\\\text{of }\left(  E,<_{1}\right)  \text{ are precisely
the}\\\text{linear extensions of }\pi\text{)}}}F_{w}\left(  \mathbf{x}\right)
\\
&  =\sum_{\substack{w\text{ is a linear}\\\text{extension of }\pi}%
}F_{w}\left(  \mathbf{x}\right)  .
\end{align*}
Now, (\ref{add.p9.FGamma.pf.F=Gamma}) yields%
\begin{equation}
\Gamma\left(  \pi\right)  =F_{\left(  E,<_{1}\right)  }=F_{\left(
E,<_{1}\right)  }\left(  \mathbf{x}\right)  =\sum_{\substack{w\text{ is a
linear}\\\text{extension of }\pi}}F_{w}\left(  \mathbf{x}\right)  .
\label{add.p9.FGamma.pf.Gamma1}%
\end{equation}


On the other hand, $L\left(  \pi\right)  $ is the sum of all linear extensions
of $\pi$ (according to the definition of $L$). In other words,%
\[
L\left(  \pi\right)  =\sum_{\substack{\sigma\text{ is a linear}%
\\\text{extension of }\pi}}\sigma.
\]
Applying the map $F$ to both sides of this equality, we obtain%
\begin{align}
F\left(  L\left(  \pi\right)  \right)   &  =F\left(  \sum_{\substack{\sigma
\text{ is a linear}\\\text{extension of }\pi}}\sigma\right)  =\sum
_{\substack{\sigma\text{ is a linear}\\\text{extension of }\pi}%
}\underbrace{F\left(  \sigma\right)  }_{\substack{=F_{C\left(  \sigma\right)
}\\\text{(by the definition of }F\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{ is }\mathbb{Z}%
\text{-linear}\right) \nonumber\\
&  =\sum_{\substack{\sigma\text{ is a linear}\\\text{extension of }\pi
}}F_{C\left(  \sigma\right)  }=\sum_{\substack{w\text{ is a linear}%
\\\text{extension of }\pi}}F_{C\left(  w\right)  } \label{add.p9.FGamma.pf.FL}%
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\sigma\text{ as }w\right)  .\nonumber
\end{align}


Now, let us fix a linear extension $w$ of $\pi$. We recall that $w$ is thus
identified with a word and with a permutation in $S_{n}$. Proposition
\ref{prop.C(sigma)} \textbf{(b)} (applied to $I=\operatorname*{Des}w$) yields
that there exists a unique composition $\alpha$ of $n$ such that $D\left(
\alpha\right)  =\operatorname*{Des}w$. Consider this $\alpha$. Then,
$\alpha=C\left(  w\right)  $ (according to Proposition \ref{prop.C(sigma)}
\textbf{(c)}, applied to $\sigma=w$). Notice that the fundamental
quasisymmetric function $F_{\alpha}$ is denoted by $L_{\alpha}$ in
\cite[\S 5.2]{GriRei15}.

Now, in \cite[Proposition 5.2.10]{GriRei15}, it is shown that $F_{w}\left(
\mathbf{x}\right)  $ equals the fundamental quasisymmetric function
$L_{\alpha}$. Since $F_{\alpha}$ is denoted by $L_{\alpha}$ in \cite[\S 5.2]%
{GriRei15}, this result rewrites as follows: $F_{w}\left(  \mathbf{x}\right)
$ equals the fundamental quasisymmetric function $F_{\alpha}$. In other words,
$F_{w}\left(  \mathbf{x}\right)  =F_{\alpha}=F_{C\left(  w\right)  }$ (since
$\alpha=C\left(  w\right)  $).

Let us now forget that we fixed $w$. We thus have shown that $F_{w}\left(
\mathbf{x}\right)  =F_{C\left(  w\right)  }$ for every linear extension $w$ of
$\pi$. Thus, (\ref{add.p9.FGamma.pf.Gamma1}) becomes%
\begin{align*}
\Gamma\left(  \pi\right)   &  =\sum_{\substack{w\text{ is a linear}%
\\\text{extension of }\pi}}\underbrace{F_{w}\left(  \mathbf{x}\right)
}_{=F_{C\left(  w\right)  }}=\sum_{\substack{w\text{ is a linear}%
\\\text{extension of }\pi}}F_{C\left(  w\right)  }=F\left(  L\left(
\pi\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{add.p9.FGamma.pf.FL})}\right) \\
&  =\left(  F\circ L\right)  \left(  \pi\right)  .
\end{align*}


Let us now forget that we fixed $\pi$. We thus have proven that $\Gamma\left(
\pi\right)  =\left(  F\circ L\right)  \left(  \pi\right)  $ for every special
double poset $\pi$. Thus, $\left(  \Gamma\mid_{\mathbb{Z}\mathbf{DS}}\right)
\left(  \pi\right)  =\Gamma\left(  \pi\right)  =\left(  F\circ L\right)
\left(  \pi\right)  $ for every special double poset $\pi$. In other words,
the two maps $\Gamma\mid_{\mathbb{Z}\mathbf{DS}}$ and $F\circ L$ are equal to
each other on the basis of the $\mathbb{Z}$-module $\mathbb{Z}\mathbf{DS}$
consisting of the (isomorphism classes of) special double posets. Since these
two maps $\Gamma\mid_{\mathbb{Z}\mathbf{DS}}$ and $F\circ L$ are $\mathbb{Z}%
$-linear, this shows that these two maps $\Gamma\mid_{\mathbb{Z}\mathbf{DS}}$
and $F\circ L$ are identical (because if two $\mathbb{Z}$-linear maps from the
same domain are equal to each other on a basis of the domain, then these two
maps must be identical). In other words, $\Gamma\mid_{\mathbb{Z}\mathbf{DS}%
}=F\circ L$. This proves Corollary 3.3.

Let me also prove the claim that \textquotedblleft the linear function
$F:\mathbb{Z}S\rightarrow\mathbf{QSym}$ defined by $\sigma\mapsto F_{C\left(
\sigma\right)  }$ is a homomorphism of bialgebras\textquotedblright. You refer
to \textquotedblleft\lbrack MR] Th.3.3\textquotedblright, but as far as I can
tell this fact is never explicitly stated in [MR]; it takes some work to
derive it from the results of [MR]. Here is a simple way to derive it from
\cite[Corollary 8.1.14]{GriRei15}:

\textit{Proof of the fact that the linear function }$F:\mathbb{Z}%
S\rightarrow\mathbf{QSym}$\textit{ defined by }$\sigma\mapsto F_{C\left(
\sigma\right)  }$\textit{ is a homomorphism of bialgebras:} For every
$n\in\mathbb{N}$ and $w\in S_{n}$, we let $\gamma\left(  w\right)  $ denote
the unique composition $\alpha$ of $n$ such that $D\left(  \alpha\right)
=\operatorname*{Des}w$ (where $\operatorname*{Des}w$ is defined as in
Proposition \ref{prop.C(sigma)} \textbf{(c)}). (This unique composition exists
according to Proposition \ref{prop.C(sigma)} \textbf{(b)} (applied to
$\sigma=w$ and $I=\operatorname*{Des}w$).)

Part of \cite[Corollary 8.1.14(a)]{GriRei15} (applied to $\mathbf{k}%
=\mathbb{Z}$) states (in the notations of \cite[\S 8]{GriRei15}) that the
$\mathbb{Z}$-linear map%
\begin{align*}
\operatorname*{FQSym}  &  \rightarrow\operatorname*{QSym},\\
F_{w}  &  \mapsto L_{\gamma\left(  w\right)  }%
\end{align*}
is a Hopf morphism. Translating this into our notations, we obtain the
following: The $\mathbb{Z}$-linear map%
\begin{align*}
\mathbb{Z}S  &  \rightarrow\mathbf{QSym},\\
w  &  \mapsto F_{\gamma\left(  w\right)  }%
\end{align*}
is a Hopf morphism\footnote{In our translation, we used the following
dictionary:
\par
\begin{itemize}
\item What is called $\operatorname*{FQSym}$ in \cite{GriRei15} is what we
call $\mathbb{Z}S$.
\par
\item What is called $\operatorname*{QSym}$ in \cite{GriRei15} is what we call
$\mathbf{QSym}$.
\par
\item What is called $F_{w}$ in \cite{GriRei15} (for $w$ being a permutation)
is what we call $w$.
\par
\item What is called $L_{\alpha}$ in \cite{GriRei15} (for $\alpha$ being a
composition) is what we call $F_{\alpha}$.
\end{itemize}
}. Renaming the index $w$ as $\sigma$ in this fact, we rewrite it as follows:
The $\mathbb{Z}$-linear map%
\begin{align*}
\mathbb{Z}S  &  \rightarrow\mathbf{QSym},\\
\sigma &  \mapsto F_{\gamma\left(  \sigma\right)  }%
\end{align*}
is a Hopf morphism. In other words, the $\mathbb{Z}$-linear map%
\begin{align*}
\mathbb{Z}S  &  \rightarrow\mathbf{QSym},\\
\sigma &  \mapsto F_{C\left(  \sigma\right)  }%
\end{align*}
is a Hopf morphism (because every $n\in\mathbb{N}$ and every $\sigma\in S_{n}$
satisfy $\gamma\left(  \sigma\right)  =C\left(  \sigma\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $n\in\mathbb{N}$ and $\sigma\in S_{n}$.
Proposition \ref{prop.C(sigma)} \textbf{(c)} (applied to $\alpha=\gamma\left(
\sigma\right)  $) yields that $\gamma\left(  \sigma\right)  =C\left(
\sigma\right)  $ (since $\gamma\left(  \sigma\right)  $ is the unique
composition $\alpha$ of $n$ such that $D\left(  \alpha\right)
=\operatorname*{Des}\sigma$ (due to the definition of $\gamma\left(
\sigma\right)  $)). Qed.}). In particular, this map is a homomorphism of
bialgebras. This completes our proof that the linear function $F:\mathbb{Z}%
S\rightarrow\mathbf{QSym}$ defined by $\sigma\mapsto F_{C\left(
\sigma\right)  }$ is a homomorphism of bialgebras.

\subsection{Page 10, Proposition 3.1: properties of standard permutations}

Let me give a self-contained proof of Proposition 3.1.

First, I am going to give a formal definition of the standard permutation of a
word. Let me start with the following simple fact:

\begin{proposition}
\label{prop.st.constr}Let $P$ be a poset. Let $w=a_{1}a_{2}\cdots a_{n}$ be a
word over the poset $P$ (that is, a finite list of elements of $P$). We define
a binary relation $\prec_{w}$ on the set $\left\{  1,2,\ldots,n\right\}  $ as
follows: For any $i\in\left\{  1,2,\ldots,n\right\}  $ and $j\in\left\{
1,2,\ldots,n\right\}  $, we set $i\prec_{w}j$ if and only if%
\[
\left(  \text{either }a_{i}<a_{j}\text{ or }\left(  a_{i}=a_{j}\text{ and
}i<j\right)  \right)  .
\]


\textbf{(a)} This binary relation $\prec_{w}$ is the smaller relation of a
partial order on the set $\left\{  1,2,\ldots,n\right\}  $.

\textbf{(b)} Assume that the poset $P$ is totally ordered. Then, the binary
relation $\prec_{w}$ is the smaller relation of a total order on the set
$\left\{  1,2,\ldots,n\right\}  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.st.constr}.]\textbf{(a)} Any two elements $p$
and $q$ of $\left\{  1,2,\ldots,n\right\}  $ satisfying $p\prec_{w}q$ must
satisfy%
\begin{equation}
a_{p}\leq a_{q}. \label{pf.prop.st.constr.a.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.st.constr.a.1}):} Let $p$ and $q$ be
two elements of $\left\{  1,2,\ldots,n\right\}  $ satisfying $p\prec_{w}q$.
\par
We have $p\prec_{w}q$ if and only if $\left(  \text{either }a_{p}<a_{q}\text{
or }\left(  a_{p}=a_{q}\text{ and }p<q\right)  \right)  $ (according to the
definition of the relation $\prec_{w}$). Hence, we must have $\left(
\text{either }a_{p}<a_{q}\text{ or }\left(  a_{p}=a_{q}\text{ and }p<q\right)
\right)  $ (since we have $p\prec_{w}q$). In other words, we are in one of the
following two cases:
\par
\textit{Case 1:} We have $a_{p}<a_{q}$.
\par
\textit{Case 2:} We have $\left(  a_{p}=a_{q}\text{ and }p<q\right)  $.
\par
Let us first consider Case 1. In this case, we have $a_{p}<a_{q}$. Thus,
$a_{p}\leq a_{q}$. Hence, (\ref{pf.prop.st.constr.a.1}) is proven in Case 1.
\par
Let us now consider Case 1. In this case, we have $\left(  a_{p}=a_{q}\text{
and }p<q\right)  $. Thus, $a_{p}=a_{q}$, so that $a_{p}\leq a_{q}$. Hence,
(\ref{pf.prop.st.constr.a.1}) is proven in Case 2.
\par
We have now proven (\ref{pf.prop.st.constr.a.1}) in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this yields that
(\ref{pf.prop.st.constr.a.1}) always holds. Qed.}

The binary relation $\prec_{w}$ is irreflexive\footnote{\textit{Proof.} Let
$p$ be an element of $\left\{  1,2,\ldots,n\right\}  $ such that $p\prec_{w}%
p$.
\par
We have $p\prec_{w}p$ if and only if $\left(  \text{either }a_{p}<a_{p}\text{
or }\left(  a_{p}=a_{p}\text{ and }p<p\right)  \right)  $ (according to the
definition of the relation $\prec_{w}$). Hence, we must have $\left(
\text{either }a_{p}<a_{p}\text{ or }\left(  a_{p}=a_{p}\text{ and }p<p\right)
\right)  $ (since we have $p\prec_{w}p$). Since $a_{p}<a_{p}$ is impossible,
we thus have $\left(  a_{p}=a_{p}\text{ and }p<p\right)  $. Thus, $p<p$, which
is absurd.
\par
Now, let us forget that we fixed $p$. We thus have obtained a contradiction
for every $p\in\left\{  1,2,\ldots,n\right\}  $ satisfying $p\prec_{w}p$.
Hence, there exists no $p\in\left\{  1,2,\ldots,n\right\}  $ satisfying
$p\prec_{w}p$. In other words, the relation $\prec_{w}$ is irreflexive, qed.},
transitive\footnote{\textit{Proof.} Let $p$, $q$ and $r$ be three elements of
$\left\{  1,2,\ldots,n\right\}  $ such that $p\prec_{w}q$ and $q\prec_{w}r$.
\par
We are going to show that $p\prec_{w}r$.
\par
We have $p\prec_{w}q$ and thus $a_{p}\leq a_{q}$ (by
(\ref{pf.prop.st.constr.a.1})). Also, we have $q\prec_{w}r$ and thus
$a_{q}\leq a_{r}$ (by (\ref{pf.prop.st.constr.a.1}), applied to $q$ and $r$
instead of $p$ and $q$). Thus, $a_{p}\leq a_{q}\leq a_{r}$. Hence, either
$a_{p}=a_{r}$ or $a_{p}<a_{r}$. In other words, we are in one of the following
two cases:
\par
\textit{Case 1:} We have $a_{p}=a_{r}$.
\par
\textit{Case 2:} We have $a_{p}<a_{r}$.
\par
Let us first consider Case 1. In this case, we have $a_{p}=a_{r}$. Thus,
$a_{r}=a_{p}\leq a_{q}$. Combined with $a_{q}\leq a_{r}$, this yields
$a_{r}=a_{q}$, so that $a_{q}=a_{r}$.
\par
We have $p\prec_{w}q$ if and only if $\left(  \text{either }a_{p}<a_{q}\text{
or }\left(  a_{p}=a_{q}\text{ and }p<q\right)  \right)  $ (according to the
definition of the relation $\prec_{w}$). Hence, we must have $\left(
\text{either }a_{p}<a_{q}\text{ or }\left(  a_{p}=a_{q}\text{ and }p<q\right)
\right)  $ (since we have $p\prec_{w}q$). Since $a_{p}<a_{q}$ is impossible
(because $a_{p}=a_{r}=a_{q}$), this yields that we have $\left(  a_{p}%
=a_{q}\text{ and }p<q\right)  $. Hence, $p<q$.
\par
We have $q\prec_{w}r$ if and only if $\left(  \text{either }a_{q}<a_{r}\text{
or }\left(  a_{q}=a_{r}\text{ and }q<r\right)  \right)  $ (according to the
definition of the relation $\prec_{w}$). Hence, we must have $\left(
\text{either }a_{q}<a_{r}\text{ or }\left(  a_{q}=a_{r}\text{ and }q<r\right)
\right)  $ (since we have $q\prec_{w}r$). Since $a_{q}<a_{r}$ is impossible
(since $a_{q}=a_{r}$), this shows that we have $\left(  a_{q}=a_{r}\text{ and
}q<r\right)  $. Hence, $q<r$.
\par
Now, $a_{p}=a_{r}$ and $p<q<r$. Hence, $\left(  a_{p}=a_{r}\text{ and
}p<r\right)  $, and therefore $\left(  \text{either }a_{p}<a_{r}\text{ or
}\left(  a_{p}=a_{r}\text{ and }p<r\right)  \right)  $.
\par
Now, we have $p\prec_{w}r$ if and only if $\left(  \text{either }a_{p}%
<a_{r}\text{ or }\left(  a_{p}=a_{r}\text{ and }p<r\right)  \right)  $
(according to the definition of the relation $\prec_{w}$). Hence, we must have
$p\prec_{w}r$ (since we have $\left(  \text{either }a_{p}<a_{r}\text{ or
}\left(  a_{p}=a_{r}\text{ and }p<r\right)  \right)  $). Hence, $p\prec_{w}r$
is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $a_{p}<a_{r}$. Thus,
$\left(  \text{either }a_{p}<a_{r}\text{ or }\left(  a_{p}=a_{r}\text{ and
}p<r\right)  \right)  $.
\par
Now, we have $p\prec_{w}r$ if and only if $\left(  \text{either }a_{p}%
<a_{r}\text{ or }\left(  a_{p}=a_{r}\text{ and }p<r\right)  \right)  $
(according to the definition of the relation $\prec_{w}$). Hence, we must have
$p\prec_{w}r$ (since we have $\left(  \text{either }a_{p}<a_{r}\text{ or
}\left(  a_{p}=a_{r}\text{ and }p<r\right)  \right)  $). Hence, $p\prec_{w}r$
is proven in Case 2.
\par
Now, we have proven $p\prec_{w}r$ in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, we can thus conclude that $p\prec
_{w}r$ always holds.
\par
Now, let us forget that we fixed $p$, $q$ and $r$. We thus have shown that if
$p$, $q$ and $r$ are any three elements of $\left\{  1,2,\ldots,n\right\}  $
such that $p\prec_{w}q$ and $q\prec_{w}r$, then $p\prec_{w}r$. In other words,
the relation $\prec_{w}$ is transitive. Qed.} and
antisymmetric\footnote{\textit{Proof.} It is known that any irreflexive
transitive binary relation is antisymmetric. Thus, the binary relation
$\prec_{w}$ is antisymmetric (since it is irreflexive and transitive). Qed.}.
Therefore, this binary relation $\prec_{w}$ is the smaller relation of a
partial order on the set $\left\{  1,2,\ldots,n\right\}  $ (since every
irreflexive, transitive and antisymmetric relation on a set $S$ must be the
smaller relation of a partial order on $S$). This proves Proposition
\ref{prop.st.constr} \textbf{(a)}.

