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\begin{document}

\title{\textbf{On the logarithm of the identity on connected filtered bialgebras}}
\author{\textit{Darij Grinberg}}
\date{Version 0.19, 18 June 2021}
\maketitle
\tableofcontents

\section*{***}

The purpose of this document is to prove several properties of coalgebras,
bialgebras and Hopf algebras. The proofs given here are mostly not new, and
often not optimal; however, they are very detailed and don't use Sweedler's notation.

\begin{quote}
\textbf{Remark (2017):}

This \textquotedblleft lab notebook\textquotedblright\ has been mostly written
in 2011--2013 (when I was a graduate student), and collects various results I
have encountered while exploring the theory of Hopf algebras (e.g., variants
of the Cartier-Milnor-Moore and Leray theorems; properties of the Eulerian
idempotent; facts like the invertibility of the antipode in a connected
filtered Hopf algebra). I expect few of these results to be new; the best I
can claim is that they are stated more explicitly here than in the available
literature. Unfortunately, this notebook is rather disorganized, and the
results are written down more or less in the order in which I have found them.
I have tried to give detailed proofs of all statements (if only to make sure
that they are correct); these proofs should be \textquotedblleft
technically\textquotedblright\ readable but in practice you might have an
easier time skimming them for their main ideas (which are, unfortunately,
sometimes hidden well) and reconstructing the rest yourself. Needless to say,
the notations used in this notebook are also not the best.
\end{quote}

\textbf{Acknowledgments:} Thanks to Philipp Var\v{s}o for finding an error in
the proof of Proposition~\ref{Proposition13} (the statement of
Corollary~\ref{Corollary15} was insufficiently general).

\section{\label{S0a}Notations and definitions}

In this document, we shall use some standard terminology from the theory of
Hopf algebras (see, for example, \cite{Schneider-Hopf} or \cite{1}) along with
the following notations:

\begin{convention}
In the following, the symbol $\mathbb{N}$ always denotes the set $\left\{
0,1,2,\ldots\right\}  $.
\end{convention}

\begin{convention}
A \textquotedblleft ring\textquotedblright\ shall always mean an associative
ring with unity.\footnote{We will sometimes say \textquotedblleft ring with
unity\textquotedblright\ to additionally stress this, but even if we do not
say \textquotedblleft with unity\textquotedblright\ we still mean unital rings
only.}
\end{convention}

\begin{definition}
\label{Definition2} Let $k$ be a field, and $U$ and $V$ be two $k$-vector
spaces. Then, $\mathcal{L}\left(  U,V\right)  $ denotes the vector space of
all $k$-linear maps $U\rightarrow V$. (This vector space is commonly denoted
by $\operatorname{Hom}_{k}\left(  U,V\right)  $.)
\end{definition}

\begin{convention}
\label{Convention3a} In the following, whenever a commutative ring $k$ exists
in the context\footnote{Most of the time we will be working over a ground
field, which will be denoted by $k$. Occasionally (e.g., in \S \ref{S11}), we
will be working over a ground ring, but it will also be denoted by $k$.}, the
$\otimes$ sign will always mean $\otimes_{k}$.
\end{convention}

\begin{convention}
\label{Convention3bs} If $Q$ is any set and $n$ is any nonnegative integer,
then we shall denote the Cartesian product $\underbrace{Q\times Q\times
\cdots\times Q}_{n\text{ times}}$ by $Q^{\times n}$. This Cartesian product is
usually denoted by $Q^{n}$ in the literature, but we shall instead reserve the
notation $Q^{n}$ for another meaning (which will be introduced in
Convention~\ref{Convention100}).
\end{convention}

\begin{convention}
\label{Convention3c} Whenever $k$ is a field, $A$, $B$, $C$ and $D$ are four
$k$-vector spaces, and $f:A\rightarrow B$ and $g:C\rightarrow D$ are two
$k$-linear maps, then the notation $f\otimes g$ can mean two different things:
On the one hand, it can mean the $k$-linear map $f\otimes g:A\otimes
C\rightarrow B\otimes D$ (which maps $a\otimes c$ to $f\left(  a\right)
\otimes g\left(  c\right)  $ for every $a\in A$ and $c\in C$). On the other
hand, it can mean the tensor $f\otimes g\in\mathcal{L}\left(  A,B\right)
\otimes\mathcal{L}\left(  C,D\right)  $ (since $f\in\mathcal{L}\left(
A,B\right)  $ and $g\in\mathcal{L}\left(  C,D\right)  $). Let us agree that in
the following, whenever a term like $f\otimes g$ (with $f$ and $g$ being two
$k$-linear maps) occurs, it will mean the first thing (i. e., the $k$-linear
map $f\otimes g:A\otimes C\rightarrow B\otimes D$) and not the second one (i.
e., the tensor $f\otimes g\in\mathcal{L}\left(  A,B\right)  \otimes
\mathcal{L}\left(  C,D\right)  $).
\end{convention}

\begin{definition}
\label{Definition4} Let $k$ be a field and $A$ be a $k$-algebra. Then,
$\mu_{A}$ will always denote the multiplication map $A\otimes A\rightarrow A$
of the $k$-algebra $A$, and $\eta_{A}$ will always denote the unity map
$k\rightarrow A$ of the $k$-algebra $A$. When it is clear which algebra we are
talking about, we will abbreviate $\mu_{A}$ and $\eta_{A}$ as $\mu$ and $\eta
$, respectively.
\end{definition}

\begin{definition}
\label{Definition5} Let $k$ be a field and $C$ be a $k$-coalgebra. Then,
$\Delta_{C}$ will always denote the comultiplication map $C\rightarrow
C\otimes C$ of the $k$-coalgebra $C$, and $\varepsilon_{C}$ will always denote
the counit map $C\rightarrow k$ of the $k$-coalgebra $C$. When it is clear
which coalgebra we are talking about, we will abbreviate $\Delta_{C}$ and
$\varepsilon_{C}$ as $\Delta$ and $\varepsilon$, respectively.
\end{definition}

\begin{definition}
\label{Definition6} Let $k$ be a field, let $A$ be a $k$-algebra, and let $C$
be a $k$-coalgebra. Then, the $k$-vector space $\mathcal{L}\left(  C,A\right)
$ becomes a $k$-algebra $\left(  \mathcal{L}\left(  C,A\right)  ,\ast\right)
$ by setting%
\begin{equation}
f\ast g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
\ \ \ \ \ \ \ \ \ \ \text{ for any }f\in\mathcal{L}\left(  C,A\right)  \text{
and }g\in\mathcal{L}\left(  C,A\right)  . \label{*}%
\end{equation}
This $k$-algebra $\left(  \mathcal{L}\left(  C,A\right)  ,\ast\right)  $ has
unity $\eta_{A}\circ\varepsilon_{C}$ and is called the \textit{convolution
algebra} of $C$ and $A$. In the following, we will simply refer to this
algebra as $\mathcal{L}\left(  C,A\right)  $.

The binary operation $\ast$ defined in (\ref{*}) is called
\textit{convolution}. In particular, for any $f\in\mathcal{L}\left(
C,A\right)  $ and $g\in\mathcal{L}\left(  C,A\right)  $, we will refer to
$f\ast g$ as the \textit{convolution} of the maps $f$ and $g$.
\end{definition}

\begin{convention}
\label{Convention7} Let $k$ be a field, let $A$ be a $k$-algebra, and let $C$
be a $k$-coalgebra. For any $n\in\mathbb{N}$ and any $f\in\mathcal{L}\left(
C,A\right)  $, we will denote by $f^{\ast n}$ the $n$-th power of the element
$f$ in the convolution algebra $\mathcal{L}\left(  C,A\right)  $.

If an element $f\in\mathcal{L}\left(  C,A\right)  $ has a multiplicative
inverse in the $k$-algebra $\mathcal{L}\left(  C,A\right)  $, then this
inverse is denoted by $f^{\ast\left(  -1\right)  }$ and called the $\ast
$\textit{-inverse} of $f$.
\end{convention}

\begin{definition}
Let $k$ be a field. Let $H$ be a $k$-Hopf algebra. Then, $S_{H}$ will always
denote the antipode of the $k$-Hopf algebra $S$ (that is, the $\ast$-inverse
of the identity map $\operatorname*{id}\nolimits_{H}\in\mathcal{L}\left(
H,H\right)  $).
\end{definition}

\begin{definition}
\label{def.eHA}Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $A$ be a
$k$-algebra. Then, $e_{C,A}$ shall denote the map $\eta_{A}\circ
\varepsilon_{C}:C\rightarrow A$. This map $e_{C,A}$ is the unity of the
convolution algebra $\mathcal{L}\left(  C,A\right)  $.
\end{definition}

Next, let us define some concepts related to \textit{filtrations} on vector
spaces. Different authors define the words \textquotedblleft
filtered\textquotedblright\ and \textquotedblleft filtration\textquotedblright%
\ in different (often non-equivalent) ways, so some care should be taken when
consulting the literature.

\begin{definition}
\label{Definition2b} Let $k$ be a field. A \textit{filtered }$k$%
\textit{-vector space} means a $k$-vector space $V$ equipped with a family
$\left(  V_{\leq\ell}\right)  _{\ell\geq0}$ of $k$-vector subspaces of $V$
satisfying $V_{\leq0}\subseteq V_{\leq1}\subseteq V_{\leq2}\subseteq\cdots$
and $V=\bigcup_{\ell\geq0}V_{\leq\ell}$. Such a filtered $k$-vector space will
often be denoted simply by $V$ (that is, the family $\left(  V_{\leq\ell
}\right)  _{\ell\geq0}$ will not be explicitly mentioned). The family $\left(
V_{\leq\ell}\right)  _{\ell\geq0}$ is called the \textit{filtration} of this
filtered $k$-vector space. For each $m\in\mathbb{N}$, the $k$-vector subspace
$V_{\leq m}$ of $V$ is called the $m$\textit{-th part of the filtration}
$\left(  V_{\leq\ell}\right)  _{\ell\geq0}$.
\end{definition}

\begin{convention}
\label{Convention3b} In the following, whenever $V$ is a filtered vector
space, we will denote the filtration on $V$ by $\left(  V_{\leq\ell}\right)
_{\ell\geq0}$. (This is a general convention, so it does not only pertain to
filtered vector spaces called $V$, but pertains to any filtered vector space.
For instance, if we have a filtered vector space called $C$, then this
convention yields that the filtration on $C$ is denoted by $\left(
C_{\leq\ell}\right)  _{\ell\geq0}$.)

Furthermore, whenever $V$ is a filtered vector space and $\ell$ is a negative
integer, we define $V_{\leq\ell}$ to mean the $k$-vector subspace $0$ of $V$.
(Thus, $V_{\leq\ell}$ is defined for each $\ell\in\mathbb{Z}$, not only for
$\ell\in\mathbb{N}$.)
\end{convention}

\begin{definition}
Let $k$ be a field. A \textit{filtered }$k$\textit{-coalgebra} means a
$k$-coalgebra $C$ that is simultaneously a filtered $k$-vector space (i.e.,
that is equipped with a family $\left(  C_{\leq\ell}\right)  _{\ell\geq0}$ of
$k$-vector subspaces of $C$ satisfying $C_{\leq0}\subseteq C_{\leq1}\subseteq
C_{\leq2}\subseteq\cdots$ and $C=\bigcup_{\ell\geq0}C_{\leq\ell}$) and has the
property that
\[
\text{each }n\in\mathbb{N}\text{ satisfies }\Delta\left(  C_{\leq n}\right)
\subseteq\sum_{u=0}^{n}C_{\leq u}\otimes C_{\leq n-u}.
\]

\end{definition}

\begin{definition}
\label{Definition7a} Let $k$ be a field, and let $C$ be a filtered
$k$-coalgebra. We say that the filtered $k$-coalgebra $C$ is
\textit{connected} if and only if the map $\varepsilon_{C}\mid_{C_{\leq0}%
}:C_{\leq0}\rightarrow k$ is a $k$-vector space isomorphism\footnote{Recall
that $\left(  C_{\leq\ell}\right)  _{\ell\geq0}$ denotes the filtration of the
filtered $k$-coalgebra $C$ (according to Convention~\ref{Convention3b}).}.
\end{definition}

\section{Unital coalgebras}

Next we shall define the notion of a \textit{unital coalgebra}. This notion is
an intermediate step between the (rather well-known) notions of a coalgebra
and of a bialgebra; it also is an intermediate step between the notions of a
coalgebra and of a filtered connected coalgebra. Its definition is very simple:

\begin{definition}
\label{Definition7b} Let $k$ be a field. Let $C$ be a $k$-coalgebra, and let
$i$ be an element of $C$. Then, $\left(  C,i\right)  $ is said to be a
\textit{unital coalgebra}\footnote{To be completely honest, we would have to
call this ``unital $k$-coalgebra'' to make clear that this notion depends on
the field $k$. However, since we are not going to change the field $k$ anytime
soon, we leave the $k$ out of the notation and simply speak of ``unital
coalgebras''.} if $\Delta_{C}\left(  i\right)  =i\otimes i$ and $\varepsilon
_{C}\left(  i\right)  =1$.

When $\left(  C,i\right)  $ is a unital coalgebra, we will denote the element
$i$ by $1_{\left(  C,i\right)  }$ and call it the \textit{unity} of the unital
coalgebra $\left(  C,i\right)  $.
\end{definition}

When $C$ is a $k$-coalgebra, there may be several elements $i\in C$ for which
$\left(  C,i\right)  $ is a unital coalgebra\footnote{Such elements are called
\textit{grouplike elements} of $C$. Thus, a unital coalgebra is a pair
$\left(  C,i\right)  $ of a coalgebra $C$ and a grouplike element $i$ of $C$.}
(but there also may be no such elements). Hence, a unital coalgebra $\left(
C,i\right)  $ is not uniquely determined by the coalgebra $C$. However, there
are many cases where we have some additional structure on $C$ (like a
$k$-bialgebra structure or a connected filtered $k$-coalgebra structure) which
gives rise to one preferred canonical $i$. First we consider the case when $C$
is a $k$-bialgebra. In this case, we have:

\begin{proposition}
\label{Proposition7c} Let $k$ be a field. Let $C$ be a $k$-bialgebra. Then,
$\left(  C,1_{C}\right)  $ is a unital coalgebra. (Here, $1_{C}$ denotes the
unity of the $k$-algebra $C$, as usual.)
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition7c}.]By the axioms of a $k$-bialgebra,
we have $\Delta_{C}\left(  1_{C}\right)  =1_{C}\otimes1_{C}$ and
$\varepsilon_{C}\left(  1_{C}\right)  =1$. By the definition of a unital
coalgebra, this means that $\left(  C,1_{C}\right)  $ is a unital coalgebra.
Proposition~\ref{Proposition7c} is thus proven.
\end{proof}

Note that Proposition~\ref{Proposition7c} really needs the condition that $C$
is a $k$-bialgebra, and not just some $k$-vector space with a $k$-algebra
structure and a $k$-coalgebra structure.

Now we consider the case of connected filtered coalgebras:

\begin{proposition}
\label{Proposition7d} Let $k$ be a field. Let $C$ be a connected filtered
$k$-coalgebra. Then, $\left(  C,\left(  \varepsilon_{C}\mid_{C_{\leq0}%
}\right)  ^{-1}\left(  1\right)  \right)  $ is a unital coalgebra.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition7d}.]Since $C$ is connected, the map
$\varepsilon_{C}\mid_{C_{\leq0}}:C_{\leq0}\rightarrow k$ is a $k$-vector space
isomorphism. Hence, the map $\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
^{-1}:k\rightarrow C_{\leq0}$ is well-defined.

Let us define an element $i\in C$ by $i=\left(  \varepsilon_{C}\mid_{C_{\leq
0}}\right)  ^{-1}\left(  1\right)  $. We are going to show that $\left(
C,i\right)  $ is a unital coalgebra.

Since $\varepsilon_{C}\mid_{C_{\leq0}}:C_{\leq0}\rightarrow k$ is an
isomorphism, we have%
\[
C_{\leq0}=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(
\underbrace{k}_{=k\cdot1}\right)  =\left(  \varepsilon_{C}\mid_{C_{\leq0}%
}\right)  ^{-1}\left(  k\cdot1\right)  =k\cdot\underbrace{\left(
\varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  }_{=i}=k\cdot
i.
\]


Since $i=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(
1\right)  $, we have $1=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
\left(  i\right)  =\underbrace{\varepsilon_{C}}_{=\varepsilon}\left(
i\right)  =\varepsilon\left(  i\right)  $. Now, $i\in C_{\leq0}$, so that
$\Delta\left(  i\right)  \in\Delta\left(  C_{\leq0}\right)  \subseteq
\sum_{u=0}^{0}C_{\leq u}\otimes C_{\leq0-u}$ (since $C$ is a filtered
coalgebra). Since%
\begin{align*}
\sum_{u=0}^{0}C_{\leq u}\otimes C_{\leq0-u}  &  =C_{\leq0}\otimes
\underbrace{C_{\leq0-0}}_{=C_{\leq0}}=C_{\leq0}\otimes C_{\leq0}=\left(
k\cdot i\right)  \otimes\left(  k\cdot i\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }C_{\leq0}=k\cdot i\right) \\
&  =k\cdot\left(  i\otimes i\right)  ,
\end{align*}
this rewrites as $\Delta\left(  i\right)  \in k\cdot\left(  i\otimes i\right)
$. Thus, there exists some $\lambda\in k$ such that $\Delta\left(  i\right)
=\lambda\cdot\left(  i\otimes i\right)  $. Consider this $\lambda$.

Let $\operatorname*{can}:C\otimes k\rightarrow C$ be the canonical $k$-module
isomorphism (sending $c\otimes x$ to $cx$ for all $c\in C$ and $x\in k$).
Then, by the axioms of a coalgebra, we have $\operatorname*{can}\circ\left(
\operatorname*{id}\otimes\varepsilon\right)  \circ\Delta=\operatorname*{id}$.
But%
\begin{align*}
\left(  \operatorname*{can}\circ\left(  \operatorname*{id}\otimes
\varepsilon\right)  \circ\Delta\right)  \left(  i\right)   &
=\operatorname*{can}\left(  \left(  \operatorname*{id}\otimes\varepsilon
\right)  \left(  \underbrace{\Delta\left(  i\right)  }_{=\lambda\cdot\left(
i\otimes i\right)  }\right)  \right)  =\operatorname*{can}\left(
\underbrace{\left(  \operatorname*{id}\otimes\varepsilon\right)  \left(
\lambda\cdot\left(  i\otimes i\right)  \right)  }_{=\lambda\cdot
\operatorname*{id}\left(  i\right)  \otimes\varepsilon\left(  i\right)
}\right) \\
&  =\operatorname*{can}\left(  \lambda\cdot\operatorname*{id}\left(  i\right)
\otimes\varepsilon\left(  i\right)  \right)  =\lambda\cdot
\underbrace{\operatorname*{id}\left(  i\right)  }_{=i}\cdot
\underbrace{\varepsilon\left(  i\right)  }_{=1}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\operatorname*{can}\right) \\
&  =\lambda i,
\end{align*}
so that%
\[
\lambda i=\underbrace{\left(  \operatorname*{can}\circ\left(
\operatorname*{id}\otimes\varepsilon\right)  \circ\Delta\right)
}_{=\operatorname*{id}}\left(  i\right)  =\operatorname*{id}\left(  i\right)
=i.
\]
Now, $\Delta\left(  i\right)  =\lambda\cdot\left(  i\otimes i\right)
=\underbrace{\lambda i}_{=i}\otimes i=i\otimes i$.

So we have $\Delta_{C}\left(  i\right)  =\Delta\left(  i\right)  =i\otimes i$
and $\varepsilon_{C}\left(  i\right)  =1$. By the definition of a unital
coalgebra, this shows that $\left(  C,i\right)  $ is a unital coalgebra. Since
$i=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  $,
this means that $\left(  C,\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
^{-1}\left(  1\right)  \right)  $ is a unital coalgebra.
Proposition~\ref{Proposition7d} is proven.
\end{proof}

Proposition~\ref{Proposition7c} gives us a unital coalgebra when we start with
a $k$-bialgebra. Proposition~\ref{Proposition7d} gives us a unital coalgebra
when we start with a connected filtered $k$-coalgebra. One might wonder what
happens if we start with a connected filtered $k$-bialgebra: In this case,
each of Propositions~\ref{Proposition7c} and~\ref{Proposition7d} gives us a
unital coalgebra. Are these two unital coalgebras the same? The answer is yes,
as the following proposition shows:

\begin{proposition}
\label{Proposition7e} Let $k$ be a field, and let $C$ be a connected
filtered\footnote{Of course, when we say that the filtered $k$-bialgebra $C$
is connected, we mean that the filtered $k$-coalgebra $C$ is connected.}
$k$-bialgebra. Then, $1_{C}=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
^{-1}\left(  1\right)  $ (where $1_{C}$ denotes the unity of the $k$-algebra
$C$). As a consequence, the unital coalgebras $\left(  C,1_{C}\right)  $ and
$\left(  C,\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(
1\right)  \right)  $ are identic.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition7e}.]By the definition of a connected
filtered $k$-coalgebra, the map $\varepsilon_{C}\mid_{C_{\leq0}}:C_{\leq
0}\rightarrow k$ is a $k$-vector space isomorphism (since $C$ is connected).
Moreover, $\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  \left(
1_{C}\right)  =\varepsilon_{C}\left(  1_{C}\right)  =1$, so that
$1_{C}=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)
$. This proves Proposition~\ref{Proposition7e}.
\end{proof}

Since it is cumbersome to explicitly mention the unity every time we are
referring to a unital coalgebra, we make the following convention:

\begin{convention}
\label{Convention7f} Let $k$ be a field. Let $\left(  C,i\right)  $ be a
unital coalgebra. Then, we will often abbreviate the ``unital coalgebra
$\left(  C,i\right)  $'' as ``unital coalgebra $C$''. This abbreviation is an
abuse of notation, since a unital coalgebra $\left(  C,i\right)  $ is not
uniquely determined by the coalgebra $C$; but we will only use this
abbreviation when it is clear what $i$ we mean. In particular, we will use
this abbreviation when there is a canonical unital coalgebra structure on $C$
obtained from either Definition~\ref{Definition7g} or
Definition~\ref{Definition7h} (below), or when the unital coalgebra is just
being defined\footnote{For instance, when we write ``Let $C$ be a unital
coalgebra'', we are using this abbreviation; this is okay, because there are
no elements $i$ of $C$ defined yet which we can confuse.}.
\end{convention}

\begin{definition}
\label{Definition7g} Let $k$ be a field. Let $C$ be a $k$-bialgebra. Then,
according to Proposition~\ref{Proposition7c}, the unital coalgebra $\left(
C,1_{C}\right)  $ is well-defined (where $1_{C}$ denotes the unity of the
$k$-algebra $C$). This unital coalgebra $\left(  C,1_{C}\right)  $ is called
the \textit{unital coalgebra canonically induced by the }$k$%
\textit{-bialgebra} $C$. Whenever we just speak of ``the unital coalgebra
$C$'' (where $C$ is a $k$-bialgebra), we will always mean this unital
coalgebra $\left(  C,1_{C}\right)  $.
\end{definition}

\begin{definition}
\label{Definition7h} Let $k$ be a field. Let $C$ be a connected filtered
$k$-coalgebra. Then, the unital coalgebra $\left(  C,\left(  \varepsilon
_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  \right)  $ (this is
well-defined according to Proposition~\ref{Proposition7d}) is called the
\textit{unital coalgebra canonically induced by the connected filtered }%
$k$\textit{-coalgebra} $C$. Whenever we just speak of ``the unital coalgebra
$C$'' (where $C$ is a connected filtered $k$-coalgebra), we will always mean
this unital coalgebra $\left(  C,\left(  \varepsilon_{C}\mid_{C_{\leq0}%
}\right)  ^{-1}\left(  1\right)  \right)  $.
\end{definition}

\begin{remark}
\label{Remark7i} Let $k$ be a field. Let $C$ be a connected filtered
$k$-bialgebra. Then, the ``unital coalgebra $C$'' as understood according to
Definition~\ref{Definition7g} is identic with the ``unital coalgebra $C$'' as
understood according to Definition~\ref{Definition7h}. (This is just a
rewording of Proposition~\ref{Proposition7e}.) Thus, the notations introduced
in Definitions~\ref{Definition7g} and~\ref{Definition7h} don't conflict with
each other.\footnote{This only pertains to the case when $C$ is a connected
filtered $k$-bialgebra. If $C$ would be a vector space with a connected
filtered $k$-coalgebra structure on one hand and a (totally unrelated!)
$k$-bialgebra structure on the other, then, of course, then the ``unital
coalgebra $C$'' in the sense of Definition~\ref{Definition7g} would not
necessarily be identic to the ``unital coalgebra $C$'' in the sense of
Definition~\ref{Definition7h}, so the notations introduced in
Definitions~\ref{Definition7g} and~\ref{Definition7h} could conflict. However,
in such a case, it would already be a very bad idea to speak of the
``coalgebra $C$'', since this could mean any of two different coalgebra
structures, so in such a case we would have to be careful with our notations
anyway.}
\end{remark}

\begin{remark}
\label{Remark7j} Let $k$ be a field. Let $C$ be a unital coalgebra. Then, the
unity of this unital coalgebra $C$ is denoted by $1_{C}$. This is not a new
notation we introduce; it is just a consequence of our
Definition~\ref{Definition7b} (where we stipulated that the unity of a unital
coalgebra $\left(  C,i\right)  $ will be denoted by $1_{\left(  C,i\right)  }%
$) because we abbreviate $\left(  C,i\right)  $ by $C$.

This notation can conflict with the notation $1_{A}$ for the unity of a
$k$-algebra $A$: In fact, if we have a vector space $V$ which happens to be a
$k$-algebra and a unital coalgebra at the same time, then in general it might
occur that the unity of the unital coalgebra $V$ is not the same as the unity
of the $k$-algebra $V$, so the notation $1_{V}$ would be ambiguous (it could
mean each of these two unities). However, when we have a $k$-bialgebra $C$,
then the unity of the unital coalgebra $C$ (which is understood according to
Definition~\ref{Definition7g}) \textit{is} the same as the unity of the
$k$-algebra $C$\ \ \ \ \footnote{\textit{Proof.} Let $C$ be a $k$-bialgebra.
According to Definition~\ref{Definition7g}, the unital coalgebra $C$ is
$\left(  C,1_{C}\right)  $, where $1_{C}$ denotes the unity of the $k$-algebra
$C$. Hence, the unity of the unital coalgebra $C$ is $1_{C}$, where $1_{C}$
denotes the unity of the $k$-algebra $C$. In other words, the unity of the
unital coalgebra $C$ is the same as the unity of the $k$-algebra $C$, qed.},
so there is no conflict in this case. Fortunately, we are going to have this
case all of the time, so we won't have to care about possible conflicts
between these notations.
\end{remark}

\begin{remark}
\label{Remark7k} Let $k$ be a field, and let $C$ be a connected filtered
$k$-coalgebra. Then, $C$ is a unital coalgebra (according to
Definition~\ref{Definition7h}) with unity $1_{C}=\left(  \varepsilon_{C}%
\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  $.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark7k}.]The unital coalgebra $C$ (as defined in
Definition~\ref{Definition7h}) is $\left(  C,\left(  \varepsilon_{C}%
\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  \right)  $. Hence, the unity
of this coalgebra is $\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
^{-1}\left(  1\right)  $. Since we denote the unity of the unital coalgebra
$C$ by $1_{C}$, this means that $1_{C}=\left(  \varepsilon_{C}\mid_{C_{\leq0}%
}\right)  ^{-1}\left(  1\right)  $. Remark~\ref{Remark7k} is proven.
\end{proof}

\begin{remark}
\label{Remark7l} Let $k$ be a field, and let $C$ be a connected filtered
$k$-coalgebra. Then, $C_{\leq0}=k\cdot1_{C}$. (Here, as usual, $1_{C}$ denotes
the unity of the unital coalgebra $C$, which unital coalgebra is defined as in
Definition~\ref{Definition7h}.)
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark7l}.]In the proof of
Proposition~\ref{Proposition7d}, we showed that $C_{\leq0}=k\cdot i$, where
$i=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  $.
But this $i$ is equal to $1_{C}$ (since $i=\left(  \varepsilon_{C}%
\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  =1_{C}$ by
Remark~\ref{Remark7k}), so that $C_{\leq0}=k\cdot i$ rewrites as $C_{\leq
0}=k\cdot1_{C}$. Remark~\ref{Remark7l} is now proven.
\end{proof}

\begin{remark}
\label{Remark7m} Let $k$ be a field, and let $H$ be a filtered $k$-bialgebra.
Then, $H$ is connected if and only if $H_{\leq0}=k\cdot1_{H}$ (where $1_{H}$
denotes the unity of the $k$-algebra $H$).
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark7m}.]\textbf{a)} Let us prove that if $H$ is
connected, then $H_{\leq0}=k\cdot1_{H}$.

\textit{Proof.} Assume that $H$ is connected. Then, $H$ becomes a unital
coalgebra according to Definition~\ref{Definition7h}. Thus, the notation
$1_{H}$ can mean two different things: On the one hand, it can mean the unity
of the $k$-algebra $H$, but on the other hand, it can mean the unity of the
unital coalgebra $H$ (defined by Definition~\ref{Definition7h}). Fortunately,
this does not yield a conflict because these two things are the same (by
Remark~\ref{Remark7j}, since $H$ is a $k$-bialgebra). Remark~\ref{Remark7l}
(applied to $C=H$) now yields $H_{\leq0}=k\cdot1_{H}$.

We thus have proven that if $H$ is connected, then $H_{\leq0}=k\cdot1_{H}$.

\textbf{b)} Let us prove that if $H_{\leq0}=k\cdot1_{H}$, then $H$ is connected.

\textit{Proof.} Assume that $H_{\leq0}=k\cdot1_{H}$. Then, every $\alpha\in
H_{\leq0}$ satisfying $\left(  \varepsilon_{H}\mid_{H_{\leq0}}\right)  \left(
\alpha\right)  =0$ must satisfy $\alpha=0\ \ \ \ $\footnote{\textit{Proof.}
Let $\alpha\in H_{\leq0}$ satisfying $\left(  \varepsilon_{H}\mid_{H_{\leq0}%
}\right)  \left(  \alpha\right)  =0$ be arbitrary. Then, $\varepsilon\left(
\alpha\right)  =\left(  \varepsilon_{H}\mid_{H_{\leq0}}\right)  \left(
\alpha\right)  =0$. On the other hand, $\alpha\in H_{\leq0}=k\cdot1_{H}$, so
that there exists some $\lambda\in k$ such that $\alpha=\lambda\cdot1_{H}$.
Consider this $\lambda$. Then, $\varepsilon\left(  \alpha\right)
=\varepsilon\left(  \lambda\cdot1_{H}\right)  =\lambda\underbrace{\varepsilon
\left(  1_{H}\right)  }_{=1}=\lambda$. Thus, $\varepsilon\left(
\alpha\right)  =0$ becomes $\lambda=0$, so that $\alpha=\underbrace{\lambda
}_{=0}\cdot1_{H}=0$, qed.}. Hence, the $k$-linear map $\varepsilon_{H}%
\mid_{H_{\leq0}}$ must be injective. Since this map is also
surjective\footnote{In fact, every $\beta\in k$ satisfies $\beta\in\left(
\varepsilon_{H}\mid_{H_{\leq0}}\right)  \left(  H_{\leq0}\right)  $ (because
$\beta\cdot\underbrace{1_{H}}_{\in H_{\leq0}}\in H_{\leq0}$ and $\left(
\varepsilon_{H}\mid_{H_{\leq0}}\right)  \left(  \beta\cdot1_{H}\right)
=\varepsilon\left(  \beta\cdot1_{H}\right)  =\beta\underbrace{\varepsilon
\left(  1_{H}\right)  }_{=1}=\beta$, so that $\beta=\left(  \varepsilon
_{H}\mid_{H_{\leq0}}\right)  \left(  \beta\cdot1_{H}\right)  \in\left(
\varepsilon_{H}\mid_{H_{\leq0}}\right)  \left(  H_{\leq0}\right)  $).}, it
thus follows that the map $\varepsilon_{H}\mid_{H_{\leq0}}$ is bijective.
Hence, $\varepsilon_{H}\mid_{H_{\leq0}}$ is an isomorphism. By the definition
of ``connected'', this yields that $H$ is connected.

We have thus proven that if $H_{\leq0}=k\cdot1_{H}$, then $H$ is connected.

Combining the above points \textbf{a)} and \textbf{b)}, we obtain
Remark~\ref{Remark7m}.
\end{proof}

\textit{Remark:} The above Remark~\ref{Remark7m} is often used as an
alternative definition of the notion of a connected filtered $k$-bialgebra.
However, we prefer Definition~\ref{Definition7a}, since it works for filtered
$k$-coalgebras as well (and not only for filtered $k$-bialgebras).

\begin{definition}
\label{Definition7n} Let $k$ be a field, and let $C$ be a unital coalgebra.
Then, we denote by $\eta_{C}$ the map $k\rightarrow C$ which sends every
$\lambda\in k$ to $\lambda\cdot1_{C}\in C$. This map is called the
\textit{unity map} of the unital coalgebra $C$.

This notation could sometimes conflict with the notation $\eta_{A}$ for the
unity map of a $k$-algebra $A$. In fact, such a conflict might emerge when we
have a $k$-vector space $H$ which is both a unital coalgebra and a $k$-algebra
at the same time; in this case, $\eta_{H}$ might mean two different things
(namely, the unity map of the $k$-algebra $H$ on the one hand, and the unity
map of the unital coalgebra $H$ on the other), just as $1_{H}$ might mean two
different things. However, when $H$ is a $k$-bialgebra, both meanings of
$1_{H}$ are the same, and therefore both meanings of $\eta_{H}$ are the
same\footnote{\textit{Proof.} Let $H$ be a $k$-bialgebra. Then, the notation
$\eta_{H}$ might mean two different things: namely, the unity map of the
$k$-algebra $H$ on the one hand, and the unity map of the unital coalgebra $H$
on the other. However, these two things are the same, since
\begin{align*}
&  \left(  \text{the unity map of the unital coalgebra }H\right) \\
&  =\left(  \text{the map }k\rightarrow H\text{ which sends every }\lambda\in
k\text{ to }\lambda\cdot1_{H}\text{, where }1_{H}\text{ denotes the unity of
the unital coalgebra }H\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because this is how the unity map of the
unital coalgebra }H\text{ was defined}\right) \\
&  =\left(  \text{the map }k\rightarrow H\text{ which sends every }\lambda\in
k\text{ to }\lambda\cdot1_{H}\text{, where }1_{H}\text{ denotes the unity of
the }k\text{-algebra }H\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Remark~\ref{Remark7j} (applied to }C=H\text{) says that the unity
of the unital coalgebra }H\text{ is the same}\\
\text{as the unity of the }k\text{-algebra }H
\end{array}
\right) \\
&  =\left(  \text{the unity map of the }k\text{-algebra }H\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because this is how the unity map of the
}k\text{-algebra }H\text{ was defined}\right)  .
\end{align*}
Hence, both meanings of $\eta_{H}$ are the same, qed.}. Thus, no conflict can
occur as long as $H$ is a $k$-bialgebra.
\end{definition}

Now that we have defined the notion of unital coalgebras and cleared up some
possible and impossible confusions, let us continue introducing notation.
First, we observe that if we replace the words \textquotedblleft%
$k$-algebra\textquotedblright\ by \textquotedblleft unital
coalgebra\textquotedblright\ in the definition of $e_{H,A}$
(Definition~\ref{def.eHA}), then this definition still makes sense (because
the expression $1_{A}$ makes sense not only when $A$ is a $k$-algebra, but
also when $A$ is a unital coalgebra), although of course the map $e_{H,A}$ it
defines is no longer the unity of the convolution algebra $\mathcal{L}\left(
H,A\right)  $ (since this convolution algebra does not exist in this
situation). Thus, we obtain the following definition:

\begin{definition}
\label{def.eHA2}Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $A$ be a
unital coalgebra. Then, $e_{C,A}$ shall denote the map $\eta_{A}%
\circ\varepsilon_{C}:C\rightarrow A$.
\end{definition}

Definition~\ref{def.eHA2} does not conflict with Definition~\ref{def.eHA} in
the case when $A$ is a $k$-bialgebra (because both interpretations of the
expression $1_{A}$ mean the same thing in this case); these definitions also
do not conflict in the case when $A$ is a connected filtered coalgebra (for
the same reason). Of course, if $A$ is simultaneously a $k$-algebra and a
unital coalgebra with two completely unrelated unities, then the two
definitions can conflict.

\section{Logarithms and exponentials in convolution algebras}

\begin{definition}
\label{Definition8} Let $k$ be a field, let $A$ be a $k$-algebra, and let $H$
be a connected filtered $k$-coalgebra.\footnote{We denote this $k$-coalgebra
by $H$ rather than by the (more appropriate) letter $C$ because this is how it
is often called in this context in standard literature.}

\textbf{(a)} We denote by $\mathfrak{g}\left(  H,A\right)  $ the subspace
$\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(  1_{H}\right)
=0\right\}  $ of $\mathcal{L}\left(  H,A\right)  $. (Here, $1_{H}$ denotes the
unity of the unital coalgebra $H$, which is defined according to
Definition~\ref{Definition7h}.)

\textbf{(b)} For every $n\in\mathbb{N}$, we denote by $\mathcal{L}^{n}\left(
H,A\right)  $ the subspace $\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\mid_{H_{\leq n-1}}=0\right\}  $\ \ \ \ \footnote{Recall that
$\left(  H_{\leq\ell}\right)  _{\ell\geq0}$ denotes the filtration of the
filtered $k$-coalgebra $H$ (according to Convention~\ref{Convention3b}).}.
Then,%
\[
\mathcal{L}^{0}\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ f\mid_{H_{\leq0-1}}=0\right\}  =\mathcal{L}\left(
H,A\right)
\]
(since every $f\in\mathcal{L}\left(  H,A\right)  $ satisfies $f\mid
_{H_{\leq0-1}}=0$ (because $H_{\leq0-1}=H_{\leq-1}=0$)) and%
\begin{align*}
\mathcal{L}^{1}\left(  H,A\right)   &  =\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ f\mid_{H_{\leq1-1}}=0\right\}  =\left\{  f\in
\mathcal{L}\left(  H,A\right)  \ \mid\ f\mid_{\left\{  \lambda\cdot1_{H}%
\mid\lambda\in k\right\}  }=0\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }H_{\leq1-1}=H_{\leq0}=k\cdot1_{H}\text{ (by Remark~\ref{Remark7l}%
, applied to }C=H\text{)}\\
\text{and thus }H_{\leq1-1}=k\cdot1_{H}=\left\{  \lambda\cdot1_{H}\mid
\lambda\in k\right\}
\end{array}
\right) \\
&  =\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ \underbrace{f\left(
\lambda\cdot1_{H}\right)  }_{=\lambda f\left(  1_{H}\right)  }=0\text{ for
every }\lambda\in k\right\} \\
&  =\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ \underbrace{\lambda
f\left(  1_{H}\right)  =0\text{ for every }\lambda\in k}_{\text{this is
equivalent to }f\left(  1_{H}\right)  =0}\right\} \\
&  =\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(  1_{H}\right)
=0\right\}  =\mathfrak{g}\left(  H,A\right)  .
\end{align*}


\textbf{(c)} Let us denote by $G\left(  H,A\right)  $ the subset $\left\{
f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(  1_{H}\right)
=1_{A}\right\}  $ of $\mathcal{L}\left(  H,A\right)  $. (Here, $1_{A}$ denotes
the unity of $A$.)
\end{definition}

\begin{definition}
\label{Definition8b} \textbf{(a)} In Definition~\ref{Definition8}
\textbf{(a)}, we have defined $\mathfrak{g}\left(  H,A\right)  $ when $H$ is a
connected filtered $k$-coalgebra and $A$ is a $k$-algebra. In the same way
(that is, by the formula%
\[
\mathfrak{g}\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\left(  1_{H}\right)  =0\right\}
\]
), we can define $\mathfrak{g}\left(  H,A\right)  $ when $H$ is a unital
coalgebra and $A$ is any $k$-vector space. In particular, $\mathfrak{g}\left(
H,A\right)  $ is thus defined when $H$ is a $k$-bialgebra and $A$ is any
$k$-vector space (because according to Definition~\ref{Definition7g}, when $H$
is a $k$-bialgebra, $H$ canonically becomes a unital coalgebra).

\textbf{(b)} In Definition~\ref{Definition8} \textbf{(c)}, we have defined
$G\left(  H,A\right)  $ when $H$ is a connected filtered $k$-coalgebra and $A$
is a $k$-algebra. In the same way (that is, by the formula
\[
G\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\left(  1_{H}\right)  =1_{A}\right\}
\]
), we can define the set $G\left(  H,A\right)  $ when $H$ is a unital
coalgebra and $A$ is a $k$-algebra. Moreover, in the same way (i.e., by the
same formula), we can define the set $G\left(  H,A\right)  $ when $H$ is a
unital coalgebra and $A$ is a unital coalgebra (because the notation $1_{A}$
makes sense whenever $A$ is a unital coalgebra).
\end{definition}

\begin{remark}
\label{rmk.eHAinG}Let $k$ be a field. Let $A$ be a $k$-algebra. Let $H$ be a
unital coalgebra. Recall (from Definition~\ref{def.eHA}) that $e_{H,A}$
denotes the map $\eta_{A}\circ\varepsilon_{H}:H\rightarrow A$. Recall the sets
$\mathfrak{g}\left(  H,A\right)  $ and $G\left(  H,A\right)  $ defined in
Definition~\ref{Definition8b}.

\textbf{(a)} We have $e_{H,A}\in G\left(  H,A\right)  $.

\textbf{(b)} We have $G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(
H,A\right)  $.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{rmk.eHAinG}.]We know that $\left(  H,1_{H}\right)  $ is
a unital coalgebra. In other words, $H$ is a $k$-coalgebra and $1_{H}$ is an
element of $H$ satisfying $\Delta_{H}\left(  1_{H}\right)  =1_{H}\otimes1_{H}$
and $\varepsilon_{H}\left(  1_{H}\right)  =1$ (by the definition of a
\textquotedblleft unital coalgebra\textquotedblright).

We have $e_{H,A}=\eta_{A}\circ\varepsilon_{H}$. Thus,%
\begin{align*}
e_{H,A}\left(  1_{H}\right)   &  =\left(  \eta_{A}\circ\varepsilon_{H}\right)
\left(  1_{H}\right)  =\eta_{A}\left(  \underbrace{\varepsilon_{H}\left(
1_{H}\right)  }_{=1}\right)  =\eta_{A}\left(  1\right) \\
&  =1\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map
}\eta_{A}\right) \\
&  =1_{A}.
\end{align*}


Recall that%
\begin{equation}
\mathfrak{g}\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\left(  1_{H}\right)  =0\right\}  \label{pf.rmk.eHAinG.1}%
\end{equation}
and%
\begin{equation}
G\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\left(  1_{H}\right)  =1_{A}\right\}  . \label{pf.rmk.eHAinG.2}%
\end{equation}


\textbf{(a)} Now, $e_{H,A}$ is an element of $\mathcal{L}\left(  H,A\right)  $
satisfying $e_{H,A}\left(  1_{H}\right)  =1_{A}$. In other words, $e_{H,A}$ is
an $f\in\mathcal{L}\left(  H,A\right)  $ satisfying $f\left(  1_{H}\right)
=1_{A}$. In other words, $e_{H,A}\in\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ f\left(  1_{H}\right)  =1_{A}\right\}  $. In view of
(\ref{pf.rmk.eHAinG.2}), this rewrites as $e_{H,A}\in G\left(  H,A\right)  $.
This proves Remark~\ref{rmk.eHAinG} \textbf{(a)}.

\textbf{(b)} Let $g\in G\left(  H,A\right)  $. Thus, $g\in G\left(
H,A\right)  =\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(
1_{H}\right)  =1_{A}\right\}  $. In other words, $g$ is an element
$f\in\mathcal{L}\left(  H,A\right)  $ satisfying $f\left(  1_{H}\right)
=1_{A}$. In other words, $g$ is an element of $\mathcal{L}\left(  H,A\right)
$ and satisfies $g\left(  1_{H}\right)  =1_{A}$. Now, $\left(  g-e_{H,A}%
\right)  \left(  1_{H}\right)  =\underbrace{g\left(  1_{H}\right)  }_{=1_{A}%
}-\underbrace{e_{H,A}\left(  1_{H}\right)  }_{=1_{A}}=1_{A}-1_{A}=0$. Thus,
$g-e_{H,A}$ is an element of $\mathcal{L}\left(  H,A\right)  $ and satisfies
$\left(  g-e_{H,A}\right)  \left(  1_{H}\right)  =0$. In other words,
$g-e_{H,A}$ is an $f\in\mathcal{L}\left(  H,A\right)  $ satisfying $f\left(
1_{H}\right)  =0$. In other words, $g-e_{H,A}\in\left\{  f\in\mathcal{L}%
\left(  H,A\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}  $. In view of
(\ref{pf.rmk.eHAinG.1}), this rewrites as $g-e_{H,A}\in\mathfrak{g}\left(
H,A\right)  $. Thus, $g\in e_{H,A}+\mathfrak{g}\left(  H,A\right)  $.

Now, forget that we fixed $g$. We thus have shown that $g\in e_{H,A}%
+\mathfrak{g}\left(  H,A\right)  $ for each $g\in G\left(  H,A\right)  $. In
other words,%
\begin{equation}
G\left(  H,A\right)  \subseteq e_{H,A}+\mathfrak{g}\left(  H,A\right)  .
\label{pf.rmk.eHAinG.b.1}%
\end{equation}


On the other hand, let $h\in e_{H,A}+\mathfrak{g}\left(  H,A\right)  $. Thus,
$h\in\mathcal{L}\left(  H,A\right)  $ and $h-e_{H,A}\in\mathfrak{g}\left(
H,A\right)  $. We have $h-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  =\left\{
f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}
$. In other words, $h-e_{H,A}$ is an element $f$ of $\mathcal{L}\left(
H,A\right)  $ satisfying $f\left(  1_{H}\right)  =0$. In other words,
$h-e_{H,A}$ is an element of $\mathcal{L}\left(  H,A\right)  $ and satisfies
$\left(  h-e_{H,A}\right)  \left(  1_{H}\right)  =0$. Comparing $\left(
h-e_{H,A}\right)  \left(  1_{H}\right)  =0$ with $\left(  h-e_{H,A}\right)
\left(  1_{H}\right)  =h\left(  1_{H}\right)  -\underbrace{e_{H,A}\left(
1_{H}\right)  }_{=1_{A}}=h\left(  1_{H}\right)  -1_{A}$, we obtain $h\left(
1_{H}\right)  -1_{A}=0$. In other words, $h\left(  1_{H}\right)  =1_{A}$. Now,
$h$ is an element of $\mathcal{L}\left(  H,A\right)  $ satisfying $h\left(
1_{H}\right)  =1_{A}$. In other words, $h$ is an $f\in\mathcal{L}\left(
H,A\right)  $ satisfying $f\left(  1_{H}\right)  =1_{A}$. In other words,
$h\in\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(
1_{H}\right)  =1_{A}\right\}  $. In view of (\ref{pf.rmk.eHAinG.2}), this
rewrites as $h\in G\left(  H,A\right)  $.

Now, forget that we fixed $h$. We thus have shown that $h\in G\left(
H,A\right)  $ for each $h\in e_{H,A}+\mathfrak{g}\left(  H,A\right)  $. In
other words, $e_{H,A}+\mathfrak{g}\left(  H,A\right)  \subseteq G\left(
H,A\right)  $. Combining this with (\ref{pf.rmk.eHAinG.b.1}), we obtain
$G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(  H,A\right)  $. This proves
Remark~\ref{rmk.eHAinG} \textbf{(b)}.
\end{proof}

\begin{remark}
\label{rmk.eHAinG2}If we replace the word \textquotedblleft$k$%
-algebra\textquotedblright\ by \textquotedblleft unital
coalgebra\textquotedblright\ in Remark~\ref{rmk.eHAinG}, then
Remark~\ref{rmk.eHAinG} still holds. In fact, the same proof given above still
applies in this situation.
\end{remark}

\begin{remark}
\label{Remark9} Let $k$ be a field, let $A$ be a $k$-algebra, and let $H$ be a
connected filtered $k$-coalgebra. Every $i\in\mathbb{N}$, $n\in\mathbb{N}$ and
$f\in\mathfrak{g}\left(  H,A\right)  $ such that $i>n$ satisfy $f^{\ast
i}\left(  H_{\leq n}\right)  =0$.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark9}.]We prove Remark~\ref{Remark9} by induction
over $i$:

\textit{Induction base:} For $i=0$, Remark~\ref{Remark9} is vacuously true
(because $i>n$ cannot hold (since $i=0$ and $n\in\mathbb{N}$)).

\textit{Induction step:} Let $j\in\mathbb{N}$ be arbitrary. Assume that
Remark~\ref{Remark9} holds for $i=j$. Now, we must prove that
Remark~\ref{Remark9} also holds for $i=j+1$.

Let $n\in\mathbb{N}$ and $f\in\mathfrak{g}\left(  H,A\right)  $ be such that
$j+1>n$. Then, $\Delta_{H}\left(  H_{\leq n}\right)  \subseteq\sum_{u=0}%
^{n}H_{\leq u}\otimes H_{\leq n-u}$ (since $H$ is a filtered coalgebra) and%
\begin{align}
&  \underbrace{f^{\ast\left(  j+1\right)  }}_{=f^{\ast j}\ast f}\left(
H_{\leq n}\right) \nonumber\\
&  =\left(  f^{\ast j}\ast f\right)  \left(  H_{\leq n}\right)  =\left(
\mu_{A}\circ\left(  f^{\ast j}\otimes f\right)  \circ\Delta_{H}\right)
\left(  H_{\leq n}\right) \nonumber\\
&  =\mu_{A}\left(  \left(  f^{\ast j}\otimes f\right)  \left(
\underbrace{\Delta_{H}\left(  H_{\leq n}\right)  }_{\subseteq\sum_{u=0}%
^{n}H_{\leq u}\otimes H_{\leq n-u}}\right)  \right) \nonumber\\
&  \subseteq\mu_{A}\left(  \left(  f^{\ast j}\otimes f\right)  \left(
\sum_{u=0}^{n}H_{\leq u}\otimes H_{\leq n-u}\right)  \right) \nonumber\\
&  =\sum_{u=0}^{n}\mu_{A}\left(  \underbrace{\left(  f^{\ast j}\otimes
f\right)  \left(  H_{\leq u}\otimes H_{\leq n-u}\right)  }_{\subseteq f^{\ast
j}\left(  H_{\leq u}\right)  \otimes f\left(  H_{\leq n-u}\right)  }\right)
\nonumber\\
&  \subseteq\sum_{u=0}^{n}\mu_{A}\left(  f^{\ast j}\left(  H_{\leq u}\right)
\otimes f\left(  H_{\leq n-u}\right)  \right) \nonumber\\
&  =\sum_{u=0}^{n-1}\mu_{A}\left(  f^{\ast j}\left(  H_{\leq u}\right)
\otimes f\left(  H_{\leq n-u}\right)  \right)  +\mu_{A}\left(  f^{\ast
j}\left(  H_{\leq n}\right)  \otimes f\left(  H_{\leq n-n}\right)  \right)  .
\label{pf.9.3}%
\end{align}
Now, every $u\in\left\{  0,1,\ldots,n-1\right\}  $ satisfies $j>u$ (since
$j+1>n=\underbrace{\left(  n-1\right)  }_{\geq u}+1\geq u+1$). Thus, for every
$u\in\left\{  0,1,\ldots,n-1\right\}  $, we can apply Remark~\ref{Remark9} to
$j$ and $u$ instead of $i$ and $n$ (since we assumed that Remark~\ref{Remark9}
holds for $i=j$), and obtain $f^{\ast j}\left(  H_{\leq u}\right)  =0$.
Besides, $f\in\mathfrak{g}\left(  H,A\right)  =\mathcal{L}^{1}\left(
H,A\right)  $ yields $f\mid_{H_{\leq1-1}}=0$ (by the definition of
$\mathcal{L}^{1}\left(  H,A\right)  $), so that $f\left(  H_{\leq1-1}\right)
=0$. Thus, $f\left(  H_{\leq n-n}\right)  =f\left(  H_{\leq0}\right)
=f\left(  H_{\leq1-1}\right)  =0$. Now, (\ref{pf.9.3}) becomes%
\begin{align*}
f^{\ast\left(  j+1\right)  }\left(  H_{\leq n}\right)   &  \subseteq\sum
_{u=0}^{n-1}\mu_{A}\left(  \underbrace{f^{\ast j}\left(  H_{\leq u}\right)
}_{=0}\otimes f\left(  H_{\leq n-u}\right)  \right)  +\mu_{A}\left(  f^{\ast
j}\left(  H_{\leq n}\right)  \otimes\underbrace{f\left(  H_{\leq n-n}\right)
}_{=0}\right) \\
&  =\underbrace{\sum_{u=0}^{n-1}\mu_{A}\left(  0\otimes f\left(  H_{\leq
n-u}\right)  \right)  }_{=0}+\underbrace{\mu_{A}\left(  f^{\ast j}\left(
H_{\leq n}\right)  \otimes0\right)  }_{=0}=0+0=0.
\end{align*}
In other words, $f^{\ast\left(  j+1\right)  }\left(  H_{\leq n}\right)  =0$.
We have thus proven that Remark~\ref{Remark9} also holds for $i=j+1$. This
completes the induction step.

We thus have completed the induction proof of Remark~\ref{Remark9}.
\end{proof}

\begin{definition}
\label{Definition10} Let $k$ be a field of characteristic $0$, let $A$ be a
$k$-algebra, and let $H$ be a connected filtered $k$-coalgebra. For every
$f\in\mathfrak{g}\left(  H,A\right)  $, let us define a map $e^{\ast
f}:H\rightarrow A$ by the formula%
\begin{equation}
\left(  e^{\ast f}\left(  x\right)  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(
x\right)  }{i!}\ \ \ \ \ \ \ \ \ \ \text{for every }x\in H\right)  .
\label{exp}%
\end{equation}
This map $e^{\ast f}$ is well-defined, because for every $x\in H$ the infinite
sum $\sum_{i\geq0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}$ converges with
respect to the discrete topology\footnote{\textit{Proof.} Let $x\in H$. Then,
there exists some $n\in\mathbb{N}$ such that $x\in H_{\leq n}$ (since $H$ is
filtered). Consider such an $n$. Then, every integer $i>n$ satisfies $f^{\ast
i}\left(  x\right)  \in f^{\ast i}\left(  H_{\leq n}\right)  =0$ (by
Remark~\ref{Remark9}) and thus $f^{\ast i}\left(  x\right)  =0$. Hence, for
every integer $i>n$, the $i$-th addend of the infinite sum $\sum_{i\geq
0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}$ is zero. Hence, this infinite sum
$\sum_{i\geq0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}$ has only finitely many
nonzero addends. Thus, this sum converges with respect to the discrete
topology.}. Besides, $e^{\ast f}$ is a $k$-linear map\footnote{\textit{Proof.}
Let $\alpha\in k$, $\beta\in k$, $x\in H$ and $y\in H$ be arbitrary. Then,
(\ref{exp}) (applied to $y$ instead of $x$) yields $e^{\ast f}\left(
y\right)  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(  y\right)  }{i!}$. But
(\ref{exp}) (applied to $\alpha x+\beta y$ instead of $x$) yields%
\begin{align*}
e^{\ast f}\left(  \alpha x+\beta y\right)   &  =\sum_{i\geq0}\dfrac{f^{\ast
i}\left(  \alpha x+\beta y\right)  }{i!}=\sum_{i\geq0}\underbrace{\dfrac
{\alpha f^{\ast i}\left(  x\right)  +\beta f^{\ast i}\left(  y\right)  }{i!}%
}_{=\alpha\dfrac{f^{\ast i}\left(  x\right)  }{i!}+\beta\dfrac{f^{\ast
i}\left(  y\right)  }{i!}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since for every }i\in\mathbb{N}\text{, we have }f^{\ast i}\left(  \alpha
x+\beta y\right)  =\alpha f^{\ast i}\left(  x\right)  +\beta f^{\ast i}\left(
y\right) \\
\text{(because }f^{\ast i}\text{ is a }k\text{-linear map)}%
\end{array}
\right) \\
&  =\alpha\underbrace{\sum_{i\geq0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}%
}_{=e^{\ast f}\left(  x\right)  }+\beta\underbrace{\sum_{i\geq0}\dfrac{f^{\ast
i}\left(  y\right)  }{i!}}_{=e^{\ast f}\left(  y\right)  }=\alpha e^{\ast
f}\left(  x\right)  +\beta e^{\ast f}\left(  y\right)  .
\end{align*}
Since this holds for all $\alpha\in k$, $\beta\in k$, $x\in H$ and $y\in H$,
we thus see that $e^{\ast f}$ is $k$-linear, qed.}, so that $e^{\ast f}%
\in\mathcal{L}\left(  H,A\right)  $. More precisely, $e^{\ast f}\in G\left(
H,A\right)  $.\ \ \ \ \footnote{\textit{Proof.} Every integer $i>0$ satisfies
$f^{\ast i}\left(  H_{\leq0}\right)  =0$ (by Remark~\ref{Remark9}, applied to
$n=0$) and thus $f^{\ast i}\left(  \underbrace{1_{H}}_{\in H_{\leq0}}\right)
\in f^{\ast i}\left(  H_{\leq0}\right)  =0$, so that $f^{\ast i}\left(
1_{H}\right)  =0$. Hence, every integer $i>0$ satisfies%
\begin{equation}
\dfrac{f^{\ast i}\left(  1_{H}\right)  }{i!}=\dfrac{0}{i!}=0.
\label{pf.Definition10.fn3.1}%
\end{equation}
\par
But applying (\ref{exp}) to $x=1_{H}$, we get%
\begin{align*}
e^{\ast f}\left(  1_{H}\right)   &  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(
1_{H}\right)  }{i!}=\underbrace{\dfrac{f^{\ast0}\left(  1_{H}\right)  }{0!}%
}_{=\dfrac{f^{\ast0}\left(  1_{H}\right)  }{1}=f^{\ast0}\left(  1_{H}\right)
}+\sum_{i>0}\underbrace{\dfrac{f^{\ast i}\left(  1_{H}\right)  }{i!}%
}_{\substack{=0\\\text{(by (\ref{pf.Definition10.fn3.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=0\text{ from the sum}\right) \\
&  =f^{\ast0}\left(  1_{H}\right)  +\underbrace{\sum_{i>0}0}_{=0}%
=\underbrace{f^{\ast0}}_{=e_{H,A}}\left(  1_{H}\right)  =e_{H,A}\left(
1_{H}\right)  =1_{A}.
\end{align*}
Thus, $e^{\ast f}\in G\left(  H,A\right)  $ (by the definition of $G\left(
H,A\right)  $).}
\end{definition}

\textit{Remark.} The $e$ in the notation $e^{\ast f}$ has nothing to do with
the $e$ in the notation $e_{H,A}$. The $e$ in the notation $e^{\ast f}$ is a
pure symbol (in particular, $e^{\ast f}$ is not an ``$f$-th power'' of any $e$
(whatever this $e$ would be) with respect to convolution) which has been
chosen to suggest similarity with the exponential function known from
analysis; despite this similarity, there is (in general) no ``Euler number''
$e\approx2.718\ldots$ in $\mathcal{L}\left(  H,A\right)  $. The $e$ in the
notation $e_{H,A}$ simply stands for ``neutral element'' (just as the neutral
element of a group is often denoted by $e$).

\begin{definition}
\label{Definition11} Let $k$ be a field of characteristic $0$, let $A$ be a
$k$-algebra, and let $H$ be a connected filtered $k$-coalgebra. For every
$f\in\mathfrak{g}\left(  H,A\right)  $, let us define a map
$\operatorname{Log}_{1}f:H\rightarrow A$ by the formula%
\begin{equation}
\left(  \left(  \operatorname{Log}_{1}f\right)  \left(  x\right)  =\sum
_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in H\right)  . \label{log}%
\end{equation}
This map $\operatorname{Log}_{1}f$ is well-defined, because for every $x\in H$
the infinite sum $\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  $ converges with respect to the discrete
topology\footnote{\textit{Proof.} Let $x\in H$. Then, there exists some
$n\in\mathbb{N}$ such that $x\in H_{\leq n}$ (since $H$ is filtered). Consider
this $n$. Then, every integer $i>n$ satisfies $f^{\ast i}\left(  x\right)  =0$
(this is proven just as in Definition~\ref{Definition10}). Therefore, every
integer $i>n$ satisfies $\dfrac{\left(  -1\right)  ^{i-1}}{i}%
\underbrace{f^{\ast i}\left(  x\right)  }_{=0}=0$. In other words, for every
integer $i>n$, the $i$-th addend of the infinite sum $\sum_{i\geq1}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)  $ is zero.
Hence, this infinite sum $\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}f^{\ast i}\left(  x\right)  $ has only finitely many nonzero addends. Thus,
this sum converges with respect to the discrete topology, qed.}. Besides,
$\operatorname{Log}_{1}f$ is a $k$-linear map\footnote{\textit{Proof.} Let
$\alpha\in k$, $\beta\in k$, $x\in H$ and $y\in H$ be arbitrary. Then,
(\ref{log}) (applied to $y$ instead of $x$) yields $\left(  \operatorname{Log}%
_{1}f\right)  \left(  y\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}f^{\ast i}\left(  y\right)  $. But (\ref{log}) (applied to $\alpha
x+\beta y$ instead of $x$) yields%
\begin{align*}
\left(  \operatorname{Log}_{1}f\right)  \left(  \alpha x+\beta y\right)   &
=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{f^{\ast
i}\left(  \alpha x+\beta y\right)  }_{\substack{=\alpha f^{\ast i}\left(
x\right)  +\beta f^{\ast i}\left(  y\right)  \\\text{(since }f^{\ast i}\text{
is a }k\text{-linear map)}}}=\sum_{i\geq1}\underbrace{\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  \alpha f^{\ast i}\left(  x\right)  +\beta f^{\ast
i}\left(  y\right)  \right)  }_{=\alpha\dfrac{\left(  -1\right)  ^{i-1}}%
{i}f^{\ast i}\left(  x\right)  +\beta\dfrac{\left(  -1\right)  ^{i-1}}%
{i}f^{\ast i}\left(  y\right)  }\\
&  =\alpha\underbrace{\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  }_{=\left(  \operatorname{Log}_{1}f\right)  \left(
x\right)  }+\beta\underbrace{\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}f^{\ast i}\left(  y\right)  }_{=\left(  \operatorname{Log}_{1}f\right)
\left(  y\right)  }=\alpha\left(  \operatorname{Log}_{1}f\right)  \left(
x\right)  +\beta\left(  \operatorname{Log}_{1}f\right)  \left(  y\right)  .
\end{align*}
Since this holds for all $\alpha\in k$, $\beta\in k$, $x\in H$ and $y\in H$,
we thus see that $\operatorname{Log}_{1}f$ is $k$-linear, qed.}, so that
$\operatorname{Log}_{1}f\in\mathcal{L}\left(  H,A\right)  $. More precisely,
$\operatorname{Log}_{1}f\in\mathfrak{g}\left(  H,A\right)  $%
.\ \ \ \ \footnote{\textit{Proof.} Every integer $i>0$ satisfies $f^{\ast
i}\left(  H_{\leq0}\right)  =0$ (by Remark~\ref{Remark9}, applied to $n=0$)
and thus $f^{\ast i}\left(  \underbrace{1_{H}}_{\in H_{\leq0}}\right)  \in
f^{\ast i}\left(  H_{\leq0}\right)  =0$, so that $f^{\ast i}\left(
1_{H}\right)  =0$. But applying (\ref{log}) to $x=1_{H}$, we get%
\[
\left(  \operatorname{Log}_{1}f\right)  \left(  1_{H}\right)  =\sum_{i\geq
1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{f^{\ast i}\left(
1_{H}\right)  }_{=0}=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}0=0.
\]
Thus, $\operatorname{Log}_{1}f\in\mathfrak{g}\left(  H,A\right)  $ (by the
definition of $\mathfrak{g}\left(  H,A\right)  $).}
\end{definition}

\begin{definition}
\label{Definition12} Let $k$ be a field of characteristic $0$, let $A$ be a
$k$-algebra, and let $H$ be a connected filtered $k$-coalgebra. For every
$F\in G\left(  H,A\right)  $, let us define an element $\operatorname*{Log}%
F\in\mathfrak{g}\left(  H,A\right)  $ by $\operatorname*{Log}%
F=\operatorname{Log}_{1}\left(  F-e_{H,A}\right)  $.\ \ \ \ \footnote{This is
well-defined since%
\[
\underbrace{F}_{\in G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(
H,A\right)  }-e_{H,A}\in e_{H,A}+\mathfrak{g}\left(  H,A\right)
-e_{H,A}=\mathfrak{g}\left(  H,A\right)  .
\]
}
\end{definition}

\section{\label{S1}$\operatorname*{Log}\operatorname*{id}$ in a connected
filtered cocommutative bialgebra}

We are now ready to state a first interesting result:

\begin{theorem}
\label{Theorem1} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected filtered cocommutative bialgebra over $k$. Consider the convolution
algebra $\mathcal{L}\left(  H,H\right)  $. The map $\operatorname*{Log}%
\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $ is a projection from $H$
to the subspace $\operatorname*{Prim}H$ of all primitive elements of $H$.
\end{theorem}

The map $\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(
H,H\right)  $ defined in Theorem~\ref{Theorem1} is known as the
\textit{Eulerian idempotent} of $H$.

Theorem~\ref{Theorem1} is a classical fact about the Eulerian idempotent in a
bialgebra. In particular, it has been used in \cite{2} (more precisely, in the
Example in \S 2 of \cite{2}).\footnote{Actually, our Theorem~\ref{Theorem1} is
stronger than the fact used in the Example in \S 2 of \cite{2}, because
\cite{2} considers only connected \textit{graded} cocommutative bialgebras,
whereas our Theorem~\ref{Theorem1} is stated (and proven) for any connected
\textit{filtered} cocommutative bialgebra.
\par
Note that the $\mathcal{L}\left(  H,H\right)  $ in our Theorem~\ref{Theorem1}
is \textit{not} the $\mathcal{L}\left(  H\right)  $ of \cite{2} in the case
when $H$ is graded. In fact, the $\mathcal{L}\left(  H\right)  $ of \cite{2}
contains only the graded $k$-linear maps, while our $\mathcal{L}\left(
H,H\right)  $ contains all $k$-linear maps $H\rightarrow H$.}

\section{\label{S2}Basic properties of $\operatorname*{Log}$ and $e^{\ast}$}

Before we start proving Theorem~\ref{Theorem1}, we shall study the concepts of
exponentiation and logarithm closer -- first, as formal power series, but then
as operators on the space $\mathcal{L}\left(  H,A\right)  $ of linear maps
from a coalgebra $H$ to an algebra $A$. (To be precise, they do not act on the
full space $\mathcal{L}\left(  H,A\right)  $; but we will make everything
precise when we state the results.)

\subsection{$\log$ and $\exp$ as power series}

Let us first study the logarithm and the exponentials as they act on formal
power series:

\begin{definition}
\label{def.exp-log.powser}Let $k$ be a field of characteristic $0$. For every
power series $P\in k\left[  \left[  X\right]  \right]  $ whose coefficient
before $X^{0}$ is $0$, let $\exp P$ denote the power series in $k\left[
\left[  X\right]  \right]  $ defined by $\exp P=\sum_{i\geq0}\dfrac{P^{i}}%
{i!}$. For every power series $Q\in k\left[  \left[  X\right]  \right]  $
whose coefficient before $X^{0}$ is $1$, let $\log Q$ denote the power series
in $k\left[  \left[  X\right]  \right]  $ defined by $\log Q=\sum_{i\geq
1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  Q-1\right)  ^{i}$.
\end{definition}

\begin{theorem}
\label{Theorem14} Let $k$ be a field of characteristic $0$.

\textbf{(a)} Every power series $P\in k\left[  \left[  X\right]  \right]  $
whose coefficient before $X^{0}$ is $0$ satisfies $\log\left(  \exp P\right)
=P$.

\textbf{(b)} Every power series $Q\in k\left[  \left[  X\right]  \right]  $
whose coefficient before $X^{0}$ is $1$ satisfies $\exp\left(  \log Q\right)
=Q$.
\end{theorem}

Theorem~\ref{Theorem14} is an important result in mathematics; in particular,
it is frequently applied in combinatorics (e.g., in dealing with generating
functions) and in algebra. Often it is derived from the analogous fact from
complex analysis (where $P$ and $Q$, rather than being formal power series,
are required to be holomorphic functions). Let us instead give an elementary
proof. The proof will be based on some basic properties of derivatives of
power series:

\begin{lemma}
\label{lem.deriv-uniq-det0}Let $k$ be a field of characteristic $0$. Let $P\in
k\left[  \left[  X\right]  \right]  $ be a power series such that $\dfrac
{d}{dX}P=0$. Assume furthermore that the coefficient of $P$ before $X^{0}$ is
$0$. Then, $P=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.deriv-uniq-det0}.]Write the power series $P\in
k\left[  \left[  X\right]  \right]  $ in the form $P=\sum_{n\in\mathbb{N}%
}p_{n}X^{n}$ for some sequence $\left(  p_{0},p_{1},p_{2},\ldots\right)  \in
k^{\mathbb{N}}$ of elements of $k$. Then,
\begin{align*}
\dfrac{d}{dX}\underbrace{P}_{=\sum_{n\in\mathbb{N}}p_{n}X^{n}}  &  =\dfrac
{d}{dX}\left(  \sum_{n\in\mathbb{N}}p_{n}X^{n}\right) \\
&  =\sum_{\substack{n\in\mathbb{N};\\n\geq1}}np_{n}X^{n-1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the operator }\dfrac
{d}{dX}\right) \\
&  =\sum_{n\in\mathbb{N}}\left(  n+1\right)  p_{n+1}\underbrace{X^{\left(
n+1\right)  -1}}_{\substack{=X^{n}\\\text{(since }\left(  n+1\right)
-1=n\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n+1\text{ for
}n\text{ in the sum}\right) \\
&  =\sum_{n\in\mathbb{N}}\left(  n+1\right)  p_{n+1}X^{n}.
\end{align*}
Thus, $\sum_{n\in\mathbb{N}}\left(  n+1\right)  p_{n+1}X^{n}=\dfrac{d}{dX}%
P=0$. Comparing coefficients on both sides of this equality, we conclude that%
\begin{equation}
\left(  n+1\right)  p_{n+1}=0\ \ \ \ \ \ \ \ \ \ \text{for each }%
n\in\mathbb{N}. \label{pf.lem.deriv-uniq-det0.1}%
\end{equation}


On the other hand, the coefficient of $P$ before $X^{0}$ is $0$. Thus,%
\begin{align*}
0  &  =\left(  \text{the coefficient of }\underbrace{P}_{=\sum_{n\in
\mathbb{N}}p_{n}X^{n}}\text{ before }X^{0}\right) \\
&  =\left(  \text{the coefficient of }\sum_{n\in\mathbb{N}}p_{n}X^{n}\text{
before }X^{0}\right)  =p_{0}.
\end{align*}
In other words, $p_{0}=0$.

Now, we have%
\begin{equation}
p_{n}=0\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}
\label{pf.lem.deriv-uniq-det0.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.deriv-uniq-det0.4}):} Let
$n\in\mathbb{N}$. We must prove (\ref{pf.lem.deriv-uniq-det0.4}).
\par
If $n=0$, then clearly $p_{n}=p_{0}=0$. Hence, (\ref{pf.lem.deriv-uniq-det0.4}%
) is proven in the case when $n=0$. Thus, for the rest of the proof of
(\ref{pf.lem.deriv-uniq-det0.4}), we can WLOG assume that we don't have $n=0$.
Assume this.
\par
We have $n\neq0$ (since we don't have $n=0$). Combined with $n\in\mathbb{N}$,
this yields $n\in\mathbb{N}\setminus\left\{  0\right\}  =\left\{
1,2,3,\ldots\right\}  $. Thus, $n-1\in\mathbb{N}$. Hence,
(\ref{pf.lem.deriv-uniq-det0.1}) (applied to $n-1$ instead of $n$) yields
$\left(  \left(  n-1\right)  +1\right)  p_{\left(  n-1\right)  +1}=0$. Since
$\left(  n-1\right)  +1=n$, this rewrites as $np_{n}=0$. But $n\in\left\{
1,2,3,\ldots\right\}  $; thus, $n$ is invertible in $k$ (since $k$ is a field
of characteristic $0$). Hence, we can cancel $n$ from the equality $np_{n}=0$.
We thus obtain $p_{n}=0$. This proves (\ref{pf.lem.deriv-uniq-det0.4}).}.
Thus, $P=\sum_{n\in\mathbb{N}}\underbrace{p_{n}}_{\substack{=0\\\text{(by
(\ref{pf.lem.deriv-uniq-det0.4}))}}}X^{n}=\sum_{n\in\mathbb{N}}0X^{n}=0$. This
proves Lemma~\ref{lem.deriv-uniq-det0}.
\end{proof}

\begin{proposition}
\label{prop.deriv-uniq-det}Let $k$ be a field of characteristic $0$. Let $U\in
k\left[  \left[  X\right]  \right]  $ and $V\in k\left[  \left[  X\right]
\right]  $ be two power series such that $\dfrac{d}{dX}U=\dfrac{d}{dX}V$.
Assume furthermore that the coefficient of $U$ before $X^{0}$ equals the
coefficient of $V$ before $X^{0}$. Then, $U=V$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{prop.deriv-uniq-det}.]We have
\begin{align*}
&  \left(  \text{the coefficient of }U-V\text{ before }X^{0}\right) \\
&  =\underbrace{\left(  \text{the coefficient of }U\text{ before }%
X^{0}\right)  }_{\substack{=\left(  \text{the coefficient of }V\text{ before
}X^{0}\right)  \\\text{(since the coefficient of }U\text{ before }%
X^{0}\\\text{equals the coefficient of }V\text{ before }X^{0}\text{)}%
}}-\left(  \text{the coefficient of }V\text{ before }X^{0}\right) \\
&  =\left(  \text{the coefficient of }V\text{ before }X^{0}\right)  -\left(
\text{the coefficient of }V\text{ before }X^{0}\right)  =0.
\end{align*}
In other words, the coefficient of $V$ before $X^{0}$ is $0$. Moreover,
$\dfrac{d}{dX}\left(  U-V\right)  =\dfrac{d}{dX}U-\dfrac{d}{dX}V=0$ (since
$\dfrac{d}{dX}U=\dfrac{d}{dX}V$). Hence, Lemma~\ref{lem.deriv-uniq-det0}
(applied to $P=U-V$) yields $U-V=0$. In other words, $U=V$. This proves
Proposition~\ref{prop.deriv-uniq-det}.
\end{proof}

\begin{proposition}
\label{prop.deriv-lei}Let $k$ be a field. Any two power series $U\in k\left[
\left[  X\right]  \right]  $ and $V\in k\left[  \left[  X\right]  \right]  $
satisfy%
\begin{equation}
\dfrac{d}{dX}\left(  UV\right)  =\left(  \dfrac{d}{dX}U\right)  \cdot
V+U\cdot\dfrac{d}{dX}V. \label{pf.prop.deriv-inverse.lei}%
\end{equation}

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{prop.deriv-lei}.]The equality
(\ref{pf.prop.deriv-inverse.lei}) is just the well-known Leibniz identity for
the derivation $\dfrac{d}{dX}$.
\end{proof}

\begin{proposition}
\label{prop.deriv-inverse}Let $k$ be a field. Let $P\in k\left[  \left[
X\right]  \right]  $ be a power series such that the coefficient of $P$ before
$X^{0}$ is $1$. Then, $\dfrac{d}{dX}\left(  P^{-1}\right)  =-\dfrac{1}{P^{2}%
}\cdot\dfrac{d}{dX}P$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{prop.deriv-inverse}.]The multiplicative inverse
$P^{-1}$ of the power series $P$ is well-defined, since the coefficient of $P$
before $X^{0}$ is $1$.