\textbf{(b)} Proposition \ref{prop.st.constr} \textbf{(a)} shows that the
binary relation $\prec_{w}$ is the smaller relation of a partial order on the
set $\left\{  1,2,\ldots,n\right\}  $. In order to prove Proposition
\ref{prop.st.constr} \textbf{(b)}, it is clearly enough to show that this
order is total. In other words, it is clearly enough to prove that if $p$ and
$q$ are two distinct elements of $\left\{  1,2,\ldots,n\right\}  $, then
either $p\prec_{w}q$ or $q\prec_{w}p$. Let us prove this now.

Let $p$ and $q$ be two distinct elements of $\left\{  1,2,\ldots,n\right\}  $.
We need to show that either $p\prec_{w}q$ or $q\prec_{w}p$.

We can WLOG assume that $p\leq q$ (since our situation is symmetric in $p$ and
$q$, and thus we can interchange $p$ with $q$ to obtain $p\leq q$). Assume
this. Combining $p\leq q$ with $p\neq q$ (since $p$ and $q$ are distinct), we
obtain $p<q$.

The poset $P$ is totally ordered. Hence, we have either $a_{p}<a_{q}$ or
$a_{p}=a_{q}$ or $a_{p}>a_{q}$. In other words, we must be in one of the
following three cases:

\textit{Case 1:} We have $a_{p}<a_{q}$.

\textit{Case 2:} We have $a_{p}=a_{q}$.

\textit{Case 3:} We have $a_{p}>a_{q}$.

Let us first consider Case 1. In this case, we have $a_{p}<a_{q}$. Hence,
\newline$\left(  \text{either }a_{p}<a_{q}\text{ or }\left(  a_{p}=a_{q}\text{
and }p<q\right)  \right)  $.

Now, we have $p\prec_{w}q$ if and only if $\left(  \text{either }a_{p}%
<a_{q}\text{ or }\left(  a_{p}=a_{q}\text{ and }p<q\right)  \right)  $
(according to the definition of the relation $\prec_{w}$). Hence, we must have
$p\prec_{w}q$ (since we have $\left(  \text{either }a_{p}<a_{q}\text{ or
}\left(  a_{p}=a_{q}\text{ and }p<q\right)  \right)  $). Consequently, we have
either $p\prec_{w}q$ or $q\prec_{w}p$. Thus, $\left(  \text{either }p\prec
_{w}q\text{ or }q\prec_{w}p\right)  $ is proven in Case 1.

Let us now consider Case 2. In this case, we have $a_{p}=a_{q}$. Hence,
$\left(  a_{p}=a_{q}\text{ and }p<q\right)  $, so that $\left(  \text{either
}a_{p}<a_{q}\text{ or }\left(  a_{p}=a_{q}\text{ and }p<q\right)  \right)  $.

Now, we have $p\prec_{w}q$ if and only if $\left(  \text{either }a_{p}%
<a_{q}\text{ or }\left(  a_{p}=a_{q}\text{ and }p<q\right)  \right)  $
(according to the definition of the relation $\prec_{w}$). Hence, we must have
$p\prec_{w}q$ (since we have $\left(  \text{either }a_{p}<a_{q}\text{ or
}\left(  a_{p}=a_{q}\text{ and }p<q\right)  \right)  $). Consequently, we have
either $p\prec_{w}q$ or $q\prec_{w}p$. Thus, $\left(  \text{either }p\prec
_{w}q\text{ or }q\prec_{w}p\right)  $ is proven in Case 2.

Let us finally consider Case 3. In this case, we have $a_{p}>a_{q}$. Hence,
$a_{q}<a_{p}$, so that $\left(  \text{either }a_{q}<a_{p}\text{ or }\left(
a_{q}=a_{p}\text{ and }q<p\right)  \right)  $.

Now, we have $q\prec_{w}p$ if and only if $\left(  \text{either }a_{q}%
<a_{p}\text{ or }\left(  a_{q}=a_{p}\text{ and }q<p\right)  \right)  $
(according to the definition of the relation $\prec_{w}$). Hence, we must have
$q\prec_{w}p$ (since we have $\left(  \text{either }a_{q}<a_{p}\text{ or
}\left(  a_{q}=a_{p}\text{ and }q<p\right)  \right)  $). Consequently, we have
either $p\prec_{w}q$ or $q\prec_{w}p$. Thus, $\left(  \text{either }p\prec
_{w}q\text{ or }q\prec_{w}p\right)  $ is proven in Case 3.

Thus, $\left(  \text{either }p\prec_{w}q\text{ or }q\prec_{w}p\right)  $ is
proven in each of the three Cases 1, 2 and 3. Since these three Cases cover
all possibilities, this yields that $\left(  \text{either }p\prec_{w}q\text{
or }q\prec_{w}p\right)  $ always holds.

Now, let us forget that we fixed $p$ and $q$. We thus have shown that if $p$
and $q$ are any two distinct elements of $\left\{  1,2,\ldots,n\right\}  $,
then either $p\prec_{w}q$ or $q\prec_{w}p$. Thus, the binary relation
$\prec_{w}$ is the smaller relation of a total order on the set $\left\{
1,2,\ldots,n\right\}  $ (since we already know that the binary relation
$\prec_{w}$ is the smaller relation of a partial order on the set $\left\{
1,2,\ldots,n\right\}  $). This proves Proposition \ref{prop.st.constr}
\textbf{(b)}.
\end{proof}

\begin{definition}
\label{def.st}Let $P$ be a totally ordered poset. Let $w=a_{1}a_{2}\cdots
a_{n}$ be a word over the poset $P$. We define the \textit{standardization} of
the word $w$ as follows:

Construct a binary relation $\prec_{w}$ on the set $\left\{  1,2,\ldots
,n\right\}  $ as in Proposition \ref{prop.st.constr}. Then, Proposition
\ref{prop.st.constr} \textbf{(b)} shows that this relation is the smaller
relation of a total order on the set $\left\{  1,2,\ldots,n\right\}  $. In
other words, $\left(  \left\{  1,2,\ldots,n\right\}  ,\prec_{w}\right)  $ is a
totally ordered set. Thus, there is a unique order isomorphism $\left(
\left\{  1,2,\ldots,n\right\}  ,\prec_{w}\right)  \rightarrow\left(  \left\{
1,2,\ldots,n\right\}  ,<_{\mathbb{Z}}\right)  $. We define the
\textit{standard permutation} of $w$ as this order isomorphism, regarded as a
permutation of $\left\{  1,2,\ldots,n\right\}  $. This standard permutation is
also denoted as the \textit{standardization} of $w$.
\end{definition}

\begin{example}
Let $P$ be a three-element totally ordered poset $\left\{  x,y,z\right\}  $
with $x<y<z$. Let $w$ be the word $yzxy$ over the poset $P$. Then, the binary
relation $\prec_{w}$ is the total order given by $3\prec_{w}1\prec_{w}%
4\prec_{w}2$. Hence, the unique order isomorphism $\left(  \left\{
1,2,\ldots,n\right\}  ,\prec_{w}\right)  \rightarrow\left(  \left\{
1,2,\ldots,n\right\}  ,<_{\mathbb{Z}}\right)  $ sends $3$ to $1$, sends $1$ to
$2$, sends $4$ to $3$, and sends $2$ to $4$. In other words, this unique order
isomorphism is the permutation $\left(  2,4,1,3\right)  $ (written in one-line
notation). Thus, the standard permutation of the word $w$ is this permutation
$\left(  2,4,1,3\right)  $.
\end{example}

\begin{proposition}
\label{prop.st.1}Let $P$ be a totally ordered poset. Let $w=a_{1}a_{2}\cdots
a_{n}$ be a word over the poset $P$. Let $s$ be the standard permutation of
$w$. Let $i\in\left\{  1,2,\ldots,n\right\}  $ and $j\in\left\{
1,2,\ldots,n\right\}  $.

\textbf{(a)} We have the following logical equivalence:%
\[
\left(  s\left(  i\right)  <s\left(  j\right)  \right)  \ \Longleftrightarrow
\ \left(  \text{either }a_{i}<a_{j}\text{ or }\left(  a_{i}=a_{j}\text{ and
}i<j\right)  \right)  .
\]


\textbf{(b)} If $i<j$, then we have the following logical equivalence:%
\[
\left(  s\left(  i\right)  <s\left(  j\right)  \right)  \ \Longleftrightarrow
\ \left(  a_{i}\leq a_{j}\right)  .
\]


\textbf{(c)} If $i\geq j$, then we have the following logical equivalence:%
\[
\left(  s\left(  i\right)  <s\left(  j\right)  \right)  \ \Longleftrightarrow
\ \left(  a_{i}<a_{j}\right)  .
\]


\textbf{(d)} If $s\left(  i\right)  <s\left(  j\right)  $, then $a_{i}\leq
a_{j}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.st.1}.]Recall a very basic property of order
isomorphisms: If $\left(  U,<_{U}\right)  $ and $\left(  V,<_{V}\right)  $ are
two posets, if $f$ is an order isomorphism $\left(  U,<_{U}\right)
\rightarrow\left(  V,<_{V}\right)  $, and if $p$ and $q$ are two elements of
$U$, then we have the following logical equivalence:%
\begin{equation}
\left(  p<_{U}q\right)  \ \Longleftrightarrow\ \left(  f\left(  p\right)
<_{V}f\left(  q\right)  \right)  . \label{pf.prop.st.1.ordiso}%
\end{equation}


\textbf{(a)} Construct a binary relation $\prec_{w}$ on the set $\left\{
1,2,\ldots,n\right\}  $ as in Proposition \ref{prop.st.constr}. We recall that
the standard permutation of $w$ is the unique order isomorphism $\left(
\left\{  1,2,\ldots,n\right\}  ,\prec_{w}\right)  \rightarrow\left(  \left\{
1,2,\ldots,n\right\}  ,<_{\mathbb{Z}}\right)  $ (indeed, this is how it was
defined). In other words, $s$ is the unique order isomorphism $\left(
\left\{  1,2,\ldots,n\right\}  ,\prec_{w}\right)  \rightarrow\left(  \left\{
1,2,\ldots,n\right\}  ,<_{\mathbb{Z}}\right)  $ (since $s$ is the standard
permutation of $w$). In particular, $s$ is an order isomorphism $\left(
\left\{  1,2,\ldots,n\right\}  ,\prec_{w}\right)  \rightarrow\left(  \left\{
1,2,\ldots,n\right\}  ,<_{\mathbb{Z}}\right)  $. Hence, for every two elements
$p$ and $q$ of $\left\{  1,2,\ldots,n\right\}  $, we have the following
logical equivalence:%
\begin{equation}
\left(  p\prec_{w}q\right)  \ \Longleftrightarrow\ \left(  s\left(  p\right)
<_{\mathbb{Z}}s\left(  q\right)  \right)  . \label{pf.prop.st.1.a.ordiso}%
\end{equation}
(Indeed, this follows from (\ref{pf.prop.st.1.ordiso}), applied to $\left(
U,<_{U}\right)  =\left(  \left\{  1,2,\ldots,n\right\}  ,\prec_{w}\right)  $,
$\left(  V,<_{V}\right)  =\left(  \left\{  1,2,\ldots,n\right\}
,<_{\mathbb{Z}}\right)  $ and $f=s$).

Applying (\ref{pf.prop.st.1.a.ordiso}) to $p=i$ and $q=j$, we obtain the
logical equivalence
\begin{align*}
\left(  i\prec_{w}j\right)  \  &  \Longleftrightarrow\ \left(  s\left(
i\right)  <_{\mathbb{Z}}s\left(  j\right)  \right)  \ \Longleftrightarrow
\ \left(  s\left(  i\right)  <s\left(  j\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the relation }<_{\mathbb{Z}}\text{ on }\left\{  1,2,\ldots
,n\right\} \\
\text{is a restriction of the relation }<\text{ on }\mathbb{Z}%
\end{array}
\right)  .
\end{align*}
Hence, we have the logical equivalence%
\[
\left(  s\left(  i\right)  <s\left(  j\right)  \right)  \ \Longleftrightarrow
\ \left(  i\prec_{w}j\right)  \ \Longleftrightarrow\ \left(  \text{either
}a_{i}<a_{j}\text{ or }\left(  a_{i}=a_{j}\text{ and }i<j\right)  \right)
\]
(because $i\prec_{w}j$ is equivalent to $\left(  \text{either }a_{i}%
<a_{j}\text{ or }\left(  a_{i}=a_{j}\text{ and }i<j\right)  \right)  $ (by the
definition of the relation $\prec_{w}$)). This proves Proposition
\ref{prop.st.1} \textbf{(a)}.

\textbf{(b)} Assume that $i<j$. Proposition \ref{prop.st.1} \textbf{(a)} shows
that we have the following logical equivalence:%
\begin{align*}
\left(  s\left(  i\right)  <s\left(  j\right)  \right)  \  &
\Longleftrightarrow\ \left(  \text{either }a_{i}<a_{j}\text{ or }%
\underbrace{\left(  a_{i}=a_{j}\text{ and }i<j\right)  }_{\substack{\text{this
is equivalent to }\left(  a_{i}=a_{j}\right)  \\\text{(since }i<j\text{ is
true)}}}\right) \\
&  \Longleftrightarrow\ \left(  \text{either }a_{i}<a_{j}\text{ or }%
a_{i}=a_{j}\right)  \ \Longleftrightarrow\ \left(  a_{i}\leq a_{j}\right)  .
\end{align*}
This proves Proposition \ref{prop.st.1} \textbf{(b)}.

\textbf{(c)} Assume that $i\geq j$. Then, $i<j$ is false. Proposition
\ref{prop.st.1} \textbf{(a)} shows that we have the following logical
equivalence:%
\begin{align*}
\left(  s\left(  i\right)  <s\left(  j\right)  \right)  \  &
\Longleftrightarrow\ \left(  \text{either }a_{i}<a_{j}\text{ or }%
\underbrace{\left(  a_{i}=a_{j}\text{ and }i<j\right)  }_{\substack{\text{this
is false}\\\text{(since }i<j\text{ is false)}}}\right) \\
&  \Longleftrightarrow\ \left(  \text{either }a_{i}<a_{j}\text{ or }\left(
\text{false}\right)  \right)  \ \Longleftrightarrow\ \left(  a_{i}%
<a_{j}\right)  .
\end{align*}
This proves Proposition \ref{prop.st.1} \textbf{(c)}.

\textbf{(d)} We have the following logical implication:%
\begin{align*}
\left(  s\left(  i\right)  <s\left(  j\right)  \right)  \  &
\Longleftrightarrow\ \left(  \text{either }a_{i}<a_{j}\text{ or }%
\underbrace{\left(  a_{i}=a_{j}\text{ and }i<j\right)  }_{\text{this implies
}a_{i}=a_{j}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.st.1}
\textbf{(a)}}\right) \\
&  \Longrightarrow\ \left(  \text{either }a_{i}<a_{j}\text{ or }a_{i}%
=a_{j}\right)  \ \Longleftrightarrow\ \left(  a_{i}\leq a_{j}\right)  .
\end{align*}
In other words, if $s\left(  i\right)  <s\left(  j\right)  $, then $a_{i}\leq
a_{j}$. This proves Proposition \ref{prop.st.1} \textbf{(d)}.
\end{proof}

Next, let us show another simple property of standardization:

\begin{proposition}
\label{prop.st.2}Let $P$ be a totally ordered poset. Let $w=a_{1}a_{2}\cdots
a_{n}$ be a word over the poset $P$. Let $s$ be the standard permutation of
$w$. We regard the permutation $s$ as a word over the poset $\left\{
1,2,\ldots,n\right\}  $ (where the partial order on this poset is just the
usual order inherited from $\mathbb{Z}$) by writing $s$ in one-line notation.
Then, the standard permutation of this word $s$ is $s$ again.
\end{proposition}

Before we prove this, let us prove a basic fact:

\begin{lemma}
\label{lem.st.2.inv}Let $n\in\mathbb{N}$. Let $s$ and $t$ be two permutations
of $\left\{  1,2,\ldots,n\right\}  $. Assume that every two elements $i$ and
$j$ of $\left\{  1,2,\ldots,n\right\}  $ satisfying $i<j$ satisfy the
following logical equivalence:%
\begin{equation}
\left(  s\left(  i\right)  <s\left(  j\right)  \right)  \ \Longleftrightarrow
\ \left(  t\left(  i\right)  <t\left(  j\right)  \right)  .
\label{eq.lem.st.2.inv.1}%
\end{equation}
Then, $s=t$.
\end{lemma}

Lemma \ref{lem.st.2.inv} is a restatement of the well-known fact that any
permutation of $\left\{  1,2,\ldots,n\right\}  $ is uniquely determined by its
set of inversions. However, for the sake of completeness, let us give a
self-contained proof of Lemma \ref{lem.st.2.inv}.