Applying Proposition~\ref{prop.deriv-lei} to $U=P^{-1}$ and $V=P$, we obtain%
\[
\dfrac{d}{dX}\left(  P^{-1}P\right)  =\left(  \dfrac{d}{dX}\left(
P^{-1}\right)  \right)  \cdot P+P^{-1}\cdot\dfrac{d}{dX}P.
\]
Comparing this with $\dfrac{d}{dX}\left(  \underbrace{P^{-1}P}_{=1}\right)
=\dfrac{d}{dX}1=0$, we obtain%
\[
\left(  \dfrac{d}{dX}\left(  P^{-1}\right)  \right)  \cdot P+P^{-1}\cdot
\dfrac{d}{dX}P=0.
\]
Hence, $\left(  \dfrac{d}{dX}P^{-1}\right)  \cdot P=-P^{-1}\cdot\dfrac{d}%
{dX}P$. Multiplying both sides of this equality by $P^{-1}$, we obtain%
\[
\left(  \dfrac{d}{dX}\left(  P^{-1}\right)  \right)  \cdot PP^{-1}%
=-P^{-1}\cdot\left(  \dfrac{d}{dX}P\right)  \cdot P^{-1}=-\dfrac{1}{P^{2}%
}\cdot\dfrac{d}{dX}P.
\]
Thus,%
\[
-\dfrac{1}{P^{2}}\cdot\dfrac{d}{dX}P=\left(  \dfrac{d}{dX}\left(
P^{-1}\right)  \right)  \cdot\underbrace{PP^{-1}}_{=1}=\dfrac{d}{dX}\left(
P^{-1}\right)  .
\]
This proves Proposition~\ref{prop.deriv-inverse}.
\end{proof}

\begin{proposition}
\label{prop.lderiv-uniq-det}Let $k$ be a field of characteristic $0$. Let
$U\in k\left[  \left[  X\right]  \right]  $ and $V\in k\left[  \left[
X\right]  \right]  $ be two power series satisfying $V\dfrac{d}{dX}%
U=U\dfrac{d}{dX}V$. Assume that the coefficient of $U$ before $X^{0}$ is $1$.
Assume that the coefficient of $V$ before $X^{0}$ is $1$. Then, $U=V$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{prop.lderiv-uniq-det}.]The multiplicative inverse
$V^{-1}$ of the power series $V$ is well-defined, since the coefficient of $V$
before $X^{0}$ is $1$.

Let $w_{0}$ be the coefficient of the power series $UV^{-1}$ before $X^{0}$.
Then, we can regard $w_{0}$ itself as a constant power series. Of course, the
coefficient of this constant power series $w_{0}$ before $X^{0}$ is $w_{0}$
itself. Thus, the coefficient of $UV^{-1}$ before $X^{0}$ equals the
coefficient of $w_{0}$ before $X^{0}$ (since both of these coefficients equal
$w_{0}$). We have $\dfrac{d}{dX}w_{0}=0$ (since $w_{0}$ is a constant power series).

Proposition~\ref{prop.deriv-lei} (applied to $V^{-1}$ instead of $V$) yields%
\begin{align*}
\dfrac{d}{dX}\left(  UV^{-1}\right)   &  =\left(  \dfrac{d}{dX}U\right)  \cdot
V^{-1}+U\cdot\underbrace{\dfrac{d}{dX}\left(  V^{-1}\right)  }%
_{\substack{=\dfrac{-1}{V^{2}}\cdot\dfrac{d}{dX}V\\\text{(by
Proposition~\ref{prop.deriv-inverse}}\\\text{(applied to }P=V\text{))}}}\\
&  =\left(  \dfrac{d}{dX}U\right)  \cdot V^{-1}+U\cdot\dfrac{-1}{V^{2}}%
\cdot\dfrac{d}{dX}V=\dfrac{1}{V}\left(  \dfrac{d}{dX}U\right)  -\dfrac
{U}{V^{2}}\cdot\dfrac{d}{dX}V\\
&  =\dfrac{1}{V^{2}}\underbrace{\left(  V\left(  \dfrac{d}{dX}U\right)
-U\left(  \dfrac{d}{dX}V\right)  \right)  }_{\substack{=0\\\text{(since
}V\dfrac{d}{dX}U=U\dfrac{d}{dX}V\text{)}}}=\dfrac{1}{V^{2}}0=0\\
&  =\dfrac{d}{dX}w_{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\dfrac{d}%
{dX}w_{0}=0\right)  .
\end{align*}
Hence, Proposition~\ref{prop.deriv-uniq-det} (applied to $UV^{-1}$ and $w_{0}$
instead of $U$ and $V$) yields $UV^{-1}=w_{0}$. Thus, $U=w_{0}V$. Hence,%
\begin{align*}
&  \left(  \text{the coefficient of }\underbrace{U}_{=w_{0}V}\text{ before
}X^{0}\right) \\
&  =\left(  \text{the coefficient of }w_{0}V\text{ before }X^{0}\right)
=w_{0}\underbrace{\left(  \text{the coefficient of }V\text{ before }%
X^{0}\right)  }_{\substack{=1\\\text{(since the coefficient of }V\text{ before
}X^{0}\text{ is }1\text{)}}}=w_{0},
\end{align*}
so that%
\[
w_{0}=\left(  \text{the coefficient of }U\text{ before }X^{0}\right)  =1
\]
(since the coefficient of $U$ before $X^{0}$ is $1$). Now,
$U=\underbrace{w_{0}}_{=1}V=V$. This proves
Proposition~\ref{prop.lderiv-uniq-det}.
\end{proof}

\begin{proposition}
\label{prop.deriv-leibnizPn}Let $k$ be a field. Let $P\in k\left[  \left[
X\right]  \right]  $ be a power series. Then, every positive integer $n$
satisfies%
\begin{equation}
\dfrac{d}{dX}\left(  P^{n}\right)  =nP^{n-1}\dfrac{d}{dX}P.
\label{eq.prop.deriv-leibnizPn.1}%
\end{equation}

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{prop.deriv-leibnizPn}.]We shall prove
(\ref{eq.prop.deriv-leibnizPn.1}) by induction over $n$:

\textit{Induction base:} We have $1\underbrace{P^{1-1}}_{=P^{0}=1}\dfrac
{d}{dX}P=1\dfrac{d}{dX}P=\dfrac{d}{dX}P$. Comparing this with $\dfrac{d}%
{dX}\left(  \underbrace{P^{1}}_{=P}\right)  =\dfrac{d}{dX}P$, we conclude
$\dfrac{d}{dX}\left(  P^{1}\right)  =1P^{1-1}\dfrac{d}{dX}P$. In other words,
(\ref{eq.prop.deriv-leibnizPn.1}) holds for $n=1$. This completes the
induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that
(\ref{eq.prop.deriv-leibnizPn.1}) holds for $n=N$. We must now show that
(\ref{eq.prop.deriv-leibnizPn.1}) holds for $n=N+1$.

We have assumed that (\ref{eq.prop.deriv-leibnizPn.1}) holds for $n=N$. In
other words, we have%
\[
\dfrac{d}{dX}\left(  P^{N}\right)  =NP^{N-1}\dfrac{d}{dX}P.
\]
Now,
\begin{align*}
\dfrac{d}{dX}\left(  \underbrace{P^{N+1}}_{=PP^{N}}\right)   &  =\dfrac{d}%
{dX}\left(  PP^{N}\right)  =\underbrace{\left(  \dfrac{d}{dX}P\right)  \cdot
P^{N}}_{=P^{N}\dfrac{d}{dX}P}+P\cdot\underbrace{\dfrac{d}{dX}\left(
P^{N}\right)  }_{=NP^{N-1}\dfrac{d}{dX}P}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition~\ref{prop.deriv-lei}
(applied to }U=P\text{ and }V=P^{N}\text{)}\right) \\
&  =P^{N}\dfrac{d}{dX}P+\underbrace{P\cdot NP^{N-1}}_{=NP^{N}}\dfrac{d}%
{dX}P=P^{N}\dfrac{d}{dX}P+NP^{N}\dfrac{d}{dX}P\\
&  =\underbrace{\left(  1+N\right)  }_{=N+1}\underbrace{P^{N}}%
_{\substack{=P^{\left(  N+1\right)  -1}\\\text{(since }N=\left(  N+1\right)
-1\text{)}}}\dfrac{d}{dX}P=\left(  N+1\right)  P^{\left(  N+1\right)
-1}\dfrac{d}{dX}P.
\end{align*}
In other words, (\ref{eq.prop.deriv-leibnizPn.1}) holds for $n=N+1$. This
completes the induction step. Thus, the induction proof of
(\ref{eq.prop.deriv-leibnizPn.1}) is complete.

This proves Proposition~\ref{prop.deriv-leibnizPn}.
\end{proof}

\begin{proposition}
\label{prop.deriv-exp}Let $k$ be a field of characteristic $0$. Let $P\in
k\left[  \left[  X\right]  \right]  $ be a power series whose coefficient
before $X^{0}$ is $0$. Then,%
\[
\dfrac{d}{dX}\left(  \exp P\right)  =\left(  \dfrac{d}{dX}P\right)  \cdot\exp
P.
\]

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{prop.deriv-exp}.]Every positive integer $n$
satisfies%
\begin{equation}
\dfrac{1}{n!}n=\dfrac{1}{\left(  n-1\right)  !} \label{pf.prop.deriv-exp.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.deriv-exp.1}):} Let $n$ be a positive
integer. The recursive definition of $n!$ yields $n!=n\cdot\left(  n-1\right)
!$. Hence, $\dfrac{1}{n!}n=\dfrac{1}{n\cdot\left(  n-1\right)  !}n=\dfrac
{1}{\left(  n-1\right)  !}$. Qed.}.

The definition of $\exp P$ yields $\exp P=\sum_{i\geq0}\dfrac{P^{i}}{i!}%
=\sum_{i\geq0}\dfrac{1}{i!}P^{i}=\sum_{n\geq0}\dfrac{1}{n!}P^{n}$ (here, we
have renamed the summation index $i$ as $n$). Hence,%
\begin{align*}
\dfrac{d}{dX}\underbrace{\left(  \exp P\right)  }_{=\sum_{n\geq0}\dfrac{1}%
{n!}P^{n}}  &  =\dfrac{d}{dX}\left(  \sum_{n\geq0}\dfrac{1}{n!}P^{n}\right)
=\sum_{n\geq0}\dfrac{1}{n!}\dfrac{d}{dX}\left(  P^{n}\right) \\
&  =\dfrac{1}{0!}\dfrac{d}{dX}\left(  \underbrace{P^{0}}_{=1}\right)
+\sum_{n\geq1}\dfrac{1}{n!}\underbrace{\dfrac{d}{dX}\left(  P^{n}\right)
}_{\substack{=nP^{n-1}\dfrac{d}{dX}P\\\text{(by
Proposition~\ref{prop.deriv-leibnizPn})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=0\text{ from the sum}\right) \\
&  =\dfrac{1}{0!}\underbrace{\dfrac{d}{dX}1}_{=0}+\sum_{n\geq1}%
\underbrace{\dfrac{1}{n!}n}_{\substack{=\dfrac{1}{\left(  n-1\right)
!}\\\text{(by (\ref{pf.prop.deriv-exp.1}))}}}P^{n-1}\dfrac{d}{dX}P\\
&  =\underbrace{\dfrac{1}{0!}0}_{=0}+\sum_{n\geq1}\dfrac{1}{\left(
n-1\right)  !}P^{n-1}\dfrac{d}{dX}P=\sum_{n\geq1}\dfrac{1}{\left(  n-1\right)
!}P^{n-1}\dfrac{d}{dX}P\\
&  =\sum_{n\geq0}\dfrac{1}{n!}P^{n}\dfrac{d}{dX}P\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }n\text{ for }n-1\\
\text{in the sum}%
\end{array}
\right)  .
\end{align*}
Comparing this with%
\[
\left(  \dfrac{d}{dX}P\right)  \cdot\underbrace{\exp P}_{=\sum_{n\geq0}%
\dfrac{1}{n!}P^{n}}=\left(  \dfrac{d}{dX}P\right)  \cdot\sum_{n\geq0}\dfrac
{1}{n!}P^{n}=\sum_{n\geq0}\underbrace{\left(  \dfrac{d}{dX}P\right)
\cdot\dfrac{1}{n!}P^{n}}_{=\dfrac{1}{n!}P^{n}\dfrac{d}{dX}P}=\sum_{n\geq
0}\dfrac{1}{n!}P^{n}\dfrac{d}{dX}P,
\]
we obtain $\dfrac{d}{dX}\left(  \exp P\right)  =\left(  \dfrac{d}{dX}P\right)
\cdot\exp P$. This proves Proposition~\ref{prop.deriv-exp}.
\end{proof}

\begin{proposition}
\label{prop.deriv-log}Let $k$ be a field. Let $Q\in k\left[  \left[  X\right]
\right]  $ be a power series whose coefficient before $X^{0}$ is $1$. Then:

\textbf{(a)} The power series $Q$ has a multiplicative inverse $Q^{-1}$.

\textbf{(b)} Assume that $k$ is a field of characteristic $0$. Then, the
multiplicative inverse $Q^{-1}$ of $Q$ satisfies
\[
\dfrac{d}{dX}\left(  \log Q\right)  =Q^{-1}\cdot\dfrac{d}{dX}Q.
\]

\end{proposition}

I call Proposition~\ref{prop.deriv-log} \textbf{(b)} the \textit{logarithmic
derivative theorem}, as it gives the derivative of the logarithm of a power
series (provided the logarithm is well-defined).

\begin{proof}
[Proof of Proposition~\ref{prop.deriv-log}.]Define a power series $R\in
k\left[  \left[  X\right]  \right]  $ by $R=Q-1$. Thus,
\begin{equation}
\dfrac{d}{dX}\underbrace{R}_{=Q-1}=\dfrac{d}{dX}\left(  Q-1\right)  =\dfrac
{d}{dX}Q-\underbrace{\dfrac{d}{dX}1}_{=0}=\dfrac{d}{dX}Q.
\label{pf.prop.deriv-log.0}%
\end{equation}


Also,%
\begin{align*}
&  \left(  \text{the coefficient of }\underbrace{R}_{=Q-1}\text{ before }%
X^{0}\right) \\
&  =\left(  \text{the coefficient of }Q-1\text{ before }X^{0}\right) \\
&  =\underbrace{\left(  \text{the coefficient of }Q\text{ before }%
X^{0}\right)  }_{\substack{=1\\\text{(since the coefficient of }Q\text{ before
}X^{0}\text{ is }1\text{)}}}-\underbrace{\left(  \text{the coefficient of
}1\text{ before }X^{0}\right)  }_{=1}\\
&  =1-1=0.
\end{align*}
In other words, the coefficient of $R$ before $X^{0}$ is $0$. Hence, the sum
$\sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}$ converges in $k\left[  \left[
X\right]  \right]  $. The power series $\sum_{n\geq0}\left(  -1\right)
^{n}R^{n}$ is a multiplicative inverse of $Q$ in the commutative ring
$k\left[  \left[  X\right]  \right]  $\ \ \ \ \footnote{\textit{Proof.} We
have%
\begin{align*}
\left(  \sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}\right)  R  &  =\sum_{n\geq
0}\left(  -1\right)  ^{n}\underbrace{R^{n}R}_{=R^{n+1}}=\sum_{n\geq0}\left(
-1\right)  ^{n}R^{n+1}=\sum_{n\geq1}\underbrace{\left(  -1\right)  ^{n-1}%
}_{=-\left(  -1\right)  ^{n}}\underbrace{R^{\left(  n-1\right)  +1}}_{=R^{n}%
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n-1\text{ for
}n\text{ in the sum}\right) \\
&  =\sum_{n\geq1}\left(  -\left(  -1\right)  ^{n}\right)  R^{n}=-\sum_{n\geq
1}\left(  -1\right)  ^{n}R^{n}.
\end{align*}
On the other hand,%
\begin{align*}
\sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}  &  =\underbrace{\left(  -1\right)
^{0}}_{=1}\underbrace{R^{0}}_{=1}+\sum_{n\geq1}\left(  -1\right)  ^{n}%
R^{n}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend}\\
\text{for }n=0\text{ from the sum}%
\end{array}
\right) \\
&  =1+\sum_{n\geq1}\left(  -1\right)  ^{n}R^{n}.
\end{align*}
Now,%
\begin{align*}
&  \left(  \sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}\right)  \underbrace{Q}%
_{\substack{=R+1\\\text{(since }R=Q-1\text{)}}}\\
&  =\left(  \sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}\right)  \left(
R+1\right)  =\underbrace{\left(  \sum_{n\geq0}\left(  -1\right)  ^{n}%
R^{n}\right)  R}_{=-\sum_{n\geq1}\left(  -1\right)  ^{n}R^{n}}%
+\underbrace{\sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}}_{=1+\sum_{n\geq
1}\left(  -1\right)  ^{n}R^{n}}\\
&  =-\sum_{n\geq1}\left(  -1\right)  ^{n}R^{n}+\left(  1+\sum_{n\geq1}\left(
-1\right)  ^{n}R^{n}\right)  =1.
\end{align*}
Thus, $\sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}$ is a multiplicative inverse
of $Q$ in the commutative ring $k\left[  \left[  X\right]  \right]  $.}. Thus,
the power series $Q$ has a multiplicative inverse $Q^{-1}$. This proves
Proposition~\ref{prop.deriv-log} \textbf{(a)}.

Notice that the multiplicative inverse $Q^{-1}$ of $Q$ must be equal to
$\sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}$ (since $\sum_{n\geq0}\left(
-1\right)  ^{n}R^{n}$ is a multiplicative inverse of $Q$). In other words,
\begin{equation}
Q^{-1}=\sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}. \label{pf.prop.deriv-log.1}%
\end{equation}


\textbf{(b)} The definition of $\log Q$ yields%
\[
\log Q=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
\underbrace{Q-1}_{=R}\right)  ^{i}=\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}R^{i}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}R^{n}%
\]
(here, we have renamed the summation index $i$ as $n$). Thus,%
\begin{align*}
\dfrac{d}{dX}\underbrace{\left(  \log Q\right)  }_{=\sum_{n\geq1}%
\dfrac{\left(  -1\right)  ^{n-1}}{n}R^{n}}  &  =\dfrac{d}{dX}\left(
\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}R^{n}\right)  =\sum_{n\geq
1}\dfrac{\left(  -1\right)  ^{n-1}}{n}\cdot\underbrace{\dfrac{d}{dX}\left(
R^{n}\right)  }_{\substack{=nR^{n-1}\dfrac{d}{dX}R\\\text{(by
Proposition~\ref{prop.deriv-leibnizPn}}\\\text{(applied to }P=R\text{))}}}\\
&  =\sum_{n\geq1}\underbrace{\dfrac{\left(  -1\right)  ^{n-1}}{n}\cdot
nR^{n-1}}_{=\left(  -1\right)  ^{n-1}R^{n-1}}\dfrac{d}{dX}R\\
&  =\sum_{n\geq1}\left(  -1\right)  ^{n-1}R^{n-1}\dfrac{d}{dX}R=\sum_{n\geq
0}\left(  -1\right)  ^{n}R^{n}\dfrac{d}{dX}R
\end{align*}
(here, we have substituted $n$ for $n-1$ in the sum). Comparing this with%
\[
\underbrace{Q^{-1}}_{\substack{=\sum_{n\geq0}\left(  -1\right)  ^{n}%
R^{n}\\\text{(by (\ref{pf.prop.deriv-log.1}))}}}\cdot\underbrace{\dfrac{d}%
{dX}Q}_{\substack{=\dfrac{d}{dX}R\\\text{(by (\ref{pf.prop.deriv-log.0}))}%
}}=\left(  \sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}\right)  \cdot\dfrac
{d}{dX}R=\sum_{n\geq0}\left(  -1\right)  ^{n}R^{n}\dfrac{d}{dX}R,
\]
we obtain $\dfrac{d}{dX}\left(  \log Q\right)  =Q^{-1}\cdot\dfrac{d}{dX}Q$.
This proves Proposition~\ref{prop.deriv-log} \textbf{(b)}.
\end{proof}

\begin{lemma}
\label{lem.powser.cst-term-prod}Let $k$ be a field. Let $U\in k\left[  \left[
X\right]  \right]  $ and $V\in k\left[  \left[  X\right]  \right]  $ be two
power series such that the coefficient of $U$ before $X^{0}$ is $0$. Then, the
coefficient of $UV$ before $X^{0}$ is $0$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.powser.cst-term-prod}.]We have $\left(  \text{the
coefficient of }U\text{ before }X^{0}\text{ is }0\right)  =0$ (since the
coefficient of $U$ before $X^{0}$ is $0$). Now, the definition of the product
of two power series shows that%
\begin{align*}
&  \left(  \text{the coefficient of }UV\text{ before }X^{n}\right) \\
&  =\sum_{g=0}^{n}\left(  \text{the coefficient of }U\text{ before }%
X^{g}\right)  \cdot\left(  \text{the coefficient of }V\text{ before }%
X^{n-g}\right)
\end{align*}
for each $n\in\mathbb{N}$. Applying this to $n=0$, we find%
\begin{align*}
&  \left(  \text{the coefficient of }UV\text{ before }X^{0}\right) \\
&  =\sum_{g=0}^{0}\left(  \text{the coefficient of }U\text{ before }%
X^{g}\right)  \cdot\left(  \text{the coefficient of }V\text{ before }%
X^{0-g}\right) \\
&  =\underbrace{\left(  \text{the coefficient of }U\text{ before }%
X^{0}\right)  }_{=0}\cdot\left(  \text{the coefficient of }V\text{ before
}X^{0-0}\right) \\
&  =0\cdot\left(  \text{the coefficient of }V\text{ before }X^{0-0}\right)
=0.
\end{align*}
In other words, the coefficient of $UV$ before $X^{0}$ is $0$. This proves
Lemma~\ref{lem.powser.cst-term-prod}.
\end{proof}

\begin{proposition}
\label{Proposition1009} Let $k$ be a field of characteristic $0$.

\textbf{(a)} If $P\in k\left[  \left[  X\right]  \right]  $ is a power series
whose coefficient before $X^{0}$ is $0$, then $\exp P$ is a power series whose
coefficient before $X^{0}$ is $1$.

\textbf{(b)} If $Q\in k\left[  \left[  X\right]  \right]  $ is a power series
whose coefficient before $X^{0}$ is $1$, then $\log Q$ is a power series whose
coefficient before $X^{0}$ is $0$.
\end{proposition}

Note that Proposition~\ref{Proposition1009} shows that the terms $\log\left(
\exp P\right)  $ and $\exp\left(  \log Q\right)  $ in Theorem~\ref{Theorem14}
make any sense at all.

\begin{proof}
[Proof of Proposition~\ref{Proposition1009}.]Let us first prove a simple fact:
If $R\in k\left[  \left[  X\right]  \right]  $ is a power series whose
coefficient before $X^{0}$ is $0$, and if $i$ is a positive integer, then%
\begin{equation}
\left(  \text{the coefficient of }R^{i}\text{ before }X^{0}\right)  =0
\label{pf.Proposition1009.RiX0}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.Proposition1009.RiX0}):} Let $R\in
k\left[  \left[  X\right]  \right]  $ is a power series whose coefficient
before $X^{0}$ is $0$. Let $i$ be a positive integer. Thus, $R^{i-1}$ is
well-defined. But Lemma~\ref{lem.powser.cst-term-prod} (applied to $U=R$ and
$V=R^{i-1}$) yields that the coefficient of $RR^{i-1}$ before $X^{0}$ is $0$.
In other words, $\left(  \text{the coefficient of }RR^{i-1}\text{ before
}X^{0}\right)  =0$. This rewrites as $\left(  \text{the coefficient of }%
R^{i}\text{ before }X^{0}\right)  =0$ (since $RR^{i-1}=R^{i}$). This proves
(\ref{pf.Proposition1009.RiX0}).}.

\textbf{(a)} Let $P\in k\left[  \left[  X\right]  \right]  $ is a power series
whose coefficient before $X^{0}$ is $0$. The definition of $\exp P$ yields
$\exp P=\sum_{i\geq0}\dfrac{P^{i}}{i!}=\sum_{i\geq0}\dfrac{1}{i!}P^{i}$.
Hence,%
\begin{align*}
&  \left(  \text{the coefficient of }\underbrace{\exp P}_{=\sum_{i\geq0}%
\dfrac{1}{i!}P^{i}}\text{ before }X^{0}\right) \\
&  =\left(  \text{the coefficient of }\sum_{i\geq0}\dfrac{1}{i!}P^{i}\text{
before }X^{0}\right) \\
&  =\sum_{i\geq0}\dfrac{1}{i!}\left(  \text{the coefficient of }P^{i}\text{
before }X^{0}\right) \\
&  =\underbrace{\dfrac{1}{0!}}_{=\dfrac{1}{1}=1}\left(  \text{the coefficient
of }\underbrace{P^{0}}_{=1}\text{ before }X^{0}\right)  +\sum_{i\geq1}%
\dfrac{1}{i!}\underbrace{\left(  \text{the coefficient of }P^{i}\text{ before
}X^{0}\right)  }_{\substack{=0\\\text{(by (\ref{pf.Proposition1009.RiX0})
(applied to }R=P\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=0\text{ from the sum}\right) \\
&  =\underbrace{\left(  \text{the coefficient of }1\text{ before }%
X^{0}\right)  }_{=1}+\underbrace{\sum_{i\geq1}\dfrac{1}{i!}0}_{=0}=1+0=1.
\end{align*}
In other words, $\exp P$ is a power series whose coefficient before $X^{0}$ is
$1$. This proves Proposition~\ref{Proposition1009} \textbf{(a)}.

\textbf{(b)} Let $Q\in k\left[  \left[  X\right]  \right]  $ be a power series
whose coefficient before $X^{0}$ is $1$. Define a power series $R\in k\left[
\left[  X\right]  \right]  $ by $R=Q-1$. Then, the coefficient of $R$ before
$X^{0}$ is $0$\ \ \ \ \footnote{This has already been proven in the proof of
Proposition~\ref{prop.deriv-log}.}. Thus, every positive integer $i$ satisfies
(\ref{pf.Proposition1009.RiX0}).

Now, the definition of $\log Q$ yields%
\[
\log Q=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
\underbrace{Q-1}_{=R}\right)  ^{i}=\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}R^{i}.
\]
Thus,%
\begin{align*}
&  \left(  \text{the coefficient of }\underbrace{\log Q}_{=\sum_{i\geq1}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}R^{i}}\text{ before }X^{0}\right) \\
&  =\left(  \text{the coefficient of }\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}R^{i}\text{ before }X^{0}\right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{\left(
\text{the coefficient of }R^{i}\text{ before }X^{0}\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.Proposition1009.RiX0}))}}}\\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}0=0.
\end{align*}
In other words, $\log Q$ is a power series whose coefficient before $X^{0}$ is
$0$. This proves Proposition~\ref{Proposition1009} \textbf{(b)}.
\end{proof}

Now, we can finally come to the proof of Theorem~\ref{Theorem14}:

\begin{proof}
[Proof of Theorem~\ref{Theorem14}.]\textbf{(a)} Let $P\in k\left[  \left[
X\right]  \right]  $ is a power series whose coefficient before $X^{0}$ is
$0$. Then, $\exp P$ is a power series whose coefficient before $X^{0}$ is $1$
(by Proposition~\ref{Proposition1009} \textbf{(a)}). Hence, $\log\left(  \exp
P\right)  $ is a well-defined power series in $k\left[  \left[  X\right]
\right]  $. Furthermore, Proposition~\ref{Proposition1009} \textbf{(b)}
(applied to $Q=\exp P$) shows that $\log\left(  \exp P\right)  $ is a power
series whose coefficient before $X^{0}$ is $0$. Thus, the coefficient of
$\log\left(  \exp P\right)  $ before $X^{0}$ is $0$. On the other hand, the
coefficient of $P$ before $X^{0}$ is $0$ (by the definition of $P$). Thus, the
coefficient of $U$ before $\log\left(  \exp P\right)  $ equals the coefficient
of $P$ before $X^{0}$ (since both of these coefficients are $0$).

Also, Proposition~\ref{prop.deriv-log} \textbf{(b)} (applied to $Q=\exp P$)
yields%
\begin{align*}
\dfrac{d}{dX}\left(  \log\left(  \exp P\right)  \right)   &  =\left(  \exp
P\right)  ^{-1}\cdot\underbrace{\dfrac{d}{dX}\left(  \exp P\right)
}_{\substack{=\left(  \dfrac{d}{dX}P\right)  \cdot\exp P\\\text{(by
Proposition~\ref{prop.deriv-exp})}}}=\left(  \exp P\right)  ^{-1}\cdot\left(
\dfrac{d}{dX}P\right)  \cdot\exp P\\
&  =\dfrac{d}{dX}P.
\end{align*}
Thus, Proposition~\ref{prop.deriv-uniq-det} (applied to $U=\log\left(  \exp
P\right)  $ and $V=P$) shows that $\log\left(  \exp P\right)  =P$. This proves
Theorem~\ref{Theorem14} \textbf{(a)}.

\textbf{(b)} Let $Q\in k\left[  \left[  X\right]  \right]  $ is a power series
whose coefficient before $X^{0}$ is $1$. Then, $\log Q$ is a power series
whose coefficient before $X^{0}$ is $0$ (by Proposition~\ref{Proposition1009}
\textbf{(b)}). Hence, $\exp\left(  \log Q\right)  $ is a well-defined power
series in $k\left[  \left[  X\right]  \right]  $. Furthermore,
Proposition~\ref{Proposition1009} \textbf{(a)} (applied to $P=\log Q$) shows
that $\exp\left(  \log Q\right)  $ is a power series whose coefficient before
$X^{0}$ is $1$. Thus, the coefficient of $\exp\left(  \log Q\right)  $ before
$X^{0}$ is $1$. On the other hand, the coefficient of $Q$ before $X^{0}$ is
$1$ (by the definition of $Q$).

Also, Proposition~\ref{prop.deriv-exp} (applied to $P=\log Q$) yields%
\begin{align*}
\dfrac{d}{dX}\left(  \exp\left(  \log Q\right)  \right)   &
=\underbrace{\left(  \dfrac{d}{dX}\left(  \log Q\right)  \right)
}_{\substack{=Q^{-1}\cdot\dfrac{d}{dX}Q\\\text{(by
Proposition~\ref{prop.deriv-log} \textbf{(b)})}}}\cdot\exp\left(  \log
Q\right) \\
&  =Q^{-1}\cdot\left(  \dfrac{d}{dX}Q\right)  \cdot\exp\left(  \log Q\right)
.
\end{align*}
Multiplying both sides of this equality by $Q$, we obtain%
\[
Q\dfrac{d}{dX}\left(  \exp\left(  \log Q\right)  \right)  =\left(  \dfrac
{d}{dX}Q\right)  \cdot\exp\left(  \log Q\right)  =\exp\left(  \log Q\right)
\cdot\dfrac{d}{dX}Q.
\]
Thus, Proposition~\ref{prop.lderiv-uniq-det} (applied to $U=\exp\left(  \log
Q\right)  $ and $V=Q$) shows that $\exp\left(  \log Q\right)  =Q$. This proves
Theorem~\ref{Theorem14} \textbf{(b)}.
\end{proof}

\subsection{$\operatorname*{Log}:G\left(  H,A\right)  \rightarrow
\mathfrak{g}\left(  H,A\right)  $ and $\mathfrak{g}\left(  H,A\right)
\rightarrow G\left(  H,A\right)  $}

Next, let us prove a basic fact about the map $\operatorname*{Log}:G\left(
H,A\right)  \rightarrow\mathfrak{g}\left(  H,A\right)  $ and the map
$\mathfrak{g}\left(  H,A\right)  \rightarrow G\left(  H,A\right)  $ which
sends every $f$ to $e^{\ast f}$ (when $H$ is a connected filtered
$k$-coalgebra and $A$ is a $k$-algebra, where $k$ is a field of characteristic
$0$): namely, that these two maps are mutually inverse. Here is how we state
this fact:

\begin{proposition}
\label{Proposition13} Let $k$ be a field of characteristic $0$, let $A$ be a
$k$-algebra, and let $H$ be a connected filtered $k$-coalgebra.

\textbf{(a)} Every map $f\in\mathfrak{g}\left(  H,A\right)  $ satisfies
$\operatorname*{Log}\left(  e^{\ast f}\right)  =f$.

\textbf{(b)} Every map $F\in G\left(  H,A\right)  $ satisfies $e^{\ast\left(
\operatorname*{Log}F\right)  }=F$.
\end{proposition}

It is easy to prove Proposition~\ref{Proposition13} using
Theorem~\ref{Theorem14} and some topology to make sense of infinite sums like
$\sum_{i\geq0}\dfrac{f^{\ast i}}{i!}$ and $\sum_{i\geq1}\dfrac{\left(
-1\right)  ^{i-1}}{i}f^{\ast i}$ in $\mathcal{L}\left(  H,A\right)  $. Here is
a more elementary version of this proof:

We start with a ``finite version'' of Theorem~\ref{Theorem14}:

\begin{corollary}
\label{Corollary15} Let $k$ be a field of characteristic $0$. Let
$n\in\mathbb{N}$.

\textbf{(a)} Let $a$ be an element of a $k$-algebra such that $a^{n+1}=0$.
Then,%
\[
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum_{j=1}^{n}%
\dfrac{a^{j}}{j!}\right)  ^{i}=a.
\]


\textbf{(b)} Let $b$ be an element of a $k$-algebra such that $b^{n+1}=0$.
Then,%
\[
\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}b^{i}\right)  ^{j}}{j!}=b.
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary15}.]\textbf{(a)} Consider the ring of power
series $k\left[  \left[  X\right]  \right]  $. Clearly, $X$ is a power series
whose coefficient before $X^{0}$ is $0$. Thus, applying
Theorem~\ref{Theorem14} \textbf{(a)} to $P=X$, we obtain $\log\left(  \exp
X\right)  =X$. Now, in the ring $k\left[  \left[  X\right]  \right]  $, we
have%
\begin{align*}
\underbrace{\exp X}_{=\sum_{j=0}^{\infty}\dfrac{X^{j}}{j!}=\dfrac{X^{0}}%
{0!}+\sum_{j=1}^{\infty}\dfrac{X^{j}}{j!}}-1  &  =\underbrace{\dfrac{X^{0}%
}{0!}}_{=1}+\sum_{j=1}^{\infty}\dfrac{X^{j}}{j!}-1=\sum_{j=1}^{\infty}%
\dfrac{X^{j}}{j!}=\sum_{j=1}^{n}\dfrac{X^{j}}{j!}+\sum_{j=n+1}^{\infty
}\underbrace{\dfrac{X^{j}}{j!}}_{\substack{\equiv0\operatorname{mod}%
X^{n+1}k\left[  \left[  X\right]  \right]  \\\text{(since }j\geq n+1\text{
leads to}\\X^{j}\equiv0\operatorname{mod}X^{n+1}k\left[  \left[  X\right]
\right]  \text{)}}}\\
&  \equiv\sum_{j=1}^{n}\dfrac{X^{j}}{j!}+\underbrace{\sum_{j=n+1}^{\infty}%
0}_{=0}=\sum_{j=1}^{n}\dfrac{X^{j}}{j!}\operatorname{mod}X^{n+1}k\left[
\left[  X\right]  \right]  .
\end{align*}
Thus,
\[
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \underbrace{\sum
_{j=1}^{n}\dfrac{X^{j}}{j!}}_{\equiv\exp X-1\operatorname{mod}X^{n+1}k\left[
\left[  X\right]  \right]  }\right)  ^{i}\equiv\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  \exp X-1\right)  ^{i}\operatorname{mod}%
X^{n+1}k\left[  \left[  X\right]  \right]  .
\]
On the other hand, $X\mid\exp X-1$ in $k\left[  \left[  X\right]  \right]  $
(since $\exp X-1=\sum_{j=1}^{\infty}\dfrac{X^{j}}{j!}=\sum_{j=1}^{\infty
}\dfrac{XX^{j-1}}{j!}=X\sum_{j=1}^{\infty}\dfrac{X^{j-1}}{j!}$ is divisible by
$X$), and thus $X^{i}\mid\left(  \exp X-1\right)  ^{i}$ in $k\left[  \left[
X\right]  \right]  $ for every $i\in\mathbb{N}$. Thus,
\begin{equation}
\left(  \exp X-1\right)  ^{i}\equiv0\operatorname{mod}X^{n+1}k\left[  \left[
X\right]  \right]  \label{pf.Corollary15.a.=0}%
\end{equation}
for every $i\in\mathbb{N}$ satisfying $i\geq n+1$ (because $i\geq n+1$ leads
to $X^{n+1}\mid X^{i}\mid\left(  \exp X-1\right)  ^{i}$). Now,%
\begin{align*}
X  &  =\log\left(  \exp X\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}\left(  \exp X-1\right)  ^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\log\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \underbrace{\exp
X-1}_{\equiv\sum_{j=1}^{n}\dfrac{X^{j}}{j!}\operatorname{mod}X^{n+1}k\left[
\left[  X\right]  \right]  }\right)  ^{i}+\sum_{i=n+1}^{\infty}\dfrac{\left(
-1\right)  ^{i-1}}{i}\underbrace{\left(  \exp X-1\right)  ^{i}}%
_{\substack{\equiv0\operatorname{mod}X^{n+1}k\left[  \left[  X\right]
\right]  \\\text{(by (\ref{pf.Corollary15.a.=0}))}}}\\
&  \equiv\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum
_{j=1}^{n}\dfrac{X^{j}}{j!}\right)  ^{i}+\underbrace{\sum_{i=n+1}^{\infty
}\dfrac{\left(  -1\right)  ^{i-1}}{i}0}_{=0}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum_{j=1}%
^{n}\dfrac{X^{j}}{j!}\right)  ^{i}\operatorname{mod}X^{n+1}k\left[  \left[
X\right]  \right]  .
\end{align*}
Thus, $X^{n+1}\mid X-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
\sum_{j=1}^{n}\dfrac{X^{j}}{j!}\right)  ^{i}$ in $k\left[  \left[  X\right]
\right]  $. This means that the coefficient of the power series $X-\sum
_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum_{j=1}^{n}%
\dfrac{X^{j}}{j!}\right)  ^{i}$ before $X^{\lambda}$ is $0$ for every
$\lambda\in\left\{  0,1,\ldots,n\right\}  $. But the power series
$X-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum_{j=1}%
^{n}\dfrac{X^{j}}{j!}\right)  ^{i}$ is actually a polynomial, so this rewrites
as follows: The coefficient of the polynomial $X-\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  \sum_{j=1}^{n}\dfrac{X^{j}}{j!}\right)  ^{i}$
before $X^{\lambda}$ is $0$ for every $\lambda\in\left\{  0,1,\ldots
,n\right\}  $. In other words, $X^{n+1}\mid X-\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  \sum_{j=1}^{n}\dfrac{X^{j}}{j!}\right)  ^{i}$ in
$k\left[  X\right]  $. Hence, there exists a polynomial $\mathbf{P}\in
k\left[  X\right]  $ such that $X-\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}\left(  \sum_{j=1}^{n}\dfrac{X^{j}}{j!}\right)  ^{i}=X^{n+1}%
\mathbf{P}$. Consider this polynomial $\mathbf{P}$.

Applying the polynomial identity $X-\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}\left(  \sum_{j=1}^{n}\dfrac{X^{j}}{j!}\right)  ^{i}=X^{n+1}%
\mathbf{P}$ to $a$ instead of $X$, we get%
\[
a-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum_{j=1}%
^{n}\dfrac{a^{j}}{j!}\right)  ^{i}=\underbrace{a^{n+1}}_{=0}\mathbf{P}\left(
a\right)  =0,
\]
so that $\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum
_{j=1}^{n}\dfrac{a^{j}}{j!}\right)  ^{i}=a$. This proves
Corollary~\ref{Corollary15} \textbf{(a)}.

\textbf{(b) }Consider the ring of power series $k\left[  \left[  X\right]
\right]  $. Clearly, $1+X$ is a power series whose coefficient before $X^{0}$
is $1$. Thus, applying Theorem~\ref{Theorem14} \textbf{(b)} to $P=1+X$, we
obtain $\exp\left(  \log\left(  1+X\right)  \right)  =1+X$. Now, in the ring
$k\left[  \left[  X\right]  \right]  $, we have%
\begin{align*}
\log\left(  1+X\right)   &  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  \underbrace{\left(  1+X\right)  -1}_{=X}\right)  ^{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\log\right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}=\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}+\sum_{i=n+1}^{\infty
}\underbrace{\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}}_{\substack{\equiv
0\operatorname{mod}X^{n+1}k\left[  \left[  X\right]  \right]  \\\text{(since
}i\geq n+1\text{ leads to}\\X^{i}\equiv0\operatorname{mod}X^{n+1}k\left[
\left[  X\right]  \right]  \text{)}}}\\
&  \equiv\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}%
+\underbrace{\sum_{i=n+1}^{\infty}0}_{=0}=\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}X^{i}\operatorname{mod}X^{n+1}k\left[  \left[  X\right]
\right]  .
\end{align*}
On the other hand,%
\begin{align*}
1+X  &  =\exp\left(  \log\left(  1+X\right)  \right)  =\sum_{j=0}^{\infty
}\dfrac{\left(  \log\left(  1+X\right)  \right)  ^{j}}{j!}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\exp\right) \\
&  =\underbrace{\dfrac{\left(  \log\left(  1+X\right)  \right)  ^{0}}{0!}%
}_{=\dfrac{1}{1}=1}+\sum_{j=1}^{\infty}\dfrac{\left(  \log\left(  1+X\right)
\right)  ^{j}}{j!}=1+\sum_{j=1}^{\infty}\dfrac{\left(  \log\left(  1+X\right)
\right)  ^{j}}{j!}.
\end{align*}
Subtracting $1$ from this yields%
\[
X=\sum_{j=1}^{\infty}\dfrac{\left(  \log\left(  1+X\right)  \right)  ^{j}}%
{j!}.
\]
Since $X\mid\log\left(  1+X\right)  $ in $k\left[  \left[  X\right]  \right]
$ (because $\log\left(  1+X\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}\underbrace{X^{i}}_{=XX^{i-1}}=X\sum_{i\geq1}\dfrac{\left(
-1\right)  ^{i-1}}{i}X^{i-1}$ is divisible by $X$), we have $X^{j}\mid\left(
\log\left(  1+X\right)  \right)  ^{j}$ for every $j\in\mathbb{N}$. Thus every
$j\in\mathbb{N}$ such that $j\geq n+1$ satisfies
\begin{equation}
\left(  \log\left(  1+X\right)  \right)  ^{j}\equiv0\operatorname{mod}%
X^{n+1}k\left[  \left[  X\right]  \right]  \label{pf.Corollary15.b.=0}%
\end{equation}
(since $j\geq n+1$ leads to $X^{n+1}\mid X^{j}\mid\left(  \log\left(
1+X\right)  \right)  ^{j}$). Now,%
\begin{align*}
X  &  =\sum_{j=1}^{\infty}\dfrac{\left(  \log\left(  1+X\right)  \right)
^{j}}{j!}=\sum_{j=1}^{n}\dfrac{\left(  \log\left(  1+X\right)  \right)  ^{j}%
}{j!}+\sum_{j=n+1}^{\infty}\underbrace{\dfrac{\left(  \log\left(  1+X\right)
\right)  ^{j}}{j!}}_{\substack{\equiv0\operatorname{mod}X^{n+1}k\left[
\left[  X\right]  \right]  \\\text{(since }j\geq n+1\text{ and thus}\\\left(
\log\left(  1+X\right)  \right)  ^{j}\equiv0\operatorname{mod}X^{n+1}k\left[
\left[  X\right]  \right]  \\\text{(by (\ref{pf.Corollary15.b.=0})))}}}\\
&  \equiv\sum_{j=1}^{n}\dfrac{\left(  \log\left(  1+X\right)  \right)  ^{j}%
}{j!}+\underbrace{\sum_{j=n+1}^{\infty}0}_{=0}=\sum_{j=1}^{n}\dfrac{\left(
\log\left(  1+X\right)  \right)  ^{j}}{j!}\\
&  \equiv\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}X^{i}\right)  ^{j}}{j!}\operatorname{mod}X^{n+1}k\left[  \left[
X\right]  \right]
\end{align*}
(since $\log\left(  1+X\right)  \equiv\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}X^{i}\operatorname{mod}X^{n+1}k\left[  \left[  X\right]  \right]
$). Thus, $X^{n+1}\mid X-\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$ in $k\left[
\left[  X\right]  \right]  $. This means that the coefficient of the power
series $X-\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$ before $X^{\lambda}$ is $0$ for every
$\lambda\in\left\{  0,1,\ldots,n\right\}  $. But the power series
$X-\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$ is actually a polynomial, so this rewrites
as follows: The coefficient of the polynomial $X-\sum_{j=1}^{n}\dfrac{\left(
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$
before $X^{\lambda}$ is $0$ for every $\lambda\in\left\{  0,1,\ldots
,n\right\}  $. In other words, $X^{n+1}\mid X-\sum_{j=1}^{n}\dfrac{\left(
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$ in
$k\left[  X\right]  $. Hence, there exists a polynomial $\mathbf{Q}\in
k\left[  X\right]  $ such that $X-\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}=X^{n+1}%
\mathbf{Q}$. Consider this polynomial $\mathbf{Q}$.

Applying the polynomial identity $X-\sum_{j=1}^{n}\dfrac{\left(  \sum
_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}%
=X^{n+1}\mathbf{Q}$ to $b$ instead of $X$, we get%
\[
b-\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}b^{i}\right)  ^{j}}{j!}=\underbrace{b^{n+1}}_{=0}\mathbf{Q}\left(
b\right)  =0,
\]
so that $\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}b^{i}\right)  ^{j}}{j!}=b$. This proves Corollary~\ref{Corollary15}
\textbf{(b)}.
\end{proof}

Next, we shall show a very easy fact:

\begin{proposition}
\label{Proposition16} Let $k$ be a field. Let $H$ be a filtered $k$-coalgebra.
Then, $H_{\leq n}$ is a subcoalgebra of $H$ for every $n\in\mathbb{N}$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition16}.]Let $n\in\mathbb{N}$. Then, since
$H$ is a filtered $k$-coalgebra, we have%
\[
\Delta_{H}\left(  H_{\leq n}\right)  \subseteq\sum_{u=0}^{n}%
\underbrace{H_{\leq u}}_{\substack{\subseteq H_{\leq n}\\\text{(since }u\leq
n\text{)}}}\otimes\underbrace{H_{\leq n-u}}_{\substack{\subseteq H_{\leq
n}\\\text{(since }n-u\leq n\text{)}}}\subseteq\sum_{u=0}^{n}H_{\leq n}\otimes
H_{\leq n}\subseteq H_{\leq n}\otimes H_{\leq n}%
\]
(since $H_{\leq n}\otimes H_{\leq n}$ is a $k$-vector space). Thus, $H_{\leq
n}$ is a subcoalgebra of $H$. This proves Proposition~\ref{Proposition16}.
\end{proof}

Now some triviality:

\begin{proposition}
\label{Proposition17} Let $k$ be a field. Let $H$ be a $k$-coalgebra. Let $J$
be a subcoalgebra of $H$. Let $A$ be a $k$-algebra.

\textbf{(a)} Then, any $f\in\mathcal{L}\left(  H,A\right)  $ and any
$g\in\mathcal{L}\left(  H,A\right)  $ satisfy $\left(  f\mid_{J}\right)
\ast\left(  g\mid_{J}\right)  =\left(  f\ast g\right)  \mid_{J}$.

\textbf{(b)} Also, any $f\in\mathcal{L}\left(  H,A\right)  $ and any
$i\in\mathbb{N}$ satisfy $\left(  f\mid_{J}\right)  ^{\ast i}=f^{\ast i}%
\mid_{J}$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition17}.]\textbf{(a)} Any $f\in
\mathcal{L}\left(  H,A\right)  $ and any $g\in\mathcal{L}\left(  H,A\right)  $
satisfy%
\begin{align*}
\left(  f\mid_{J}\right)  \ast\left(  g\mid_{J}\right)   &  =\mu_{A}%
\circ\underbrace{\left(  \left(  f\mid_{J}\right)  \otimes\left(  g\mid
_{J}\right)  \right)  }_{=\left(  f\otimes g\right)  \mid_{J\otimes J}}%
\circ\underbrace{\Delta_{J}}_{=\Delta_{H}\mid_{J}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of convolution}\right) \\
&  =\mu_{A}\circ\left(  \left(  f\otimes g\right)  \mid_{J\otimes J}\right)
\circ\left(  \Delta_{H}\mid_{J}\right)  =\underbrace{\left(  \mu_{A}%
\circ\left(  f\otimes g\right)  \circ\Delta_{H}\right)  }_{=f\ast g}\mid
_{J}=\left(  f\ast g\right)  \mid_{J}.
\end{align*}
This proves Proposition~\ref{Proposition17} \textbf{(a)}.

\textbf{(b)} Proposition~\ref{Proposition17} \textbf{(b)} easily follows from
Proposition~\ref{Proposition17} \textbf{(a)} by induction over $i$.

This completes the proof of Proposition~\ref{Proposition17}.
\end{proof}

Now let us give a proof of Proposition~\ref{Proposition13}. (We will give
another proof of Proposition~\ref{Proposition13} (and even of a more general
fact: Proposition~\ref{Proposition82}) in \S \ref{S11}; it will avoid the use
of Propositions~\ref{Proposition16} and~\ref{Proposition17}.)

\begin{proof}
[Proof of Proposition~\ref{Proposition13}.]\textbf{(a)} Let $f\in
\mathfrak{g}\left(  H,A\right)  $. Let $n\in\mathbb{N}$.
Proposition~\ref{Proposition16} yields that $H_{\leq n}$ is a subcoalgebra of
$H$.

Let $g=e^{\ast f}-e_{H,A}$. Then, $g\in\mathfrak{g}\left(  H,A\right)  $
(since $e^{\ast f}\in G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(
H,A\right)  $, so that $e^{\ast f}-e_{H,A}\in\mathfrak{g}\left(  H,A\right)
$). Hence, Remark~\ref{Remark9} (applied to $g$ instead of $f$) yields
$g^{\ast i}\left(  H_{\leq n}\right)  =0$ for every $i>n$. Also,
Remark~\ref{Remark9} yields $f^{\ast i}\left(  H_{\leq n}\right)  =0$ for
every $i>n$.

We have
\begin{align*}
\operatorname*{Log}\left(  e^{\ast f}\right)   &  =\operatorname{Log}%
_{1}\underbrace{\left(  e^{\ast f}-e_{H,A}\right)  }_{=g}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Log}%
\right) \\
&  =\operatorname{Log}_{1}g.
\end{align*}
Thus, every $x\in H_{\leq n}$ satisfies%
\begin{align*}
\left(  \operatorname*{Log}\left(  e^{\ast f}\right)  \right)  \left(
x\right)   &  =\left(  \operatorname{Log}_{1}g\right)  \left(  x\right)
=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname{Log}%
_{1}\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  +\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{g^{\ast i}\left(  x\right)  }_{\substack{=0\text{ (since
}\\x\in H_{\leq n}\text{ and thus}\\g^{\ast i}\left(  x\right)  \in g^{\ast
i}\left(  H_{\leq n}\right)  =0\\\text{(since }i>n\text{))}}}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  +\underbrace{\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}0}_{=0}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\left(  x\right)  .
\end{align*}
In other words,%
\[
\operatorname*{Log}\left(  e^{\ast f}\right)  \mid_{H_{\leq n}}=\left(
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\right)
\mid_{H_{\leq n}}.
\]
Since%
\[
\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\right)
\mid_{H_{\leq n}}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{\left(  g^{\ast i}\mid_{H_{\leq n}}\right)  }%
_{\substack{=\left(  g\mid_{H_{\leq n}}\right)  ^{\ast i}\\\text{(because
Proposition~\ref{Proposition17} \textbf{(b)}}\\\text{(applied to }H_{\leq
n}\text{ and }g\text{ instead of}\\J\text{ and }f\text{) yields }\left(
g\mid_{H_{\leq n}}\right)  ^{\ast i}=g^{\ast i}\mid_{H_{\leq n}}\text{)}%
}}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  g\mid_{H_{\leq
n}}\right)  ^{\ast i},
\]
this rewrites as%
\[
\operatorname*{Log}\left(  e^{\ast f}\right)  \mid_{H_{\leq n}}=\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  g\mid_{H_{\leq n}}\right)
^{\ast i}.
\]


But every $x\in H_{\leq n}$ satisfies%
\begin{align*}
e^{\ast f}\left(  x\right)   &  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(
x\right)  }{i!}=\sum_{i=0}^{n}\dfrac{f^{\ast i}\left(  x\right)  }{i!}%
+\sum_{i=n+1}^{\infty}\underbrace{\dfrac{f^{\ast i}\left(  x\right)  }{i!}%
}_{\substack{=0\text{ (since }\\x\in H_{\leq n}\text{ and thus}\\f^{\ast
i}\left(  x\right)  \in f^{\ast i}\left(  H_{\leq n}\right)  =0\\\text{(since
}i>n\text{), so that}\\f^{\ast i}\left(  x\right)  =0\text{)}}}=\sum_{i=0}%
^{n}\dfrac{f^{\ast i}\left(  x\right)  }{i!}+\underbrace{\sum_{i=n+1}^{\infty
}0}_{=0}\\
&  =\sum_{i=0}^{n}\dfrac{f^{\ast i}\left(  x\right)  }{i!}=\underbrace{\dfrac
{f^{\ast0}\left(  x\right)  }{0!}}_{=\dfrac{e_{H,A}\left(  x\right)  }%
{1}=e_{H,A}\left(  x\right)  }+\sum_{i=1}^{n}\dfrac{f^{\ast i}\left(
x\right)  }{i!}=e_{H,A}\left(  x\right)  +\sum_{i=1}^{n}\dfrac{f^{\ast
i}\left(  x\right)  }{i!}%
\end{align*}
and thus%
\begin{align*}
\underbrace{g}_{=e^{\ast f}-e_{H,A}}\left(  x\right)   &  =\left(  e^{\ast
f}-e_{H,A}\right)  \left(  x\right)  =\underbrace{e^{\ast f}\left(  x\right)
}_{=e_{H,A}\left(  x\right)  +\sum_{i=1}^{n}\dfrac{f^{\ast i}\left(  x\right)
}{i!}}-e_{H,A}\left(  x\right)  =\sum_{i=1}^{n}\dfrac{f^{\ast i}\left(
x\right)  }{i!}=\sum_{i=1}^{n}\dfrac{f^{\ast i}}{i!}\left(  x\right) \\
&  =\sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for }i\text{
in the sum}\right)  .
\end{align*}
In other words,
\begin{align*}
g\mid_{H_{\leq n}}  &  =\left(  \sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\right)
\mid_{H_{\leq n}}=\sum_{j=1}^{n}\dfrac{f^{\ast j}\mid_{H_{\leq n}}}{j!}\\
&  =\sum_{j=1}^{n}\dfrac{\left(  f\mid_{H_{\leq n}}\right)  ^{\ast j}}{j!}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Proposition~\ref{Proposition17} \textbf{(b)} (applied to }H_{\leq
n}\text{ and }j\text{ instead of }J\text{ and }i\text{)}\\
\text{yields }\left(  f\mid_{H_{\leq n}}\right)  ^{\ast j}=f^{\ast j}%
\mid_{H_{\leq n}}\text{, so that }f^{\ast j}\mid_{H_{\leq n}}=\left(
f\mid_{H_{\leq n}}\right)  ^{\ast j}%
\end{array}
\right)  .
\end{align*}


Now,%
\begin{equation}
\operatorname*{Log}\left(  e^{\ast f}\right)  \mid_{H_{\leq n}}=\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \underbrace{g\mid_{H_{\leq n}%
}}_{=\sum_{j=1}^{n}\dfrac{\left(  f\mid_{H_{\leq n}}\right)  ^{\ast j}}{j!}%
}\right)  ^{\ast i}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
\sum_{j=1}^{n}\dfrac{\left(  f\mid_{H_{\leq n}}\right)  ^{\ast j}}{j!}\right)
^{\ast i}. \label{13.pf.1}%
\end{equation}


But $n+1>n$. Hence, Remark~\ref{Remark9} (applied to $i=n+1$) yields
$f^{\ast\left(  n+1\right)  }\left(  H_{\leq n}\right)  =0$. Since%
\begin{align*}
\left(  f\mid_{H_{\leq n}}\right)  ^{\ast\left(  n+1\right)  }  &
=f^{\ast\left(  n+1\right)  }\mid_{H_{\leq n}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition~\ref{Proposition17} \textbf{(b)}}\\
\text{(applied to }H_{\leq n}\text{ and }n+1\text{ instead of }J\text{ and
}i\text{)}%
\end{array}
\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f^{\ast\left(  n+1\right)
}\left(  H_{\leq n}\right)  =0\right)  ,
\end{align*}
we can apply Corollary~\ref{Corollary15} \textbf{(a)} to $a=f\mid_{H_{\leq n}%
}$ and obtain $\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
\sum_{j=1}^{n}\dfrac{\left(  f\mid_{H_{\leq n}}\right)  ^{\ast j}}{j!}\right)
^{\ast i}=f\mid_{H_{\leq n}}$. Thus, (\ref{13.pf.1}) becomes%
\[
\operatorname*{Log}\left(  e^{\ast f}\right)  \mid_{H_{\leq n}}=f\mid_{H_{\leq
n}}.
\]
We have thus proven this for every $n\in\mathbb{N}$.

Now, let $x\in H$ be arbitrary. Since $H$ is filtered, there must exist some
$n\in\mathbb{N}$ such that $x\in H_{\leq n}$. Consider this $n$. From $x \in
H_{\leq n}$, we obtain
\[
\left(  \operatorname*{Log}\left(  e^{\ast f}\right)  \right)  \left(
x\right)  =\underbrace{\left(  \operatorname*{Log}\left(  e^{\ast f}\right)
\mid_{H_{\leq n}}\right)  }_{=f\mid_{H_{\leq n}}}\left(  x\right)  =\left(
f\mid_{H_{\leq n}}\right)  \left(  x\right)  =f\left(  x\right)  .
\]
Since this holds for every $x\in H$, we can now conclude that
$\operatorname*{Log}\left(  e^{\ast f}\right)  =f$.

This proves Proposition~\ref{Proposition13} \textbf{(a)}.

\textbf{(b)} Let $F\in G\left(  H,A\right)  $. Let $n\in\mathbb{N}$.
Proposition~\ref{Proposition16} yields that $H_{\leq n}$ is a subcoalgebra of
$H$.

Let $g=F-e_{H,A}$. Then, $g\in\mathfrak{g}\left(  H,A\right)  $ (since $F\in
G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(  H,A\right)  $, so that
$F-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  $). Hence, Remark~\ref{Remark9}
(applied to $g$ instead of $f$) yields $g^{\ast i}\left(  H_{\leq n}\right)
=0$ for every $i>n$.

Let $\varphi=\operatorname*{Log}F$. Then, $\varphi\in\mathfrak{g}\left(
H,A\right)  $, so that Remark~\ref{Remark9} (applied to $\varphi$ instead of
$f$) yields $\varphi^{\ast i}\left(  H_{\leq n}\right)  =0$ for every $i>n$.

By the definition of $\operatorname*{Log}$, we have $\operatorname*{Log}%
F=\operatorname{Log}_{1}\underbrace{\left(  F-e_{H,A}\right)  }_{=g}%
=\operatorname{Log}_{1}g$. Hence, $\varphi=\operatorname*{Log}%
f=\operatorname{Log}_{1}g$.

Thus, every $x\in H_{\leq n}$ satisfies%
\begin{align*}
\varphi\left(  x\right)   &  =\left(  \operatorname{Log}_{1}g\right)  \left(
x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname{Log}_{1}\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  +\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{g^{\ast i}\left(  x\right)  }_{\substack{=0\text{ (since
}\\x\in H_{\leq n}\text{ and thus}\\g^{\ast i}\left(  x\right)  \in g^{\ast
i}\left(  H_{\leq n}\right)  =0\\\text{(since }i>n\text{))}}}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  +\underbrace{\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}0}_{=0}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  .
\end{align*}
In other words,%
\[
\varphi\mid_{H_{\leq n}}=\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}.
\]
Since%
\[
\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\right)
\mid_{H_{\leq n}}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{\left(  g^{\ast i}\mid_{H_{\leq n}}\right)  }%
_{\substack{=\left(  g\mid_{H_{\leq n}}\right)  ^{\ast i}\text{ (because
Proposition~\ref{Proposition17} \textbf{(b)}}\\\text{(applied to }H_{\leq
n}\text{ and }g\text{ instead of}\\J\text{ and }f\text{) yields }\left(
g\mid_{H_{\leq n}}\right)  ^{\ast i}=g^{\ast i}\mid_{H_{\leq n}}\text{)}%
}}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  g\mid_{H_{\leq
n}}\right)  ^{\ast i},
\]
this rewrites as%
\[
\varphi\mid_{H_{\leq n}}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  g\mid_{H_{\leq n}}\right)  ^{\ast i}.
\]


But every $x\in H_{\leq n}$ satisfies%
\begin{align*}
e^{\ast\varphi}\left(  x\right)   &  =\sum_{i\geq0}\dfrac{\varphi^{\ast
i}\left(  x\right)  }{i!}=\sum_{i=0}^{n}\dfrac{\varphi^{\ast i}\left(
x\right)  }{i!}+\sum_{i=n+1}^{\infty}\underbrace{\dfrac{\varphi^{\ast
i}\left(  x\right)  }{i!}}_{\substack{=0\text{ (since }\\x\in H_{\leq n}\text{
and thus}\\\varphi^{\ast i}\left(  x\right)  \in\varphi^{\ast i}\left(
H_{\leq n}\right)  =0\\\text{(since }i>n\text{), so that}\\\varphi^{\ast
i}\left(  x\right)  =0\text{)}}}=\sum_{i=0}^{n}\dfrac{\varphi^{\ast i}\left(
x\right)  }{i!}+\underbrace{\sum_{i=n+1}^{\infty}0}_{=0}=\sum_{i=0}^{n}%
\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}\\
&  =\underbrace{\dfrac{\varphi^{\ast0}\left(  x\right)  }{0!}}_{=\dfrac
{e_{H,A}\left(  x\right)  }{1}=e_{H,A}\left(  x\right)  }+\sum_{i=1}^{n}%
\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}=e_{H,A}\left(  x\right)
+\sum_{i=1}^{n}\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}%
\end{align*}
and thus%
\begin{align*}
\left(  e^{\ast\varphi}-e_{H,A}\right)  \left(  x\right)   &
=\underbrace{e^{\ast\varphi}\left(  x\right)  }_{=e_{H,A}\left(  x\right)
+\sum_{i=1}^{n}\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}}-e_{H,A}\left(
x\right)  =\sum_{i=1}^{n}\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}%
=\sum_{i=1}^{n}\dfrac{\varphi^{\ast i}}{i!}\left(  x\right) \\
&  =\sum_{j=1}^{n}\dfrac{\varphi^{\ast j}}{j!}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for }i\text{
in the sum}\right)  .
\end{align*}
In other words,
\begin{align}
&  \left(  e^{\ast\varphi}-e_{H,A}\right)  \mid_{H_{\leq n}}\nonumber\\
&  =\left(  \sum_{j=1}^{n}\dfrac{\varphi^{\ast j}}{j!}\right)  \mid_{H_{\leq
n}}=\sum_{j=1}^{n}\dfrac{\varphi^{\ast j}\mid_{H_{\leq n}}}{j!}\nonumber\\
&  =\sum_{j=1}^{n}\dfrac{\left(  \varphi\mid_{H_{\leq n}}\right)  ^{\ast j}%
}{j!}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Proposition~\ref{Proposition17} \textbf{(b)} (applied to }%
\varphi\text{, }H_{\leq n}\text{ and }j\text{ instead of }f\text{, }J\text{
and }i\text{)}\\
\text{yields }\left(  \varphi\mid_{H_{\leq n}}\right)  ^{\ast j}=\varphi^{\ast
j}\mid_{H_{\leq n}}\text{, so that }\varphi^{\ast j}\mid_{H_{\leq n}}=\left(
\varphi\mid_{H_{\leq n}}\right)  ^{\ast j}%
\end{array}
\right) \nonumber\\
&  =\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}\left(  g\mid_{H_{\leq n}}\right)  ^{\ast i}\right)  ^{\ast j}}%
{j!}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\mid_{H_{\leq n}}%
=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  g\mid_{H_{\leq n}%
}\right)  ^{\ast i}\right)  . \label{13.pf.2}%
\end{align}


But Remark~\ref{Remark9} (applied to $n+1$ and $g$ instead of $i$ and $f$)
yields $g^{\ast\left(  n+1\right)  }\left(  H_{\leq n}\right)  =0$ (since
$n+1>n$). Since%
\begin{align*}
&  \left(  g\mid_{H_{\leq n}}\right)  ^{\ast\left(  n+1\right)  }\\
&  =g^{\ast\left(  n+1\right)  }\mid_{H_{\leq n}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition~\ref{Proposition17} \textbf{(b)}}\\
\text{(applied to }g\text{, }H_{\leq n}\text{ and }n+1\text{ instead of
}f\text{, }J\text{ and }i\text{)}%
\end{array}
\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }g^{\ast\left(  n+1\right)
}\left(  H_{\leq n}\right)  =0\right)  ,
\end{align*}
we can apply Corollary~\ref{Corollary15} \textbf{(b)} to $b=g\mid_{H_{\leq n}%
}$ and obtain $\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  g\mid_{H_{\leq n}}\right)  ^{\ast i}\right)
^{\ast j}}{j!}=g\mid_{H_{\leq n}}$. Thus, (\ref{13.pf.2}) becomes%
\begin{align*}
\left(  e^{\ast\varphi}-e_{H,A}\right)  \mid_{H_{\leq n}}  &  =g\mid_{H_{\leq
n}}=\left(  F-e_{H,A}\right)  \mid_{H_{\leq n}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }g=F-e_{H,A}\right) \\
&  =F\mid_{H_{\leq n}}-e_{H,A}\mid_{H_{\leq n}}.
\end{align*}
Since $\left(  e^{\ast\varphi}-e_{H,A}\right)  \mid_{H_{\leq n}}%
=e^{\ast\varphi}\mid_{H_{\leq n}}-e_{H,A}\mid_{H_{\leq n}}$, this rewrites as
$e^{\ast\varphi}\mid_{H_{\leq n}}-e_{H,A}\mid_{H_{\leq n}}=F\mid_{H_{\leq n}%
}-e_{H,A}\mid_{H_{\leq n}}$. Thus, $e^{\ast\varphi}\mid_{H_{\leq n}}%
=F\mid_{H_{\leq n}}$. Since $\varphi=\operatorname*{Log}F$, this becomes
$e^{\ast\left(  \operatorname*{Log}F\right)  }\mid_{H_{\leq n}}=F\mid_{H_{\leq
n}}$. We have thus proven this for every $n\in\mathbb{N}$.

Now, let $x\in H$ be arbitrary. Since $H$ is filtered, there must exist some
$n\in\mathbb{N}$ such that $x\in H_{\leq n}$. Consider this $n$. From $x \in
H_{\leq n}$, we obtain
\[
e^{\ast\left(  \operatorname*{Log}F\right)  }\left(  x\right)
=\underbrace{\left(  e^{\ast\left(  \operatorname*{Log}F\right)  }%
\mid_{H_{\leq n}}\right)  }_{=F\mid_{H_{\leq n}}}\left(  x\right)  =\left(
F\mid_{H_{\leq n}}\right)  \left(  x\right)  =F\left(  x\right)  .
\]
Since this holds for every $x\in H$, we can now conclude that $e^{\ast\left(
\operatorname*{Log}F\right)  }=F$.

This proves Proposition~\ref{Proposition13} \textbf{(b)}.
\end{proof}

\section{\label{S3}Some properties of primitive elements}

Next let us recall the definition of a primitive element:

\begin{definition}
\label{Definition18} Let $k$ be a field. Let $H$ be a unital coalgebra. Let
$x\in H$ be an element. The element $x$ is said to be \textit{primitive} if
$\Delta\left(  x\right)  =x\otimes1_{H}+1_{H}\otimes x$. Here, $1_{H}\in H$
denotes the unity of the unital coalgebra $H$ (as defined in
Definition~\ref{Definition7b}).
\end{definition}

\textit{Remark.} Some authors define the notion of primitivity slightly
differently: they define an element $x\in H$ to be \textit{primitive} if
$\Delta\left(  x\right)  =x\otimes1_{H}+1_{H}\otimes x$ and $\varepsilon
\left(  x\right)  =0$. However, this definition of ``primitive'' turns out to
be equivalent to our Definition~\ref{Definition18}. This is because every
element $x\in H$ which is primitive in the sense of
Definition~\ref{Definition18} satisfies $\Delta\left(  x\right)
=x\otimes1_{H}+1_{H}\otimes x$ and $\varepsilon\left(  x\right)  =0$ (by
Remark~\ref{Remark21}), and conversely, every element $x\in H$ satisfying
$\Delta\left(  x\right)  =x\otimes1_{H}+1_{H}\otimes x$ and $\varepsilon
\left(  x\right)  =0$ is clearly primitive in the sense of
Definition~\ref{Definition18}.

Next we show:

\begin{proposition}
\label{Proposition19} Let $k$ be a field, let $A$ be a $k$-algebra, and let
$H$ be a$\ $unital coalgebra. Let $\operatorname*{Prim}H$ denote the set of
all primitive elements of $H$.

\textbf{(a)} Any $f\in\mathfrak{g}\left(  H,A\right)  $ and any $g\in
\mathfrak{g}\left(  H,A\right)  $ satisfy $f\ast g\in\mathfrak{g}\left(
H,A\right)  $ and $\left(  f\ast g\right)  \left(  \operatorname*{Prim}%
H\right)  =0$.

\textbf{(b)} Every $f\in\mathfrak{g}\left(  H,A\right)  $ satisfies $f^{\ast
i}\in\mathfrak{g}\left(  H,A\right)  $ and $f^{\ast i}\left(
\operatorname*{Prim}H\right)  =0$ for every integer $i>1$.

\textbf{(c)} Assume that the field $k$ has characteristic $0$, and that $H$ is
a connected filtered $k$-coalgebra. Then, every $F\in G\left(  H,A\right)  $
satisfies $\left(  \operatorname*{Log}F\right)  \mid_{\operatorname*{Prim}%
H}=F\mid_{\operatorname*{Prim}H}$.
\end{proposition}

Before we prove this, let us show a very easy (and classical) observation
about primitive elements:

\begin{remark}
\label{Remark21} Let $k$ be a field. Let $H$ be a unital coalgebra. Then,
$\varepsilon\left(  x\right)  =0$ for every $x\in\operatorname*{Prim}H$.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark21}.]Recall that the unity of the unital coalgebra
$H$ is denoted by $1_{H}$. Thus, $\left(  H,1_{H}\right)  $ is a unital
coalgebra. Hence, by the definition of a unital coalgebra, we have $\Delta
_{H}\left(  1_{H}\right)  =1_{H}\otimes1_{H}$ and $\varepsilon_{H}\left(
1_{H}\right)  =1$. Since we abbreviate $\varepsilon_{H}$ as $\varepsilon$, we
have $\varepsilon\left(  1_{H}\right)  =\varepsilon_{H}\left(  1_{H}\right)
=1$.

Let $x\in\operatorname*{Prim}H$. Then, $x$ is primitive (by the definition of
$\operatorname*{Prim}H$). In other words, $\Delta\left(  x\right)
=x\otimes1_{H}+1_{H}\otimes x$.

Now, let $\operatorname*{can}:H\otimes k\rightarrow H$ be the canonical
$k$-module isomorphism (sending $c\otimes x$ to $cx$ for all $c\in H$ and
$x\in k$). Then, by the axioms of a coalgebra, we have $\operatorname*{can}%
\circ\left(  \operatorname*{id}\otimes\varepsilon\right)  \circ\Delta
=\operatorname*{id}$. But%
\begin{align*}
\left(  \operatorname*{can}\circ\left(  \operatorname*{id}\otimes
\varepsilon\right)  \circ\Delta\right)  \left(  x\right)   &
=\operatorname*{can}\left(  \left(  \operatorname*{id}\otimes\varepsilon
\right)  \left(  \underbrace{\Delta\left(  x\right)  }_{=x\otimes1_{H}%
+1_{H}\otimes x}\right)  \right)  =\operatorname*{can}\left(
\underbrace{\left(  \operatorname*{id}\otimes\varepsilon\right)  \left(
x\otimes1_{H}+1_{H}\otimes x\right)  }_{=\operatorname*{id}\left(  x\right)
\otimes\varepsilon\left(  1_{H}\right)  +\operatorname*{id}\left(
1_{H}\right)  \otimes\varepsilon\left(  x\right)  }\right) \\
&  =\operatorname*{can}\left(  \operatorname*{id}\left(  x\right)
\otimes\varepsilon\left(  1_{H}\right)  +\operatorname*{id}\left(
1_{H}\right)  \otimes\varepsilon\left(  x\right)  \right) \\
&  =\underbrace{\operatorname*{id}\left(  x\right)  }_{=x}\cdot
\underbrace{\varepsilon\left(  1_{H}\right)  }_{=1}%
+\underbrace{\operatorname*{id}\left(  1_{H}\right)  }_{=1_{H}}\cdot
\varepsilon\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\operatorname*{can}\right) \\
&  =x+1_{H}\cdot\varepsilon\left(  x\right)  .
\end{align*}
Comparing this with $\left(  \underbrace{\operatorname*{can}\circ\left(
\operatorname*{id}\otimes\varepsilon\right)  \circ\Delta}_{=\operatorname*{id}%
}\right)  \left(  x\right)  =\operatorname*{id}\left(  x\right)  =x$, we
obtain $x+1_{H}\cdot\varepsilon\left(  x\right)  =x$. Thus, $1_{H}%
\cdot\varepsilon\left(  x\right)  =0$. Hence, $\varepsilon\left(  1_{H}%
\cdot\varepsilon\left(  x\right)  \right)  =0$. Since $\varepsilon\left(
1_{H}\cdot\varepsilon\left(  x\right)  \right)  =\underbrace{\varepsilon
\left(  1_{H}\right)  }_{=1}\cdot\varepsilon\left(  x\right)  =\varepsilon
\left(  x\right)  $, this rewrites as $\varepsilon\left(  x\right)  =0$. This
proves Remark~\ref{Remark21}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition19}.]Recall that the unity of the unital
coalgebra $H$ is denoted by $1_{H}$. Thus, $\left(  H,1_{H}\right)  $ is a
unital coalgebra. Hence, by the definition of a unital coalgebra, we have
$\Delta_{H}\left(  1_{H}\right)  =1_{H}\otimes1_{H}$ and $\varepsilon
_{H}\left(  1_{H}\right)  =1$. Since we abbreviate $\Delta_{H}$ by $\Delta$,
we have $\Delta\left(  1_{H}\right)  =\Delta_{H}\left(  1_{H}\right)
=1_{H}\otimes1_{H}$.