\begin{proof}
[Proof of Lemma \ref{lem.st.2.inv}.]Let $u$ be the permutation $s\circ t^{-1}$
of $\left\{  1,2,\ldots,n\right\}  $. Then, $u$ is a bijection (since $u$ is a
permutation) and therefore an injection. It is easy to see that every two
elements $i$ and $j$ of $\left\{  1,2,\ldots,n\right\}  $ satisfy the
following logical equivalence:%
\begin{equation}
\left(  i<j\right)  \ \Longleftrightarrow\ \left(  u\left(  i\right)
<u\left(  j\right)  \right)  . \label{pf.lem.st.2.inv.u}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.st.2.inv.u}):} Let $i$ and $j$ be two
elements of $\left\{  1,2,\ldots,n\right\}  $. We want to prove the
equivalence (\ref{pf.lem.st.2.inv.u}).
\par
We must be in one of the following three cases:
\par
\textit{Case 1:} We have $t^{-1}\left(  i\right)  <t^{-1}\left(  j\right)  $.
\par
\textit{Case 2:} We have $t^{-1}\left(  i\right)  =t^{-1}\left(  j\right)  $.
\par
\textit{Case 3:} We have $t^{-1}\left(  i\right)  >t^{-1}\left(  j\right)  $.
\par
Let us first consider Case 1. In this case, we have $t^{-1}\left(  i\right)
<t^{-1}\left(  j\right)  $. Thus, (\ref{eq.lem.st.2.inv.1}) (applied to
$t^{-1}\left(  i\right)  $ and $t^{-1}\left(  j\right)  $ instead of $i$ and
$j$) yields that we have the following logical equivalence: $\left(  s\left(
t^{-1}\left(  i\right)  \right)  <s\left(  t^{-1}\left(  j\right)  \right)
\right)  \ \Longleftrightarrow\ \left(  t\left(  t^{-1}\left(  i\right)
\right)  <t\left(  t^{-1}\left(  j\right)  \right)  \right)  $. Thus, we have
the following logical equivalence:%
\[
\left(  s\left(  t^{-1}\left(  i\right)  \right)  <s\left(  t^{-1}\left(
j\right)  \right)  \right)  \ \Longleftrightarrow\ \left(
\underbrace{t\left(  t^{-1}\left(  i\right)  \right)  }_{=i}%
<\underbrace{t\left(  t^{-1}\left(  j\right)  \right)  }_{=j}\right)
\ \Longleftrightarrow\ \left(  i<j\right)  .
\]
Hence, we have the following logical equivalence:%
\begin{align*}
\left(  i<j\right)  \  &  \Longleftrightarrow\ \left(  \underbrace{s\left(
t^{-1}\left(  i\right)  \right)  }_{=\left(  s\circ t^{-1}\right)  \left(
i\right)  }<\underbrace{s\left(  t^{-1}\left(  j\right)  \right)  }_{=\left(
s\circ t^{-1}\right)  \left(  j\right)  }\right)  \ \Longleftrightarrow
\ \left(  \underbrace{\left(  s\circ t^{-1}\right)  }_{=u}\left(  i\right)
<\underbrace{\left(  s\circ t^{-1}\right)  }_{=u}\left(  j\right)  \right) \\
&  \Longleftrightarrow\ \left(  u\left(  i\right)  <u\left(  j\right)
\right)  .
\end{align*}
Thus, (\ref{pf.lem.st.2.inv.u}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $t^{-1}\left(  i\right)
=t^{-1}\left(  j\right)  $. Hence, $i=t\left(  \underbrace{t^{-1}\left(
i\right)  }_{=t^{-1}\left(  j\right)  }\right)  =t\left(  t^{-1}\left(
j\right)  \right)  =j$. Therefore, we do not have $i<j$. Also, $u\left(
\underbrace{i}_{=j}\right)  =u\left(  j\right)  $. Thus, we do not have
$u\left(  i\right)  <u\left(  j\right)  $. Hence, we have neither $i<j$ nor
$u\left(  i\right)  <u\left(  j\right)  $. Therefore, we have the logical
equivalence $\left(  i<j\right)  \ \Longleftrightarrow\ \left(  u\left(
i\right)  <u\left(  j\right)  \right)  $. Thus, (\ref{pf.lem.st.2.inv.u}) is
proven in Case 2.
\par
Let us first consider Case 3. In this case, we have $t^{-1}\left(  i\right)
>t^{-1}\left(  j\right)  $. Hence, if we had $i=j$, then we would have
$t^{-1}\left(  \underbrace{i}_{=j}\right)  =t^{-1}\left(  j\right)  $, which
would contradict $t^{-1}\left(  i\right)  >t^{-1}\left(  j\right)  $. Thus, we
cannot have $i=j$. Moreover, $u$ is an injection. Hence, if we had $u\left(
i\right)  =u\left(  j\right)  $, then we would have $i=j$, which would
contradict the fact that we cannot have $i=j$. Hence, we cannot have $u\left(
i\right)  =u\left(  j\right)  $.
\par
From $t^{-1}\left(  i\right)  >t^{-1}\left(  j\right)  $, we obtain
$t^{-1}\left(  j\right)  <t^{-1}\left(  i\right)  $. Consequently,
(\ref{eq.lem.st.2.inv.1}) (applied to $t^{-1}\left(  j\right)  $ and
$t^{-1}\left(  i\right)  $ instead of $i$ and $j$) yields that we have the
following logical equivalence: $\left(  s\left(  t^{-1}\left(  j\right)
\right)  <s\left(  t^{-1}\left(  i\right)  \right)  \right)
\ \Longleftrightarrow\ \left(  t\left(  t^{-1}\left(  j\right)  \right)
<t\left(  t^{-1}\left(  i\right)  \right)  \right)  $. Thus, we have the
following logical equivalence:%
\[
\left(  s\left(  t^{-1}\left(  j\right)  \right)  <s\left(  t^{-1}\left(
i\right)  \right)  \right)  \ \Longleftrightarrow\ \left(
\underbrace{t\left(  t^{-1}\left(  j\right)  \right)  }_{=j}%
<\underbrace{t\left(  t^{-1}\left(  i\right)  \right)  }_{=i}\right)
\ \Longleftrightarrow\ \left(  j<i\right)  .
\]
Hence, we have the following logical equivalence:%
\begin{align*}
\left(  j<i\right)  \  &  \Longleftrightarrow\ \left(  \underbrace{s\left(
t^{-1}\left(  j\right)  \right)  }_{=\left(  s\circ t^{-1}\right)  \left(
j\right)  }<\underbrace{s\left(  t^{-1}\left(  i\right)  \right)  }_{=\left(
s\circ t^{-1}\right)  \left(  i\right)  }\right)  \ \Longleftrightarrow
\ \left(  \underbrace{\left(  s\circ t^{-1}\right)  }_{=u}\left(  j\right)
<\underbrace{\left(  s\circ t^{-1}\right)  }_{=u}\left(  i\right)  \right) \\
&  \Longleftrightarrow\ \left(  u\left(  j\right)  <u\left(  i\right)
\right)  .
\end{align*}
Hence, we have the following logical equivalence:%
\begin{align*}
&  \ \left(  \text{not }\underbrace{j<i}_{\text{this is equivalent
to}\ \left(  u\left(  j\right)  <u\left(  i\right)  \right)  }\right) \\
&  \Longleftrightarrow\ \left(  \text{not }u\left(  j\right)  <u\left(
i\right)  \right)  \ \Longleftrightarrow\ \left(  u\left(  j\right)  \geq
u\left(  i\right)  \right) \\
&  \Longleftrightarrow\ \left(  u\left(  i\right)  \leq u\left(  j\right)
\right)  \ \Longleftrightarrow\ \left(  u\left(  i\right)  <u\left(  j\right)
\text{ or }\underbrace{u\left(  i\right)  =u\left(  j\right)  }%
_{\substack{\text{this is equivalent to }\left(  \text{false}\right)
\\\text{(since we cannot have }u\left(  i\right)  =u\left(  j\right)
\text{)}}}\right) \\
&  \Longleftrightarrow\ \left(  u\left(  i\right)  <u\left(  j\right)  \text{
or }\left(  \text{false}\right)  \right)  \ \Longleftrightarrow\ \left(
u\left(  i\right)  <u\left(  j\right)  \right)  .
\end{align*}
Thus, we have the following logical equivalence:%
\begin{align*}
&  \ \left(  u\left(  i\right)  <u\left(  j\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{not }j<i\right)  \ \Longleftrightarrow
\ \left(  j\geq i\right)  \ \Longleftrightarrow\ \left(  i\leq j\right) \\
&  \Longleftrightarrow\ \left(  i<j\text{ or }\underbrace{i=j}%
_{\substack{\text{this is equivalent to }\left(  \text{false}\right)
\\\text{(since we cannot have }i=j\text{)}}}\right)  \ \Longleftrightarrow
\ \left(  i<j\text{ or }\left(  \text{false}\right)  \right)
\ \Longleftrightarrow\ \left(  i<j\right)  .
\end{align*}
In other words, we have the logical equivalence $\left(  i<j\right)
\ \Longleftrightarrow\ \left(  u\left(  i\right)  <u\left(  j\right)  \right)
$. Thus, (\ref{pf.lem.st.2.inv.u}) is proven in Case 3.
\par
We have now proven (\ref{pf.lem.st.2.inv.u}) in each of the three Cases 1, 2
and 3. Since these three Cases cover all possibilities, this yields that
(\ref{pf.lem.st.2.inv.u}) always holds. Qed.}

Now, we are going to prove that $u\left(  k\right)  =k$ for every
$k\in\left\{  1,2,\ldots,n\right\}  $.

Indeed, assume the contrary. Then, we don't have \newline$\left(  u\left(
k\right)  =k\text{ for every }k\in\left\{  1,2,\ldots,n\right\}  \right)  $.
Thus, there exists a $k\in\left\{  1,2,\ldots,n\right\}  $ such that $u\left(
k\right)  \neq k$. Let $p$ be the smallest such $k$.

We know that $p$ is the \textbf{smallest} $k\in\left\{  1,2,\ldots,n\right\}
$ such that $u\left(  k\right)  \neq k$. Hence, if $k\in\left\{
1,2,\ldots,n\right\}  $ is such that $u\left(  k\right)  \neq k$, then%
\begin{equation}
k\geq p. \label{pf.lem.st.2.inv.min}%
\end{equation}


On the other hand, $p$ is the smallest $k\in\left\{  1,2,\ldots,n\right\}  $
such that $u\left(  k\right)  \neq k$. Hence, $p$ is a $k\in\left\{
1,2,\ldots,n\right\}  $ such that $u\left(  k\right)  \neq k$. In other words,
$p\in\left\{  1,2,\ldots,n\right\}  $ and $u\left(  p\right)  \neq p$. Let
$q=u^{-1}\left(  p\right)  $. Then, $u\left(  q\right)  =p$. Also, $p\neq
q$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $p=q$, so that
$u\left(  \underbrace{p}_{=q}\right)  =u\left(  q\right)  =p$. This
contradicts $u\left(  p\right)  \neq p$. This contradiction shows that our
assumption was wrong, qed.}. Thus, $u\left(  q\right)  =p\neq q$. Hence,
$q\geq p$ (by (\ref{pf.lem.st.2.inv.min}), applied to $k=q$). Combined with
$q\neq p$, this yields $q>p$. Thus, $p<q$. But (\ref{pf.lem.st.2.inv.u})
(applied to $i=p$ and $j=q$) yields that we have the following logical
equivalence: $\left(  p<q\right)  \ \Longleftrightarrow\ \left(  u\left(
p\right)  <u\left(  q\right)  \right)  $. Thus, $u\left(  p\right)  <u\left(
q\right)  $ holds (since $p<q$ holds). Hence, $u\left(  p\right)  <u\left(
q\right)  =p$.

If we had $u\left(  u\left(  p\right)  \right)  \neq u\left(  p\right)  $,
then we would have $u\left(  p\right)  \geq p$ (by (\ref{pf.lem.st.2.inv.min}%
), applied to $k=u\left(  p\right)  $), which would contradict $u\left(
p\right)  <p$. Hence, we do not have $u\left(  u\left(  p\right)  \right)
\neq u\left(  p\right)  $. In other words, we have $u\left(  u\left(
p\right)  \right)  =u\left(  p\right)  $. Since $u$ is an injection, this
yields $u\left(  p\right)  =p$. This contradicts $u\left(  p\right)  <p$.

This contradiction proves that our assumption was wrong. Hence, we have shown
that $u\left(  k\right)  =k$ for every $k\in\left\{  1,2,\ldots,n\right\}  $.
Thus, every $k\in\left\{  1,2,\ldots,n\right\}  $ satisfies $u\left(
k\right)  =k=\operatorname*{id}\left(  k\right)  $. Hence,
$u=\operatorname*{id}$. Since $u=s\circ t^{-1}$, this rewrites as $s\circ
t^{-1}=\operatorname*{id}$. Thus, $s=t$. Lemma \ref{lem.st.2.inv} is thus proven.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.st.2}.]Let $t$ denote the standard permutation
of $s$ (where we regard $s$ as a word by writing $s$ in one-line notation).
Regarded as a word, the permutation $s$ has the form $s=s\left(  1\right)
s\left(  2\right)  \cdots s\left(  n\right)  $.

Let $i$ and $j$ be two elements of $\left\{  1,2,\ldots,n\right\}  $ such that
$i<j$. Then, Proposition \ref{prop.st.1} \textbf{(b)} (applied to $\left\{
1,2,\ldots,n\right\}  $, $s$, $s\left(  k\right)  $ and $t$ instead of $P$,
$w$, $a_{k}$ and $s$) yields that we have the following logical equivalence:%
\[
\left(  t\left(  i\right)  <t\left(  j\right)  \right)  \ \Longleftrightarrow
\ \left(  s\left(  i\right)  <s\left(  j\right)  \right)  .
\]


Now, let us forget that we fixed $i$ and $j$. We thus have proven that every
two elements $i$ and $j$ of $\left\{  1,2,\ldots,n\right\}  $ satisfying $i<j$
satisfy the following logical equivalence:%
\[
\left(  t\left(  i\right)  <t\left(  j\right)  \right)  \ \Longleftrightarrow
\ \left(  s\left(  i\right)  <s\left(  j\right)  \right)  .
\]
Lemma \ref{lem.st.2.inv} (applied to $t$ and $s$ instead of $s$ and $t$) thus
shows that $t=s$. In other words, $t$ is $s$. In other words, the standard
permutation of $s$ is $s$ (since $t$ is the standard permutation of $s$). This
proves Proposition \ref{prop.st.2}.
\end{proof}

We can now prove Proposition 3.1. Indeed, let us prove a slightly stronger claim:

\begin{proposition}
\label{prop.st.3.1.extended}Let $\pi$ be a special double poset. Let $n$ be
the size of the special double poset $\pi$ (that is, the size of the
underlying set of $\pi$). Let $w=a_{1}a_{2}\cdots a_{n}$ be a word whose
letters belong to a totally ordered alphabet $A$.

\textbf{(a)} The word $w$ fits into $\pi$ if and only if the standard
permutation of $w$ fits into $\pi$.

\textbf{(b)} Write the special double poset $\pi$ in the form $\left(
E,<_{1},<_{2}\right)  $. Let $\omega$ be the labelling of the special double
poset $\pi$. Let $s$ be the standard permutation of $w$. The word $w$ fits
into $\pi$ if and only if the map $s\circ\omega:E\rightarrow\left\{
1,2,\ldots,n\right\}  $ is an order homomorphism $\left(  E,<_{1}\right)
\rightarrow\left\{  1,2,\ldots,n\right\}  $. (Here, $\left\{  1,2,\ldots
,n\right\}  $ denotes the poset $\left(  \left\{  1,2,\ldots,n\right\}
,<\right)  $, where the relation $<$ is the smaller relation of $\mathbb{Z}$.
In other words, $\left\{  1,2,\ldots,n\right\}  =\left(  \left\{
1,2,\ldots,n\right\}  ,<_{\mathbb{Z}}\right)  $.)
\end{proposition}

Proposition \ref{prop.st.3.1.extended} \textbf{(a)} is precisely Proposition 3.1.

\begin{proof}
[Proof of Proposition \ref{prop.st.3.1.extended}.]Write the special double
poset $\pi$ in the form $\left(  E,<_{1},<_{2}\right)  $. Let $\omega$ be the
labelling of the special double poset $\pi$.

The map $\omega$ is the labelling of the special double poset $\pi=\left(
E,<_{1},<_{2}\right)  $. In other words, $\omega$ is the unique order
isomorphism $\left(  E,<_{2}\right)  \rightarrow\left\{  1,2,\ldots,n\right\}
$ (because the labelling of the special double poset $\left(  E,<_{1}%
,<_{2}\right)  $ is the unique order isomorphism $\left(  E,<_{2}\right)
\rightarrow\left\{  1,2,\ldots,n\right\}  $ (by the definition of the
labelling of the special double poset $\left(  E,<_{1},<_{2}\right)  $)). In
particular, $\omega$ is an order isomorphism $\left(  E,<_{2}\right)
\rightarrow\left\{  1,2,\ldots,n\right\}  $. In other words, $\omega$ is an
order isomorphism $\left(  E,<_{2}\right)  \rightarrow\left(  \left\{
1,2,\ldots,n\right\}  ,<\right)  $. Hence, for any two elements $p$ and $q$ of
$E$, we have the following logical equivalence:%
\begin{equation}
\left(  p<_{2}q\right)  \ \Longleftrightarrow\ \left(  \omega\left(  p\right)
<\omega\left(  q\right)  \right)  \label{pf.prop.st.3.1.extended.labelling}%
\end{equation}
(by (\ref{pf.prop.st.1.ordiso}), applied to $\left(  U,<_{U}\right)  =\left(
E,<_{2}\right)  $, $\left(  V,<_{V}\right)  =\left(  \left\{  1,2,\ldots
,n\right\}  ,<\right)  $ and $f=\omega$).

The map $\omega$ is an order isomorphism, thus a bijection, and therefore an injection.

Let $s$ be the standard permutation of $w$. Then, $s$ is a permutation, hence
a bijection. Thus, the map $s\circ\omega$ is a bijection (since it is the
composition of the two bijections $s$ and $\omega$). In particular, this
yields that the map $s\circ\omega$ is an injection.

Let $\eta$ denote the map%
\[
E\rightarrow A,\ \ \ \ \ \ \ \ \ \ e\mapsto a_{\omega\left(  e\right)  }.
\]


The definition of the notion of \textquotedblleft the word $a_{1}a_{2}\cdots
a_{n}$ fits into $\pi$\textquotedblright\ yields the following: The word
$a_{1}a_{2}\cdots a_{n}$ fits into $\pi$ if and only if the map
\[
E\rightarrow A,\ \ \ \ \ \ \ \ \ \ e\mapsto a_{\omega\left(  e\right)  }%
\]
is a $\pi$-partition. In other words, the word $a_{1}a_{2}\cdots a_{n}$ fits
into $\pi$ if and only if the map $\eta$ is a $\pi$-partition\footnote{since
the map%
\[
E\rightarrow A,\ \ \ \ \ \ \ \ \ \ e\mapsto a_{\omega\left(  e\right)  }%
\]
is the map $\eta$}. In other words, the word $w$ fits into $\pi$ if and only
if the map $\eta$ is a $\pi$-partition (since $w=a_{1}a_{2}\cdots a_{n}$).

Definition \ref{def.pi-par} (applied to $A$ and $\eta$ instead of $X$ and $x$)
yields that $\eta$ is a $\pi$-partition if and only if any two elements $e$
and $e^{\prime}$ of $E$ satisfy%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ implies }\eta\left(  e\right)  \leq\eta\left(
e^{\prime}\right)  \right)  \label{pf.prop.st.3.1.extended.p-par.1}%
\end{equation}
and%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ and }e\geq_{2}e^{\prime}\text{ implies }%
\eta\left(  e\right)  <\eta\left(  e^{\prime}\right)  \right)  .
\label{pf.prop.st.3.1.extended.p-par.2}%
\end{equation}


\textbf{(b)} We are going to prove the following two logical implications:%
\begin{align}
&  \ \left(  \text{the word }w\text{ fits into }\pi\right) \nonumber\\
&  \Longrightarrow\ \left(  \text{the map }s\circ\omega\text{ is an order
homomorphism }\left(  E,<_{1}\right)  \rightarrow\left\{  1,2,\ldots
,n\right\}  \right)  \label{pf.prop.st.3.1.extended.b.=>}%
\end{align}
and%
\begin{align}
&  \ \left(  \text{the map }s\circ\omega\text{ is an order homomorphism
}\left(  E,<_{1}\right)  \rightarrow\left\{  1,2,\ldots,n\right\}  \right)
\nonumber\\
&  \Longrightarrow\ \left(  \text{the word }w\text{ fits into }\pi\right)  .
\label{pf.prop.st.3.1.extended.b.<=}%
\end{align}


\textit{Proof of (\ref{pf.prop.st.3.1.extended.b.=>}):} We assume that the
word $w$ fits into $\pi$. We then want to show that the map $s\circ\omega$ is
an order homomorphism $\left(  E,<_{1}\right)  \rightarrow\left\{
1,2,\ldots,n\right\}  $.

Recall that the word $w$ fits into $\pi$ if and only if the map $\eta$ is a
$\pi$-partition. Hence, the map $\eta$ is a $\pi$-partition (since we know
that the word $w$ fits into $\pi$). In other words, any two elements $e$ and
$e^{\prime}$ of $E$ satisfy (\ref{pf.prop.st.3.1.extended.p-par.1}) and
(\ref{pf.prop.st.3.1.extended.p-par.2})\footnote{since we know that $\eta$ is
a $\pi$-partition if and only if any two elements $e$ and $e^{\prime}$ of $E$
satisfy (\ref{pf.prop.st.3.1.extended.p-par.1}) and
(\ref{pf.prop.st.3.1.extended.p-par.2})}.

Now, let $e$ and $e^{\prime}$ be two elements of $E$ such that $e\leq
_{1}e^{\prime}$. We are going to show that $\left(  s\circ\omega\right)
\left(  e\right)  \leq\left(  s\circ\omega\right)  \left(  e^{\prime}\right)
$. This is obvious when $e=e^{\prime}$ (because when $e=e^{\prime}$, then
$\left(  s\circ\omega\right)  \left(  \underbrace{e}_{=e^{\prime}}\right)
=\left(  s\circ\omega\right)  \left(  e^{\prime}\right)  $). Hence, for the
rest of this proof, we can WLOG assume that we don't have $e=e^{\prime}$.
Assume this. Then, $e\neq e^{\prime}$ (since not $e=e^{\prime}$) and thus
$e<_{1}e^{\prime}$ (since $e\leq_{1}e^{\prime}$ and $e\neq e^{\prime}$) and
thus $\eta\left(  e\right)  \leq\eta\left(  e^{\prime}\right)  $ (by
(\ref{pf.prop.st.3.1.extended.p-par.1})). The definition of $\eta$ yields
$\eta\left(  e\right)  =a_{\omega\left(  e\right)  }$ and $\eta\left(
e^{\prime}\right)  =a_{\omega\left(  e^{\prime}\right)  }$. Thus,
$a_{\omega\left(  e\right)  }=\eta\left(  e\right)  \leq\eta\left(  e^{\prime
}\right)  =a_{\omega\left(  e^{\prime}\right)  }$.

We must be in one of the following two cases:

\textit{Case 1:} We have $\omega\left(  e\right)  <\omega\left(  e^{\prime
}\right)  $.

\textit{Case 2:} We have $\omega\left(  e\right)  \geq\omega\left(  e^{\prime
}\right)  $.

Let us first consider Case 1. In this case, we have $\omega\left(  e\right)
<\omega\left(  e^{\prime}\right)  $. Hence, Proposition \ref{prop.st.1}
\textbf{(b)} (applied to $i=\omega\left(  e\right)  $ and $j=\omega\left(
e^{\prime}\right)  $) yields that we have the following logical equivalence:%
\[
\left(  s\left(  \omega\left(  e\right)  \right)  <s\left(  \omega\left(
e^{\prime}\right)  \right)  \right)  \ \Longleftrightarrow\ \left(
a_{\omega\left(  e\right)  }\leq a_{\omega\left(  e^{\prime}\right)  }\right)
.
\]
Thus, we have $s\left(  \omega\left(  e\right)  \right)  <s\left(
\omega\left(  e^{\prime}\right)  \right)  $ (since we know that we have
$a_{\omega\left(  e\right)  }\leq a_{\omega\left(  e^{\prime}\right)  }$).
Hence, $\left(  s\circ\omega\right)  \left(  e\right)  =s\left(  \omega\left(
e\right)  \right)  <s\left(  \omega\left(  e^{\prime}\right)  \right)
=\left(  s\circ\omega\right)  \left(  e^{\prime}\right)  $. Thus, $\left(
s\circ\omega\right)  \left(  e\right)  \leq\left(  s\circ\omega\right)
\left(  e^{\prime}\right)  $. We have thus proven $\left(  s\circ
\omega\right)  \left(  e\right)  \leq\left(  s\circ\omega\right)  \left(
e^{\prime}\right)  $ in Case 1.

Let us now consider Case 2. In this case, we have $\omega\left(  e\right)
\geq\omega\left(  e^{\prime}\right)  $. If we had $\omega\left(  e\right)
=\omega\left(  e^{\prime}\right)  $, then we would have $e=e^{\prime}$ (since
the map $\omega$ is an injection), which would contradict $e\neq e^{\prime}$.
Hence, we cannot have $\omega\left(  e\right)  =\omega\left(  e^{\prime
}\right)  $. We thus have $\omega\left(  e\right)  \neq\omega\left(
e^{\prime}\right)  $. Combined with $\omega\left(  e\right)  \geq\omega\left(
e^{\prime}\right)  $, this yields $\omega\left(  e\right)  >\omega\left(
e^{\prime}\right)  $. Thus, $\omega\left(  e^{\prime}\right)  <\omega\left(
e\right)  $. But (\ref{pf.prop.st.3.1.extended.labelling}) (applied to
$p=e^{\prime}$ and $q=e$) yields that we have the following logical
equivalence: $\left(  e^{\prime}<_{2}e\right)  \ \Longleftrightarrow\ \left(
\omega\left(  e^{\prime}\right)  <\omega\left(  e\right)  \right)  $. Hence,
$e^{\prime}<_{2}e$ holds (since $\omega\left(  e^{\prime}\right)
<\omega\left(  e\right)  $ holds). Thus, $e>_{2}e^{\prime}$, so that
$e\geq_{2}e^{\prime}$. Therefore, (\ref{pf.prop.st.3.1.extended.p-par.2})
yields $\eta\left(  e\right)  <\eta\left(  e^{\prime}\right)  $. Hence,
$a_{\omega\left(  e\right)  }=\eta\left(  e\right)  <\eta\left(  e^{\prime
}\right)  =a_{\omega\left(  e^{\prime}\right)  }$.

But $\omega\left(  e\right)  \geq\omega\left(  e^{\prime}\right)  $. Hence,
Proposition \ref{prop.st.1} \textbf{(c)} (applied to $i=\omega\left(
e\right)  $ and $j=\omega\left(  e^{\prime}\right)  $) yields that we have the
following logical equivalence:%
\[
\left(  s\left(  \omega\left(  e\right)  \right)  <s\left(  \omega\left(
e^{\prime}\right)  \right)  \right)  \ \Longleftrightarrow\ \left(
a_{\omega\left(  e\right)  }<a_{\omega\left(  e^{\prime}\right)  }\right)  .
\]
Thus, we have $s\left(  \omega\left(  e\right)  \right)  <s\left(
\omega\left(  e^{\prime}\right)  \right)  $ (since we know that we have
$a_{\omega\left(  e\right)  }<a_{\omega\left(  e^{\prime}\right)  }$). Hence,
$\left(  s\circ\omega\right)  \left(  e\right)  =s\left(  \omega\left(
e\right)  \right)  <s\left(  \omega\left(  e^{\prime}\right)  \right)
=\left(  s\circ\omega\right)  \left(  e^{\prime}\right)  $. Thus, $\left(
s\circ\omega\right)  \left(  e\right)  \leq\left(  s\circ\omega\right)
\left(  e^{\prime}\right)  $. We have thus proven $\left(  s\circ
\omega\right)  \left(  e\right)  \leq\left(  s\circ\omega\right)  \left(
e^{\prime}\right)  $ in Case 2.

Hence, we have proven $\left(  s\circ\omega\right)  \left(  e\right)
\leq\left(  s\circ\omega\right)  \left(  e^{\prime}\right)  $ in each of the
two Cases 1 and 2. Since these two Cases cover all possibilities, we thus
conclude that $\left(  s\circ\omega\right)  \left(  e\right)  \leq\left(
s\circ\omega\right)  \left(  e^{\prime}\right)  $ always holds.

Now, let us forget that we fixed $e$ and $e^{\prime}$. We thus have proven
that if $e$ and $e^{\prime}$ are any two elements of $E$ such that $e\leq
_{1}e^{\prime}$, then $\left(  s\circ\omega\right)  \left(  e\right)
\leq\left(  s\circ\omega\right)  \left(  e^{\prime}\right)  $. In other words,
the map $s\circ\omega$ is an order homomorphism $\left(  E,<_{1}\right)
\rightarrow\left\{  1,2,\ldots,n\right\}  $. Thus,
(\ref{pf.prop.st.3.1.extended.b.=>}) is proven.

\textit{Proof of (\ref{pf.prop.st.3.1.extended.b.<=}):} We assume that the map
$s\circ\omega$ is an order homomorphism $\left(  E,<_{1}\right)
\rightarrow\left\{  1,2,\ldots,n\right\}  $. We are going to show that the
word $w$ fits into $\pi$.

Let now $e$ and $e^{\prime}$ be any two elements of $E$. We will prove
(\ref{pf.prop.st.3.1.extended.p-par.1}) and
(\ref{pf.prop.st.3.1.extended.p-par.2}).

\textit{Proof of (\ref{pf.prop.st.3.1.extended.p-par.1}):} Assume that
$e<_{1}e^{\prime}$. We are going to show that $\eta\left(  e\right)  \leq
\eta\left(  e^{\prime}\right)  $.

We have $e<_{1}e^{\prime}$, so that $e\leq_{1}e^{\prime}$. Thus, $\left(
s\circ\omega\right)  \left(  e\right)  \leq\left(  s\circ\omega\right)
\left(  e^{\prime}\right)  $ (since the map $s\circ\omega$ is an order
homomorphism $\left(  E,<_{1}\right)  \rightarrow\left\{  1,2,\ldots
,n\right\}  $). But if we had $\left(  s\circ\omega\right)  \left(  e\right)
=\left(  s\circ\omega\right)  \left(  e^{\prime}\right)  $, then we would have
$e=e^{\prime}$ (since the map $s\circ\omega$ is an injection), which would
contradict $e<_{1}e^{\prime}$. Thus, we do not have $\left(  s\circ
\omega\right)  \left(  e\right)  =\left(  s\circ\omega\right)  \left(
e^{\prime}\right)  $. In other words, we have $\left(  s\circ\omega\right)
\left(  e\right)  \neq\left(  s\circ\omega\right)  \left(  e^{\prime}\right)
$. Combined with $\left(  s\circ\omega\right)  \left(  e\right)  \leq\left(
s\circ\omega\right)  \left(  e^{\prime}\right)  $, this yields $\left(
s\circ\omega\right)  \left(  e\right)  <\left(  s\circ\omega\right)  \left(
e^{\prime}\right)  $. Thus,
\begin{equation}
s\left(  \omega\left(  e\right)  \right)  =\left(  s\circ\omega\right)
\left(  e\right)  <\left(  s\circ\omega\right)  \left(  e^{\prime}\right)
=s\left(  \omega\left(  e^{\prime}\right)  \right)  .
\label{pf.prop.st.3.1.extended.b.<=.pf.U}%
\end{equation}
Thus, Proposition \ref{prop.st.1} \textbf{(d)} (applied to $i=\omega\left(
e\right)  $ and $j=\omega\left(  e^{\prime}\right)  $) yields $a_{\omega
\left(  e\right)  }\leq a_{\omega\left(  e^{\prime}\right)  }$. But the
definition of $\eta$ yields $\eta\left(  e\right)  =a_{\omega\left(  e\right)
}$ and $\eta\left(  e^{\prime}\right)  =a_{\omega\left(  e^{\prime}\right)  }%
$. Hence, $\eta\left(  e\right)  =a_{\omega\left(  e\right)  }\leq
a_{\omega\left(  e^{\prime}\right)  }=\eta\left(  e^{\prime}\right)  $. Hence,
$\eta\left(  e\right)  \leq\eta\left(  e^{\prime}\right)  $ is shown. Thus, we
have proven (\ref{pf.prop.st.3.1.extended.p-par.1}).

\textit{Proof of (\ref{pf.prop.st.3.1.extended.p-par.2}):} Assume that
$e<_{1}e^{\prime}$ and $e\geq_{2}e^{\prime}$. We are going to show that
$\eta\left(  e\right)  <\eta\left(  e^{\prime}\right)  $.

We have $e<_{1}e^{\prime}$. Hence, we can prove
(\ref{pf.prop.st.3.1.extended.b.<=.pf.U}) (in the same way as we have proved
it in our above proof of (\ref{pf.prop.st.3.1.extended.p-par.1})). Thus, we
have $s\left(  \omega\left(  e\right)  \right)  <s\left(  \omega\left(
e^{\prime}\right)  \right)  $.

We have $e\neq e^{\prime}$ (since $e<_{1}e^{\prime}$). Combined with
$e\geq_{2}e^{\prime}$, this yields $e>_{2}e^{\prime}$, so that $e^{\prime
}<_{2}e$. But (\ref{pf.prop.st.3.1.extended.labelling}) (applied to
$p=e^{\prime}$ and $q=e$) yields that we have the following logical
equivalence: $\left(  e^{\prime}<_{2}e\right)  \ \Longleftrightarrow\ \left(
\omega\left(  e^{\prime}\right)  <\omega\left(  e\right)  \right)  $. Thus, we
have $\omega\left(  e^{\prime}\right)  <\omega\left(  e\right)  $ (since we
have $e^{\prime}<_{2}e$). Hence, $\omega\left(  e\right)  >\omega\left(
e^{\prime}\right)  $, so that $\omega\left(  e\right)  \geq\omega\left(
e^{\prime}\right)  $. Thus, Proposition \ref{prop.st.1} \textbf{(c)} (applied
to $i=\omega\left(  e\right)  $ and $j=\omega\left(  e^{\prime}\right)  $)
yields that we have the following logical equivalence:%
\[
\left(  s\left(  \omega\left(  e\right)  \right)  <s\left(  \omega\left(
e^{\prime}\right)  \right)  \right)  \ \Longleftrightarrow\ \left(
a_{\omega\left(  e\right)  }<a_{\omega\left(  e^{\prime}\right)  }\right)  .
\]
Hence, we have $a_{\omega\left(  e\right)  }<a_{\omega\left(  e^{\prime
}\right)  }$ (since we have $s\left(  \omega\left(  e\right)  \right)
<s\left(  \omega\left(  e^{\prime}\right)  \right)  $). But the definition of
$\eta$ yields $\eta\left(  e\right)  =a_{\omega\left(  e\right)  }$ and
$\eta\left(  e^{\prime}\right)  =a_{\omega\left(  e^{\prime}\right)  }$.
Hence, $\eta\left(  e\right)  =a_{\omega\left(  e\right)  }<a_{\omega\left(
e^{\prime}\right)  }=\eta\left(  e^{\prime}\right)  $. Hence, $\eta\left(
e\right)  <\eta\left(  e^{\prime}\right)  $ is shown. Thus, we have proven
(\ref{pf.prop.st.3.1.extended.p-par.2}).

Now, we have proven both (\ref{pf.prop.st.3.1.extended.p-par.1}) and
(\ref{pf.prop.st.3.1.extended.p-par.2}).

Let us now forget that we fixed $e$ and $e^{\prime}$. We thus have shown that
any two elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{pf.prop.st.3.1.extended.p-par.1}) and
(\ref{pf.prop.st.3.1.extended.p-par.2}). In other words, the map $\eta$ is a
$\pi$-partition\footnote{since we know that $\eta$ is a $\pi$-partition if and
only if any two elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{pf.prop.st.3.1.extended.p-par.1}) and
(\ref{pf.prop.st.3.1.extended.p-par.2})}. In other words, the word $w$ fits
into $\pi$ (since we know that the word $w$ fits into $\pi$ if and only if the
map $\eta$ is a $\pi$-partition). This proves
(\ref{pf.prop.st.3.1.extended.b.<=}).

We now have proven both implications (\ref{pf.prop.st.3.1.extended.b.=>}) and
(\ref{pf.prop.st.3.1.extended.b.<=}). Combining these two implications, we
obtain the logical equivalence%
\begin{align*}
&  \ \left(  \text{the word }w\text{ fits into }\pi\right) \\
&  \Longleftrightarrow\ \left(  \text{the map }s\circ\omega\text{ is an order
homomorphism }\left(  E,<_{1}\right)  \rightarrow\left\{  1,2,\ldots
,n\right\}  \right)  .
\end{align*}
In other words, Proposition \ref{prop.st.3.1.extended} \textbf{(b)} is proven.