\textbf{(a)} Let $x\in\operatorname*{Prim}H$. Then, $x$ is a primitive element
of $H$ (since $\operatorname*{Prim}H$ is the set of all primitive elements of
$H$). In other words, $\Delta\left(  x\right)  =x\otimes1_{H}+1_{H}\otimes x$.
But by the definition of convolution, $f\ast g=\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta$. Thus,%
\begin{align*}
\left(  f\ast g\right)  \left(  x\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta\right)  \left(  x\right)  =\mu_{A}\left(
\left(  f\otimes g\right)  \underbrace{\left(  \Delta\left(  x\right)
\right)  }_{=x\otimes1_{H}+1_{H}\otimes x}\right) \\
&  =\mu_{A}\left(  \underbrace{\left(  f\otimes g\right)  \left(
x\otimes1_{H}+1_{H}\otimes x\right)  }_{=f\left(  x\right)  \otimes g\left(
1_{H}\right)  +f\left(  1_{H}\right)  \otimes g\left(  x\right)  }\right) \\
&  =\mu_{A}\left(  f\left(  x\right)  \otimes\underbrace{g\left(
1_{H}\right)  }_{\substack{=0\\\text{(since }g\in\mathfrak{g}\left(
H,A\right)  \text{)}}}+\underbrace{f\left(  1_{H}\right)  }%
_{\substack{=0\\\text{(since }f\in\mathfrak{g}\left(  H,A\right)  \text{)}%
}}\otimes g\left(  x\right)  \right) \\
&  =\mu_{A}\left(  \underbrace{f\left(  x\right)  \otimes0+0\otimes g\left(
x\right)  }_{=0}\right)  =0.
\end{align*}
Since this holds for every $x\in\operatorname*{Prim}H$, we thus get $\left(
f\ast g\right)  \left(  \operatorname*{Prim}H\right)  =0$.

Besides, $f\ast g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta$ leads to%
\begin{align*}
\left(  f\ast g\right)  \left(  1_{H}\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta\right)  \left(  1_{H}\right)  =\mu_{A}\left(
\left(  f\otimes g\right)  \underbrace{\left(  \Delta\left(  1_{H}\right)
\right)  }_{=1_{H}\otimes1_{H}}\right)  =\mu_{A}\left(  \underbrace{\left(
f\otimes g\right)  \left(  1_{H}\otimes1_{H}\right)  }_{=f\left(
1_{H}\right)  \otimes g\left(  1_{H}\right)  }\right) \\
&  =\mu_{A}\left(  \underbrace{f\left(  1_{H}\right)  }%
_{\substack{=0\\\text{(since }f\in\mathfrak{g}\left(  H,A\right)  \text{)}%
}}\otimes\underbrace{g\left(  1_{H}\right)  }_{\substack{=0\\\text{(since
}g\in\mathfrak{g}\left(  H,A\right)  \text{)}}}\right)  =\mu_{A}\left(
0\otimes0\right)  =0.
\end{align*}
Thus, $f\ast g\in\mathfrak{g}\left(  H,A\right)  $ (by the definition of
$\mathfrak{g}\left(  H,A\right)  $).

This proves Proposition~\ref{Proposition19} \textbf{(a)}.

\textbf{(b)} We will prove Proposition~\ref{Proposition19} \textbf{(b)} by
induction over $i$:

\textit{Induction base:} Proposition~\ref{Proposition19} \textbf{(a)} (applied
to $g=f$) yields $f\ast f\in\mathfrak{g}\left(  H,A\right)  $ and $\left(
f\ast f\right)  \left(  \operatorname*{Prim}H\right)  =0$. Since $f\ast
f=f^{\ast2}$, this rewrites as follows: We have $f^{\ast2}\in\mathfrak{g}%
\left(  H,A\right)  $ and $f^{\ast2}\left(  \operatorname*{Prim}H\right)  =0$.
This proves Proposition~\ref{Proposition19} \textbf{(b)} for $i=2$. The
induction base is thus complete.

\textit{Induction step:} Let $j>1$ be an integer. Assume that
Proposition~\ref{Proposition19} \textbf{(b)} holds for $i=j$. We must then
prove Proposition~\ref{Proposition19} \textbf{(b)} for $i=j+1$.

Since Proposition~\ref{Proposition19} \textbf{(b)} holds for $i=j$, we have
$f^{\ast j}\in\mathfrak{g}\left(  H,A\right)  $ and $f^{\ast j}\left(
\operatorname*{Prim}H\right)  =0$. Now, Proposition~\ref{Proposition19}
\textbf{(a)} (applied to $g=f^{\ast j}$) yields $f\ast f^{\ast j}%
\in\mathfrak{g}\left(  H,A\right)  $ and $\left(  f\ast f^{\ast j}\right)
\left(  \operatorname*{Prim}H\right)  =0$. Since $f\ast f^{\ast j}%
=f^{\ast\left(  j+1\right)  }$, this rewrites as follows: We have
$f^{\ast\left(  j+1\right)  }\in\mathfrak{g}\left(  H,A\right)  $ and
$f^{\ast\left(  j+1\right)  }\left(  \operatorname*{Prim}H\right)  =0$. In
other words, Proposition~\ref{Proposition19} \textbf{(b)} holds for $i=j+1$.
The induction step is thus complete.

This completes the induction proof of Proposition~\ref{Proposition19}
\textbf{(b)}.

\textbf{(c)} Let $F\in G\left(  H,A\right)  $.

Let $f=F-e_{H,A}$. Then, $f\in\mathfrak{g}\left(  H,A\right)  $ (since $F\in
G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(  H,A\right)  $, so that
$F-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  $).
Proposition~\ref{Proposition19} \textbf{(b)} thus yields $f^{\ast i}\left(
\operatorname*{Prim}H\right)  =0$ for every integer $i>1$.

Now, the definition of $\operatorname*{Log}$ says that $\operatorname*{Log}%
F=\operatorname{Log}_{1}\underbrace{\left(  F-e_{H,A}\right)  }_{=f}%
=\operatorname{Log}_{1}f$.

Let $x\in\operatorname*{Prim}H$. Then, $f^{\ast i}\left(  x\right)  \in
f^{\ast i}\left(  \operatorname*{Prim}H\right)  =0$ for every integer $i>1$.
In other words, $f^{\ast i}\left(  x\right)  =0$ for every integer $i>1$. On
the other hand, $\varepsilon\left(  x\right)  =0$ (by Remark~\ref{Remark21}),
so that $\underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{H}}\left(  x\right)
=\left(  \eta_{A}\circ\varepsilon_{H}\right)  \left(  x\right)  =\eta
_{A}\left(  \underbrace{\varepsilon_{H}\left(  x\right)  }_{=\varepsilon
\left(  x\right)  =0}\right)  =0$. Thus, $F\left(  x\right)  -e_{H,A}\left(
x\right)  =F\left(  x\right)  $. Since $F\left(  x\right)  -e_{H,A}\left(
x\right)  =\underbrace{\left(  F-e_{H,A}\right)  }_{=f}\left(  x\right)
=f\left(  x\right)  $, this rewrites as $f\left(  x\right)  =F\left(
x\right)  $. Now,
\begin{align*}
\underbrace{\left(  \operatorname*{Log}F\right)  }_{=\operatorname{Log}_{1}%
f}\left(  x\right)   &  =\left(  \operatorname{Log}_{1}f\right)  \left(
x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
x\right)  =\underbrace{\dfrac{\left(  -1\right)  ^{1-1}}{1}}_{=1}f^{\ast
1}\left(  x\right)  +\sum_{i>1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{f^{\ast i}\left(  x\right)  }_{=0}\\
&  =\underbrace{1f^{\ast1}\left(  x\right)  }_{=f^{\ast1}\left(  x\right)
=f\left(  x\right)  }+\underbrace{\sum_{i>1}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}0}_{=0}=f\left(  x\right)  =F\left(  x\right)  .
\end{align*}
Since this holds for every $x\in\operatorname*{Prim}H$, we thus conclude that
$\left(  \operatorname*{Log}F\right)  \mid_{\operatorname*{Prim}H}%
=F\mid_{\operatorname*{Prim}H}$. This proves Proposition~\ref{Proposition19}
\textbf{(c)}.
\end{proof}

\section{\label{S4}$\left(  \varepsilon,\varepsilon\right)  $-coderivations}

We now introduce the notion of an $\left(  \varepsilon,\varepsilon\right)
$\textit{-coderivation}. This notion can be defined in two ways; we are going
to take the one that is more similar to the notion of an $\left(
\varepsilon,\varepsilon\right)  $-derivation\footnote{The notion of an
$\left(  \varepsilon,\varepsilon\right)  $-derivation is a known one; we
recall its definition in \S \ref{S12} (Definitions~\ref{Definition108}
and~\ref{Definition109}). Note that we speak of ``$\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivations'' instead of ``$\left(
\varepsilon,\varepsilon\right)  $-derivations'' for reasons of pedantry.} as
definition, and then prove the equivalence to the other one as a theorem
(Theorem~\ref{Theorem23}).

\begin{definition}
\label{Definition22} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $H$
be a unital coalgebra. Let $f:C\rightarrow H$ be a $k$-linear map. Then, $f$
is said to be an $\left(  \varepsilon,\varepsilon\right)  $%
\textit{-coderivation} if and only if $\Delta_{H}\circ f=\left(  f\otimes
e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta_{C}$. Here, the map $e_{C,H}$ is
defined to be the map $\eta_{H}\circ\varepsilon_{C}:C\rightarrow H$ (this
definition of the map $e_{C,H}$ is identical with the definition of the map
$e_{H,A}$ in Definition~\ref{def.eHA2}).
\end{definition}

The following theorem shows how we can actually think of coderivations:

\begin{theorem}
\label{Theorem23} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $H$ be a
unital coalgebra. We denote by $\operatorname*{Prim}H$ the set of all
primitive elements of $H$. Let $f:C\rightarrow H$ be a $k$-linear map. Then,
$f$ is an $\left(  \varepsilon,\varepsilon\right)  $-coderivation if and only
if $f\left(  C\right)  \subseteq\operatorname*{Prim}H$.
\end{theorem}

Before we prove this, we recall a familiar fact from basic algebra:%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{Any six }k\text{-vector spaces }U\text{, }V\text{, }W\text{, }U^{\prime
}\text{, }V^{\prime}\text{ and }W^{\prime}\\
\text{and any four }k\text{-linear maps }\alpha:U\rightarrow V\text{, }%
\beta:V\rightarrow W\text{, }\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime
}\\
\text{and }\beta^{\prime}:V^{\prime}\rightarrow W^{\prime}\text{ satisfy
}\left(  \beta\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ
\alpha^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)
\circ\left(  \alpha\otimes\alpha^{\prime}\right)  .
\end{array}
\right)  \label{(X)(X)}%
\end{equation}


\begin{proof}
[Proof of Theorem~\ref{Theorem23}.]\textbf{a)} First, let us show that every
$x\in C$ satisfies%
\begin{equation}
\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta
_{C}\right)  \left(  x\right)  =f\left(  x\right)  \otimes1_{H}+1_{H}\otimes
f\left(  x\right)  . \label{23.pf.1}%
\end{equation}
(This holds no matter whether $f$ is an $\left(  \varepsilon,\varepsilon
\right)  $-coderivation or not, and whether $f\left(  C\right)  \subseteq
\operatorname*{Prim}H$ or not.)

\textit{Proof.} First it is clear that every $x\in C$ satisfies $\left(
\varepsilon_{C}\otimes\operatorname*{id}\right)  \left(  \Delta_{C}\left(
x\right)  \right)  =1\otimes x$\ \ \ \ \footnote{\textit{Proof.} Let $x\in C$.
Let $\operatorname*{kan}:C\rightarrow k\otimes C$ be the canonical isomorphism
which sends every $y\in C$ to $1\otimes y$. Then, by the axioms of a
coalgebra, we have $\left(  \varepsilon_{C}\otimes\operatorname*{id}\right)
\circ\Delta_{C}=\operatorname*{kan}$ (since $C$ is a coalgebra), and $\left(
\varepsilon_{C}\otimes\operatorname*{id}\right)  \left(  \Delta_{C}\left(
x\right)  \right)  =\underbrace{\left(  \left(  \varepsilon_{C}\otimes
\operatorname*{id}\right)  \circ\Delta_{C}\right)  }_{=\operatorname*{kan}%
}\left(  x\right)  =\operatorname*{kan}\left(  x\right)  =1\otimes x$ (by the
definition of $\operatorname*{kan}$), qed.}.

Since $e_{C,H}=\eta_{H}\circ\varepsilon_{C}$ and $f=f\circ\operatorname*{id}$,
we have
\begin{align*}
e_{C,H}\otimes f  &  =\left(  \eta_{H}\circ\varepsilon_{C}\right)
\otimes\left(  f\circ\operatorname*{id}\right)  =\left(  \eta_{H}\otimes
f\right)  \circ\left(  \varepsilon_{C}\otimes\operatorname*{id}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by an application of (\ref{(X)(X)}%
)}\right)  ,
\end{align*}
so that every $x\in C$ satisfies%
\begin{align*}
\left(  e_{C,H}\otimes f\right)  \left(  \Delta_{C}\left(  x\right)  \right)
&  =\left(  \left(  \eta_{H}\otimes f\right)  \circ\left(  \varepsilon
_{C}\otimes\operatorname*{id}\right)  \right)  \left(  \Delta_{C}\left(
x\right)  \right)  =\left(  \eta_{H}\otimes f\right)  \left(
\underbrace{\left(  \varepsilon_{C}\otimes\operatorname*{id}\right)  \left(
\Delta_{C}\left(  x\right)  \right)  }_{=1\otimes x}\right) \\
&  =\left(  \eta_{H}\otimes f\right)  \left(  1\otimes x\right)
=\underbrace{\eta_{H}\left(  1\right)  }_{\substack{=1\cdot1_{H}\\\text{(by
the definition of }\eta_{H}\text{)}}}\otimes f\left(  x\right)
=\underbrace{1\cdot1_{H}}_{=1_{H}}\otimes f\left(  x\right)  =1_{H}\otimes
f\left(  x\right)  .
\end{align*}
The same argument (but with permuted tensorands) shows that every $x\in C$
satisfies%
\[
\left(  f\otimes e_{C,H}\right)  \left(  \Delta_{C}\left(  x\right)  \right)
=f\left(  x\right)  \otimes1_{H}.
\]
Now, every $x\in C$ satisfies%
\begin{align*}
\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta
_{C}\right)  \left(  x\right)   &  =\left(  f\otimes e_{C,H}+e_{C,H}\otimes
f\right)  \left(  \Delta_{C}\left(  x\right)  \right) \\
&  =\underbrace{\left(  f\otimes e_{C,H}\right)  \left(  \Delta_{C}\left(
x\right)  \right)  }_{=f\left(  x\right)  \otimes1_{H}}+\underbrace{\left(
e_{C,H}\otimes f\right)  \left(  \Delta_{C}\left(  x\right)  \right)
}_{=1_{H}\otimes f\left(  x\right)  }\\
&  =f\left(  x\right)  \otimes1_{H}+1_{H}\otimes f\left(  x\right)  .
\end{align*}
This proves (\ref{23.pf.1}). Thus, \textbf{a)} is proven.

\textbf{b) }Now, let us prove that if $f$ is an $\left(  \varepsilon
,\varepsilon\right)  $-coderivation, then $f\left(  C\right)  \subseteq
\operatorname*{Prim}H$.

\textit{Proof.} Assume that $f$ is an $\left(  \varepsilon,\varepsilon\right)
$-coderivation. Then, $\Delta_{H}\circ f=\left(  f\otimes e_{C,H}%
+e_{C,H}\otimes f\right)  \circ\Delta_{C}$ (by the definition of an $\left(
\varepsilon,\varepsilon\right)  $-coderivation, since $f$ is an $\left(
\varepsilon,\varepsilon\right)  $-coderivation).

Using (\ref{23.pf.1}), it is easy to see that every $y\in f\left(  C\right)  $
satisfies $y\in\operatorname*{Prim}H$.\ \ \ \ \footnote{\textit{Proof.} Let
$y\in f\left(  C\right)  $. Then, there exists some $x\in C$ such that
$y=f\left(  x\right)  $. Consider this $x$. Then,%
\begin{align*}
\Delta_{H}\left(  y\right)   &  =\Delta_{H}\left(  f\left(  x\right)  \right)
=\underbrace{\left(  \Delta_{H}\circ f\right)  }_{=\left(  f\otimes
e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta_{C}}\left(  x\right)  =\left(
\left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta_{C}\right)
\left(  x\right) \\
&  =\underbrace{f\left(  x\right)  }_{=y}\otimes1_{H}+1_{H}\otimes
\underbrace{f\left(  x\right)  }_{=y}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{23.pf.1})}\right) \\
&  =y\otimes1_{H}+1_{H}\otimes y.
\end{align*}
But this yields that $y$ is primitive (because a primitive element of $H$ was
defined as an element $z\in H$ satisfying $\Delta_{H}\left(  z\right)
=z\otimes1_{H}+1_{H}\otimes z$). In other words, $y\in\operatorname*{Prim}H$
(since $\operatorname*{Prim}H$ is the set of all primitive elements of $H$),
qed.} In other words, $f\left(  C\right)  \subseteq\operatorname*{Prim}H$.
This proves \textbf{b)}.

\textbf{c)} Now, let us prove that if $f\left(  C\right)  \subseteq
\operatorname*{Prim}H$, then $f$ is an $\left(  \varepsilon,\varepsilon
\right)  $-coderivation.

\textit{Proof.} Assume that $f\left(  C\right)  \subseteq\operatorname*{Prim}%
H$. Then, for every $x\in C$, the element $f\left(  x\right)  \in H$ is
primitive (since $x\in C$ and thus $f\left(  x\right)  \in f\left(  C\right)
\subseteq\operatorname*{Prim}H=\left(  \text{the set of all primitive elements
of }H\right)  $). Now, for every $x\in C$, we have%
\begin{align*}
&  \left(  \Delta_{H}\circ f\right)  \left(  x\right) \\
&  =\Delta_{H}\left(  f\left(  x\right)  \right)  =f\left(  x\right)
\otimes1_{H}+1_{H}\otimes f\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }f\left(  x\right)  \text{ is primitive, and since a primitive
element of }H\text{ was defined}\\
\text{as an element }z\in H\text{ satisfying }\Delta_{H}\left(  z\right)
=z\otimes1_{H}+1_{H}\otimes z
\end{array}
\right) \\
&  =\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta
_{C}\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{23.pf.1})}\right)  .
\end{align*}
Thus, $\Delta_{H}\circ f=\left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)
\circ\Delta_{C}$. In other words, $f$ is an $\left(  \varepsilon
,\varepsilon\right)  $-coderivation (because this is how $\left(
\varepsilon,\varepsilon\right)  $-coderivations were defined). This proves
\textbf{c)}.

\textbf{d)} Combining the results of \textbf{b)} and \textbf{c)}, we see that
$f$ is an $\left(  \varepsilon,\varepsilon\right)  $-coderivation if and only
if $f\left(  C\right)  \subseteq\operatorname*{Prim}H$.
Theorem~\ref{Theorem23} is thus proven.
\end{proof}

\section{\label{S5}The exponent-logarithm bijection between $\left(
\varepsilon,\varepsilon\right)  $-coderivations and coalgebra homomorphisms}

We are now ready to formulate a fact which will give us a major part of
Theorem~\ref{Theorem1}, and is of significant interest in its own:

\begin{theorem}
\label{Theorem30} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered cocommutative $k$-coalgebra. Let $H$ be a $k$-bialgebra.
Let $f\in\mathfrak{g}\left(  C,H\right)  $. Then, $f$ is an $\left(
\varepsilon,\varepsilon\right)  $-coderivation if and only if $e^{\ast f}$ is
a $k$-coalgebra homomorphism.
\end{theorem}

Being an ``if and only if'' statement, the assertion of this theorem splits
into two parts, which we will now formulate as two independent lemmas:

\begin{lemma}
\label{Lemma31} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered cocommutative $k$-coalgebra. Let $H$ be a $k$-bialgebra.
Let $f\in\mathfrak{g}\left(  C,H\right)  $. If $f$ is an $\left(
\varepsilon,\varepsilon\right)  $-coderivation, then $e^{\ast f}$ is a
$k$-coalgebra homomorphism.
\end{lemma}

\begin{lemma}
\label{Lemma32} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered cocommutative $k$-coalgebra. Let $H$ be a $k$-bialgebra.
Let $f\in\mathfrak{g}\left(  C,H\right)  $. If $e^{\ast f}$ is a $k$-coalgebra
homomorphism, then $f$ is an $\left(  \varepsilon,\varepsilon\right)  $-coderivation.
\end{lemma}

Of these two lemmas, only one will be used in the proof of
Theorem~\ref{Theorem1} - namely, Lemma~\ref{Lemma32}. However, both of them
are interesting and we will prove both for the sake of completeness.

\section{\label{S6}The ``$\left(  \varepsilon,\varepsilon\right)
$-coderivation $\Longrightarrow$ coalgebra homomorphism'' direction}

First let us prepare some auxiliary results for proving Lemma~\ref{Lemma31}.
We start with a triviality:

\begin{lemma}
\label{Lemma33} Let $k$ be a field. Let $C$ and $D$ be $k$-coalgebras. Let $A$
and $B$ be $k$-algebras. Let $p:C\rightarrow A$, $q:C\rightarrow A$,
$r:D\rightarrow B$ and $s:D\rightarrow B$ be four $k$-linear maps. Then,%
\[
\left(  \mu_{A}\otimes\mu_{B}\right)  \circ\left(  p\otimes q\otimes r\otimes
s\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  =\left(  p\ast
q\right)  \otimes\left(  r\ast s\right)
\]
(this is an equality between two maps $C\otimes D\rightarrow A\otimes B$).
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma33}.]By (\ref{(X)(X)}) (applied to $U=C$,
$V=C\otimes C$, $W=A\otimes A$, $U^{\prime}=D$, $V^{\prime}=D\otimes D$,
$W^{\prime}=B\otimes B$, $\alpha=\Delta_{C}$, $\beta=p\otimes q$,
$\alpha^{\prime}=\Delta_{D}$ and $\beta^{\prime}=r\otimes s$), we have%
\[
\left(  \left(  p\otimes q\right)  \circ\Delta_{C}\right)  \otimes\left(
\left(  r\otimes s\right)  \circ\Delta_{D}\right)  =\left(  p\otimes q\otimes
r\otimes s\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  .
\]


By (\ref{(X)(X)}) (applied to $U=C$, $V=A\otimes A$, $W=A$, $U^{\prime}=D$,
$V^{\prime}=B\otimes B$, $W^{\prime}=B$, $\alpha=\left(  p\otimes q\right)
\circ\Delta_{C}$, $\beta=\mu_{A}$, $\alpha^{\prime}=\left(  r\otimes s\right)
\circ\Delta_{D}$ and $\beta^{\prime}=\mu_{B}$), we have%
\begin{align*}
&  \left(  \mu_{A}\circ\left(  p\otimes q\right)  \circ\Delta_{C}\right)
\otimes\left(  \mu_{B}\circ\left(  r\otimes s\right)  \circ\Delta_{D}\right)
\\
&  =\left(  \mu_{A}\otimes\mu_{B}\right)  \circ\underbrace{\left(  \left(
\left(  p\otimes q\right)  \circ\Delta_{C}\right)  \otimes\left(  \left(
r\otimes s\right)  \circ\Delta_{D}\right)  \right)  }_{=\left(  p\otimes
q\otimes r\otimes s\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)
}\\
&  =\left(  \mu_{A}\otimes\mu_{B}\right)  \circ\left(  p\otimes q\otimes
r\otimes s\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  .
\end{align*}
Now,%
\begin{align*}
&  \underbrace{\left(  p\ast q\right)  }_{\substack{=\mu_{A}\circ\left(
p\otimes q\right)  \circ\Delta_{C}\\\text{(by the definition of convolution)}%
}}\otimes\underbrace{\left(  r\ast s\right)  }_{\substack{=\mu_{B}\circ\left(
r\otimes s\right)  \circ\Delta_{D}\\\text{(by the definition of convolution)}%
}}\\
&  =\left(  \mu_{A}\circ\left(  p\otimes q\right)  \circ\Delta_{C}\right)
\otimes\left(  \mu_{B}\circ\left(  r\otimes s\right)  \circ\Delta_{D}\right)
\\
&  =\left(  \mu_{A}\otimes\mu_{B}\right)  \circ\left(  p\otimes q\otimes
r\otimes s\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  ,
\end{align*}
so that Lemma~\ref{Lemma33} is proven.
\end{proof}

Next, a definition:

\begin{definition}
\label{Definition34} Let $k$ be a field. Let $V$ and $W$ be two $k$-vector
spaces. Then, $\tau_{V,W}$ will denote the $\left(  V,W\right)  $%
\textit{-flip}; this is the $k$-linear map $V\otimes W\rightarrow W\otimes V$
which, for every $v\in V$ and $w\in W$, sends the element $v\otimes w\in
V\otimes W$ to the element $w\otimes v\in W\otimes V$.
\end{definition}

First let us show a trivial property of these flips:

\begin{proposition}
\label{Proposition35} Let $k$ be a field.

\textbf{(a)} Let $V$, $W$, $V^{\prime}$ and $W^{\prime}$ be four $k$-vector
spaces, and $f:V\rightarrow V^{\prime}$ and $g:W\rightarrow W^{\prime}$ be two
$k$-linear maps. Then, $\left(  g\otimes f\right)  \circ\tau_{V,W}%
=\tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes g\right)  $.

\textbf{(b)} Let $U$, $V$, $W$, $T$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$
and $T^{\prime}$ be eight $k$-vector spaces, and $e:U\rightarrow U^{\prime}$,
$f:V\rightarrow V^{\prime}$, $g:W\rightarrow W^{\prime}$ and $h:T\rightarrow
T^{\prime}$ be four $k$-linear maps. Then,%
\[
\left(  \operatorname{id}_{U^{\prime}}\otimes\tau_{V^{\prime},W^{\prime}%
}\otimes\operatorname{id}_{T^{\prime}}\right)  \circ\left(  e\otimes f\otimes
g\otimes h\right)  =\left(  e\otimes g\otimes f\otimes h\right)  \circ\left(
\operatorname{id}_{U}\otimes\tau_{V,W}\otimes\operatorname{id}_{T}\right)
\]
(this is an equality between $k$-linear maps from $U\otimes V\otimes W\otimes
T$ to $U^{\prime}\otimes W^{\prime}\otimes V^{\prime}\otimes T^{\prime}$).
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition35}.]\textbf{(a)} Every $v\in V$ and
$w\in W$ satisfy%
\begin{align*}
\left(  \left(  g\otimes f\right)  \circ\tau_{V,W}\right)  \left(  v\otimes
w\right)   &  =\left(  g\otimes f\right)  \left(  \underbrace{\tau
_{V,W}\left(  v\otimes w\right)  }_{\substack{=w\otimes v\\\text{(by the
definition of }\tau_{V,W}\text{)}}}\right) \\
&  =\left(  g\otimes f\right)  \left(  w\otimes v\right)  =g\left(  w\right)
\otimes f\left(  v\right)
\end{align*}
and%
\begin{align*}
\left(  \tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes g\right)  \right)
\left(  v\otimes w\right)   &  =\tau_{V^{\prime},W^{\prime}}\left(
\underbrace{\left(  f\otimes g\right)  \left(  v\otimes w\right)  }_{=f\left(
v\right)  \otimes g\left(  w\right)  }\right)  =\tau_{V^{\prime},W^{\prime}%
}\left(  f\left(  v\right)  \otimes g\left(  w\right)  \right) \\
&  =g\left(  w\right)  \otimes f\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\tau_{V^{\prime},W^{\prime}}\right)  .
\end{align*}
Hence, every $v\in V$ and $w\in W$ satisfy
\[
\left(  \left(  g\otimes f\right)  \circ\tau_{V,W}\right)  \left(  v\otimes
w\right)  =g\left(  w\right)  \otimes f\left(  v\right)  =\left(
\tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes g\right)  \right)  \left(
v\otimes w\right)  .
\]
In other words, the two maps $\left(  g\otimes f\right)  \circ\tau_{V,W}$ and
$\tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes g\right)  $ are equal on
every pure tensor. But since these two maps are $k$-linear, this yields that
these maps must be identic (because whenever two $k$-linear maps from a tensor
product are equal on every pure tensor, they must be identic). In other words,
$\left(  g\otimes f\right)  \circ\tau_{V,W}=\tau_{V^{\prime},W^{\prime}}%
\circ\left(  f\otimes g\right)  $. This proves Proposition~\ref{Proposition35}
\textbf{(a)}.

\textbf{(b)} Applying (\ref{(X)(X)}) to $V\otimes W$, $W\otimes V$,
$W^{\prime}\otimes V^{\prime}$, $T$, $T$, $T^{\prime}$, $\tau_{V,W}$,
$g\otimes f$, $\operatorname{id}_{T}$ and $h$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$
and $\beta^{\prime}$, we obtain%
\[
\left(  \left(  g\otimes f\right)  \circ\tau_{V,W}\right)  \otimes\left(
h\circ\operatorname{id}_{T}\right)  =\left(  g\otimes f\otimes h\right)
\circ\left(  \tau_{V,W}\otimes\operatorname{id}_{T}\right)  .
\]
Applying (\ref{(X)(X)}) to $U$, $U$, $U^{\prime}$, $V\otimes W\otimes T$,
$W\otimes V\otimes T$, $W^{\prime}\otimes V^{\prime}\otimes T^{\prime}$,
$\operatorname{id}_{U}$, $e$, $\tau_{V,W}\otimes\operatorname{id}_{T}$ and
$g\otimes f\otimes h$ instead of $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$,
$W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$ and $\beta^{\prime}$, we
obtain%
\[
\left(  e\circ\operatorname{id}_{U}\right)  \otimes\left(  \left(  g\otimes
f\otimes h\right)  \circ\left(  \tau_{V,W}\otimes\operatorname{id}_{T}\right)
\right)  =\left(  e\otimes g\otimes f\otimes h\right)  \circ\left(
\operatorname{id}_{U}\otimes\tau_{V,W}\otimes\operatorname{id}_{T}\right)  .
\]
Thus,%
\begin{align}
\left(  e\otimes g\otimes f\otimes h\right)  \circ\left(  \operatorname{id}%
_{U}\otimes\tau_{V,W}\otimes\operatorname{id}_{T}\right)   &
=\underbrace{\left(  e\circ\operatorname{id}_{U}\right)  }_{=e}\otimes
\underbrace{\left(  \left(  g\otimes f\otimes h\right)  \circ\left(
\tau_{V,W}\otimes\operatorname{id}_{T}\right)  \right)  }_{=\left(  \left(
g\otimes f\right)  \circ\tau_{V,W}\right)  \otimes\left(  h\circ
\operatorname{id}_{T}\right)  }\nonumber\\
&  =e\otimes\underbrace{\left(  \left(  g\otimes f\right)  \circ\tau
_{V,W}\right)  }_{\substack{=\tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes
g\right)  \\\text{(by Proposition~\ref{Proposition35} \textbf{(a)})}}%
}\otimes\underbrace{\left(  h\circ\operatorname{id}_{T}\right)  }%
_{=h}\nonumber\\
&  =e\otimes\left(  \tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes
g\right)  \right)  \otimes h. \label{35.pf.b.1}%
\end{align}


On the other hand, applying (\ref{(X)(X)}) to $V\otimes W$, $V^{\prime}\otimes
W^{\prime}$, $W^{\prime}\otimes V^{\prime}$, $T$, $T^{\prime}$, $T^{\prime}$,
$f\otimes g$, $\tau_{V^{\prime},W^{\prime}}$, $h$ and $\operatorname{id}%
_{T^{\prime}}$ instead of $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$,
$W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$ and $\beta^{\prime}$, we
obtain%
\[
\left(  \tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes g\right)  \right)
\otimes\left(  \operatorname{id}_{T^{\prime}}\circ h\right)  =\left(
\tau_{V^{\prime},W^{\prime}}\otimes\operatorname{id}_{T^{\prime}}\right)
\circ\left(  f\otimes g\otimes h\right)  .
\]
Applying (\ref{(X)(X)}) to $U$, $U^{\prime}$, $U^{\prime}$, $V\otimes W\otimes
T$, $V^{\prime}\otimes W^{\prime}\otimes T^{\prime}$, $W^{\prime}\otimes
V^{\prime}\otimes T^{\prime}$, $e$, $\operatorname{id}_{U^{\prime}}$,
$f\otimes g\otimes h$ and $\tau_{V^{\prime},W^{\prime}}\otimes
\operatorname{id}_{T^{\prime}}$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$ and
$\beta^{\prime}$, we obtain%
\[
\left(  \operatorname{id}_{U^{\prime}}\circ e\right)  \otimes\left(  \left(
\tau_{V^{\prime},W^{\prime}}\otimes\operatorname{id}_{T^{\prime}}\right)
\circ\left(  f\otimes g\otimes h\right)  \right)  =\left(  \operatorname{id}%
_{U^{\prime}}\otimes\tau_{V^{\prime},W^{\prime}}\otimes\operatorname{id}%
_{T^{\prime}}\right)  \circ\left(  e\otimes f\otimes g\otimes h\right)  .
\]
Thus,%
\begin{align*}
\left(  \operatorname{id}_{U^{\prime}}\otimes\tau_{V^{\prime},W^{\prime}%
}\otimes\operatorname{id}_{T^{\prime}}\right)  \circ\left(  e\otimes f\otimes
g\otimes h\right)   &  =\underbrace{\left(  \operatorname{id}_{U^{\prime}%
}\circ e\right)  }_{=e}\otimes\underbrace{\left(  \left(  \tau_{V^{\prime
},W^{\prime}}\otimes\operatorname{id}_{T^{\prime}}\right)  \circ\left(
f\otimes g\otimes h\right)  \right)  }_{=\left(  \tau_{V^{\prime},W^{\prime}%
}\circ\left(  f\otimes g\right)  \right)  \otimes\left(  \operatorname{id}%
_{T^{\prime}}\circ h\right)  }\\
&  =e\otimes\left(  \tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes
g\right)  \right)  \otimes\underbrace{\left(  \operatorname{id}_{T^{\prime}%
}\circ h\right)  }_{=h}\\
&  =e\otimes\left(  \tau_{V^{\prime},W^{\prime}}\circ\left(  f\otimes
g\right)  \right)  \otimes h.
\end{align*}
Compared with (\ref{35.pf.b.1}), this yields%
\[
\left(  e\otimes g\otimes f\otimes h\right)  \circ\left(  \operatorname{id}%
_{U}\otimes\tau_{V,W}\otimes\operatorname{id}_{T}\right)  =\left(
\operatorname{id}_{U^{\prime}}\otimes\tau_{V^{\prime},W^{\prime}}%
\otimes\operatorname{id}_{T^{\prime}}\right)  \circ\left(  e\otimes f\otimes
g\otimes h\right)  .
\]
This proves Proposition~\ref{Proposition35} \textbf{(b)}.
\end{proof}

Next, we shall show a general property of coalgebras:

\begin{lemma}
\label{Lemma554} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Then,%
\[
\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}=\left(
\operatorname{id}_{C}\otimes\Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma554}.]By the axioms of a coalgebra, we have
\[
\left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}=\left(
\Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}%
\]
(since $C$ is a coalgebra).

On the other hand,%
\[
\underbrace{\Delta_{C}}_{=\Delta_{C}\circ\operatorname{id}_{C}}\otimes
\underbrace{\Delta_{C}}_{=\operatorname{id}_{C\otimes C}\circ\Delta_{C}%
}=\left(  \Delta_{C}\circ\operatorname{id}_{C}\right)  \otimes\left(
\operatorname{id}_{C\otimes C}\circ\Delta_{C}\right)  =\left(  \Delta
_{C}\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right)
\]
(by (\ref{(X)(X)}) (applied to $U=C$, $V=C$, $W=C\otimes C$, $U^{\prime}=C$,
$V^{\prime}=C\otimes C$, $W^{\prime}=C\otimes C$, $\alpha=\operatorname*{id}%
\nolimits_{C}$, $\beta=\Delta_{C}$, $\alpha^{\prime}=\Delta_{C}$ and
$\beta^{\prime}=\operatorname*{id}\nolimits_{C\otimes C}$)), and thus%
\begin{align}
\underbrace{\left(  \Delta_{C}\otimes\Delta_{C}\right)  }_{=\left(  \Delta
_{C}\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right)  }\circ\Delta_{C}  &  =\left(
\Delta_{C}\otimes\underbrace{\operatorname{id}_{C\otimes C}}%
_{=\operatorname{id}_{C}\otimes\operatorname{id}_{C}}\right)  \circ
\underbrace{\left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)  \circ
\Delta_{C}}_{=\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\Delta_{C}}\nonumber\\
&  =\left(  \Delta_{C}\otimes\operatorname{id}_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\Delta_{C}. \label{36.pf.1}%
\end{align}
But (\ref{(X)(X)}) (applied to $U=C$, $V=C\otimes C$, $W=C\otimes C\otimes C$,
$U^{\prime}=C$, $V^{\prime}=C$, $W^{\prime}=C$, $\alpha=\Delta_{C}$,
$\beta=\Delta_{C}\otimes\operatorname{id}_{C}$, $\alpha^{\prime}%
=\operatorname{id}_{C}$ and $\beta^{\prime}=\operatorname{id}_{C}$) yields%
\[
\left(  \left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta
_{C}\right)  \otimes\left(  \operatorname{id}_{C}\circ\operatorname{id}%
_{C}\right)  =\left(  \Delta_{C}\otimes\operatorname{id}_{C}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}%
_{C}\right)  ,
\]
so that%
\begin{align*}
\left(  \Delta_{C}\otimes\operatorname{id}_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)   &
=\underbrace{\left(  \left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\Delta_{C}\right)  }_{=\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right)  \circ\Delta_{C}}\otimes\left(  \operatorname{id}_{C}%
\circ\operatorname{id}_{C}\right) \\
&  =\left(  \left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)  \circ
\Delta_{C}\right)  \otimes\left(  \operatorname{id}_{C}\circ\operatorname{id}%
_{C}\right) \\
&  =\left(  \operatorname{id}_{C}\otimes\Delta_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)
\end{align*}
(by (\ref{(X)(X)}), applied to $U=C$, $V=C\otimes C$, $W=C\otimes C\otimes C$,
$U^{\prime}=C$, $V^{\prime}=C$, $W^{\prime}=C$, $\alpha=\Delta_{C}$,
$\beta=\operatorname{id}_{C}\otimes\Delta_{C}$, $\alpha^{\prime}%
=\operatorname{id}_{C}$ and $\beta^{\prime}=\operatorname{id}_{C}$). Thus,
(\ref{36.pf.1}) becomes%
\begin{align*}
\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}  &
=\underbrace{\left(  \Delta_{C}\otimes\operatorname{id}_{C}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}%
_{C}\right)  }_{=\left(  \operatorname{id}_{C}\otimes\Delta_{C}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}%
_{C}\right)  }\circ\Delta_{C}\\
&  =\left(  \operatorname{id}_{C}\otimes\Delta_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\Delta_{C}.
\end{align*}
This proves Lemma~\ref{Lemma554}.
\end{proof}

Now, we turn to a basic property of cocommutative coalgebras:

\begin{lemma}
\label{Lemma36} Let $k$ be a field. Let $C$ be a cocommutative $k$-coalgebra.
Then, the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{coversion of (ac)(bd) = (ab)(cd)}{\xymatrixcolsep{7pc}
%\xymatrix{
%C \ar[r]^{\Delta_C} \ar[d]_{\Delta_C} & C\otimes C \ar[rd]^{\Delta
%_C\otimes\Delta_C} & \\
%C\otimes C \ar[r]_{\Delta_C\otimes\Delta_C} & C\otimes C\otimes C\otimes
%C \ar[r]_{\id_C\otimes\tau_{C,C}\otimes\id_C} & C\otimes C\otimes C\otimes C
%}} }%
%BeginExpansion
\xymatrixcolsep{7pc}
\xymatrix{
C \ar[r]^{\Delta_C} \ar[d]_{\Delta_C} & C\otimes C \ar[rd]^{\Delta
_C\otimes\Delta_C} & \\
C\otimes C \ar[r]_{\Delta_C\otimes\Delta_C} & C\otimes C\otimes C\otimes
C \ar[r]_{\id_C\otimes\tau_{C,C}\otimes\id_C} & C\otimes C\otimes C\otimes C
}
%EndExpansion
\label{36.diag}%
\end{equation}
commutes. In other words,%
\begin{equation}
\left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta
_{C}=\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}.
\label{36.equation}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma36}.]Lemma~\ref{Lemma554} yields%
\begin{equation}
\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}=\left(
\operatorname{id}_{C}\otimes\Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}.
\label{36.pf.2}%
\end{equation}


The equality (\ref{(X)(X)}) (applied to $U=C$, $V=C$, $W=C$, $U^{\prime
}=C\otimes C$, $V^{\prime}=C\otimes C\otimes C$, $W^{\prime}=C\otimes C\otimes
C$, $\alpha=\operatorname{id}_{C}$, $\beta=\operatorname{id}_{C}$,
$\alpha^{\prime}=\Delta_{C}\otimes\operatorname{id}_{C}$ and $\beta^{\prime
}=\tau_{C,C}\otimes\operatorname{id}_{C}$) yields%
\begin{equation}
\left(  \operatorname{id}_{C}\circ\operatorname{id}_{C}\right)  \otimes\left(
\left(  \tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta
_{C}\otimes\operatorname{id}_{C}\right)  \right)  =\left(  \operatorname{id}%
_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\otimes\operatorname{id}_{C}\right)  .
\label{36.pf.5}%
\end{equation}
But (\ref{(X)(X)}) (applied to $U=C$, $V=C\otimes C$, $W=C\otimes C$,
$U^{\prime}=C$, $V^{\prime}=C$, $W^{\prime}=C$, $\alpha=\Delta_{C}$,
$\beta=\tau_{C,C}$, $\alpha^{\prime}=\operatorname{id}_{C}$ and $\beta
^{\prime}=\operatorname{id}_{C}$) yields $\left(  \tau_{C,C}\circ\Delta
_{C}\right)  \otimes\left(  \operatorname{id}_{C}\circ\operatorname{id}%
_{C}\right)  =\left(  \tau_{C,C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)  $, so that
\begin{align}
\left(  \tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta
_{C}\otimes\operatorname{id}_{C}\right)   &  =\left(  \tau_{C,C}\circ
\Delta_{C}\right)  \otimes\underbrace{\left(  \operatorname{id}_{C}%
\circ\operatorname{id}_{C}\right)  }_{=\operatorname{id}_{C}}%
=\underbrace{\left(  \tau_{C,C}\circ\Delta_{C}\right)  }_{\substack{=\Delta
_{C}\\\text{(since }C\text{ is cocommutative)}}}\otimes\operatorname{id}%
_{C}\nonumber\\
&  =\Delta_{C}\otimes\operatorname{id}_{C}. \label{36.pf.6}%
\end{align}
Now, (\ref{36.pf.5}) becomes%
\begin{align}
&  \left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta_{C}%
\otimes\operatorname{id}_{C}\right) \nonumber\\
&  =\underbrace{\left(  \operatorname{id}_{C}\circ\operatorname{id}%
_{C}\right)  }_{=\operatorname{id}_{C}}\otimes\underbrace{\left(  \left(
\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\operatorname{id}_{C}\right)  \right)  }_{\substack{=\Delta_{C}%
\otimes\operatorname{id}_{C}\\\text{(by (\ref{36.pf.6}))}}}=\operatorname{id}%
_{C}\otimes\Delta_{C}\otimes\operatorname{id}_{C}. \label{36.pf.7}%
\end{align}
Now,%
\begin{align*}
&  \left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  \circ\underbrace{\left(  \Delta_{C}\otimes\Delta_{C}\right)
\circ\Delta_{C}}_{\substack{=\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}\\\text{(by (\ref{36.pf.2}%
))}}}\\
&  =\underbrace{\left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}\otimes
\Delta_{C}\otimes\operatorname{id}_{C}\right)  }_{=\operatorname{id}%
_{C}\otimes\Delta_{C}\otimes\operatorname{id}_{C}}\circ\left(  \Delta
_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}\\
&  =\left(  \operatorname{id}_{C}\otimes\Delta_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\Delta_{C}=\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}%
\end{align*}
(by (\ref{36.pf.2})). Hence, (\ref{36.equation}) holds. In other words, the
diagram (\ref{36.diag}) is commutative. This proves Lemma~\ref{Lemma36}.
\end{proof}

\textit{Remark:} Actually, many elementary properties of coalgebras (in fact,
all purely diagrammatic properties) are ``duals'' of analogous properties of
algebras. For instance, Lemma~\ref{Lemma36} is a ``dual'' of the fact that for
any four elements $a$, $b$, $c$, $d$ of a commutative $k$-algebra, we have
$\left(  ac\right)  \cdot\left(  bd\right)  =\left(  ab\right)  \cdot\left(
cd\right)  $. This fact is, of course, trivial. This gives us two ways to
prove Lemma~\ref{Lemma36}:

\begin{itemize}
\item Either invoke the theorem that the ``dual'' of a true purely
diagrammatic property of $k$-algebras must be a true property of
$k$-coalgebras. However, this would require us to prove this theorem.

\item Or reformulate the property of $k$-algebras (that for any four elements
$a$, $b$, $c$, $d$ of a commutative $k$-algebra, we have $\left(  ac\right)
\cdot\left(  bd\right)  =\left(  ab\right)  \cdot\left(  cd\right)  $) in a
purely diagrammatic form (i. e., in form of a commutative diagram with no
concrete elements occurring) and \textit{prove it purely by diagram chasing}
(again, without using elements), and then reverse all the arrows in this
proof. As a result you get a proof of Lemma~\ref{Lemma36}. This is how I
constructed the above proof of Lemma~\ref{Lemma36}.\footnote{Of course, if you
are comparing different proofs, you need to count the proof of
Lemma~\ref{Lemma554} as a part of the above proof of Lemma~\ref{Lemma36}.}
\end{itemize}

So when Lemma~\ref{Lemma36} is just a \textquotedblleft dual\textquotedblright%
\ of a fact about $k$-algebras, then why is the fact about $k$-algebras
trivial while Lemma~\ref{Lemma36} took us so long to prove? This is an example
of how working with coalgebras is a lot more cumbersome than working with
algebras, because working with algebras is easily done elementwise, while
working with coalgebras usually requires transforming equations into
commutative diagrams and chasing diagrams. There \textit{is} a second way to
efficiently work with coalgebras: namely, by using Sweedler's notation;
however, this would require us to introduce Sweedler's notation\footnote{To my
knowledge, there is not a single book about Hopf algebras that cares to
formally and correctly introduce Sweedler's notation. However, I have not read
too many books, so this needs not say much.}, which I don't want to do here.
Using Sweedler's notation, we could give a proof of Lemma~\ref{Lemma36}
shorter than the one given above, but still not as short as the obvious proof
of the fact about $k$-algebras (that for any four elements $a$, $b$, $c$, $d$
of a commutative $k$-algebra, we have $\left(  ac\right)  \cdot\left(
bd\right)  =\left(  ab\right)  \cdot\left(  cd\right)  $).

Next, let us show another simple property of arbitrary coalgebras:

\begin{lemma}
\label{lem.coalg.epsi-hom}Let $k$ be a field. Let $C$ be a $k$-coalgebra.

\textbf{(a)} Every $x\in C$ satisfies $\left(  \left(  \varepsilon_{C}%
\otimes\varepsilon_{C}\right)  \circ\Delta_{C}\right)  \left(  x\right)
=\varepsilon_{C}\left(  x\right)  1\otimes1$.

\textbf{(b)} Consider the obvious canonical $k$-coalgebra structure on $k$
(with $\Delta_{k}$ being the canonical isomorphism $k\rightarrow k\otimes k$,
and $\varepsilon_{k}$ being the identity map). The map $\varepsilon_{C}$ is a
$k$-coalgebra homomorphism from $C$ to $k$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.coalg.epsi-hom}.]\textbf{(a)} Let
$\operatorname*{kan}$ be the canonical isomorphism $C\rightarrow C\otimes k$
which maps every $c\in C$ to $c\otimes1$. Since $C$ is a coalgebra, we have
$\left(  \operatorname*{id}\otimes\varepsilon_{C}\right)  \circ\Delta
_{C}=\operatorname*{kan}$ (by the axioms of a coalgebra).

We have $\underbrace{\varepsilon_{C}}_{=\varepsilon_{C}\circ\operatorname*{id}%
}\otimes\underbrace{\varepsilon_{C}}_{=\operatorname*{id}\circ\varepsilon_{C}%
}=\left(  \varepsilon_{C}\circ\operatorname*{id}\right)  \otimes\left(
\operatorname*{id}\circ\varepsilon_{C}\right)  =\left(  \varepsilon_{C}%
\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes\varepsilon_{C}\right)  $ (by an application of (\ref{(X)(X)})), so
that%
\[
\left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ\Delta_{C}=\left(
\varepsilon_{C}\otimes\operatorname*{id}\right)  \circ\underbrace{\left(
\operatorname*{id}\otimes\varepsilon_{C}\right)  \circ\Delta_{C}%
}_{=\operatorname*{kan}}=\left(  \varepsilon_{C}\otimes\operatorname*{id}%
\right)  \circ\operatorname*{kan}.
\]
Hence, for every $x\in C$, we have%
\begin{align*}
\left(  \left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ\Delta
_{C}\right)  \left(  x\right)   &  =\left(  \left(  \varepsilon_{C}%
\otimes\operatorname*{id}\right)  \circ\operatorname*{kan}\right)  \left(
x\right)  =\left(  \varepsilon_{C}\otimes\operatorname*{id}\right)
\underbrace{\left(  \operatorname*{kan}x\right)  }_{\substack{=x\otimes
1\\\text{(by the definition of }\operatorname*{kan}\text{)}}}=\left(
\varepsilon_{C}\otimes\operatorname*{id}\right)  \left(  x\otimes1\right) \\
&  =\varepsilon_{C}\left(  x\right)  \otimes\underbrace{\operatorname*{id}%
\left(  1\right)  }_{=1}=\varepsilon_{C}\left(  x\right)  \otimes
1=\varepsilon_{C}\left(  x\right)  1\otimes1.
\end{align*}
This proves Lemma~\ref{lem.coalg.epsi-hom} \textbf{(a)}.

\textbf{(b)} Let $\operatorname*{kan}\nolimits_{k}:k\rightarrow k\otimes k$ be
the canonical isomorphism which sends every $\lambda\in k$ to $\lambda
1\otimes1=\lambda\otimes1=1\otimes\lambda\in k\otimes k$. Then, $\Delta
_{k}=\operatorname*{kan}\nolimits_{k}$ (by the definition of the coalgebra
structure on $k$). Also, $\varepsilon_{k}=\operatorname*{id}$ (by the
definition of the coalgebra structure on $k$).

Now, every $x\in C$ satisfies
\begin{align*}
&  \left(  \left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ
\Delta_{C}\right)  \left(  x\right) \\
&  =\varepsilon_{C}\left(  x\right)  1\otimes1\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma~\ref{lem.coalg.epsi-hom} \textbf{(a)}}\right) \\
&  =\operatorname*{kan}\nolimits_{k}\left(  \varepsilon_{C}\left(  x\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{because }\operatorname*{kan}%
\nolimits_{k}\left(  \varepsilon_{C}\left(  x\right)  \right)  =\varepsilon
_{C}\left(  x\right)  1\otimes1\text{ (by the definition of }%
\operatorname*{kan}\nolimits_{k}\text{)}\right) \\
&  =\left(  \operatorname*{kan}\nolimits_{k}\circ\varepsilon_{C}\right)
\left(  x\right)  .
\end{align*}
Thus, $\left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ\Delta
_{C}=\underbrace{\operatorname*{kan}\nolimits_{k}}_{=\Delta_{k}}%
\circ\varepsilon_{C}=\Delta_{k}\circ\varepsilon_{C}$. Combined with
$\underbrace{\varepsilon_{k}}_{=\operatorname*{id}}\circ\varepsilon
_{C}=\varepsilon_{C}$, this yields that $\varepsilon_{C}$ is a $k$-coalgebra
homomorphism from $C$ to $k$. This proves Lemma~\ref{lem.coalg.epsi-hom}
\textbf{(b)}.
\end{proof}

Now we will show a less trivial fact, much closer to Lemma~\ref{Lemma31}:

\begin{lemma}
\label{Lemma37} Let $k$ be a field. Let $C$ be a cocommutative $k$-coalgebra.
Let $H$ be a $k$-bialgebra. Let $f\in\mathcal{L}\left(  C,H\right)  $ be an
$\left(  \varepsilon,\varepsilon\right)  $-coderivation. Then, for every
$n\in\mathbb{N}$, we have%
\begin{equation}
\Delta_{H}\circ f^{\ast n}=\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\ast
i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}. \label{37.eq}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma37}.]We are going to prove Lemma~\ref{Lemma37} by
induction over $n$.

\textit{Induction base:} Recall that $H$ is a bialgebra, so that $\Delta
_{H}\left(  1_{H}\right)  =1_{H}\otimes1_{H}$ by the axioms of a bialgebra.

We have $f^{\ast0}=e_{C,H}$ and%
\begin{align}
\sum_{i=0}^{0}\dbinom{0}{i}\left(  f^{\ast i}\otimes f^{\ast\left(
0-i\right)  }\right)  \circ\Delta_{C}  &  =\underbrace{\dbinom{0}{0}}%
_{=1}\left(  \underbrace{f^{\ast0}}_{=e_{C,H}}\otimes\underbrace{f^{\ast0}%
}_{=e_{C,H}}\right)  \circ\Delta_{C}\nonumber\\
&  =\left(  e_{C,H}\otimes e_{C,H}\right)  \circ\Delta_{C}. \label{37.pf.0}%
\end{align}
Since $e_{C,H}=\eta_{H}\circ\varepsilon_{C}$, we have%
\[
e_{C,H}\otimes e_{C,H}=\left(  \eta_{H}\circ\varepsilon_{C}\right)
\otimes\left(  \eta_{H}\circ\varepsilon_{C}\right)  =\left(  \eta_{H}%
\otimes\eta_{H}\right)  \circ\left(  \varepsilon_{C}\otimes\varepsilon
_{C}\right)
\]
(by an application of (\ref{(X)(X)})). Hence, every $x\in C$ satisfies%
\begin{align*}
\left(  \underbrace{\left(  e_{C,H}\otimes e_{C,H}\right)  }_{=\left(
\eta_{H}\otimes\eta_{H}\right)  \circ\left(  \varepsilon_{C}\otimes
\varepsilon_{C}\right)  }\circ\Delta_{C}\right)  \left(  x\right)   &
=\left(  \left(  \eta_{H}\otimes\eta_{H}\right)  \circ\left(  \varepsilon
_{C}\otimes\varepsilon_{C}\right)  \circ\Delta_{C}\right)  \left(  x\right) \\
&  =\left(  \eta_{H}\otimes\eta_{H}\right)  \left(  \underbrace{\left(
\left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ\Delta_{C}\right)
\left(  x\right)  }_{\substack{=\varepsilon_{C}\left(  x\right)
1\otimes1\\\text{(by Lemma~\ref{lem.coalg.epsi-hom} \textbf{(a)})}}}\right) \\
&  =\left(  \eta_{H}\otimes\eta_{H}\right)  \left(  \varepsilon_{C}\left(
x\right)  1\otimes1\right)  =\varepsilon_{C}\left(  x\right)  \eta_{H}\left(
1\right)  \otimes\eta_{H}\left(  1\right) \\
&  =\varepsilon_{C}\left(  x\right)  \underbrace{1_{H}\otimes1_{H}}%
_{=\Delta_{H}\left(  1_{H}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the definition of }\eta_{H}\text{
yields }\eta_{H}\left(  1\right)  =1\cdot1_{H}=1_{H}\right)
\end{align*}
and%
\begin{align*}
\left(  \Delta_{H}\circ\underbrace{e_{C,H}}_{=\eta_{H}\circ\varepsilon_{C}%
}\right)  \left(  x\right)   &  =\left(  \Delta_{H}\circ\eta_{H}%
\circ\varepsilon_{C}\right)  \left(  x\right)  =\Delta_{H}\left(
\underbrace{\eta_{H}\left(  \varepsilon_{C}\left(  x\right)  \right)
}_{\substack{=\varepsilon_{C}\left(  x\right)  1_{H}\\\text{(by the definition
of }\eta_{H}\text{)}}}\right) \\
&  =\Delta_{H}\left(  \varepsilon_{C}\left(  x\right)  1_{H}\right)
=\varepsilon_{C}\left(  x\right)  \underbrace{\Delta_{H}\left(  1_{H}\right)
}_{=1_{H}\otimes1_{H}}=\varepsilon_{C}\left(  x\right)  1_{H}\otimes1_{H}.
\end{align*}
Hence, every $x\in C$ satisfies%
\[
\left(  \left(  e_{C,H}\otimes e_{C,H}\right)  \circ\Delta_{C}\right)  \left(
x\right)  =\left(  \Delta_{H}\circ e_{C,H}\right)  \left(  x\right)  .
\]
In other words, $\left(  e_{C,H}\otimes e_{C,H}\right)  \circ\Delta_{C}%
=\Delta_{H}\circ e_{C,H}$. Now,%
\[
\Delta_{H}\circ\underbrace{f^{\ast0}}_{=e_{C,H}}=\Delta_{H}\circ
e_{C,H}=\left(  e_{C,H}\otimes e_{C,H}\right)  \circ\Delta_{C}=\sum_{i=0}%
^{0}\dbinom{0}{i}\left(  f^{\ast i}\otimes f^{\ast\left(  0-i\right)
}\right)  \circ\Delta_{C}%
\]
(by (\ref{37.pf.0})). In other words, Lemma~\ref{Lemma37} holds for $n=0$.
This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that Lemma~\ref{Lemma37}
holds for $n=N$. To complete the induction, we must show that
Lemma~\ref{Lemma37} also holds for $n=N+1$.

Since Lemma~\ref{Lemma37} holds for $n=N$, we have%
\begin{equation}
\Delta_{H}\circ f^{\ast N}=\sum_{i=0}^{N}\dbinom{N}{i}\left(  f^{\ast
i}\otimes f^{\ast\left(  N-i\right)  }\right)  \circ\Delta_{C}.
\label{37.pf.indass}%
\end{equation}
Denote the $k$-linear map $\sum_{i=0}^{N}\dbinom{N}{i}\left(  f^{\ast
i}\otimes f^{\ast\left(  N-i\right)  }\right)  :C\otimes C\rightarrow H\otimes
H$ by $\Phi$. Then, (\ref{37.pf.indass}) becomes%
\begin{align}
\Delta_{H}\circ f^{\ast N}  &  =\sum_{i=0}^{N}\dbinom{N}{i}\left(  f^{\ast
i}\otimes f^{\ast\left(  N-i\right)  }\right)  \circ\Delta_{C}%
=\underbrace{\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  f^{\ast i}\otimes
f^{\ast\left(  N-i\right)  }\right)  \right)  }_{=\Phi}\circ\Delta
_{C}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right) \nonumber\\
&  =\Phi\circ\Delta_{C}. \label{37.pf.1}%
\end{align}


Now,%
\[
f^{\ast\left(  N+1\right)  }=f\ast f^{\ast N}=\mu_{H}\circ\left(  f\otimes
f^{\ast N}\right)  \circ\Delta_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of convolution}\right)  ,
\]
so that%
\begin{align}
\Delta_{H}\circ f^{\ast\left(  N+1\right)  }  &  =\underbrace{\Delta_{H}%
\circ\mu_{H}}_{\substack{=\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(
\operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)
\circ\left(  \Delta_{H}\otimes\Delta_{H}\right)  \\\text{(by the axioms of a
bialgebra, since }H\text{ is a bialgebra)}}}\circ\left(  f\otimes f^{\ast
N}\right)  \circ\Delta_{C}\nonumber\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  \operatorname{id}%
_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)  \circ\left(
\Delta_{H}\otimes\Delta_{H}\right)  \circ\left(  f\otimes f^{\ast N}\right)
\circ\Delta_{C}. \label{37.pf.4}%
\end{align}
But an application of (\ref{(X)(X)}) yields $\left(  \Delta_{H}\circ f\right)
\otimes\left(  \Delta_{H}\circ f^{\ast N}\right)  =\left(  \Delta_{H}%
\otimes\Delta_{H}\right)  \circ\left(  f\otimes f^{\ast N}\right)  $, so that%
\begin{align*}
&  \left(  \Delta_{H}\otimes\Delta_{H}\right)  \circ\left(  f\otimes f^{\ast
N}\right) \\
&  =\underbrace{\left(  \Delta_{H}\circ f\right)  }_{\substack{=\left(
f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta_{C}\\\text{(since
}f\text{ is an }\left(  \varepsilon,\varepsilon\right)  \text{-coderivation)}%
}}\otimes\underbrace{\left(  \Delta_{H}\circ f^{\ast N}\right)  }%
_{\substack{=\Phi\circ\Delta_{C}\\\text{(by (\ref{37.pf.1}))}}}\\
&  =\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta
_{C}\right)  \otimes\left(  \Phi\circ\Delta_{C}\right) \\
&  =\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \otimes
\Phi\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}), applied to }U=C\text{, }V=C\otimes C\text{,
}W=H\otimes H\text{, }U^{\prime}=C\text{, }V^{\prime}=C\otimes C\text{, }\\
W^{\prime}=H\otimes H\text{, }\alpha=\Delta_{C}\text{, }\beta=\left(  f\otimes
e_{C,H}+e_{C,H}\otimes f\right)  \text{, }\alpha^{\prime}=\Delta_{C}\text{ and
}\beta^{\prime}=\Phi
\end{array}
\right)  .
\end{align*}
Hence, (\ref{37.pf.4}) becomes%
\begin{align}
&  \Delta_{H}\circ f^{\ast\left(  N+1\right)  }\nonumber\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  \operatorname{id}%
_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)  \circ
\underbrace{\left(  \Delta_{H}\otimes\Delta_{H}\right)  \circ\left(  f\otimes
f^{\ast N}\right)  }_{=\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes
f\right)  \otimes\Phi\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
}\circ\Delta_{C}\nonumber\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  \operatorname{id}%
_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)  \circ\left(  \left(
f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \otimes\Phi\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}. \label{37.pf.5}%
\end{align}


But since tensoring of $k$-linear maps is distributive, we have%
\begin{align*}
&  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \otimes\Phi\\
&  =\left(  f\otimes e_{C,H}\right)  \otimes\Phi+\left(  e_{C,H}\otimes
f\right)  \otimes\Phi\\
&  =\underbrace{\left(  f\otimes e_{C,H}\right)  \otimes\left(  \sum_{i=0}%
^{N}\dbinom{N}{i}\left(  f^{\ast i}\otimes f^{\ast\left(  N-i\right)
}\right)  \right)  }_{\substack{=\sum_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes
e_{C,H}\right)  \otimes\left(  f^{\ast i}\otimes f^{\ast\left(  N-i\right)
}\right)  \\\text{(since tensoring of }k\text{-linear maps is }%
k\text{-bilinear)}}}+\underbrace{\left(  e_{C,H}\otimes f\right)
\otimes\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  f^{\ast i}\otimes
f^{\ast\left(  N-i\right)  }\right)  \right)  }_{\substack{=\sum_{i=0}%
^{N}\dbinom{N}{i}\left(  e_{C,H}\otimes f\right)  \otimes\left(  f^{\ast
i}\otimes f^{\ast\left(  N-i\right)  }\right)  \\\text{(since tensoring of
}k\text{-linear maps is }k\text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since tensoring of }k\text{-linear maps
is distributive}\right) \\
&  =\sum_{i=0}^{N}\dbinom{N}{i}\underbrace{\left(  f\otimes e_{C,H}\right)
\otimes\left(  f^{\ast i}\otimes f^{\ast\left(  N-i\right)  }\right)
}_{=f\otimes e_{C,H}\otimes f^{\ast i}\otimes f^{\ast\left(  N-i\right)  }%
}+\sum_{i=0}^{N}\dbinom{N}{i}\underbrace{\left(  e_{C,H}\otimes f\right)
\otimes\left(  f^{\ast i}\otimes f^{\ast\left(  N-i\right)  }\right)
}_{=e_{C,H}\otimes f\otimes f^{\ast i}\otimes f^{\ast\left(  N-i\right)  }}\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}f\otimes e_{C,H}\otimes f^{\ast i}\otimes
f^{\ast\left(  N-i\right)  }+\sum_{i=0}^{N}\dbinom{N}{i}e_{C,H}\otimes
f\otimes f^{\ast i}\otimes f^{\ast\left(  N-i\right)  },
\end{align*}
and thus%
\begin{align*}
&  \left(  \operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}%
_{H}\right)  \circ\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)
\otimes\Phi\right) \\
&  =\left(  \operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}%
_{H}\right)  \circ\left(  \sum_{i=0}^{N}\dbinom{N}{i}f\otimes e_{C,H}\otimes
f^{\ast i}\otimes f^{\ast\left(  N-i\right)  }+\sum_{i=0}^{N}\dbinom{N}%
{i}e_{C,H}\otimes f\otimes f^{\ast i}\otimes f^{\ast\left(  N-i\right)
}\right) \\
&  =\underbrace{\left(  \operatorname{id}_{H}\otimes\tau_{H,H}\otimes
\operatorname{id}_{H}\right)  \circ\left(  \sum_{i=0}^{N}\dbinom{N}{i}f\otimes
e_{C,H}\otimes f^{\ast i}\otimes f^{\ast\left(  N-i\right)  }\right)
}_{\substack{=\sum_{i=0}^{N}\dbinom{N}{i}\left(  \operatorname{id}_{H}%
\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)  \circ\left(  f\otimes
e_{C,H}\otimes f^{\ast i}\otimes f^{\ast\left(  N-i\right)  }\right)
\\\text{(since composition of }k\text{-linear maps is }k\text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \operatorname{id}_{H}\otimes
\tau_{H,H}\otimes\operatorname{id}_{H}\right)  \circ\left(  \sum_{i=0}%
^{N}\dbinom{N}{i}e_{C,H}\otimes f\otimes f^{\ast i}\otimes f^{\ast\left(
N-i\right)  }\right)  }_{\substack{=\sum_{i=0}^{N}\dbinom{N}{i}\left(
\operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)
\circ\left(  e_{C,H}\otimes f\otimes f^{\ast i}\otimes f^{\ast\left(
N-i\right)  }\right)  \\\text{(since composition of }k\text{-linear maps is
}k\text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right) \\
&  =\sum_{i=0}^{N}\dbinom{N}{i}\underbrace{\left(  \operatorname{id}%
_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)  \circ\left(
f\otimes e_{C,H}\otimes f^{\ast i}\otimes f^{\ast\left(  N-i\right)  }\right)
}_{\substack{=\left(  f\otimes f^{\ast i}\otimes e_{C,H}\otimes f^{\ast\left(
N-i\right)  }\right)  \circ\left(  \operatorname{id}_{C}\otimes\tau
_{C,C}\otimes\operatorname{id}_{C}\right)  \\\text{(by
Proposition~\ref{Proposition35} \textbf{(b)}, applied to }C\text{, }C\text{,
}C\text{, }C\text{,}\\H\text{, }H\text{, }H\text{, }H\text{, }f\text{,
}e_{C,H}\text{, }f^{\ast i}\text{, }f^{\ast\left(  N-i\right)  }\text{ instead
of}\\U\text{, }V\text{, }W\text{, }T\text{, }U^{\prime}\text{, }V^{\prime
}\text{, }W^{\prime}\text{, }T^{\prime}\text{, }e\text{, }f\text{, }g\text{,
}h\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=0}^{N}\dbinom{N}{i}\underbrace{\left(
\operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)
\circ\left(  e_{C,H}\otimes f\otimes f^{\ast i}\otimes f^{\ast\left(
N-i\right)  }\right)  }_{\substack{=\left(  e_{C,H}\otimes f^{\ast i}\otimes
f\otimes f^{\ast\left(  N-i\right)  }\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \\\text{(by
Proposition~\ref{Proposition35} \textbf{(b)}, applied to }C\text{, }C\text{,
}C\text{, }C\text{,}\\H\text{, }H\text{, }H\text{, }H\text{, }e_{C,H}\text{,
}f\text{, }f^{\ast i}\text{, }f^{\ast\left(  N-i\right)  }\text{ instead
of}\\U\text{, }V\text{, }W\text{, }T\text{, }U^{\prime}\text{, }V^{\prime
}\text{, }W^{\prime}\text{, }T^{\prime}\text{, }e\text{, }f\text{, }g\text{,
}h\text{)}}}\\
&  =\underbrace{\sum_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes f^{\ast i}\otimes
e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)  \circ\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
}_{\substack{=\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes f^{\ast
i}\otimes e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)  \right)
\circ\left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  \\\text{(since composition of }k\text{-linear maps is
}k\text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{i=0}^{N}\dbinom{N}{i}\left(
e_{C,H}\otimes f^{\ast i}\otimes f\otimes f^{\ast\left(  N-i\right)  }\right)
\circ\left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  }_{\substack{=\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(
e_{C,H}\otimes f^{\ast i}\otimes f\otimes f^{\ast\left(  N-i\right)  }\right)
\right)  \circ\left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes
\operatorname{id}_{C}\right)  \\\text{(since composition of }k\text{-linear
maps is }k\text{-bilinear)}}}
\end{align*}%
\begin{align}
&  =\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes f^{\ast i}\otimes
e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)  \right)  \circ\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(
e_{C,H}\otimes f^{\ast i}\otimes f\otimes f^{\ast\left(  N-i\right)  }\right)
\right)  \circ\left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes
\operatorname{id}_{C}\right) \nonumber\\
&  =\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes f^{\ast i}\otimes
e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)  +\sum_{i=0}^{N}\dbinom
{N}{i}\left(  e_{C,H}\otimes f^{\ast i}\otimes f\otimes f^{\ast\left(
N-i\right)  }\right)  \right)  \circ\left(  \operatorname{id}_{C}\otimes
\tau_{C,C}\otimes\operatorname{id}_{C}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is }k\text{-bilinear}\right)  . \label{37.pf.6}%
\end{align}
Let us denote by $\Psi$ the $k$-linear map
\begin{align*}
&  \sum_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes f^{\ast i}\otimes
e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)  +\sum_{i=0}^{N}\dbinom
{N}{i}\left(  e_{C,H}\otimes f^{\ast i}\otimes f\otimes f^{\ast\left(
N-i\right)  }\right) \\
&  :C\otimes C\otimes C\otimes C\rightarrow H\otimes H\otimes H\otimes H.
\end{align*}
Then, (\ref{37.pf.6}) becomes%
\begin{align}
&  \left(  \operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}%
_{H}\right)  \circ\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)
\otimes\Phi\right) \nonumber\\
&  =\underbrace{\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes f^{\ast
i}\otimes e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)  +\sum_{i=0}%
^{N}\dbinom{N}{i}\left(  e_{C,H}\otimes f^{\ast i}\otimes f\otimes
f^{\ast\left(  N-i\right)  }\right)  \right)  }_{=\Psi}\circ\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
\nonumber\\
&  =\Psi\circ\left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes
\operatorname{id}_{C}\right)  . \label{37.pf.7}%
\end{align}
Hence, (\ref{37.pf.5}) becomes%
\begin{align}
&  \Delta_{H}\circ f^{\ast\left(  N+1\right)  }\nonumber\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\underbrace{\left(
\operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)
\circ\left(  \left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \otimes
\Phi\right)  }_{=\Psi\circ\left(  \operatorname{id}_{C}\otimes\tau
_{C,C}\otimes\operatorname{id}_{C}\right)  }\circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \circ\Delta_{C}\nonumber\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\Psi\circ\underbrace{\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}%
}_{\substack{=\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta
_{C}\\\text{(by (\ref{36.equation}))}}}\nonumber\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\Psi\circ\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}. \label{37.pf.8}%
\end{align}


But since%
\[
\Psi=\sum_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes f^{\ast i}\otimes
e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)  +\sum_{i=0}^{N}\dbinom
{N}{i}\left(  e_{C,H}\otimes f^{\ast i}\otimes f\otimes f^{\ast\left(
N-i\right)  }\right)  ,
\]
we have%
\begin{align*}
&  \left(  \mu_{H}\otimes\mu_{H}\right)  \circ\Psi\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  \sum_{i=0}^{N}%
\dbinom{N}{i}\left(  f\otimes f^{\ast i}\otimes e_{C,H}\otimes f^{\ast\left(
N-i\right)  }\right)  +\sum_{i=0}^{N}\dbinom{N}{i}\left(  e_{C,H}\otimes
f^{\ast i}\otimes f\otimes f^{\ast\left(  N-i\right)  }\right)  \right) \\
&  =\underbrace{\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  \sum
_{i=0}^{N}\dbinom{N}{i}\left(  f\otimes f^{\ast i}\otimes e_{C,H}\otimes
f^{\ast\left(  N-i\right)  }\right)  \right)  }_{\substack{=\sum_{i=0}%
^{N}\dbinom{N}{i}\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  f\otimes
f^{\ast i}\otimes e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)
\\\text{(since composition of }k\text{-linear maps is }k\text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \mu_{H}\otimes\mu_{H}\right)
\circ\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  e_{C,H}\otimes f^{\ast
i}\otimes f\otimes f^{\ast\left(  N-i\right)  }\right)  \right)
}_{\substack{=\sum_{i=0}^{N}\dbinom{N}{i}\left(  \mu_{H}\otimes\mu_{H}\right)
\circ\left(  e_{C,H}\otimes f^{\ast i}\otimes f\otimes f^{\ast\left(
N-i\right)  }\right)  \\\text{(since composition of }k\text{-linear maps is
}k\text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right) \\
&  =\sum_{i=0}^{N}\dbinom{N}{i}\left(  \mu_{H}\otimes\mu_{H}\right)
\circ\left(  f\otimes f^{\ast i}\otimes e_{C,H}\otimes f^{\ast\left(
N-i\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=0}^{N}\dbinom{N}{i}\left(  \mu_{H}\otimes
\mu_{H}\right)  \circ\left(  e_{C,H}\otimes f^{\ast i}\otimes f\otimes
f^{\ast\left(  N-i\right)  }\right)  ,
\end{align*}
so that%
\begin{align*}
&  \left(  \mu_{H}\otimes\mu_{H}\right)  \circ\Psi\circ\left(  \Delta
_{C}\otimes\Delta_{C}\right) \\
&  =\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  \mu_{H}\otimes\mu_{H}\right)
\circ\left(  f\otimes f^{\ast i}\otimes e_{C,H}\otimes f^{\ast\left(
N-i\right)  }\right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  +\sum_{i=0}^{N}\dbinom{N}{i}\left(  \mu
_{H}\otimes\mu_{H}\right)  \circ\left(  e_{C,H}\otimes f^{\ast i}\otimes
f\otimes f^{\ast\left(  N-i\right)  }\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =\underbrace{\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(  \mu_{H}\otimes
\mu_{H}\right)  \circ\left(  f\otimes f^{\ast i}\otimes e_{C,H}\otimes
f^{\ast\left(  N-i\right)  }\right)  \right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  }_{\substack{=\sum_{i=0}^{N}\dbinom{N}{i}\left(
\mu_{H}\otimes\mu_{H}\right)  \circ\left(  f\otimes f^{\ast i}\otimes
e_{C,H}\otimes f^{\ast\left(  N-i\right)  }\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \\\text{(since composition of }k\text{-linear
maps is }k\text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \sum_{i=0}^{N}\dbinom{N}{i}\left(
\mu_{H}\otimes\mu_{H}\right)  \circ\left(  e_{C,H}\otimes f^{\ast i}\otimes
f\otimes f^{\ast\left(  N-i\right)  }\right)  \right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right)  }_{\substack{=\sum_{i=0}^{N}\dbinom{N}{i}\left(
\mu_{H}\otimes\mu_{H}\right)  \circ\left(  e_{C,H}\otimes f^{\ast i}\otimes
f\otimes f^{\ast\left(  N-i\right)  }\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \\\text{(since composition of }k\text{-linear maps
is }k\text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right) \\
&  =\sum_{i=0}^{N}\dbinom{N}{i}\underbrace{\left(  \mu_{H}\otimes\mu
_{H}\right)  \circ\left(  f\otimes f^{\ast i}\otimes e_{C,H}\otimes
f^{\ast\left(  N-i\right)  }\right)  \circ\left(  \Delta_{C}\otimes\Delta
_{C}\right)  }_{\substack{=\left(  f\ast f^{\ast i}\right)  \otimes\left(
e_{C,H}\ast f^{\ast\left(  N-i\right)  }\right)  \\\text{(by
Lemma~\ref{Lemma33}, applied to }A=H\text{, }B=H\text{, }D=C\text{,
}p=f\text{, }q=f^{\ast i}\text{, }r=e_{C,H}\text{ and }s=f^{\ast\left(
N-i\right)  }\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=0}^{N}\dbinom{N}{i}\underbrace{\left(  \mu
_{H}\otimes\mu_{H}\right)  \circ\left(  e_{C,H}\otimes f^{\ast i}\otimes
f\otimes f^{\ast\left(  N-i\right)  }\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  }_{\substack{=\left(  e_{C,H}\ast f^{\ast i}\right)
\otimes\left(  f\ast f^{\ast\left(  N-i\right)  }\right)  \\\text{(by
Lemma~\ref{Lemma33}, applied to }A=H\text{, }B=H\text{, }D=C\text{, }
p=e_{C,H}\text{, }q=f^{\ast i}\text{, }r=f\text{ and }s=f^{\ast\left(
N-i\right)  }\text{)}}}\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}\underbrace{\left(  f\ast f^{\ast i}\right)
}_{=f^{\ast\left(  i+1\right)  }}\otimes\underbrace{\left(  e_{C,H}\ast
f^{\ast\left(  N-i\right)  }\right)  }_{=f^{\ast\left(  N-i\right)  }}%
+\sum_{i=0}^{N}\dbinom{N}{i}\underbrace{\left(  e_{C,H}\ast f^{\ast i}\right)
}_{=f^{\ast i}}\otimes\underbrace{\left(  f\ast f^{\ast\left(  N-i\right)
}\right)  }_{=f^{\ast\left(  \left(  N-i\right)  +1\right)  }=f^{\ast\left(
N+1-i\right)  }}\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast\left(  i+1\right)  }\otimes
f^{\ast\left(  N-i\right)  }+\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\otimes
f^{\ast\left(  N+1-i\right)  }.
\end{align*}
Compared with%
\begin{align*}
&  \sum_{i=0}^{N+1}\dbinom{N+1}{i}f^{\ast i}\otimes f^{\ast\left(
N+1-i\right)  }\\
&  =\sum_{i=0}^{N+1}\left(  \dbinom{N}{i-1}+\dbinom{N}{i}\right)  f^{\ast
i}\otimes f^{\ast\left(  N+1-i\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\dbinom{N+1}{i}=\dbinom{N}%
{i-1}+\dbinom{N}{i}\text{ by the recursion of the binomial coefficients}%
\right) \\
&  =\underbrace{\sum_{i=0}^{N+1}\dbinom{N}{i-1}f^{\ast i}\otimes
f^{\ast\left(  N+1-i\right)  }}_{=\dbinom{N}{0-1}f^{\ast0}\otimes
f^{\ast\left(  N+1-0\right)  }+\sum_{i=1}^{N+1}\dbinom{N}{i-1}f^{\ast
i}\otimes f^{\ast\left(  N+1-i\right)  }}+\underbrace{\sum_{i=0}^{N+1}%
\dbinom{N}{i}f^{\ast i}\otimes f^{\ast\left(  N+1-i\right)  }}_{=\sum
_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\otimes f^{\ast\left(  N+1-i\right)
}+\dbinom{N}{N+1}f^{\ast\left(  N+1\right)  }\otimes f^{\ast\left(
N+1-\left(  N+1\right)  \right)  }}\\
&  =\underbrace{\dbinom{N}{0-1}}_{=0}f^{\ast0}\otimes f^{\ast\left(
N+1-0\right)  }+\underbrace{\sum_{i=1}^{N+1}\dbinom{N}{i-1}f^{\ast i}\otimes
f^{\ast\left(  N+1-i\right)  }}_{\substack{=\sum_{i=0}^{N}\dbinom{N}%
{i+1-1}f^{\ast\left(  i+1\right)  }\otimes f^{\ast\left(  N+1-\left(
i+1\right)  \right)  }\\\text{(here, we substituted }i+1\text{ for }i\text{ in
the sum)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\otimes
f^{\ast\left(  N+1-i\right)  }+\underbrace{\dbinom{N}{N+1}}_{=0}f^{\ast\left(
N+1\right)  }\otimes f^{\ast\left(  N+1-\left(  N+1\right)  \right)  }\\
&  =\underbrace{0f^{\ast0}\otimes f^{\ast\left(  N+1-0\right)  }}_{=0}%
+\sum_{i=0}^{N}\underbrace{\dbinom{N}{i+1-1}}_{=\dbinom{N}{i}}f^{\ast\left(
i+1\right)  }\otimes\underbrace{f^{\ast\left(  N+1-\left(  i+1\right)
\right)  }}_{=f^{\ast\left(  N-i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\otimes
f^{\ast\left(  N+1-i\right)  }+\underbrace{0f^{\ast\left(  N+1\right)
}\otimes f^{\ast\left(  N+1-\left(  N+1\right)  \right)  }}_{=0}\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast\left(  i+1\right)  }\otimes
f^{\ast\left(  N-i\right)  }+\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\otimes
f^{\ast\left(  N+1-i\right)  },
\end{align*}
this yields%
\[
\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\Psi\circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  =\sum_{i=0}^{N+1}\dbinom{N+1}{i}f^{\ast i}\otimes
f^{\ast\left(  N+1-i\right)  }.
\]
Hence, (\ref{37.pf.8}) becomes
\begin{align*}
\Delta_{H}\circ f^{\ast\left(  N+1\right)  }  &  =\underbrace{\left(  \mu
_{H}\otimes\mu_{H}\right)  \circ\Psi\circ\left(  \Delta_{C}\otimes\Delta
_{C}\right)  }_{=\sum_{i=0}^{N+1}\dbinom{N+1}{i}f^{\ast i}\otimes
f^{\ast\left(  N+1-i\right)  }}\circ\Delta_{C}\\
&  =\sum_{i=0}^{N+1}\dbinom{N+1}{i}\left(  f^{\ast i}\otimes f^{\ast\left(
N+1-i\right)  }\right)  \circ\Delta_{C}.
\end{align*}
In other words, Lemma~\ref{Lemma37} holds for $n=N+1$. This completes the
induction step. The induction proof of Lemma~\ref{Lemma37} is thus complete.
\end{proof}

Here is a little brother of Lemma~\ref{Lemma37} (a similar property of
$\varepsilon_{C}$, much easier to prove):

\begin{lemma}
\label{Lemma38} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $H$ be a
$k$-bialgebra. Let $f\in\mathcal{L}\left(  C,H\right)  $ be an $\left(
\varepsilon,\varepsilon\right)  $-coderivation. Then, for every positive
$n\in\mathbb{N}$, we have $\varepsilon_{H}\circ f^{\ast n}=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma38}.]Since $n$ is positive, we have $f^{\ast
n}=f\ast f^{\ast\left(  n-1\right)  }=\mu_{H}\circ\left(  f\otimes
f^{\ast\left(  n-1\right)  }\right)  \circ\Delta_{C}$ (by the definition of
convolution). Since $H$ is a $k$-bialgebra, we have $\varepsilon_{H}\circ
\mu_{H}=\mu_{k}\circ\left(  \varepsilon_{H}\otimes\varepsilon_{H}\right)  $
(by the axioms of a bialgebra), where $\mu_{k}$ is the canonical $k$-module
isomorphism $k\otimes k\rightarrow k$.