\textbf{(a)} Proposition \ref{prop.st.3.1.extended} \textbf{(b)} yields that
the word $w$ fits into $\pi$ if and only if the map $s\circ\omega
:E\rightarrow\left\{  1,2,\ldots,n\right\}  $ is an order homomorphism
$\left(  E,<_{1}\right)  \rightarrow\left\{  1,2,\ldots,n\right\}  $. In other
words, we have the following logical equivalence:%
\begin{align}
&  \ \left(  \text{the word }w\text{ fits into }\pi\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the map }s\circ\omega\text{ is an order
homomorphism }\left(  E,<_{1}\right)  \rightarrow\left\{  1,2,\ldots
,n\right\}  \right)  . \label{pf.prop.st.3.1.extended.a.1}%
\end{align}


But Proposition \ref{prop.st.2} shows that the standard permutation of the
word $s$ is $s$ again. Hence, we can also apply
(\ref{pf.prop.st.3.1.extended.a.1}) to $\left\{  1,2,\ldots,n\right\}  $, $s$
and $s\left(  k\right)  $ instead of $A$, $w$ and $a_{k}$. As a result, we
obtain the following logical equivalence:%
\begin{align}
&  \ \left(  \text{the word }s\text{ fits into }\pi\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the map }s\circ\omega\text{ is an order
homomorphism }\left(  E,<_{1}\right)  \rightarrow\left\{  1,2,\ldots
,n\right\}  \right)  . \label{pf.prop.st.3.1.extended.a.2}%
\end{align}


Hence, we have the following logical equivalence:%
\begin{align*}
&  \ \left(  \text{the word }w\text{ fits into }\pi\right) \\
&  \Longleftrightarrow\ \left(  \text{the map }s\circ\omega\text{ is an order
homomorphism }\left(  E,<_{1}\right)  \rightarrow\left\{  1,2,\ldots
,n\right\}  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.st.3.1.extended.a.1}%
)}\right) \\
&  \Longleftrightarrow\ \left(  \text{the word }s\text{ fits into }\pi\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.st.3.1.extended.a.2}%
)}\right) \\
&  \Longleftrightarrow\ \left(  \text{the standard permutation of }w\text{
fits into }\pi\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }s\text{ is the standard
permutation of }w\right)  .
\end{align*}
In other words, Proposition \ref{prop.st.3.1.extended} \textbf{(a)} is proven.
\end{proof}

Now that we are already studying $\pi$-partitions, let us also prove Theorem 1
in [G]. Let us first give our own version of it (whose equivalence to Theorem
1 in [G], however, is rather obvious, unlike that of Proposition 3.1):

\begin{theorem}
\label{thm.gessel-thm1}Let $X$ be a totally ordered alphabet. For every double
poset $D$, we let $\mathcal{A}\left(  D\right)  $ denote the set of all
$D$-partitions into $X$.

Let $\pi=\left(  E,<_{1},<_{2}\right)  $ be a special double poset. Then, the
sets $\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  $, where
$\prec$ runs over all linear extensions of $\pi$, are pairwise disjoint. The
union of these sets is $\mathcal{A}\left(  \pi\right)  $.
\end{theorem}

Theorem \ref{thm.gessel-thm1} is not only easily seen to be equivalent to
Theorem 1 in [G], but also quickly yields \cite[Theorem 5.2.11]{GriRei15}.

Before we prove Theorem \ref{thm.gessel-thm1}, let us state an auxiliary fact
which itself might be of some interest:

\begin{proposition}
\label{prop.gessel-thm1.subdiv}Let $X$ be a totally ordered alphabet. For
every double poset $D$, we let $\mathcal{A}\left(  D\right)  $ denote the set
of all $D$-partitions into $X$.

Let $\pi=\left(  E,<_{1},<_{2}\right)  $ be a special double poset. Let $n$ be
the size of the special double poset $\pi$ (that is, the size of the
underlying set $E$ of $\pi$). Let $\omega$ be the labelling of the special
double poset $\pi$. Thus, $\omega$ is the unique order isomorphism $\left(
E,<_{2}\right)  \rightarrow\left\{  1,2,\ldots,n\right\}  $ (by the definition
of the labelling of the special double poset $\left(  E,<_{1},<_{2}\right)  $).

For every $f:E\rightarrow X$, we let $\mathbf{w}\left(  f\right)  $ denote the
word $f\left(  \omega^{-1}\left(  1\right)  \right)  f\left(  \omega
^{-1}\left(  2\right)  \right)  \cdots f\left(  \omega^{-1}\left(  n\right)
\right)  $. This word $\mathbf{w}\left(  f\right)  $ gives rise to a binary
relation $\prec_{\mathbf{w}\left(  f\right)  }$ on the set $\left\{
1,2,\ldots,n\right\}  $ (defined according to Definition \ref{def.st}), and
this relation, in turn, gives rise to a binary relation $\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}$ on the set $E$.

\textbf{(a)} If $f\in\mathcal{A}\left(  \pi\right)  $, then the binary
relation $\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}$ is a
linear extension of $\pi$.

\textbf{(b)} If $\prec$ is any linear extension of $\pi$, then%
\[
\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  =\left\{
f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation }\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals }%
\prec\right\}  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.gessel-thm1.subdiv}.]We know that the map
$\omega$ is an order isomorphism $\left(  E,<_{2}\right)  \rightarrow\left(
\left\{  1,2,\ldots,n\right\}  ,<\right)  $. Hence, for any two elements $p$
and $q$ of $E$, we have the following logical equivalence:%
\begin{equation}
\left(  p<_{2}q\right)  \ \Longleftrightarrow\ \left(  \omega\left(  p\right)
<\omega\left(  q\right)  \right)  \label{pf.prop.gessel-thm1.subdiv.ordiso}%
\end{equation}
(by (\ref{pf.prop.st.1.ordiso}), applied to $\left(  U,<_{U}\right)  =\left(
E,<_{2}\right)  $, $\left(  V,<_{V}\right)  =\left(  \left\{  1,2,\ldots
,n\right\}  ,<\right)  $ and $f=\omega$).

The map $\omega$ is an order isomorphism, thus a bijection, thus an injection.

\textbf{(a)} Let $f\in\mathcal{A}\left(  \pi\right)  $. Then, $f$ is an
element of $\mathcal{A}\left(  \pi\right)  $. In other words, $f$ is a $\pi
$-partition into $X$ (since $\mathcal{A}\left(  \pi\right)  $ is the set of
all $\pi$-partitions into $X$ (by the definition of $\mathcal{A}\left(
\pi\right)  $)).

Definition \ref{def.pi-par} (applied to $x=f$) yields that the map $f$ is a
$\pi$-partition if and only if any two elements $e$ and $e^{\prime}$ of $E$
satisfy%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ implies }f\left(  e\right)  \leq f\left(
e^{\prime}\right)  \right)  \label{pf.prop.gessel-thm1.subdiv.1}%
\end{equation}
and%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ and }e\geq_{2}e^{\prime}\text{ implies
}f\left(  e\right)  <f\left(  e^{\prime}\right)  \right)  .
\label{pf.prop.gessel-thm1.subdiv.2}%
\end{equation}
Hence, any two elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{pf.prop.gessel-thm1.subdiv.1}) and (\ref{pf.prop.gessel-thm1.subdiv.2})
(since $f$ is a $\pi$-partition).

Proposition \ref{prop.st.constr} \textbf{(b)} (applied to $\mathbf{w}\left(
f\right)  $, $f\left(  \omega^{-1}\left(  k\right)  \right)  $ and $X$ instead
of $w$, $a_{k}$ and $P$) yields that the binary relation $\prec_{\mathbf{w}%
\left(  f\right)  }$ is the smaller relation of a total order on the set
$\left\{  1,2,\ldots,n\right\}  $. Thus, Proposition \ref{prop.assoc.AB.total}
(applied to $E$, $\left\{  1,2,\ldots,n\right\}  $, $\prec_{\mathbf{w}\left(
f\right)  }$ and $\omega$ instead of $A$, $B$, $<_{B}$ and $f$) yields that
the relation $\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}$
is the smaller relation of a total order on $E$.

We shall now show that this relation $\left(  \prec_{\mathbf{w}\left(
f\right)  }\right)  ^{\omega}$ extends the first order $<_{1}$ on $E$.

For any two elements $e$ and $e^{\prime}$ of $E$, we have the following
logical equivalence:%
\begin{align}
&  \ \left(  e\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega
}e^{\prime}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{either }f\left(  e\right)  <f\left(
e^{\prime}\right)  \text{ or }\left(  f\left(  e\right)  =f\left(  e^{\prime
}\right)  \text{ and }\omega\left(  e\right)  <\omega\left(  e^{\prime
}\right)  \right)  \right)  . \label{pf.prop.gessel-thm1.subdiv.5}%
\end{align}
\footnote{\textit{Proof of (\ref{pf.prop.gessel-thm1.subdiv.5}):} Let $e$ and
$e^{\prime}$ be two elements of $E$. We need to prove the logical equivalence
(\ref{pf.prop.gessel-thm1.subdiv.5}).
\par
Define $n$ elements $a_{1}$, $a_{2}$, $\ldots$, $a_{n}$ of $X$ by $\left(
a_{k}=f\left(  \omega^{-1}\left(  k\right)  \right)  \text{ for every }%
k\in\left\{  1,2,\ldots,n\right\}  \right)  $. Then, $a_{1}a_{2}\cdots
a_{n}=f\left(  \omega^{-1}\left(  1\right)  \right)  f\left(  \omega
^{-1}\left(  2\right)  \right)  \cdots f\left(  \omega^{-1}\left(  n\right)
\right)  =\mathbf{w}\left(  f\right)  $.
\par
Let $i=\omega\left(  e\right)  $ and $j=\omega\left(  e^{\prime}\right)  $.
Then, the definition of $a_{i}$ yields $a_{i}=f\left(  \underbrace{\omega
^{-1}\left(  i\right)  }_{\substack{=e\\\text{(since }i=\omega\left(
e\right)  \text{)}}}\right)  =f\left(  e\right)  $, and the definition of
$a_{j}$ yields $a_{j}=f\left(  \underbrace{\omega^{-1}\left(  j\right)
}_{\substack{=e^{\prime}\\\text{(since }j=\omega\left(  e^{\prime}\right)
\text{)}}}\right)  =f\left(  e^{\prime}\right)  $.
\par
Recall that $\mathbf{w}\left(  f\right)  =a_{1}a_{2}\cdots a_{n}$. Hence, we
have $i\prec_{\mathbf{w}\left(  f\right)  }j$ if and only if $\left(
\text{either }a_{i}<a_{j}\text{ or }\left(  a_{i}=a_{j}\text{ and }i<j\right)
\right)  $ (due to the definition of \textquotedblleft$i\prec_{\mathbf{w}%
\left(  f\right)  }j$\textquotedblright). In other words, we have the
following logical equivalence:%
\[
\left(  i\prec_{\mathbf{w}\left(  f\right)  }j\right)  \ \Longleftrightarrow
\ \left(  \text{either }a_{i}<a_{j}\text{ or }\left(  a_{i}=a_{j}\text{ and
}i<j\right)  \right)  .
\]
\par
We have $e\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega
}e^{\prime}$ if and only if $\omega\left(  e\right)  \prec_{\mathbf{w}\left(
f\right)  }\omega\left(  e^{\prime}\right)  $ (due to the definition of
\textquotedblleft$e\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)
^{\omega}e^{\prime}$\textquotedblright). Hence, we have the following logical
equivalence:%
\begin{align*}
&  \ \left(  e\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega
}e^{\prime}\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\omega\left(  e\right)  }%
_{=i}\prec_{\mathbf{w}\left(  f\right)  }\underbrace{\omega\left(  e^{\prime
}\right)  }_{=j}\right)  \ \Longleftrightarrow\ \left(  i\prec_{\mathbf{w}%
\left(  f\right)  }j\right) \\
&  \Longleftrightarrow\ \left(  \text{either }\underbrace{a_{i}}_{=f\left(
e\right)  }<\underbrace{a_{j}}_{=f\left(  e^{\prime}\right)  }\text{ or
}\left(  \underbrace{a_{i}}_{=f\left(  e\right)  }=\underbrace{a_{j}%
}_{=f\left(  e^{\prime}\right)  }\text{ and }\underbrace{i}_{=\omega\left(
e\right)  }<\underbrace{j}_{=\omega\left(  e^{\prime}\right)  }\right)
\right) \\
&  \Longleftrightarrow\ \left(  \text{either }f\left(  e\right)  <f\left(
e^{\prime}\right)  \text{ or }\left(  f\left(  e\right)  =f\left(  e^{\prime
}\right)  \text{ and }\omega\left(  e\right)  <\omega\left(  e^{\prime
}\right)  \right)  \right)  .
\end{align*}
Thus, (\ref{pf.prop.gessel-thm1.subdiv.5}) is proven.}

Now, let $e$ and $e^{\prime}$ be two elements of $E$ such that $e<_{1}%
e^{\prime}$. We are going to show that $e\left(  \prec_{\mathbf{w}\left(
f\right)  }\right)  ^{\omega}e^{\prime}$.

We now recall that $<_{2}$ is a total order (since $\left(  E,<_{1}%
,<_{2}\right)  $ is a special double poset). Hence, we have either
$e<_{2}e^{\prime}$ or $e\geq_{2}e^{\prime}$. In other words, we must be in one
of the following two cases:

\textit{Case 1:} We have $e<_{2}e^{\prime}$.

\textit{Case 2:} We have $e\geq_{2}e^{\prime}$.

Let us first consider Case 1. In this case, we have $e<_{2}e^{\prime}$.
Applying (\ref{pf.prop.gessel-thm1.subdiv.ordiso}) to $p=e$ and $q=e^{\prime}%
$, we obtain the following logical equivalence: $\left(  e<_{2}e^{\prime
}\right)  \ \Longleftrightarrow\ \left(  \omega\left(  e\right)
<\omega\left(  e^{\prime}\right)  \right)  $. Hence, we have $\omega\left(
e\right)  <\omega\left(  e^{\prime}\right)  $ (since $e<_{2}e^{\prime}$). But
(\ref{pf.prop.gessel-thm1.subdiv.1}) shows that $f\left(  e\right)  \leq
f\left(  e^{\prime}\right)  $ (since $e<_{1}e^{\prime}$). Thus, either
$f\left(  e\right)  <f\left(  e^{\prime}\right)  $ or $f\left(  e\right)
=f\left(  e^{\prime}\right)  $. Hence, either $f\left(  e\right)  <f\left(
e^{\prime}\right)  $ or $\left(  f\left(  e\right)  =f\left(  e^{\prime
}\right)  \text{ and }\omega\left(  e\right)  <\omega\left(  e^{\prime
}\right)  \right)  $ (because we have $i<j$). Thus, we have $e\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}e^{\prime}$ (because of
the equivalence (\ref{pf.prop.gessel-thm1.subdiv.5})). Hence, $e\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}e^{\prime}$ is proven in
Case 1.

Let us now consider Case 2. In this case, we have $e\geq_{2}e^{\prime}$.
Hence, (\ref{pf.prop.gessel-thm1.subdiv.2}) shows that $f\left(  e\right)
<f\left(  e^{\prime}\right)  $ (since $e<_{1}e^{\prime}$). Hence, either
$f\left(  e\right)  <f\left(  e^{\prime}\right)  $ or \newline$\left(
f\left(  e\right)  =f\left(  e^{\prime}\right)  \text{ and }\omega\left(
e\right)  <\omega\left(  e^{\prime}\right)  \right)  $. Thus, we have
$e\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}e^{\prime}$
(because of the equivalence (\ref{pf.prop.gessel-thm1.subdiv.5})). Hence,
$e\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}e^{\prime}$ is
proven in Case 2.

Thus, $e\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega
}e^{\prime}$ is proven in each of the two Cases 1 and 2. Since these two Cases
cover all possibilities, this yields that $e\left(  \prec_{\mathbf{w}\left(
f\right)  }\right)  ^{\omega}e^{\prime}$ always holds.

Let us now forget that we fixed $e$ and $e^{\prime}$. We thus have shown that
if $e$ and $e^{\prime}$ are any two elements of $E$ such that $e<_{1}%
e^{\prime}$, then $e\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)
^{\omega}e^{\prime}$. Hence, the relation $\left(  \prec_{\mathbf{w}\left(
f\right)  }\right)  ^{\omega}$ extends the relation $<_{1}$. Combining this
with the fact that $\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)
^{\omega}$ is the smaller relation of a total order on $E$, we conclude that
$\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}$ is a total
order on $E$ which extends the first order $<_{1}$ of $E$. In other words,
$\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}$ is a linear
extension of $\pi$ (because this is how a linear extension of $\pi$ is
defined). This proves Proposition \ref{prop.gessel-thm1.subdiv} \textbf{(a)}.

\textbf{(b)} Let $\prec$ be any linear extension of $\pi$. In other words,
$\prec$ is a total order on $E$ which extends the first order $<_{1}$ of $E$
(because this is how a linear extension of $\pi$ is defined).

We are going to prove that%
\begin{equation}
\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  \subseteq\left\{
f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation }\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals }%
\prec\right\}  \label{pf.prop.gessel-thm1.subdiv.b.left}%
\end{equation}
and%
\begin{equation}
\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation
}\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals
}\prec\right\}  \subseteq\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)
\right)  . \label{pf.prop.gessel-thm1.subdiv.b.right}%
\end{equation}


\textit{Proof of (\ref{pf.prop.gessel-thm1.subdiv.b.left}):} Let
$g\in\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  $. We shall
show that $g\in\mathcal{A}\left(  \pi\right)  $ and that the binary relation
$\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}$ equals
$\prec$.

We know that $g$ is an element of $\mathcal{A}\left(  \left(  E,\prec
,<_{2}\right)  \right)  $. In other words, $f$ is a $\left(  E,\prec
,<_{2}\right)  $-partition into $X$ (since $\mathcal{A}\left(  \left(
E,\prec,<_{2}\right)  \right)  $ is the set of all $\left(  E,\prec
,<_{2}\right)  $-partitions into $X$ (by the definition of $\mathcal{A}\left(
\left(  E,\prec,<_{2}\right)  \right)  $)).

Definition \ref{def.pi-par} (applied to $g$, $\prec$ and $\left(
E,\prec,<_{2}\right)  $ instead of $x$, $<_{1}$ and $\pi$) yields that the map
$g$ is a $\left(  E,\prec,<_{2}\right)  $-partition if and only if any two
elements $e$ and $e^{\prime}$ of $E$ satisfy%
\begin{equation}
\left(  e\prec e^{\prime}\text{ implies }g\left(  e\right)  \leq g\left(
e^{\prime}\right)  \right)  \label{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1}%
\end{equation}
and%
\begin{equation}
\left(  e\prec e^{\prime}\text{ and }e\geq_{2}e^{\prime}\text{ implies
}g\left(  e\right)  <g\left(  e^{\prime}\right)  \right)  .
\label{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2}%
\end{equation}
Hence, any two elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1}) and
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2}) (since $g$ is a $\left(
E,\prec,<_{2}\right)  $-partition). Now, it is easy to see that any two
elements $e$ and $e^{\prime}$ of $E$ satisfy%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ implies }g\left(  e\right)  \leq g\left(
e^{\prime}\right)  \right)
\label{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1'}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1'}%
):} Let $e$ and $e^{\prime}$ be two elements of $E$. We need to prove
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1'}). Assume that
$e<_{1}e^{\prime}$. Then, $e\prec e^{\prime}$ (since the relation $\prec$
extends the first order $<_{1}$ of $E$). Hence, $g\left(  e\right)  \leq
g\left(  e^{\prime}\right)  $ (according to
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1})). This proves
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1'}).} and%
\begin{equation}
\left(  e<_{1}e^{\prime}\text{ and }e\geq_{2}e^{\prime}\text{ implies
}g\left(  e\right)  <g\left(  e^{\prime}\right)  \right)
\label{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2'}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2'}%
):} Let $e$ and $e^{\prime}$ be two elements of $E$. We need to prove
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2'}). Assume that
$e<_{1}e^{\prime}$ and $e\geq_{2}e^{\prime}$. We have $e<_{1}e^{\prime}$;
therefore, $e\prec e^{\prime}$ (since the relation $\prec$ extends the first
order $<_{1}$ of $E$). Hence, $g\left(  e\right)  <g\left(  e^{\prime}\right)
$ (according to (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2})). This
proves (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2'}).}. But Definition
\ref{def.pi-par} (applied to $x=g$) yields that the map $g$ is a $\pi
$-partition if and only if any two elements $e$ and $e^{\prime}$ of $E$
satisfy (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1'}) and
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2'}). Hence, the map $g$ is a
$\pi$-partition (since any two elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1'}) and
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2'})). More precisely, $g$ is a
$\pi$-partition into $X$. In other words, $g$ is an element of $\mathcal{A}%
\left(  \pi\right)  $ (since $\mathcal{A}\left(  \pi\right)  $ is the set of
all $\pi$-partitions into $X$ (by the definition of $\mathcal{A}\left(
\pi\right)  $)). That is, we have $g\in\mathcal{A}\left(  \pi\right)  $.

It is easy to see that any two elements $e$ and $e^{\prime}$ of $E$ satisfying
$e\prec e^{\prime}$ satisfy%
\begin{equation}
\left(  \text{either }g\left(  e\right)  <g\left(  e^{\prime}\right)  \text{
or }\left(  g\left(  e\right)  =g\left(  e^{\prime}\right)  \text{ and }%
\omega\left(  e\right)  <\omega\left(  e^{\prime}\right)  \right)  \right)
\label{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}%
):} Let $e$ and $e^{\prime}$ be two elements of $E$ satisfying $e\prec
e^{\prime}$. We need to prove
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}).
\par
We distinguish between two cases:
\par
\textit{Case 1:} We have $\omega\left(  e\right)  \leq\omega\left(  e^{\prime
}\right)  $.
\par
\textit{Case 2:} We don't have $\omega\left(  e\right)  \leq\omega\left(
e^{\prime}\right)  $.
\par
Let us first consider Case 1. In this case, we have $\omega\left(  e\right)
\leq\omega\left(  e^{\prime}\right)  $. If we have $\omega\left(  e\right)
=\omega\left(  e^{\prime}\right)  $, then we have $e=e^{\prime}$ (since
$\omega$ is an injection), which contradicts $e\prec e^{\prime}$. Hence, we
cannot have $\omega\left(  e\right)  =\omega\left(  e^{\prime}\right)  $. We
thus have $\omega\left(  e\right)  \neq\omega\left(  e^{\prime}\right)  $.
Combined with $\omega\left(  e\right)  \leq\omega\left(  e^{\prime}\right)  $,
this yields $\omega\left(  e\right)  <\omega\left(  e^{\prime}\right)  $.
Also, $e\prec e^{\prime}$, and thus $g\left(  e\right)  \leq g\left(
e^{\prime}\right)  $ (due to (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1}%
)). In other words, either $g\left(  e\right)  <g\left(  e^{\prime}\right)  $
or $g\left(  e\right)  =g\left(  e^{\prime}\right)  $. Hence, either $g\left(
e\right)  <g\left(  e^{\prime}\right)  $ or $\left(  g\left(  e\right)
=g\left(  e^{\prime}\right)  \text{ and }\omega\left(  e\right)
<\omega\left(  e^{\prime}\right)  \right)  $ (because we have $\omega\left(
e\right)  <\omega\left(  e^{\prime}\right)  $). Thus,
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we don't have $\omega\left(
e\right)  \leq\omega\left(  e^{\prime}\right)  $. In other words, we have
$\omega\left(  e\right)  >\omega\left(  e^{\prime}\right)  $. In other words,
$\omega\left(  e^{\prime}\right)  <\omega\left(  e\right)  $. Applying
(\ref{pf.prop.gessel-thm1.subdiv.ordiso}) to $p=e^{\prime}$ and $q=e$, we
obtain the logical equivalence $\left(  e^{\prime}<_{2}e\right)
\ \Longleftrightarrow\ \left(  \omega\left(  e^{\prime}\right)  <\omega\left(
e\right)  \right)  $. Thus, we have $e^{\prime}<_{2}e$ (since we have
$\omega\left(  e^{\prime}\right)  <\omega\left(  e\right)  $). In other words,
$e>_{2}e^{\prime}$, so that $e\geq_{2}e^{\prime}$. Recall also that $e\prec
e^{\prime}$. Thus, (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2}) shows
that $g\left(  e\right)  <g\left(  e^{\prime}\right)  $. Hence, either
$g\left(  e\right)  <g\left(  e^{\prime}\right)  $ or $\left(  g\left(
e\right)  =g\left(  e^{\prime}\right)  \text{ and }\omega\left(  e\right)
<\omega\left(  e^{\prime}\right)  \right)  $. Thus,
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}) is proven in Case 2.
\par
We have now proven (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}) in each
of the two Cases 1 and 2. Since these two Cases cover all possibilities, this
shows that (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}) always holds,
qed.}.

Now, let $i$ and $j$ be any two elements of $E$. We are going to prove that we
have the logical equivalence $\left(  i\left(  \prec_{\mathbf{w}\left(
g\right)  }\right)  ^{\omega}j\right)  \ \Longleftrightarrow\ \left(  i\prec
j\right)  $.

In the proof of Proposition \ref{prop.gessel-thm1.subdiv} \textbf{(a)}, we
have shown that for every $f\in\mathcal{A}\left(  \pi\right)  $ and every two
elements $e$ and $e^{\prime}$ of $E$, the logical equivalence
(\ref{pf.prop.gessel-thm1.subdiv.5}) holds. Thus,
(\ref{pf.prop.gessel-thm1.subdiv.5}) (applied to $f=g$, $e=i$ and $e^{\prime
}=j$) shows that we have the following logical equivalence:%
\begin{align}
&  \ \left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega
}j\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{either }g\left(  i\right)  <g\left(
j\right)  \text{ or }\left(  g\left(  i\right)  =g\left(  j\right)  \text{ and
}\omega\left(  i\right)  <\omega\left(  j\right)  \right)  \right)
\label{pf.prop.gessel-thm1.subdiv.b.left.subdiv.left.equiv}%
\end{align}
(since we have $g\in\mathcal{A}\left(  \pi\right)  $). Also,
(\ref{pf.prop.gessel-thm1.subdiv.5}) (applied to $f=g$, $e=j$ and $e^{\prime
}=i$) shows that we have the following logical equivalence:%
\begin{align}
&  \ \left(  j\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega
}i\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{either }g\left(  j\right)  <g\left(
i\right)  \text{ or }\left(  g\left(  j\right)  =g\left(  i\right)  \text{ and
}\omega\left(  j\right)  <\omega\left(  i\right)  \right)  \right)  .
\label{pf.prop.gessel-thm1.subdiv.b.left.subdiv.left.equiv'}%
\end{align}


Now, let us recall that $\prec$ is a total order. Thus, we have either $i\prec
j$ or $i=j$ or $j\prec i$. In other words, we must be in one of the following
three cases:

\textit{Case 1:} We have $i\prec j$.

\textit{Case 2:} We have $i=j$.

\textit{Case 3:} We have $j\prec i$.

Let us first consider Case 1. In this case, we have $i\prec j$. Hence,
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}) (applied to $e=i$ and
$e^{\prime}=j$) yields%
\[
\left(  \text{either }g\left(  i\right)  <g\left(  j\right)  \text{ or
}\left(  g\left(  i\right)  =g\left(  j\right)  \text{ and }\omega\left(
i\right)  <\omega\left(  j\right)  \right)  \right)  .
\]
Therefore, we have $i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}j$ (according to the equivalence
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.left.equiv})). Thus, both
statements $\left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}j\right)  $ and $\left(  i\prec j\right)  $ hold. Therefore, we have
the equivalence $\left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}j\right)  \ \Longleftrightarrow\ \left(  i\prec j\right)  $. Hence,
the equivalence $\left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}j\right)  \ \Longleftrightarrow\ \left(  i\prec j\right)  $ is proven
in Case 1.

Let us now consider Case 2. In this case, we have $i=j$. Hence, we have
neither $\left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}j\right)  $ nor $\left(  i\prec j\right)  $ (because both $\left(
\prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}$ and $\prec$ are total
orders). Therefore, we have the equivalence $\left(  i\left(  \prec
_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}j\right)
\ \Longleftrightarrow\ \left(  i\prec j\right)  $. Hence, the equivalence
$\left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega
}j\right)  \ \Longleftrightarrow\ \left(  i\prec j\right)  $ is proven in Case 2.

Let us finally consider Case 3. In this case, we have $j\prec i$. Hence,
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.3}) (applied to $e=j$ and
$e^{\prime}=i$) yields%
\[
\left(  \text{either }g\left(  j\right)  <g\left(  i\right)  \text{ or
}\left(  g\left(  j\right)  =g\left(  i\right)  \text{ and }\omega\left(
j\right)  <\omega\left(  i\right)  \right)  \right)  .
\]
Thus, we have $j\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega
}i$ (according to the equivalence
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.left.equiv'})). Therefore, we
do not have $i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}j$
(since the relation $\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}$ is asymmetric (because $\left(  \prec_{\mathbf{w}\left(  g\right)
}\right)  ^{\omega}$ is a total order)). Also, we have $j\prec i$, and thus we
do not have $i\prec j$ (since the relation $\prec$ is asymmetric (because
$\prec$ is a total order)). Hence, we have neither $\left(  i\left(
\prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}j\right)  $ nor $\left(
i\prec j\right)  $ (because both $\left(  \prec_{\mathbf{w}\left(  g\right)
}\right)  ^{\omega}$ and $\prec$ are total orders). Therefore, we have the
equivalence $\left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}j\right)  \ \Longleftrightarrow\ \left(  i\prec j\right)  $. Hence,
the equivalence $\left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}j\right)  \ \Longleftrightarrow\ \left(  i\prec j\right)  $ is proven
in Case 3.

We thus have proven the equivalence $\left(  i\left(  \prec_{\mathbf{w}\left(
g\right)  }\right)  ^{\omega}j\right)  \ \Longleftrightarrow\ \left(  i\prec
j\right)  $ in each of the three Cases 1, 2 and 3. Since these three Cases
cover all possibilities, this yields that the equivalence $\left(  i\left(
\prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}j\right)
\ \Longleftrightarrow\ \left(  i\prec j\right)  $ always holds.

Now, let us forget that we fixed $i$ and $j$. We thus have proven the logical
equivalence $\left(  i\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}j\right)  \ \Longleftrightarrow\ \left(  i\prec j\right)  $ for all
$i\in E$ and $j\in E$. In other words, the binary relation $\left(
\prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}$ equals $\prec$.

So we know that $g\in\mathcal{A}\left(  \pi\right)  $, and that the binary
relation $\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}$
equals $\prec$. In other words, $g$ is an element $f$ of $\mathcal{A}\left(
\pi\right)  $ such that the binary relation $\left(  \prec_{\mathbf{w}\left(
f\right)  }\right)  ^{\omega}$ equals $\prec$. In other words,%
\[
g\in\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\prec\right\}  .
\]


Let us now forget that we fixed $g$. We thus have shown that \newline%
$g\in\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\prec\right\}  $ for every\newline$g\in\mathcal{A}\left(  \left(
E,\prec,<_{2}\right)  \right)  $. In other words,%
\[
\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  \subseteq\left\{
f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation }\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals }%
\prec\right\}  .
\]
This proves (\ref{pf.prop.gessel-thm1.subdiv.b.left}).

\textit{Proof of (\ref{pf.prop.gessel-thm1.subdiv.b.right}):} Let
$g\in\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\prec\right\}  $. Then, $g$ is an element $f$ of $\mathcal{A}\left(
\pi\right)  $ such that the binary relation $\left(  \prec_{\mathbf{w}\left(
f\right)  }\right)  ^{\omega}$ equals $\prec$. In other words, $g$ is an
element of $\mathcal{A}\left(  \pi\right)  $, and the binary relation $\left(
\prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}$ equals $\prec$.

In the proof of Proposition \ref{prop.gessel-thm1.subdiv} \textbf{(a)}, we
have shown that for every $f\in\mathcal{A}\left(  \pi\right)  $ and every two
elements $e$ and $e^{\prime}$ of $E$, the logical equivalence
(\ref{pf.prop.gessel-thm1.subdiv.5}) holds. We are going to use this in the following.

We are going to prove that $g\in\mathcal{A}\left(  \left(  E,\prec
,<_{2}\right)  \right)  $. In order to do so, we shall show that any two
elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1}) and
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2}).

Let $e$ and $e^{\prime}$ be two elements of $E$. Let us first prove
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1}).

Indeed, we assume that $e\prec e^{\prime}$. Then, we have $e\prec e^{\prime}$.
In other words, $e\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}e^{\prime}$ (since the binary relation $\left(  \prec_{\mathbf{w}%
\left(  g\right)  }\right)  ^{\omega}$ equals $\prec$). But recall that
$g\in\mathcal{A}\left(  \pi\right)  $. Thus, we can apply
(\ref{pf.prop.gessel-thm1.subdiv.5}) to $f=g$. As a result, we conclude that
the following logical equivalence holds:%
\begin{align*}
&  \ \left(  e\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega
}e^{\prime}\right) \\
&  \Longleftrightarrow\ \left(  \text{either }g\left(  e\right)  <g\left(
e^{\prime}\right)  \text{ or }\left(  g\left(  e\right)  =g\left(  e^{\prime
}\right)  \text{ and }\omega\left(  e\right)  <\omega\left(  e^{\prime
}\right)  \right)  \right)  .
\end{align*}
Therefore, we have%
\[
\left(  \text{either }g\left(  e\right)  <g\left(  e^{\prime}\right)  \text{
or }\left(  g\left(  e\right)  =g\left(  e^{\prime}\right)  \text{ and }%
\omega\left(  e\right)  <\omega\left(  e^{\prime}\right)  \right)  \right)
\]
(since we have $e\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}e^{\prime}$). Consequently, we have
\[
\left(  \text{either }g\left(  e\right)  <g\left(  e^{\prime}\right)  \text{
or }g\left(  e\right)  =g\left(  e^{\prime}\right)  \right)
\]
(because of the logical implication \newline$\left(  g\left(  e\right)
=g\left(  e^{\prime}\right)  \text{ and }\omega\left(  e\right)
<\omega\left(  e^{\prime}\right)  \right)  \ \Longrightarrow\ \left(  g\left(
e\right)  =g\left(  e^{\prime}\right)  \right)  $). In other words, $g\left(
e\right)  \leq g\left(  e^{\prime}\right)  $. This proves
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1}).

Let us next prove (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2}).

Indeed, we assume that $e\prec e^{\prime}$ and $e\geq_{2}e^{\prime}$. Then, it
is easy to see that $\omega\left(  e\right)  <\omega\left(  e^{\prime}\right)
$ is false\footnote{\textit{Proof.} Assume the contrary. Then, we have
$\omega\left(  e\right)  <\omega\left(  e^{\prime}\right)  $. Applying
(\ref{pf.prop.gessel-thm1.subdiv.ordiso}) to $p=e$ and $q=e^{\prime}$, we
obtain the logical equivalence $\left(  e<_{2}e^{\prime}\right)
\ \Longleftrightarrow\ \left(  \omega\left(  e\right)  <\omega\left(
e^{\prime}\right)  \right)  $. Hence, we have $e<_{2}e^{\prime}$ (since we
have $\omega\left(  e\right)  <\omega\left(  e^{\prime}\right)  $). But this
contradicts $e\geq_{2}e^{\prime}$. This contradiction shows that our
assumption was wrong, qed.}. Now, recall that $e\prec e^{\prime}$. In other
words, $e\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega
}e^{\prime}$ (since the binary relation $\left(  \prec_{\mathbf{w}\left(
g\right)  }\right)  ^{\omega}$ equals $\prec$). But recall that $g\in
\mathcal{A}\left(  \pi\right)  $. Thus, we can apply
(\ref{pf.prop.gessel-thm1.subdiv.5}) to $f=g$. As a result, we conclude that
the following logical equivalence holds:%
\begin{align*}
&  \ \left(  e\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega
}e^{\prime}\right) \\
&  \Longleftrightarrow\ \left(  \text{either }g\left(  e\right)  <g\left(
e^{\prime}\right)  \text{ or }\left(  g\left(  e\right)  =g\left(  e^{\prime
}\right)  \text{ and }\underbrace{\omega\left(  e\right)  <\omega\left(
e^{\prime}\right)  }_{\substack{\text{this is false}\\\text{(since }%
\omega\left(  e\right)  <\omega\left(  e^{\prime}\right)  \text{ is false)}%
}}\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{either }g\left(  e\right)  <g\left(
e^{\prime}\right)  \text{ or }\underbrace{\left(  g\left(  e\right)  =g\left(
e^{\prime}\right)  \text{ and }\left(  \text{false}\right)  \right)
}_{\text{this is false}}\right) \\
&  \Longleftrightarrow\ \left(  \text{either }g\left(  e\right)  <g\left(
e^{\prime}\right)  \text{ or }\left(  \text{false}\right)  \right)
\ \Longleftrightarrow\ \left(  g\left(  e\right)  <g\left(  e^{\prime}\right)
\right)  .
\end{align*}
Hence, we have $g\left(  e\right)  <g\left(  e^{\prime}\right)  $ (since we
have $e\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}%
e^{\prime}$). This proves (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2}).