Since $f$ is an $\left(  \varepsilon,\varepsilon\right)  $-coderivation, we
have $f\left(  C\right)  \subseteq\operatorname*{Prim}H$ (since by
Theorem~\ref{Theorem23}, we know that $f$ is an $\left(  \varepsilon
,\varepsilon\right)  $-coderivation if and only if $f\left(  C\right)
\subseteq\operatorname*{Prim}H$). Thus, every $x\in C$ satisfies $f\left(
x\right)  \in f\left(  C\right)  \subseteq\operatorname*{Prim}H$, so that
$\varepsilon\left(  f\left(  x\right)  \right)  =0$ (by Remark~\ref{Remark21},
applied to $f\left(  x\right)  $ instead of $x$) and therefore%
\[
\left(  \varepsilon_{H}\circ f\right)  \left(  x\right)  =\varepsilon
_{H}\left(  f\left(  x\right)  \right)  =\varepsilon\left(  f\left(  x\right)
\right)  =0.
\]
Since this holds for every $x\in C$, we thus conclude that $\varepsilon
_{H}\circ f=0$.

Now, for every positive $n\in\mathbb{N}$, we have%
\begin{align*}
\varepsilon_{H}\circ\underbrace{f^{\ast n}}_{=\mu_{H}\circ\left(  f\otimes
f^{\ast\left(  n-1\right)  }\right)  \circ\Delta_{C}}  &
=\underbrace{\varepsilon_{H}\circ\mu_{H}}_{=\mu_{k}\circ\left(  \varepsilon
_{H}\otimes\varepsilon_{H}\right)  }\circ\left(  f\otimes f^{\ast\left(
n-1\right)  }\right)  \circ\Delta_{C}\\
&  =\mu_{k}\circ\left(  \varepsilon_{H}\otimes\varepsilon_{H}\right)
\circ\left(  f\otimes f^{\ast\left(  n-1\right)  }\right)  \circ\Delta_{C}.
\end{align*}
Since an application of (\ref{(X)(X)}) yields $\left(  \varepsilon_{H}\circ
f\right)  \otimes\left(  \varepsilon_{H}\circ f^{\ast\left(  n-1\right)
}\right)  =\left(  \varepsilon_{H}\otimes\varepsilon_{H}\right)  \circ\left(
f\otimes f^{\ast\left(  n-1\right)  }\right)  $, this becomes%
\begin{align*}
\varepsilon_{H}\circ f^{\ast n}  &  =\mu_{k}\circ\underbrace{\left(
\varepsilon_{H}\otimes\varepsilon_{H}\right)  \circ\left(  f\otimes
f^{\ast\left(  n-1\right)  }\right)  }_{=\left(  \varepsilon_{H}\circ
f\right)  \otimes\left(  \varepsilon_{H}\circ f^{\ast\left(  n-1\right)
}\right)  }\circ\Delta_{C}\\
&  =\mu_{k}\circ\left(  \underbrace{\left(  \varepsilon_{H}\circ f\right)
}_{=0}\otimes\left(  \varepsilon_{H}\circ f^{\ast\left(  n-1\right)  }\right)
\right)  \circ\Delta_{C}=\mu_{k}\circ\underbrace{\left(  0\otimes\left(
\varepsilon_{H}\circ f^{\ast\left(  n-1\right)  }\right)  \right)  }_{=0}%
\circ\Delta_{C}\\
&  =\mu_{k}\circ0\circ\Delta_{C}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mu_{k}\text{ is }k\text{-linear}\right)  .
\end{align*}
This proves Lemma~\ref{Lemma38}.
\end{proof}

We can now move on to the proof of Lemma~\ref{Lemma31}:

\begin{proof}
[Proof of Lemma~\ref{Lemma31}.]Assume that $f$ is an $\left(  \varepsilon
,\varepsilon\right)  $-coderivation. Let $x\in C$.

Since $\Delta_{C}\left(  x\right)  $ is a tensor in $C\otimes C$, we can write
$\Delta_{C}\left(  x\right)  $ as $\Delta_{C}\left(  x\right)  =\sum_{j=1}%
^{m}\lambda_{j}a_{j}\otimes b_{j}$ for some $m\in\mathbb{N}$, some elements
$a_{1}$, $a_{2}$, $\ldots$, $a_{m}$ of $C$, and some elements $b_{1}$, $b_{2}%
$, $\ldots$, $b_{m}$ of $C$. Consider this $m$, these $a_{1}$, $a_{2}$,
$\ldots$, $a_{m}$, and these $b_{1}$, $b_{2}$, $\ldots$, $b_{m}$.

Every $n\in\mathbb{N}$ and every $i\in\left\{  0,1,\ldots,n\right\}  $ satisfy%
\begin{align}
&  \dfrac{1}{n!}\dbinom{n}{i}\underbrace{\left(  \left(  f^{\ast i}\otimes
f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right)
}_{=\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \left(
\Delta_{C}\left(  x\right)  \right)  }\nonumber\\
&  =\dfrac{1}{n!}\underbrace{\dbinom{n}{i}}_{=\dfrac{n!}{i!\left(  n-i\right)
!}}\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \left(
\underbrace{\Delta_{C}\left(  x\right)  }_{=\sum_{j=1}^{m}\lambda_{j}%
a_{j}\otimes b_{j}}\right) \nonumber\\
&  =\underbrace{\dfrac{1}{n!}\cdot\dfrac{n!}{i!\left(  n-i\right)  !}%
}_{=\dfrac{1}{i!}\cdot\dfrac{1}{\left(  n-i\right)  !}}\underbrace{\left(
f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \left(  \sum_{j=1}%
^{m}\lambda_{j}a_{j}\otimes b_{j}\right)  }_{=\sum_{j=1}^{m}\lambda_{j}f^{\ast
i}\left(  a_{j}\right)  \otimes f^{\ast\left(  n-i\right)  }\left(
b_{j}\right)  }\nonumber\\
&  =\dfrac{1}{i!}\cdot\dfrac{1}{\left(  n-i\right)  !}\cdot\sum_{j=1}%
^{m}\lambda_{j}f^{\ast i}\left(  a_{j}\right)  \otimes f^{\ast\left(
n-i\right)  }\left(  b_{j}\right)  =\sum_{j=1}^{m}\lambda_{j}\dfrac{f^{\ast
i}\left(  a_{j}\right)  }{i!}\otimes\dfrac{f^{\ast\left(  n-i\right)  }\left(
b_{j}\right)  }{\left(  n-i\right)  !}. \label{31.pf.0}%
\end{align}


We have%
\begin{align*}
&  \left(  \Delta_{H}\circ e^{\ast f}\right)  \left(  x\right) \\
&  =\Delta_{H}\left(  e^{\ast f}\left(  x\right)  \right)  =\Delta_{H}\left(
\sum_{i\geq0}\underbrace{\dfrac{f^{\ast i}\left(  x\right)  }{i!}}_{=\dfrac
{1}{i!}f^{\ast i}\left(  x\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{exp})}\right) \\
&  =\Delta_{H}\left(  \sum_{i\geq0}\dfrac{1}{i!}f^{\ast i}\left(  x\right)
\right)  =\sum_{i\geq0}\dfrac{1}{i!}\underbrace{\Delta_{H}\left(  f^{\ast
i}\left(  x\right)  \right)  }_{=\left(  \Delta_{H}\circ f^{\ast i}\right)
\left(  x\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{H}\text{
is }k\text{-linear}\right) \\
&  =\sum_{i\geq0}\dfrac{1}{i!}\left(  \Delta_{H}\circ f^{\ast i}\right)
\left(  x\right)  =\sum_{n\geq0}\dfrac{1}{n!}\underbrace{\left(  \Delta
_{H}\circ f^{\ast n}\right)  }_{\substack{=\sum_{i=0}^{n}\dbinom{n}{i}\left(
f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta
_{C}\\\text{(by Lemma~\ref{Lemma37})}}}\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }i\text{ as
}n\text{ in the sum}\right) \\
&  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(
f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)
\left(  x\right)  .
\end{align*}
Compared to\footnote{The following computation is a manipulation with infinite
sums. Such manipulations may be dangerous, since infinite sums (even when they
converge) may fail to satisfy some of the rules one would expect infinite sums
to satisfy: For example, switching two summation signs might not always
preserve the sum. However, the specific computation that we are going to do is
safe from such troubles, for the following reasons:
\par
\begin{itemize}
\item Whenever a sum appears in the computation, it has the property that all
but finitely many of its addends are zero.
\par
\item Whenever two summation signs get switched in the computation, they have
the property that all but finitely many addends of the resulting
\textbf{double sum} are zero. (For example, we can transform the
\textquotedblleft$\sum_{j=1}^{m}\sum_{n\geq0}\sum_{\substack{i\geq0;\\i\leq
n}}$\textquotedblright\ into \textquotedblleft$\sum_{n\geq0}\sum
_{\substack{i\geq0;\\i\leq n}}\sum_{j=1}^{m}$\textquotedblright\ in the triple
sum $\sum_{j=1}^{m}\sum_{n\geq0}\sum_{\substack{i\geq0;\\i\leq n}}\lambda
_{j}\dfrac{f^{\ast i}\left(  a_{j}\right)  }{i!}\otimes\dfrac{f^{\ast\left(
n-i\right)  }\left(  b_{j}\right)  }{\left(  n-i\right)  !}$, because all but
finitely many \textbf{triples} $\left(  j,n,i\right)  \in\left\{
1,2,\ldots,m\right\}  \times\mathbb{N}\times\mathbb{N}$ satisfying $i\leq n$
satisfy $\lambda_{j}\dfrac{f^{\ast i}\left(  a_{j}\right)  }{i!}\otimes
\dfrac{f^{\ast\left(  n-i\right)  }\left(  b_{j}\right)  }{\left(  n-i\right)
!}=0$.)
\end{itemize}
}%
\begin{align*}
&  \left(  \left(  e^{\ast f}\otimes e^{\ast f}\right)  \circ\Delta
_{C}\right)  \left(  x\right) \\
&  =\left(  e^{\ast f}\otimes e^{\ast f}\right)  \left(  \underbrace{\Delta
_{C}\left(  x\right)  }_{=\sum_{j=1}^{m}\lambda_{j}a_{j}\otimes b_{j}}\right)
=\left(  e^{\ast f}\otimes e^{\ast f}\right)  \left(  \sum_{j=1}^{m}%
\lambda_{j}a_{j}\otimes b_{j}\right) \\
&  =\sum_{j=1}^{m}\lambda_{j}\underbrace{e^{\ast f}\left(  a_{j}\right)
}_{\substack{=\sum_{i\geq0}\dfrac{f^{\ast i}\left(  a_{j}\right)  }%
{i!}\\\text{(by (\ref{exp}), applied to }a_{j}\\\text{instead of }x\text{)}%
}}\otimes\underbrace{e^{\ast f}\left(  b_{j}\right)  }_{\substack{=\sum
_{i\geq0}\dfrac{f^{\ast i}\left(  b_{j}\right)  }{i!}\\\text{(by (\ref{exp}),
applied to }b_{j}\\\text{instead of }x\text{)}}}\\
&  =\sum_{j=1}^{m}\lambda_{j}\left(  \sum_{i\geq0}\dfrac{f^{\ast i}\left(
a_{j}\right)  }{i!}\right)  \otimes\left(  \sum_{i\geq0}\dfrac{f^{\ast
i}\left(  b_{j}\right)  }{i!}\right)  =\sum_{j=1}^{m}\lambda_{j}\left(
\sum_{i\geq0}\dfrac{f^{\ast i}\left(  a_{j}\right)  }{i!}\right)
\otimes\left(  \sum_{\ell\geq0}\dfrac{f^{\ast\ell}\left(  b_{j}\right)  }%
{\ell!}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }i\text{ as
}\ell\text{ in the third sum}\right) \\
&  =\sum_{j=1}^{m}\lambda_{j}\sum_{i\geq0}\sum_{\ell\geq0}\dfrac{f^{\ast
i}\left(  a_{j}\right)  }{i!}\otimes\dfrac{f^{\ast\ell}\left(  b_{j}\right)
}{\ell!}=\sum_{j=1}^{m}\lambda_{j}\underbrace{\sum_{i\geq0}\sum
_{\substack{n\geq0;\\i\leq n}}}_{=\sum_{n\geq0}\sum_{\substack{i\geq0;\\i\leq
n}}}\dfrac{f^{\ast i}\left(  a_{j}\right)  }{i!}\otimes\dfrac{f^{\ast\left(
n-i\right)  }\left(  b_{j}\right)  }{\left(  n-i\right)  !}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }n\text{ for }%
i+\ell\text{ in the third sum}\right)
\end{align*}
\begin{align*}
&  =\sum_{j=1}^{m}\lambda_{j}\sum_{n\geq0}\sum_{\substack{i\geq0;\\i\leq
n}}\dfrac{f^{\ast i}\left(  a_{j}\right)  }{i!}\otimes\dfrac{f^{\ast\left(
n-i\right)  }\left(  b_{j}\right)  }{\left(  n-i\right)  !}=\sum_{j=1}^{m}%
\sum_{n\geq0}\sum_{\substack{i\geq0;\\i\leq n}}\lambda_{j}\dfrac{f^{\ast
i}\left(  a_{j}\right)  }{i!}\otimes\dfrac{f^{\ast\left(  n-i\right)  }\left(
b_{j}\right)  }{\left(  n-i\right)  !}\\
&  =\sum_{n\geq0}\underbrace{\sum_{\substack{i\geq0;\\i\leq n}}}_{=\sum
_{i=0}^{n}}\underbrace{\sum_{j=1}^{m}\lambda_{j}\dfrac{f^{\ast i}\left(
a_{j}\right)  }{i!}\otimes\dfrac{f^{\ast\left(  n-i\right)  }\left(
b_{j}\right)  }{\left(  n-i\right)  !}}_{\substack{=\dfrac{1}{n!}\dbinom{n}%
{i}\left(  \left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)
\circ\Delta_{C}\right)  \left(  x\right)  \\\text{(by (\ref{31.pf.0}))}}}\\
&  =\sum_{n\geq0}\sum_{i=0}^{n}\dfrac{1}{n!}\dbinom{n}{i}\left(  \left(
f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)
\left(  x\right)  =\sum_{n\geq0}\dfrac{1}{n!}\sum_{i=0}^{n}\dbinom{n}%
{i}\left(  \left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)
\circ\Delta_{C}\right)  \left(  x\right) \\
&  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(
f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)
\left(  x\right)  ,
\end{align*}
this yields%
\[
\left(  \Delta_{H}\circ e^{\ast f}\right)  \left(  x\right)  =\left(  \left(
e^{\ast f}\otimes e^{\ast f}\right)  \circ\Delta_{C}\right)  \left(  x\right)
.
\]
Since this holds for every $x\in C$, we thus conclude that $\Delta_{H}\circ
e^{\ast f}=\left(  e^{\ast f}\otimes e^{\ast f}\right)  \circ\Delta_{C}$.

On the other hand, every $x\in C$ satisfies%
\begin{align*}
\left(  \varepsilon_{H}\circ e^{\ast f}\right)  \left(  x\right)   &
=\varepsilon_{H}\left(  e^{\ast f}\left(  x\right)  \right)  =\varepsilon
_{H}\left(  \sum_{i\geq0}\underbrace{\dfrac{f^{\ast i}\left(  x\right)  }{i!}%
}_{=\dfrac{1}{i!}f^{\ast i}\left(  x\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{exp})}\right) \\
&  =\varepsilon_{H}\left(  \sum_{i\geq0}\dfrac{1}{i!}f^{\ast i}\left(
x\right)  \right)  =\sum_{i\geq0}\dfrac{1}{i!}\varepsilon_{H}\left(  f^{\ast
i}\left(  x\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\varepsilon_{H}\text{ is }k\text{-linear}\right) \\
&  =\underbrace{\dfrac{1}{0!}}_{=\dfrac{1}{1}=1}\varepsilon_{H}\left(
\underbrace{f^{\ast0}}_{=e_{C,H}=\eta_{H}\circ\varepsilon_{C}}\left(
x\right)  \right)  +\sum_{i\geq1}\dfrac{1}{i!}\underbrace{\varepsilon
_{H}\left(  f^{\ast i}\left(  x\right)  \right)  }_{=\left(  \varepsilon
_{H}\circ f^{\ast i}\right)  \left(  x\right)  }\\
&  =\underbrace{\varepsilon_{H}\left(  \left(  \eta_{H}\circ\varepsilon
_{C}\right)  \left(  x\right)  \right)  }_{=\left(  \varepsilon_{H}\circ
\eta_{H}\circ\varepsilon_{C}\right)  \left(  x\right)  }+\sum_{i\geq1}%
\dfrac{1}{i!}\underbrace{\left(  \varepsilon_{H}\circ f^{\ast i}\right)
}_{\substack{=0\\\text{(by Lemma~\ref{Lemma38},}\\\text{applied to
}n=i\text{)}}}\left(  x\right) \\
&  =\left(  \underbrace{\varepsilon_{H}\circ\eta_{H}}%
_{\substack{=\operatorname*{id}\\\text{(by the axioms of a bialgebra,}%
\\\text{since }H\text{ is a bialgebra)}}}\circ\varepsilon_{C}\right)  \left(
x\right)  +\sum_{i\geq1}\dfrac{1}{i!}\underbrace{0\left(  x\right)  }%
_{=0}=\varepsilon_{C}\left(  x\right)  +\underbrace{\sum_{i\geq1}\dfrac{1}%
{i!}0}_{=0}=\varepsilon_{C}\left(  x\right)  .
\end{align*}
Thus, $\varepsilon_{H}\circ e^{\ast f}=\varepsilon_{C}$. Combined with
$\Delta_{H}\circ e^{\ast f}=\left(  e^{\ast f}\otimes e^{\ast f}\right)
\circ\Delta_{C}$, this yields that $e^{\ast f}$ is a $k$-coalgebra
homomorphism. This proves Lemma~\ref{Lemma31}.
\end{proof}

Now that we have proved Lemma~\ref{Lemma31}, let us formulate three easy
corollaries of Lemma~\ref{Lemma33} and Proposition~\ref{Proposition35}
\textbf{(b)}. We will not use them in this paper until \S \ref{S19}, but they
are commonly used elsewhere:

\begin{corollary}
\label{Corollary40} Let $k$ be a field. Let $C$ and $D$ be $k$-coalgebras. Let
$A$ and $B$ be $k$-algebras. Let $p:C\rightarrow A$, $q:C\rightarrow A$,
$r:D\rightarrow B$ and $s:D\rightarrow B$ be four $k$-linear maps. Then,%
\[
\left(  p\otimes r\right)  \ast\left(  q\otimes s\right)  =\left(  p\ast
q\right)  \otimes\left(  r\ast s\right)
\]
(this is an equality between two maps $C\otimes D\rightarrow A\otimes B$).
Here, $\left(  p\otimes r\right)  \ast\left(  q\otimes s\right)  $ denotes the
convolution of the two $k$-linear maps $p\otimes r:C\otimes D\rightarrow
A\otimes B$ and $q\otimes s:C\otimes D\rightarrow A\otimes B$. (This
convolution is well-defined, since $C\otimes D$ is a $k$-coalgebra and
$A\otimes B$ is a $k$-algebra.)
\end{corollary}

\begin{corollary}
\label{Corollary41} Let $k$ be a field. Let $C$ and $D$ be $k$-coalgebras. Let
$A$ and $B$ be $k$-algebras. Let $f:C\rightarrow A$ and $g:D\rightarrow B$ be
two $k$-linear maps. Let maps $e_{C,A}:C\rightarrow A$ and $e_{D,B}%
:D\rightarrow B$ be defined as in Definition~\ref{def.eHA}. Then,%
\[
\left(  f\otimes e_{D,B}\right)  \ast\left(  e_{C,A}\otimes g\right)
=f\otimes g=\left(  e_{C,A}\otimes g\right)  \ast\left(  f\otimes
e_{D,B}\right)
\]
(this is an equality between two maps $C\otimes D\rightarrow A\otimes B$).
Here, $\ast$ denotes the convolution of the $k$-linear maps $C\otimes
D\rightarrow A\otimes B$. (This convolution is well-defined, since $C\otimes
D$ is a $k$-coalgebra and $A\otimes B$ is a $k$-algebra.)
\end{corollary}

\begin{corollary}
\label{Corollary42} Let $k$ be a field. Let $C$ and $D$ be $k$-coalgebras. Let
$A$ and $B$ be $k$-algebras. Let a map $e_{D,B}:D\rightarrow B$ be defined as
in Definition~\ref{def.eHA}. Recall that $C\otimes D$ is a $k$-coalgebra and
$A\otimes B$ is a $k$-algebra; hence, $\mathcal{L}\left(  C\otimes D,A\otimes
B\right)  $ becomes a $k$-algebra with respect to convolution.

\textbf{(a)} For any two $k$-linear maps $f:C\rightarrow A$ and
$g:C\rightarrow A$, we have%
\[
\left(  f\otimes e_{D,B}\right)  \ast\left(  g\otimes e_{D,B}\right)  =\left(
f\ast g\right)  \otimes e_{D,B}%
\]
(this is an equality between two maps $C\otimes D\rightarrow A\otimes B$).

\textbf{(b)} We have $e_{C,A}\otimes e_{D,B}=e_{C\otimes D,A\otimes B}$.

\textbf{(c)} For any $k$-linear map $f:C\rightarrow A$ and any $i\in
\mathbb{N}$, we have $\left(  f\otimes e_{D,B}\right)  ^{\ast i}=f^{\ast
i}\otimes e_{D,B}$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary40}.]By the definition of the $k$-algebra
$A\otimes B$, we have $\mu_{A\otimes B}=\left(  \mu_{A}\otimes\mu_{B}\right)
\circ\left(  \operatorname{id}_{A}\otimes\tau_{B,A}\otimes\operatorname{id}%
_{B}\right)  $, where $\tau_{B,A}$ is defined according to
Definition~\ref{Definition34}.

By the definition of the $k$-coalgebra $C\otimes D$, we have $\Delta_{C\otimes
D}=\left(  \operatorname{id}_{C}\otimes\tau_{C,D}\otimes\operatorname{id}%
_{D}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  $, where
$\tau_{C,D}$ is defined according to Definition~\ref{Definition34}.

Now, by the definition of convolution,%
\begin{align}
&  \left(  p\otimes r\right)  \ast\left(  q\otimes s\right) \nonumber\\
&  =\underbrace{\mu_{A\otimes B}}_{=\left(  \mu_{A}\otimes\mu_{B}\right)
\circ\left(  \operatorname{id}_{A}\otimes\tau_{B,A}\otimes\operatorname{id}%
_{B}\right)  }\circ\underbrace{\left(  \left(  p\otimes r\right)
\otimes\left(  q\otimes s\right)  \right)  }_{=p\otimes r\otimes q\otimes
s}\circ\underbrace{\Delta_{C\otimes D}}_{=\left(  \operatorname{id}_{C}%
\otimes\tau_{C,D}\otimes\operatorname{id}_{D}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{D}\right)  }\nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{B}\right)  \circ\underbrace{\left(
\operatorname{id}_{A}\otimes\tau_{B,A}\otimes\operatorname{id}_{B}\right)
\circ\left(  p\otimes r\otimes q\otimes s\right)  }_{\substack{=\left(
p\otimes q\otimes r\otimes s\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\tau_{D,C}\otimes\operatorname{id}_{D}\right)  \\\text{(by
Proposition~\ref{Proposition35} \textbf{(b)}, applied to}\\U=C\text{,
}V=D\text{, }W=C\text{, }T=D\text{, }U^{\prime}=A\text{, }V^{\prime}=B\text{,
}W^{\prime}=A\text{, }T^{\prime}=B\text{,}\\e=p\text{, }f=r\text{, }g=q\text{
and }h=s\text{)}}}\circ\left(  \operatorname{id}_{C}\otimes\tau_{C,D}%
\otimes\operatorname{id}_{D}\right)  \circ\left(  \Delta_{C}\otimes\Delta
_{D}\right) \nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{B}\right)  \circ\left(  p\otimes q\otimes
r\otimes s\right)  \circ\left(  \operatorname{id}_{C}\otimes\tau_{D,C}%
\otimes\operatorname{id}_{D}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\tau_{C,D}\otimes\operatorname{id}_{D}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{D}\right)  . \label{40.pf.1}%
\end{align}


Recall that any two $k$-vector spaces $V$ and $W$ satisfy $\tau_{V,W}\circ
\tau_{W,V}=\operatorname{id}_{W\otimes V}$. Applied to $V=D$ and $W=C$, this
yields $\tau_{D,C}\circ\tau_{C,D}=\operatorname{id}_{C\otimes D}$. Hence,
$\underbrace{\left(  \tau_{D,C}\circ\tau_{C,D}\right)  }_{=\operatorname{id}%
_{C\otimes D}}\otimes\operatorname{id}_{D}=\operatorname{id}_{C\otimes
D}\otimes\operatorname{id}_{D}=\operatorname{id}_{C\otimes D\otimes D}$. But
since%
\[
\left(  \tau_{D,C}\circ\tau_{C,D}\right)  \otimes\underbrace{\operatorname{id}%
_{D}}_{=\operatorname{id}_{D}\circ\operatorname{id}_{D}}=\left(  \tau
_{D,C}\circ\tau_{C,D}\right)  \otimes\left(  \operatorname{id}_{D}%
\circ\operatorname{id}_{D}\right)  =\left(  \tau_{D,C}\otimes\operatorname{id}%
_{D}\right)  \circ\left(  \tau_{C,D}\otimes\operatorname{id}_{D}\right)
\]
(by (\ref{(X)(X)}), applied to $U=C\otimes D$, $V=D\otimes C$, $W=C\otimes D$,
$U^{\prime}=D$, $V^{\prime}=D$, $W^{\prime}=D$, $\alpha=\tau_{C,D}$,
$\beta=\tau_{D,C}$, $\alpha^{\prime}=\operatorname{id}_{D}$, $\beta^{\prime
}=\operatorname{id}_{D}$), this becomes%
\[
\left(  \tau_{D,C}\otimes\operatorname{id}_{D}\right)  \circ\left(  \tau
_{C,D}\otimes\operatorname{id}_{D}\right)  =\operatorname{id}_{C\otimes
D\otimes D}.
\]
Hence,%
\[
\operatorname{id}_{C}\otimes\underbrace{\left(  \left(  \tau_{D,C}%
\otimes\operatorname{id}_{D}\right)  \circ\left(  \tau_{C,D}\otimes
\operatorname{id}_{D}\right)  \right)  }_{=\operatorname{id}_{C\otimes
D\otimes D}}=\operatorname{id}_{C}\otimes\operatorname{id}_{C\otimes D\otimes
D}=\operatorname{id}_{C\otimes C\otimes D\otimes D}.
\]
But since%
\begin{align*}
\underbrace{\operatorname{id}_{C}}_{=\operatorname{id}_{C}\circ
\operatorname{id}_{C}}\otimes\left(  \left(  \tau_{D,C}\otimes
\operatorname{id}_{D}\right)  \circ\left(  \tau_{C,D}\otimes\operatorname{id}%
_{D}\right)  \right)   &  =\left(  \operatorname{id}_{C}\circ\operatorname{id}%
_{C}\right)  \otimes\left(  \left(  \tau_{D,C}\otimes\operatorname{id}%
_{D}\right)  \circ\left(  \tau_{C,D}\otimes\operatorname{id}_{D}\right)
\right) \\
&  =\left(  \operatorname{id}_{C}\otimes\tau_{D,C}\otimes\operatorname{id}%
_{D}\right)  \circ\left(  \operatorname{id}_{C}\otimes\tau_{C,D}%
\otimes\operatorname{id}_{D}\right)
\end{align*}
(by (\ref{(X)(X)}), applied to $U=C$, $V=C$, $W=C$, $U^{\prime}=C\otimes
D\otimes D$, $V^{\prime}=D\otimes C\otimes D$, $W^{\prime}=C\otimes D\otimes
D$, $\alpha=\operatorname{id}_{C}$, $\beta=\operatorname{id}_{C}$,
$\alpha^{\prime}=\tau_{C,D}\otimes\operatorname{id}_{D}$, $\beta^{\prime}%
=\tau_{D,C}\otimes\operatorname{id}_{D}$), this becomes%
\[
\left(  \operatorname{id}_{C}\otimes\tau_{D,C}\otimes\operatorname{id}%
_{D}\right)  \circ\left(  \operatorname{id}_{C}\otimes\tau_{C,D}%
\otimes\operatorname{id}_{D}\right)  =\operatorname{id}_{C\otimes C\otimes
D\otimes D}.
\]
Hence, (\ref{40.pf.1}) becomes%
\begin{align*}
&  \left(  p\otimes r\right)  \ast\left(  q\otimes s\right) \\
&  =\left(  \mu_{A}\otimes\mu_{B}\right)  \circ\left(  p\otimes q\otimes
r\otimes s\right)  \circ\underbrace{\left(  \operatorname{id}_{C}\otimes
\tau_{D,C}\otimes\operatorname{id}_{D}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\tau_{C,D}\otimes\operatorname{id}_{D}\right)  }%
_{=\operatorname{id}_{C\otimes C\otimes D\otimes D}}\circ\left(  \Delta
_{C}\otimes\Delta_{D}\right) \\
&  =\left(  \mu_{A}\otimes\mu_{B}\right)  \circ\left(  p\otimes q\otimes
r\otimes s\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  =\left(
p\ast q\right)  \otimes\left(  r\ast s\right)
\end{align*}
(by Lemma~\ref{Lemma33}). This proves Corollary~\ref{Corollary40}.
\end{proof}

\begin{proof}
[Proof of Corollary~\ref{Corollary41}.]Applying Corollary~\ref{Corollary40} to
$p=f$, $q=e_{C,A}$, $r=e_{D,B}$ and $s=g$, we obtain%
\[
\left(  f\otimes e_{D,B}\right)  \ast\left(  e_{C,A}\otimes g\right)
=\underbrace{\left(  f\ast e_{C,A}\right)  }_{=f}\otimes\underbrace{\left(
e_{D,B}\ast g\right)  }_{=g}=f\ast g.
\]


Applying Corollary~\ref{Corollary40} to $p=e_{C,A}$, $q=f$, $r=g$ and
$s=e_{D,B}$, we obtain%
\[
\left(  e_{C,A}\otimes g\right)  \ast\left(  f\otimes e_{D,B}\right)
=\underbrace{\left(  e_{C,A}\ast f\right)  }_{=f}\otimes\underbrace{\left(
g\ast e_{D,B}\right)  }_{=g}=f\ast g.
\]


Thus, we have shown that $\left(  f\otimes e_{D,B}\right)  \ast\left(
e_{C,A}\otimes g\right)  =f\ast g=\left(  e_{C,A}\otimes g\right)  \ast\left(
f\otimes e_{D,B}\right)  $. This proves Corollary~\ref{Corollary41}.
\end{proof}

\begin{proof}
[Proof of Corollary~\ref{Corollary42}.]\textbf{(a)} Let $f:C\rightarrow A$ and
$g:C\rightarrow A$ be two $k$-linear maps. Applying
Corollary~\ref{Corollary40} to $p=f$, $q=g$, $r=e_{D,B}$ and $s=e_{D,B}$, we
obtain%
\[
\left(  f\otimes e_{D,B}\right)  \ast\left(  g\otimes e_{D,B}\right)  =\left(
f\ast g\right)  \otimes\underbrace{\left(  e_{D,B}\ast e_{D,B}\right)
}_{=e_{D,B}}=\left(  f\ast g\right)  \otimes e_{D,B}.
\]
This proves Corollary~\ref{Corollary42} \textbf{(a)}.

\textbf{(b)} We have%
\begin{align*}
e_{C\otimes D,A\otimes B}  &  =\underbrace{\eta_{A\otimes B}}_{\substack{=\eta
_{A}\otimes\eta_{B}\\\text{(by the definition of }\\\text{the }k\text{-algebra
}A\otimes B\text{)}}}\circ\underbrace{\varepsilon_{C\otimes D}}%
_{\substack{=\varepsilon_{C}\otimes\varepsilon_{D}\\\text{(by the definition
of }\\\text{the }k\text{-coalgebra }C\otimes D\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }e_{C\otimes
D,A\otimes B}\right) \\
&  =\left(  \eta_{A}\otimes\eta_{B}\right)  \circ\left(  \varepsilon
_{C}\otimes\varepsilon_{D}\right)
\end{align*}
and%
\[
\underbrace{e_{C,A}}_{\substack{=\eta_{A}\circ\varepsilon_{C}\\\text{(by the
definition of }e_{C,A}\text{)}}}\otimes\underbrace{e_{D,B}}_{\substack{=\eta
_{B}\circ\varepsilon_{D}\\\text{(by the definition of }e_{D,B}\text{)}%
}}=\left(  \eta_{A}\circ\varepsilon_{C}\right)  \otimes\left(  \eta_{B}%
\circ\varepsilon_{D}\right)  =\left(  \eta_{A}\otimes\eta_{B}\right)
\circ\left(  \varepsilon_{C}\otimes\varepsilon_{D}\right)
\]
(by (\ref{(X)(X)}), applied to $U=C$, $V=k$, $W=A$, $U^{\prime}=D$,
$V^{\prime}=k$, $W^{\prime}=B$, $\alpha=\varepsilon_{C}$, $\beta=\eta_{A}$,
$\alpha^{\prime}=\varepsilon_{D}$ and $\beta^{\prime}=\eta_{B}$). Thus,
$e_{C\otimes D,A\otimes B}=\left(  \eta_{A}\otimes\eta_{B}\right)
\circ\left(  \varepsilon_{C}\otimes\varepsilon_{D}\right)  =e_{C,A}\otimes
e_{D,B}$. This proves Corollary~\ref{Corollary42} \textbf{(b)}.

\textbf{(c)} We are going to prove Corollary~\ref{Corollary42} \textbf{(c)} by
induction over $i$:

\textit{Induction base:} For every $k$-linear map $f:C\rightarrow A$, we have%
\begin{align*}
\left(  f\otimes e_{D,B}\right)  ^{\ast0}  &  =e_{C\otimes D,A\otimes
B}=\underbrace{e_{C,A}}_{=f^{\ast0}}\otimes e_{D,B}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Corollary~\ref{Corollary42} \textbf{(b)}}\right) \\
&  =f^{\ast0}\otimes e_{D,B}.
\end{align*}
In other words, Corollary~\ref{Corollary42} \textbf{(c)} holds for $i=0$. This
completes the induction base.

\textit{Induction step:} Let $n\in\mathbb{N}$. Assume that
Corollary~\ref{Corollary42} \textbf{(c)} holds for $i=n$. Now we will prove
that Corollary~\ref{Corollary42} \textbf{(c)} also holds for $i=n+1$.

Let $f:C\rightarrow A$ be a $k$-linear map. Then, $\left(  f\otimes
e_{D,B}\right)  ^{\ast n}=f^{\ast n}\otimes e_{D,B}$ (since
Corollary~\ref{Corollary42} \textbf{(c)} holds for $i=n$). Now,%
\begin{align*}
\left(  f\otimes e_{D,B}\right)  ^{\ast\left(  n+1\right)  }  &  =\left(
f\otimes e_{D,B}\right)  \ast\underbrace{\left(  f\otimes e_{D,B}\right)
^{\ast n}}_{=f^{\ast n}\otimes e_{D,B}}=\left(  f\otimes e_{D,B}\right)
\ast\left(  f^{\ast n}\otimes e_{D,B}\right) \\
&  =\underbrace{\left(  f\ast f^{\ast n}\right)  }_{=f^{\ast\left(
n+1\right)  }}\otimes e_{D,B}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Corollary~\ref{Corollary42} \textbf{(a)}, applied to }g=f^{\ast n}\right) \\
&  =f^{\ast\left(  n+1\right)  }\otimes e_{D,B}.
\end{align*}
Thus, Corollary~\ref{Corollary42} \textbf{(c)} holds for $i=n+1$. We thus have
completed the induction step. The induction proof of
Corollary~\ref{Corollary42} \textbf{(c)} is thus complete.
\end{proof}

\section{\label{S7}The product of coalgebra homomorphisms}

The next result will be used in our proof of Lemma~\ref{Lemma32}, but is
actually much more fundamental and important than Lemma~\ref{Lemma32}:

\begin{proposition}
\label{Proposition45} Let $k$ be a field. Let $C$ be a cocommutative
$k$-coalgebra. Let $H$ be a $k$-bialgebra. Let $f:C\rightarrow H$ and
$g:C\rightarrow H$ be two $k$-coalgebra homomorphisms. Then, $f\ast
g:C\rightarrow H$ is also a $k$-coalgebra homomorphism.
\end{proposition}

The proof of this proposition is similar to (but much simpler than!) the
induction step in the proof of Lemma~\ref{Lemma37} above. (It is simpler
because we don't have to work with sums, so we do not need distributivity and
$k$-bilinearity of tensoring and composition.) Here are the details that any
reader should be able to see on his own anyway:

\begin{proof}
[Proof of Proposition~\ref{Proposition45}.]Since $f$ is a $k$-coalgebra
homomorphism, we have $\Delta_{H}\circ f=\left(  f\otimes f\right)
\circ\Delta_{C}$ and $\varepsilon_{H}\circ f=\varepsilon_{C}$. Since $g$ is a
$k$-coalgebra homomorphism, we have $\Delta_{H}\circ g=\left(  g\otimes
g\right)  \circ\Delta_{C}$ and $\varepsilon_{H}\circ g=\varepsilon_{C}$.

By the definition of convolution, $f\ast g=\mu_{H}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$, so that%
\begin{align}
\Delta_{H}\circ\left(  f\ast g\right)   &  =\underbrace{\Delta_{H}\circ\mu
_{H}}_{\substack{=\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(
\operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)
\circ\left(  \Delta_{H}\otimes\Delta_{H}\right)  \\\text{(by the axioms of a
bialgebra, since }H\text{ is a bialgebra)}}}\circ\left(  f\otimes g\right)
\circ\Delta_{C}\nonumber\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  \operatorname{id}%
_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)  \circ\left(
\Delta_{H}\otimes\Delta_{H}\right)  \circ\left(  f\otimes g\right)
\circ\Delta_{C}. \label{45.pf.4}%
\end{align}
But an application of (\ref{(X)(X)}) yields $\left(  \Delta_{H}\circ f\right)
\otimes\left(  \Delta_{H}\circ g\right)  =\left(  \Delta_{H}\otimes\Delta
_{H}\right)  \circ\left(  f\otimes g\right)  $, so that%
\begin{align*}
&  \left(  \Delta_{H}\otimes\Delta_{H}\right)  \circ\left(  f\otimes g\right)
\\
&  =\underbrace{\left(  \Delta_{H}\circ f\right)  }_{=\left(  f\otimes
f\right)  \circ\Delta_{C}}\otimes\underbrace{\left(  \Delta_{H}\circ g\right)
}_{=\left(  g\otimes g\right)  \circ\Delta_{C}}\\
&  =\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)  \otimes\left(
\left(  g\otimes g\right)  \circ\Delta_{C}\right) \\
&  =\left(  f\otimes f\otimes g\otimes g\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}), applied to }U=C\text{, }V=C\otimes C\text{,
}W=H\otimes H\text{, }U^{\prime}=C\text{, }V^{\prime}=C\otimes C\text{, }\\
W^{\prime}=H\otimes H\text{, }\alpha=\Delta_{C}\text{, }\beta=f\otimes
f\text{, }\alpha^{\prime}=\Delta_{C}\text{ and }\beta^{\prime}=g\otimes g
\end{array}
\right)  .
\end{align*}
Hence, (\ref{45.pf.4}) becomes%
\begin{align*}
&  \Delta_{H}\circ\left(  f\ast g\right) \\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  \operatorname{id}%
_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)  \circ
\underbrace{\left(  \Delta_{H}\otimes\Delta_{H}\right)  \circ\left(  f\otimes
g\right)  }_{=\left(  f\otimes f\otimes g\otimes g\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right)  }\circ\Delta_{C}\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\underbrace{\left(
\operatorname{id}_{H}\otimes\tau_{H,H}\otimes\operatorname{id}_{H}\right)
\circ\left(  f\otimes f\otimes g\otimes g\right)  }_{\substack{=\left(
f\otimes g\otimes f\otimes g\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \\\text{(by
Proposition~\ref{Proposition35} \textbf{(b)}, applied to }C\text{, }C\text{,
}C\text{, }C\text{,}\\H\text{, }H\text{, }H\text{, }H\text{, }f\text{,
}f\text{, }g\text{, }g\text{ instead of}\\U\text{, }V\text{, }W\text{,
}T\text{, }U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }T^{\prime
}\text{, }e\text{, }f\text{, }g\text{, }h\text{)}}}\circ\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\\
&  =\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  f\otimes g\otimes
f\otimes g\right)  \circ\underbrace{\left(  \operatorname{id}_{C}\otimes
\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \circ\Delta_{C}}_{\substack{=\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\\\text{(by (\ref{36.equation}%
))}}}\\
&  =\underbrace{\left(  \mu_{H}\otimes\mu_{H}\right)  \circ\left(  f\otimes
g\otimes f\otimes g\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
}_{\substack{=\left(  f\ast g\right)  \otimes\left(  f\ast g\right)
\\\text{(by Lemma~\ref{Lemma33}, applied to }A=H\text{, }B=H\text{,
}D=C\text{, }p=f\text{, }q=g\text{, }r=f\text{ and }s=g\text{)}}}\circ
\Delta_{C}\\
&  =\left(  \left(  f\ast g\right)  \otimes\left(  f\ast g\right)  \right)
\circ\Delta_{C}.
\end{align*}


Also, let us denote by $\mu_{k}$ the canonical isomorphism $k\otimes
k\rightarrow k$. Then, $\mu_{k}$ is the multiplication map of the $k$-algebra
$k$. By the axioms of a bialgebra, $\varepsilon_{H}$ is a $k$-algebra
homomorphism (since $H$ is a bialgebra); thus, $\varepsilon_{H}\circ\mu
_{H}=\mu_{k}\circ\left(  \varepsilon_{H}\otimes\varepsilon_{H}\right)  $. On
the other hand, $\left(  \varepsilon_{H}\circ f\right)  \otimes\left(
\varepsilon_{H}\circ g\right)  =\left(  \varepsilon_{H}\otimes\varepsilon
_{H}\right)  \circ\left(  f\otimes g\right)  $ (by an application of
(\ref{(X)(X)})). Using these equalities, we have%
\begin{align*}
\varepsilon_{H}\circ\underbrace{\left(  f\ast g\right)  }_{=\mu_{H}%
\circ\left(  f\otimes g\right)  \circ\Delta_{C}}  &  =\underbrace{\varepsilon
_{H}\circ\mu_{H}}_{=\mu_{k}\circ\left(  \varepsilon_{H}\otimes\varepsilon
_{H}\right)  }\circ\left(  f\otimes g\right)  \circ\Delta_{C}=\mu_{k}%
\circ\underbrace{\left(  \varepsilon_{H}\otimes\varepsilon_{H}\right)
\circ\left(  f\otimes g\right)  }_{=\left(  \varepsilon_{H}\circ f\right)
\otimes\left(  \varepsilon_{H}\circ g\right)  }\circ\Delta_{C}\\
&  =\mu_{k}\circ\left(  \underbrace{\left(  \varepsilon_{H}\circ f\right)
}_{=\varepsilon_{C}}\otimes\underbrace{\left(  \varepsilon_{H}\circ g\right)
}_{=\varepsilon_{C}}\right)  \circ\Delta_{C}=\mu_{k}\circ\left(
\varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ\Delta_{C}.
\end{align*}
But from the axioms of a coalgebra, it is easy to see that $\mu_{k}%
\circ\left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ\Delta
_{C}=\varepsilon_{C}$\ \ \ \ \footnote{\textit{Proof.} Every $x\in C$
satisfies $\left(  \left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)
\circ\Delta_{C}\right)  \left(  x\right)  =\varepsilon_{C}\left(  x\right)
1\otimes1$ (by Lemma~\ref{lem.coalg.epsi-hom} \textbf{(a)}) and thus%
\begin{align*}
\left(  \mu_{k}\circ\left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)
\circ\Delta_{C}\right)  \left(  x\right)   &  =\mu_{k}\left(
\underbrace{\left(  \left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)
\circ\Delta_{C}\right)  \left(  x\right)  }_{=\varepsilon_{C}\left(  x\right)
1\otimes1}\right)  =\mu_{k}\left(  \varepsilon_{C}\left(  x\right)
1\otimes1\right) \\
&  =\varepsilon_{C}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mu_{k}\right)  .
\end{align*}
Hence, $\mu_{k}\circ\left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)
\circ\Delta_{C}=\varepsilon_{C}$, qed.}. Altogether, we thus have%
\[
\varepsilon_{H}\circ\left(  f\ast g\right)  =\mu_{k}\circ\left(
\varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ\Delta_{C}=\varepsilon
_{C}.
\]
Combined with $\Delta_{H}\circ\left(  f\ast g\right)  =\left(  \left(  f\ast
g\right)  \otimes\left(  f\ast g\right)  \right)  \circ\Delta_{C}$, this
yields that $f\ast g$ is a $k$-coalgebra homomorphism. This proves
Proposition~\ref{Proposition45}.
\end{proof}

As a consequence, we have:

\begin{corollary}
\label{Corollary46} Let $k$ be a field. Let $C$ be a cocommutative
$k$-coalgebra. Let $H$ be a $k$-bialgebra. Let $f:C\rightarrow H$ be a
$k$-coalgebra homomorphism. Let $n\in\mathbb{N}$. Then, $f^{\ast
n}:C\rightarrow H$ is also a $k$-coalgebra homomorphism.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary46}.]We are going to prove
Corollary~\ref{Corollary46} by induction over $n$.

\textit{Induction base:} Consider the obvious canonical $k$-coalgebra
structure on $k$ (with $\Delta_{k}$ being the canonical isomorphism
$k\rightarrow k\otimes k$, and $\varepsilon_{k}$ being the identity map).
Lemma~\ref{lem.coalg.epsi-hom} \textbf{(b)} shows that $\varepsilon
_{C}:C\rightarrow k$ is a $k$-coalgebra homomorphism. Combined with the fact
that $\eta_{H}:k\rightarrow H$ is a $k$-coalgebra homomorphism (because $H$ is
a $k$-bialgebra), this yields that $\eta_{H}\circ\varepsilon_{C}$ is a
$k$-coalgebra homomorphism (since the composition of two $k$-coalgebra
homomorphisms is a $k$-coalgebra homomorphism). Since $f^{\ast0}=e_{C,H}%
=\eta_{H}\circ\varepsilon_{C}$ (by the definition of $e_{C,H}$), this yields
that $f^{\ast0}$ is a $k$-coalgebra homomorphism. In other words,
Corollary~\ref{Corollary46} holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
Corollary~\ref{Corollary46} holds for $n=N$. We must now prove that
Corollary~\ref{Corollary46} also holds for $n=N+1$.

Since Corollary~\ref{Corollary46} holds for $n=N$, we know that $f^{\ast N}$
is a $k$-coalgebra homomorphism. Proposition~\ref{Proposition45} (applied to
$g=f^{\ast N}$) now yields that $f\ast f^{\ast N}$ is a $k$-coalgebra
homomorphism. Since $f\ast f^{\ast N}=f^{\ast\left(  N+1\right)  }$, this
yields that $f^{\ast\left(  N+1\right)  }$ is a $k$-coalgebra homomorphism. In
other words, Corollary~\ref{Corollary46} holds for $n=N+1$. This completes the
induction step. The induction proof of Corollary~\ref{Corollary46} is thus complete.
\end{proof}

\section{\label{S8}The addition-to-multiplication property of the exponent}

One more thing we need about exponentiation (the map $\mathfrak{g}\left(
C,H\right)  \rightarrow G\left(  C,H\right)  $, $f\mapsto e^{\ast f}$) is the
following property (which gives us the moral claim to call it exponentiation!):

\begin{proposition}
\label{Proposition47} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered $k$-coalgebra. Let $H$ be a $k$-algebra. Let $f\in
\mathfrak{g}\left(  C,H\right)  $ and $g\in\mathfrak{g}\left(  C,H\right)  $
be such that $f\ast g=g\ast f$. Then, $e^{\ast\left(  f+g\right)  }=e^{\ast
f}\ast e^{\ast g}$.
\end{proposition}

Note that the condition $f\ast g=g\ast f$ in this proposition can be replaced
by the stronger condition that $C$ be cocommutative and $H$ be commutative: In
fact, whenever $C$ is cocommutative and $H$ is commutative, it is easy to see
that any two $k$-linear maps $f:C\rightarrow H$ and $g:C\rightarrow H$ satisfy
$f\ast g=g\ast f$. We are not going to use and prove this, though.

Before we prove Proposition~\ref{Proposition47}, let us state a basic fact in
noncommutative algebra:

\begin{proposition}
\label{Proposition47a} Let $k$ be a field. Let $A$ be a $k$-algebra. Let $S$
be a subset of $A$ such that the $k$-algebra $A$ is generated by $S$. Assume
that any two elements of $S$ commute. Then, the $k$-algebra $A$ is commutative.
\end{proposition}

Proposition~\ref{Proposition47a} is usually stated in the form
\textquotedblleft a $k$-algebra generated by pairwise commuting generators
must be commutative\textquotedblright.

\begin{proof}
[Proof of Proposition~\ref{Proposition47a}.]For every $a\in A$, let $Z\left(
a\right)  $ denote the subset
\[
\left\{  b\in A\ \mid\ ab=ba\right\}
\]
of $A$. Then, for every $a\in A$, the subset $Z\left(  a\right)  $ of $A$ is a
$k$-subalgebra of $A$\ \ \ \ \footnote{\textit{Proof.} Let $a\in A$. The
definition of $Z\left(  a\right)  $ yields $Z\left(  a\right)  =\left\{  b\in
A\ \mid\ ab=ba\right\}  $.
\par
We know that $0$ is an element of $A$ and satisfies $a\cdot0=0\cdot a$ (since
$a\cdot0=0=0\cdot a$). In other words, $0\in\left\{  b\in A\ \mid
\ ab=ba\right\}  =Z\left(  a\right)  $.
\par
We know that $1$ is an element of $A$ and satisfies $a\cdot1=1\cdot a$ (since
$a\cdot1=a=1\cdot a$). In other words, $1\in\left\{  b\in A\ \mid
\ ab=ba\right\}  =Z\left(  a\right)  $.
\par
Let $c\in Z\left(  a\right)  $ and $d\in Z\left(  a\right)  $. Then, $c\in
Z\left(  a\right)  =\left\{  b\in A\ \mid\ ab=ba\right\}  $. In other words,
$c$ is an element of $A$ and satisfies $ac=ca$. Also, $d\in Z\left(  a\right)
=\left\{  b\in A\ \mid\ ab=ba\right\}  $. In other words, $d$ is an element of
$A$ and satisfies $ad=da$. Now, $a\left(  c+d\right)  =\underbrace{ac}%
_{=ca}+\underbrace{ad}_{=da}=ca+da=\left(  c+d\right)  a$ and $\underbrace{ac}%
_{=ca}d=c\underbrace{ad}_{=da}=cda$.
\par
So we know that $c+d$ is an element of $A$ and satisfies $a\left(  c+d\right)
=\left(  c+d\right)  a$. In other words, $c+d\in\left\{  b\in A\ \mid
\ ab=ba\right\}  =Z\left(  a\right)  $. Also, $cd$ is an element of $A$ and
satisfies $acd=cda$. In other words, $cd\in\left\{  b\in A\ \mid
\ ab=ba\right\}  =Z\left(  a\right)  $.
\par
Now, forget that we fixed $c$ and $d$. We thus have shown that every $c\in
Z\left(  a\right)  $ and $d\in Z\left(  a\right)  $ satisfy $c+d\in Z\left(
a\right)  $ and $cd\in Z\left(  a\right)  $.
\par
Next, let $\lambda\in k$ and $c\in Z\left(  a\right)  $. Then, $c\in Z\left(
a\right)  =\left\{  b\in A\ \mid\ ab=ba\right\}  $. In other words, $c$ is an
element of $A$ and satisfies $ac=ca$. Now, $\lambda c$ is an element of $A$
and satisfies $a\left(  \lambda c\right)  =\left(  \lambda c\right)  a$ (since
$a\left(  \lambda c\right)  =\lambda\underbrace{ac}_{=ca}=\lambda ca=\left(
\lambda c\right)  a$). In other words, $\lambda c\in\left\{  b\in
A\ \mid\ ab=ba\right\}  =Z\left(  a\right)  $. So we know that $\lambda c$ is
an element of $A$ and satisfies $\lambda c\in Z\left(  a\right)  $. Combined
with the fact that every $c\in Z\left(  a\right)  $ and $d\in Z\left(
a\right)  $ satisfy $c+d\in Z\left(  a\right)  $, and combined with the fact
that $0\in Z\left(  a\right)  $, this yields that $Z\left(  a\right)  $ is a
$k$-vector subspace of $A$. Combined with the fact that $1\in Z\left(
a\right)  $, and combined with the fact that $c\in Z\left(  a\right)  $ and
$d\in Z\left(  a\right)  $ satisfy $cd\in Z\left(  a\right)  $, this yields
that $Z\left(  a\right)  $ is a $k$-subalgebra of $A$, qed.}.

The $k$-algebra $A$ is generated by $S$. In other words, the $k$-subalgebra of
$A$ generated by $S$ is $A$. In other words, $A$ is the $k$-subalgebra of $A$
generated by $S$. In other words, $A$ is the smallest $k$-subalgebra of $A$
which contains $S$ as a subset (because the $k$-subalgebra of $A$ generated by
$S$ is the smallest $k$-subalgebra of $A$ which contains $S$ as a subset).
Hence,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{whenever }U\text{ is a }k\text{-subalgebra of }A\text{ which contains
}S\text{ as}\\
\text{a subset, we must necessarily have }A\subseteq U
\end{array}
\right)  . \label{47a.pf.1}%
\end{equation}


Now, fix an $s\in S$. We know that for every $a\in A$, the subset $Z\left(
a\right)  $ of $A$ is a $k$-subalgebra of $A$. Applying this to $a=s$, we
obtain the following: The subset $Z\left(  s\right)  $ of $A$ is a
$k$-subalgebra of $A$ (since we know that for every $a\in A$, the subset
$Z\left(  a\right)  $ of $A$ is a $k$-subalgebra of $A$). But we have
$S\subseteq Z\left(  s\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $t\in
S$. Then, $t$ is an element of $S$. Also, the elements $s$ and $t$ of $S$
commute (since any two elements of $S$ commute). Thus, $st=ts$. Now, the
definition of $Z\left(  s\right)  $ yields $Z\left(  s\right)  =\left\{  b\in
A\ \mid\ sb=bs\right\}  $. But $t$ is an element of $A$ and satisfying
$st=ts$. In other words, $t\in\left\{  b\in A\ \mid\ sb=bs\right\}  =Z\left(
s\right)  $.
\par
Now, forget that we fixed $t$. We thus have proven that every $t\in S$
satisfies $t\in Z\left(  s\right)  $. In other words, $S\subseteq Z\left(
s\right)  $, qed.}. Hence, $Z\left(  s\right)  $ contains $S$ as a subset.

Now, we know that $Z\left(  s\right)  $ is a $k$-subalgebra of $A$ which
contains $S$ as a subset. Hence, (\ref{47a.pf.1}) (applied to $U=Z\left(
s\right)  $) yields $Z\left(  s\right)  =A$. Thus, $A=Z\left(  s\right)
=\left\{  b\in A\ \mid\ sb=bs\right\}  $ (by the definition of $Z\left(
s\right)  $).

Now, forget that we fixed $s\in S$. We thus have shown that every $s\in S$
satisfies $A=\left\{  b\in A\ \mid\ sb=bs\right\}  $.

Now, fix some $a\in A$. Recall that the subset $Z\left(  a\right)  $ of $A$ is
a $k$-subalgebra of $A$. Moreover, $S\subseteq Z\left(  a\right)
$.\ \ \ \ \footnote{\textit{Proof.} Let $s\in S$. We have $a\in A=\left\{
b\in A\ \mid\ sb=bs\right\}  $. In other words, $a$ is an element of $A$ and
satisfies $sa=as$. Thus, $as=sa$.
\par
But recall that the definition of $Z\left(  a\right)  $ yields $Z\left(
a\right)  =\left\{  b\in A\ \mid\ ab=ba\right\}  $. Now, $s$ is an element of
$a$ and satisfies $as=sa$. In other words, $s\in\left\{  b\in A\ \mid
\ ab=ba\right\}  =Z\left(  a\right)  $.
\par
Now, forget that we fixed $s$. We thus have proven that every $s\in S$
satisfies $s\in Z\left(  a\right)  $. In other words, $S\subseteq Z\left(
a\right)  $, qed.} In other words, $Z\left(  a\right)  $ contains $S$ as a subset.

Now, we know that $Z\left(  a\right)  $ is a $k$-subalgebra of $A$ which
contains $A$ as a subset. Hence, (\ref{47a.pf.1}) (applied to $U=Z\left(
a\right)  $) yields $Z\left(  a\right)  =A$. Hence, $A=Z\left(  a\right)  $.

Now let $c\in A$ be arbitrary. Then, $c\in A=Z\left(  a\right)  =\left\{  b\in
A\ \mid\ ab=ba\right\}  $ (by the definition of $Z\left(  a\right)  $). In
other words, $c$ is an element of $A$ and satisfies $ac=ca$.

Now, forget that we fixed $a$ and $c$. We thus have proven that every $a\in A$
and $c\in A$ satisfy $ac=ca$. In other words, the $k$-algebra $A$ is
commutative. Proposition~\ref{Proposition47a} is proven.
\end{proof}

We record a corollary of Proposition~\ref{Proposition47a} for easy reference:

\begin{corollary}
\label{Corollary47b} Let $k$ be a field. Let $A$ be a $k$-algebra. Let $f$ and
$g$ be two elements of $A$. Assume that $f$ and $g$ commute. Let
$\mathfrak{H}$ be the $k$-subalgebra of $A$ generated by $f$ and $g$. Then,
the $k$-algebra $\mathfrak{H}$ is commutative.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary47b}.]Let $S=\left\{  f,g\right\}  $. Recall
that $\mathfrak{H}$ is the $k$-subalgebra of $A$ generated by $f$ and $g$. In
other words,%
\begin{align*}
\mathfrak{H}  &  =\left(  \text{the }k\text{-subalgebra of }A\text{ generated
by }f\text{ and }g\right) \\
&  =\left(  \text{the }k\text{-subalgebra of }A\text{ generated by
}\underbrace{\left\{  f,g\right\}  }_{=S}\right) \\
&  =\left(  \text{the }k\text{-subalgebra of }A\text{ generated by }S\right)
\supseteq S.
\end{align*}
Thus, $S$ is a subset of $\mathfrak{H}$. Moreover, the $k$-algebra
$\mathfrak{H}$ is generated by $S$ (because $\mathfrak{H}=\left(  \text{the
}k\text{-subalgebra of }A\text{ generated by }S\right)  $). Finally, any two
elements of $S$ commute.\footnote{\textit{Proof.} Let $u$ and $v$ be two
elements of $S$. We are going to prove that $uv=vu$.
\par
We have $u\in S=\left\{  f,g\right\}  $. Hence, we must be in one of the
following two cases:
\par
\textit{Case 1:} We have $u=f$.
\par
\textit{Case 2:} We have $u=g$.
\par
Let us consider Case 1 first. In this case, we have $u=f$.
\par
We have $v\in S=\left\{  f,g\right\}  $. Hence, we must be in one of the
following two subcases:
\par
\textit{Subcase 1.1:} We have $v=f$.
\par
\textit{Subcase 1.2:} We have $v=g$.
\par
Let us first consider Subcase 1.1. In this subcase, we have $v=f$. Now,
$\underbrace{u}_{=f}\underbrace{v}_{=f}=ff$ and $\underbrace{v}_{=f}%
\underbrace{u}_{=f}=ff$. Hence, $uv=ff=vu$. Thus, $uv=vu$ is proven in Subcase
1.1.
\par
Let us next consider Subcase 1.2. In this subcase, we have $v=g$. Now,
$\underbrace{u}_{=f}\underbrace{v}_{=g}=fg$ and $\underbrace{v}_{=g}%
\underbrace{u}_{=f}=gf$. Hence, $uv=fg=gf=vu$. Thus, $uv=vu$ is proven in
Subcase 1.2.
\par
We now have proven $uv=vu$ in each of the two Subcases 1.1 and 1.2. Since
these two Subcases cover the whole Case 1, this yields that $uv=vu$ always
holds in Case 1.
\par
Let us now consider Case 2. In this case, we must have $u=g$.
\par
We have $v\in S=\left\{  f,g\right\}  $. Hence, we must be in one of the
following two subcases:
\par
\textit{Subcase 2.1:} We have $v=f$.
\par
\textit{Subcase 2.2:} We have $v=g$.
\par
Let us first consider Subcase 2.1. In this subcase, we have $v=f$. Now,
$\underbrace{u}_{=g}\underbrace{v}_{=f}=gf$ and $\underbrace{v}_{=f}%
\underbrace{u}_{=g}=fg=gf$. Hence, $uv=gf=vu$. Thus, $uv=vu$ is proven in
Subcase 2.1.
\par
Let us next consider Subcase 2.2. In this subcase, we have $v=g$. Now,
$\underbrace{u}_{=g}\underbrace{v}_{=g}=gg$ and $\underbrace{v}_{=g}%
\underbrace{u}_{=g}=gg$. Hence, $uv=gg=vu$. Thus, $uv=vu$ is proven in Subcase
2.2.
\par
We now have proven $uv=vu$ in each of the two Subcases 2.1 and 2.2. Since
these two Subcases cover the whole Case 2, this yields that $uv=vu$ always
holds in Case 2.
\par
We now have proven $uv=vu$ in each of the two Cases 1 and 2. Since these two
Cases cover all possibilities, this yields that $uv=vu$ always holds. In other
words, $u$ and $v$ commute.
\par
Now, forget that we fixed $u$ and $v$. We thus have shown that $u$ and $v$
commute for any two elements $u$ and $v$ of $S$. In other words, any two
elements of $S$ commute, qed.} Hence, Proposition~\ref{Proposition47a}
(applied to $\mathfrak{H}$ instead of $A$) yields that the $k$-algebra
$\mathfrak{H}$ is commutative. This proves Corollary~\ref{Corollary47b}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition47}.]Let $x\in C$. By (\ref{exp})
(applied to $f\ast g$ instead of $f$), we have%
\begin{equation}
e^{\ast\left(  f+g\right)  }\left(  x\right)  =\sum_{i\geq0}\dfrac{\left(
f+g\right)  ^{\ast i}\left(  x\right)  }{i!}. \label{47.pf.1}%
\end{equation}


We have $f\ast g=g\ast f$. In other words, $f$ and $g$ commute (as elements of
$\mathcal{L}\left(  C,H\right)  $). Now, let $\mathfrak{H}$ be the
$k$-subalgebra of $\mathcal{L}\left(  C,H\right)  $ generated by $f$ and $g$.
Then, the $k$-algebra $\mathfrak{H}$ is commutative (by
Corollary~\ref{Corollary47b}, applied to $A=\mathcal{L}\left(  C,H\right)  $).