Now, let us forget that we fixed $e$ and $e^{\prime}$. We thus have proven
that any two elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1}) and
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2}).

But Definition \ref{def.pi-par} (applied to $g$, $\prec$ and $\left(
E,\prec,<_{2}\right)  $ instead of $x$, $<_{1}$ and $\pi$) yields that the map
$g$ is a $\left(  E,\prec,<_{2}\right)  $-partition if and only if any two
elements $e$ and $e^{\prime}$ of $E$ satisfy
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1}) and
(\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2}). Hence, $g$ is a $\left(
E,\prec,<_{2}\right)  $-partition (since we know that any two elements $e$ and
$e^{\prime}$ of $E$ satisfy (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.1})
and (\ref{pf.prop.gessel-thm1.subdiv.b.left.subdiv.2})). More precisely, $g$
is a $\left(  E,\prec,<_{2}\right)  $-partition into $X$. In other words, $g$
is an element of $\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  $
(since $\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  $ is the set
of all $\left(  E,\prec,<_{2}\right)  $-partitions into $X$ (by the definition
of $\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  $)). Thus,
$g\in\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  $.

Let us now forget that we fixed $g$. We thus have shown that $g\in
\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  $ for every
\newline$g\in\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the
binary relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)
^{\omega}\text{ equals }\prec\right\}  $. In other words,%
\[
\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation
}\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals
}\prec\right\}  \subseteq\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)
\right)  .
\]
This proves (\ref{pf.prop.gessel-thm1.subdiv.b.right}).

We thus have proven the inclusions (\ref{pf.prop.gessel-thm1.subdiv.b.left})
and (\ref{pf.prop.gessel-thm1.subdiv.b.right}). Combining these two
inclusions, we obtain%
\[
\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  =\left\{
f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation }\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals }%
\prec\right\}  .
\]
This proves Proposition \ref{prop.gessel-thm1.subdiv} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.gessel-thm1}.]We have%
\[
\mathcal{A}\left(  \left(  E,\prec,<_{2}\right)  \right)  =\left\{
f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation }\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals }%
\prec\right\}
\]
for every linear extension $\prec$ of $\pi$ (due to Proposition
\ref{prop.gessel-thm1.subdiv} \textbf{(b)}).

The sets $\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\prec\right\}  $, where $\prec$ runs over all linear extensions of
$\pi$, are pairwise disjoint (since the binary relation $\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}$ is uniquely determined
by $f$). In other words, the sets $\mathcal{A}\left(  \left(  E,\prec
,<_{2}\right)  \right)  $, where $\prec$ runs over all linear extensions of
$\pi$, are pairwise disjoint (because $\mathcal{A}\left(  \left(
E,\prec,<_{2}\right)  \right)  =\left\{  f\in\mathcal{A}\left(  \pi\right)
\ \mid\ \text{the binary relation }\left(  \prec_{\mathbf{w}\left(  f\right)
}\right)  ^{\omega}\text{ equals }\prec\right\}  $ for every linear extension
$\prec$ of $\pi$). Thus, in order to prove Theorem \ref{thm.gessel-thm1}, it
remains to show that the union of these sets is $\mathcal{A}\left(
\pi\right)  $. In other words, it remains to show that%
\begin{align}
&  \bigcup_{\substack{\prec\text{ is a linear}\\\text{extension of }\pi
}}\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\prec\right\} \nonumber\\
&  =\mathcal{A}\left(  \pi\right)  . \label{pf.thm.gessel-thm.part2}%
\end{align}


But we have%
\[
\mathcal{A}\left(  \pi\right)  \subseteq\bigcup_{\substack{\prec\text{ is a
linear}\\\text{extension of }\pi}}\left\{  f\in\mathcal{A}\left(  \pi\right)
\ \mid\ \text{the binary relation }\left(  \prec_{\mathbf{w}\left(  f\right)
}\right)  ^{\omega}\text{ equals }\prec\right\}
\]
\footnote{\textit{Proof.} Let $g\in\mathcal{A}\left(  \pi\right)  $. Then, the
binary relation $\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}$ is a linear extension of $\pi$ (according to Proposition
\ref{prop.gessel-thm1.subdiv} \textbf{(a)}, applied to $f=g$). Thus, $\left\{
f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation }\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals }\left(
\prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}\right\}  $ is a summand
in the union $\bigcup_{\substack{\prec\text{ is a linear}\\\text{extension of
}\pi}}\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\prec\right\}  $. As a consequence,%
\begin{align*}
&  \left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}\right\}
\\
&  \subseteq\bigcup_{\substack{\prec\text{ is a linear}\\\text{extension of
}\pi}}\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\prec\right\}  .
\end{align*}
But $g$ is an element $f$ of $\mathcal{A}\left(  \pi\right)  $ such that the
binary relation $\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)
^{\omega}$ equals $\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)
^{\omega}$ (since $g$ is an element of $\mathcal{A}\left(  \pi\right)  $, and
since the binary relation $\left(  \prec_{\mathbf{w}\left(  g\right)
}\right)  ^{\omega}$ equals $\left(  \prec_{\mathbf{w}\left(  g\right)
}\right)  ^{\omega}$). In other words,%
\begin{align*}
g  &  \in\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\left(  \prec_{\mathbf{w}\left(  g\right)  }\right)  ^{\omega}\right\}
\\
&  \subseteq\bigcup_{\substack{\prec\text{ is a linear}\\\text{extension of
}\pi}}\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary
relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{
equals }\prec\right\}  .
\end{align*}
\par
Let us now forget that we fixed $g$. We thus have shown that every
$g\in\mathcal{A}\left(  \pi\right)  $ satisfies $g\in\bigcup_{\substack{\prec
\text{ is a linear}\\\text{extension of }\pi}}\left\{  f\in\mathcal{A}\left(
\pi\right)  \ \mid\ \text{the binary relation }\left(  \prec_{\mathbf{w}%
\left(  f\right)  }\right)  ^{\omega}\text{ equals }\prec\right\}  $. In other
words,%
\[
\mathcal{A}\left(  \pi\right)  \subseteq\bigcup_{\substack{\prec\text{ is a
linear}\\\text{extension of }\pi}}\left\{  f\in\mathcal{A}\left(  \pi\right)
\ \mid\ \text{the binary relation }\left(  \prec_{\mathbf{w}\left(  f\right)
}\right)  ^{\omega}\text{ equals }\prec\right\}  ,
\]
qed.}. Combining this with%
\begin{align*}
&  \bigcup_{\substack{\prec\text{ is a linear}\\\text{extension of }\pi
}}\underbrace{\left\{  f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the
binary relation }\left(  \prec_{\mathbf{w}\left(  f\right)  }\right)
^{\omega}\text{ equals }\prec\right\}  }_{\subseteq\mathcal{A}\left(
\pi\right)  }\\
&  \subseteq\bigcup_{\substack{\prec\text{ is a linear}\\\text{extension of
}\pi}}\mathcal{A}\left(  \pi\right)  \subseteq\mathcal{A}\left(  \pi\right)  ,
\end{align*}
we obtain%
\begin{equation}
\bigcup_{\substack{\prec\text{ is a linear}\\\text{extension of }\pi}}\left\{
f\in\mathcal{A}\left(  \pi\right)  \ \mid\ \text{the binary relation }\left(
\prec_{\mathbf{w}\left(  f\right)  }\right)  ^{\omega}\text{ equals }%
\prec\right\}  =\mathcal{A}\left(  \pi\right)  .\nonumber
\end{equation}
This proves (\ref{pf.thm.gessel-thm.part2}). Thus, the proof of Theorem
\ref{thm.gessel-thm1} is complete.
\end{proof}

\section{Some brief remarks}

\subsection{How nondegenerate is the form?}

As mentioned in the errata above, the bilinear form $\left\langle \cdot
,\cdot\right\rangle $ on $\mathbb{Z}\mathbf{D}$ gives rise to a Hopf algebra
homomorphism $\mathbb{Z}\mathbf{D}\rightarrow\left(  \mathbb{Z}\mathbf{D}%
\right)  ^{\ast}$ which is \textbf{not} bijective. However, \cite[Theorem 36
1.]{Foissy11} shows that this homomorphism is at least injective; it becomes
bijective upon base change from $\mathbb{Z}$ to a characteristic-$0$ field.

We can in fact say a bit more. For any double poset $E=\left(  E,<_{1}%
,<_{2}\right)  $, we define its \textit{relational weight}
$\operatorname*{rlw}\left(  E\right)  $ to be the integer
\begin{align*}
\operatorname*{rlw}\left(  E\right)   &  :=\left(  \text{\# of pairs }\left(
i,j\right)  \in E\times E\text{ such that }i<_{1}j\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \text{\# of pairs }\left(  i,j\right)  \in
E\times E\text{ such that }i<_{2}j\right)  .
\end{align*}
It is easy to see the following (essentially a result of Foissy in \cite[proof
of Theorem 36 1.]{Foissy11}):

\begin{proposition}
\label{prop.rlw.1}Let $E$ and $F$ be two double posets. Let $\widetilde{F}$ be
the double poset obtained from $F$ by swapping its two orders (i.e.: if
$F=\left(  F,<_{1},<_{2}\right)  $, then $\widetilde{F}=\left(  F,<_{2}%
,<_{1}\right)  $). Then:

\textbf{(a)} If $\operatorname*{rlw}\left(  E\right)  >\operatorname*{rlw}%
\left(  \widetilde{F}\right)  $, then there exists no picture from $E$ to $F$,
and thus we have $\left\langle E,F\right\rangle =0$.

\textbf{(b)} If $\operatorname*{rlw}\left(  E\right)  =\operatorname*{rlw}%
\left(  \widetilde{F}\right)  $, then the pictures from $E$ to $F$ are
precisely the double poset isomorphisms from $E$ to $\widetilde{F}$.

\textbf{(c)} If $\operatorname*{rlw}\left(  E\right)  =\operatorname*{rlw}%
\left(  \widetilde{F}\right)  $ but not $E\cong\widetilde{F}$ as double
posets, then $\left\langle E,F\right\rangle =0$.

\textbf{(d)} If $E\cong\widetilde{F}$ as double posets, then $\left\langle
E,F\right\rangle =\left\vert \operatorname*{Aut}E\right\vert $, where
$\operatorname*{Aut}E$ is the group of automorphisms of the double poset $E$.
\end{proposition}

\begin{proof}
[Proof sketch.]Write the double posets $E$ and $F$ as $E=\left(  E,<_{1}%
,<_{2}\right)  $ and $F=\left(  F,<_{1},<_{2}\right)  $. (Do not worry about
the double meaning of the notations $<_{1}$ and $<_{2}$; the context will
always make clear on which poset we are working on.)

\textbf{(a)} Let $\phi:E\rightarrow F$ be a picture. Consider the map%
\begin{align*}
&  \text{from }\left\{  \text{pairs }\left(  i,j\right)  \in E\times E\text{
such that }i<_{1}j\right\} \\
&  \text{to }\left\{  \text{pairs }\left(  u,v\right)  \in F\times F\text{
such that }u<_{2}v\right\}
\end{align*}
that sends each pair $\left(  i,j\right)  $ to $\left(  \phi\left(  i\right)
,\phi\left(  j\right)  \right)  $. This map is well-defined (since $\phi$ is a
picture, so that $i<_{1}j$ entails $\phi\left(  i\right)  <_{2}\phi\left(
j\right)  $) and injective (since $\phi$ is bijective). Hence,%
\begin{align*}
&  \left(  \text{\# of pairs }\left(  i,j\right)  \in E\times E\text{ such
that }i<_{1}j\right) \\
&  \leq\left(  \text{\# of pairs }\left(  u,v\right)  \in F\times F\text{ such
that }u<_{2}v\right)  .
\end{align*}
Similarly, using $\phi^{-1}$ instead of $\phi$, we obtain%
\begin{align*}
&  \left(  \text{\# of pairs }\left(  i,j\right)  \in F\times F\text{ such
that }i<_{1}j\right) \\
&  \leq\left(  \text{\# of pairs }\left(  u,v\right)  \in E\times E\text{ such
that }u<_{2}v\right)  .
\end{align*}
Adding these two inequalities together, we find%
\begin{align*}
&  \left(  \text{\# of pairs }\left(  i,j\right)  \in E\times E\text{ such
that }i<_{1}j\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{\# of pairs }\left(  i,j\right)  \in
F\times F\text{ such that }i<_{1}j\right) \\
&  \leq\left(  \text{\# of pairs }\left(  u,v\right)  \in F\times F\text{ such
that }u<_{2}v\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{\# of pairs }\left(  u,v\right)  \in
E\times E\text{ such that }u<_{2}v\right)  .
\end{align*}
In other words,%
\begin{align*}
&  \left(  \text{\# of pairs }\left(  i,j\right)  \in E\times E\text{ such
that }i<_{1}j\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \text{\# of pairs }\left(  u,v\right)  \in
E\times E\text{ such that }u<_{2}v\right) \\
&  \leq\left(  \text{\# of pairs }\left(  u,v\right)  \in F\times F\text{ such
that }u<_{2}v\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \text{\# of pairs }\left(  i,j\right)  \in
F\times F\text{ such that }i<_{1}j\right)  .
\end{align*}
This inequality can be rewritten as $\operatorname*{rlw}\left(  E\right)
\leq\operatorname*{rlw}\left(  \widetilde{F}\right)  $ (since its left hand
side is $\operatorname*{rlw}\left(  E\right)  $, while its right hand side is
$\operatorname*{rlw}\left(  \widetilde{F}\right)  $).

Forget that we fixed $\phi$. We thus have shown that if $\phi:E\rightarrow F$
is any picture, then $\operatorname*{rlw}\left(  E\right)  \leq
\operatorname*{rlw}\left(  \widetilde{F}\right)  $. Hence, if
$\operatorname*{rlw}\left(  E\right)  >\operatorname*{rlw}\left(
\widetilde{F}\right)  $, then there exists no picture from $E$ to $F$, and
thus we have $\left\langle E,F\right\rangle =0$ by the definition of the
pairing $\left\langle ,\right\rangle $. This proves Proposition
\ref{prop.rlw.1} \textbf{(a)}.

\textbf{(b)} Assume that $\operatorname*{rlw}\left(  E\right)
=\operatorname*{rlw}\left(  \widetilde{F}\right)  $. Let $\phi:E\rightarrow F$
be a picture. Revisit the above proof of Proposition \ref{prop.rlw.1}
\textbf{(a)}, and note that the new assumption $\operatorname*{rlw}\left(
E\right)  =\operatorname*{rlw}\left(  \widetilde{F}\right)  $ means that
equality must hold in all relevant inequalities. Hence, in particular, the map%
\begin{align*}
&  \text{from }\left\{  \text{pairs }\left(  i,j\right)  \in E\times E\text{
such that }i<_{1}j\right\} \\
&  \text{to }\left\{  \text{pairs }\left(  u,v\right)  \in F\times F\text{
such that }u<_{2}v\right\}
\end{align*}
that sends each pair $\left(  i,j\right)  $ to $\left(  \phi\left(  i\right)
,\phi\left(  j\right)  \right)  $ must not only be injective, but also be
surjective (otherwise, the inequality that it leads to would be a strict
inequality), and hence bijective. In other words, for any pair $\left(
i,j\right)  \in E\times E$, we have $i<_{1}j$ if and only if $\phi\left(
i\right)  <_{2}\phi\left(  j\right)  $ (since $\phi$ is a bijection). In other
words, the map $\phi$ is an order isomorphism from $\left(  E,<_{1}\right)  $
to $\left(  F,<_{2}\right)  $. Similarly, the map $\phi^{-1}$ is an order
isomorphism from $\left(  F,<_{1}\right)  $ to $\left(  E,<_{2}\right)  $. In
other words, $\phi$ is an order isomorphism from $\left(  E,<_{2}\right)  $ to
$\left(  F,<_{1}\right)  $.

Now we know that $\phi$ is both an order isomorphism from $\left(
E,<_{1}\right)  $ to $\left(  F,<_{2}\right)  $ and an order isomorphism from
$\left(  E,<_{2}\right)  $ to $\left(  F,<_{1}\right)  $. In other words,
$\phi$ is a double poset isomorphism from $\left(  E,<_{1},<_{2}\right)  $ to
$\left(  F,<_{2},<_{1}\right)  $. In other words, $\phi$ is a double poset
isomorphism from $E$ to $\widetilde{F}$ (since $E=\left(  E,<_{1}%
,<_{2}\right)  $ and $\widetilde{F}=\left(  F,<_{2},<_{1}\right)  $).

Forget that we fixed $\phi$. We thus have shown that each picture
$\phi:E\rightarrow F$ is a double poset isomorphism from $E$ to $\widetilde{F}%
$. Conversely, it is easy to see that any double poset isomorphism from $E$ to
$\widetilde{F}$ is a picture $\phi:E\rightarrow F$. Combining these two facts,
we conclude that the pictures $\phi:E\rightarrow F$ are precisely the double
poset isomorphisms from $E$ to $\widetilde{F}$. This proves Proposition
\ref{prop.rlw.1} \textbf{(b)}.

\textbf{(c)} Assume that $\operatorname*{rlw}\left(  E\right)
=\operatorname*{rlw}\left(  \widetilde{F}\right)  $ but not $E\cong%
\widetilde{F}$ as double posets. As we recall, $\left\langle E,F\right\rangle
$ is defined as the number of all pictures $\phi:E\rightarrow F$. But these
pictures are precisely the double poset isomorphisms from $E$ to
$\widetilde{F}$ (by Proposition \ref{prop.rlw.1} \textbf{(b)}), and the number
of these isomorphisms is $0$ (since we don't have $E\cong\widetilde{F}$ as
double posets). Hence, we conclude that $\left\langle E,F\right\rangle =0$.
This proves Proposition \ref{prop.rlw.1} \textbf{(c)}.

\textbf{(d)} Assume that $E\cong\widetilde{F}$ as double posets. Then, of
course, $\operatorname*{rlw}\left(  E\right)  =\operatorname*{rlw}\left(
\widetilde{F}\right)  $. As we recall, $\left\langle E,F\right\rangle $ is
defined as the number of all pictures $\phi:E\rightarrow F$. But these
pictures are precisely the double poset isomorphisms from $E$ to
$\widetilde{F}$ (by Proposition \ref{prop.rlw.1} \textbf{(b)}), and these are
clearly in bijection with the double poset isomorphisms from $E$ to $E$ (since
$E\cong\widetilde{F}$ as double posets), i.e., with the automorphisms of the
double poset $E$. Hence, we conclude that $\left\langle E,F\right\rangle $ is
the number of all automorphisms of the double poset $E$. In other words,
$\left\langle E,F\right\rangle =\left\vert \operatorname*{Aut}E\right\vert $,
where $\operatorname*{Aut}E$ is the group of automorphisms of the double poset
$E$. This proves Proposition \ref{prop.rlw.1} \textbf{(d)}.
\end{proof}

Proposition \ref{prop.rlw.1} tells us something about the Gram
matrix\footnote{If $f:V\times V\rightarrow\mathbf{k}$ is a bilinear form on a
$\mathbf{k}$-module $V$, then its \textit{Gram matrix} on a given basis
$\left(  b_{1},b_{2},\ldots,b_{m}\right)  $ of $V$ is the matrix $\left(
f\left(  b_{i},b_{j}\right)  \right)  _{i,j\in\left\{  1,2,\ldots,m\right\}
}\in\mathbf{k}^{m\times m}$. Of course, this definition applies (with the
obvious changes) to infinite bases as well. One of the main uses of Gram
matrices is in determining whether a bilinear form is nondegenerate: A
bilinear form $f:V\times V\rightarrow\mathbf{k}$ on a finite-dimensional
$\mathbf{k}$-vector space $V$ over a field $\mathbf{k}$ is nondegenerate if
and only if its Gram matrix (on any given basis) is invertible. We shall use a
variant of this result here, in which $V$ is infinite-dimensional but graded,
and we use the graded version of nondegeneracy.} of the bilinear form
$\left\langle ,\right\rangle $ on the basis $\mathbf{D}$ of $\mathbb{Z}%
\mathbf{D}$: Namely, this matrix is upper-triangular up to a permutation of
its rows (or columns), and its diagonal entries (after this permutation) are
the numbers $\left\vert \operatorname*{Aut}E\right\vert $ for all double
posets $E$. Since these numbers are not all invertible in $\mathbb{Z}$ (in
fact, some of them are larger than $1$), we conclude that the form
$\left\langle ,\right\rangle $ fails to be nondegenerate; however, it becomes
nondegenerate if we change our base ring from $\mathbb{Z}$ to a
characteristic-$0$ field.

We can actually rescale the form $\left\langle ,\right\rangle $ to make it
nondegenerate: Namely, we define a new bilinear form $\left\langle
,\right\rangle ^{\prime}:\mathbb{Z}\mathbf{D}\times\mathbb{Z}\mathbf{D}%
\rightarrow\mathbb{Z}$ by%
\begin{equation}
\left\langle E,F\right\rangle ^{\prime}:=\dfrac{\left\vert \left\{
\text{pictures }\alpha:E\rightarrow F\right\}  \right\vert }{\left\vert
\operatorname*{Aut}E\right\vert } \label{eq.rlw.newform.def}%
\end{equation}
for all double posets $E$ and $F$. We claim the following:

\begin{proposition}
\label{prop.rlw.newform}\textbf{(a)} This form $\left\langle ,\right\rangle
^{\prime}$ is well-defined (i.e., the fraction on the right hand side of
(\ref{eq.rlw.newform.def}) really is an integer).

\textbf{(b)} However, this new bilinear form $\left\langle ,\right\rangle
^{\prime}$ is no longer symmetric.

\textbf{(c)} Nevertheless, it still satisfies%
\[
\left\langle EF,G\right\rangle ^{\prime}=\left\langle E\otimes F,\delta
G\right\rangle ^{\prime}%
\]
for any three double posets $E,F,G$.

\textbf{(d)} But it does \textbf{not} satisfy $\left\langle G,EF\right\rangle
^{\prime}=\left\langle \delta G,E\otimes F\right\rangle ^{\prime}$, so it does
not make the Hopf algebra $\mathbb{Z}\mathbf{D}$ self-dual.
\end{proposition}

\begin{proof}
[Proof idea.]\textbf{(a)} Let $E$ and $F$ be two double posets. Then, the
automorphism group $\operatorname*{Aut}E$ of the double poset $E$ acts freely
on the set of all pictures $\alpha:E\rightarrow F$. (The action is a right
action by composition: $\alpha\varphi=\alpha\circ\varphi$ for any picture
$\alpha:E\rightarrow F$ and any automorphism $\varphi\in\operatorname*{Aut}%
E$.) Hence, $\left\vert \left\{  \text{pictures }\alpha:E\rightarrow
F\right\}  \right\vert $ is a multiple of $\left\vert \operatorname*{Aut}%
E\right\vert $. Thus, the fraction on the right hand side of
(\ref{eq.rlw.newform.def}) really is an integer. This proves Proposition
\ref{prop.rlw.newform} \textbf{(a)}.

\textbf{(b)} This is easy to see: Let $E=\left(  \left\{  1,2\right\}
,<_{\varnothing},<_{\varnothing}\right)  $ and $F=\left(  \left\{
1,2\right\}  ,<_{\varnothing},<_{\mathbb{Z}}\right)  $, where $<_{\varnothing
}$ is the antichain partial order (i.e., no two distinct elements are
comparable). Then, $\left\vert \operatorname*{Aut}E\right\vert =2$ but
$\left\vert \operatorname*{Aut}F\right\vert =1$ (since the second order of $F$
distinguishes the two elements). But $\left\langle E,F\right\rangle =2$
(indeed, since the first orders on both $E$ and $F$ are antichains, it is
clear that the pictures from $E$ to $F$ are just the bijections from $E$ to
$F$). Hence, $\left\langle E,F\right\rangle ^{\prime}=\dfrac{2}{2}$ but
$\left\langle F,E\right\rangle ^{\prime}=\dfrac{2}{1}$, which of course is not
the same. Thus, the form $\left\langle ,\right\rangle ^{\prime}$ is not
symmetric. This proves Proposition \ref{prop.rlw.newform} \textbf{(b)}.

\textbf{(c)} This follows from $\left\langle EF,G\right\rangle =\left\langle
E\otimes F,\delta G\right\rangle $, which is proved on page 3, once we show
that $\operatorname*{Aut}\left(  EF\right)  \cong\operatorname*{Aut}%
E\times\operatorname*{Aut}F$. But the latter is a nice exercise (show that the
second order on $EF$ forces any automorphism of $EF$ to send $E$ to $E$ and
$F$ to $F$).

\textbf{(d)} Easy counterexample omitted.
\end{proof}

\subsection{What is the kernel of $\Gamma$ ?}

What is the kernel of the morphism $\Gamma:\mathbb{Z}\mathbf{D}\rightarrow
\mathbf{QSym}$ from Theorem 2.2?

\subsection{The kernel of $L$}

Note that it is not hard to describe the kernel of the map $L:\mathbb{Z}%
\mathbf{DS}\rightarrow\mathbb{Z}S$ (defined on page 7). Namely:

\begin{proposition}
\label{prop.kerL}The kernel of the map $L:\mathbb{Z}\mathbf{DS}\rightarrow
\mathbb{Z}S$ is spanned by all elements of the form
\begin{equation}
\left[  E+\left(  a<_{1}b\right)  \right]  +\left[  E+\left(  b<_{1}a\right)
\right]  -E \label{eq.prop.kerL.1}%
\end{equation}
where $E=\left(  E,<_{1},<_{2}\right)  $ is a special double poset and $a$ and
$b$ are two incomparable elements of $\left(  E,<_{1}\right)  $. Here, the
notation \textquotedblleft$\left[  E+\left(  a<_{1}b\right)  \right]
$\textquotedblright\ means the double poset $E$ with its first order $<_{1}$
extended so that $a$ becomes smaller than $b$ (and thus all elements that are
$\leq_{1}a$ become smaller than all elements that are $\geq_{1}b$). Likewise,
the notation \textquotedblleft$\left[  E+\left(  b<_{1}a\right)  \right]
$\textquotedblright\ is to be understood. (The second orders of all three
posets are the same.)
\end{proposition}

\begin{proof}
[Proof idea.]It is easy to see that any element of the form
(\ref{eq.prop.kerL.1}) lies in the kernel of $L$, since any linear extension
of $E$ is either a linear extension of $\left[  E+\left(  a<_{1}b\right)
\right]  $ or a linear extension of $\left[  E+\left(  b<_{1}a\right)
\right]  $ (but not both).

It remains to show that these elements span $\operatorname*{Ker}L$. For this,
it suffices to show that any isomorphism class $E$ of a special double poset
can be reduced to a linear combination of double total orders (i.e., double
posets whose \textbf{both} orders are total) by adding a few elements of the
form (\ref{eq.prop.kerL.1}). This can be proved by induction on the number of
incomparable pairs in the first order of $E$ (that is, pairs $\left(
a,b\right)  \in E\times E$ that satisfy neither $a<_{1}b$ nor $b<_{1}a$). In
fact, each of the two double posets $\left[  E+\left(  a<_{1}b\right)
\right]  $ and $\left[  E+\left(  b<_{1}a\right)  \right]  $ has fewer
incomparable pairs in its first order than $E$ does, and so we can apply the
induction hypothesis to them.
\end{proof}

The elements of the form (\ref{eq.prop.kerL.1}) have another peculiar property:

\begin{proposition}
\label{prop.EabEbaE}Let $E=\left(  E,<_{1},<_{2}\right)  $ be a double poset,
and let $F$ be a special double poset. Let $a$ and $b$ be two elements of $E$
that are incomparable in the poset $\left(  E,<_{1}\right)  $. Then,%
\[
\left(  \left[  E+\left(  a<_{1}b\right)  \right]  +\left[  E+\left(
b<_{1}a\right)  \right]  -E\right)  \circ F=0.
\]
(See Proposition \ref{prop.kerL} for the meaning of $\left[  E+\left(
a<_{1}b\right)  \right]  $ and $\left[  E+\left(  b<_{1}a\right)  \right]  $.)
\end{proposition}

\begin{proof}
[Proof idea.]Write $E$ and $F$ as $E=\left(  E,<_{E1},<_{E2}\right)  $ and
$F=\left(  F,<_{F1},<_{F2}\right)  $. Then, Lemma \ref{lem.assoc.fiber}
\textbf{(d)} yields
\begin{align*}
E\circ F  &  =\sum_{\substack{\phi\text{ is an increasing}\\\text{bijection
}\left(  E,<_{E1}\right)  \rightarrow\left(  F,<_{F2}\right)  }}\left(
E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)  ;\\
\left[  E+\left(  a<_{1}b\right)  \right]  \circ F  &  =\sum_{\substack{\phi
\text{ is an increasing}\\\text{bijection }\left(  E,<_{\left[  E+\left(
a<_{1}b\right)  \right]  }\right)  \rightarrow\left(  F,<_{F2}\right)
}}\left(  E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)  ;\\
\left[  E+\left(  b<_{1}a\right)  \right]  \circ F  &  =\sum_{\substack{\phi
\text{ is an increasing}\\\text{bijection }\left(  E,<_{\left[  E+\left(
b<_{1}a\right)  \right]  }\right)  \rightarrow\left(  F,<_{F2}\right)
}}\left(  E,\left(  <_{F1}\right)  ^{\phi},<_{E2}\right)  .
\end{align*}
But an increasing bijection $\phi:\left(  E,<_{E1}\right)  \rightarrow\left(
F,<_{F2}\right)  $ is either an increasing bijection from $\left(  \left[
E+\left(  a<_{1}b\right)  \right]  ,<_{1}\right)  $ to $\left(  F,<_{F2}%
\right)  $ or an increasing bijection from $\left(  \left[  E+\left(
b<_{1}a\right)  \right]  ,<_{1}\right)  $ to $\left(  F,<_{F2}\right)  $
(depending on whether it satisfies $\phi\left(  a\right)  <_{F2}\phi\left(
b\right)  $ or satisfies $\phi\left(  b\right)  <_{F2}\phi\left(  a\right)  $
(in fact, one of these two inequalities must hold, since $F$ is special)).
Hence, the above equalities yield%
\[
\left[  E+\left(  a<_{1}b\right)  \right]  \circ F+\left[  E+\left(
b<_{1}a\right)  \right]  \circ F=E\circ F.
\]
This proves Proposition \ref{prop.EabEbaE}.
\end{proof}

\begin{corollary}
\label{cor.kerL.rad}The kernel of the map $L:\mathbb{Z}\mathbf{DS}%
\rightarrow\mathbb{Z}S$ is the Jacobson radical of the nonunital algebra
$\left(  \mathbb{Z}\mathbf{DS},\circ\right)  $.
\end{corollary}

\begin{proof}
[Proof idea.]Consider the internal multiplication $\circ$ on $\mathbb{Z}S$
given by multiplying permutations of the same size: i.e., by%
\[
\sigma\circ\tau=%
\begin{cases}
\sigma\tau, & \text{if }\sigma,\tau\in S_{n}\text{ for the same }n;\\
0, & \text{if }\sigma\in S_{n}\text{ and }\tau\in S_{m}\text{ for }n\neq m.
\end{cases}
\]
The algebra $\left(  \mathbb{Z}S,\circ\right)  $ is just the direct product of
the group rings of the symmetric groups $S_{n}$ for all $n\geq0$. As such, it
is \textquotedblleft almost unital\textquotedblright\ (its unity would be the
sum of all identity permutations, if this sum was not infinite); it is a
nonunital Frobenius algebra (with the J\"{o}llenbeck scalar product being the
Frobenius form), and its Jacobson radical is $0$ (by Maschke's theorem). The
map $L:\mathbb{Z}\mathbf{DS}\rightarrow\mathbb{Z}S$ is a surjective ring
morphism from the nonunital ring $\left(  \mathbb{Z}\mathbf{DS},\circ\right)
$ to the nonunital ring $\left(  \mathbb{Z}S,\circ\right)  $.

If $a\in\operatorname*{Ker}L$, then Proposition \ref{prop.kerL} shows that $a$
is a linear combination of elements of the form (\ref{eq.prop.kerL.1}), and
therefore Proposition \ref{prop.EabEbaE} shows that $a\circ F=0$ for each
special double poset $F$; hence, $a\circ b=0$ for each $b\in\mathbb{Z}%
\mathbf{DS}$. In particular, we thus have $a\circ b=0$ for each $a\in
\operatorname*{Ker}L$. This shows that the ideal $\operatorname*{Ker}L$ of
$\mathbb{Z}\mathbf{DS}$ is nilpotent. Hence, $\operatorname*{Ker}L$ is
contained in the Jacobson radical of $\left(  \mathbb{Z}\mathbf{DS}%
,\circ\right)  $.

Conversely, any element of the Jacobson radical of $\left(  \mathbb{Z}%
\mathbf{DS},\circ\right)  $ must be mapped to an element of the Jacobson
radical of $\left(  \mathbb{Z}S,\circ\right)  $ under the map $L$ (since $L$
is a surjective ring morphism). But the Jacobson radical of $\left(
\mathbb{Z}S,\circ\right)  $ is $0$. Thus, any element of the Jacobson radical
of $\left(  \mathbb{Z}\mathbf{DS},\circ\right)  $ must be mapped to $0$ under
the map $L$. In other words, the Jacobson radical of $\left(  \mathbb{Z}%
\mathbf{DS},\circ\right)  $ is contained in $\operatorname*{Ker}L$.
Altogether, Corollary \ref{cor.kerL.rad} is proved.
\end{proof}

Corollary \ref{cor.kerL.rad} is somewhat reminiscent of the projection map
$\operatorname*{NSym}\rightarrow\operatorname*{Sym}$, whose kernel is (at
least in characteristic $0$) the Jacobson radical of the nonunital algebra
that is $\operatorname*{NSym}$ under the internal product. (This was
originally proved by Solomon in the language of descent algebras; it served as
a foundation for the Garsia-Reutenauer study of the descent algebra
\cite[Theorem 1.1]{GarReu89}.) Can we do the Garsia-Reutenauer descent-algebra
theory for $\mathbb{Z}\mathbf{DS}$ instead of the descent algebra?

\subsection{Weirder questions}

Here are two open-ended questions that crossed my mind:

\begin{itemize}
\item Can double posets be seen as bimodules over some sort of algebra, and
the internal product correspond to tensoring these bimodules? Or what other
categorical reason is there for the associativity of the internal product?

\item When do two double posets $\left(  E,<_{1},<_{2}\right)  $ and $\left(
F,<_{1},<_{2}\right)  $ satisfy
\begin{align*}
&  \left(  \text{number of pictures from }\left(  E,<_{1},<_{2}\right)  \text{
to }\left(  F,<_{1},<_{2}\right)  \right) \\
&  =\left(  \text{number of pictures from }\left(  E,<_{1},>_{2}\right)
\text{ to }\left(  F,<_{1},>_{2}\right)  \right)  \text{ ?}%
\end{align*}
This would generalize a symmetry of the Littlewood--Richardson coefficients.
\end{itemize}

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[Foissy11]{Foissy11}%
\href{http://www.arxiv.org/abs/1101.5231v4}{Lo\"{\i}c Foissy,
\textit{Algebraic structures on double and plane posets}, arXiv preprint
arXiv:1101.5231v4}.

\bibitem[GarReu89]{GarReu89}%
\href{https://doi.org/10.1016/0001-8708(89)90020-0}{A. M. Garsia, C.
Reutenauer, \textit{A decomposition of Solomon's descent algebra}, Advances in
Mathematics \textbf{77}, Issue 2, October 1989, pp. 189--262.}

\bibitem[Gashar98]{Gashar98}%
\href{https://doi.org/10.1006/eujc.1998.0212}{Vesselin Gasharov, \textit{A
Short Proof of the Littlewood--Richardson Rule}, Europ. J. Combinatorics
(1998) \textbf{19}, pp. 451--453.}

\bibitem[GriRei15]{GriRei15}\href{https://arxiv.org/abs/1409.8356v7}{Darij
Grinberg, Victor Reiner, \textit{Hopf algebras in Combinatorics},
arXiv:1409.8356v7, version of 28 July 2020.}
\end{thebibliography}


\end{document}