Thus, $\mathfrak{H}$ is a commutative $k$-algebra. Hence, we can calculate
inside $\mathfrak{H}$ as in any commutative algebra; in particular, we thus
obtain $\left(  f+g\right)  ^{\ast i}=\sum_{j=0}^{i}\dbinom{i}{j}f^{\ast
j}\ast g^{\ast\left(  i-j\right)  }$ (by the binomial formula) for every
$i\in\mathbb{N}$. Hence, (\ref{47.pf.1}) becomes%
\begin{align}
e^{\ast\left(  f+g\right)  }\left(  x\right)   &  =\sum_{i\geq0}\dfrac{\left(
\sum_{j=0}^{i}\dbinom{i}{j}f^{\ast j}\ast g^{\ast\left(  i-j\right)  }\right)
\left(  x\right)  }{i!}=\sum_{i\geq0}\dfrac{\sum_{j=0}^{i}\dbinom{i}{j}\left(
f^{\ast j}\ast g^{\ast\left(  i-j\right)  }\right)  \left(  x\right)  }%
{i!}\nonumber\\
&  =\sum_{i\geq0}\dfrac{1}{i!}\sum_{j=0}^{i}\dbinom{i}{j}\left(  f^{\ast
j}\ast g^{\ast\left(  i-j\right)  }\right)  \left(  x\right) \nonumber\\
&  =\sum_{i\geq0}\underbrace{\sum_{j=0}^{i}}_{=\sum_{\substack{j\geq0;\\j\leq
i}}}\underbrace{\dfrac{1}{i!}\dbinom{i}{j}}_{\substack{=\dfrac{1}{j!\left(
i-j\right)  !}\\\text{(since }\dbinom{i}{j}=\dfrac{i!}{j!\left(  i-j\right)
!}\text{)}}}\left(  f^{\ast j}\ast g^{\ast\left(  i-j\right)  }\right)
\left(  x\right) \nonumber\\
&  =\sum_{i\geq0}\sum_{\substack{j\geq0;\\j\leq i}}\dfrac{1}{j!\left(
i-j\right)  !}\left(  f^{\ast j}\ast g^{\ast\left(  i-j\right)  }\right)
\left(  x\right)  . \label{47.pf.5a}%
\end{align}


Since $\Delta_{C}\left(  x\right)  $ is a tensor in $C\otimes C$, we can write
$\Delta_{C}\left(  x\right)  $ as $\Delta_{C}\left(  x\right)  =\sum_{\ell
=1}^{m}\lambda_{\ell}a_{\ell}\otimes b_{\ell}$ for some $m\in\mathbb{N}$, some
elements $a_{1}$, $a_{2}$, $\ldots$, $a_{m}$ of $C$, and some elements $b_{1}%
$, $b_{2}$, $\ldots$, $b_{m}$ of $C$. Consider this $m$, these $a_{1}$,
$a_{2}$, $\ldots$, $a_{m}$, and these $b_{1}$, $b_{2}$, $\ldots$, $b_{m}$.
Then, every $i\in\mathbb{N}$ and $j\in\mathbb{N}$ satisfy%
\begin{align}
&  \underbrace{\left(  f^{\ast j}\ast g^{\ast i}\right)  }_{\substack{=\mu
_{H}\circ\left(  f^{\ast j}\otimes g^{\ast i}\right)  \circ\Delta
_{C}\\\text{(by the definition of convolution)}}}\left(  x\right) \nonumber\\
&  =\left(  \mu_{H}\circ\left(  f^{\ast j}\otimes g^{\ast i}\right)
\circ\Delta_{C}\right)  \left(  x\right)  =\mu_{H}\left(  \left(  f^{\ast
j}\otimes g^{\ast i}\right)  \left(  \underbrace{\Delta_{C}\left(  x\right)
}_{=\sum_{\ell=1}^{m}\lambda_{\ell}a_{\ell}\otimes b_{\ell}}\right)  \right)
\nonumber\\
&  =\mu_{H}\left(  \underbrace{\left(  f^{\ast j}\otimes g^{\ast i}\right)
\left(  \sum_{\ell=1}^{m}\lambda_{\ell}a_{\ell}\otimes b_{\ell}\right)
}_{=\sum_{\ell=1}^{m}\lambda_{\ell}f^{\ast j}\left(  a_{\ell}\right)  \otimes
g^{\ast i}\left(  b_{\ell}\right)  }\right)  =\mu_{H}\left(  \sum_{\ell=1}%
^{m}\lambda_{\ell}f^{\ast j}\left(  a_{\ell}\right)  \otimes g^{\ast i}\left(
b_{\ell}\right)  \right) \nonumber\\
&  =\sum_{\ell=1}^{m}\lambda_{\ell}f^{\ast j}\left(  a_{\ell}\right)  g^{\ast
i}\left(  b_{\ell}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu
_{H}\text{ is the multiplication map}\right)  . \label{47.pf.6}%
\end{align}


On the other hand,\footnote{The following computation is a manipulation with
infinite sums. Such manipulations may be dangerous, since infinite sums (even
when they converge) may fail to satisfy some of the rules one would expect
infinite sums to satisfy: For example, switching two summation signs might not
always preserve the sum. However, the specific computation that we are going
to do is safe from such troubles, for the following reasons:
\par
\begin{itemize}
\item Whenever a sum appears in the computation, it has the property that all
but finitely many of its addends are zero.
\par
\item Whenever two summation signs get switched in the computation, they have
the property that all but finitely many addends of the resulting
\textbf{double sum} are zero. (For example, we can transform the
\textquotedblleft$\sum_{j\geq0}\sum_{\substack{i\geq0;\\j\leq i}%
}$\textquotedblright\ into \textquotedblleft$\sum_{i\geq0}\sum
_{\substack{j\geq0;\\j\leq i}}$\textquotedblright\ in the double sum
$\sum_{j\geq0}\sum_{\substack{i\geq0;\\j\leq i}}\dfrac{1}{j!\left(
i-j\right)  !}\left(  f^{\ast j}\ast g^{\ast\left(  i-j\right)  }\right)
\left(  x\right)  $, because all but finitely many \textbf{pairs} $\left(
i,j\right)  \in\mathbb{N}^{\times2}$ satisfying $j\leq i$ satisfy $\dfrac
{1}{j!\left(  i-j\right)  !}\left(  f^{\ast j}\ast g^{\ast\left(  i-j\right)
}\right)  \left(  x\right)  =0$.)
\end{itemize}
}%
\begin{align*}
&  \underbrace{\left(  e^{\ast f}\ast e^{\ast g}\right)  }_{\substack{=\mu
_{H}\circ\left(  e^{\ast f}\otimes e^{\ast g}\right)  \circ\Delta
_{C}\\\text{(by the definition of convolution)}}}\left(  x\right) \\
&  =\left(  \mu_{H}\circ\left(  e^{\ast f}\otimes e^{\ast g}\right)
\circ\Delta_{C}\right)  \left(  x\right)  =\mu_{H}\left(  \left(  e^{\ast
f}\otimes e^{\ast g}\right)  \left(  \underbrace{\Delta_{C}\left(  x\right)
}_{=\sum_{\ell=1}^{m}\lambda_{\ell}a_{\ell}\otimes b_{\ell}}\right)  \right)
\\
&  =\mu_{H}\left(  \underbrace{\left(  e^{\ast f}\otimes e^{\ast g}\right)
\left(  \sum_{\ell=1}^{m}\lambda_{\ell}a_{\ell}\otimes b_{\ell}\right)
}_{=\sum_{\ell=1}^{m}\lambda_{\ell}e^{\ast f}\left(  a_{\ell}\right)  \otimes
e^{\ast g}\left(  b_{\ell}\right)  }\right) \\
&  =\mu_{H}\left(  \sum_{\ell=1}^{m}\lambda_{\ell}e^{\ast f}\left(  a_{\ell
}\right)  \otimes e^{\ast g}\left(  b_{\ell}\right)  \right)  =\sum_{\ell
=1}^{m}\lambda_{\ell}\underbrace{e^{\ast f}\left(  a_{\ell}\right)
}_{\substack{=\sum_{i\geq0}\dfrac{f^{\ast i}\left(  a_{\ell}\right)  }%
{i!}\\\text{(by (\ref{exp}), applied}\\\text{to }a_{\ell}\text{ instead of
}x\text{)}}}\underbrace{e^{\ast g}\left(  b_{\ell}\right)  }_{\substack{=\sum
_{i\geq0}\dfrac{g^{\ast i}\left(  b_{\ell}\right)  }{i!}\\\text{(by
(\ref{exp}), applied}\\\text{to }b_{\ell}\text{ and }g\text{ instead of
}x\text{ and }f\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because }\mu_{H}\text{ is the
multiplication map}\right)
\end{align*}%
\begin{align*}
&  =\sum_{\ell=1}^{m}\lambda_{\ell}\left(  \sum_{i\geq0}\dfrac{f^{\ast
i}\left(  a_{\ell}\right)  }{i!}\right)  \left(  \sum_{i\geq0}\dfrac{g^{\ast
i}\left(  b_{\ell}\right)  }{i!}\right)  =\sum_{\ell=1}^{m}\lambda_{\ell
}\left(  \sum_{j\geq0}\dfrac{f^{\ast j}\left(  a_{\ell}\right)  }{j!}\right)
\left(  \sum_{i\geq0}\dfrac{g^{\ast i}\left(  b_{\ell}\right)  }{i!}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }i\text{ as
}j\text{ in the second sum}\right) \\
&  =\sum_{\ell=1}^{m}\lambda_{\ell}\sum_{j\geq0}\sum_{i\geq0}%
\underbrace{\dfrac{f^{\ast j}\left(  a_{\ell}\right)  }{j!}\dfrac{g^{\ast
i}\left(  b_{\ell}\right)  }{i!}}_{=\dfrac{1}{j!i!}f^{\ast j}\left(  a_{\ell
}\right)  g^{\ast i}\left(  b_{\ell}\right)  }=\sum_{\ell=1}^{m}\lambda_{\ell
}\sum_{j\geq0}\sum_{i\geq0}\dfrac{1}{j!i!}f^{\ast j}\left(  a_{\ell}\right)
g^{\ast i}\left(  b_{\ell}\right) \\
&  =\sum_{j\geq0}\sum_{i\geq0}\dfrac{1}{j!i!}\underbrace{\sum_{\ell=1}%
^{m}\lambda_{\ell}f^{\ast j}\left(  a_{\ell}\right)  g^{\ast i}\left(
b_{\ell}\right)  }_{\substack{=\left(  f^{\ast j}\ast g^{\ast i}\right)
\left(  x\right)  \\\text{(by (\ref{47.pf.6}))}}}=\sum_{j\geq0}\sum_{i\geq
0}\dfrac{1}{j!i!}\left(  f^{\ast j}\ast g^{\ast i}\right)  \left(  x\right) \\
&  =\sum_{j\geq0}\underbrace{\sum_{i\geq j}}_{=\sum_{\substack{i\geq0;\\i\geq
j}}\sum_{\substack{i\geq0;\\j\leq i}}}\dfrac{1}{j!\left(  i-j\right)
!}\left(  f^{\ast j}\ast g^{\ast\left(  i-j\right)  }\right)  \left(  x\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i-j\text{ for
}i\text{ in the second sum}\right) \\
&  =\underbrace{\sum_{j\geq0}\sum_{\substack{i\geq0;\\j\leq i}}}_{=\sum
_{i\geq0}\sum_{\substack{j\geq0;\\j\leq i}}}\dfrac{1}{j!\left(  i-j\right)
!}\left(  f^{\ast j}\ast g^{\ast\left(  i-j\right)  }\right)  \left(
x\right)  =\sum_{i\geq0}\sum_{\substack{j\geq0;\\j\leq i}}\dfrac{1}{j!\left(
i-j\right)  !}\left(  f^{\ast j}\ast g^{\ast\left(  i-j\right)  }\right)
\left(  x\right)  .
\end{align*}
Compared to (\ref{47.pf.5a}), this yields $\left(  e^{\ast f}\ast e^{\ast
g}\right)  \left(  x\right)  =e^{\ast\left(  f+g\right)  }\left(  x\right)  $.
Since this is proven for all $x\in C$, we can conclude that $e^{\ast f}\ast
e^{\ast g}=e^{\ast\left(  f+g\right)  }$. Proposition~\ref{Proposition47} is
thus proven.
\end{proof}

As a consequence, we can describe the natural powers of exponentials:

\begin{corollary}
\label{Corollary48} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered $k$-coalgebra. Let $H$ be a $k$-algebra. Let $f\in
\mathfrak{g}\left(  C,H\right)  $ and $n\in\mathbb{N}$. Then, $e^{\ast\left(
nf\right)  }=\left(  e^{\ast f}\right)  ^{\ast n}$.
\end{corollary}

We could extend this corollary to hold for all $n\in\mathbb{Z}$ (it is not a
priori clear that $e^{\ast f}$ is invertible as an element of $\mathcal{L}%
\left(  C,H\right)  $, but it is true), but we won't need this extension, so
we don't prove it.

\begin{proof}
[Proof of Corollary~\ref{Corollary48}.]We are going to prove
Corollary~\ref{Corollary48} by induction over $n$:

\textit{Induction base:} We have $e^{\ast0}=e_{C,H}$%
\ \ \ \ \footnote{\textit{Proof.} Every $x\in C$ satisfies%
\begin{align*}
e^{\ast0}\left(  x\right)   &  =\sum_{i\geq0}\dfrac{0^{\ast i}\left(
x\right)  }{i!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{exp}), applied to
}0\text{ instead of }f\right) \\
&  =\underbrace{\dfrac{0^{\ast0}\left(  x\right)  }{0!}}_{=\dfrac{0^{\ast
0}\left(  x\right)  }{1}=0^{\ast0}\left(  x\right)  }+\sum_{i\geq
1}\underbrace{\dfrac{0^{\ast i}\left(  x\right)  }{i!}}_{\substack{=0\text{
(since }0^{\ast i}\left(  x\right)  =0\\\text{(due to }i\geq1\text{))}}}\\
&  =\underbrace{0^{\ast0}}_{=e_{C,H}}\left(  x\right)  +\underbrace{\sum
_{i\geq1}0}_{=0}=e_{C,H}\left(  x\right)  .
\end{align*}
Thus, $e^{\ast0}=e_{C,H}$, qed.}. Thus, $e^{\ast\left(  0f\right)  }=e^{\ast
0}=e_{C,H}=\left(  e^{\ast f}\right)  ^{\ast0}$. In other words,
Corollary~\ref{Corollary48} holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$ be arbitrary. Assume that
Corollary~\ref{Corollary48} holds for $n=N$. We now must prove that
Corollary~\ref{Corollary48} also holds for $n=N+1$.

Since Corollary~\ref{Corollary48} holds for $n=N$, we have $e^{\ast\left(
Nf\right)  }=\left(  e^{\ast f}\right)  ^{\ast N}$. Since $f\ast\left(
Nf\right)  =N\left(  f\ast f\right)  =\left(  Nf\right)  \ast f$, we can apply
Proposition~\ref{Proposition47} to $g=Nf$, and conclude that $e^{\ast\left(
f+Nf\right)  }=e^{\ast f}\ast e^{\ast\left(  Nf\right)  }$. Since $\left(
N+1\right)  f=f+Nf$, we now have%
\[
e^{\ast\left(  \left(  N+1\right)  f\right)  }=e^{\ast\left(  f+Nf\right)
}=e^{\ast f}\ast\underbrace{e^{\ast\left(  Nf\right)  }}_{=\left(  e^{\ast
f}\right)  ^{\ast N}}=e^{\ast f}\ast\left(  e^{\ast f}\right)  ^{\ast
N}=\left(  e^{\ast f}\right)  ^{\ast\left(  N+1\right)  }.
\]
In other words, Corollary~\ref{Corollary48} holds for $n=N+1$. This completes
the induction step. Thus, the induction proof of Corollary~\ref{Corollary48}
is done.
\end{proof}

\section{\label{S9}The ``coalgebra homomorphism $\Longrightarrow$ $\left(
\varepsilon,\varepsilon\right)  $-coderivation'' direction}

Our proof of Lemma~\ref{Lemma32} will further need the following, purely
linear-algebraic fact about filtered vector spaces over a field of
characteristic $0$.

\begin{proposition}
\label{Proposition49} Let $k$ be a field of characteristic $0$. Let $V$ be a
filtered $k$-vector space. Let $W$ be a $k$-vector space. For every
$n\in\mathbb{N}$, let $h_{n}:V\rightarrow W$ be a $k$-linear map such that
$h_{n}\left(  V_{\leq n-1}\right)  =0$.

\textbf{(a)} Then, for every $x\in V$ and every $t\in\mathbb{Z}$, the element
$\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  \in W$ is well-defined, i. e., the
infinite sum $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  $ converges with
respect to the discrete topology.

\textbf{(b)} Assume that every $x\in V$ and every $t\in\mathbb{N}$ satisfy
$\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$. Then, $h_{n}=0$ for every
$n\in\mathbb{N}$.
\end{proposition}

To get an intuition for this proposition, we should note that part
\textbf{(a)} is more or less trivial (we have to use that every $x\in V$ lies
in $V_{n}$ for some high enough $n$), and part \textbf{(b)} is an assertion of
the kind ``if a polynomial in one variable vanishes on every $t\in\mathbb{N}$,
then it must be the zero polynomial''. Note that part \textbf{(b)} is the only
part which uses the condition that $k$ be of characteristic $0$. There are two
easy ways to prove part \textbf{(b)}: one is by using Vandermonde
determinants, the other by finite differences. We are going to follow the
second way here. The proof will need the following theorem as an auxiliary result:

\begin{theorem}
\label{Theorem50} Let $k$ be a commutative ring with unity. Let $N\in
\mathbb{N}$. Then, the equalities%
\begin{equation}
\sum_{t=0}^{N}\left(  -1\right)  ^{t}\dbinom{N}{t}t^{\ell}%
=0\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\left\{  0,1,\ldots,N-1\right\}
\label{50-1}%
\end{equation}
and%
\begin{equation}
\sum_{t=0}^{N}\left(  -1\right)  ^{t}\dbinom{N}{t}t^{N}=\left(  -1\right)
^{N}N! \label{50-2}%
\end{equation}
are satisfied in $k$.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem50}.]Theorem~\ref{Theorem50} is identical with
Theorem 1 of \cite{3}, with the only difference that the $R$ and the $k$ from
Theorem 1 of \cite{3} have been renamed as $k$ and $t$ in
Theorem~\ref{Theorem50}. Since Theorem 1 of \cite{3} is proven in \cite{3}, we
thus don't need to prove Theorem~\ref{Theorem50} here.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition49}.]First we notice that
\begin{equation}
h_{n}\left(  V_{\leq i}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for any }%
n\in\mathbb{N}\text{ and }i\in\mathbb{N}\text{ satisfying }i<n \label{49.pf.1}%
\end{equation}
\footnote{\textit{Proof.} Let $n\in\mathbb{N}$ and $i\in\mathbb{N}$ satisfy
$i<n$. Then, $i\leq n-1$ (since $i$ and $n$ are integers) and thus $V_{\leq
i}\subseteq V_{\leq n-1}$ (since $\left(  V_{\leq\ell}\right)  _{\ell\geq0}$
is a filtration). Hence, $h_{n}\left(  V_{\leq i}\right)  \subseteq
h_{n}\left(  V_{\leq n-1}\right)  =0$. In other words, $h_{n}\left(  V_{\leq
i}\right)  =0$. This proves (\ref{49.pf.1}).}.

\textbf{(a)} Let $x\in V$ and $t\in\mathbb{Z}$. Since $V$ is filtered, there
exists some $n\in\mathbb{N}$ such that $x\in V_{\leq n}$. Consider such an $n$.

Every integer $i>n$ satisfies $h_{i}\left(  V_{\leq n}\right)  =0$ (by
(\ref{49.pf.1}), applied to $i$ and $n$ instead of $n$ and $i$) and thus
$h_{i}\left(  x\right)  =0$ (since $x\in V_{\leq n}$ and therefore
$h_{i}\left(  x\right)  \in h_{i}\left(  V_{\leq n}\right)  =0$, so that
$h_{i}\left(  x\right)  =0$), so that $t^{i}h_{i}\left(  x\right)  =0$. Hence,
for every integer $i>n$, the $i$-th addend of the infinite sum $\sum_{i\geq
0}t^{i}h_{i}\left(  x\right)  $ is zero. Hence, this infinite sum $\sum
_{i\geq0}t^{i}h_{i}\left(  x\right)  $ has only finitely many nonzero addends.
Thus, this sum converges with respect to the discrete topology.
Proposition~\ref{Proposition49} \textbf{(a)} is thus proven.

\textbf{(b)} Assume that every $x\in V$ and every $t\in\mathbb{N}$ satisfy
$\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$. We are now going to show that%
\begin{equation}
\text{every }x\in V\text{ and every }n\in\mathbb{N}\text{ satisfy }%
h_{n}\left(  x\right)  =0. \label{49.pf.2}%
\end{equation}


\textit{Proof of (\ref{49.pf.2}).} Fix some $x\in V$. We then must prove that
$h_{n}\left(  x\right)  =0$ for every $n\in\mathbb{N}$.

Since $V$ is filtered, there exists some $j\in\mathbb{N}$ such that $x\in
V_{\leq j}$. Consider such a $j$.

Every integer $n>j$ satisfies $h_{n}\left(  x\right)  \in h_{n}\left(  V_{\leq
j}\right)  $ (since $x\in V_{\leq j}$) and $h_{n}\left(  V_{\leq j}\right)
=0$ (by (\ref{49.pf.1}), applied to $j$ instead of $i$). Hence, every integer
$n>j$ satisfies $h_{n}\left(  x\right)  \in h_{n}\left(  V_{\leq j}\right)
=0$. In other words, every integer $n>j$ satisfies $h_{n}\left(  x\right)  =0$.

Now we are going to prove that%
\begin{equation}
\text{for every }\ell\in\left\{  0,1,\ldots,j+1\right\}  \text{, every integer
}n>j-\ell\text{ satisfies }h_{n}\left(  x\right)  =0. \label{49.pf.4}%
\end{equation}
\footnote{Note that for $\ell=0$, this is pretty much obvious, whereas for
$\ell=j+1$, this is our goal.}

\textit{Proof of (\ref{49.pf.4}).} We are going to prove (\ref{49.pf.4}) by
induction over $\ell$:

\textit{Induction base:} For $\ell=0$, every integer $n>j-\ell$ satisfies
$h_{n}\left(  x\right)  =0$.\ \ \ $\ $\footnote{\textit{Proof.} Let $\ell=0$.
We know that every integer $n>j$ satisfies $h_{n}\left(  x\right)  =0$. Since
$j=j-\underbrace{0}_{=\ell}=j-\ell$, this rewrites as follows: Every integer
$n>j-\ell$ satisfies $h_{n}\left(  x\right)  =0$, qed.} In other words,
(\ref{49.pf.4}) holds for $\ell=0$. This completes the induction base.

\textit{Induction step:} Let $L\in\left\{  0,1,\ldots,j\right\}  $. Assume
that (\ref{49.pf.4}) holds for $\ell=L$. In order to complete the induction
step, we must then prove that (\ref{49.pf.4}) also holds for $\ell=L+1$.

We know that%
\begin{equation}
\text{every integer }n>j-L\text{ satisfies }h_{n}\left(  x\right)  =0
\label{49.pf.5}%
\end{equation}
(because (\ref{49.pf.4}) holds for $\ell=L$). Now, let $n$ be an integer such
that $n>j-\left(  L+1\right)  $. We are going to prove that $h_{n}\left(
x\right)  =0$.

\textit{Proof of }$h_{n}\left(  x\right)  =0$\textit{:} First we notice that
$n>j-\left(  L+1\right)  =j-L-1$. Since $n$ and $j-L$ are integers, this
yields $n\geq\left(  j-L-1\right)  +1=j-L$. Hence, two cases are possible:

\textit{Case 1:} We have $n>j-L$.

\textit{Case 2:} We have $n=j-L$.

In Case 1, we trivially have $h_{n}\left(  x\right)  =0$ (by (\ref{49.pf.5})).

Now consider Case 2. In this case, $n=j-L$. Notice that $n=j-L\in\mathbb{N}$
(since $L\in\left\{  0,1,\ldots,j\right\}  $). For every $t\in\mathbb{N}$, we
have%
\begin{align}
0  &  =\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by our assumption}\right) \nonumber\\
&  =\sum_{i=0}^{j-L}t^{i}h_{i}\left(  x\right)  +\sum_{i>j-L}t^{i}%
\underbrace{h_{i}\left(  x\right)  }_{\substack{=0\\\text{(by (\ref{49.pf.5}),
applied to }i\text{ instead of }n\text{)}}}=\sum_{i=0}^{j-L}t^{i}h_{i}\left(
x\right)  +\underbrace{\sum_{i>j-L}t^{i}0}_{=0}=\sum_{i=0}^{j-L}t^{i}%
h_{i}\left(  x\right)  . \label{49.pf.6}%
\end{align}
Clearly,%
\[
\sum_{t=0}^{N}\left(  -1\right)  ^{t}\dbinom{j-L}{t}\underbrace{\left(
\sum_{i=0}^{j-L}t^{i}h_{i}\left(  x\right)  \right)  }_{=0\text{ (by
(\ref{49.pf.6}))}}=\sum_{t=0}^{N}\left(  -1\right)  ^{t}\dbinom{j-L}{t}0=0.
\]
Compared to%
\begin{align*}
&  \sum_{t=0}^{j-L}\left(  -1\right)  ^{t}\dbinom{j-L}{t}\left(  \sum
_{i=0}^{j-L}t^{i}h_{i}\left(  x\right)  \right) \\
&  =\sum_{i=0}^{j-L}\left(  \sum_{t=0}^{j-L}\left(  -1\right)  ^{t}%
\dbinom{j-L}{t}t^{i}\right)  h_{i}\left(  x\right) \\
&  =\sum_{i=0}^{j-L-1}\underbrace{\left(  \sum_{t=0}^{j-L}\left(  -1\right)
^{t}\dbinom{j-L}{t}t^{i}\right)  }_{\substack{=0\\\text{(by (\ref{50-1}),
applied to }N=j-L\text{ and }\ell=i\text{)}}}h_{i}\left(  x\right)
+\underbrace{\left(  \sum_{t=0}^{j-L}\left(  -1\right)  ^{t}\dbinom{j-L}%
{t}t^{j-L}\right)  }_{\substack{=\left(  -1\right)  ^{j-L}\left(  j-L\right)
!\\\text{(by (\ref{50-2}), applied to }N=j-L\text{)}}}h_{j-L}\left(  x\right)
\\
&  =\underbrace{\sum_{i=0}^{j-L-1}0h_{i}\left(  x\right)  }_{=0}+\left(
-1\right)  ^{j-L}\left(  j-L\right)  !h_{j-L}\left(  x\right)  =\left(
-1\right)  ^{j-L}\left(  j-L\right)  !h_{j-L}\left(  x\right)  ,
\end{align*}
this yields%
\[
\left(  -1\right)  ^{j-L}\left(  j-L\right)  !h_{j-L}\left(  x\right)  =0.
\]
Since $\left(  -1\right)  ^{j-L}\left(  j-L\right)  !$ is invertible in $k$
(because $k$ has characteristic $0$), this simplifies to $h_{j-L}\left(
x\right)  =0$. Since $j-L=n$, this becomes $h_{n}\left(  x\right)  =0$. We
thus have proven $h_{n}\left(  x\right)  =0$ in Case 2.

Hence, $h_{n}\left(  x\right)  =0$ is proven in both possible cases 1 and 2.
This completes the proof of $h_{n}\left(  x\right)  =0$.

We have thus shown that%
\[
\text{every integer }n>j-\left(  L+1\right)  \text{ satisfies }h_{n}\left(
x\right)  =0.
\]
In other words, we have shown that (\ref{49.pf.4}) holds for $\ell=L+1$. This
completes the induction step (of the proof of (\ref{49.pf.4})). Hence, the
induction proof of (\ref{49.pf.4}) is complete.

Now we can apply (\ref{49.pf.4}) to $\ell=j+1$, and conclude that
\begin{equation}
\text{every integer }n>j-\left(  j+1\right)  \text{ satisfies }h_{n}\left(
x\right)  =0. \label{49.pf.9}%
\end{equation}
Thus, every $n\in\mathbb{N}$ satisfies $h_{n}\left(  x\right)  =0$ (because
$n\in\mathbb{N}$ yields $n>-1=j-\left(  j+1\right)  $, and thus $h_{n}\left(
x\right)  =0$ by (\ref{49.pf.9})).

So we have proven that for every $n\in\mathbb{N}$, every $x\in V$ satisfies
$h_{n}\left(  x\right)  =0$. Thus, for every $n\in\mathbb{N}$, we have
$h_{n}=0$ (because $h_{n}\left(  x\right)  =0$ for every $x\in V$). This
proves Proposition~\ref{Proposition49} \textbf{(b)}.
\end{proof}

Our next result is a distillate of the proof of Lemma~\ref{Lemma31}:

\begin{lemma}
\label{Lemma53} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered $k$-coalgebra. Let $H$ be a $k$-bialgebra. Let
$f\in\mathfrak{g}\left(  C,H\right)  $. Let $x\in C$. Then,%
\begin{equation}
\left(  \Delta_{H}\circ e^{\ast f}\right)  \left(  x\right)  =\sum_{n\geq
0}\dfrac{1}{n!}\left(  \Delta_{H}\circ f^{\ast n}\right)  \left(  x\right)
\label{53.1}%
\end{equation}
and%
\begin{equation}
\left(  \left(  e^{\ast f}\otimes e^{\ast f}\right)  \circ\Delta_{C}\right)
\left(  x\right)  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom
{n}{i}\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)
\circ\Delta_{C}\right)  \left(  x\right)  . \label{53.2}%
\end{equation}
(In particular, the infinite sums $\sum_{n\geq0}\dfrac{1}{n!}\left(
\Delta_{H}\circ f^{\ast n}\right)  \left(  x\right)  $ and \newline%
$\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\ast
i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(
x\right)  $ converge with respect to the discrete topology, i. e., each of
these sums has only finitely many nonzero addends.)
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma53}.]The equalities (\ref{53.1}) and (\ref{53.2})
have been proven during our proof of Lemma~\ref{Lemma31}, \textit{without
using the assumptions that }$C$ \textit{be cocommutative and }$f$ \textit{be
an }$\left(  \varepsilon,\varepsilon\right)  $\textit{-coderivation}. Hence,
these equalities are true. Lemma~\ref{Lemma53} is thus true.
\end{proof}

We now proceed to proving Lemma~\ref{Lemma32}:

\begin{proof}
[Proof of Lemma~\ref{Lemma32}.]Assume that $e^{\ast f}$ is a $k$-coalgebra homomorphism.

For every $n\in\mathbb{N}$, define a $k$-linear map $h_{n}:C\rightarrow H$ by%
\begin{equation}
h_{n}=\dfrac{1}{n!}\Delta_{H}\circ f^{\ast n}-\dfrac{1}{n!}\sum_{i=0}%
^{n}\dbinom{n}{i}\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)
}\right)  \circ\Delta_{C}. \label{32.pf.h_n}%
\end{equation}


\textbf{a)} We now will prove that every $n\in\mathbb{N}$ satisfies
$h_{n}\left(  C_{\leq n-1}\right)  =0$.

\textit{Proof.} Fix some $n\in\mathbb{N}$. If $n=0$, then $h_{n}\left(
C_{\leq n-1}\right)  =0$ is obviously true (since if $n=0$, then $C_{\leq
n-1}=C_{\leq0-1}=0$, so that $h\left(  C_{\leq n-1}\right)  =h\left(
0\right)  =0$). Hence, we can WLOG assume (for the rest of the proof of
\textbf{a)}) that $n\neq0$. Then, $n\geq1$. Thus, $n-1\in\mathbb{N}$.

We have $f^{\ast n}\left(  C_{\leq n-1}\right)  =0$ (by Remark~\ref{Remark9},
applied to $C$, $H$, $n$ and $n-1$ instead of $H$, $A$, $i$ and $n$). On the
other hand, since $C$ is a filtered coalgebra, every $m\in\mathbb{N}$
satisfies $\Delta_{C}\left(  C_{\leq m}\right)  \subseteq\sum_{u=0}^{m}C_{\leq
u}\otimes C_{\leq m-u}$. Applied to $m=n-1$, this yields $\Delta_{C}\left(
C_{\leq n-1}\right)  \subseteq\sum_{u=0}^{n-1}C_{\leq u}\otimes C_{\leq
n-1-u}$. Thus, for every $i\in\left\{  0,1,\ldots,n\right\}  $, we have%
\begin{align}
&  \left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \left(
\Delta_{C}\left(  C_{\leq n-1}\right)  \right) \nonumber\\
&  \subseteq\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)
\left(  \sum_{u=0}^{n-1}C_{\leq u}\otimes C_{\leq n-1-u}\right) \nonumber\\
&  =\sum_{u=0}^{n-1}\underbrace{\left(  f^{\ast i}\otimes f^{\ast\left(
n-i\right)  }\right)  \left(  C_{\leq u}\otimes C_{\leq n-1-u}\right)
}_{\subseteq f^{\ast i}\left(  C_{\leq u}\right)  \otimes f^{\ast\left(
n-i\right)  }\left(  C_{\leq n-1-u}\right)  }\subseteq\sum_{u=0}^{n-1}f^{\ast
i}\left(  C_{\leq u}\right)  \otimes f^{\ast\left(  n-i\right)  }\left(
C_{\leq n-1-u}\right) \nonumber\\
&  =\sum_{u=0}^{i-1}\underbrace{f^{\ast i}\left(  C_{\leq u}\right)
}_{\substack{=0\text{ (by Remark~\ref{Remark9}}\\\text{(applied to }C\text{,
}H\text{, }i\text{ and }u\\\text{instead of }H\text{, }A\text{, }i\text{ and
}n\text{),}\\\text{since }i>u\text{)}}}\otimes f^{\ast\left(  n-i\right)
}\left(  C_{\leq n-1-u}\right)  +\sum_{u=i}^{n-1}f^{\ast i}\left(  C_{\leq
u}\right)  \otimes\underbrace{f^{\ast\left(  n-i\right)  }\left(  C_{\leq
n-1-u}\right)  }_{\substack{=0\text{ (by Remark~\ref{Remark9}}\\\text{(applied
to }C\text{, }H\text{, }n-i\text{ and }n-1-u\\\text{instead of }H\text{,
}A\text{, }i\text{ and }n\text{),}\\\text{since }n-i>n-1-u\text{
(because}\\i\leq u\text{, so that }n-i\geq n-u>n-1-u\text{))}}}\nonumber\\
&  =\sum_{u=0}^{i-1}\underbrace{0\otimes f^{\ast\left(  n-i\right)  }\left(
C_{\leq n-1-u}\right)  }_{=0}+\sum_{u=i}^{n-1}\underbrace{f^{\ast i}\left(
C_{\leq u}\right)  \otimes0}_{=0}=\underbrace{\sum_{u=0}^{i-1}0}%
_{=0}+\underbrace{\sum_{u=i}^{n-1}0}_{=0}=0+0=0. \label{32.pf.1}%
\end{align}
Now, (\ref{32.pf.h_n}) yields%
\begin{align*}
h_{n}\left(  C_{\leq n-1}\right)   &  =\left(  \dfrac{1}{n!}\Delta_{H}\circ
f^{\ast n}-\dfrac{1}{n!}\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\ast i}\otimes
f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(  C_{\leq
n-1}\right) \\
&  \subseteq\underbrace{\left(  \dfrac{1}{n!}\Delta_{H}\circ f^{\ast
n}\right)  \left(  C_{\leq n-1}\right)  }_{=\dfrac{1}{n!}\left(  \Delta
_{H}\circ f^{\ast n}\right)  \left(  C_{\leq n-1}\right)  }%
-\underbrace{\left(  \dfrac{1}{n!}\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\ast
i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(
C_{\leq n-1}\right)  }_{=\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}%
{i}\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ
\Delta_{C}\right)  \left(  C_{\leq n-1}\right)  }\\
&  =\dfrac{1}{n!}\underbrace{\left(  \Delta_{H}\circ f^{\ast n}\right)
\left(  C_{\leq n-1}\right)  }_{=\Delta_{H}\left(  f^{\ast n}\left(  C_{\leq
n-1}\right)  \right)  }-\dfrac{1}{n!}\underbrace{\left(  \sum_{i=0}^{n}%
\dbinom{n}{i}\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)
\circ\Delta_{C}\right)  \left(  C_{\leq n-1}\right)  }_{\subseteq\sum
_{i=0}^{n}\dbinom{n}{i}\left(  \left(  f^{\ast i}\otimes f^{\ast\left(
n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(  C_{\leq n-1}\right)  }\\
&  =\dfrac{1}{n!}\Delta_{H}\left(  \underbrace{f^{\ast n}\left(  C_{\leq
n-1}\right)  }_{=0}\right)  -\dfrac{1}{n!}\sum_{i=0}^{n}\dbinom{n}%
{i}\underbrace{\left(  \left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)
}\right)  \circ\Delta_{C}\right)  \left(  C_{\leq n-1}\right)  }%
_{\substack{=\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)
\left(  \Delta_{C}\left(  C_{\leq n-1}\right)  \right)  =0\\\text{(by
(\ref{32.pf.1}))}}}\\
&  =\dfrac{1}{n!}\underbrace{\Delta_{H}\left(  0\right)  }_{=0}-\dfrac{1}%
{n!}\underbrace{\sum_{i=0}^{n}\dbinom{n}{i}0}_{=0}=\dfrac{1}{n!}0-\dfrac
{1}{n!}0=0-0=0.
\end{align*}
Thus we have proven that every $n\in\mathbb{N}$ satisfies $h_{n}\left(
C_{\leq n-1}\right)  =0$. In other words, \textbf{a)} is proven.

\textbf{b)} Now, we will show that every $x\in C$ and every $t\in\mathbb{N}$
satisfy $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$.\ \ \ \ \footnote{Note
that the element $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  $ is well-defined
due to Proposition~\ref{Proposition49} \textbf{(a)} (applied to $V=C$ and
$W=H$).}

\textit{Proof.} Let $x\in C$ and $t\in\mathbb{N}$ be arbitrary. By
Corollary~\ref{Corollary46} (applied to $t$ and $e^{\ast f}$ instead of $n$
and $f$), we see that $\left(  e^{\ast f}\right)  ^{\ast t}$ is a
$k$-coalgebra homomorphism. Since $e^{\ast\left(  tf\right)  }=\left(  e^{\ast
f}\right)  ^{\ast t}$ (by Corollary~\ref{Corollary48}, applied to $t$ instead
of $n$), this rewrites as follows: The map $e^{\ast\left(  tf\right)  }$ is a
$k$-coalgebra homomorphism. In other words,
\begin{equation}
\Delta_{H}\circ e^{\ast\left(  tf\right)  }=\left(  e^{\ast\left(  tf\right)
}\otimes e^{\ast\left(  tf\right)  }\right)  \circ\Delta_{C} \label{32.pf.3}%
\end{equation}
(by the definition of a $k$-coalgebra homomorphism).

But applying Lemma~\ref{Lemma53} to $tf$ instead of $f$ (this is allowed since
$f\in\mathfrak{g}\left(  C,H\right)  $ yields $tf\in\mathfrak{g}\left(
C,H\right)  $), we obtain%
\begin{equation}
\left(  \Delta_{H}\circ e^{\ast\left(  tf\right)  }\right)  \left(  x\right)
=\sum_{n\geq0}\dfrac{1}{n!}\left(  \Delta_{H}\circ\left(  tf\right)  ^{\ast
n}\right)  \left(  x\right)  \label{32.pf.4}%
\end{equation}
and%
\begin{equation}
\left(  \left(  e^{\ast\left(  tf\right)  }\otimes e^{\ast\left(  tf\right)
}\right)  \circ\Delta_{C}\right)  \left(  x\right)  =\sum_{n\geq0}\dfrac
{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(  \left(  tf\right)  ^{\ast
i}\otimes\left(  tf\right)  ^{\ast\left(  n-i\right)  }\right)  \circ
\Delta_{C}\right)  \left(  x\right)  . \label{32.pf.5}%
\end{equation}


We are now going to rewrite these two equalities by taking $t$ out of the brackets.

Since%
\begin{align*}
\sum_{n\geq0}\dfrac{1}{n!}\left(  \Delta_{H}\circ\underbrace{\left(
tf\right)  ^{\ast n}}_{=t^{n}f^{\ast n}}\right)  \left(  x\right)   &
=\sum_{n\geq0}\dfrac{1}{n!}\left(  \underbrace{\Delta_{H}\circ\left(
t^{n}f^{\ast n}\right)  }_{\substack{=t^{n}\cdot\left(  \Delta_{H}\circ
f^{\ast n}\right)  \\\text{(since composition of }\\k\text{-linear maps is
}k\text{-bilinear)}}}\right)  \left(  x\right)  =\sum_{n\geq0}\dfrac{1}%
{n!}\underbrace{\left(  t^{n}\cdot\left(  \Delta_{H}\circ f^{\ast n}\right)
\right)  \left(  x\right)  }_{=t^{n}\cdot\left(  \Delta_{H}\circ f^{\ast
n}\right)  \left(  x\right)  }\\
&  =\sum_{n\geq0}\dfrac{1}{n!}t^{n}\cdot\left(  \Delta_{H}\circ f^{\ast
n}\right)  \left(  x\right)  =\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(
\Delta_{H}\circ f^{\ast n}\right)  \left(  x\right)  ,
\end{align*}
the equality (\ref{32.pf.4}) rewrites as%
\begin{equation}
\left(  \Delta_{H}\circ e^{\ast\left(  tf\right)  }\right)  \left(  x\right)
=\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(  \Delta_{H}\circ f^{\ast n}\right)
\left(  x\right)  . \label{32.pf.4'}%
\end{equation}


Since%
\begin{align*}
&  \sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(
\underbrace{\left(  tf\right)  ^{\ast i}}_{=t^{i}f^{\ast i}}\otimes
\underbrace{\left(  tf\right)  ^{\ast\left(  n-i\right)  }}_{=t^{n-i}%
f^{\ast\left(  n-i\right)  }}\right)  \circ\Delta_{C}\right)  \left(  x\right)
\\
&  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(
\underbrace{t^{i}f^{\ast i}\otimes t^{n-i}f^{\ast\left(  n-i\right)  }%
}_{\substack{=t^{i}t^{n-i}\cdot\left(  f^{\ast i}\otimes f^{\ast\left(
n-i\right)  }\right)  \\\text{(since tensoring of }k\text{-linear
maps}\\\text{is }k\text{-bilinear)}}}\right)  \circ\Delta_{C}\right)  \left(
x\right) \\
&  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(
\underbrace{t^{i}t^{n-i}}_{=t^{n}}\cdot\left(  f^{\ast i}\otimes
f^{\ast\left(  n-i\right)  }\right)  \right)  \circ\Delta_{C}\right)  \left(
x\right) \\
&  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}%
\underbrace{\left(  t^{n}\cdot\left(  f^{\ast i}\otimes f^{\ast\left(
n-i\right)  }\right)  \right)  \circ\Delta_{C}}_{\substack{=t^{n}\cdot\left(
\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta
_{C}\right)  \\\text{(since composition of }k\text{-linear}\\\text{maps is
}k\text{-bilinear)}}}\right)  \left(  x\right) \\
&  =\sum_{n\geq0}\dfrac{1}{n!}\underbrace{\left(  \sum_{i=0}^{n}\dbinom{n}%
{i}t^{n}\cdot\left(  \left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)
}\right)  \circ\Delta_{C}\right)  \right)  \left(  x\right)  }_{=\sum
_{i=0}^{n}\dbinom{n}{i}t^{n}\cdot\left(  \left(  f^{\ast i}\otimes
f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right)
}\\
&  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}t^{n}%
\cdot\left(  \left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)
\circ\Delta_{C}\right)  \left(  x\right)  \right)  =\sum_{n\geq0}t^{n}%
\dfrac{1}{n!}\underbrace{\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(  \left(
f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)
\left(  x\right)  \right)  }_{=\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(
f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)
\left(  x\right)  }\\
&  =\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(
f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)
\left(  x\right)  ,
\end{align*}
the equality (\ref{32.pf.5}) rewrites as%
\begin{equation}
\left(  \left(  e^{\ast\left(  tf\right)  }\otimes e^{\ast\left(  tf\right)
}\right)  \circ\Delta_{C}\right)  \left(  x\right)  =\sum_{n\geq0}t^{n}%
\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\ast i}\otimes
f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right)
. \label{32.pf.5'}%
\end{equation}


Now,
\begin{align*}
\sum_{i\geq0}t^{i}h_{i}\left(  x\right)   &  =\sum_{n\geq0}t^{n}h_{n}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index
}i\text{ as }n\text{ in the sum}\right) \\
&  =\sum_{n\geq0}t^{n}\underbrace{\left(  \dfrac{1}{n!}\Delta_{H}\circ f^{\ast
n}-\dfrac{1}{n!}\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\ast i}\otimes
f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right)
}_{=\dfrac{1}{n!}\left(  \Delta_{H}\circ f^{\ast n}\right)  \left(  x\right)
-\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\ast i}\otimes
f^{\ast\left(  n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{32.pf.h_n})}\right) \\
&  =\sum_{n\geq0}t^{n}\left(  \dfrac{1}{n!}\left(  \Delta_{H}\circ f^{\ast
n}\right)  \left(  x\right)  -\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}%
{i}\left(  f^{\ast i}\otimes f^{\ast\left(  n-i\right)  }\right)  \circ
\Delta_{C}\right)  \left(  x\right)  \right) \\
&  =\underbrace{\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(  \Delta_{H}\circ f^{\ast
n}\right)  \left(  x\right)  }_{\substack{=\left(  \Delta_{H}\circ
e^{\ast\left(  tf\right)  }\right)  \left(  x\right)  \\\text{(by
(\ref{32.pf.4'}))}}}-\underbrace{\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(
\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\ast i}\otimes f^{\ast\left(
n-i\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right)  }%
_{\substack{=\left(  \left(  e^{\ast\left(  tf\right)  }\otimes e^{\ast\left(
tf\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right)  \\\text{(by
(\ref{32.pf.5'}))}}}\\
&  =\underbrace{\left(  \Delta_{H}\circ e^{\ast\left(  tf\right)  }\right)
}_{\substack{=\left(  e^{\ast\left(  tf\right)  }\otimes e^{\ast\left(
tf\right)  }\right)  \circ\Delta_{C}\\\text{(by (\ref{32.pf.3}))}}}\left(
x\right)  -\left(  \left(  e^{\ast\left(  tf\right)  }\otimes e^{\ast\left(
tf\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right) \\
&  =\left(  \left(  e^{\ast\left(  tf\right)  }\otimes e^{\ast\left(
tf\right)  }\right)  \circ\Delta_{C}\right)  \left(  x\right)  -\left(
\left(  e^{\ast\left(  tf\right)  }\otimes e^{\ast\left(  tf\right)  }\right)
\circ\Delta_{C}\right)  \left(  x\right)  =0.
\end{align*}


We thus have proven that every $x\in C$ and every $t\in\mathbb{N}$ satisfy
$\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$. In other words, part
\textbf{b)} of the proof is done.

\textbf{c)} We know that every $n\in\mathbb{N}$ satisfies $h_{n}\left(
C_{\leq n-1}\right)  =0$ (by part \textbf{a)}), and that every $x\in C$ and
every $t\in\mathbb{N}$ satisfy $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$
(by part \textbf{b)}). Hence, Proposition~\ref{Proposition49} \textbf{(b)}
(applied to $C$ and $H$ instead of $V$ and $W$) yields that $h_{n}=0$ for
every $n\in\mathbb{N}$. Applied to $n=1$, this yields $h_{1}=0$. But
(\ref{32.pf.h_n}) (applied to $n=1$) yields%
\begin{align*}
h_{1}  &  =\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}=1}\Delta_{H}%
\circ\underbrace{f^{\ast1}}_{=f}-\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}%
=1}\underbrace{\sum_{i=0}^{1}\dbinom{1}{i}\left(  f^{\ast i}\otimes
f^{\ast\left(  1-i\right)  }\right)  \circ\Delta_{C}}_{=\dbinom{1}{0}\left(
f^{\ast0}\otimes f^{\ast\left(  1-0\right)  }\right)  \circ\Delta_{C}%
+\dbinom{1}{1}\left(  f^{\ast1}\otimes f^{\ast\left(  1-1\right)  }\right)
\circ\Delta_{C}}\\
&  =\Delta_{H}\circ f-\left(  \underbrace{\dbinom{1}{0}}_{=1}\left(
\underbrace{f^{\ast0}}_{=e_{C,H}}\otimes\underbrace{f^{\ast\left(  1-0\right)
}}_{=f^{\ast1}=f}\right)  \circ\Delta_{C}+\underbrace{\dbinom{1}{1}}%
_{=1}\left(  \underbrace{f^{\ast1}}_{=f}\otimes\underbrace{f^{\ast\left(
1-1\right)  }}_{=f^{\ast0}=e_{C,H}}\right)  \circ\Delta_{C}\right) \\
&  =\Delta_{H}\circ f-\left(  \left(  e_{C,H}\otimes f\right)  \circ\Delta
_{C}+\left(  f\otimes e_{C,H}\right)  \circ\Delta_{C}\right)  .
\end{align*}
Since $h_{1}=0$, this rewrites as%
\[
0=\Delta_{H}\circ f-\left(  \left(  e_{C,H}\otimes f\right)  \circ\Delta
_{C}+\left(  f\otimes e_{C,H}\right)  \circ\Delta_{C}\right)  ,
\]
so that%
\begin{align*}
\Delta_{H}\circ f  &  =\left(  e_{C,H}\otimes f\right)  \circ\Delta
_{C}+\left(  f\otimes e_{C,H}\right)  \circ\Delta_{C}=\left(  f\otimes
e_{C,H}\right)  \circ\Delta_{C}+\left(  e_{C,H}\otimes f\right)  \circ
\Delta_{C}\\
&  =\left(  f\otimes e_{C,H}+e_{C,H}\otimes f\right)  \circ\Delta_{C}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right)  .
\end{align*}
By the definition of ``$\left(  \varepsilon,\varepsilon\right)  $%
-coderivation'', this means that $f$ is an $\left(  \varepsilon,\varepsilon
\right)  $-coderivation.

We thus have proven that, under the assumption that $e^{\ast f}$ is a
$k$-coalgebra homomorphism, the map $f$ is an $\left(  \varepsilon
,\varepsilon\right)  $-coderivation. Thus, Lemma~\ref{Lemma32} is proven.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem30}.]The assertion of Theorem~\ref{Theorem30} is
an ``if and only if'' assertion. Its ``if'' part was proven in
Lemma~\ref{Lemma32}, and its ``only if'' part was proven in
Lemma~\ref{Lemma31}. Hence, both parts of the assertion of
Theorem~\ref{Theorem30} are proven. This finally completes the proof of
Theorem~\ref{Theorem30}.
\end{proof}

\section{\label{S10}Proof of Theorem~\ref{Theorem1}}

Theorem~\ref{Theorem1} will now be merely a trivial consequence of the facts
proven above.

\begin{proof}
[Proof of Theorem~\ref{Theorem1}.]Let $f$ be the map $\operatorname*{Log}%
\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $. Then,
$f=\operatorname*{Log}\operatorname*{id}\in\mathfrak{g}\left(  H,H\right)  $
(because $\operatorname*{Log}F\in\mathfrak{g}\left(  H,H\right)  $ for every
$F\in G\left(  H,H\right)  $). Besides, $f=\operatorname*{Log}%
\operatorname*{id}$ yields $e^{\ast f}=e^{\ast\left(  \operatorname*{Log}%
\operatorname*{id}\right)  }=\operatorname*{id}$ (by
Proposition~\ref{Proposition13} \textbf{(b)}, applied to $F=\operatorname*{id}%
$ and $A=H$). Hence, $e^{\ast f}$ is a $k$-coalgebra homomorphism. By
Lemma~\ref{Lemma32} (applied to $C=H$), this yields that $f$ is an $\left(
\varepsilon,\varepsilon\right)  $-coderivation.

But Theorem~\ref{Theorem23} (applied to $C=H$) tells us that $f$ is an
$\left(  \varepsilon,\varepsilon\right)  $-coderivation if and only if
$f\left(  H\right)  \subseteq\operatorname*{Prim}H$. Hence, $f\left(
H\right)  \subseteq\operatorname*{Prim}H$ (since we know that $f$ is an
$\left(  \varepsilon,\varepsilon\right)  $-coderivation).

Proposition~\ref{Proposition19} \textbf{(c)} (applied to $F=\operatorname*{id}%
$ and $A=H$) yields $\left(  \operatorname*{Log}\operatorname*{id}\right)
\mid_{\operatorname*{Prim}H}=\operatorname*{id}\mid_{\operatorname*{Prim}%
H}=\operatorname{id}_{\operatorname*{Prim}H}$. Since $\operatorname*{Log}%
\operatorname*{id}=f$, this rewrites as $f\mid_{\operatorname*{Prim}%
H}=\operatorname{id}_{\operatorname*{Prim}H}$.

So we know that $f$ is a $k$-linear map satisfying $f\left(  H\right)
\subseteq\operatorname*{Prim}H$ and $f\mid_{\operatorname*{Prim}%
H}=\operatorname{id}_{\operatorname*{Prim}H}$. In other words, $f$ is a
projection from $H$ to the subspace $\operatorname*{Prim}H$. Since
$f=\operatorname*{Log}\operatorname*{id}$, this rewrites as follows: The map
$\operatorname*{Log}\operatorname*{id}$ is a projection from $H$ to the
subspace $\operatorname*{Prim}H$. Theorem~\ref{Theorem1} is therefore proven.
\end{proof}

\section{\label{S11}On the case of $k$ being a ring}

The above results have been formulated for $k$ being a field (except of
Theorem~\ref{Theorem50}). However, all of them, apart from
Propositions~\ref{Proposition16} and~\ref{Proposition17}, still hold if $k$ is
just a commutative ring with unity, as long as the following replacements are made:

\begin{itemize}
\item Any occurrence of ``$k$-vector space'' must be replaced by ``$k$-module''.

\item Any requirement that $k$ be a field of characteristic $0$ must be
replaced by a requirement that $k$ be a commutative $\mathbb{Q}$-algebra.

\item The definition of the notion of a ``filtered $k$-coalgebra'' has to be
replaced by the following Definition~\ref{Definition80} (which is adjusted to
the case of $k$ being a commutative ring with unity):
\end{itemize}

\begin{definition}
\label{Definition80} Let $k$ be a commutative ring with unity. Let $C$ be a
$k$-coalgebra and, at the same time, a filtered $k$-vector space. Then, $C$ is
said to be a \textit{filtered }$k$\textit{-coalgebra} if and only if every
$n\in\mathbb{N}$ satisfies%
\[
\Delta_{C}\left(  C_{\leq n}\right)  \subseteq\sum_{u=0}^{n}\left(  \iota
_{u}\otimes\iota_{n-u}\right)  \left(  C_{\leq u}\otimes C_{\leq n-u}\right)
.
\]
Here, for every $v\in\mathbb{N}$, we denote by $\iota_{v}$ the canonical
inclusion $C_{\leq v}\rightarrow C$.
\end{definition}

Note that this Definition~\ref{Definition80} is not the only possible way to
define a ``filtered $k$-coalgebra'' in the case of $k$ being a commutative
ring with unity. There might be the other definitions around, and possibly
even non-equivalent ones. However, Definition~\ref{Definition80} makes all of
our results (except of Propositions~\ref{Proposition16}
and~\ref{Proposition17}) valid in the case of $k$ being a commutative ring
with unity. (I suspect that the other possible definitions also make them
valid, but I am not sure, since there might always be some definition I
haven't thought of.)

Here is the reason why Propositions~\ref{Proposition16}
and~\ref{Proposition17} do not hold when $k$ is a commutative ring with unity:
If $k$ is just a commutative ring with unity, then tensor products might
behave strangely; in particular, if $U$, $V$, $U^{\prime}$, $V^{\prime}$ are
$k$-modules such that $U\subseteq U^{\prime}$ and $V\subseteq V^{\prime}$,
then we cannot in general identify $U\otimes V$ with a $k$-submodule of
$U^{\prime}\otimes V^{\prime}$. (We still have a canonical map $U\otimes
V\rightarrow U^{\prime}\otimes V^{\prime}$ obtained by tensoring the inclusion
maps $U\rightarrow U^{\prime}$ and $V\rightarrow V^{\prime}$, but this map can
fail to be injective.) Hence, the notion of a \textquotedblleft
subcoalgebra\textquotedblright\ becomes ambiguous: If we define a subcoalgebra
of a coalgebra $C$ to be a $k$-submodule $D$ of $C$ satisfying $\Delta
_{C}\left(  D\right)  \subseteq\left(  \iota\otimes\iota\right)  \left(
D\otimes D\right)  $ (with $\iota$ being the inclusion map $D\rightarrow C$),
then it is not true in general that a subcoalgebra $D$ of a coalgebra $C$ is a
coalgebra, so that Proposition~\ref{Proposition17} makes no sense anymore! On
the other hand, if we define a subcoalgebra of a coalgebra $C$ to be a
coalgebra $D$ equipped with an injective coalgebra homomorphism $D\rightarrow
C$, then Proposition~\ref{Proposition17} remains true, but I am not sure
whether Proposition~\ref{Proposition16} still holds (at least our proof
becomes hopelessly wrong).

While all of our results except of Propositions~\ref{Proposition16}
and~\ref{Proposition17} still hold if $k$ is just a commutative ring with
unity, the same cannot be said about the proofs. Instead, we must take some
more care. What is true is that all of our above proofs, except of the proofs
of Propositions~\ref{Proposition13},~\ref{Proposition16}
and~\ref{Proposition17}, still hold if $k$ is just a commutative ring with
unity, as long as we make the following replacements:

\begin{itemize}
\item Any occurrence of ``$k$-vector space'' must be replaced by ``$k$-module''.

\item Any requirement that $k$ be a field of characteristic $0$ must be
replaced by a requirement that $k$ be a commutative $\mathbb{Q}$-algebra.

\item The definition of the notion of a ``filtered $k$-coalgebra'' has to be
replaced by Definition~\ref{Definition80}.

\item In some situations, inclusion maps have to be made more explicit. This
means the following: Whenever we have four $k$-modules $U$, $V$, $U^{\prime}$,
$V^{\prime}$ (they need not actually be called $U$, $V$, $U^{\prime}$,
$V^{\prime}$) such that $U\subseteq U^{\prime}$ and $V\subseteq V^{\prime}$,
then the assumption that $k$ be a field allows us to identify $U\otimes V$
with a $k$-submodule of $U^{\prime}\otimes V^{\prime}$ (by abuse of notation).
This identification has been done several times in our above proofs (because
we worked with the assumption that $k$ be a field), in order to save space. In
order to adjust the proofs to the case of $k$ being just a commutative ring
with unity, we have to get rid of this identification (because we can no
longer identify $U\otimes V$ with a $k$-submodule of $U^{\prime}\otimes
V^{\prime}$ when $k$ is just a commutative ring with unity). This means that
every time the notation $U\otimes V$ is used to mean a $k$-submodule of
$U^{\prime}\otimes V^{\prime}$, we have to replace it by $\left(  \iota
_{U}\otimes\iota_{V}\right)  \left(  U\otimes V\right)  $, where $\iota_{U}$
is the canonical inclusion $U^{\prime}\rightarrow U$ and $\iota_{V}$ is the
canonical inclusion $V^{\prime}\rightarrow V$. (But when the notation
$U\otimes V$ is used to mean just the $k$-module $U\otimes V$ itself, then it
should stay a $U\otimes V$.) These replacements are straightforward and we are
not going to perform them en detail. (Note that Definition~\ref{Definition80}
is exactly what comes out if we take the standard definition of a filtered
$k$-coalgebra over a field $k$, and perform these replacements on that definition!)
\end{itemize}

However, even these replacement do not salvage our proof of
Proposition~\ref{Proposition13} in the case when $k$ is just a commutative
ring with unity. In fact, this proof made use of
Propositions~\ref{Proposition16} and~\ref{Proposition17}, which both require
$k$ to be a field and are not valid otherwise (at least not in general).
Hence, in the case when $k$ is just a commutative ring with unity, we need a
new proof of Proposition~\ref{Proposition13}. We are going to provide one such
proof. First, let us show an auxiliary result:

\begin{proposition}
\label{Proposition81} Let $k$ be a commutative ring with unity, let $A$ be a
$k$-algebra, and let $H$ be a filtered $k$-coalgebra. For every $n\in
\mathbb{N}$, the subset $\mathcal{L}^{n}\left(  H,A\right)  $ of
$\mathcal{L}\left(  H,A\right)  $ is an ideal of the $k$-algebra
$\mathcal{L}\left(  H,A\right)  $. (For the definition of $\mathcal{L}%
^{n}\left(  H,A\right)  $, see Definition~\ref{Definition8} \textbf{(b)}.)
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition81}.]In Definition~\ref{Definition8}
\textbf{(b)}, we defined $\mathcal{L}^{n}\left(  H,A\right)  $ by%
\[
\mathcal{L}^{n}\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ f\mid_{H_{\leq n-1}}=0\right\}  .
\]
Thus it is easy to see that $\mathcal{L}^{n}\left(  H,A\right)  $ is a
$k$-submodule of $\mathcal{L}\left(  H,A\right)  $%
.\ \ \ \ \footnote{\textit{Proof.} Let $\alpha\in k$, $\beta\in k$,
$g\in\mathcal{L}^{n}\left(  H,A\right)  $ and $h\in\mathcal{L}^{n}\left(
H,A\right)  $ be arbitrary. Since $g\in\mathcal{L}^{n}\left(  H,A\right)
=\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\mid_{H_{\leq n-1}%
}=0\right\}  $, we have $g\mid_{H_{\leq n-1}}=0$. Similarly, $h\mid_{H_{\leq
n-1}}=0$. Now,%
\[
\left(  \alpha g+\beta h\right)  \mid_{H_{\leq n-1}}=\alpha\underbrace{g\mid
_{H_{\leq n-1}}}_{=0}+\beta\underbrace{h\mid_{H_{\leq n-1}}}_{=0}%
=\alpha0+\beta0=0,
\]
so that $\alpha g+\beta h\in\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\mid_{H_{\leq n-1}}=0\right\}  =\mathcal{L}^{n}\left(  H,A\right)  $.
\par
Hence, we have proven that for every $\alpha\in k$, $\beta\in k$,
$g\in\mathcal{L}^{n}\left(  H,A\right)  $ and $h\in\mathcal{L}^{n}\left(
H,A\right)  $, we have $\alpha g+\beta h\in\mathcal{L}^{n}\left(  H,A\right)
$. In other words, $\mathcal{L}^{n}\left(  H,A\right)  $ is a $k$-submodule of
$\mathcal{L}\left(  H,A\right)  $.} Besides, $\mathcal{L}^{n}\left(
H,A\right)  $ is a left ideal of $\mathcal{L}\left(  H,A\right)
\ \ \ \ $\footnote{\textit{Proof.} Let $g\in\mathcal{L}\left(  H,A\right)  $
and $h\in\mathcal{L}^{n}\left(  H,A\right)  $. Then, we are going to prove
that $g\ast h\in\mathcal{L}^{n}\left(  H,A\right)  $.
\par
If $n=0$, then this is obvious (because $\mathcal{L}^{0}\left(  H,A\right)  $
is the whole $k$-module $\mathcal{L}\left(  H,A\right)  $), so let us now WLOG
assume that $n\neq0$. Then, $n\geq1$, so that $n-1\in\mathbb{N}$.
\par
We have $h\in\mathcal{L}^{n}\left(  H,A\right)  =\left\{  f\in\mathcal{L}%
\left(  H,A\right)  \ \mid\ f\mid_{H_{\leq n-1}}=0\right\}  $, so that
$h\mid_{H_{\leq n-1}}=0$. We now get $h\left(  H_{\leq n-1}\right)
=\underbrace{\left(  h\mid_{H_{\leq n-1}}\right)  }_{=0}\left(  H_{\leq
n-1}\right)  =0\left(  H_{\leq n-1}\right)  =0$.
\par
Now, for every $v\in\mathbb{N}$, let $\iota_{v}$ denote the canonical
inclusion map $H_{\leq v}\rightarrow H$. Then, by the definition of a filtered
$k$-coalgebra (Definition~\ref{Definition80}), we have%
\[
\Delta_{H}\left(  H_{\leq m}\right)  \subseteq\sum_{u=0}^{m}\left(  \iota
_{u}\otimes\iota_{m-u}\right)  \left(  H_{\leq u}\otimes H_{\leq m-u}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}%
\]
(since $H$ is a filtered $k$-coalgebra). Applied to $m=n-1$, this yields%
\[
\Delta_{H}\left(  H_{\leq n-1}\right)  \subseteq\sum_{u=0}^{n-1}\left(
\iota_{u}\otimes\iota_{n-1-u}\right)  \left(  H_{\leq u}\otimes H_{\leq
n-1-u}\right)  .
\]
\par
On the other hand, for every $u\in\left\{  0,1,\ldots,n-1\right\}  $, we have
$h\circ\iota_{n-1-u}=0$ (because every $x\in H_{\leq n-1-u}$ satisfies%
\begin{align*}
\left(  h\circ\iota_{n-1-u}\right)  \left(  x\right)   &  =h\left(
\underbrace{\iota_{n-1-u}\left(  x\right)  }_{=x\text{ (since }\iota
_{n-1-u}\text{ is just an inclusion map)}}\right)  =h\left(  \underbrace{x}%
_{\in H_{\leq n-1-u}}\right)  \in h\left(  H_{\leq n-1-u}\right)  \subseteq
h\left(  H_{\leq n-1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  H_{\leq\ell}\right)
_{\ell\geq0}\text{ is a filtration, and thus }H_{\leq n-1-u}\subseteq H_{\leq
n-1}\text{ (since }n-1-u\leq n-1\text{)}\right) \\
&  =0
\end{align*}
and thus $\left(  h\circ\iota_{n-1-u}\right)  \left(  x\right)  =0$) and%
\begin{align}
\left(  g\otimes h\right)  \circ\left(  \iota_{u}\otimes\iota_{n-1-u}\right)
&  =\left(  g\circ\iota_{u}\right)  \otimes\left(  h\circ\iota_{n-1-u}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because an application of (\ref{(X)(X)}) yields}\\
\left(  g\circ\iota_{u}\right)  \otimes\left(  h\circ\iota_{n-1-u}\right)
=\left(  \left(  g\otimes h\right)  \circ\left(  \iota_{u}\otimes\iota
_{n-1-u}\right)  \right)
\end{array}
\right) \nonumber\\
&  =\left(  g\circ\iota_{u}\right)  \otimes\underbrace{\left(  h\circ
\iota_{n-1-u}\right)  }_{=0}=\left(  g\circ\iota_{u}\right)  \otimes0=0.
\label{81.pf.1}%
\end{align}
\par
By the definition of convolution, $g\ast h=\mu_{H}\circ\left(  g\otimes
h\right)  \circ\Delta_{H}$, so that%
\begin{align*}
\left(  g\ast h\right)  \left(  H_{\leq n-1}\right)   &  =\left(  \mu_{H}%
\circ\left(  g\otimes h\right)  \circ\Delta_{H}\right)  \left(  H_{\leq
n-1}\right)  =\mu_{H}\left(  \left(  g\otimes h\right)  \underbrace{\left(
\Delta_{H}\left(  H_{\leq n-1}\right)  \right)  }_{\subseteq\sum_{u=0}%
^{n-1}\left(  \iota_{u}\otimes\iota_{n-1-u}\right)  \left(  H_{\leq u}\otimes
H_{\leq n-1-u}\right)  }\right) \\
&  \subseteq\mu_{H}\left(  \underbrace{\left(  g\otimes h\right)  \left(
\sum_{u=0}^{n-1}\left(  \iota_{u}\otimes\iota_{n-1-u}\right)  \left(  H_{\leq
u}\otimes H_{\leq n-1-u}\right)  \right)  }_{\substack{\subseteq\sum
_{u=0}^{n-1}\left(  g\otimes h\right)  \left(  \left(  \iota_{u}\otimes
\iota_{n-1-u}\right)  \left(  H_{\leq u}\otimes H_{\leq n-1-u}\right)
\right)  \\\text{(since }g\otimes h\text{ is }k\text{-linear)}}}\right) \\
&  =\mu_{H}\left(  \sum_{u=0}^{n-1}\underbrace{\left(  g\otimes h\right)
\left(  \left(  \iota_{u}\otimes\iota_{n-1-u}\right)  \left(  H_{\leq
u}\otimes H_{\leq n-1-u}\right)  \right)  }_{=\left(  \left(  g\otimes
h\right)  \circ\left(  \iota_{u}\otimes\iota_{n-1-u}\right)  \right)  \left(
H_{\leq u}\otimes H_{\leq n-1-u}\right)  }\right) \\
&  =\mu_{H}\left(  \sum_{u=0}^{n-1}\underbrace{\left(  \left(  g\otimes
h\right)  \circ\left(  \iota_{u}\otimes\iota_{n-1-u}\right)  \right)
}_{\substack{=0\\\text{(by (\ref{81.pf.1}))}}}\left(  H_{\leq u}\otimes
H_{\leq n-1-u}\right)  \right) \\
&  =\mu_{H}\left(  \sum_{u=0}^{n-1}\underbrace{0\left(  H_{\leq u}\otimes
H_{\leq n-1-u}\right)  }_{=0}\right)  =\mu_{H}\left(  \underbrace{\sum
_{u=0}^{n-1}0}_{=0}\right)  =\mu_{H}\left(  0\right)  =0
\end{align*}
and therefore $g\ast h\in\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\mid_{H_{\leq n-1}}=0\right\}  \subseteq\mathcal{L}^{n}\left(
H,A\right)  $.
\par
We thus have shown that every $g\in\mathcal{L}\left(  H,A\right)  $ and
$h\in\mathcal{L}^{n}\left(  H,A\right)  $ satisfy $g\ast h\in\mathcal{L}%
^{n}\left(  H,A\right)  $. In other words, $\mathcal{L}^{n}\left(  H,A\right)
$ is a left ideal of $\mathcal{L}\left(  H,A\right)  $, qed.} and a right
ideal of $\mathcal{L}\left(  H,A\right)  $ (similarly). Hence, $\mathcal{L}%
^{n}\left(  H,A\right)  $ is an ideal of $\mathcal{L}\left(  H,A\right)  $.
Proposition~\ref{Proposition81} is thus proven.
\end{proof}

Now we can finally prove the generalization of Proposition~\ref{Proposition13}
to the case of $k$ being a commutative ring with unity. Our proof will be
similar to the one we gave above in the case of $k$ being a field, but instead
of restricting maps to $H_{\leq n}$ we will now consider their equivalence
classes modulo the ideal $\mathcal{L}^{n+1}\left(  H,A\right)  $.

First we formulate the generalization of Proposition~\ref{Proposition13} to
the case of $k$ being a commutative ring with unity:

\begin{proposition}
\label{Proposition82} Let $k$ be a commutative $\mathbb{Q}$-algebra. Let $A$
be a $k$-algebra, and let $H$ be a connected filtered $k$-coalgebra.

\textbf{(a)} Every map $f\in\mathfrak{g}\left(  H,A\right)  $ satisfies
$\operatorname*{Log}\left(  e^{\ast f}\right)  =f$.

\textbf{(b)} Every map $F\in G\left(  H,A\right)  $ satisfies $e^{\ast\left(
\operatorname*{Log}F\right)  }=F$.
\end{proposition}

Note that the following proof of Proposition~\ref{Proposition82} automatically
gives us a second proof of Proposition~\ref{Proposition13}, since
Proposition~\ref{Proposition82} generalizes Proposition~\ref{Proposition13}.

\begin{proof}
[Proof of Proposition~\ref{Proposition82}.]Due to how we defined
$\mathcal{L}^{n+1}\left(  H,A\right)  $ (in Definition~\ref{Definition8}
\textbf{(b)}), we have%
\begin{align}
\mathcal{L}^{n+1}\left(  H,A\right)   &  =\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ \underbrace{f\mid_{H_{\leq n+1-1}}}_{=f\mid_{H_{\leq n}}%
}=0\right\}  =\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid
\ f\mid_{H_{\leq n}}=0\right\} \nonumber\\
&  =\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid_{H_{\leq n}%
}=0\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }f\text{ as
}h\right)  \label{82.pf.0}%
\end{align}
for every $n\in\mathbb{N}$.

\textbf{(a)} Let $f\in\mathfrak{g}\left(  H,A\right)  $. Let $n\in\mathbb{N}$.

According to Proposition~\ref{Proposition81} (applied to $n+1$ instead of
$n$), the subset $\mathcal{L}^{n+1}\left(  H,A\right)  $ of $\mathcal{L}%
\left(  H,A\right)  $ is an ideal of $\mathcal{L}\left(  H,A\right)  $. Thus,
there is a factor algebra $\mathcal{L}\left(  H,A\right)  \diagup
\mathcal{L}^{n+1}\left(  H,A\right)  $. For every $p\in\mathcal{L}\left(
H,A\right)  $, we are going to denote by $\overline{p}$ the projection of $p$
to this factor algebra (i. e., the residue class of $p$ modulo the ideal
$\mathcal{L}^{n+1}\left(  H,A\right)  $). We are going to denote the
multiplication in the factor algebra $\mathcal{L}\left(  H,A\right)
\diagup\mathcal{L}^{n+1}\left(  H,A\right)  $ by the $\ast$ sign, and we are
going to write $q^{\ast i}$ for the $i$-th power of $q$ (in the factor algebra
$\mathcal{L}\left(  H,A\right)  \diagup\mathcal{L}^{n+1}\left(  H,A\right)  $)
whenever $q\in\mathcal{L}\left(  H,A\right)  \diagup\mathcal{L}^{n+1}\left(
H,A\right)  $ and $i\in\mathbb{N}$.

Let $g=e^{\ast f}-e_{H,A}$. Then, $g\in\mathfrak{g}\left(  H,A\right)  $
(since $e^{\ast f}\in G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(
H,A\right)  $, so that $e^{\ast f}-e_{H,A}\in\mathfrak{g}\left(  H,A\right)
$). Hence, Remark~\ref{Remark9} (applied to $g$ instead of $f$) yields
$g^{\ast i}\left(  H_{\leq n}\right)  =0$ for every $i>n$. Also,
Remark~\ref{Remark9} yields $f^{\ast i}\left(  H_{\leq n}\right)  =0$ for
every $i>n$.

We have
\begin{align*}
\operatorname*{Log}\left(  e^{\ast f}\right)   &  =\operatorname{Log}%
_{1}\underbrace{\left(  e^{\ast f}-e_{H,A}\right)  }_{=g}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Log}%
\right) \\
&  =\operatorname{Log}_{1}g.
\end{align*}
Thus, every $x\in H_{\leq n}$ satisfies%
\begin{align*}
\left(  \operatorname*{Log}\left(  e^{\ast f}\right)  \right)  \left(
x\right)   &  =\left(  \operatorname{Log}_{1}g\right)  \left(  x\right)
=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname{Log}%
_{1}\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  +\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{g^{\ast i}\left(  x\right)  }_{\substack{=0\text{ (since
}\\x\in H_{\leq n}\text{ and thus}\\g^{\ast i}\left(  x\right)  \in g^{\ast
i}\left(  H_{\leq n}\right)  =0\\\text{(since }i>n\text{))}}}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  +\underbrace{\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}0}_{=0}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  .
\end{align*}
In other words,%
\[
\operatorname*{Log}\left(  e^{\ast f}\right)  \mid_{H_{\leq n}}=\left(
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\right)
\mid_{H_{\leq n}},
\]
so that%
\begin{align*}
\left(  \operatorname*{Log}\left(  e^{\ast f}\right)  -\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}  &
=\underbrace{\operatorname*{Log}\left(  e^{\ast f}\right)  \mid_{H_{\leq n}}%
}_{=\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\right)  \mid_{H_{\leq n}}}-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}g^{\ast i}\mid_{H_{\leq n}}\\
&  =\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\right)  \mid_{H_{\leq n}}-\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}=0
\end{align*}
and thus%
\[
\operatorname*{Log}\left(  e^{\ast f}\right)  -\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}g^{\ast i}\in\left\{  h\in\mathcal{L}\left(  H,A\right)
\ \mid\ h\mid_{H_{\leq n}}=0\right\}  =\mathcal{L}^{n+1}\left(  H,A\right)
\]
(by (\ref{82.pf.0})). In other words,%
\begin{equation}
\operatorname*{Log}\left(  e^{\ast f}\right)  \equiv\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\operatorname{mod}%
\mathcal{L}^{n+1}\left(  H,A\right)  . \label{82.pf.1}%
\end{equation}


But every $x\in H_{\leq n}$ satisfies%
\begin{align*}
e^{\ast f}\left(  x\right)   &  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(
x\right)  }{i!}=\sum_{i=0}^{n}\dfrac{f^{\ast i}\left(  x\right)  }{i!}%
+\sum_{i=n+1}^{\infty}\underbrace{\dfrac{f^{\ast i}\left(  x\right)  }{i!}%
}_{\substack{=0\text{ (since }\\x\in H_{\leq n}\text{ and thus}\\f^{\ast
i}\left(  x\right)  \in f^{\ast i}\left(  H_{\leq n}\right)  =0\\\text{(since
}i>n\text{), so that}\\f^{\ast i}\left(  x\right)  =0\text{)}}}=\sum_{i=0}%
^{n}\dfrac{f^{\ast i}\left(  x\right)  }{i!}+\underbrace{\sum_{i=n+1}^{\infty
}0}_{=0}=\sum_{i=0}^{n}\dfrac{f^{\ast i}\left(  x\right)  }{i!}\\
&  =\underbrace{\dfrac{f^{\ast0}\left(  x\right)  }{0!}}_{=\dfrac
{e_{H,A}\left(  x\right)  }{1}=e_{H,A}\left(  x\right)  }+\sum_{i=1}^{n}%
\dfrac{f^{\ast i}\left(  x\right)  }{i!}=e_{H,A}\left(  x\right)  +\sum
_{i=1}^{n}\dfrac{f^{\ast i}\left(  x\right)  }{i!}%
\end{align*}
and thus%
\begin{align*}
\underbrace{g}_{=e^{\ast f}-e_{H,A}}\left(  x\right)   &  =\left(  e^{\ast
f}-e_{H,A}\right)  \left(  x\right)  =\underbrace{e^{\ast f}\left(  x\right)
}_{=e_{H,A}\left(  x\right)  +\sum_{i=1}^{n}\dfrac{f^{\ast i}\left(  x\right)
}{i!}}-e_{H,A}\left(  x\right)  =\sum_{i=1}^{n}\dfrac{f^{\ast i}\left(
x\right)  }{i!}=\sum_{i=1}^{n}\dfrac{f^{\ast i}}{i!}\left(  x\right) \\
&  =\sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for }i\text{
in the sum}\right)  .
\end{align*}
In other words,
\[
g\mid_{H_{\leq n}}=\left(  \sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\right)
\mid_{H_{\leq n}}.
\]
Hence,%
\begin{align*}
\left(  g-\sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\right)  \mid_{H_{\leq n}}  &
=\underbrace{g\mid_{H_{\leq n}}}_{=\left(  \sum_{j=1}^{n}\dfrac{f^{\ast j}%
}{j!}\right)  \mid_{H_{\leq n}}}-\left(  \sum_{j=1}^{n}\dfrac{f^{\ast j}}%
{j!}\right)  \mid_{H_{\leq n}}\\
&  =\left(  \sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\right)  \mid_{H_{\leq n}%
}-\left(  \sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\right)  \mid_{H_{\leq n}}=0.
\end{align*}
This yields%
\[
g-\sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\in\left\{  h\in\mathcal{L}\left(
H,A\right)  \ \mid\ h\mid_{H_{\leq n}}=0\right\}  =\mathcal{L}^{n+1}\left(
H,A\right)
\]
(by (\ref{82.pf.0})). In other words,%
\[
g\equiv\sum_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\operatorname{mod}\mathcal{L}%
^{n+1}\left(  H,A\right)  .
\]
Thus, (\ref{82.pf.1}) becomes%
\begin{align}
\operatorname*{Log}\left(  e^{\ast f}\right)   &  \equiv\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \underbrace{g}_{\equiv\sum
_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\operatorname{mod}\mathcal{L}^{n+1}\left(
H,A\right)  }\right)  ^{\ast i}\nonumber\\
&  \equiv\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum
_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\right)  ^{\ast i}\operatorname{mod}%
\mathcal{L}^{n+1}\left(  H,A\right)  . \label{82.pf.3}%
\end{align}


But since $f^{\ast\left(  n+1\right)  }\mid_{H_{\leq n}}=0$ (since
Remark~\ref{Remark9} (applied to $i=n+1$) yields $f^{\ast\left(  n+1\right)
}\left(  H_{\leq n}\right)  =0$), we have $f^{\ast\left(  n+1\right)  }%
\in\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid_{H_{\leq n}%
}=0\right\}  =\mathcal{L}^{n+1}\left(  H,A\right)  $ (by (\ref{82.pf.0})). In
other words, $f^{\ast\left(  n+1\right)  }\equiv0\operatorname{mod}%
\mathcal{L}^{n+1}\left(  H,A\right)  $. In other words, $\overline
{f^{\ast\left(  n+1\right)  }}=0$. Thus, $\overline{f}^{\ast\left(
n+1\right)  }=\overline{f^{\ast\left(  n+1\right)  }}=0$. Hence, we can apply
Corollary~\ref{Corollary15} \textbf{(a)} to $a=\overline{f}$ and obtain
$\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum_{j=1}%
^{n}\dfrac{\overline{f}^{\ast j}}{j!}\right)  ^{\ast i}=\overline{f}$. Hence,%
\[
\overline{\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum
_{j=1}^{n}\dfrac{f^{\ast j}}{j!}\right)  ^{\ast i}}=\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum_{j=1}^{n}\dfrac{\overline
{f}^{\ast j}}{j!}\right)  ^{\ast i}=\overline{f}.
\]
In other words,%
\[
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \sum_{j=1}^{n}%
\dfrac{f^{\ast j}}{j!}\right)  ^{\ast i}\equiv f\operatorname{mod}%
\mathcal{L}^{n+1}\left(  H,A\right)  .
\]
Combined with (\ref{82.pf.3}), this yields $\operatorname*{Log}\left(  e^{\ast
f}\right)  \equiv f\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  $.
In other words,%
\[
\operatorname*{Log}\left(  e^{\ast f}\right)  -f\in\mathcal{L}^{n+1}\left(
H,A\right)  =\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid
_{H_{\leq n}}=0\right\}
\]
(by (\ref{82.pf.0})). In other words, $\left(  \operatorname*{Log}\left(
e^{\ast f}\right)  -f\right)  \mid_{H_{\leq n}}=0$. Hence,%
\[
0=\left(  \operatorname*{Log}\left(  e^{\ast f}\right)  -f\right)
\mid_{H_{\leq n}}=\left(  \operatorname*{Log}\left(  e^{\ast f}\right)
\right)  \mid_{H_{\leq n}}-f\mid_{H_{\leq n}}.
\]
Hence,
\begin{equation}
\left(  \operatorname*{Log}\left(  e^{\ast f}\right)  \right)  \mid_{H_{\leq
n}}=f\mid_{H_{\leq n}}. \label{82.pf.end}%
\end{equation}
We have thus proven this for every $n\in\mathbb{N}$.

Now, let $x\in H$ be arbitrary. Since $H$ is filtered, there must exist some
$n\in\mathbb{N}$ such that $x\in H_{\leq n}$. Consider this $n$. From $x \in
H_{\leq n}$, we obtain
\[
\left(  \operatorname*{Log}\left(  e^{\ast f}\right)  \right)  \left(
x\right)  =\underbrace{\left(  \operatorname*{Log}\left(  e^{\ast f}\right)
\mid_{H_{\leq n}}\right)  }_{\substack{=f\mid_{H_{\leq n}}\\\text{(by
(\ref{82.pf.end}))}}}\left(  x\right)  =\left(  f\mid_{H_{\leq n}}\right)
\left(  x\right)  =f\left(  x\right)  .
\]
Since this holds for every $x\in H$, we can now conclude that
$\operatorname*{Log}\left(  e^{\ast f}\right)  =f$.

This proves Proposition~\ref{Proposition82} \textbf{(a)}.

\textbf{(b)} Let $F\in G\left(  H,A\right)  $. Let $n\in\mathbb{N}$.

According to Proposition~\ref{Proposition81} (applied to $n+1$ instead of
$n$), the subset $\mathcal{L}^{n+1}\left(  H,A\right)  $ of $\mathcal{L}%
\left(  H,A\right)  $ is an ideal of $\mathcal{L}\left(  H,A\right)  $. Thus,
there is a factor algebra $\mathcal{L}\left(  H,A\right)  \diagup
\mathcal{L}^{n+1}\left(  H,A\right)  $. For every $p\in\mathcal{L}\left(
H,A\right)  $, we are going to denote by $\overline{p}$ the projection of $p$
to this factor algebra (i. e., the residue class of $p$ modulo the ideal
$\mathcal{L}^{n+1}\left(  H,A\right)  $). We are going to denote the
multiplication in the factor algebra $\mathcal{L}\left(  H,A\right)
\diagup\mathcal{L}^{n+1}\left(  H,A\right)  $ by the $\ast$ sign, and we are
going to write $q^{\ast i}$ for the $i$-th power of $q$ (in the factor algebra
$\mathcal{L}\left(  H,A\right)  \diagup\mathcal{L}^{n+1}\left(  H,A\right)  $)
whenever $q\in\mathcal{L}\left(  H,A\right)  \diagup\mathcal{L}^{n+1}\left(
H,A\right)  $ and $i\in\mathbb{N}$.

Let $g=F-e_{H,A}$. Then, $g\in\mathfrak{g}\left(  H,A\right)  $ (since $F\in
G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(  H,A\right)  $, so that
$F-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  $). Hence, Remark~\ref{Remark9}
(applied to $g$ instead of $f$) yields $g^{\ast i}\left(  H_{\leq n}\right)
=0$ for every $i>n$.

Let $\varphi=\operatorname*{Log}F$. Then, $\varphi\in\mathfrak{g}\left(
H,A\right)  $, so that Remark~\ref{Remark9} (applied to $\varphi$ instead of
$f$) yields $\varphi^{\ast i}\left(  H_{\leq n}\right)  =0$ for every $i>n$.

By the definition of $\operatorname*{Log}$, we have $\operatorname*{Log}%
F=\operatorname{Log}_{1}\underbrace{\left(  F-e_{H,A}\right)  }_{=g}%
=\operatorname{Log}_{1}g$. Hence, $\varphi=\operatorname*{Log}%
f=\operatorname{Log}_{1}g$.

Thus, every $x\in H_{\leq n}$ satisfies%
\begin{align*}
\varphi\left(  x\right)   &  =\left(  \operatorname{Log}_{1}g\right)  \left(
x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname{Log}_{1}\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  +\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{g^{\ast i}\left(  x\right)  }_{\substack{=0\text{ (since
}\\x\in H_{\leq n}\text{ and thus}\\g^{\ast i}\left(  x\right)  \in g^{\ast
i}\left(  H_{\leq n}\right)  =0\\\text{(since }i>n\text{))}}}=\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(  x\right)
+\underbrace{\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}}{i}0}%
_{=0}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  .
\end{align*}
In other words,%
\[
\varphi\mid_{H_{\leq n}}=\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}.
\]
Now,%
\begin{align*}
\left(  \varphi-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\right)  \mid_{H_{\leq n}}  &  =\underbrace{\varphi\mid_{H_{\leq n}}%
}_{=\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\right)  \mid_{H_{\leq n}}}-\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}\\
&  =\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\right)  \mid_{H_{\leq n}}-\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}=0.
\end{align*}
In other words,%
\[
\varphi-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}%
\in\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid_{H_{\leq n}%
}=0\right\}  =\mathcal{L}^{n+1}\left(  H,A\right)
\]
(by (\ref{82.pf.0})). This rewrites as%
\begin{equation}
\varphi\equiv\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  . \label{82.pf.6}%
\end{equation}


But every $x\in H_{\leq n}$ satisfies%
\begin{align*}
e^{\ast\varphi}\left(  x\right)   &  =\sum_{i\geq0}\dfrac{\varphi^{\ast
i}\left(  x\right)  }{i!}=\sum_{i=0}^{n}\dfrac{\varphi^{\ast i}\left(
x\right)  }{i!}+\sum_{i=n+1}^{\infty}\underbrace{\dfrac{\varphi^{\ast
i}\left(  x\right)  }{i!}}_{\substack{=0\text{ (since }\\x\in H_{\leq n}\text{
and thus}\\\varphi^{\ast i}\left(  x\right)  \in\varphi^{\ast i}\left(
H_{\leq n}\right)  =0\\\text{(since }i>n\text{), so that}\\\varphi^{\ast
i}\left(  x\right)  =0\text{)}}}=\sum_{i=0}^{n}\dfrac{\varphi^{\ast i}\left(
x\right)  }{i!}+\underbrace{\sum_{i=n+1}^{\infty}0}_{=0}=\sum_{i=0}^{n}%
\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}\\
&  =\underbrace{\dfrac{\varphi^{\ast0}\left(  x\right)  }{0!}}_{=\dfrac
{e_{H,A}\left(  x\right)  }{1}=e_{H,A}\left(  x\right)  }+\sum_{i=1}^{n}%
\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}=e_{H,A}\left(  x\right)
+\sum_{i=1}^{n}\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}%
\end{align*}
and thus%
\begin{align*}
\left(  e^{\ast\varphi}-e_{H,A}\right)  \left(  x\right)   &
=\underbrace{e^{\ast\varphi}\left(  x\right)  }_{=e_{H,A}\left(  x\right)
+\sum_{i=1}^{n}\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}}-e_{H,A}\left(
x\right)  =\sum_{i=1}^{n}\dfrac{\varphi^{\ast i}\left(  x\right)  }{i!}%
=\sum_{i=1}^{n}\dfrac{\varphi^{\ast i}}{i!}\left(  x\right) \\
&  =\sum_{j=1}^{n}\dfrac{\varphi^{\ast j}}{j!}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for }i\text{
in the sum}\right)  .
\end{align*}
In other words,
\[
\left(  e^{\ast\varphi}-e_{H,A}\right)  \mid_{H_{\leq n}}=\left(  \sum
_{j=1}^{n}\dfrac{\varphi^{\ast j}}{j!}\right)  \mid_{H_{\leq n}}.
\]
But now,%
\begin{align*}
&  \left(  \left(  e^{\ast\varphi}-e_{H,A}\right)  -\sum_{j=1}^{n}%
\dfrac{\varphi^{\ast j}}{j!}\right)  \mid_{H_{\leq n}}\\
&  =\underbrace{\left(  e^{\ast\varphi}-e_{H,A}\right)  \mid_{H_{\leq n}}%
}_{=\left(  \sum_{j=1}^{n}\dfrac{\varphi^{\ast j}}{j!}\right)  \mid_{H_{\leq
n}}}-\left(  \sum_{j=1}^{n}\dfrac{\varphi^{\ast j}}{j!}\right)  \mid_{H_{\leq
n}}\\
&  =\left(  \sum_{j=1}^{n}\dfrac{\varphi^{\ast j}}{j!}\right)  \mid_{H_{\leq
n}}-\left(  \sum_{j=1}^{n}\dfrac{\varphi^{\ast j}}{j!}\right)  \mid_{H_{\leq
n}}=0.
\end{align*}
In other words,%
\[
\left(  e^{\ast\varphi}-e_{H,A}\right)  -\sum_{j=1}^{n}\dfrac{\varphi^{\ast
j}}{j!}\in\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid_{H_{\leq
n}}=0\right\}  =\mathcal{L}^{n+1}\left(  H,A\right)
\]
(by (\ref{82.pf.0})). This rewrites as%
\[
e^{\ast\varphi}-e_{H,A}\equiv\sum_{j=1}^{n}\dfrac{\varphi^{\ast j}}%
{j!}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  .
\]
Substituting (\ref{82.pf.6}) into the right hand side of this congruence, we
get%
\begin{equation}
e^{\ast\varphi}-e_{H,A}\equiv\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\right)  ^{\ast j}}%
{j!}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  . \label{82.pf.8}%
\end{equation}


But since $g^{\ast\left(  n+1\right)  }\mid_{H_{\leq n}}=0$ (since
Remark~\ref{Remark9} (applied to $n+1$ and $g$ instead of $i$ and $f$) yields
$g^{\ast\left(  n+1\right)  }\left(  H_{\leq n}\right)  =0$), we have
$g^{\ast\left(  n+1\right)  }\in\left\{  h\in\mathcal{L}\left(  H,A\right)
\ \mid\ h\mid_{H_{\leq n}}=0\right\}  =\mathcal{L}^{n+1}\left(  H,A\right)  $
(by (\ref{82.pf.0})). In other words, $g^{\ast\left(  n+1\right)  }%
\equiv0\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  $, so that
$\overline{g^{\ast\left(  n+1\right)  }}=0$. Thus, $\overline{g}^{\ast\left(
n+1\right)  }=\overline{g^{\ast\left(  n+1\right)  }}=0$. Hence, we can apply
Corollary~\ref{Corollary15} \textbf{(b)} to $b=\overline{g}$ and obtain
$\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}\overline{g}^{\ast i}\right)  ^{\ast j}}{j!}=\overline{g}$. Thus,%
\[
\overline{\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  ^{\ast j}}{j!}}=\sum_{j=1}^{n}\dfrac{\left(
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\overline{g}^{\ast
i}\right)  ^{\ast j}}{j!}=\overline{g}.
\]
In other words,%
\[
\sum_{j=1}^{n}\dfrac{\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}g^{\ast i}\right)  ^{\ast j}}{j!}\equiv g\operatorname{mod}\mathcal{L}%
^{n+1}\left(  H,A\right)  .
\]
Combined with (\ref{82.pf.8}), this yields $e^{\ast\varphi}-e_{H,A}\equiv
g\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  $. Since $g=F-e_{H,A}%
$, this rewrites as $e^{\ast\varphi}-e_{H,A}\equiv F-e_{H,A}\operatorname{mod}%
\mathcal{L}^{n+1}\left(  H,A\right)  $. This simplifies to $e^{\ast\varphi
}\equiv F\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  $. In other
words,%
\[
e^{\ast\varphi}-F\in\mathcal{L}^{n+1}\left(  H,A\right)  =\left\{
h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid_{H_{\leq n}}=0\right\}
\]
(by (\ref{82.pf.0})). In other words, $\left(  e^{\ast\varphi}-F\right)
\mid_{H_{\leq n}}=0$. Thus,%
\[
0=\left(  e^{\ast\varphi}-F\right)  \mid_{H_{\leq n}}=e^{\ast\varphi}%
\mid_{H_{\leq n}}-F\mid_{H_{\leq n}},
\]
so that $e^{\ast\varphi}\mid_{H_{\leq n}}=F\mid_{H_{\leq n}}$. Since
$\varphi=\operatorname*{Log}F$, this becomes
\begin{equation}
e^{\ast\left(  \operatorname*{Log}F\right)  }\mid_{H_{\leq n}}=F\mid_{H_{\leq
n}}. \label{82.pf.fnd}%
\end{equation}
We have thus proven this for every $n\in\mathbb{N}$.

Now, let $x\in H$ be arbitrary. Since $H$ is filtered, there must exist some
$n\in\mathbb{N}$ such that $x\in H_{\leq n}$. Consider this $n$. From $x \in
H_{\leq n}$, we obtain
\[
e^{\ast\left(  \operatorname*{Log}F\right)  }\left(  x\right)
=\underbrace{\left(  e^{\ast\left(  \operatorname*{Log}F\right)  }%
\mid_{H_{\leq n}}\right)  }_{\substack{=F\mid_{H_{\leq n}}\\\text{(by
(\ref{82.pf.fnd}))}}}\left(  x\right)  =\left(  F\mid_{H_{\leq n}}\right)
\left(  x\right)  =F\left(  x\right)  .
\]
Since this holds for every $x\in H$, we can now conclude that $e^{\ast\left(
\operatorname*{Log}F\right)  }=F$.

This proves Proposition~\ref{Proposition82} \textbf{(b)}.

We are thus done proving Proposition~\ref{Proposition82}.
\end{proof}

\section{\label{S12}The dual theorem}

Before we proceed, let us introduce two pieces of notation:

\begin{convention}
\label{Convention99} Let $k$ be a field. Let $V$ be a $k$-vector space. Let
$S$ be a subset of $V$. Then, $\left\langle S\right\rangle $ will denote the
$k$-vector subspace of $V$ generated by the subset $S$. Note that
$\left\langle S\right\rangle $ is the set of all $k$-linear combinations of
elements of $S$. We have $S\subseteq\left\langle S\right\rangle $.
\end{convention}

\begin{convention}
\label{Convention100} Let $k$ be a field. Let $A$ be a $k$-algebra.

If $U$ and $V$ are two $k$-vector subspaces of $A$, then we let $UV$ denote
the $k$-vector subspace%
\[
\left\langle uv\ \mid\ \left(  u,v\right)  \in U\times V\right\rangle
\]
of $A$. This subspace is called the \textit{product} of the subspaces $U$ and
$V$ of $A$; we will also denote it by $U\cdot V$. (Note that it depends on the
base field $k$.)

This definition makes the set of all $k$-vector subspaces of $A$ into a
multiplicative monoid, with neutral element $k\cdot1_{A}$ (the subspace of all
scalar multiples of the unity $1_{A}$). Any $k$-vector subspace $V$ of $A$
satisfies $\left(  k\cdot1_{A}\right)  \cdot V=V\cdot\left(  k\cdot
1_{A}\right)  =V$. Any three $k$-vector subspaces $U$, $V$ and $W$ of $A$
satisfy $\left(  UV\right)  W=U\left(  VW\right)  $. We will use the standard
notations that are used in multiplicative monoids (such as writing $UVW$ for
the product $\left(  UV\right)  W=U\left(  VW\right)  $). In particular, if
$V$ is a $k$-vector subspace of $A$ and $n$ is a nonnegative integer, then
$V^{n}$ will denote the $n$-th power of $V$ in the multiplicative monoid of
all $k$-vector subspaces of $A$ (that is, the subspace $\underbrace{VV\cdots
V}_{n\text{ times}}$ of $A$).

Three remarks about this are in order. Firstly, the notation $V^{n}$ cannot be
misunderstood for the Cartesian product $\underbrace{V\times V\times
\cdots\times V}_{n\text{ times}}$, because we are denoting the latter
Cartesian product by $V^{\times n}$ and not by $V^{n}$ (see
Convention~\ref{Convention3bs}).

Secondly, it is obvious that standard power rules that hold in monoids hold in
the multiplicative monoid of all $k$-vector subspaces of $A$. For instance,
$V^{n}V^{m}=V^{n+m}$ for any $k$-vector subspace $V$ of $A$ and every
$n\in\mathbb{N}$ and $m\in\mathbb{N}$.

Furthermore, if $n$ is a nonnegative integer, and $V_{1}$, $V_{2}$, $\ldots$,
$V_{n}$ are $n$ arbitrary $k$-vector subspaces of a $k$-algebra $A$, then%
\begin{equation}
V_{1}V_{2}\cdots V_{n}=\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V_{1}\times V_{2}\times\cdots\times
V_{n}\right\rangle . \label{100.V1V2...Vn}%
\end{equation}
In particular, if $n$ is a nonnegative integer, and $V$ is a $k$-vector
subspace of a $k$-algebra $A$, then%
\begin{equation}
V^{n}=\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\rangle . \label{100.Vn}%
\end{equation}


Thirdly, some authors define the notion of an $n$-th power of an ideal of a
ring. When the ring is $A$, their definition of the $n$-th power of an ideal
$I$ of $A$ differs from our definition of $I^{n}$ in that they define it as an
$A$-linear span whereas we define it as a $k$-linear span. This difference is
insubstantial if $n>0$ (that is, the reader can easily check that the two
definitions are equivalent if $n>0$). But if $n=0$, then their definition
gives a different result than ours. Namely, their definition leads to
$I^{0}=A$ while ours leads to $I^{0}=k\cdot1_{A}$. While their definition is
more suitable for studying ideals (in particular, it ensures that
$I^{0}\supseteq I^{1}\supseteq I^{2}\supseteq\cdots$ for any ideal $I$,
whereas our definition only ensures that ideals $I$ satisfy $I^{1}\supseteq
I^{2}\supseteq I^{3}\supseteq\cdots$ but usually not $I^{0}\supseteq I^{1}$),
our definition is more useful for arbitrary $k$-vector subspaces.
\end{convention}

Now we are going to state and prove a result which is, in some sense, dual to
Theorem~\ref{Theorem1}:

\begin{theorem}
\label{Theorem101} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected filtered commutative bialgebra over $k$. Consider the convolution
algebra $\mathcal{L}\left(  H,H\right)  $. The map $\operatorname*{Log}%
\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $ is a projection such
that $\operatorname*{Ker}\left(  \operatorname*{Log}\operatorname*{id}\right)
=\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}$.\ \ \ \ \footnote{Recall that $\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}$ is to be understood according
to Convention~\ref{Convention100}. Thus, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of $H$.}
\end{theorem}

Theorem~\ref{Theorem101} is not directly obtainable from
Theorem~\ref{Theorem1} by dualization: First of all, the dual of a bialgebra
is not necessarily a bialgebra (the dual of a coalgebra is always an algebra,
but the dual of an algebra needs not be a coalgebra unless the algebra is
finite-dimensional), but even this technical hurdle aside, the dual of a
coalgebra filtration is not a filtration but something reasonably called a
``cofiltration'' (a decreasing filtration with intersection $0$). Given these
difficulties, one might be surprised that Theorem~\ref{Theorem101} is correct
at all. Fortunately, the filtration is not really an important part of
Theorems~\ref{Theorem1} and~\ref{Theorem101} - its main purpose is to make
$\operatorname*{Log}\operatorname*{id}$ (and, more generally,
$\operatorname*{Log}f$ for $f\in\mathfrak{g}\left(  H,H\right)  $)
well-defined. The main core of Theorem~\ref{Theorem101} \textit{is} dual to
that of Theorem~\ref{Theorem1}. We will make this clear by giving a proof of
Theorem~\ref{Theorem101} which mirrors the above proof of
Theorem~\ref{Theorem1} - at least it will have the same large-scale structure,
using lemmas which are duals of respective lemmas used to prove
Theorem~\ref{Theorem1} (though some of the lemmas will not have to be modified
at all). As far as the details of the proof are concerned (the proofs of these
lemmas), they will be simpler than those in the proof of
Theorem~\ref{Theorem1}, since we will be able to replace many complicated
computations with maps by simpler computations with elements.

First let us formulate a proposition, which is in some sense a dual of
Proposition~\ref{Proposition19}:

\begin{proposition}
\label{Proposition104} Let $k$ be a field of characteristic $0$. Let $A$ be a
$k$-algebra, and let $H$ be a$\ $connected filtered $k$-coalgebra. Let
$\varepsilon_{A}:A\rightarrow k$ be a $k$-algebra homomorphism. Then:

\textbf{(a)} Any $f\in\mathcal{L}\left(  H,A\right)  $ and any $g\in
\mathcal{L}\left(  H,A\right)  $ such that $\varepsilon_{A}\circ f=0$ and
$\varepsilon_{A}\circ g=0$ satisfy $\varepsilon_{A}\circ\left(  f\ast
g\right)  =0$ and $\left(  f\ast g\right)  \left(  H\right)  \subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}%
$.\ \ \ \ \footnote{Recall that $\left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  ^{2}$ is defined according to
Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  $ of $A$.}

\textbf{(b)} Every $f\in\mathcal{L}\left(  H,A\right)  $ such that
$\varepsilon_{A}\circ f=0$ and every integer $i>1$ satisfy $\varepsilon
_{A}\circ f^{\ast i}=0$ and $f^{\ast i}\left(  H\right)  \subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}$.

\textbf{(c)} Every $F\in G\left(  H,A\right)  $ satisfying $\varepsilon
_{A}\circ F=\varepsilon_{H}$ satisfies $\left(  \operatorname*{Log}F-F\right)
\left(  H\right)  \subseteq\left(  \operatorname*{Ker}\left(  \varepsilon
_{A}\right)  \right)  ^{2}+k\cdot1_{A}$.
\end{proposition}

Note that the duality between Proposition~\ref{Proposition19} and
Proposition~\ref{Proposition104} is not perfect. The condition $\varepsilon
_{A}\circ F=\varepsilon_{H}$ of Proposition~\ref{Proposition104} \textbf{(c)}
is dual to the condition $F\left(  1_{H}\right)  =1_{A}$ (which is implicit in
the condition $F\in G\left(  H,A\right)  $) of Proposition~\ref{Proposition19}
\textbf{(c)}, but the condition $F\in G\left(  H,A\right)  $ of
Proposition~\ref{Proposition104} \textbf{(c)} is not dual to anything required
in Proposition~\ref{Proposition19} \textbf{(c)} - it just is there to make
sure that $\operatorname*{Log}F$ is well-defined.

Before we start proving Proposition~\ref{Proposition104}, let us show a simple lemma:

\begin{lemma}
\label{lem.alg.muUV}Let $k$ be a field. Let $A$ be a $k$-algebra. Then,%
\begin{equation}
\text{any two }k\text{-vector subspaces }U\text{ and }V\text{ of }A\text{
satisfy }\mu_{A}\left(  U\otimes V\right)  =UV \label{104.UV}%
\end{equation}
(where we consider $U\otimes V$ as a $k$-vector subspace of $A\otimes A$ by
tensoring the inclusion maps $U\rightarrow A$ and $V\rightarrow A$).
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.alg.muUV}.]Let $U$ and $V$ be two $k$-vector
subspaces of $A$. Let $\operatorname*{tensor}:U\times V\rightarrow U\otimes V$
be the map defined by%
\[
\left(  \operatorname*{tensor}\left(  u,v\right)  =u\otimes v\text{ for every
}\left(  u,v\right)  \in U\times V\right)  .
\]


Recall that for every $k$-vector space $M$ and every subset $S$ of $M$, we
denote by $\left\langle S\right\rangle $ the $k$-vector subspace of $M$
generated by the elements of $S$.

For every $k$-vector space $M$, every set $\Phi$ and every map $P:\Phi
\rightarrow M$, let us denote by $\left\langle P\left(  v\right)  \ \mid
\ v\in\Phi\right\rangle $ the subspace $\left\langle \left\{  P\left(
v\right)  \ \mid\ v\in\Phi\right\}  \right\rangle $ of $M$ (this is the
$k$-vector subspace of $M$ generated by all the elements $P\left(  v\right)  $
with $v\in\Phi$).

It is a known fact that any two $k$-vector spaces $M$ and $R$, any $k$-linear
map $\phi:M\rightarrow R$ and every subset $S$ of $M$ satisfy $\phi\left(
\left\langle S\right\rangle \right)  =\left\langle \phi\left(  S\right)
\right\rangle $. Applied to $M=A\otimes A$, $R=A$, $\phi=\mu_{A}$ and
$S=\left\{  \operatorname*{tensor}\left(  u,v\right)  \ \mid\ \left(
u,v\right)  \in U\times V\right\}  $, this yields%
\[
\mu_{A}\left(  \left\langle \left\{  \operatorname*{tensor}\left(  u,v\right)
\ \mid\ \left(  u,v\right)  \in U\times V\right\}  \right\rangle \right)
=\left\langle \mu_{A}\left(  \left\{  \operatorname*{tensor}\left(
u,v\right)  \ \mid\ \left(  u,v\right)  \in U\times V\right\}  \right)
\right\rangle .
\]
Since $\operatorname*{tensor}\left(  u,v\right)  =u\otimes v$, this rewrites
as%
\[
\mu_{A}\left(  \left\langle \left\{  u\otimes v\ \mid\ \left(  u,v\right)  \in
U\times V\right\}  \right\rangle \right)  =\left\langle \mu_{A}\left(
\left\{  u\otimes v\ \mid\ \left(  u,v\right)  \in U\times V\right\}  \right)
\right\rangle .
\]


Now, since the tensor product $U\otimes V$ is generated by pure tensors, we
have%
\[
U\otimes V=\left\langle u\otimes v\ \mid\ \left(  u,v\right)  \in U\times
V\right\rangle =\left\langle \left\{  u\otimes v\ \mid\ \left(  u,v\right)
\in U\times V\right\}  \right\rangle ,
\]
so that%
\begin{align*}
\mu_{A}\left(  U\otimes V\right)   &  =\mu_{A}\left(  \left\langle \left\{
u\otimes v\ \mid\ \left(  u,v\right)  \in U\times V\right\}  \right\rangle
\right)  =\left\langle \underbrace{\mu_{A}\left(  \left\{  u\otimes
v\ \mid\ \left(  u,v\right)  \in U\times V\right\}  \right)  }_{=\left\{
\mu_{A}\left(  u\otimes v\right)  \ \mid\ \left(  u,v\right)  \in U\times
V\right\}  }\right\rangle \\
&  =\left\langle \left\{  \underbrace{\mu_{A}\left(  u\otimes v\right)
}_{\substack{=uv\\\text{(since }\mu_{A}\text{ is the multiplication map)}%
}}\ \mid\ \left(  u,v\right)  \in U\times V\right\}  \right\rangle \\
&  =\left\langle \left\{  uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}
\right\rangle =\left\langle uv\ \mid\ \left(  u,v\right)  \in U\times
V\right\rangle =UV.
\end{align*}
This proves Lemma~\ref{lem.alg.muUV}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition104}.]\textbf{(a)} We have
$\varepsilon_{A}\left(  f\left(  H\right)  \right)  =\underbrace{\left(
\varepsilon_{A}\circ f\right)  }_{=0}\left(  H\right)  =0\left(  H\right)  =0$
and thus $f\left(  H\right)  \subseteq\operatorname*{Ker}\left(
\varepsilon_{A}\right)  $. Similarly, $g\left(  H\right)  \subseteq
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  $.

By the definition of convolution, $f\ast g=\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{H}$. Thus,%
\begin{align*}
\left(  f\ast g\right)  \left(  H\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{H}\right)  \left(  H\right)  =\mu_{A}\left(
\left(  f\otimes g\right)  \underbrace{\left(  \Delta_{H}\left(  H\right)
\right)  }_{\subseteq H\otimes H}\right)  \subseteq\mu_{A}\left(
\underbrace{\left(  f\otimes g\right)  \left(  H\otimes H\right)  }_{=f\left(
H\right)  \otimes g\left(  H\right)  }\right) \\
&  =\mu_{A}\left(  \underbrace{f\left(  H\right)  }_{\subseteq
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  }\otimes
\underbrace{g\left(  H\right)  }_{\subseteq\operatorname*{Ker}\left(
\varepsilon_{A}\right)  }\right)  \subseteq\mu_{A}\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  \otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  \right)  .
\end{align*}


But (\ref{104.UV}) (applied to $U=\operatorname*{Ker}\left(  \varepsilon
_{A}\right)  $ and $V=\operatorname*{Ker}\left(  \varepsilon_{A}\right)  $)
yields $\mu_{A}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{A}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(  \varepsilon
_{A}\right)  \right)  \right)  =\left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  \left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  =\left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  ^{2}$. Thus, $\left(  f\ast g\right)  \left(
H\right)  \subseteq\mu_{A}\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  \right)  $ rewrites as $\left(  f\ast
g\right)  \left(  H\right)  \subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  ^{2}$. Hence,%
\begin{align*}
\varepsilon_{A}\left(  \left(  f\ast g\right)  \left(  H\right)  \right)   &
\subseteq\varepsilon_{A}\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  ^{2}\right)  \subseteq\left(
\underbrace{\varepsilon_{A}\left(  \operatorname*{Ker}\left(  \varepsilon
_{A}\right)  \right)  }_{=0}\right)  ^{2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon_{A}\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  =0^{2}=0,
\end{align*}
so that $\left(  \varepsilon_{A}\circ\left(  f\ast g\right)  \right)  \left(
H\right)  =\varepsilon_{A}\left(  \left(  f\ast g\right)  \left(  H\right)
\right)  =0$. In other words, $\varepsilon_{A}\circ\left(  f\ast g\right)
=0$. This completes the proof of Proposition~\ref{Proposition104} \textbf{(a)}.

\textbf{(b)} We will prove Proposition~\ref{Proposition104} \textbf{(b)} by
induction over $i$:

\textit{Induction base:} Proposition~\ref{Proposition104} \textbf{(a)}
(applied to $g=f$) yields $\varepsilon_{A}\circ\left(  f\ast f\right)  =0$ and
$\left(  f\ast f\right)  \left(  H\right)  \subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}$. Since
$f\ast f=f^{\ast2}$, this rewrites as follows: We have $\varepsilon_{A}\circ
f^{\ast2}=0$ and $f^{\ast2}\left(  H\right)  \subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}$. This proves
Proposition~\ref{Proposition104} \textbf{(b)} for $i=2$. The induction base is
thus complete.

\textit{Induction step:} Let $j>1$ be an integer. Assume that
Proposition~\ref{Proposition104} \textbf{(b)} holds for $i=j$. We must then
prove Proposition~\ref{Proposition104} \textbf{(b)} for $i=j+1$.

Since Proposition~\ref{Proposition104} \textbf{(b)} holds for $i=j$, we have
$\varepsilon_{A}\circ f^{\ast j}=0$ and $f^{\ast j}\left(  H\right)
\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)
^{2}$. Now, Proposition~\ref{Proposition104} \textbf{(a)} (applied to
$g=f^{\ast j}$) yields $\varepsilon_{A}\circ\left(  f\ast f^{\ast j}\right)
=0$ and $\left(  f\ast f^{\ast j}\right)  \left(  H\right)  \subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}$. Since
$f\ast f^{\ast j}=f^{\ast\left(  j+1\right)  }$, this rewrites as follows: We
have $\varepsilon_{A}\circ f^{\ast\left(  j+1\right)  }=0$ and $f^{\ast\left(
j+1\right)  }\left(  H\right)  \subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{A}\right)  \right)  ^{2}$. In other words,
Proposition~\ref{Proposition104} \textbf{(b)} holds for $i=j+1$. The induction
step is thus complete.

This completes the induction proof of Proposition~\ref{Proposition104}
\textbf{(b)}.

\textbf{(c)} Let $F\in G\left(  H,A\right)  $ satisfy $\varepsilon_{A}\circ
F=\varepsilon_{H}$.

Since $\varepsilon_{A}$ is a $k$-algebra homomorphism, we have $\varepsilon
_{A}\circ\eta_{A}=\eta_{k}=\operatorname{id}_{k}$.

Let $f=F-e_{H,A}$. Then,%
\begin{align*}
\varepsilon_{A}\circ f  &  =\varepsilon_{A}\circ\left(  F-e_{H,A}\right)
=\underbrace{\varepsilon_{A}\circ F}_{=\varepsilon_{H}}-\varepsilon_{A}%
\circ\underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{H}}=\varepsilon
_{H}-\underbrace{\varepsilon_{A}\circ\eta_{A}}_{=\operatorname{id}_{k}}%
\circ\varepsilon_{H}\\
&  =\varepsilon_{H}-\varepsilon_{H}=0
\end{align*}
and $f\in\mathfrak{g}\left(  H,A\right)  $ (since $F\in G\left(  H,A\right)
=e_{H,A}+\mathfrak{g}\left(  H,A\right)  $, so that $F-e_{H,A}\in
\mathfrak{g}\left(  H,A\right)  $).

Now, the definition of $\operatorname*{Log}$ says that $\operatorname*{Log}%
F=\operatorname{Log}_{1}\underbrace{\left(  F-e_{H,A}\right)  }_{=f}%
=\operatorname{Log}_{1}f$.

Let $x\in H$. Then,%
\begin{align*}
\underbrace{\left(  \operatorname*{Log}F\right)  }_{=\operatorname{Log}_{1}%
f}\left(  x\right)   &  =\left(  \operatorname{Log}_{1}f\right)  \left(
x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
x\right)  =\underbrace{\dfrac{\left(  -1\right)  ^{1-1}}{1}}_{=1}f^{\ast
1}\left(  x\right)  +\sum_{i>1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right) \\
&  =\underbrace{1f^{\ast1}\left(  x\right)  }_{=f^{\ast1}\left(  x\right)
=f\left(  x\right)  }+\sum_{i>1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  =f\left(  x\right)  +\sum_{i>1}\dfrac{\left(  -1\right)
^{i-1}}{i}f^{\ast i}\left(  x\right)  .
\end{align*}
On the other hand, $f=F-e_{H,A}$ yields
\begin{align*}
f\left(  x\right)   &  =\left(  F-e_{H,A}\right)  \left(  x\right)  =F\left(
x\right)  -\underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{E}}\left(
x\right)  =F\left(  x\right)  -\left(  \eta_{A}\circ\varepsilon_{H}\right)
\left(  x\right) \\
&  =F\left(  x\right)  -\underbrace{\eta_{A}\left(  \varepsilon_{H}\left(
x\right)  \right)  }_{\substack{=\varepsilon_{H}\left(  x\right)  \cdot
1_{A}\\\text{(by the definition of }\eta_{A}\text{)}}}=F\left(  x\right)
-\varepsilon_{H}\left(  x\right)  \cdot1_{A},
\end{align*}
so that $F\left(  x\right)  =f\left(  x\right)  +\varepsilon_{H}\left(
x\right)  \cdot1_{A}$. Thus,%
\begin{align}
\left(  \operatorname*{Log}F-F\right)  \left(  x\right)   &
=\underbrace{\left(  \operatorname*{Log}F\right)  \left(  x\right)
}_{=f\left(  x\right)  +\sum_{i>1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  }-\underbrace{F\left(  x\right)  }_{=f\left(  x\right)
+\varepsilon_{H}\left(  x\right)  \cdot1_{A}}\nonumber\\
&  =\left(  f\left(  x\right)  +\sum_{i>1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}f^{\ast i}\left(  x\right)  \right)  -\left(  f\left(  x\right)
+\varepsilon_{H}\left(  x\right)  \cdot1_{A}\right) \nonumber\\
&  =\sum_{i>1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)
+\varepsilon_{H}\left(  x\right)  \cdot1_{A}. \label{104.pf.2}%
\end{align}
But every integer $i>1$ satisfies $f^{\ast i}\left(  H\right)  \subseteq
\left(  \operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}$ (by
Proposition~\ref{Proposition104} \textbf{(b)}) and thus $f^{\ast i}\left(
x\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{A}\right)
\right)  ^{2}$ (since $x\in H$ and thus $f^{\ast i}\left(  x\right)  \in
f^{\ast i}\left(  H\right)  $). Hence, $\sum_{i>1}\dfrac{\left(  -1\right)
^{i-1}}{i}f^{\ast i}\left(  x\right)  $ is a finite $k$-linear combination of
elements of $\left(  \operatorname*{Ker}\left(  \varepsilon_{A}\right)
\right)  ^{2}$ (it is finite because we know that only finitely many addends
of the infinite sum $\sum_{i>1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  $ are nonzero). Therefore, $\sum_{i>1}\dfrac{\left(
-1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)  $ is an element of $\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}$ (because
$\left(  \operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}$ is
a $k$-vector space, so that every finite $k$-linear combination of elements of
$\left(  \operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}$
must be an element of $\left(  \operatorname*{Ker}\left(  \varepsilon
_{A}\right)  \right)  ^{2}$ itself). Thus, (\ref{104.pf.2}) becomes%
\[
\left(  \operatorname*{Log}F-F\right)  \left(  x\right)  =\underbrace{\sum
_{i>1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)  }%
_{\in\left(  \operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}%
}+\underbrace{\varepsilon_{H}\left(  x\right)  \cdot1_{A}}_{\in k\cdot1_{A}%
}\in\left(  \operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)
^{2}+k\cdot1_{A}.
\]
Since this holds for every $x\in H$, we thus have proven that $\left(
\operatorname*{Log}F-F\right)  \left(  H\right)  \subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{A}\right)  \right)  ^{2}+k\cdot1_{A}$.
This proves Proposition~\ref{Proposition104} \textbf{(c)}.
\end{proof}

Next, just as we defined the notion of $\left(  \varepsilon,\varepsilon
\right)  $-coderivations in Definition~\ref{Definition22}, let us define
$\left(  \varepsilon,\varepsilon\right)  $-derivations (which is, in fact, a
well-known notion, after which the notion of $\left(  \varepsilon
,\varepsilon\right)  $-coderivations was modelled after):

\begin{definition}
\label{Definition108} Let $k$ be a field. Let $H$ be a $k$-algebra, and let
$\varepsilon_{H}:H\rightarrow k$ be a $k$-algebra homomorphism. Let $A$ be a
$k$-algebra. Let $f:H\rightarrow A$ be a $k$-linear map. Then, $f$ is said to
be an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $\textit{-derivation}
if and only if $f\circ\mu_{H}=\mu_{A}\circ\left(  f\otimes e_{H,A}%
+e_{H,A}\otimes f\right)  $. Here, the map $e_{H,A}$ is defined to be the map
$\eta_{A}\circ\varepsilon_{H}:H\rightarrow A$ (this definition of the map
$e_{H,A}$ is identical with the definition of the map $e_{H,A}$ in
Definition~\ref{def.eHA}).
\end{definition}

This definition can be very easily equivalently rewritten as follows:

\begin{definition}
\label{Definition109} Let $k$ be a field. Let $H$ be a $k$-algebra, and let
$\varepsilon_{H}:H\rightarrow k$ be a $k$-algebra homomorphism. Let $A$ be a
$k$-algebra. Let $f:H\rightarrow A$ be a $k$-linear map. Then, $f$ is said to
be an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $\textit{-derivation}
if and only if every $\left(  a,b\right)  \in H\times H$ satisfies $f\left(
ab\right)  =f\left(  a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon
_{H}\left(  a\right)  f\left(  b\right)  $.
\end{definition}

We will be able to use both Definitions~\ref{Definition108}
and~\ref{Definition109} in parallel as soon as we have shown the following proposition:

\begin{proposition}
\label{Proposition110} Definition~\ref{Definition108} and
Definition~\ref{Definition109} are equivalent.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition110}.]Let $k$ be a field. Let $H$ be a
$k$-algebra, and let $\varepsilon_{H}:H\rightarrow k$ be a $k$-algebra
homomorphism. Let $A$ be a $k$-algebra. Let $f:H\rightarrow A$ be a $k$-linear map.

It is well-known that two $k$-linear maps from a tensor product are equal if
and only if they are equal on each pure tensor. Applying this fact to the two
$k$-linear maps $f\circ\mu_{H}:H\otimes H\rightarrow A$ and $\mu_{A}%
\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)  :H\otimes H\rightarrow
A$, we conclude that we have the following equivalence:%
\begin{align}
&  \ \left(  \text{the two maps }f\circ\mu_{H}\text{ and }\mu_{A}\circ\left(
f\otimes e_{H,A}+e_{H,A}\otimes f\right)  \text{ are equal}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the two maps }f\circ\mu_{H}\text{ and
}\mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)  \text{ are
equal on each pure tensor}\right)  . \label{110.pf.1}%
\end{align}
Since pure tensors in $H\otimes H$ are tensors of the form $a\otimes b$ with
$\left(  a,b\right)  \in H\times H$, we have the equivalence%
\begin{align}
&  \ \left(  \text{the two maps }f\circ\mu_{H}\text{ and }\mu_{A}\circ\left(
f\otimes e_{H,A}+e_{H,A}\otimes f\right)  \text{ are equal on each pure
tensor}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the two maps }f\circ\mu_{H}\text{ and
}\mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)  \text{
are}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  \text{equal on }a\otimes
b\text{ for each }\left(  a,b\right)  \in H\times H\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  f\circ\mu_{H}\right)  \left(  a\otimes
b\right)  =\left(  \mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes
f\right)  \right)  \left(  a\otimes b\right)  \text{ for each }\left(
a,b\right)  \in H\times H\right)  . \label{110.pf.3}%
\end{align}
But every $\left(  a,b\right)  \in H\times H$ satisfies%
\[
\left(  f\circ\mu_{H}\right)  \left(  a\otimes b\right)  =f\left(
\underbrace{\mu_{H}\left(  a\otimes b\right)  }_{\substack{=ab\\\text{(since
}\mu_{H}\text{ is the multiplication map)}}}\right)  =f\left(  ab\right)
\]
and%
\begin{align*}
&  \left(  \mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)
\right)  \left(  a\otimes b\right) \\
&  =\mu_{A}\left(  \underbrace{\left(  f\otimes e_{H,A}+e_{H,A}\otimes
f\right)  \left(  a\otimes b\right)  }_{=\left(  f\otimes e_{H,A}\right)
\left(  a\otimes b\right)  +\left(  e_{H,A}\otimes f\right)  \left(  a\otimes
b\right)  }\right) \\
&  =\mu_{A}\left(  \underbrace{\left(  f\otimes e_{H,A}\right)  \left(
a\otimes b\right)  }_{=f\left(  a\right)  \otimes e_{H,A}\left(  b\right)
}+\underbrace{\left(  e_{H,A}\otimes f\right)  \left(  a\otimes b\right)
}_{=e_{H,A}\left(  a\right)  \otimes f\left(  b\right)  }\right) \\
&  =\mu_{A}\left(  f\left(  a\right)  \otimes e_{H,A}\left(  b\right)
+e_{H,A}\left(  a\right)  \otimes f\left(  b\right)  \right) \\
&  =f\left(  a\right)  \underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{H}%
}\left(  b\right)  +\underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{H}%
}\left(  a\right)  f\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mu_{A}\text{ is the multiplication map}\right) \\
&  =f\left(  a\right)  \underbrace{\left(  \eta_{A}\circ\varepsilon
_{H}\right)  \left(  b\right)  }_{\substack{=\eta_{A}\left(  \varepsilon
_{H}\left(  b\right)  \right)  =\varepsilon_{H}\left(  b\right)  \cdot
1_{A}\\\text{(by the definition of }\eta_{A}\text{)}}}+\underbrace{\left(
\eta_{A}\circ\varepsilon_{H}\right)  \left(  a\right)  }_{\substack{=\eta
_{A}\left(  \varepsilon_{H}\left(  a\right)  \right)  =\varepsilon_{H}\left(
a\right)  \cdot1_{A}\\\text{(by the definition of }\eta_{A}\text{)}}}f\left(
b\right) \\
&  =f\left(  a\right)  \varepsilon_{H}\left(  b\right)  \cdot1_{A}%
+\varepsilon_{H}\left(  a\right)  \cdot1_{A}f\left(  b\right)  =f\left(
a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)
f\left(  b\right)  .
\end{align*}


Now, we have the following chain of equivalences:%
\begin{align*}
&  \ \left(  \text{the map }f\text{ is an }\left(  \varepsilon_{H}%
,\varepsilon_{H}\right)  \text{-derivation in the sense of
Definition~\ref{Definition108}}\right) \\
&  \Longleftrightarrow\ \left(  f\circ\mu_{H}=\mu_{A}\circ\left(  f\otimes
e_{H,A}+e_{H,A}\otimes f\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Definition~\ref{Definition108}}\right) \\
&  \Longleftrightarrow\ \left(  \text{the two maps }f\circ\mu_{H}\text{ and
}\mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)  \text{ are
equal}\right) \\
&  \Longleftrightarrow\ \left(  \text{the two maps }f\circ\mu_{H}\text{ and
}\mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)  \text{ are
equal on each pure tensor}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{110.pf.1})}\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  f\circ\mu_{H}\right)
\left(  a\otimes b\right)  }_{=f\left(  ab\right)  }=\underbrace{\left(
\mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)  \right)  \left(
a\otimes b\right)  }_{=f\left(  a\right)  \varepsilon_{H}\left(  b\right)
+\varepsilon_{H}\left(  a\right)  f\left(  b\right)  }\text{ for each }\left(
a,b\right)  \in H\times H\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{110.pf.3})}\right) \\
&  \Longleftrightarrow\ \left(  f\left(  ab\right)  =f\left(  a\right)
\varepsilon_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)  f\left(
b\right)  \text{ for each }\left(  a,b\right)  \in H\times H\right) \\
&  \Longleftrightarrow\ \left(  \text{the map }f\text{ is an }\left(
\varepsilon_{H},\varepsilon_{H}\right)  \text{-derivation in the sense of
Definition~\ref{Definition109}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Definition~\ref{Definition109}%
}\right)  .
\end{align*}
In other words, Definition~\ref{Definition108} and
Definition~\ref{Definition109} are equivalent. This proves
Proposition~\ref{Proposition110}.
\end{proof}

There is another way of thinking about $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivations. It is given by a result (in some way) dual to
Theorem~\ref{Theorem23}:

\begin{theorem}
\label{Theorem115} Let $k$ be a field. Let $H$ be a $k$-algebra, and let
$\varepsilon_{H}:H\rightarrow k$ be a $k$-algebra homomorphism. Let $A$ be a
$k$-algebra. Let $f:H\rightarrow A$ be a $k$-linear map. Then, $f$ is an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and only if
$f\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  =0$. \ \ \ \ \footnote{Recall that $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ is defined
according to Convention~\ref{Convention100}; thus, $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ means the
subspace $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of
$H$.}
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem115}.]\textbf{a)} Let us prove that if $f$ is an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation, then
\newline$f\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0$.

\textit{Proof.} Assume that $f$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation. By Definition~\ref{Definition109}, this means that%
\begin{equation}
\text{every }\left(  a,b\right)  \in H\times H\text{ satisfies }f\left(
ab\right)  =f\left(  a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon
_{H}\left(  a\right)  f\left(  b\right)  . \label{115.pf.1}%
\end{equation}


Applying (\ref{115.pf.1}) to $\left(  a,b\right)  =\left(  1_{H},1_{H}\right)
$, we get%
\begin{align*}
f\left(  1_{H}1_{H}\right)   &  =f\left(  1_{H}\right)  \varepsilon_{H}\left(
1_{H}\right)  +\varepsilon_{H}\left(  1_{H}\right)  f\left(  1_{H}\right)
=f\left(  1_{H}\right)  1+1f\left(  1_{H}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon_{H}\text{ is a
}k\text{-algebra homomorphism, and thus }\varepsilon_{H}\left(  1_{H}\right)
=1\right) \\
&  =f\left(  1_{H}\right)  +f\left(  1_{H}\right)  .
\end{align*}
Since $1_{H}1_{H}=1_{H}$, this rewrites as $f\left(  1_{H}\right)  =f\left(
1_{H}\right)  +f\left(  1_{H}\right)  $. This simplifies to $f\left(
1_{H}\right)  =0$.

On the other hand,
\begin{equation}
\text{every }\left(  a,b\right)  \in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \times\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \text{ satisfies }f\left(  ab\right)  =0.
\label{115.pf.3}%
\end{equation}
\footnote{\textit{Proof.} Let $\left(  a,b\right)  \in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \times\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $. Then,
$a\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $, so that
$\varepsilon_{H}\left(  a\right)  =0$ and similarly $\varepsilon_{H}\left(
b\right)  =0$. Now, (\ref{115.pf.1}) yields $f\left(  ab\right)  =f\left(
a\right)  \underbrace{\varepsilon_{H}\left(  b\right)  }_{=0}%
+\underbrace{\varepsilon_{H}\left(  a\right)  }_{=0}f\left(  b\right)
=f\left(  a\right)  0+0f\left(  b\right)  =0$, and thus (\ref{115.pf.3}) is
proven.}

Now it is easy to see that $f\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\right)  =0$%
.\ \ \ \ \footnote{\textit{Proof.} Let $x\in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. Then,%
\begin{align*}
x  &  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}=\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right) \\
&  =\left(  \text{the }k\text{-vector subspace of }H\text{ generated by the
terms }ab\text{ with }\left(  a,b\right)  \in\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \times\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \right) \\
&  =\left(  \text{the set of all }k\text{-linear combinations of terms of the
form }ab\text{ with }\left(  a,b\right)  \in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \times\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \right)  .
\end{align*}
Hence, $x$ is a $k$-linear combination of terms of the form $ab$ with $\left(
a,b\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \times\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  $. In other words, there exists some $n\in\mathbb{N}$, some elements
$\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$ of $k$, some elements
$a_{1}$, $a_{2}$, $\ldots$, $a_{n}$ of $\operatorname*{Ker}\left(
\varepsilon_{H}\right)  $, and some elements $b_{1}$, $b_{2}$, $\ldots$,
$b_{n}$ of $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ such that
$x=\sum_{i=1}^{n}\lambda_{i}a_{i}b_{i}$. Consider this $n$, these $\lambda
_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$, these $a_{1}$, $a_{2}$,
$\ldots$, $a_{n}$, and these $b_{1}$, $b_{2}$, $\ldots$, $b_{n}$. Then,
$x=\sum_{i=1}^{n}\lambda_{i}a_{i}b_{i}$ leads to%
\begin{align*}
f\left(  x\right)   &  =f\left(  \sum_{i=1}^{n}\lambda_{i}a_{i}b_{i}\right)
=\sum_{i=1}^{n}\lambda_{i}\underbrace{f\left(  a_{i}b_{i}\right)
}_{\substack{=0\text{ }\\\text{(by (\ref{115.pf.3}), applied to }\left(
a,b\right)  =\left(  a_{i},b_{i}\right)  \text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }f\text{ is }k\text{-linear}\right) \\
&  =\sum_{i=1}^{n}\lambda_{i}0=0.
\end{align*}
Since this is proven for every $x\in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$, we thus have $f\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$,
qed.}

But $f$ is $k$-linear, so that%
\[
f\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  =\underbrace{f\left(  \left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  }_{=0}+k\cdot
\underbrace{f\left(  1_{H}\right)  }_{=0}=0+k\cdot0=0.
\]
So we have shown that $f\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$. This proves
\textbf{a)}.

\textbf{b)} Now let us prove that if $f\left(  \left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$, then
$f$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation.

\textit{Proof.} Assume that $f\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$.

Let $\left(  a,b\right)  \in H\times H$ be arbitrary. Since%
\begin{align*}
\varepsilon_{H}\left(  a-\varepsilon_{H}\left(  a\right)  \cdot1_{H}\right)
&  =\varepsilon_{H}\left(  a\right)  -\varepsilon_{H}\left(  a\right)
\cdot1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon_{H}\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  =\varepsilon_{H}\left(  a\right)  -\varepsilon_{H}\left(  a\right)  =0,
\end{align*}
we have $a-\varepsilon_{H}\left(  a\right)  \cdot1_{H}\in\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $ and similarly $b-\varepsilon_{H}\left(
b\right)  \cdot1_{H}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $.
Hence, $\underbrace{\left(  a-\varepsilon_{H}\left(  a\right)  \cdot
1_{H}\right)  }_{\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)
}\underbrace{\left(  b-\varepsilon_{H}\left(  b\right)  \cdot1_{H}\right)
}_{\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  }\in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  =\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$. Thus,%
\[
\underbrace{\left(  a-\varepsilon_{H}\left(  a\right)  \cdot1_{H}\right)
\left(  b-\varepsilon_{H}\left(  b\right)  \cdot1_{H}\right)  }_{\in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}}+\left(
\underbrace{-\varepsilon_{H}\left(  a\right)  \varepsilon_{H}\left(  b\right)
}_{\in k}\cdot1_{H}\right)  \in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H},
\]
so that
\[
f\left(  \left(  a-\varepsilon_{H}\left(  a\right)  \cdot1_{H}\right)  \left(
b-\varepsilon_{H}\left(  b\right)  \cdot1_{H}\right)  +\left(  -\varepsilon
_{H}\left(  a\right)  \varepsilon_{H}\left(  b\right)  \cdot1_{H}\right)
\right)  \in f\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0.
\]
In other words,
\[
f\left(  \left(  a-\varepsilon_{H}\left(  a\right)  \cdot1_{H}\right)  \left(
b-\varepsilon_{H}\left(  b\right)  \cdot1_{H}\right)  +\left(  -\varepsilon
_{H}\left(  a\right)  \varepsilon_{H}\left(  b\right)  \cdot1_{H}\right)
\right)  =0.
\]
Thus,%
\begin{align*}
0  &  =f\left(  \underbrace{\left(  a-\varepsilon_{H}\left(  a\right)
\cdot1_{H}\right)  \left(  b-\varepsilon_{H}\left(  b\right)  \cdot
1_{H}\right)  }_{=ab-a\varepsilon_{H}\left(  b\right)  -\varepsilon_{H}\left(
a\right)  b+\varepsilon_{H}\left(  a\right)  \varepsilon_{H}\left(  b\right)
1_{H}}+\left(  -\varepsilon_{H}\left(  a\right)  \varepsilon_{H}\left(
b\right)  \cdot1_{H}\right)  \right) \\
&  =f\left(  ab-a\varepsilon_{H}\left(  b\right)  -\varepsilon_{H}\left(
a\right)  b+\underbrace{\varepsilon_{H}\left(  a\right)  \varepsilon
_{H}\left(  b\right)  1_{H}-\varepsilon_{H}\left(  a\right)  \varepsilon
_{H}\left(  b\right)  \cdot1_{H}}_{=0}\right) \\
&  =f\left(  ab-a\varepsilon_{H}\left(  b\right)  -\varepsilon_{H}\left(
a\right)  b\right)  =f\left(  ab\right)  -f\left(  a\right)  \varepsilon
_{H}\left(  b\right)  -\varepsilon_{H}\left(  a\right)  f\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is }k\text{-linear}\right)  .
\end{align*}
This rewrites as%
\[
f\left(  ab\right)  =f\left(  a\right)  \varepsilon_{H}\left(  b\right)
+\varepsilon_{H}\left(  a\right)  f\left(  b\right)  .
\]


Forget that we fixed $\left(  a,b\right)  $. We have proven that every
$\left(  a,b\right)  \in H\times H$ satisfies $f\left(  ab\right)  =f\left(
a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)
f\left(  b\right)  $. By Definition~\ref{Definition109}, this means that $f$
is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation. This
proves \textbf{b)}.

\textbf{c)} Combining the results of \textbf{a)} and \textbf{b)}, we see that
$f$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and
only if $f\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0$. This proves Theorem~\ref{Theorem115}.
\end{proof}

Our main thrust on Theorem~\ref{Theorem101} will come from the next theorem,
which is (in some ways, but not completely) a dual version of
Theorem~\ref{Theorem30}:

\begin{theorem}
\label{Theorem120} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $A$ be a commutative
$k$-algebra. Let $f\in\mathfrak{g}\left(  H,A\right)  $. Then, $f$ is an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and only if
$e^{\ast f}$ is a $k$-algebra homomorphism.
\end{theorem}

Being an ``if and only if'' statement, the assertion of this theorem splits
into two parts, which we will now formulate as two independent lemmas:

\begin{lemma}
\label{Lemma121} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $A$ be a commutative
$k$-algebra. Let $f\in\mathfrak{g}\left(  H,A\right)  $. If $f$ is an $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivation, then $e^{\ast f}$ is a
$k$-algebra homomorphism.
\end{lemma}

\begin{lemma}
\label{Lemma122} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $A$ be a commutative
$k$-algebra. Let $f\in\mathfrak{g}\left(  H,A\right)  $. If $e^{\ast f}$ is a
$k$-algebra homomorphism, then $f$ is an $\left(  \varepsilon_{H}%
,\varepsilon_{H}\right)  $-derivation.
\end{lemma}

In order to prove Lemma~\ref{Lemma121}, we first show an auxiliary result:

\begin{lemma}
\label{Lemma125} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be a
commutative $k$-algebra. Let $f\in\mathcal{L}\left(  H,A\right)  $ be an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation. Then, for every
$n\in\mathbb{N}$, we have%
\begin{equation}
f^{\ast n}\left(  ab\right)  =\sum_{i=0}^{n}\dbinom{n}{i}f^{\ast i}\left(
a\right)  f^{\ast\left(  n-i\right)  }\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in H\text{ and }b\in H. \label{125.fab}%
\end{equation}

\end{lemma}

Note that this Lemma~\ref{Lemma125} is the dual of Lemma~\ref{Lemma37} - as
you can see if you rewrite the equation (\ref{125.fab}) as%
\[
f^{\ast n}\circ\mu_{H}=\sum_{i=0}^{n}\dbinom{n}{i}\mu_{A}\circ\left(  f^{\ast
i}\otimes f^{\ast\left(  n-i\right)  }\right)  .
\]
This time, it is a true duality, without any filtrations that could prevent us
from dualizing things. As a consequence, one could obtain a proof of
Lemma~\ref{Lemma125} by ``reversing all arrows'' in our above proof of
Lemma~\ref{Lemma37} (but first, one would have to rewrite our above proof of
Lemma~\ref{Lemma37} in a purely categorical form). Let us here take a somewhat
different route towards proving Lemma~\ref{Lemma125} - namely, by direct
computation with elements. It will turn out to be faster and easier (although
still requiring a lot of computation since we are not using Sweedler notation).

\begin{proof}
[Proof of Lemma~\ref{Lemma125}.]We are going to prove Lemma~\ref{Lemma125} by
induction over $n$.

\textit{Induction base:} We have $f^{\ast0}=e_{H,A}=\eta_{A}\circ
\varepsilon_{H}$. Hence, every $x\in H$ satisfies%
\begin{equation}
f^{\ast0}\left(  x\right)  =\left(  \eta_{A}\circ\varepsilon_{H}\right)
\left(  x\right)  =\eta_{A}\left(  \varepsilon_{H}\left(  x\right)  \right)
=\varepsilon_{H}\left(  x\right)  1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\eta_{A}\right)  . \label{125.pf.1}%
\end{equation}


Applying (\ref{125.pf.1}) to $x=a$, we obtain $f^{\ast0}\left(  a\right)
=\varepsilon_{H}\left(  a\right)  1_{A}$. Applying (\ref{125.pf.1}) to $x=b$,
we obtain $f^{\ast0}\left(  b\right)  =\varepsilon_{H}\left(  b\right)  1_{A}%
$. Applying (\ref{125.pf.1}) to $x=ab$, we obtain $f^{\ast0}\left(  ab\right)
=\varepsilon_{H}\left(  ab\right)  1_{A}$. Since $H$ is a bialgebra, we have
$\varepsilon_{H}\left(  ab\right)  =\varepsilon_{H}\left(  a\right)
\varepsilon_{H}\left(  b\right)  $ (by the axioms of a bialgebra). Now,
comparing%
\[
f^{\ast0}\left(  ab\right)  =\underbrace{\varepsilon_{H}\left(  ab\right)
}_{=\varepsilon_{H}\left(  a\right)  \varepsilon_{H}\left(  b\right)  }%
1_{A}=\varepsilon_{H}\left(  a\right)  \varepsilon_{H}\left(  b\right)  1_{A}%
\]
with%
\begin{align*}
\sum_{i=0}^{0}\dbinom{0}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(
0-i\right)  }\left(  b\right)   &  =\underbrace{\dbinom{0}{0}}_{=1}f^{\ast
0}\left(  a\right)  \underbrace{f^{\ast\left(  0-0\right)  }}_{=f^{\ast0}%
}\left(  b\right)  =\underbrace{f^{\ast0}\left(  a\right)  }_{=\varepsilon
_{H}\left(  a\right)  1_{A}}\underbrace{f^{\ast0}\left(  b\right)
}_{=\varepsilon_{H}\left(  b\right)  1_{A}}=\varepsilon_{H}\left(  a\right)
1_{A}\cdot\varepsilon_{H}\left(  b\right)  1_{A}\\
&  =\varepsilon_{H}\left(  a\right)  \varepsilon_{H}\left(  b\right)  1_{A},
\end{align*}
we obtain $f^{\ast0}\left(  ab\right)  =\sum_{i=0}^{0}\dbinom{0}{i}f^{\ast
i}\left(  a\right)  f^{\ast\left(  0-i\right)  }\left(  b\right)  $. In other
words, Lemma~\ref{Lemma125} holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
Lemma~\ref{Lemma125} holds for $n=N$. To complete the induction, we must show
that Lemma~\ref{Lemma125} also holds for $n=N+1$.

Since Lemma~\ref{Lemma125} holds for $n=N$, we have%
\begin{equation}
f^{\ast N}\left(  ab\right)  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(
a\right)  f^{\ast\left(  N-i\right)  }\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in H\text{ and }b\in H.
\label{125.pf.indass}%
\end{equation}


On the other hand, $f$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation. By Definition~\ref{Definition109}, this means that%
\begin{equation}
f\left(  ab\right)  =f\left(  a\right)  \varepsilon_{H}\left(  b\right)
+\varepsilon_{H}\left(  a\right)  f\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }\left(  a,b\right)  \in H\times H.
\label{125.pf.deri}%
\end{equation}


Now let $a\in H$ and $b\in H$ be arbitrary.

Since $\Delta_{H}\left(  a\right)  \in H\otimes H$, we can write $\Delta
_{H}\left(  a\right)  $ in the form $\Delta_{H}\left(  a\right)  =\sum
_{j=1}^{M}\lambda_{j}a_{j}\otimes a_{j}^{\prime}$ for some $M\in\mathbb{N}$,
some elements $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{M}$ of $k$,
some elements $a_{1}$, $a_{2}$, $\ldots$, $a_{M}$ of $H$, and some elements
$a_{1}^{\prime}$, $a_{2}^{\prime}$, $\ldots$, $a_{M}^{\prime}$ of $H$.
Consider this $M$, these $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{M}%
$, these $a_{1}$, $a_{2}$, $\ldots$, $a_{M}$, and these $a_{1}^{\prime}$,
$a_{2}^{\prime}$, $\ldots$, $a_{M}^{\prime}$.

Since $\Delta_{H}\left(  b\right)  \in H\otimes H$, we can write $\Delta
_{H}\left(  b\right)  $ in the form $\Delta_{H}\left(  b\right)  =\sum
_{\ell=1}^{K}\mu_{\ell}b_{\ell}\otimes b_{\ell}^{\prime}$ for some
$K\in\mathbb{N}$, some elements $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{K}$ of
$k$, some elements $b_{1}$, $b_{2}$, $\ldots$, $b_{K}$ of $H$, and some
elements $b_{1}^{\prime}$, $b_{2}^{\prime}$, $\ldots$, $b_{K}^{\prime}$ of
$H$. Consider this $K$, these $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{K}$, these
$b_{1}$, $b_{2}$, $\ldots$, $b_{K}$, and these $b_{1}^{\prime}$,
$b_{2}^{\prime}$, $\ldots$, $b_{K}^{\prime}$.

Since $H$ is a bialgebra,%
\begin{align*}
\Delta_{H}\left(  ab\right)   &  =\underbrace{\Delta_{H}\left(  a\right)
}_{=\sum_{j=1}^{M}\lambda_{j}a_{j}\otimes a_{j}^{\prime}}\cdot
\underbrace{\Delta_{H}\left(  b\right)  }_{=\sum_{\ell=1}^{K}\mu_{\ell}%
b_{\ell}\otimes b_{\ell}^{\prime}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
axioms of a bialgebra}\right) \\
&  =\left(  \sum_{j=1}^{M}\lambda_{j}a_{j}\otimes a_{j}^{\prime}\right)
\cdot\left(  \sum_{\ell=1}^{K}\mu_{\ell}b_{\ell}\otimes b_{\ell}^{\prime
}\right)  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}%
\underbrace{\left(  a_{j}\otimes a_{j}^{\prime}\right)  \left(  b_{\ell
}\otimes b_{\ell}^{\prime}\right)  }_{=a_{j}b_{\ell}\otimes a_{j}^{\prime
}b_{\ell}^{\prime}}\\
&  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}a_{j}b_{\ell}\otimes
a_{j}^{\prime}b_{\ell}^{\prime}.
\end{align*}


Now, $f^{\ast\left(  N+1\right)  }=f\ast f^{\ast N}=\mu_{A}\circ\left(
f\otimes f^{\ast N}\right)  \circ\Delta_{H}$ (by the definition of
convolution), so that%
\begin{align}
f^{\ast\left(  N+1\right)  }\left(  ab\right)   &  =\left(  \mu_{A}%
\circ\left(  f\otimes f^{\ast N}\right)  \circ\Delta_{H}\right)  \left(
ab\right) \nonumber\\
&  =\mu_{A}\left(  \left(  f\otimes f^{\ast N}\right)  \underbrace{\left(
\Delta_{H}\left(  ab\right)  \right)  }_{=\sum_{j=1}^{M}\sum_{\ell=1}%
^{K}\lambda_{j}\mu_{\ell}a_{j}b_{\ell}\otimes a_{j}^{\prime}b_{\ell}^{\prime}%
}\right)  =\mu_{A}\left(  \underbrace{\left(  f\otimes f^{\ast N}\right)
\left(  \sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}a_{j}b_{\ell
}\otimes a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{\substack{=\sum_{j=1}%
^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}f\left(  a_{j}b_{\ell}\right)
\otimes f^{\ast N}\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  \\\text{(by
the definition of }f\otimes f^{\ast N}\text{)}}}\right) \nonumber\\
&  =\mu_{A}\left(  \sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell
}f\left(  a_{j}b_{\ell}\right)  \otimes f^{\ast N}\left(  a_{j}^{\prime
}b_{\ell}^{\prime}\right)  \right)  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}%
\lambda_{j}\mu_{\ell}\underbrace{f\left(  a_{j}b_{\ell}\right)  }%
_{\substack{=f\left(  a_{j}\right)  \varepsilon_{H}\left(  b_{\ell}\right)
+\varepsilon_{H}\left(  a_{j}\right)  f\left(  b_{\ell}\right)  \\\text{(by
(\ref{125.pf.deri}), applied to}\\\left(  a_{j},b_{\ell}\right)  \text{
instead of }\left(  a,b\right)  \text{)}}}\cdot f^{\ast N}\left(
a_{j}^{\prime}b_{\ell}^{\prime}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{A}\text{ is the
multiplication map}\right) \nonumber\\
&  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\underbrace{\lambda_{j}\mu_{\ell}\left(
f\left(  a_{j}\right)  \varepsilon_{H}\left(  b_{\ell}\right)  +\varepsilon
_{H}\left(  a_{j}\right)  f\left(  b_{\ell}\right)  \right)  \cdot f^{\ast
N}\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{=\lambda_{j}\mu_{\ell
}f\left(  a_{j}\right)  \varepsilon_{H}\left(  b_{\ell}\right)  \cdot f^{\ast
N}\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  +\lambda_{j}\mu_{\ell
}\varepsilon_{H}\left(  a_{j}\right)  f\left(  b_{\ell}\right)  \cdot f^{\ast
N}\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  }\nonumber\\
&  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\left(  \lambda_{j}\mu_{\ell}f\left(
a_{j}\right)  \varepsilon_{H}\left(  b_{\ell}\right)  \cdot f^{\ast N}\left(
a_{j}^{\prime}b_{\ell}^{\prime}\right)  +\lambda_{j}\mu_{\ell}\varepsilon
_{H}\left(  a_{j}\right)  f\left(  b_{\ell}\right)  \cdot f^{\ast N}\left(
a_{j}^{\prime}b_{\ell}^{\prime}\right)  \right) \nonumber\\
&  =\underbrace{\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}f\left(
a_{j}\right)  \varepsilon_{H}\left(  b_{\ell}\right)  \cdot f^{\ast N}\left(
a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{=\sum_{j=1}^{M}\lambda_{j}f\left(
a_{j}\right)  \sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell
}\right)  \cdot f^{\ast N}\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)
}+\underbrace{\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}%
\varepsilon_{H}\left(  a_{j}\right)  f\left(  b_{\ell}\right)  \cdot f^{\ast
N}\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{=\sum_{\ell=1}^{K}%
\mu_{\ell}f\left(  b_{\ell}\right)  \sum_{j=1}^{M}\lambda_{j}\varepsilon
_{H}\left(  a_{j}\right)  \cdot f^{\ast N}\left(  a_{j}^{\prime}b_{\ell
}^{\prime}\right)  }\nonumber\\
&  =\sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  \underbrace{\sum_{\ell
=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)  \cdot f^{\ast
N}\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{\substack{=f^{\ast
N}\left(  \sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)
a_{j}^{\prime}b_{\ell}^{\prime}\right)  \\\text{(since }f^{\ast N}\text{ is
}k\text{-linear)}}}+\sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)
\underbrace{\sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(  a_{j}\right)
\cdot f^{\ast N}\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  }%
_{\substack{=f^{\ast N}\left(  \sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(
a_{j}\right)  \cdot a_{j}^{\prime}b_{\ell}^{\prime}\right)  \\\text{(since
}f^{\ast N}\text{ is }k\text{-linear)}}}\nonumber\\
&  =\sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  f^{\ast N}\left(
\sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)
a_{j}^{\prime}b_{\ell}^{\prime}\right)  +\sum_{\ell=1}^{K}\mu_{\ell}f\left(
b_{\ell}\right)  f^{\ast N}\left(  \sum_{j=1}^{M}\lambda_{j}\varepsilon
_{H}\left(  a_{j}\right)  \cdot a_{j}^{\prime}b_{\ell}^{\prime}\right)  .
\label{125.pf.5}%
\end{align}


Now, it is easy to see (using the axioms of a coalgebra and the fact that
$\Delta_{H}\left(  b\right)  =\sum_{\ell=1}^{K}\mu_{\ell}b_{\ell}\otimes
b_{\ell}^{\prime}$) that $\sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(
b_{\ell}\right)  b_{\ell}^{\prime}=b$\ \ \ \ \ \footnote{\textit{Proof.} Let
$\operatorname*{kan}:k\otimes H\rightarrow H$ be the canonical isomorphism
(which maps $1\otimes x$ to $x$ for every $x\in H$). Then, by the axioms of a
coalgebra, $\operatorname*{kan}\circ\left(  \varepsilon_{H}\otimes
\operatorname{id}_{H}\right)  \circ\Delta_{H}=\operatorname{id}_{H}$ (since
$H$ is a coalgebra). Thus, $\left(  \operatorname*{kan}\circ\left(
\varepsilon_{H}\otimes\operatorname{id}_{H}\right)  \circ\Delta_{H}\right)
\left(  b\right)  =\operatorname{id}_{H}\left(  b\right)  $. Since%
\begin{align*}
\left(  \operatorname*{kan}\circ\left(  \varepsilon_{H}\otimes
\operatorname{id}_{H}\right)  \circ\Delta_{H}\right)  \left(  b\right)   &
=\operatorname*{kan}\left(  \left(  \varepsilon_{H}\otimes\operatorname{id}%
_{H}\right)  \underbrace{\left(  \Delta_{H}\left(  b\right)  \right)  }%
_{=\sum_{\ell=1}^{K}\mu_{\ell}b_{\ell}\otimes b_{\ell}^{\prime}}\right)
=\operatorname*{kan}\left(  \underbrace{\left(  \varepsilon_{H}\otimes
\operatorname{id}_{H}\right)  \left(  \sum_{\ell=1}^{K}\mu_{\ell}b_{\ell
}\otimes b_{\ell}^{\prime}\right)  }_{\substack{=\sum_{\ell=1}^{K}\mu_{\ell
}\varepsilon_{H}\left(  b_{\ell}\right)  \otimes\operatorname{id}_{H}\left(
b_{\ell}^{\prime}\right)  \\\text{(by the definition of }\varepsilon
_{H}\otimes\operatorname{id}_{H}\text{)}}}\right) \\
&  =\operatorname*{kan}\left(  \sum_{\ell=1}^{K}\mu_{\ell}\varepsilon
_{H}\left(  b_{\ell}\right)  \otimes\operatorname{id}_{H}\left(  b_{\ell
}^{\prime}\right)  \right)  =\sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(
b_{\ell}\right)  \underbrace{\operatorname{id}_{H}\left(  b_{\ell}^{\prime
}\right)  }_{=b_{\ell}^{\prime}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\operatorname*{kan}\right) \\
&  =\sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)
b_{\ell}^{\prime}%
\end{align*}
and $\operatorname{id}_{H}\left(  b\right)  =b$, this rewrites as $\sum
_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)  b_{\ell
}^{\prime}=b$, qed.}, and thus every $j\in\left\{  1,2,\ldots,M\right\}  $
satisfies%
\[
\sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)
a_{j}^{\prime}b_{\ell}^{\prime}=a_{j}^{\prime}\cdot\underbrace{\sum_{\ell
=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)  b_{\ell}^{\prime}%
}_{=b}=a_{j}^{\prime}b,
\]
so that%
\begin{align*}
f^{\ast N}\left(  \sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell
}\right)  a_{j}^{\prime}b_{\ell}^{\prime}\right)   &  =f^{\ast N}\left(
a_{j}^{\prime}b\right)  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(
a_{j}^{\prime}\right)  f^{\ast\left(  N-i\right)  }\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{125.pf.indass}), applied to
}a_{j}^{\prime}\text{ instead of }a\right)  .
\end{align*}
Hence,%
\begin{align}
&  \sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  \underbrace{f^{\ast
N}\left(  \sum_{\ell=1}^{K}\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)
a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{=\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast
i}\left(  a_{j}^{\prime}\right)  f^{\ast\left(  N-i\right)  }\left(  b\right)
}\nonumber\\
&  =\sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  \sum_{i=0}^{N}\dbinom
{N}{i}f^{\ast i}\left(  a_{j}^{\prime}\right)  f^{\ast\left(  N-i\right)
}\left(  b\right) \nonumber\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}\left(  \sum_{j=1}^{M}\lambda_{j}f\left(
a_{j}\right)  f^{\ast i}\left(  a_{j}^{\prime}\right)  \right)  f^{\ast\left(
N-i\right)  }\left(  b\right)  . \label{125.pf.6}%
\end{align}


Also, it is easy to see (using the axioms of a coalgebra and the fact that
$\Delta_{H}\left(  a\right)  =\sum_{j=1}^{M}\lambda_{j}a_{j}\otimes
a_{j}^{\prime}$) that $\sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(
a_{j}\right)  a_{j}^{\prime}=a$\ \ \ \ \ \footnote{\textit{Proof.} Let
$\operatorname*{kan}:k\otimes H\rightarrow H$ be the canonical isomorphism
(which maps $1\otimes x$ to $x$ for every $x\in H$). Then, by the axioms of a
coalgebra, $\operatorname*{kan}\circ\left(  \varepsilon_{H}\otimes
\operatorname{id}_{H}\right)  \circ\Delta_{H}=\operatorname{id}_{H}$ (since
$H$ is a coalgebra). Thus, $\left(  \operatorname*{kan}\circ\left(
\varepsilon_{H}\otimes\operatorname{id}_{H}\right)  \circ\Delta_{H}\right)
\left(  a\right)  =\operatorname{id}_{H}\left(  a\right)  $. Since%
\begin{align*}
&  \left(  \operatorname*{kan}\circ\left(  \varepsilon_{H}\otimes
\operatorname{id}_{H}\right)  \circ\Delta_{H}\right)  \left(  a\right) \\
&  =\operatorname*{kan}\left(  \left(  \varepsilon_{H}\otimes\operatorname{id}%
_{H}\right)  \underbrace{\left(  \Delta_{H}\left(  a\right)  \right)  }%
_{=\sum_{j=1}^{M}\lambda_{j}a_{j}\otimes a_{j}^{\prime}}\right)
=\operatorname*{kan}\left(  \underbrace{\left(  \varepsilon_{H}\otimes
\operatorname{id}_{H}\right)  \left(  \sum_{j=1}^{M}\lambda_{j}a_{j}\otimes
a_{j}^{\prime}\right)  }_{\substack{=\sum_{j=1}^{M}\lambda_{j}\varepsilon
_{H}\left(  a_{j}\right)  \otimes\operatorname{id}_{H}\left(  a_{j}^{\prime
}\right)  \\\text{(by the definition of }\varepsilon_{H}\otimes
\operatorname{id}_{H}\text{)}}}\right) \\
&  =\operatorname*{kan}\left(  \sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(
a_{j}\right)  \otimes\operatorname{id}_{H}\left(  a_{j}^{\prime}\right)
\right)  =\sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(  a_{j}\right)
\underbrace{\operatorname{id}_{H}\left(  a_{j}^{\prime}\right)  }%
_{=a_{j}^{\prime}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{kan}\right) \\
&  =\sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(  a_{j}\right)
a_{j}^{\prime}%
\end{align*}
and $\operatorname{id}_{H}\left(  a\right)  =a$, this rewrites as $\sum
_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(  a_{j}\right)  a_{j}^{\prime}=a$,
qed.}, and thus every $\ell\in\left\{  1,2,\ldots,M\right\}  $ satisfies%
\[
\sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(  a_{j}\right)  \cdot
a_{j}^{\prime}b_{\ell}^{\prime}=\underbrace{\sum_{j=1}^{M}\lambda
_{j}\varepsilon_{H}\left(  a_{j}\right)  a_{j}^{\prime}}_{=a}\cdot b_{\ell
}^{\prime}=ab_{\ell}^{\prime},
\]
so that%
\begin{align*}
f^{\ast N}\left(  \sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(
a_{j}\right)  \cdot a_{j}^{\prime}b_{\ell}^{\prime}\right)   &  =f^{\ast
N}\left(  ab_{\ell}^{\prime}\right)  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast
i}\left(  a\right)  f^{\ast\left(  N-i\right)  }\left(  b_{\ell}^{\prime
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{125.pf.indass}), applied to
}b_{\ell}^{\prime}\text{ instead of }b\right)  .
\end{align*}
Hence,%
\begin{align}
&  \sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)  \underbrace{f^{\ast
N}\left(  \sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(  a_{j}\right)  \cdot
a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{=\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast
i}\left(  a\right)  f^{\ast\left(  N-i\right)  }\left(  b_{\ell}^{\prime
}\right)  }\nonumber\\
&  =\sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)  \sum_{i=0}^{N}%
\dbinom{N}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(  N-i\right)  }\left(
b_{\ell}^{\prime}\right)  =\sum_{i=0}^{N}\dbinom{N}{i}\sum_{\ell=1}^{K}%
\mu_{\ell}\underbrace{f\left(  b_{\ell}\right)  f^{\ast i}\left(  a\right)
}_{\substack{=f^{\ast i}\left(  a\right)  f\left(  b_{\ell}\right)
\\\text{(since }A\text{ is commutative)}}}f^{\ast\left(  N-i\right)  }\left(
b_{\ell}^{\prime}\right) \nonumber\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}\sum_{\ell=1}^{K}\mu_{\ell}f^{\ast i}\left(
a\right)  f\left(  b_{\ell}\right)  f^{\ast\left(  N-i\right)  }\left(
b_{\ell}^{\prime}\right)  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(
a\right)  \left(  \sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)
f^{\ast\left(  N-i\right)  }\left(  b_{\ell}^{\prime}\right)  \right)  .
\label{125.pf.6a}%
\end{align}


Now, (\ref{125.pf.5}) becomes%
\begin{align}
f^{\ast\left(  N+1\right)  }\left(  ab\right)   &  =\underbrace{\sum_{j=1}%
^{M}\lambda_{j}f\left(  a_{j}\right)  f^{\ast N}\left(  \sum_{\ell=1}^{K}%
\mu_{\ell}\varepsilon_{H}\left(  b_{\ell}\right)  a_{j}^{\prime}b_{\ell
}^{\prime}\right)  }_{\substack{=\sum_{i=0}^{N}\dbinom{N}{i}\left(  \sum
_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  f^{\ast i}\left(  a_{j}^{\prime
}\right)  \right)  f^{\ast\left(  N-i\right)  }\left(  b\right)  \\\text{(by
(\ref{125.pf.6}))}}}+\underbrace{\sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell
}\right)  f^{\ast N}\left(  \sum_{j=1}^{M}\lambda_{j}\varepsilon_{H}\left(
a_{j}\right)  \cdot a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{\substack{=\sum
_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)  \left(  \sum_{\ell=1}%
^{K}\mu_{\ell}f\left(  b_{\ell}\right)  f^{\ast\left(  N-i\right)  }\left(
b_{\ell}^{\prime}\right)  \right)  \\\text{(by (\ref{125.pf.6a}))}%
}}\nonumber\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}\left(  \sum_{j=1}^{M}\lambda_{j}f\left(
a_{j}\right)  f^{\ast i}\left(  a_{j}^{\prime}\right)  \right)  f^{\ast\left(
N-i\right)  }\left(  b\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)
\left(  \sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)  f^{\ast\left(
N-i\right)  }\left(  b_{\ell}^{\prime}\right)  \right)  . \label{125.pf.8}%
\end{align}


But now, we notice that every $i\in\left\{  0,1,\ldots,N\right\}  $ satisfies%
\begin{align}
f^{\ast\left(  i+1\right)  }\left(  a\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes f^{\ast i}\right)  \circ\Delta_{H}\right)  \left(  a\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f^{\ast\left(  i+1\right)  }=f\ast
f^{\ast i}=\mu_{A}\circ\left(  f\otimes f^{\ast i}\right)  \circ\Delta
_{H}\text{ by the definition of convolution}\right) \nonumber\\
&  =\mu_{A}\left(  \left(  f\otimes f^{\ast i}\right)  \left(
\underbrace{\Delta_{H}\left(  a\right)  }_{=\sum_{j=1}^{M}\lambda_{j}%
a_{j}\otimes a_{j}^{\prime}}\right)  \right)  =\mu_{A}\left(
\underbrace{\left(  f\otimes f^{\ast i}\right)  \left(  \sum_{j=1}^{M}%
\lambda_{j}a_{j}\otimes a_{j}^{\prime}\right)  }_{\substack{=\sum_{j=1}%
^{M}\lambda_{j}f\left(  a_{j}\right)  \otimes f^{\ast i}\left(  a_{j}^{\prime
}\right)  \\\text{(by the definition of }f\otimes f^{\ast i}\text{)}}}\right)
\nonumber\\
&  =\mu_{A}\left(  \sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  \otimes
f^{\ast i}\left(  a_{j}^{\prime}\right)  \right)  =\sum_{j=1}^{M}\lambda
_{j}f\left(  a_{j}\right)  f^{\ast i}\left(  a_{j}^{\prime}\right)
\label{125.pf.9a}%
\end{align}
(since $\mu_{A}$ is the multiplication map) and%
\begin{align}
&  f^{\ast\left(  N-i+1\right)  }\left(  b\right) \nonumber\\
&  =\left(  \mu_{A}\circ\left(  f\otimes f^{\ast\left(  N-i\right)  }\right)
\circ\Delta_{H}\right)  \left(  b\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f^{\ast\left(  N-i+1\right)
}=f\ast f^{\ast\left(  N-i\right)  }=\mu_{A}\circ\left(  f\otimes
f^{\ast\left(  N-i\right)  }\right)  \circ\Delta_{H}\text{ by the definition
of convolution}\right) \nonumber\\
&  =\mu_{A}\left(  \left(  f\otimes f^{\ast\left(  N-i\right)  }\right)
\left(  \underbrace{\Delta_{H}\left(  b\right)  }_{=\sum_{\ell=1}^{K}\mu
_{\ell}b_{\ell}\otimes b_{\ell}^{\prime}}\right)  \right)  =\mu_{A}\left(
\underbrace{\left(  f\otimes f^{\ast\left(  N-i\right)  }\right)  \left(
\sum_{\ell=1}^{K}\mu_{\ell}b_{\ell}\otimes b_{\ell}^{\prime}\right)
}_{\substack{=\sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)  \otimes
f^{\ast\left(  N-i\right)  }\left(  b_{\ell}^{\prime}\right)  \\\text{(by the
definition of }f\otimes f^{\ast\left(  N-i\right)  }\text{)}}}\right)
\nonumber\\
&  =\mu_{A}\left(  \sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)
\otimes f^{\ast\left(  N-i\right)  }\left(  b_{\ell}^{\prime}\right)  \right)
=\sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)  f^{\ast\left(
N-i\right)  }\left(  b_{\ell}^{\prime}\right)  \label{125.pf.9b}%
\end{align}
(since $\mu_{A}$ is the multiplication map).

With the help of these identities, (\ref{125.pf.8}) becomes%
\begin{align*}
f^{\ast\left(  N+1\right)  }\left(  ab\right)   &  =\sum_{i=0}^{N}\dbinom
{N}{i}\underbrace{\left(  \sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)
f^{\ast i}\left(  a_{j}^{\prime}\right)  \right)  }_{\substack{=f^{\ast\left(
i+1\right)  }\left(  a\right)  \\\text{(by (\ref{125.pf.9a}))}}}f^{\ast\left(
N-i\right)  }\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)
\underbrace{\left(  \sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)
f^{\ast\left(  N-i\right)  }\left(  b_{\ell}^{\prime}\right)  \right)
}_{\substack{=f^{\ast\left(  N-i+1\right)  }\left(  b\right)  \\\text{(by
(\ref{125.pf.9b}))}}}\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast\left(  i+1\right)  }\left(  a\right)
f^{\ast\left(  N-i\right)  }\left(  b\right)  +\sum_{i=0}^{N}\dbinom{N}%
{i}f^{\ast i}\left(  a\right)  f^{\ast\left(  N-i+1\right)  }\left(  b\right)
\\
&  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(
N-i+1\right)  }\left(  b\right)  +\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast\left(
i+1\right)  }\left(  a\right)  f^{\ast\left(  N-i\right)  }\left(  b\right) \\
&  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)  \underbrace{f^{\ast
\left(  N-i+1\right)  }}_{=f^{\ast\left(  N+1-i\right)  }}\left(  b\right)
+\sum_{i=1}^{N+1}\dbinom{N}{i-1}\underbrace{f^{\ast\left(  \left(  i-1\right)
+1\right)  }}_{=f^{\ast i}}\left(  a\right)  \underbrace{f^{\ast\left(
N-\left(  i-1\right)  \right)  }}_{=f^{\ast\left(  N+1-i\right)  }}\left(
b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i-1\text{ for
}i\text{ in the second sum}\right) \\
&  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(
N+1-i\right)  }\left(  b\right)  +\sum_{i=1}^{N+1}\dbinom{N}{i-1}f^{\ast
i}\left(  a\right)  f^{\ast\left(  N+1-i\right)  }\left(  b\right)  .
\end{align*}
Compared to%
\begin{align*}
&  \sum_{i=0}^{N+1}\dbinom{N+1}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(
N+1-i\right)  }\left(  b\right) \\
&  =\sum_{i=0}^{N+1}\left(  \dbinom{N}{i}+\dbinom{N}{i-1}\right)  f^{\ast
i}\left(  a\right)  f^{\ast\left(  N+1-i\right)  }\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\dbinom{N+1}{i}=\dbinom{N}%
{i}+\dbinom{N}{i-1}\text{ by the recurrence of the binomial coefficients}%
\right) \\
&  =\underbrace{\sum_{i=0}^{N+1}\dbinom{N}{i}f^{\ast i}\left(  a\right)
f^{\ast\left(  N+1-i\right)  }\left(  b\right)  }_{=\sum_{i=0}^{N}\dbinom
{N}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(  N+1-i\right)  }\left(
b\right)  +\dbinom{N}{N+1}f^{\ast\left(  N+1\right)  }\left(  a\right)
f^{\ast\left(  N+1-\left(  N+1\right)  \right)  }\left(  b\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{i=0}^{N+1}\dbinom{N}{i-1}f^{\ast
i}\left(  a\right)  f^{\ast\left(  N+1-i\right)  }\left(  b\right)
}_{=\dbinom{N}{0-1}f^{\ast0}\left(  a\right)  f^{\ast\left(  N+1-0\right)
}\left(  b\right)  +\sum_{i=1}^{N+1}\dbinom{N}{i-1}f^{\ast i}\left(  a\right)
f^{\ast\left(  N+1-i\right)  }\left(  b\right)  }\\
&  =\left(  \sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)
f^{\ast\left(  N+1-i\right)  }\left(  b\right)  +\underbrace{\dbinom{N}{N+1}%
}_{=0\text{ (since }N+1>N\text{)}}f^{\ast\left(  N+1\right)  }\left(
a\right)  f^{\ast\left(  N+1-\left(  N+1\right)  \right)  }\left(  b\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \underbrace{\dbinom{N}{0-1}}_{=0\text{ (since
}0-1<0\text{)}}f^{\ast0}\left(  a\right)  f^{\ast\left(  N+1-0\right)
}\left(  b\right)  +\sum_{i=1}^{N+1}\dbinom{N}{i-1}f^{\ast i}\left(  a\right)
f^{\ast\left(  N+1-i\right)  }\left(  b\right)  \right) \\
&  =\left(  \sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)
f^{\ast\left(  N+1-i\right)  }\left(  b\right)  +\underbrace{0f^{\ast\left(
N+1\right)  }\left(  a\right)  f^{\ast\left(  N+1-\left(  N+1\right)  \right)
}\left(  b\right)  }_{=0}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \underbrace{0f^{\ast0}\left(  a\right)
f^{\ast\left(  N+1-0\right)  }\left(  b\right)  }_{=0}+\sum_{i=1}^{N+1}%
\dbinom{N}{i-1}f^{\ast i}\left(  a\right)  f^{\ast\left(  N+1-i\right)
}\left(  b\right)  \right) \\
&  =\sum_{i=0}^{N}\dbinom{N}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(
N+1-i\right)  }\left(  b\right)  +\sum_{i=1}^{N+1}\dbinom{N}{i-1}f^{\ast
i}\left(  a\right)  f^{\ast\left(  N+1-i\right)  }\left(  b\right)  ,
\end{align*}
this yields%
\[
f^{\ast\left(  N+1\right)  }\left(  ab\right)  =\sum_{i=0}^{N+1}\dbinom
{N+1}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(  N+1-i\right)  }\left(
b\right)  .
\]


Now forget that we fixed $a$ and $b$. We thus have shown that every $a\in H$
and $b\in H$ satisfy%
\[
f^{\ast\left(  N+1\right)  }\left(  ab\right)  =\sum_{i=0}^{N+1}\dbinom
{N+1}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(  N+1-i\right)  }\left(
b\right)  .
\]
In other words, Lemma~\ref{Lemma125} holds for $n=N+1$. This completes the
induction step. The induction proof of Lemma~\ref{Lemma125} is thus complete.
\end{proof}

Here is a little brother of Lemma~\ref{Lemma125} and, unsurprisingly, the dual
of Lemma~\ref{Lemma38}:

\begin{lemma}
\label{Lemma126} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be a
commutative $k$-algebra. Let $f\in\mathcal{L}\left(  H,A\right)  $ be an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation. Then, for every
positive $n\in\mathbb{N}$, we have $f^{\ast n}\left(  1_{H}\right)  =0$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma126}.]Since $H$ is a $k$-bialgebra, we have
$\Delta_{H}\left(  1_{H}\right)  =1_{H}\otimes1_{H}$ (by the axioms of a bialgebra).

Just as in part \textbf{a)} of the proof of Theorem~\ref{Theorem115}, we can
find that $f\left(  1_{H}\right)  =0$ (because $H$ is a $k$-bialgebra, and
thus $\varepsilon_{H}$ is a $k$-algebra homomorphism (by the axioms of a bialgebra)).

Since $n$ is positive, we have $f^{\ast n}=f\ast f^{\ast\left(  n-1\right)
}=\mu_{A}\circ\left(  f\otimes f^{\ast\left(  n-1\right)  }\right)
\circ\Delta_{H}$ (by the definition of convolution), and thus%
\begin{align*}
f^{\ast n}\left(  1_{H}\right)   &  =\left(  \mu_{A}\circ\left(  f\otimes
f^{\ast\left(  n-1\right)  }\right)  \circ\Delta_{H}\right)  \left(
1_{H}\right)  =\mu_{A}\left(  \left(  f\otimes f^{\ast\left(  n-1\right)
}\right)  \underbrace{\left(  \Delta_{H}\left(  1_{H}\right)  \right)
}_{=1_{H}\otimes1_{H}}\right) \\
&  =\mu_{A}\left(  \underbrace{\left(  f\otimes f^{\ast\left(  n-1\right)
}\right)  \left(  1_{H}\otimes1_{H}\right)  }_{=f\left(  1_{H}\right)  \otimes
f^{\ast\left(  n-1\right)  }\left(  1_{H}\right)  }\right)  =\mu_{A}\left(
\underbrace{f\left(  1_{H}\right)  }_{=0}\otimes f^{\ast\left(  n-1\right)
}\left(  1_{H}\right)  \right) \\
&  =\mu_{A}\underbrace{\left(  0\otimes f^{\ast\left(  n-1\right)  }\left(
1_{H}\right)  \right)  }_{=0}=\mu_{A}\left(  0\right)  =0.
\end{align*}
This proves Lemma~\ref{Lemma126}.
\end{proof}

Now, in analogy to our above proof of Lemma~\ref{Lemma31} (but, again,
slightly simpler), we can prove Lemma~\ref{Lemma121}:

\begin{proof}
[Proof of Lemma~\ref{Lemma121}.]Assume that $f$ is an $\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivation.

Let $a\in H$ and $b\in H$. Then, (\ref{exp}) (applied to $x=a$) yields
$e^{\ast f}\left(  a\right)  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(  a\right)
}{i!}$. Also, (\ref{exp}) (applied to $x=b$) yields $e^{\ast f}\left(
b\right)  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(  b\right)  }{i!}=\sum_{j\geq
0}\dfrac{f^{\ast j}\left(  b\right)  }{j!}$ (here, we renamed the index $i$ as
$j$ in the sum). Hence,%
\begin{align*}
e^{\ast f}\left(  a\right)  \cdot e^{\ast f}\left(  b\right)   &  =\left(
\sum_{i\geq0}\dfrac{f^{\ast i}\left(  a\right)  }{i!}\right)  \cdot\left(
\sum_{j\geq0}\dfrac{f^{\ast j}\left(  b\right)  }{j!}\right)  =\sum_{i\geq
0}\sum_{j\geq0}\dfrac{f^{\ast i}\left(  a\right)  }{i!}\cdot\dfrac{f^{\ast
j}\left(  b\right)  }{j!}\\
&  =\underbrace{\sum_{i\geq0}\sum_{\substack{n\geq0;\\n\geq i}}}_{=\sum
_{n\geq0}\sum_{\substack{i\geq0;\\n\geq i}}}\dfrac{f^{\ast i}\left(  a\right)
}{i!}\cdot\dfrac{f^{\ast\left(  n-i\right)  }\left(  b\right)  }{\left(
n-i\right)  !}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }n-i\text{ for
}j\text{ in the second sum}\right) \\
&  =\sum_{n\geq0}\underbrace{\sum_{\substack{i\geq0;\\n\geq i}}}_{=\sum
_{i=0}^{n}}\underbrace{\dfrac{f^{\ast i}\left(  a\right)  }{i!}\cdot
\dfrac{f^{\ast\left(  n-i\right)  }\left(  b\right)  }{\left(  n-i\right)  !}%
}_{=\dfrac{1}{i!\left(  n-i\right)  !}f^{\ast i}\left(  a\right)
f^{\ast\left(  n-i\right)  }\left(  b\right)  }\\
&  =\sum_{n\geq0}\sum_{i=0}^{n}\underbrace{\dfrac{1}{i!\left(  n-i\right)  !}%
}_{\substack{=\dfrac{1}{n!}\dbinom{n}{i}\\\text{(since }\dfrac{n!}{i!\left(
n-i\right)  !}=\dbinom{n}{i}\text{)}}}f^{\ast i}\left(  a\right)
f^{\ast\left(  n-i\right)  }\left(  b\right) \\
&  =\sum_{n\geq0}\sum_{i=0}^{n}\dfrac{1}{n!}\dbinom{n}{i}f^{\ast i}\left(
a\right)  f^{\ast\left(  n-i\right)  }\left(  b\right)  =\sum_{n\geq0}%
\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast i}\left(  a\right)
f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)  .
\end{align*}
Compared with%
\begin{align*}
e^{\ast f}\left(  ab\right)   &  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(
ab\right)  }{i!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{exp}), applied to
}x=ab\right) \\
&  =\sum_{n\geq0}\dfrac{f^{\ast n}\left(  ab\right)  }{n!}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }i\text{ as
}n\text{ in the sum}\right) \\
&  =\sum_{n\geq0}\dfrac{1}{n!}f^{\ast n}\left(  ab\right)  =\sum_{n\geq
0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast i}\left(  a\right)
f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{125.fab})}\right)  ,
\end{align*}
this yields $e^{\ast f}\left(  a\right)  \cdot e^{\ast f}\left(  b\right)
=e^{\ast f}\left(  ab\right)  $.

Now forget that we fixed $a$ and $b$. We thus have proven that all $a\in H$
and $b\in H$ satisfy $e^{\ast f}\left(  a\right)  \cdot e^{\ast f}\left(
b\right)  =e^{\ast f}\left(  ab\right)  $. Combined with the fact that%
\begin{align*}
e^{\ast f}\left(  1_{H}\right)   &  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(
1_{H}\right)  }{i!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{exp}), applied
to }x=1_{H}\right) \\
&  =\dfrac{f^{\ast0}\left(  1_{H}\right)  }{1!}+\sum_{i>0}\underbrace{\dfrac
{f^{\ast i}\left(  1_{H}\right)  }{i!}}_{\substack{=0\\\text{(because
Lemma~\ref{Lemma126}}\\\text{(applied to }n=i\text{)}\\\text{yields }f^{\ast
i}\left(  1_{H}\right)  =0\text{)}}}=\dfrac{f^{\ast0}\left(  1_{H}\right)
}{1!}+\underbrace{\sum_{i>0}0}_{=0}=\dfrac{f^{\ast0}\left(  1_{H}\right)
}{1!}=\dfrac{f^{\ast0}\left(  1_{H}\right)  }{1}\\
&  =\underbrace{f^{\ast0}}_{=e_{H,A}=\eta_{A}\circ\varepsilon_{H}}\left(
1_{H}\right)  =\left(  \eta_{A}\circ\varepsilon_{H}\right)  \left(
1_{H}\right)  =\eta_{A}\left(  \varepsilon_{H}\left(  1_{H}\right)  \right)
=\underbrace{\varepsilon_{H}\left(  1_{H}\right)  }_{\substack{=1\\\text{(by
the axioms of a bialgebra,}\\\text{since }H\text{ is a bialgebra)}}}1_{A}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\eta_{A}\right) \\
&  =1\cdot1_{A}=1_{A},
\end{align*}
this shows that $e^{\ast f}$ is a $k$-algebra homomorphism. This proves
Lemma~\ref{Lemma121}.
\end{proof}

Next let us prepare for the proof of Lemma~\ref{Lemma122}. We are not going to
need a ``dual'' Corollary~\ref{Corollary48}, since Corollary~\ref{Corollary48}
is already fine (it is as self-dual as a statement about filtered coalgebras
can get), but we need a dual of Corollary~\ref{Corollary46}, and for this we
first need a dual of Proposition~\ref{Proposition45}:

\begin{proposition}
\label{Proposition135} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a commutative $k$-algebra. Let $f:H\rightarrow A$ and $g:H\rightarrow A$ be
two $k$-algebra homomorphisms. Then, $f\ast g:H\rightarrow A$ is also a
$k$-algebra homomorphism.
\end{proposition}

One possible proof of this proposition would be obtained by reversing the
arrows in the proof of Proposition~\ref{Proposition45} (though, of course, we
would need to do the same with Lemma~\ref{Lemma36}). But let us (like in the
proof of Lemma~\ref{Lemma125}) work elementwise:

\begin{proof}
[Proof of Proposition~\ref{Proposition135}.]First we notice that $\Delta
_{H}\left(  1_{H}\right)  =1_{H}\otimes1_{H}$ (by the axioms of a bialgebra,
since $H$ is a bialgebra). Since $f\ast g=\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{H}$ (by the definition of convolution), we have%
\begin{align*}
\left(  f\ast g\right)  \left(  1_{H}\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{H}\right)  \left(  1_{H}\right)  =\mu
_{A}\left(  \left(  f\otimes g\right)  \underbrace{\left(  \Delta_{H}\left(
1_{H}\right)  \right)  }_{=1_{H}\otimes1_{H}}\right)  =\mu_{A}\left(
\underbrace{\left(  f\otimes g\right)  \left(  1_{H}\otimes1_{H}\right)
}_{=f\left(  1_{H}\right)  \otimes g\left(  1_{H}\right)  }\right) \\
&  =\mu_{A}\left(  f\left(  1_{H}\right)  \otimes g\left(  1_{H}\right)
\right)  =\underbrace{f\left(  1_{H}\right)  }_{\substack{=1_{A}\\\text{(since
}f\text{ is a }k\text{-algebra}\\\text{homomorphism)}}}\underbrace{g\left(
1_{H}\right)  }_{\substack{=1_{A}\\\text{(since }g\text{ is a }%
k\text{-algebra}\\\text{homomorphism)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{A}\text{ is the
multiplication map}\right) \\
&  =1_{A}\cdot1_{A}=1_{A}.
\end{align*}


Now, let $a\in H$ and $b\in H$ be arbitrary.

Since $\Delta_{H}\left(  a\right)  \in H\otimes H$, we can write $\Delta
_{H}\left(  a\right)  $ in the form $\Delta_{H}\left(  a\right)  =\sum
_{j=1}^{M}\lambda_{j}a_{j}\otimes a_{j}^{\prime}$ for some $M\in\mathbb{N}$,
some elements $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{M}$ of $k$,
some elements $a_{1}$, $a_{2}$, $\ldots$, $a_{M}$ of $H$, and some elements
$a_{1}^{\prime}$, $a_{2}^{\prime}$, $\ldots$, $a_{M}^{\prime}$ of $H$.
Consider this $M$, these $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{M}%
$, these $a_{1}$, $a_{2}$, $\ldots$, $a_{M}$, and these $a_{1}^{\prime}$,
$a_{2}^{\prime}$, $\ldots$, $a_{M}^{\prime}$.

Since $\Delta_{H}\left(  b\right)  \in H\otimes H$, we can write $\Delta
_{H}\left(  b\right)  $ in the form $\Delta_{H}\left(  b\right)  =\sum
_{\ell=1}^{K}\mu_{\ell}b_{\ell}\otimes b_{\ell}^{\prime}$ for some
$K\in\mathbb{N}$, some elements $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{K}$ of
$k$, some elements $b_{1}$, $b_{2}$, $\ldots$, $b_{K}$ of $H$, and some
elements $b_{1}^{\prime}$, $b_{2}^{\prime}$, $\ldots$, $b_{K}^{\prime}$ of
$H$. Consider this $K$, these $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{K}$, these
$b_{1}$, $b_{2}$, $\ldots$, $b_{K}$, and these $b_{1}^{\prime}$,
$b_{2}^{\prime}$, $\ldots$, $b_{K}^{\prime}$.

Since $H$ is a bialgebra,%
\begin{align*}
\Delta_{H}\left(  ab\right)   &  =\underbrace{\Delta_{H}\left(  a\right)
}_{=\sum_{j=1}^{M}\lambda_{j}a_{j}\otimes a_{j}^{\prime}}\cdot
\underbrace{\Delta_{H}\left(  b\right)  }_{=\sum_{\ell=1}^{K}\mu_{\ell}%
b_{\ell}\otimes b_{\ell}^{\prime}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
axioms of a bialgebra}\right) \\
&  =\left(  \sum_{j=1}^{M}\lambda_{j}a_{j}\otimes a_{j}^{\prime}\right)
\cdot\left(  \sum_{\ell=1}^{K}\mu_{\ell}b_{\ell}\otimes b_{\ell}^{\prime
}\right)  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}%
\underbrace{\left(  a_{j}\otimes a_{j}^{\prime}\right)  \left(  b_{\ell
}\otimes b_{\ell}^{\prime}\right)  }_{=a_{j}b_{\ell}\otimes a_{j}^{\prime
}b_{\ell}^{\prime}}\\
&  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}a_{j}b_{\ell}\otimes
a_{j}^{\prime}b_{\ell}^{\prime}.
\end{align*}
Since $f\ast g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{H}$, we now
have%
\begin{align}
\left(  f\ast g\right)  \left(  ab\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{H}\right)  \left(  ab\right)  =\mu_{A}\left(
\left(  f\otimes g\right)  \underbrace{\left(  \Delta_{H}\left(  ab\right)
\right)  }_{=\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}a_{j}b_{\ell
}\otimes a_{j}^{\prime}b_{\ell}^{\prime}}\right) \nonumber\\
&  =\mu_{A}\left(  \underbrace{\left(  f\otimes g\right)  \left(  \sum
_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}a_{j}b_{\ell}\otimes
a_{j}^{\prime}b_{\ell}^{\prime}\right)  }_{\substack{=\sum_{j=1}^{M}\sum
_{\ell=1}^{K}\lambda_{j}\mu_{\ell}f\left(  a_{j}b_{\ell}\right)  \otimes
g\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  \\\text{(by the definition of
}f\otimes g\text{)}}}\right) \nonumber\\
&  =\mu_{A}\left(  \sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell
}f\left(  a_{j}b_{\ell}\right)  \otimes g\left(  a_{j}^{\prime}b_{\ell
}^{\prime}\right)  \right) \nonumber\\
&  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}\underbrace{f\left(
a_{j}b_{\ell}\right)  }_{\substack{=f\left(  a_{j}\right)  f\left(  b_{\ell
}\right)  \\\text{(since }f\text{ is a }k\text{-algebra}\\\text{homomorphism)}%
}}\underbrace{g\left(  a_{j}^{\prime}b_{\ell}^{\prime}\right)  }%
_{\substack{=g\left(  a_{j}^{\prime}\right)  g\left(  b_{\ell}^{\prime
}\right)  \\\text{(since }g\text{ is a }k\text{-algebra}\\\text{homomorphism)}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{A}\text{ is the
multiplication map}\right) \nonumber\\
&  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}f\left(  a_{j}\right)
\underbrace{f\left(  b_{\ell}\right)  g\left(  a_{j}^{\prime}\right)
}_{\substack{=g\left(  a_{j}^{\prime}\right)  f\left(  b_{\ell}\right)
\\\text{(since }A\text{ is commutative)}}}g\left(  b_{\ell}^{\prime}\right)
\nonumber\\
&  =\sum_{j=1}^{M}\sum_{\ell=1}^{K}\lambda_{j}\mu_{\ell}f\left(  a_{j}\right)
g\left(  a_{j}^{\prime}\right)  f\left(  b_{\ell}\right)  g\left(  b_{\ell
}^{\prime}\right) \nonumber\\
&  =\left(  \sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  g\left(
a_{j}^{\prime}\right)  \right)  \left(  \sum_{\ell=1}^{K}\mu_{\ell}f\left(
b_{\ell}\right)  g\left(  b_{\ell}^{\prime}\right)  \right)  .
\label{135.pf.3}%
\end{align}
Since%
\begin{align*}
\left(  f\ast g\right)  \left(  a\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{H}\right)  \left(  a\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{because }f\ast g=\mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{H}\right) \\
&  =\mu_{A}\left(  \left(  f\otimes g\right)  \underbrace{\left(  \Delta
_{H}\left(  a\right)  \right)  }_{=\sum_{j=1}^{M}\lambda_{j}a_{j}\otimes
a_{j}^{\prime}}\right)  =\mu_{A}\left(  \underbrace{\left(  f\otimes g\right)
\left(  \sum_{j=1}^{M}\lambda_{j}a_{j}\otimes a_{j}^{\prime}\right)
}_{\substack{=\sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  \otimes g\left(
a_{j}^{\prime}\right)  \\\text{(by the definition of }f\otimes g\text{)}%
}}\right) \\
&  =\mu_{A}\left(  \sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  \otimes
g\left(  a_{j}^{\prime}\right)  \right) \\
&  =\sum_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  g\left(  a_{j}^{\prime
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{A}\text{ is the
multiplication map}\right)
\end{align*}
and%
\[
\left(  f\ast g\right)  \left(  b\right)  =\sum_{\ell=1}^{K}\mu_{\ell}f\left(
b_{\ell}\right)  g\left(  b_{\ell}^{\prime}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the same argument, done for }b\text{
instead of }a\right)  ,
\]
the equality (\ref{135.pf.3}) becomes%
\begin{align*}
\left(  f\ast g\right)  \left(  ab\right)   &  =\underbrace{\left(  \sum
_{j=1}^{M}\lambda_{j}f\left(  a_{j}\right)  g\left(  a_{j}^{\prime}\right)
\right)  }_{=\left(  f\ast g\right)  \left(  a\right)  }\underbrace{\left(
\sum_{\ell=1}^{K}\mu_{\ell}f\left(  b_{\ell}\right)  g\left(  b_{\ell}%
^{\prime}\right)  \right)  }_{=\left(  f\ast g\right)  \left(  b\right)  }\\
&  =\left(  f\ast g\right)  \left(  a\right)  \cdot\left(  f\ast g\right)
\left(  b\right)  .
\end{align*}


Now forget that we fixed $a$ and $b$. We have now shown that any $a\in H$ and
$b\in H$ satisfy $\left(  f\ast g\right)  \left(  ab\right)  =\left(  f\ast
g\right)  \left(  a\right)  \cdot\left(  f\ast g\right)  \left(  b\right)  $.
Combined with $\left(  f\ast g\right)  \left(  1_{H}\right)  =1_{A}$, this
yields that $f\ast g$ is a $k$-algebra homomorphism.
Proposition~\ref{Proposition135} is proven.
\end{proof}

As a consequence, we have:

\begin{corollary}
\label{Corollary136} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a commutative $k$-algebra. Let $f:H\rightarrow A$ be a $k$-algebra
homomorphism. Let $n\in\mathbb{N}$. Then, $f^{\ast n}:H\rightarrow A$ is also
a $k$-algebra homomorphism.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary136}.]We are going to prove
Corollary~\ref{Corollary136} by induction over $n$.

\textit{Induction base:} Consider the obvious canonical $k$-algebra structure
on $k$ (with $\mu_{k}$ being the canonical isomorphism $k\otimes k\rightarrow
k$, and $\eta_{k}$ being the identity map). Since $A$ is a $k$-algebra, it is
clear that $\eta_{A}:k\rightarrow A$ is a $k$-algebra
homomorphism.\footnote{\textit{Proof.} Any two $\lambda\in k$ and $\mu\in k$
satisfy%
\[
\underbrace{\eta_{A}\left(  \lambda\right)  }_{\substack{=\lambda\cdot
1_{A}\\\text{(by the definition of }\eta_{A}\text{)}}}\cdot\underbrace{\eta
_{A}\left(  \mu\right)  }_{\substack{=\mu\cdot1_{A}\\\text{(by the definition
of }\eta_{A}\text{)}}}=\left(  \lambda\cdot1_{A}\right)  \cdot\left(  \mu
\cdot1_{A}\right)  =\lambda\mu\cdot1_{A}=\eta_{A}\left(  \lambda\mu\right)
\]
(since $\eta_{A}\left(  \lambda\mu\right)  =\lambda\mu\cdot1_{A}$ by the
definition of $\eta_{A}$). Combined with the fact that (again, as the
definition of $\eta_{A}$ shows) $\eta_{A}\left(  1\right)  =1\cdot1_{A}=1_{A}%
$, this yields that $\eta_{A}$ is a $k$-algebra homomorphism, qed.} Combined
with the fact that $\varepsilon_{H}:H\rightarrow k$ is a $k$-algebra
homomorphism (because $H$ is a $k$-bialgebra), this yields that $\eta_{A}%
\circ\varepsilon_{H}$ is a $k$-algebra homomorphism (since the composition of
two $k$-algebra homomorphisms is a $k$-algebra homomorphism). Since $f^{\ast
0}=e_{H,A}=\eta_{A}\circ\varepsilon_{H}$ (by the definition of $e_{H,A}$),
this yields that $f^{\ast0}$ is a $k$-algebra homomorphism. In other words,
Corollary~\ref{Corollary136} holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
Corollary~\ref{Corollary136} holds for $n=N$. We must now prove that
Corollary~\ref{Corollary136} also holds for $n=N+1$.

Since Corollary~\ref{Corollary136} holds for $n=N$, we know that $f^{\ast N}$
is a $k$-algebra homomorphism. Proposition~\ref{Proposition135} (applied to
$g=f^{\ast N}$) now yields that $f\ast f^{\ast N}$ is a $k$-algebra
homomorphism. Since $f\ast f^{\ast N}=f^{\ast\left(  N+1\right)  }$, this
yields that $f^{\ast\left(  N+1\right)  }$ is a $k$-algebra homomorphism. In
other words, Corollary~\ref{Corollary136} holds for $n=N+1$. This completes
the induction step. The induction proof of Corollary~\ref{Corollary136} is
thus complete.
\end{proof}

Next, we dualize Lemma~\ref{Lemma53}:

\begin{lemma}
\label{Lemma141} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $A$ be a $k$-algebra. Let
$f\in\mathfrak{g}\left(  H,A\right)  $. Let $a\in H$ and $b\in H$. Then,%
\begin{equation}
e^{\ast f}\left(  ab\right)  =\sum_{n\geq0}\dfrac{1}{n!}f^{\ast n}\left(
ab\right)  \label{141.1}%
\end{equation}
and%
\begin{equation}
e^{\ast f}\left(  a\right)  \cdot e^{\ast f}\left(  b\right)  =\sum_{n\geq
0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast i}\left(  a\right)
f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)  . \label{141.2}%
\end{equation}
(In particular, the infinite sums $\sum_{n\geq0}\dfrac{1}{n!}f^{\ast n}\left(
ab\right)  $ and $\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom
{n}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(  n-i\right)  }\left(
b\right)  \right)  $ converge with respect to the discrete topology, i. e.,
each of these sums has only finitely many nonzero addends.)
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma141}.]The equalities (\ref{141.1}) and (\ref{141.2})
have been proven during our proof of Lemma~\ref{Lemma121}, \textit{without
using the assumptions that }$A$ \textit{be commutative and }$f$ \textit{be an
}$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $\textit{-derivation}.
Hence, these equalities are true. Lemma~\ref{Lemma141} is thus true.
\end{proof}

We can now prove Lemma~\ref{Lemma122}:

\begin{proof}
[Proof of Lemma~\ref{Lemma122}.]Assume that $e^{\ast f}$ is a $k$-algebra homomorphism.

Fix some $\left(  a,b\right)  \in H\times H$. Thus, $a\in H$ and $b\in H$.

We want to prove that $f\left(  ab\right)  =f\left(  a\right)  \varepsilon
_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)  f\left(  b\right)  $.

Since $H$ is filtered, there exists some $N\in\mathbb{N}$ such that $a\in
H_{\leq N}$. Consider this $N$.

Since $H$ is filtered, there exists some $M\in\mathbb{N}$ such that $b\in
H_{\leq M}$. Consider this $M$.

Since $H$ is filtered, there exists some $K\in\mathbb{N}$ such that $ab\in
H_{\leq K}$. Consider this $K$.

We define a filtration $\left(  k_{\leq\ell}\right)  _{\ell\geq0}$ on the
$k$-vector space $k$ by
\[
\left(  k_{\leq\ell}=\left\{
\begin{array}
[c]{c}%
0\text{, if }\ell\leq N+M+K;\\
k\text{, if }\ell>N+M+K
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{N}\right)  .
\]


For every $n\in\mathbb{N}$, define a $k$-linear map $h_{n}:k\rightarrow A$ by%
\begin{equation}
h_{n}=\dfrac{1}{n!}f^{\ast n}\left(  ab\right)  \eta_{A}-\dfrac{1}{n!}\left(
\sum_{i=0}^{n}\dbinom{n}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(
n-i\right)  }\left(  b\right)  \right)  \eta_{A}. \label{122.pf.h_n}%
\end{equation}


\textbf{a)} We now will prove that every $n\in\mathbb{N}$ satisfies
$h_{n}\left(  k_{\leq n-1}\right)  =0$.

\textit{Proof.} Fix some $n\in\mathbb{N}$. We distinguish between two cases:

\textit{Case 1:} We have $n\leq N+M+K+1$.

\textit{Case 2:} We have $n>N+M+K+1$.

First let us consider Case 1. In this case, $n\leq N+M+K+1$, so that $n-1\leq
N+M+K$, and thus%
\begin{align*}
k_{\leq n-1}  &  =\left\{
\begin{array}
[c]{c}%
0\text{, if }n-1\leq N+M+K;\\
k\text{, if }n-1>N+M+K
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }k_{\leq
n-1}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-1\leq N+M+K\right)  ,
\end{align*}
so that $h_{n}\left(  k_{\leq n-1}\right)  =h_{n}\left(  0\right)  =0$. Hence,
$h_{n}\left(  k_{\leq n-1}\right)  =0$ is proven in Case 1.

Now let us consider Case 2. In this case, $n>N+M+K+1>K$. Thus,
Remark~\ref{Remark9} (applied to $n$ and $K$ instead of $i$ and $n$) yields
$f^{\ast n}\left(  H_{\leq K}\right)  =0$. Since $ab\in H_{\leq K}$, this
yields $f^{\ast n}\left(  ab\right)  =0$. On the other hand,%
\begin{equation}
\text{every }i\in\left\{  0,1,\ldots,N\right\}  \text{ satisfies }%
f^{\ast\left(  n-i\right)  }\left(  b\right)  =0. \label{122.pf.1}%
\end{equation}
\footnote{\textit{Proof.} Let $i\in\left\{  0,1,\ldots,N\right\}  $. Then,
$i\leq N$, so that $\underbrace{n}_{>N+M+K+1>N+M}-\underbrace{i}_{\leq
N}>N+M-N=M$. Thus, Remark~\ref{Remark9} (applied to $n-i$ and $M$ instead of
$i$ and $n$) yields $f^{\ast\left(  n-i\right)  }\left(  H_{\leq M}\right)
=0$. Since $b\in H_{\leq M}$, this yields $f^{\ast\left(  n-i\right)  }\left(
b\right)  =0$, and thus (\ref{122.pf.1}) is proven.}

Also,%
\begin{equation}
\text{every }i\in\left\{  N+1,N+2,\ldots,n\right\}  \text{ satisfies }f^{\ast
i}\left(  a\right)  =0. \label{122.pf.2}%
\end{equation}
\footnote{\textit{Proof.} Let $i\in\left\{  N+1,N+2,\ldots,n\right\}  $. Then,
$i>N$. Hence, Remark~\ref{Remark9} (applied to $N$ instead of $n$) yields
$f^{\ast i}\left(  H_{\leq N}\right)  =0$. Since $a\in H_{\leq N}$, this
yields $f^{\ast i}\left(  a\right)  =0$, and thus (\ref{122.pf.2}) is proven.}

Keeping these facts in mind, we have%
\begin{align*}
\sum_{i=0}^{n}\dbinom{n}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(
n-i\right)  }\left(  b\right)   &  =\sum_{i=0}^{N}\dbinom{n}{i}f^{\ast
i}\left(  a\right)  \underbrace{f^{\ast\left(  n-i\right)  }\left(  b\right)
} _{\substack{=0 \\\text{(by (\ref{122.pf.1}))}}}+\sum_{i=N+1}^{n}\dbinom
{n}{i}\underbrace{f^{\ast i}\left(  a\right)  }_{\substack{=0 \\\text{(by
(\ref{122.pf.2}))}}}f^{\ast\left(  n-i\right)  }\left(  b\right) \\
&  =\underbrace{\sum_{i=0}^{N}\dbinom{n}{i}f^{\ast i}\left(  a\right)  0}%
_{=0}+\underbrace{\sum_{i=N+1}^{n}\dbinom{n}{i}0f^{\ast\left(  n-i\right)
}\left(  b\right)  }_{=0}=0+0=0.
\end{align*}


Now, by the definition of $h_{n}$, we have%
\[
h_{n}=\dfrac{1}{n!}\underbrace{f^{\ast n}\left(  ab\right)  }_{=0}\eta
_{A}-\dfrac{1}{n!}\underbrace{\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast
i}\left(  a\right)  f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)
}_{=0}\eta_{A}=\dfrac{1}{n!}0\eta_{A}-\dfrac{1}{n!}0\eta_{A}=0
\]
and thus $h_{n}\left(  k_{\leq n-1}\right)  =0$. Hence, $h_{n}\left(  k_{\leq
n-1}\right)  =0$ is proven in Case 2.

Hence, in both possible cases, we have shown that $h_{n}\left(  k_{\leq
n-1}\right)  =0$. This proves that $h_{n}\left(  k_{\leq n-1}\right)  =0$
always holds. In other words, part \textbf{a)} is proven.

\textbf{b)} Now, we will show that every $x\in k$ and every $t\in\mathbb{N}$
satisfy $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$.\ \ \ \ \footnote{Note
that the element $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  $ is well-defined
due to Proposition~\ref{Proposition49} \textbf{(a)} (applied to $V=k$ and
$W=A$), but in our case even the element $\sum_{i\geq0}t^{i}h_{i}$ itself is
well-defined (we can see that, for instance, by following the proof of
\textbf{b)}).}

\textit{Proof.} Let $x\in k$ and $t\in\mathbb{N}$ be arbitrary. By
Corollary~\ref{Corollary136} (applied to $t$ and $e^{\ast f}$ instead of $n$
and $f$), we see that $\left(  e^{\ast f}\right)  ^{\ast t}$ is a $k$-algebra
homomorphism. Since $e^{\ast\left(  tf\right)  }=\left(  e^{\ast f}\right)
^{\ast t}$ (by Corollary~\ref{Corollary48}, applied to $H$, $A$ and $t$
instead of $C$, $H$ and $n$), this rewrites as follows: The map $e^{\ast
\left(  tf\right)  }$ is a $k$-algebra homomorphism. Thus,%
\begin{equation}
e^{\ast\left(  tf\right)  }\left(  ab\right)  =e^{\ast\left(  tf\right)
}\left(  a\right)  \cdot e^{\ast\left(  tf\right)  }\left(  b\right)  .
\label{122.pf.3}%
\end{equation}


But applying Lemma~\ref{Lemma141} to $tf$ instead of $f$ (this is allowed
since $f\in\mathfrak{g}\left(  H,A\right)  $ yields $tf\in\mathfrak{g}\left(
H,A\right)  $), we obtain%
\begin{equation}
e^{\ast\left(  tf\right)  }\left(  ab\right)  =\sum_{n\geq0}\dfrac{1}%
{n!}\left(  tf\right)  ^{\ast n}\left(  ab\right)  \label{122.pf.4}%
\end{equation}
and%
\begin{equation}
e^{\ast\left(  tf\right)  }\left(  a\right)  \cdot e^{\ast\left(  tf\right)
}\left(  b\right)  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom
{n}{i}\left(  tf\right)  ^{\ast i}\left(  a\right)  \left(  tf\right)
^{\ast\left(  n-i\right)  }\left(  b\right)  \right)  . \label{122.pf.5}%
\end{equation}


We are now going to rewrite these two equalities by taking $t$ out of the brackets.

Since%
\[
\sum_{n\geq0}\dfrac{1}{n!}\underbrace{\left(  tf\right)  ^{\ast n}}%
_{=t^{n}f^{\ast n}}\left(  ab\right)  =\sum_{n\geq0}\dfrac{1}{n!}%
\underbrace{\left(  t^{n}f^{\ast n}\right)  \left(  ab\right)  }%
_{=t^{n}f^{\ast n}\left(  ab\right)  }=\sum_{n\geq0}\dfrac{1}{n!}t^{n}f^{\ast
n}\left(  ab\right)  =\sum_{n\geq0}t^{n}\dfrac{1}{n!}f^{\ast n}\left(
ab\right)  ,
\]
the equality (\ref{122.pf.3}) rewrites as%
\begin{equation}
e^{\ast\left(  tf\right)  }\left(  ab\right)  =\sum_{n\geq0}t^{n}\dfrac{1}%
{n!}f^{\ast n}\left(  ab\right)  . \label{122.pf.4'}%
\end{equation}


Since%
\begin{align*}
&  \sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}%
\underbrace{\left(  tf\right)  ^{\ast i}}_{=t^{i}f^{\ast i}}\left(  a\right)
\underbrace{\left(  tf\right)  ^{\ast\left(  n-i\right)  }}_{=t^{n-i}%
f^{\ast\left(  n-i\right)  }}\left(  b\right)  \right) \\
&  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}%
\underbrace{\left(  t^{i}f^{\ast i}\right)  \left(  a\right)  }_{=t^{i}f^{\ast
i}\left(  a\right)  }\underbrace{\left(  t^{n-i}f^{\ast\left(  n-i\right)
}\right)  \left(  b\right)  }_{=t^{n-i}f^{\ast\left(  n-i\right)  }\left(
b\right)  }\right)  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}%
\dbinom{n}{i}\underbrace{t^{i}f^{\ast i}\left(  a\right)  t^{n-i}%
f^{\ast\left(  n-i\right)  }\left(  b\right)  }_{=t^{i}t^{n-i}f^{\ast
i}\left(  a\right)  f^{\ast\left(  n-i\right)  }\left(  b\right)  }\right) \\
&  =\sum_{n\geq0}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}%
\underbrace{t^{i}t^{n-i}}_{=t^{n}}f^{\ast i}\left(  a\right)  f^{\ast\left(
n-i\right)  }\left(  b\right)  \right)  =\sum_{n\geq0}\dfrac{1}{n!}\left(
\sum_{i=0}^{n}\dbinom{n}{i}t^{n}f^{\ast i}\left(  a\right)  f^{\ast\left(
n-i\right)  }\left(  b\right)  \right) \\
&  =\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast
i}\left(  a\right)  f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)  ,
\end{align*}
the equality (\ref{122.pf.5}) rewrites as%
\begin{equation}
e^{\ast\left(  tf\right)  }\left(  a\right)  \cdot e^{\ast\left(  tf\right)
}\left(  b\right)  =\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(  \sum_{i=0}%
^{n}\dbinom{n}{i}f^{\ast i}\left(  a\right)  f^{\ast\left(  n-i\right)
}\left(  b\right)  \right)  . \label{122.pf.5'}%
\end{equation}


Now,
\begin{align*}
\sum_{i\geq0}t^{i}h_{i}  &  =\sum_{n\geq0}t^{n}h_{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }i\text{ as
}n\text{ in the sum}\right) \\
&  =\sum_{n\geq0}t^{n}\left(  \dfrac{1}{n!}f^{\ast n}\left(  ab\right)
\eta_{A}-\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast i}\left(
a\right)  f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)  \eta
_{A}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{122.pf.h_n})}\right)
\\
&  =\underbrace{\sum_{n\geq0}t^{n}\dfrac{1}{n!}f^{\ast n}\left(  ab\right)
\eta_{A}}_{=\left(  \sum_{n\geq0}t^{n}\dfrac{1}{n!}f^{\ast n}\left(
ab\right)  \right)  \eta_{A}}-\underbrace{\sum_{n\geq0}t^{n}\dfrac{1}%
{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast i}\left(  a\right)
f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)  \eta_{A}}_{=\left(
\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast
i}\left(  a\right)  f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)
\right)  \eta_{A}}\\
&  =\underbrace{\left(  \sum_{n\geq0}t^{n}\dfrac{1}{n!}f^{\ast n}\left(
ab\right)  \right)  }_{\substack{=e^{\ast\left(  tf\right)  }\left(
ab\right)  \\\text{(by (\ref{122.pf.4'}))}}}\eta_{A}-\underbrace{\left(
\sum_{n\geq0}t^{n}\dfrac{1}{n!}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\ast
i}\left(  a\right)  f^{\ast\left(  n-i\right)  }\left(  b\right)  \right)
\right)  }_{\substack{=e^{\ast\left(  tf\right)  }\left(  a\right)  \cdot
e^{\ast\left(  tf\right)  }\left(  b\right)  \\\text{(by (\ref{122.pf.5'}))}%
}}\eta_{A}\\
&  =\underbrace{e^{\ast\left(  tf\right)  }\left(  ab\right)  }%
_{\substack{=e^{\ast\left(  tf\right)  }\left(  a\right)  \cdot e^{\ast\left(
tf\right)  }\left(  b\right)  \\\text{(by (\ref{122.pf.3}))}}}\eta_{A}%
-e^{\ast\left(  tf\right)  }\left(  a\right)  \cdot e^{\ast\left(  tf\right)
}\left(  b\right)  \eta_{A}\\
&  =e^{\ast\left(  tf\right)  }\left(  a\right)  \cdot e^{\ast\left(
tf\right)  }\left(  b\right)  \eta_{A}-e^{\ast\left(  tf\right)  }\left(
a\right)  \cdot e^{\ast\left(  tf\right)  }\left(  b\right)  \eta_{A}=0,
\end{align*}
so that%
\[
\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =\underbrace{\left(  \sum_{i\geq
0}t^{i}h_{i}\right)  }_{=0}\left(  x\right)  =0\left(  x\right)  =0.
\]


We thus have proven that every $x\in k$ and every $t\in\mathbb{N}$ satisfy
$\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$. In other words, part
\textbf{b)} of the proof is done.

\textbf{c)} We know that every $n\in\mathbb{N}$ satisfies $h_{n}\left(
k_{\leq n-1}\right)  =0$ (by part \textbf{a)}), and that every $x\in k$ and
every $t\in\mathbb{N}$ satisfy $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$
(by part \textbf{b)}). Hence, Proposition~\ref{Proposition49} \textbf{(b)}
(applied to $k$ and $A$ instead of $V$ and $W$) yields that $h_{n}=0$ for
every $n\in\mathbb{N}$. Applied to $n=1$, this yields $h_{1}=0$. But
(\ref{122.pf.h_n}) (applied to $n=1$) yields%
\begin{align*}
h_{1}  &  =\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}=1}\underbrace{f^{\ast1}%
}_{=f}\left(  ab\right)  \eta_{A}-\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}%
=1}\underbrace{\left(  \sum_{i=0}^{1}\dbinom{1}{i}f^{\ast i}\left(  a\right)
f^{\ast\left(  1-i\right)  }\left(  b\right)  \right)  }_{=\dbinom{1}%
{0}f^{\ast0}\left(  a\right)  f^{\ast\left(  1-0\right)  }\left(  b\right)
+\dbinom{1}{1}f^{\ast1}\left(  a\right)  f^{\ast\left(  1-1\right)  }\left(
b\right)  }\eta_{A}\\
&  =f\left(  ab\right)  \eta_{A}-\left(  \underbrace{\dbinom{1}{0}}%
_{=1}\underbrace{f^{\ast0}}_{=e_{H,A}}\left(  a\right)  \underbrace{f^{\ast
\left(  1-0\right)  }}_{=f^{\ast1}=f}\left(  b\right)  +\underbrace{\dbinom
{1}{1}}_{=1}\underbrace{f^{\ast1}}_{=f}\left(  a\right)  \underbrace{f^{\ast
\left(  1-1\right)  }}_{=e_{H,A}}\left(  b\right)  \right)  \eta_{A}\\
&  =f\left(  ab\right)  \eta_{A}-\left(  e_{H,A}\left(  a\right)  f\left(
b\right)  +f\left(  a\right)  e_{H,A}\left(  b\right)  \right)  \eta
_{A}=\left(  f\left(  ab\right)  -\left(  e_{H,A}\left(  a\right)  f\left(
b\right)  +f\left(  a\right)  e_{H,A}\left(  b\right)  \right)  \right)
\eta_{A}.
\end{align*}
Since $h_{1}=0$, this rewrites as%
\[
0=\left(  f\left(  ab\right)  -\left(  e_{H,A}\left(  a\right)  f\left(
b\right)  +f\left(  a\right)  e_{H,A}\left(  b\right)  \right)  \right)
\eta_{A}.
\]
Hence,%
\begin{align*}
0\left(  1\right)   &  =\left(  \left(  f\left(  ab\right)  -\left(
e_{H,A}\left(  a\right)  f\left(  b\right)  +f\left(  a\right)  e_{H,A}\left(
b\right)  \right)  \right)  \eta_{A}\right)  \left(  1\right) \\
&  =\left(  f\left(  ab\right)  -\left(  e_{H,A}\left(  a\right)  f\left(
b\right)  +f\left(  a\right)  e_{H,A}\left(  b\right)  \right)  \right)
\underbrace{\left(  \eta_{A}\left(  1\right)  \right)  }_{\substack{=1\cdot
1_{A} \\\text{(by the definition of the map }\eta_{A}\text{)}}}\\
&  =\left(  f\left(  ab\right)  -\left(  e_{H,A}\left(  a\right)  f\left(
b\right)  +f\left(  a\right)  e_{H,A}\left(  b\right)  \right)  \right)
1\cdot1_{A}\\
&  =f\left(  ab\right)  -\left(  e_{H,A}\left(  a\right)  f\left(  b\right)
+f\left(  a\right)  e_{H,A}\left(  b\right)  \right)  .
\end{align*}
So we have%
\[
0=0\left(  1\right)  =f\left(  ab\right)  -\left(  e_{H,A}\left(  a\right)
f\left(  b\right)  +f\left(  a\right)  e_{H,A}\left(  b\right)  \right)  ,
\]
thus%
\[
f\left(  ab\right)  =e_{H,A}\left(  a\right)  f\left(  b\right)  +f\left(
a\right)  e_{H,A}\left(  b\right)  .
\]
Since%
\[
\underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{H}}\left(  a\right)  =\left(
\eta_{A}\circ\varepsilon_{H}\right)  \left(  a\right)  =\eta_{A}\left(
\varepsilon_{H}\left(  a\right)  \right)  =\varepsilon_{H}\left(  a\right)
\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\eta
_{A}\right)
\]
and%
\[
e_{H,A}\left(  b\right)  =\varepsilon_{H}\left(  b\right)  \cdot
1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{for similar reasons}\right)  ,
\]
this becomes%
\begin{align*}
f\left(  ab\right)   &  =\underbrace{e_{H,A}\left(  a\right)  }_{=\varepsilon
_{H}\left(  a\right)  \cdot1_{A}}f\left(  b\right)  +f\left(  a\right)
\underbrace{e_{H,A}\left(  b\right)  }_{=\varepsilon_{H}\left(  b\right)
\cdot1_{A}}=\varepsilon_{H}\left(  a\right)  \cdot1_{A}f\left(  b\right)
+f\left(  a\right)  \varepsilon_{H}\left(  b\right)  \cdot1_{A}\\
&  =\varepsilon_{H}\left(  a\right)  f\left(  b\right)  +f\left(  a\right)
\varepsilon_{H}\left(  b\right)  =f\left(  a\right)  \varepsilon_{H}\left(
b\right)  +\varepsilon_{H}\left(  a\right)  f\left(  b\right)  .
\end{align*}


\textbf{d)} Now forget that we fixed $\left(  a,b\right)  $. What we have
shown is that every $\left(  a,b\right)  \in H\times H$ satisfies $f\left(
ab\right)  =f\left(  a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon
_{H}\left(  a\right)  f\left(  b\right)  $. According to
Definition~\ref{Definition109}, this means that $f$ is an $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivation.

We have thus proven that, under the assumption that $e^{\ast f}$ is a
$k$-algebra homomorphism, the map $f$ is an $\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivation. Thus, Lemma~\ref{Lemma122} is proven.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem120}.]The assertion of Theorem~\ref{Theorem120}
is an ``if and only if'' assertion. Its ``if'' part was proven in
Lemma~\ref{Lemma122}, and its ``only if'' part was proven in
Lemma~\ref{Lemma121}. Hence, both parts of the assertion of
Theorem~\ref{Theorem120} are proven. This finally completes the proof of
Theorem~\ref{Theorem120}.
\end{proof}

Now we can finally deal with Theorem~\ref{Theorem101}:

\begin{proof}
[Proof of Theorem~\ref{Theorem101}.]Recall that $\operatorname*{id}\in
G\left(  H,H\right)  $. Thus, $\operatorname*{Log}\operatorname*{id}%
\in\mathfrak{g}\left(  H,H\right)  $ (because $\operatorname*{Log}%
F\in\mathfrak{g}\left(  H,H\right)  $ for every $F\in G\left(  H,H\right)  $).

Let $f$ be the map $\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(
H,H\right)  $. Then, $f=\operatorname*{Log}\operatorname*{id}\in
\mathfrak{g}\left(  H,H\right)  $. Besides, $f=\operatorname*{Log}%
\operatorname*{id}$ yields $e^{\ast f}=e^{\ast\left(  \operatorname*{Log}%
\operatorname*{id}\right)  }=\operatorname*{id}$ (by
Proposition~\ref{Proposition13} \textbf{(b)}, applied to $F=\operatorname*{id}%
$ and $A=H$). Hence, $e^{\ast f}$ is a $k$-algebra homomorphism (since
$\operatorname*{id}$ is a $k$-algebra homomorphism). By Lemma~\ref{Lemma122}
(applied to $A=H$), this yields that $f$ is an $\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivation.

Since $H$ is a bialgebra, $\varepsilon_{H}$ is a $k$-algebra homomorphism (by
the axioms of a bialgebra). Hence, we can apply Theorem~\ref{Theorem115} and
Proposition~\ref{Proposition104} to $A=H$.

Theorem~\ref{Theorem115} (applied to $A=H$) tells us that $f$ is an $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and only if $f\left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  =0$. Hence, $f\left(  \left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$ (since
we know that $f$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation).

We have $\varepsilon_{H}\circ\operatorname*{id}=\varepsilon_{H}$. Thus,
Proposition~\ref{Proposition104} \textbf{(c)} (applied to
$F=\operatorname*{id}$ and $A=H$) yields $\left(  \operatorname*{Log}%
\operatorname*{id}-\operatorname*{id}\right)  \left(  H\right)  \subseteq
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}$. Since $\operatorname*{Log}\operatorname*{id}=f$, this
rewrites as $\left(  f-\operatorname*{id}\right)  \left(  H\right)
\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}$. Now,%
\[
\underbrace{\left(  f^{2}-f\right)  }_{=f\circ\left(  f-\operatorname*{id}%
\right)  }\left(  H\right)  =\left(  f\circ\left(  f-\operatorname*{id}%
\right)  \right)  \left(  H\right)  =f\left(  \underbrace{\left(
f-\operatorname*{id}\right)  \left(  H\right)  }_{\subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}%
}\right)  \subseteq f\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0,
\]
so that $\left(  f^{2}-f\right)  \left(  H\right)  =0$. Thus, $f^{2}-f=0$, so
that $f^{2}=f$. In other words, $f$ is a projection. Since
$f=\operatorname*{Log}\operatorname*{id}$, this rewrites as follows: The map
$\operatorname*{Log}\operatorname*{id}$ is a projection.

Also, $f\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0$ yields $\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\subseteq
\operatorname*{Ker}f$. On the other hand, $\operatorname*{Ker}f\subseteq
\left(  f-\operatorname*{id}\right)  \left(  H\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $x\in\operatorname*{Ker}f$ be arbitrary.
Then, $f\left(  x\right)  =0$, so that $\left(  f-\operatorname*{id}\right)
\left(  x\right)  =\underbrace{f\left(  x\right)  }_{=0}%
-\underbrace{\operatorname*{id}\left(  x\right)  }_{=x}=-x$. Thus, $x=-\left(
f-\operatorname*{id}\right)  \left(  x\right)  =\left(  f-\operatorname*{id}%
\right)  \left(  -x\right)  $ (since $f-\operatorname*{id}$ is linear). Hence,
$x\in\left(  f-\operatorname*{id}\right)  \left(  H\right)  $ (since $-x\in
H$).
\par
We have thus shown that every $x\in\operatorname*{Ker}f$ satisfies
$x\in\left(  f-\operatorname*{id}\right)  \left(  H\right)  $. In other words,
$\operatorname*{Ker}f\subseteq\left(  f-\operatorname*{id}\right)  \left(
H\right)  $, qed.} combined with $\left(  f-\operatorname*{id}\right)  \left(
H\right)  \subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}$ yields $\operatorname*{Ker}f\subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}$.
Combined with $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\subseteq\operatorname*{Ker}f$, this results in
$\operatorname*{Ker}f=\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}+k\cdot1_{H}$. Since $f=\operatorname*{Log}%
\operatorname*{id}$, this rewrites as $\operatorname*{Ker}\left(
\operatorname*{Log}\operatorname*{id}\right)  =\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H} $.

This completes the proof of Theorem~\ref{Theorem101}.
\end{proof}

\section{\label{S13}Consequences for graded bialgebras}

We have hitherto been considering filtered bialgebras (and general
bialgebras), but not graded bialgebras. The reason for this is that we had no
reason for considering graded bialgebras: As long as the same statement holds
for graded and for filtered bialgebras, it is clearly enough to prove it for
filtered ones only, since graded bialgebras can always be seen as filtered
bialgebras (by Proposition~\ref{Proposition210} below), but not the other way
round. In this section \S \ref{S13}, we are going to formulate the relation
between graded and filtered bialgebras, and show some additional properties
special to graded bialgebras.

\begin{convention}
\label{Convention201} In the following, whenever $V$ is a graded vector
space\footnote{Note that a \textit{graded vector space} is defined as a vector
space $V$ along with a family $\left(  V_{\ell}\right)  _{\ell\geq0}$ of
subspaces of $V$ such that $V=\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$. There
are some more general definitions of the word ``graded'' in literature, but we
need this one. (In particular, if you are used to the notion of a ``$G$-graded
vector space'' where $G$ is a monoid, then you should notice that our notion
of ``graded vector space'' means an $\mathbb{N}$-graded vector space.)}, we
will denote the grading on $V$ by $\left(  V_{\ell}\right)  _{\ell\geq0}$.
(This is a general convention, so it does not only pertain to graded vector
spaces called $V$, but pertains to any graded vector space. For instance, if
we have a graded vector space called $C$, then this convention yields that the
grading on $C$ is denoted by $\left(  C_{\ell}\right)  _{\ell\geq0}$.)
\end{convention}

\begin{proposition}
\label{Proposition202} Let $k$ be a field. Let $V$ be a graded vector space.
For every $n\in\mathbb{N}$, define a $k$-vector subspace $V_{\leq n}$ of $V$
by $V_{\leq n}=\bigoplus_{\ell=0}^{n}V_{\ell}\ \ \ \ $\footnote{This is
well-defined, because the sum $\sum_{\ell=0}^{n}V_{\ell}$ is a direct sum (in
fact, since $V$ is a graded vector space, we have $V=\bigoplus_{\ell
\in\mathbb{N}}V_{\ell}$; now, the sum $\sum_{\ell=0}^{n}V_{\ell}$ is a partial
sum of the direct sum $\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$, and thus a
direct sum itself).}. Then, $\left(  V,\left(  V_{\leq n}\right)  _{n\geq
0}\right)  $ is a filtered $k$-vector space.
\end{proposition}

\begin{convention}
\label{Convention203} Let $k$ be a field. Let $V$ be a graded vector space.
Then, whenever we speak of ``the filtered $k$-vector space $V$'', we are going
to mean the filtered $k$-vector space $\left(  V,\left(  V_{\leq n}\right)
_{n\geq0}\right)  $ defined in Proposition~\ref{Proposition202}. In
particular, whenever we mention some $V_{\leq n}$ (for some $n\in\mathbb{N}$),
we are going to mean the $V_{\leq n}$ defined in
Proposition~\ref{Proposition202}.
\end{convention}

\begin{proof}
[Proof of Proposition~\ref{Proposition202}.]\textbf{a)} We have $V_{\leq
n}\subseteq V_{\leq n+1}$ for every $n\in\mathbb{N}$.

\textit{Proof.} Let $n\in\mathbb{N}$. By the definition of $V_{\leq n}$, we
have $V_{\leq n}=\bigoplus_{\ell=0}^{n}V_{\ell}$. By the definition of
$V_{\leq n+1}$, we have%
\[
V_{\leq n+1}=\bigoplus_{\ell=0}^{n+1}V_{\ell}=\underbrace{\bigoplus_{\ell
=0}^{n}V_{\ell}}_{=V_{\leq n}}\oplus V_{n+1}=V_{\leq n}\oplus V_{n+1}\supseteq
V_{\leq n},
\]
so that $V_{\leq n}\subseteq V_{\leq n+1}$. This proves \textbf{a)}.

\textbf{b)} We have%
\begin{equation}
V_{\leq a}\subseteq V_{\leq b}\text{ for every }a\in\mathbb{N}\text{ and }%
b\in\mathbb{N}\text{ satisfying }a\leq b. \label{202.pf.1}%
\end{equation}


\textit{Proof.} From part \textbf{a)}, we see that the sequence $\left(
V_{\leq n}\right)  _{n\geq0}$ of $k$-vector subspaces of $V$ is monotonically
increasing with respect to inclusion. This yields (\ref{202.pf.1}). Thus,
\textbf{b)} is proven.

\textbf{c)} We have $V=\bigcup_{n\in\mathbb{N}}V_{\leq n}$.

\textit{Proof.} Let $v\in V$. Then, $v\in V=\bigoplus_{\ell\in\mathbb{N}%
}V_{\ell}$ (since $V$ is a graded vector space), so there exists some family
$\left(  a_{\ell}\right)  _{\ell\geq0}\in\prod_{\ell\in\mathbb{N}}V_{\ell}$
such that $\left(  \text{all but finitely many }\ell\in\mathbb{N}\text{
satisfy }a_{\ell}=0\right)  $ and $v=\sum_{\ell\in\mathbb{N}}a_{\ell}$.
Consider this family $\left(  a_{\ell}\right)  _{\ell\geq0}$.

Notice that $a_{\ell}\in V_{\ell}$ for each $\ell\in\mathbb{N}$ (since
$\left(  a_{\ell}\right)  _{\ell\geq0}\in\prod_{\ell\in\mathbb{N}}V_{\ell}$).

There exists a finite subset $S$ of $\mathbb{N}$ such that%
\begin{equation}
\left(  \text{all }\ell\in\mathbb{N}\diagdown S\text{ satisfy }a_{\ell
}=0\right)  \label{202.pf.3}%
\end{equation}
(because all but finitely many $\ell\in\mathbb{N}$ satisfy $a_{\ell}=0$).
Consider this $S$. Then, $S$ is a finite set, and thus has a supremum in
$\mathbb{N}$ (this supremum is actually a maximum unless $S=\varnothing$). Let
$N$ be this supremum. Then, every $m\in S$ satisfies%
\begin{equation}
a_{m}\in V_{\leq N} \label{202.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{202.pf.2}):} Let $m\in S$. Recall that $N$ is
the supremum of the set $S$ in $\mathbb{N}$. Hence, $N\geq s$ for each $s\in
S$. Applying this to $s=m$, we obtain $N\geq m$. Thus, $m\leq N$. Hence,
$m\in\left\{  0,1,\ldots,N\right\}  $.
\par
The definition of $V_{\leq N}$ yields $V_{\leq N}=\bigoplus_{\ell=0}%
^{N}V_{\ell}$. But $V_{m}$ is an addend of the direct sum $\bigoplus_{\ell
=0}^{N}V_{\ell}$ (since $m\in\left\{  0,1,\ldots,N\right\}  $). Hence,
$V_{m}\subseteq\bigoplus_{\ell=0}^{N}V_{\ell}$. But recall that $a_{\ell}\in
V_{\ell}$ for each $\ell\in\mathbb{N}$. Applying this to $\ell=m$, we obtain
$a_{m}\in V_{m}\subseteq\bigoplus_{\ell=0}^{N}V_{\ell}=V_{\leq N}$. This
proves (\ref{202.pf.2}).}. Now we have%
\begin{align*}
v  &  =\sum_{\ell\in\mathbb{N}}a_{\ell}=\sum_{\ell\in S}\underbrace{a_{\ell}%
}_{\substack{\in V_{\leq N}\\\text{(by (\ref{202.pf.2}) (applied to }%
m=\ell\text{))}}}+\sum_{\ell\in\mathbb{N}\diagdown S}\underbrace{a_{\ell}%
}_{\substack{=0\\\text{(by (\ref{202.pf.3}))}}}\\
&  \in\sum_{\ell\in S}V_{\leq N}+\underbrace{\sum_{\ell\in\mathbb{N}\diagdown
S}0}_{=0}=\sum_{\ell\in S}V_{\leq N}\subseteq V_{\leq N}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }V_{\leq N}\text{ is a }k\text{-vector
subspace}\right) \\
&  \subseteq\bigcup_{n\in\mathbb{N}}V_{\leq n}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }V_{\leq N}\text{ is a term of the union }\bigcup_{n\in\mathbb{N}%
}V_{\leq n}\right)  .
\end{align*}


Now forget that we fixed $v$. We have thus shown that for every $v\in V$, we
have $v\in\bigcup_{n\in\mathbb{N}}V_{\leq n}$. In other words, $V\subseteq
\bigcup_{n\in\mathbb{N}}V_{\leq n}$. Combined with the triviality
$\bigcup_{n\in\mathbb{N}}V_{\leq n}\subseteq V$, this yields $V=\bigcup
_{n\in\mathbb{N}}V_{\leq n}$. This proves \textbf{c)}.

\textbf{d)} Combining the statements of parts \textbf{a)} and \textbf{c)}, we
conclude that $\left(  V,\left(  V_{\leq n}\right)  _{n\geq0}\right)  $ is a
filtered $k$-vector space. This proves Proposition~\ref{Proposition202}.
\end{proof}

\begin{proposition}
\label{Proposition204} Let $k$ be a field. Let $A$ be a graded $k$-algebra.
Then, $\left(  A,\left(  A_{\leq n}\right)  _{n\geq0}\right)  $ is a filtered
$k$-algebra.\footnote{Of course, the notation $A_{\leq n}$ has to be
understood here as according to Convention~\ref{Convention203}; that is,
$A_{\leq n}$ is defined by $A_{\leq n}=\bigoplus_{\ell=0}^{n}A_{\ell}$.}
\end{proposition}

\begin{convention}
\label{Convention205} Let $k$ be a field. Let $A$ be a graded $k$-algebra.
Then, whenever we speak of \textquotedblleft the filtered $k$-algebra
$A$\textquotedblright, we are going to mean the filtered $k$-algebra $\left(
A,\left(  A_{\leq n}\right)  _{n\geq0}\right)  $ defined in
Proposition~\ref{Proposition204}.
\end{convention}

\begin{proof}
[Proof of Proposition~\ref{Proposition204}.]\textbf{a)} By
Proposition~\ref{Proposition202} (applied to $V=A$), we know that $\left(
A,\left(  A_{\leq n}\right)  _{n\geq0}\right)  $ is a filtered $k$-vector space.

\textbf{b)} We have $1\in A_{\leq0}$.

\textit{Proof.} By the definition of $A_{\leq0}$, we have $A_{\leq0}%
=\bigoplus_{\ell=0}^{0}A_{\ell}=A_{0}$. Since $A$ is a graded $k$-algebra, we
have $1\in A_{0}=A_{\leq0}$. This proves \textbf{b)}.

\textbf{c)} We have
\begin{equation}
A_{u}\subseteq A_{\leq v}\ \ \ \ \ \ \ \ \ \ \text{for any }u\in
\mathbb{N}\text{ and }v\in\mathbb{N}\text{ satisfying }u\leq v.
\label{204.pf.0}%
\end{equation}


\textit{Proof.} Let $u\in\mathbb{N}$ and $v\in\mathbb{N}$ satisfy $u\leq v$.
Then, the definition of $A_{\leq v}$ says that $A_{\leq v}=\bigoplus_{\ell
=0}^{v}A_{\ell}$. Since $A_{u}$ is a summand in the direct sum $\bigoplus
_{\ell=0}^{v}A_{\ell}$ (because $u\in\mathbb{N}$ and $u\leq v$), this yields
$A_{\leq v}\supseteq A_{u}$, so that $A_{u}\subseteq A_{\leq v}$. This proves
\textbf{c)}.

\textbf{d)} We have $A_{\leq i}A_{\leq j}\subseteq A_{\leq i+j}$ for every
$i\in\mathbb{N}$ and $j\in\mathbb{N}$.

\textit{Proof.} Since $A$ is a graded $k$-algebra, we have
\begin{equation}
A_{\ell}A_{m}\subseteq A_{\ell+m}\ \ \ \ \ \ \ \ \ \ \text{for any }\ell
\in\mathbb{N}\text{ and }m\in\mathbb{N}. \label{204.pf.1}%
\end{equation}


Now, let $i\in\mathbb{N}$ and $j\in\mathbb{N}$ be arbitrary. Then, by the
definition of $A_{\leq i}$, we have $A_{\leq i}=\bigoplus_{\ell=0}^{i}A_{\ell
}=\sum_{\ell=0}^{i}A_{\ell}$ (since direct sums are sums). Also, by the
definition of $A_{\leq j}$, we have%
\begin{align*}
A_{\leq j}  &  =\bigoplus_{\ell=0}^{j}A_{\ell}=\bigoplus_{m=0}^{j}%
A_{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }\ell\text{
as }m\right) \\
&  =\sum_{m=0}^{j}A_{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are
sums}\right)  .
\end{align*}
Thus,%
\[
\underbrace{A_{\leq i}}_{=\sum_{\ell=0}^{i}A_{\ell}}\underbrace{A_{\leq j}%
}_{=\sum_{m=0}^{j}A_{m}}=\left(  \sum_{\ell=0}^{i}A_{\ell}\right)  \left(
\sum_{m=0}^{j}A_{m}\right)  =\sum_{\ell=0}^{i}\sum_{m=0}^{j}%
\underbrace{A_{\ell}A_{m}}_{\substack{\subseteq A_{\ell+m}\\\text{(by
(\ref{204.pf.1}))}}}\subseteq\sum_{\ell=0}^{i}\sum_{m=0}^{j}A_{\ell+m}.
\]
Since every $\ell\in\left\{  0,1,\ldots,i\right\}  $ and $m\in\left\{
0,1,\ldots,j\right\}  $ satisfy $A_{\ell+m}\subseteq A_{\leq i+j}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\ell\in\left\{  0,1,\ldots,i\right\}
$ and $m\in\left\{  0,1,\ldots,j\right\}  $ be arbitrary. Then, $0\leq\ell\leq
i$ and $0\leq m\leq j$. Hence, $0\leq\ell+m\leq i+j$, so that $A_{\ell
+m}\subseteq A_{\leq i+j}$ (by (\ref{204.pf.0}), applied to $u=\ell+m$ and
$v=i+j$),}, this becomes%
\[
A_{\leq i}A_{\leq j}\subseteq\sum_{\ell=0}^{i}\sum_{m=0}^{j}%
\underbrace{A_{\ell+m}}_{\subseteq A_{\leq i+j}}\subseteq\sum_{\ell=0}^{i}%
\sum_{m=0}^{j}A_{\leq i+j}\subseteq A_{\leq i+j}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }A_{\leq i+j}\text{ is a }k\text{-vector space}\right)  .
\]
This proves \textbf{d)}.

\textbf{e)} By combining the results of parts \textbf{b)} and \textbf{d)}, we
see that $\left(  A,\left(  A_{\leq n}\right)  _{n\geq0}\right)  $ is a
filtered $k$-algebra. This proves Proposition~\ref{Proposition204}.
\end{proof}

\begin{proposition}
\label{Proposition206} Let $k$ be a field. Let $C$ be a graded $k$-coalgebra.
Then, $\left(  C,\left(  C_{\leq n}\right)  _{n\geq0}\right)  $ is a filtered
$k$-coalgebra.\footnote{Of course, the notation $C_{\leq n}$ has to be
understood here as according to Convention~\ref{Convention203}; that is,
$C_{\leq n}$ is defined by $C_{\leq n}=\bigoplus_{\ell=0}^{n}C_{\ell}$.}
\end{proposition}

\begin{convention}
\label{Convention207} Let $k$ be a field. Let $C$ be a graded $k$-coalgebra.
Then, whenever we speak of ``the filtered $k$-coalgebra $C$'', we are going to
mean the filtered $k$-coalgebra $\left(  C,\left(  C_{\leq n}\right)
_{n\geq0}\right)  $ defined in Proposition~\ref{Proposition206}.
\end{convention}

\begin{proof}
[Proof of Proposition~\ref{Proposition206}.]\textbf{a)} By
Proposition~\ref{Proposition202} (applied to $V=C$), we know that $\left(
C,\left(  C_{\leq n}\right)  _{n\geq0}\right)  $ is a filtered $k$-vector space.

Since $C$ is a graded $k$-coalgebra, we have%
\begin{equation}
\Delta_{C}\left(  C_{\ell}\right)  \subseteq\bigoplus_{\substack{\left(
i,j\right)  \in\mathbb{N}^{\times2};\\i+j=\ell}}C_{i}\otimes C_{j}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{N}. \label{206.pf.1}%
\end{equation}
Here, the sum $\bigoplus_{\substack{\left(  i,j\right)  \in\mathbb{N}%
^{\times2};\\i+j=\ell}}C_{i}\otimes C_{j}$ is considered as a partial sum of
the infinite direct sum $\bigoplus_{\left(  i,j\right)  \in\mathbb{N}%
^{\times2}}C_{i}\otimes C_{j}=\underbrace{\left(  \bigoplus_{i\in\mathbb{N}%
}C_{i}\right)  }_{=C}\otimes\underbrace{\left(  \bigoplus_{j\in\mathbb{N}%
}C_{j}\right)  }_{=C}=C\otimes C$.

We can rewrite the right hand side of (\ref{206.pf.1}). In fact, for every
$\ell\in\mathbb{N}$, the map $\left\{  0,1,\ldots,\ell\right\}  \rightarrow
\left\{  \left(  i,j\right)  \in\mathbb{N}^{\times2}\ \mid\ i+j=\ell\right\}
$ which sends every $i\in\left\{  0,1,\ldots,\ell\right\}  $ to $\left(
i,\ell-i\right)  $ is a bijection (according to elementary combinatorics).
Hence, for every $\ell\in\mathbb{N}$, we have%
\begin{align*}
\bigoplus_{\substack{\left(  i,j\right)  \in\mathbb{N}^{\times2};\\i+j=\ell
}}C_{i}\otimes C_{j}  &  =\underbrace{\bigoplus_{i\in\left\{  0,1,\ldots
,\ell\right\}  }}_{=\bigoplus_{i=0}^{\ell}}C_{i}\otimes C_{\ell-i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\left(  i,\ell-i\right)  \text{ for }\left(
i,j\right)  \text{ in the direct sum,}\\
\text{because the map }\left\{  0,1,\ldots,\ell\right\}  \rightarrow\left\{
\left(  i,j\right)  \in\mathbb{N}^{\times2}\ \mid\ i+j=\ell\right\} \\
\text{which sends every }i\in\left\{  0,1,\ldots,\ell\right\}  \text{ to
}\left(  i,\ell-i\right)  \text{ is a bijection}%
\end{array}
\right) \\
&  =\bigoplus_{i=0}^{\ell}C_{i}\otimes C_{\ell-i}.
\end{align*}
Now, (\ref{206.pf.1}) becomes%
\begin{equation}
\Delta_{C}\left(  C_{\ell}\right)  \subseteq\bigoplus_{\substack{\left(
i,j\right)  \in\mathbb{N}^{\times2};\\i+j=\ell}}C_{i}\otimes C_{j}%
=\bigoplus_{i=0}^{\ell}C_{i}\otimes C_{\ell-i}=\sum_{i=0}^{\ell}C_{i}\otimes
C_{\ell-i} \label{206.pf.2}%
\end{equation}
(since direct sums are sums) for every $\ell\in\mathbb{N}$.

\textbf{b)} We have
\begin{equation}
C_{u}\subseteq C_{\leq v}\ \text{for any }u\in\mathbb{N}\text{ and }%
v\in\mathbb{N}\text{ satisfying }u\leq v. \label{206.pf.0}%
\end{equation}


\textit{Proof.} Let $u\in\mathbb{N}$ and $v\in\mathbb{N}$ satisfy $u\leq v$.
Then, the definition of $C_{\leq v}$ says that $C_{\leq v}=\bigoplus_{\ell
=0}^{v}C_{\ell}$. Since $C_{u}$ is a summand in the direct sum $\bigoplus
_{\ell=0}^{v}C_{\ell}$ (because $u\in\mathbb{N}$ and $u\leq v$), this yields
that $C_{u}$ is a direct addend of $\bigoplus_{\ell=0}^{v}C_{\ell}=C_{\leq v}%
$. Thus, $C_{u}\subseteq C_{\leq v}$. This proves \textbf{b)}.

\textbf{c)} For every $n\in\mathbb{N}$, we have%
\[
\Delta_{C}\left(  C_{\leq n}\right)  \subseteq\sum_{u=0}^{n}C_{\leq u}\otimes
C_{\leq n-u}.
\]


\textit{Proof.} Let $n\in\mathbb{N}$. Then, $C_{\leq n}=\bigoplus_{\ell=0}%
^{n}C_{\ell}=\sum_{\ell=0}^{n}C_{\ell}$ (since direct sums are sums), so that%
\begin{align*}
\Delta_{C}\left(  C_{\leq n}\right)   &  =\Delta_{C}\left(  \sum_{\ell=0}%
^{n}C_{\ell}\right)  =\sum_{\ell=0}^{n}\underbrace{\Delta_{C}\left(  C_{\ell
}\right)  }_{\substack{\subseteq\sum_{i=0}^{\ell}C_{i}\otimes C_{\ell
-i}\\\text{(by (\ref{206.pf.2}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\Delta_{C}\text{ is }k\text{-linear}\right) \\
&  =\sum_{\ell=0}^{n}\sum_{i=0}^{\ell}\underbrace{C_{i}}_{\substack{\subseteq
C_{\leq i}\\\text{(by (\ref{206.pf.0}) (applied to}\\u=i\text{ and
}v=i\text{), since }i\leq i\text{)}}}\otimes\underbrace{C_{\ell-i}%
}_{\substack{\subseteq C_{\leq n-i}\\\text{(by (\ref{206.pf.0}) (applied to
}u=\ell-i\text{ and }v=n-i\text{),}\\\text{ since }\ell-i\leq n-i\text{
(because }\ell\leq n\text{))}}}\\
&  \subseteq\sum_{\ell=0}^{n}\underbrace{\sum_{i=0}^{\ell}C_{\leq i}\otimes
C_{\leq n-i}}_{\substack{\subseteq\sum_{i=0}^{n}C_{\leq i}\otimes C_{\leq
n-i}\\\text{(because }\ell\leq n\text{, and thus the sum }\sum_{i=0}^{\ell
}C_{\leq i}\otimes C_{\leq n-i}\\\text{is a partial sum of the sum }\sum
_{i=0}^{n}C_{\leq i}\otimes C_{\leq n-i}\text{)}}}\subseteq\sum_{\ell=0}%
^{n}\sum_{i=0}^{n}C_{\leq i}\otimes C_{\leq n-i}\\
&  \subseteq\sum_{i=0}^{n}C_{\leq i}\otimes C_{\leq n-i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{i=0}^{n}C_{\leq i}\otimes
C_{\leq n-i}\text{ is a }k\text{-vector space}\right) \\
&  =\sum_{u=0}^{n}C_{\leq u}\otimes C_{\leq n-u}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we renamed }i\text{ as }u\text{ in the sum}\right)  .
\end{align*}
This proves \textbf{c)}.

\textbf{d)} From part \textbf{c)}, we immediately see that $\left(  C,\left(
C_{\leq n}\right)  _{n\geq0}\right)  $ is a filtered $k$-coalgebra. This
proves Proposition~\ref{Proposition206}.
\end{proof}

\begin{proposition}
\label{Proposition210} Let $k$ be a field. Let $H$ be a graded $k$-bialgebra.
Then, $\left(  H,\left(  H_{\leq n}\right)  _{n\geq0}\right)  $ is a filtered
$k$-bialgebra.\footnote{Of course, the notation $H_{\leq n}$ has to be
understood here as according to Convention~\ref{Convention203}; that is,
$H_{\leq n}$ is defined by $H_{\leq n}=\bigoplus_{\ell=0}^{n}H_{\ell}$.}
\end{proposition}

\begin{convention}
\label{Convention211} Let $k$ be a field. Let $H$ be a graded $k$-bialgebra.
Then, whenever we speak of ``the filtered $k$-bialgebra $H$'', we are going to
mean the filtered $k$-bialgebra $\left(  H,\left(  H_{\leq n}\right)
_{n\geq0}\right)  $ defined in Proposition~\ref{Proposition210}.
\end{convention}

\begin{proof}
[Proof of Proposition~\ref{Proposition210}.]Since $\left(  H,\left(  H_{\leq
n}\right)  _{n\geq0}\right)  $ is a filtered $k$-algebra (by
Proposition~\ref{Proposition204}, applied to $A=H$) and a filtered
$k$-coalgebra (by Proposition~\ref{Proposition206}, applied to $C=H$), we see
that $\left(  H,\left(  H_{\leq n}\right)  _{n\geq0}\right)  $ is a filtered
$k$-bialgebra. This proves Proposition~\ref{Proposition210}.
\end{proof}

Next we define the notion of a connected graded $k$-coalgebra:

\begin{definition}
\label{Definition212} Let $k$ be a field. Let $C$ be a graded $k$-coalgebra.
We say that the graded $k$-coalgebra $C$ is \textit{connected} if and only if
the map $\varepsilon_{C}\mid_{C_{0}}:C_{0}\rightarrow k$ is a $k$-vector space isomorphism.
\end{definition}

Note that such a definition of a connected graded $k$-coalgebra might bring us
into trouble: In fact, if $C$ is a filtered $k$-coalgebra, then when we just
say that ``$C$ is connected'' it is not immediately clear whether we mean that
the \textit{graded} $k$-coalgebra $C$ is connected, or whether we mean that
the \textit{filtered} $k$-coalgebra $C$ (defined in
Convention~\ref{Convention207}) is connected. Fortunately, these two meanings
are the same, because we have the following fact:

\begin{remark}
\label{Remark213} Let $k$ be a field. Let $C$ be a graded $k$-coalgebra. The
graded $k$-coalgebra $C$ is connected if and only if the filtered
$k$-coalgebra $C$ (defined in Convention~\ref{Convention207}) is connected.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark213}.]Consider the filtered $k$-coalgebra $C$
(defined in Convention~\ref{Convention207}). Then, by the definition of
$C_{\leq0}$, we have $C_{\leq0}=\bigoplus_{\ell=0}^{0}C_{\ell}=C_{0}$. Now, by
Definition~\ref{Definition212}, we have the following equivalence of
assertions:%
\begin{align*}
&  \ \left(  \text{the graded }k\text{-coalgebra }C\text{ is connected}\right)
\\
&  \Longleftrightarrow\ \left(  \text{the map }\varepsilon_{C}\mid_{C_{0}%
}:C_{0}\rightarrow k\text{ is a }k\text{-vector space isomorphism}\right) \\
&  \Longleftrightarrow\ \left(  \text{the map }\varepsilon_{C}\mid_{C_{\leq0}%
}:C_{\leq0}\rightarrow k\text{ is a }k\text{-vector space isomorphism}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }C_{0}=C_{\leq0}\right) \\
&  \Longleftrightarrow\ \left(  \text{the filtered }k\text{-coalgebra }C\text{
is connected}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Definition~\ref{Definition7a}}\right)  .
\end{align*}
This proves Remark~\ref{Remark213}.
\end{proof}

Remark~\ref{Remark213} was a triviality to prove, but it shows us an important
thing: It shows us that all properties of connected filtered bialgebras that
we showed above automatically yield properties of connected graded bialgebras.
Since connected graded bialgebras have more structure than connected filtered
bialgebras, we can expect connected graded bialgebras to also have some
additional properties that don't hold (or don't even make sense) for connected
filtered bialgebras. This Section \S \ref{S13} is devoted to some such properties.

Here is a simple consequence of Definition~\ref{Definition212}:

\begin{remark}
\label{Remark214} Let $k$ be a field. Let $C$ be a graded $k$-coalgebra. Let
$\lambda\in C_{0}$ be such that $C_{0}=k\cdot\lambda$ and $\varepsilon
_{C}\left(  \lambda\right)  =1$. Then, the graded $k$-coalgebra $C$ is connected.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark214}.]Since the map $\varepsilon_{C}$ is
$k$-linear, we have $\varepsilon_{C}\left(  k\cdot\lambda\right)
=k\cdot\underbrace{\varepsilon_{C}\left(  \lambda\right)  }_{=1}=k\cdot1=k$.
Now,%
\[
\left(  \varepsilon_{C}\mid_{C_{0}}\right)  \left(  C_{0}\right)
=\varepsilon_{C}\left(  \underbrace{C_{0}}_{=k\cdot\lambda}\right)
=\varepsilon_{C}\left(  k\cdot\lambda\right)  =k.
\]
In other words, the map $\varepsilon_{C}\mid_{C_{0}}:C_{0}\rightarrow k$ is surjective.

Let $x\in\operatorname*{Ker}\left(  \varepsilon_{C}\mid_{C_{0}}\right)  $.
Then, $x\in C_{0}$ satisfies $\left(  \varepsilon_{C}\mid_{C_{0}}\right)
\left(  x\right)  =0$. Since $x\in C_{0}=k\cdot\lambda$, there exists a
$\tau\in k$ such that $x=\tau\lambda$. Consider this $\tau$. Since%
\begin{align*}
0  &  =\left(  \varepsilon_{C}\mid_{C_{0}}\right)  \left(  x\right)
=\varepsilon_{C}\left(  \underbrace{x}_{=\tau\lambda}\right)  =\varepsilon
_{C}\left(  \tau\lambda\right)  =\tau\underbrace{\varepsilon_{C}\left(
\lambda\right)  }_{=1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon
_{C}\text{ is }k\text{-linear}\right) \\
&  =\tau,
\end{align*}
we have $x=\underbrace{\tau}_{=0}\lambda=0$. Now, forget that we fixed $x$. We
thus have shown that every $x\in\operatorname*{Ker}\left(  \varepsilon_{C}%
\mid_{C_{0}}\right)  $ satisfies $x=0$. In other words, $\operatorname*{Ker}%
\left(  \varepsilon_{C}\mid_{C_{0}}\right)  =0$. Hence, the map $\varepsilon
_{C}\mid_{C_{0}}$ is injective. Combined with the fact that the map
$\varepsilon_{C}\mid_{C_{0}}$ is surjective, this yields that the map
$\varepsilon_{C}\mid_{C_{0}}$ is bijective.

So $\varepsilon_{C}\mid_{C_{0}}:C_{0}\rightarrow k$ is a bijective $k$-linear
map. Hence, $\varepsilon_{C}\mid_{C_{0}}:C_{0}\rightarrow k$ is a $k$-vector
space isomorphism.

But Definition~\ref{Definition212} yields that the graded $k$-coalgebra $C$ is
connected if and only if the map $\varepsilon_{C}\mid_{C_{0}}:C_{0}\rightarrow
k$ is a $k$-vector space isomorphism. Thus, the graded $k$-coalgebra $C$ is
connected (because the map $\varepsilon_{C}\mid_{C_{0}}:C_{0}\rightarrow k$ is
a $k$-vector space isomorphism). Remark~\ref{Remark214} is thus proven.
\end{proof}

An analogue of Remark~\ref{Remark214} holds for filtered $k$-coalgebras:

\begin{remark}
\label{Remark215} Let $k$ be a field. Let $C$ be a filtered $k$-coalgebra. Let
$\lambda\in C_{\leq0}$ be such that $C_{\leq0}=k\cdot\lambda$ and
$\varepsilon_{C}\left(  \lambda\right)  =1$. Then, the filtered $k$-coalgebra
$C$ is connected.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark215}.]A proof of Remark~\ref{Remark215} can be
obtained by making the following replacements to the proof of
Remark~\ref{Remark214}:

\begin{itemize}
\item Any occurrence of ``$C_{0}$'' should be replaced by ``$C_{\leq0}$''.

\item Any occurrence of ``graded'' should be replaced by ``filtered''.

\item The reference to Definition~\ref{Definition212} should be replaced by a
reference to Definition~\ref{Definition7a}.
\end{itemize}

Thus, Remark~\ref{Remark215} is proven.
\end{proof}

We next repeat some basics related to graded vector spaces:

\begin{definition}
\label{Definition222} Let $k$ be a field. Let $V$ and $W$ be graded $k$-vector
spaces. Let $f:V\rightarrow W$ be a $k$-linear map. Then, the map $f$ is said
to be \textit{graded} if every $n\in\mathbb{N}$ satisfies $f\left(
V_{n}\right)  \subseteq W_{n}$.
\end{definition}

\begin{remark}
\label{Remark220} Let $k$ be a field. Let $V$ be a graded $k$-vector space.
Then, $V=\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$. For every $n\in\mathbb{N}$,
let $\pi_{n}:\bigoplus_{\ell\in\mathbb{N}}V_{\ell}\rightarrow V_{n}$ be the
canonical projection from the direct sum to its $n$-th addend, and let
$\iota_{n}:V_{n}\rightarrow\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ be the
canonical injection of the $n$-th addend into the direct sum. Since
$\bigoplus_{\ell\in\mathbb{N}}V_{\ell}=V$, the map $\pi_{n}:\bigoplus_{\ell
\in\mathbb{N}}V_{\ell}\rightarrow V_{n}$ is actually a map $V\rightarrow
V_{n}$, and the map $\iota_{n}:V_{n}\rightarrow\bigoplus_{\ell\in\mathbb{N}%
}V_{\ell}$ is actually a map $V_{n}\rightarrow V$. Thus, the composition
$\iota_{n}\circ\pi_{n}$ is a map $V\rightarrow V$. This map $\iota_{n}\circ
\pi_{n}$ maps every element $v\in V$ to the $n$-th graded component of $v$,
seen as an element of $V$.
\end{remark}

\begin{definition}
\label{Definition221} Let $k$ be a field. Let $V$ be a graded $k$-vector
space. For every $n\in\mathbb{N}$, we denote the $k$-linear map $\iota
_{n}\circ\pi_{n}$ defined in Remark~\ref{Remark220} as the $n$\textit{-th
grade identity} of $V$, and we denote it by $p_{n,V}$. As we mentioned in
Remark~\ref{Remark220}, this map $p_{n,V}$ (which we denoted by $\iota
_{n}\circ\pi_{n}$ in Remark~\ref{Remark220}) maps every element $v\in V$ to
the $n$-th graded component of $v$, seen as an element of $V$.

Let us recall some well-known properties of these maps $p_{n,V}$:

For every $n\in\mathbb{N}$, the map $p_{n,V}$ is idempotent, i. e., it
satisfies%
\begin{equation}
p_{n,V}\circ p_{n,V}=p_{n,V} \label{221.idp}%
\end{equation}
\footnote{\textit{Proof of (\ref{221.idp}).} Let $n\in\mathbb{N}$. Let us use
the notations of Remark~\ref{Remark220}. Then, $p_{n,V}=\iota_{n}\circ\pi_{n}%
$. Since $\pi_{n}:\bigoplus_{\ell\in\mathbb{N}}V_{\ell}\rightarrow V_{n}$ the
canonical projection from the direct sum to its $n$-th addend, whereas
$\iota_{n}:V_{n}\rightarrow\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ is the
canonical injection of the $n$-th addend into the direct sum, it is clear that
$\pi_{n}\circ\iota_{n}=\operatorname*{id}$. Now, $p_{n,V}=\iota_{n}\circ
\pi_{n}$ yields%
\[
p_{n,V}\circ p_{n,V}=\iota_{n}\circ\underbrace{\pi_{n}\circ\iota_{n}%
}_{=\operatorname*{id}}\circ\pi_{n}=\iota_{n}\circ\operatorname*{id}\circ
\pi_{n}=\iota_{n}\circ\pi_{n}=p_{n,V},
\]
thus proving (\ref{221.idp}).}. It is known that%
\begin{equation}
p_{n,V}\mid_{V_{n}}=\operatorname{id}_{V}\mid_{V_{n}}%
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N} \label{221.id}%
\end{equation}
\footnote{\textit{Proof of (\ref{221.id}).} Let $n\in\mathbb{N}$. Let us use
the notations of Remark~\ref{Remark220}. Then, $p_{n,V}=\iota_{n}\circ\pi_{n}%
$. Since $\pi_{n}:\bigoplus_{\ell\in\mathbb{N}}V_{\ell}\rightarrow V_{n}$ the
canonical projection from the direct sum to its $n$-th addend, it is clear
that $\pi_{n}\mid_{V_{n}}=\operatorname*{id}$. Now, $p_{n,V}=\iota_{n}\circ
\pi_{n}$ yields%
\[
p_{n,V}\mid_{V_{n}}=\left(  \iota_{n}\circ\pi_{n}\right)  \mid_{V_{n}}%
=\iota_{n}\circ\underbrace{\left(  \pi_{n}\mid_{V_{n}}\right)  }%
_{=\operatorname*{id}}=\iota_{n}\circ\operatorname*{id}=\iota_{n}%
=\operatorname{id}_{V}\mid_{V_{n}}%
\]
(since $\iota_{n}:V_{n}\rightarrow\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ is
the canonical injection of the $n$-th addend into the direct sum). This proves
(\ref{221.id}).}, and that%
\begin{equation}
p_{n,V}\mid_{V_{m}}=0\ \ \ \ \ \ \ \ \ \ \text{for any }n\in\mathbb{N}\text{
and }m\in\mathbb{N}\text{ satisfying }n\neq m \label{221.0}%
\end{equation}
\footnote{\textit{Proof of (\ref{221.0}).} Let $n\in\mathbb{N}$ and
$m\in\mathbb{N}$ satisfy $n\neq m$. Let us use the notations of
Remark~\ref{Remark220}. Then, $p_{n,V}=\iota_{n}\circ\pi_{n}$. Since $\pi
_{n}:\bigoplus_{\ell\in\mathbb{N}}V_{\ell}\rightarrow V_{n}$ the canonical
projection from the direct sum to its $n$-th addend, it is clear that $\pi
_{n}\mid_{V_{\ell}}=0$ for every $\ell\in\mathbb{N}$ such that $\ell\neq n$.
Applied to $\ell=m$, this yields that $\pi_{n}\mid_{V_{m}}=0$. Now,
$p_{n,V}=\iota_{n}\circ\pi_{n}$ yields%
\[
p_{n,V}\mid_{V_{m}}=\left(  \iota_{n}\circ\pi_{n}\right)  \mid_{V_{m}}%
=\iota_{n}\circ\underbrace{\left(  \pi_{n}\mid_{V_{m}}\right)  }_{=0}%
=\iota_{n}\circ0=0.
\]
This proves (\ref{221.0}).}. Also,%
\begin{equation}
p_{n,V}\left(  V\right)  =V_{n}\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\mathbb{N} \label{221.im}%
\end{equation}
\footnote{\textit{Proof of (\ref{221.im}).} Let $n\in\mathbb{N}$. Let us use
the notations of Remark~\ref{Remark220}. Then, $p_{n,V}=\iota_{n}\circ\pi_{n}%
$. Since $\pi_{n}$ is the canonical projection $\bigoplus_{\ell\in\mathbb{N}%
}V_{\ell}\rightarrow V_{n}$, it is clear that $\pi_{n}\left(  V\right)
=V_{n}$. Since $\iota_{n}$ is the canonical injection $V_{n}\rightarrow
\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$, we have $\iota_{n}\left(
V_{n}\right)  =V_{n}$. Thus,
\[
\underbrace{p_{n,V}}_{=\iota_{n}\circ\pi_{n}}\left(  V\right)  =\left(
\iota_{n}\circ\pi_{n}\right)  \left(  V\right)  =\iota_{n}\left(
\underbrace{\pi_{n}\left(  V\right)  }_{=V_{n}}\right)  =\iota_{n}\left(
V_{n}\right)  =V_{n},
\]
so that (\ref{221.im}) is proven.}. Also,%
\begin{equation}
\text{the map }p_{n,V}\text{ is graded for every }n\in\mathbb{N}
\label{221.gr}%
\end{equation}
\footnote{\textit{Proof of (\ref{221.gr}).} Let $n\in\mathbb{N}$. Let
$m\in\mathbb{N}$. We are going to prove that $p_{n,V}\left(  V_{m}\right)
\subseteq V_{m}$.
\par
In fact, let us distinguish between two cases:
\par
\textit{Case 1:} We have $m=n$.
\par
\textit{Case 2:} We have $m\neq n$.
\par
First let us consider Case 1. In this case, $m=n$, so that $V_{m}=V_{n}$. Now,
$V_{m}\subseteq V$, so that $p_{n,V}\left(  V_{m}\right)  \subseteq
p_{n,V}\left(  V\right)  =V_{n}$ (by (\ref{221.im})), so that $p_{n,V}\left(
V_{m}\right)  \subseteq V_{n}=V_{m}$.
\par
We thus have proven $p_{n,V}\left(  V_{m}\right)  \subseteq V_{m}$ in Case 1.
\par
Now let us consider Case 2. In this case, $m\neq n$, so that $p_{n,V}%
\mid_{V_{m}}=0$ (by (\ref{221.0})). But now, $p_{n,V}\left(  V_{m}\right)
=\underbrace{\left(  p_{n,V}\mid_{V_{m}}\right)  }_{=0}\left(  V_{m}\right)
=0\left(  V_{m}\right)  =0\subseteq V_{m}$.
\par
We thus have proven $p_{n,V}\left(  V_{m}\right)  \subseteq V_{m}$ in Case 2.
\par
Now, $p_{n,V}\left(  V_{m}\right)  \subseteq V_{m}$ is proven in both cases 1
and 2. Since these cases are the only possible cases, this means that
$p_{n,V}\left(  V_{m}\right)  \subseteq V_{m}$ always holds.
\par
Now forget that we fixed $m$. We have proved that $p_{n,V}\left(
V_{m}\right)  \subseteq V_{m}$ for every $m\in\mathbb{N}$. In other words,
$p_{n,V}$ is graded. This proves (\ref{221.gr}).}.

It is also known that every $v\in V$ satisfies%
\begin{equation}
v=\sum_{\ell\in\mathbb{N}}p_{\ell,V}\left(  v\right)  \label{221.vsum}%
\end{equation}
(where the sum $\sum_{\ell\in\mathbb{N}}p_{\ell,V}\left(  v\right)  $ is
well-defined since it has only finitely many nonzero
terms)\footnote{\textit{Proof of (\ref{221.vsum}).} Let $v\in V$.
\par
Since $v\in V=\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$, we can write $v$ in the
form $\sum_{\ell\in\mathbb{N}}v_{\ell}$, where $\left(  v_{\ell}\in V_{\ell
}\text{ for every }\ell\in\mathbb{N}\right)  $ and $\left(  v_{\ell}=0\text{
for all but finitely many }\ell\in\mathbb{N}\right)  $. So let us write $v$ in
this form.
\par
For every $n\in\mathbb{N}$, we have%
\begin{align*}
p_{n,V}\left(  v\right)   &  =p_{n,V}\left(  \sum_{\ell\in\mathbb{N}}v_{\ell
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }v=\sum_{\ell\in\mathbb{N}%
}v_{\ell}\right) \\
&  =\sum_{\ell\in\mathbb{N}}\underbrace{p_{n,V}\left(  v_{\ell}\right)
}_{\substack{=\left(  p_{n,V}\mid_{V_{\ell}}\right)  \left(  v_{\ell}\right)
\\\text{(since }v_{\ell}\in V_{\ell}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }p_{n,V}\text{ is }k\text{-linear}\right) \\
&  =\sum_{\ell\in\mathbb{N}}\left(  p_{n,V}\mid_{V_{\ell}}\right)  \left(
v_{\ell}\right)  =\underbrace{\sum_{\substack{\ell\in\mathbb{N};\\n=\ell
}}\left(  p_{n,V}\mid_{V_{\ell}}\right)  \left(  v_{\ell}\right)  }_{=\left(
p_{n,V}\mid_{V_{n}}\right)  \left(  v_{n}\right)  }+\sum_{\substack{\ell
\in\mathbb{N};\\n\neq\ell}}\left(  p_{n,V}\mid_{V_{\ell}}\right)  \left(
v_{\ell}\right) \\
&  =\underbrace{\left(  p_{n,V}\mid_{V_{n}}\right)  }%
_{\substack{=\operatorname{id}_{V}\mid_{V_{n}}\\\text{(by (\ref{221.id}))}%
}}\left(  v_{n}\right)  +\sum_{\substack{\ell\in\mathbb{N};\\n\neq\ell
}}\underbrace{\left(  p_{n,V}\mid_{V_{\ell}}\right)  }_{\substack{=0\text{ (by
(\ref{221.0}) (applied to }m=\ell\text{),}\\\text{since }n\neq\ell\text{)}%
}}\left(  v_{\ell}\right) \\
&  =\underbrace{\left(  \operatorname{id}_{V}\mid_{V_{n}}\right)  \left(
v_{n}\right)  }_{=\operatorname{id}_{V}\left(  v_{n}\right)  =v_{n}%
}+\underbrace{\sum_{\substack{\ell\in\mathbb{N};\\n\neq\ell}}0\left(  v_{\ell
}\right)  }_{=0}=v_{n}.
\end{align*}
Thus,%
\begin{align*}
\sum_{n\in\mathbb{N}}p_{n,V}\left(  v\right)   &  =\sum_{n\in\mathbb{N}}%
v_{n}=\sum_{\ell\in\mathbb{N}}v_{\ell}\ \ \ \ \ \ \ \ \ \ \left(  \text{here,
we renamed the index }n\text{ as }\ell\text{ in the sum}\right) \\
&  =v,
\end{align*}
so that%
\[
v=\sum_{n\in\mathbb{N}}p_{n,V}\left(  v\right)  =\sum_{\ell\in\mathbb{N}%
}p_{\ell,V}\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we
renamed the index }n\text{ as }\ell\text{ in the sum}\right)  .
\]
This proves (\ref{221.vsum}).}.
\end{definition}

\begin{proposition}
\label{Proposition223} Let $k$ be a field. Let $V$ and $W$ be graded
$k$-vector spaces. Let $f:V\rightarrow W$ be a graded $k$-linear map. Then,
$f\circ p_{n,V}=p_{n,W}\circ f=p_{n,W}\circ f\circ p_{n,V}$ (where $p_{n,V}$
and $p_{n,W}$ are the $n$-th grade identities of $V$ and $W$, as defined in
Definition~\ref{Definition221}) for every $n\in\mathbb{N}$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition223}.]Let $n\in\mathbb{N}$.

\textbf{a)} Every $m\in\mathbb{N}$ satisfies $\left(  f\circ p_{n,V}%
-p_{n,W}\circ f\right)  \left(  V_{m}\right)  =0$.

\textit{Proof.} Let $m\in\mathbb{N}$. We distinguish between two cases:

\textit{Case 1:} We have $m=n$.

\textit{Case 2:} We have $m\neq n$.

First, let us consider Case 1. In this case, $m=n$. Thus,%
\begin{equation}
\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(  V_{m}\right)  =\left(
f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(  V_{n}\right)  . \label{223.pf.1}%
\end{equation}


Now, let $x\in V_{n}$ be arbitrary. Then, $p_{n,V}\left(  x\right)
=\underbrace{\left(  p_{n,V}\mid_{V_{n}}\right)  }%
_{\substack{=\operatorname{id}_{V}\mid_{V_{n}}\\\text{(by (\ref{221.id}))}%
}}\left(  x\right)  =\left(  \operatorname{id}_{V}\mid_{V_{n}}\right)  \left(
x\right)  =\operatorname{id}_{V}\left(  x\right)  =x$. On the other hand,
since $x\in V_{n}$, we have $f\left(  x\right)  \in f\left(  V_{n}\right)
\subseteq W_{n}$ (since $f$ is graded), so that $p_{n,W}\left(  f\left(
x\right)  \right)  =\underbrace{\left(  p_{n,W}\mid_{W_{n}}\right)
}_{\substack{=\operatorname{id}_{W}\mid_{W_{n}}\\\text{(by (\ref{221.id}),
applied}\\\text{to }W\text{ instead of }V\text{)}}}\left(  f\left(  x\right)
\right)  =\left(  \operatorname{id}_{W}\mid_{W_{n}}\right)  \left(  f\left(
x\right)  \right)  =\operatorname{id}_{W}\left(  f\left(  x\right)  \right)
=f\left(  x\right)  $. But clearly,%
\begin{align*}
\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(  x\right)   &  =\left(
f\circ p_{n,V}\right)  \left(  x\right)  -\left(  p_{n,W}\circ f\right)
\left(  x\right) \\
&  =f\left(  \underbrace{p_{n,V}\left(  x\right)  }_{=x}\right)
-\underbrace{p_{n,W}\left(  f\left(  x\right)  \right)  }_{=f\left(  x\right)
}=f\left(  x\right)  -f\left(  x\right)  =0.
\end{align*}


Forget that we fixed $x$. We have thus proven that $\left(  f\circ
p_{n,V}-p_{n,W}\circ f\right)  \left(  x\right)  =0$ for every $x\in V_{n}$.
In other words, $\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(
V_{n}\right)  =0$. Combined with (\ref{223.pf.1}), this becomes $\left(
f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(  V_{m}\right)  =0$.

We thus have proven that $\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)
\left(  V_{m}\right)  =0$ in Case 1.

Now let us treat Case 2. In this case, $m\neq n$. Now, $f\left(  V_{m}\right)
\subseteq W_{m}$ (since $f$ is graded). We have%
\begin{align*}
\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(  V_{m}\right)   &
\subseteq\underbrace{\left(  f\circ p_{n,V}\right)  \left(  V_{m}\right)
}_{=f\left(  p_{n,V}\left(  V_{m}\right)  \right)  }-\underbrace{\left(
p_{n,W}\circ f\right)  \left(  V_{m}\right)  }_{=p_{n,W}\left(  f\left(
V_{m}\right)  \right)  }\\
&  =f\left(  p_{n,V}\left(  V_{m}\right)  \right)  -p_{n,W}\left(
\underbrace{f\left(  V_{m}\right)  }_{\subseteq W_{m}}\right)  \subseteq
f\left(  \underbrace{p_{n,V}\left(  V_{m}\right)  }_{\substack{=0\\\text{(by
(\ref{221.0}))}}}\right)  -\underbrace{p_{n,W}\left(  W_{m}\right)
}_{\substack{=0\\\text{(by (\ref{221.0}), applied}\\\text{to }W\text{ instead
of }V\text{)}}}\\
&  =\underbrace{f\left(  0\right)  }_{=0}-0=0-0=0.
\end{align*}
Hence, $\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(  V_{m}\right)
=0$. This proves $\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(
V_{m}\right)  =0$ in Case 2.

Hence, we have proven $\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)  \left(
V_{m}\right)  =0$ in each of the cases 1 and 2. Since these two cases are all
cases that can occur, this means that we have proven $\left(  f\circ
p_{n,V}-p_{n,W}\circ f\right)  \left(  V_{m}\right)  =0$ in every case. This
completes the proof of \textbf{a)}.

\textbf{b)} We have $f\circ p_{n,V}=p_{n,W}\circ f$.

\textit{Proof.} Every $m\in\mathbb{N}$ satisfies $\left(  f\circ
p_{n,V}-p_{n,W}\circ f\right)  \left(  V_{m}\right)  =0$ (according to part
\textbf{a)}). In other words, every $m\in\mathbb{N}$ satisfies $V_{m}%
\subseteq\operatorname*{Ker}\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)  $.
Thus,%
\begin{align*}
\sum_{m\in\mathbb{N}}V_{m}  &  \subseteq\sum_{m\in\mathbb{N}}%
\operatorname*{Ker}\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)
\subseteq\operatorname*{Ker}\left(  f\circ p_{n,V}-p_{n,W}\circ f\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\operatorname*{Ker}\left(  f\circ p_{n,V}-p_{n,W}\circ f\right)
\text{ is a }k\text{-vector subspace of }V\\
\text{(because }f\circ p_{n,V}-p_{n,W}\circ f\text{ is a }k\text{-linear map)}%
\end{array}
\right)  .
\end{align*}
But since $V$ is graded, we have $V=\bigoplus_{m\in\mathbb{N}}V_{m}=\sum
_{m\in\mathbb{N}}V_{m}$ (since direct sums are sums). Thus,%
\[
V=\sum_{m\in\mathbb{N}}V_{m}\subseteq\operatorname*{Ker}\left(  f\circ
p_{n,V}-p_{n,W}\circ f\right)  ,
\]
so that $f\circ p_{n,V}-p_{n,W}\circ f=0$. In other words, $f\circ
p_{n,V}=p_{n,W}\circ f$. This proves part \textbf{b)}.

\textbf{c)} By part \textbf{b)}, we know that $f\circ p_{n,V}=p_{n,W}\circ f$.
Thus, $\underbrace{p_{n,W}\circ f}_{=f\circ p_{n,V}}\circ p_{n,V}%
=f\circ\underbrace{p_{n,V}\circ p_{n,V}}_{\substack{=p_{n,V}\\\text{(by
(\ref{221.idp}))}}}=f\circ p_{n,V}$. Combined with $f\circ p_{n,V}%
=p_{n,W}\circ f$, this yields $f\circ p_{n,V}=p_{n,W}\circ f=p_{n,W}\circ
f\circ p_{n,V}$. This proves Proposition~\ref{Proposition223}.
\end{proof}

We now prove that gradedness is compatible with convolution:

\begin{proposition}
\label{Proposition224} Let $k$ be a field. Let $C$ be a graded $k$-coalgebra.
Let $A$ be a graded $k$-algebra.

\textbf{(a)} If $f\in\mathcal{L}\left(  C,A\right)  $ and $g\in\mathcal{L}%
\left(  C,A\right)  $ are two graded maps, then $f\ast g\in\mathcal{L}\left(
C,A\right)  $ is graded as well.

\textbf{(b)} The map $e_{C,A}:C\rightarrow A$ is graded.

\textbf{(c)} If $f\in\mathcal{L}\left(  C,A\right)  $ is a graded map and
$n\in\mathbb{N}$, then $f^{\ast n}\in\mathcal{L}\left(  C,A\right)  $ is
graded as well.

\textbf{(d)} Assume that $C$ is a connected filtered $k$-coalgebra, and that
the field $k$ has characteristic $0$. If $f\in\mathfrak{g}\left(  C,A\right)
$ is a graded map, then $e^{\ast f}$ is a graded map.

\textbf{(e)} Assume that $C$ is a connected filtered $k$-coalgebra, and that
the field $k$ has characteristic $0$. If $F\in G\left(  C,A\right)  $ is a
graded map, then $\operatorname*{Log}F$ is a graded map.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition224}.]\textbf{(a)} Let $f\in
\mathcal{L}\left(  C,A\right)  $ and $g\in\mathcal{L}\left(  C,A\right)  $ be
two graded maps.

By the definition of convolution, $f\ast g=\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$. Now, since $A$ is a graded $k$-algebra, its
multiplication map $\mu_{A}$ is graded (where the grading on $A\otimes A$ is
the usual one that is given to the tensor product of two graded $k$-vector
spaces). Since $C$ is a graded $k$-coalgebra, its comultiplication map
$\Delta_{C}$ is graded (where the grading on $C\otimes C$ is the usual one
that is given to the tensor product of two graded $k$-vector spaces). Also,
the map $f\otimes g$ is graded (since $f$ and $g$ are graded, and since the
tensor product of two graded maps is graded). Now recall that the composition
of graded maps is graded. Thus, the map $\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$ is graded (since $\mu_{A}$, $f\otimes g$ and
$\Delta_{C}$ are graded). Since $\mu_{A}\circ\left(  f\otimes g\right)
\circ\Delta_{C}=f\ast g$, this means that the map $f\ast g$ is graded. This
proves Proposition~\ref{Proposition224} \textbf{(a)}.

\textbf{(b)} Let us give $k$ the usual grading (the one where $k_{0}=k$ and
$k_{n}=0$ for all positive $n\in\mathbb{N}$). Since $A$ is a graded
$k$-algebra, its unity map $\eta_{A}:k\rightarrow A$ is graded. Since $C$ is a
graded $k$-coalgebra, its counity map $\varepsilon_{C}:C\rightarrow k$ is
graded. Now, the map $e_{C,A}=\eta_{A}\circ\varepsilon_{C}$ is graded (since
$\eta_{A}$ and $\varepsilon_{C}$ are graded, and since the composition of
graded maps is graded). This proves Proposition~\ref{Proposition224}
\textbf{(b)}.

\textbf{(c)} Let us prove Proposition~\ref{Proposition224} \textbf{(c)} by
induction over $n$:

\textit{Induction base:} The map $f^{\ast0}$ is graded for every
$f\in\mathcal{L}\left(  C,A\right)  $ (because for every $f\in\mathcal{L}%
\left(  C,A\right)  $, the map $f^{\ast0}$ equals $e_{C,A}$ and thus is graded
by Proposition~\ref{Proposition224} \textbf{(b)}). Thus,
Proposition~\ref{Proposition224} \textbf{(c)} holds for $n=0$. This completes
the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
Proposition~\ref{Proposition224} \textbf{(c)} holds for $n=N$. We must show
that Proposition~\ref{Proposition224} \textbf{(c)} also holds for $n=N+1$.

Let $f\in\mathcal{L}\left(  C,A\right)  $ be graded. Then, $f^{\ast N}$ is
graded as well (since we assumed that Proposition~\ref{Proposition224}
\textbf{(c)} holds for $n=N$). Now, Proposition~\ref{Proposition224}
\textbf{(a)} (applied to $g=f^{\ast N}$) yields that $f\ast f^{\ast N}$ is
graded. Since $f\ast f^{\ast N}=f^{\ast\left(  N+1\right)  }$, this means that
$f^{\ast\left(  N+1\right)  }$ is graded.

Thus we have shown that for every graded $f\in\mathcal{L}\left(  C,A\right)
$, the map $f^{\ast\left(  N+1\right)  }\in\mathcal{L}\left(  C,A\right)  $ is
graded as well. In other words, we have proven that
Proposition~\ref{Proposition224} \textbf{(c)} holds for $n=N+1$. This
completes the induction step.

Thus, the induction proof of Proposition~\ref{Proposition224} \textbf{(c)} is complete.

\textbf{(e)} Let $F\in G\left(  C,A\right)  $ be a graded map.

The map $e_{C,A}$ is graded (by Proposition~\ref{Proposition224} \textbf{(b)}).

By Definition~\ref{Definition12}, we have $\operatorname*{Log}%
F=\operatorname{Log}_{1}\left(  F-e_{C,A}\right)  $. Let $f=F-e_{C,A}$. Then,
$f=F-e_{C,A}$ is graded (because $F$ and $e_{C,A}$ are graded, and because the
difference of graded maps is graded). Hence, for every integer $i\geq1$, the
map $f^{\ast i}$ is graded (by Proposition~\ref{Proposition224} \textbf{(c)},
applied to $n=i$). Also, $f=F-e_{C,A}\in\mathfrak{g}\left(  C,A\right)  $
(since $F\in G\left(  C,A\right)  =e_{C,A}+\mathfrak{g}\left(  C,A\right)  $).
Thus, every $i\in\mathbb{N}$ and $n\in\mathbb{N}$ such that $i>n$ satisfy
$f^{\ast i}\left(  C_{\leq n}\right)  =0$ (by Remark~\ref{Remark9}, applied to
$H=C$).

Now let $n\in\mathbb{N}$. The definition of $C_{\leq n}$ says that $C_{\leq
n}=\bigoplus_{\ell=0}^{n}C_{\ell}=C_{n}\oplus\bigoplus_{\ell=0}^{n-1}C_{\ell
}\supseteq C_{n}$. Hence, $C_{n}\subseteq C_{\leq n}$.

Let $x\in C_{n}$. Then, every $i\in\mathbb{N}$ and $n\in\mathbb{N}$ such that
$i>n$ satisfy $f^{\ast i}\left(  x\right)  =0$ (since $f^{\ast i}\left(
\underbrace{x}_{\in C_{n}\subseteq C_{\leq n}}\right)  \in f^{\ast i}\left(
C_{\leq n}\right)  =0$). Thus, $\sum_{\substack{i\geq1;\\i>n}}\dfrac{\left(
-1\right)  ^{i-1}}{i}\underbrace{f^{\ast i}\left(  x\right)  }_{=0\text{
(since }i>n\text{)}}=\sum_{\substack{i\geq1;\\i>n}}\dfrac{\left(  -1\right)
^{i-1}}{i}0=0$.

For every integer $i\geq1$, we have $f^{\ast i}\left(  C_{n}\right)  \subseteq
A_{n}$ (since $f^{\ast i}$ is graded). Thus, for every integer $i\geq1$, we
have $f^{\ast i}\left(  x\right)  \in A_{n}$ (because $x\in C_{n}$ and thus
$f^{\ast i}\left(  x\right)  \in f^{\ast i}\left(  C_{n}\right)  \subseteq
A_{n}$). But%
\begin{align*}
\left(  \operatorname{Log}_{1}f\right)  \left(  x\right)   &  =\sum_{i\geq
1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{log})}\right) \\
&  =\sum_{\substack{i\geq1;\\i\leq n}}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{f^{\ast i}\left(  x\right)  }_{\in A_{n}}+\underbrace{\sum
_{\substack{i\geq1;\\i>n}}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  }_{=0}\in\sum_{\substack{i\geq1;\\i\leq n}}\dfrac{\left(
-1\right)  ^{i-1}}{i}A_{n}\subseteq A_{n}%
\end{align*}
(since $A_{n}$ is a $k$-vector space).

Now forget that we fixed $x$. We thus have proven that $\left(
\operatorname{Log}_{1}f\right)  \left(  x\right)  \in A_{n}$ for every $x\in
C_{n}$. In other words, $\left(  \operatorname{Log}_{1}f\right)  \left(
C_{n}\right)  \subseteq A_{n}$.

Now forget that we fixed $n$. We have thus proven that $\left(
\operatorname{Log}_{1}f\right)  \left(  C_{n}\right)  \subseteq A_{n}$ for
every $n\in\mathbb{N}$. In other words, the map $\operatorname{Log}_{1}f$ is graded.

The map $\operatorname*{Log}F=\operatorname{Log}_{1}\underbrace{\left(
F-e_{C,A}\right)  }_{=f}=\operatorname{Log}_{1}f$ is graded (as we just have
shown). This proves Proposition~\ref{Proposition224} \textbf{(e)}.

\textbf{(d)} The proof of Proposition~\ref{Proposition224} \textbf{(d)} is
very similar to that of Proposition~\ref{Proposition224} \textbf{(e)} (and
even easier since we do not have two maps $F$ and $f$ but only one map $f$).
We will leave it at that and let the reader fill in the details. (We won't
need Proposition~\ref{Proposition224} \textbf{(d)} anyway.)
\end{proof}

We now need a notation:

\begin{convention}
\label{Convention230} Let $k$ be a field, let $A$ be a $k$-algebra, and let
$S$ be a subset of $A$. Then, \textit{the }$k$\textit{-subalgebra of }%
$A$\textit{ generated by }$S$ can be defined in several ways. Here are four
ways to define it:

\begin{itemize}
\item It is the smallest $k$-subalgebra of $A$ which contains $S$ as a subset.
(``Smallest'' means that it is contained in every such $k$-subalgebra. Of
course, it is not immediately trivial that the smallest $k$-subalgebra of $A$
which contains $S$ exists at all.)

\item It is the intersection of all $k$-subalgebras of $A$ which contain $S$
as a subset.

\item It is the subset of $A$ formed by all elements that can be obtained by
repeated addition, multiplication and scaling with elements of $k$%
\ \ \ \ \footnote{By ``scaling with an element $\lambda\in k$\ \ \ \ '', we
mean the map $A\rightarrow A$ which sends every $a\in A$ to $\lambda a$.} from
elements of $S$. Here, ``multiplication'' doesn't only mean multiplication of
two elements, but can also mean multiplication of $n$ elements for any
$n\in\mathbb{N}$. (In particular, it can mean multiplication of $0$ elements;
this gives the unity $1_{A}$ as the result.)

\item It is the $k$-vector subspace $\left\langle S\right\rangle
^{0}+\left\langle S\right\rangle ^{1}+\left\langle S\right\rangle ^{2}%
+\cdots=\sum_{\ell\in\mathbb{N}}\left\langle S\right\rangle ^{\ell}$ of $A$
(where $\left\langle S\right\rangle $ denotes the $k$-vector subspace of $A$
generated by the subset $S$).
\end{itemize}

It is rather well-known that all of these definitions of the $k$-subalgebra of
$A$ generated by $S$ are equivalent, and that the $k$-subalgebra of $A$
generated by $S$ is really a $k$-subalgebra of $A$. We denote by
$\operatorname{AlgGen}_{k}S$ the $k$-subalgebra of $A$ generated by $S$.
\end{convention}

Some rather obvious properties of this notation will be used without explicit
mention. For example, if $S$ and $T$ are two subsets of a $k$-subalgebra $A$
satisfying $S\subseteq T$, then $\operatorname{AlgGen}_{k}S\subseteq
\operatorname{AlgGen}_{k}T$. (This can easily be derived from any of the
definitions of $\operatorname{AlgGen}_{k}S$.)

Here is another, even more obvious fact:

\begin{lemma}
\label{lem.AlgGen.Sinit}Let $k$ be a field. Let $A$ be a $k$-algebra. Let $S$
be any subset of $A$. Then,%
\begin{equation}
S\subseteq\operatorname{AlgGen}_{k}S. \label{1731.pf.triv-in}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.AlgGen.Sinit}.]Recall that $\operatorname{AlgGen}%
_{k}S$ is the $k$-subalgebra of $A$ generated by $S$; this $k$-subalgebra
clearly contains $S$ as a subset. In other words, $\operatorname{AlgGen}_{k}S$
contains $S$ as a subset. In other words, $S\subseteq\operatorname{AlgGen}%
_{k}S$. This proves Lemma~\ref{lem.AlgGen.Sinit}.
\end{proof}

Our next goal is the following theorem:

\begin{theorem}
\label{Theorem231} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected graded $k$-bialgebra. Let $\zeta$ be the map $\operatorname*{Log}%
\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $.

\textbf{(a)} The map $\zeta$ is graded.

\textbf{(b)} For every $n\in\mathbb{N}$, let $p_{n}$ denote the map $p_{n,H}$
(defined according to Definition~\ref{Definition221}), and let $\zeta_{n}$
denote the map $\zeta\circ p_{n}$. Then,%
\begin{equation}
\zeta_{n}=\zeta\circ p_{n}=p_{n}\circ\zeta=p_{n}\circ\zeta\circ p_{n}%
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{.} \label{231.comm}%
\end{equation}
Besides,%
\begin{equation}
\operatorname{AlgGen}_{k}\left\{  \zeta_{1},\zeta_{2},\zeta_{3},\ldots
\right\}  =\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},p_{3},\ldots\right\}
. \label{231.gen}%
\end{equation}


\textit{Remark.} In Theorem~\ref{Theorem231} and in the following
Proposition~\ref{Proposition232}, we are considering the algebra
$\mathcal{L}\left(  H,H\right)  $ whose multiplication is convolution,
\textit{not} the algebra $\operatorname*{End}H$ whose multiplication is
composition of mappings. Thus, for example, $\operatorname{AlgGen}_{k}\left\{
\zeta_{1},\zeta_{2},\zeta_{3},\ldots\right\}  $ denotes the $k$-subalgebra of
$\mathcal{L}\left(  H,H\right)  $ (not of $\operatorname*{End}H$) generated by
the subset $\left\{  \zeta_{1},\zeta_{2},\zeta_{3},\ldots\right\}  $.
\end{theorem}

Before we begin proving Theorem~\ref{Theorem231}, let us notice that its part
\textbf{(a)} is a trivial consequence of Proposition~\ref{Proposition224}
\textbf{(e)}, and the equation (\ref{231.comm}) in part \textbf{(b)} follows
immediately from part \textbf{(a)} due to Proposition~\ref{Proposition223}.
The main part of Theorem~\ref{Theorem231} is the equality (\ref{231.gen}).
This equality was mentioned in the Example in \S 2 of \cite{2}\footnote{More
precisely, the notations $F$, $A$, $e$, $e_{n}$, $\mathcal{D}\left(  A\right)
$ and $\mathcal{D}_{e}$ of \cite{2} correspond to the notations $k$, $H$,
$\zeta$, $\zeta_{n}$, $\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2}%
,p_{3},\ldots\right\}  $ and $\operatorname{AlgGen}_{k}\left\{  \zeta
_{1},\zeta_{2},\zeta_{3},\ldots\right\}  $ in our text (but we do not require
$H$ to be cocommutative here). Hence, our equation (\ref{231.gen}) rewrites as
$\mathcal{D}\left(  A\right)  =\mathcal{D}_{e}$ using the notations of
\cite{2}. This is mentioned in the Example in \S 2 of \cite{2}, and used
afterwards in the proof of Theorem 5.1 of \cite{2}.}.

First let us prove the easy part of Theorem~\ref{Theorem231}:

\begin{proof}
[Proof of Theorem~\ref{Theorem231}, first part.]\textbf{(a)} Applying
Proposition~\ref{Proposition224} \textbf{(e)} to $C=H$, $A=H$ and
$f=\operatorname*{id}$, we conclude that $\operatorname*{Log}%
\operatorname*{id}$ is graded (since $\operatorname*{id}\in G\left(
H,H\right)  $ is graded). Since $\zeta=\operatorname*{Log}\operatorname*{id}$,
this rewrites as follows: The map $\zeta$ is graded. This proves
Theorem~\ref{Theorem231} \textbf{(a)}.

\textbf{(b)} Let $n\in\mathbb{N}$. Since the map $\zeta$ is graded (by
Theorem~\ref{Theorem231} \textbf{(a)}), we have%
\[
\zeta\circ p_{n,H}=p_{n,H}\circ\zeta=p_{n,H}\circ\zeta\circ p_{n,H}%
\]
(by Proposition~\ref{Proposition223}, applied to $V=H$, $W=H$ and $f=\zeta$).
Since $p_{n,H}=p_{n}$, this rewrites as $\zeta\circ p_{n}=p_{n}\circ
\zeta=p_{n}\circ\zeta\circ p_{n}$. Combined with $\zeta_{n}=\zeta\circ p_{n}$,
this yields (\ref{231.comm}).

We thus have proven (\ref{231.comm}). We are going to prove (\ref{231.gen})
later. For the time being, we leave the proof of Theorem~\ref{Theorem231}.
\end{proof}

Here is an assertion which will turn out to be somewhat stronger than
(\ref{231.gen}) (we are later going to derive (\ref{231.gen}) from it):

\begin{proposition}
\label{Proposition232} Let $k$ be a field of characteristic $0$, and let $H$
be a connected graded $k$-bialgebra. Let $\zeta$ be the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $. We
are going to use the notations of Theorem~\ref{Theorem231} \textbf{(b)}.

For every $n\in\mathbb{N}$, we have%
\[
p_{n}-\zeta_{n}\in\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots
,p_{n-1}\right\}  .
\]

\end{proposition}

And here is a ``finite'' version of (\ref{231.gen}):

\begin{proposition}
\label{Proposition233} Let $k$ be a field of characteristic $0$, and let $H$
be a connected graded $k$-bialgebra. Let $\zeta$ be the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $. We
are going to use the notations of Theorem~\ref{Theorem231} \textbf{(b)}.

For every $n\in\mathbb{N}$, we have%
\[
\operatorname{AlgGen}_{k}\left\{  \zeta_{1},\zeta_{2},\ldots,\zeta
_{n}\right\}  =\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots
,p_{n}\right\}  .
\]

\end{proposition}

We will prove these two propositions before showing Theorem~\ref{Theorem231}.
But first, we introduce another notation, and show some of its properties.
First, the notation:

\begin{definition}
\label{Definition235} Let $k$ be a field. Let $H$ be a graded $k$-vector
space. Let $f\in\mathcal{L}\left(  H,H\right)  $ be a graded map. Let
$n\in\mathbb{N}$. Just as in Theorem~\ref{Theorem231} \textbf{(b)}, let
$p_{n}$ denote the map $p_{n,H}$ (defined according to
Definition~\ref{Definition221}). Then, we say that $f$ is \textit{concentrated
in degree }$n$ if and only if $f=p_{n}\circ f$.
\end{definition}

It is easily seen that (in the context of Definition~\ref{Definition235}) we
have the following chain of equivalences:%
\begin{align*}
\  &  \left(  f\text{ is concentrated in degree }n\right)  \ \\
\Longleftrightarrow &  \ \left(  f=p_{n}\circ f\right)  \ \Longleftrightarrow
\ \left(  f=f\circ p_{n}\right)  \ \Longleftrightarrow\ \left(  f=p_{n}\circ
f\circ p_{n}\right) \\
\Longleftrightarrow &  \ \left(  f\left(  H\right)  \subseteq H_{n}\right)
\ \Longleftrightarrow\ \left(  f\left(  H_{m}\right)  =0\text{ for all }%
m\in\mathbb{N}\text{ satisfying }m\neq n\right)  .
\end{align*}
However, we are not going to use these equivalences (at least not explicitly).

The following properties of being concentrated in degree $n$ will be used:

\begin{proposition}
\label{Proposition236} Let $k$ be a field. Let $H$ be a graded $k$-bialgebra.

\textbf{(a)} The map $e_{H,H}$ is graded and concentrated in degree $0$.

\textbf{(b)} Let $a\in\mathbb{N}$, $b\in\mathbb{N}$, $f\in\mathcal{L}\left(
H,H\right)  $ and $g\in\mathcal{L}\left(  H,H\right)  $. Assume that $f$ is
graded and concentrated in degree $a$. Assume that $g$ is graded and
concentrated in degree $b$. Then, $f\ast g$ is graded and concentrated in
degree $a+b$.

\textbf{(c)} Let $\ell\in\mathbb{N}$. Let $a_{i}$ be a nonnegative integer for
every $i\in\left\{  1,2,\ldots,\ell\right\}  $. Let $f_{i}$ be an element of
$\mathcal{L}\left(  H,H\right)  $ for every $i\in\left\{  1,2,\ldots
,\ell\right\}  $. Assume that for every $i\in\left\{  1,2,\ldots,\ell\right\}
$, the map $f_{i}$ is graded and concentrated in degree $a_{i}$. Then,
$f_{1}\ast f_{2}\ast\cdots\ast f_{\ell}$ is graded and concentrated in degree
$a_{1}+a_{2}+\cdots+a_{\ell}$.

\textbf{(d)} For every $n\in\mathbb{N}$, the map $p_{n}$ (defined in
Theorem~\ref{Theorem231} \textbf{(b)}) is graded and concentrated in degree
$n$.

\textbf{(e)} If the graded $k$-bialgebra $H$ is connected, then $e_{H,H}%
=p_{0}$ (where the map $p_{0}$ is defined as $p_{0,H}$, just as in
Theorem~\ref{Theorem231} \textbf{(b)}).
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition236}.]Just as in
Theorem~\ref{Theorem231} \textbf{(b)}, let $p_{n}$ denote the map $p_{n,H}$
for every $n\in\mathbb{N}$ (defined according to
Definition~\ref{Definition221}).

\textbf{(a)} We have $e_{H,H}=\eta_{H}\circ\varepsilon_{H}$. But since $H$ is
a graded $k$-bialgebra, we have $1_{H}\in H_{0}$.

By the definition of $p_{0}$, we have $p_{0}=p_{0,H}$. Thus, $p_{0}\mid
_{H_{0}}=p_{0,H}\mid_{H_{0}}=\operatorname{id}_{H}\mid_{H_{0}}$ (by
(\ref{221.id}), applied to $V=H$ and $n=0$).

Every $x\in H$ satisfies%
\[
\underbrace{e_{H,H}}_{=\eta_{H}\circ\varepsilon_{H}}\left(  x\right)  =\left(
\eta_{H}\circ\varepsilon_{H}\right)  \left(  x\right)  =\eta_{H}\left(
\varepsilon_{H}\left(  x\right)  \right)  =\varepsilon_{H}\left(  x\right)
\cdot1_{H}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\eta
_{H}\right)
\]
and now%
\begin{align*}
&  \left(  p_{0}\circ e_{H,H}\right)  \left(  x\right) \\
&  =p_{0}\left(  \underbrace{e_{H,H}\left(  x\right)  }_{=\varepsilon
_{H}\left(  x\right)  \cdot1_{H}}\right)  =p_{0}\left(  \varepsilon_{H}\left(
x\right)  \cdot1_{H}\right) \\
&  =\underbrace{\left(  p_{0}\mid_{H_{0}}\right)  }_{=\operatorname{id}%
_{H}\mid_{H_{0}}}\left(  \varepsilon_{H}\left(  x\right)  \cdot1_{H}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon_{H}\left(  x\right)
\cdot\underbrace{1_{H}}_{\in H_{0}}\in H_{0}\text{ (since }H_{0}\text{ is a
}k\text{-vector space)}\right) \\
&  =\left(  \operatorname{id}_{H}\mid_{H_{0}}\right)  \left(  \varepsilon
_{H}\left(  x\right)  \cdot1_{H}\right)  =\varepsilon_{H}\left(  x\right)
\cdot1_{H}=e_{H,H}\left(  x\right)  .
\end{align*}
Thus, $p_{0}\circ e_{H,H}=e_{H,H}$. In other words, $e_{H,H}=p_{0}\circ
e_{H,H}$. Combined with the fact that $e_{H,H}$ is graded (by
Proposition~\ref{Proposition224} \textbf{(b)}, applied to $C=H$ and $A=H$),
this yields that $e_{H,H}$ is concentrated in degree $0$ (by
Definition~\ref{Definition235}). This proves Proposition~\ref{Proposition236}
\textbf{(a)}.

\textbf{(b)} Since $f$ is graded and concentrated in degree $a$, we have
$f=p_{a}\circ f$ (by Definition~\ref{Definition235}). Thus,
\begin{align*}
f\left(  H\right)   &  =\left(  p_{a}\circ f\right)  \left(  H\right)
=\underbrace{p_{a}}_{\substack{=p_{a,H}\\\text{(by the definition of }%
p_{a}\text{)}}}\left(  \underbrace{f\left(  H\right)  }_{\subseteq H}\right)
\subseteq p_{a,H}\left(  H\right) \\
&  =H_{a}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{221.im}), applied to
}n=a\text{ and }V=H\right)  .
\end{align*}
Similarly, $g\left(  H\right)  \subseteq H_{b}$.

Since $H$ is a graded $k$-bialgebra, its multiplication map $\mu_{H}:H\otimes
H\rightarrow H$ is graded, where the grading on $H\otimes H$ is the usual
grading on the tensor product of two graded $k$-vector spaces. Now, this
grading is defined by $\left(  H\otimes H\right)  _{n}=\sum_{\ell=0}%
^{n}H_{\ell}\otimes H_{n-\ell}$ for every $n\in\mathbb{N}$. Hence, for every
$a\in\mathbb{N}$ and $b\in\mathbb{N}$, we have
\begin{align*}
\left(  H\otimes H\right)  _{a+b}  &  =\sum_{\ell=0}^{a+b}H_{\ell}\otimes
H_{a+b-\ell}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{due to }\left(  H\otimes H\right)
_{n}=\sum_{\ell=0}^{n}H_{\ell}\otimes H_{n-\ell}\text{, applied to
}n=a+b\right) \\
&  \supseteq H_{a}\otimes\underbrace{H_{a+b-a}}_{=H_{b}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }H_{a}\otimes H_{a+b-a}\text{ is an addend in the sum }\sum
_{\ell=0}^{a+b}H_{\ell}\otimes H_{a+b-\ell}\\
\text{(namely, the addend for }\ell=a\text{)}%
\end{array}
\right) \\
&  =H_{a}\otimes H_{b}.
\end{align*}
Thus, for every $a\in\mathbb{N}$ and $b\in\mathbb{N}$, we have $H_{a}\otimes
H_{b}\subseteq\left(  H\otimes H\right)  _{a+b}$, so that%
\[
\mu_{H}\left(  H_{a}\otimes H_{b}\right)  \subseteq\mu_{H}\left(  \left(
H\otimes H\right)  _{a+b}\right)  \subseteq H_{a+b}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mu_{H}\text{ is graded}\right)  .
\]


Now, by the definition of convolution, $f\ast g=\mu_{H}\circ\left(  f\otimes
g\right)  \circ\Delta_{H}$, so that%
\begin{align*}
\left(  f\ast g\right)  \left(  H\right)   &  =\left(  \mu_{H}\circ\left(
f\otimes g\right)  \circ\Delta_{H}\right)  \left(  H\right)  =\mu_{H}\left(
\left(  f\otimes g\right)  \underbrace{\left(  \Delta_{H}\left(  H\right)
\right)  }_{\subseteq H\otimes H}\right)  \subseteq\mu_{H}\left(
\underbrace{\left(  f\otimes g\right)  \left(  H\otimes H\right)  }_{=f\left(
H\right)  \otimes g\left(  H\right)  }\right) \\
&  =\mu_{H}\left(  \underbrace{f\left(  H\right)  }_{\subseteq H_{a}}%
\otimes\underbrace{g\left(  H\right)  }_{\subseteq H_{b}}\right)  \subseteq
\mu_{H}\left(  H_{a}\otimes H_{b}\right)  \subseteq H_{a+b}.
\end{align*}
Thus, every $x\in H$ satisfies $\left(  f\ast g\right)  \left(  x\right)  \in
H_{a+b}$. Hence, every $x\in H$ satisfies%
\begin{align*}
\left(  p_{a+b}\circ\left(  f\ast g\right)  \right)  \left(  x\right)   &
=p_{a+b}\left(  \left(  f\ast g\right)  \left(  x\right)  \right)  =\left(
\underbrace{p_{a+b}}_{\substack{=p_{a+b,H}\\\text{(by the definition of
}p_{a+b}\text{)}}}\mid_{H_{a+b}}\right)  \left(  \left(  f\ast g\right)
\left(  x\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  f\ast g\right)  \left(
x\right)  \in H_{a+b}\right) \\
&  =\underbrace{\left(  p_{a+b,H}\mid_{H_{a+b}}\right)  }%
_{\substack{=\operatorname{id}_{H}\mid_{H_{a+b}}\\\text{(by (\ref{221.id}),
applied to}\\V=H\text{ and }n=a+b\text{)}}}\left(  \left(  f\ast g\right)
\left(  x\right)  \right)  =\left(  \operatorname{id}_{H}\mid_{H_{a+b}%
}\right)  \left(  \left(  f\ast g\right)  \left(  x\right)  \right) \\
&  =\left(  f\ast g\right)  \left(  x\right)  .
\end{align*}
In other words, $p_{a+b}\circ\left(  f\ast g\right)  =f\ast g$.

Hence, $f\ast g=p_{a+b}\circ\left(  f\ast g\right)  $. Combined with the fact
that $f\ast g$ is graded (by Proposition~\ref{Proposition224} \textbf{(a)},
applied to $C=H$ and $A=H$), this yields that $f\ast g$ is concentrated in
degree $a+b$ (by Definition~\ref{Definition235}). This proves
Proposition~\ref{Proposition236} \textbf{(b)}.

\textbf{(c)} We are going to prove that for every $j\in\left\{  0,1,\ldots
,\ell\right\}  $,%
\begin{equation}
\text{the map }f_{1}\ast f_{2}\ast\cdots\ast f_{j}\text{ is graded and
concentrated in degree }a_{1}+a_{2}+\cdots+a_{j}\text{.} \label{236.pf.c}%
\end{equation}


\textit{Proof of (\ref{236.pf.c}).} We will prove (\ref{236.pf.c}) by
induction over $j$:

\textit{Induction base:} The map $f_{1}\ast f_{2}\ast\cdots\ast f_{0}=e_{H,H}$
is graded and concentrated in degree $0$ (by Proposition~\ref{Proposition236}
\textbf{(a)}). Since $0=a_{1}+a_{2}+\cdots+a_{0}$, we thus have shown that the
map $f_{1}\ast f_{2}\ast\cdots\ast f_{0}$ is graded and concentrated in degree
$a_{1}+a_{2}+\cdots+a_{0}$. In other words, we have shown that (\ref{236.pf.c}%
) holds for $j=0$. This completes the induction base.

\textit{Induction step:} Let $J\in\left\{  0,1,\ldots,\ell-1\right\}  $.
Assume that (\ref{236.pf.c}) holds for $j=J$. We now must prove that
(\ref{236.pf.c}) holds for $j=J+1$.

We know that for every $i\in\left\{  1,2,\ldots,\ell\right\}  $, the map
$f_{i}$ is graded and concentrated in degree $a_{i}$. Hence, the map $f_{J+1}$
is graded and concentrated in degree $a_{J+1}$.

Since (\ref{236.pf.c}) holds for $j=J$, the map $f_{1}\ast f_{2}\ast\cdots\ast
f_{J}$ is graded and concentrated in degree $a_{1}+a_{2}+\cdots+a_{J}$.
Applying Proposition~\ref{Proposition236} \textbf{(b)} to $a=a_{1}%
+a_{2}+\cdots+a_{J}$, $b=a_{J+1}$, $f=f_{1}\ast f_{2}\ast\cdots\ast f_{J}$ and
$g=f_{J+1}$, we now see that the map $\left(  f_{1}\ast f_{2}\ast\cdots\ast
f_{J}\right)  \ast f_{J+1}$ is graded and concentrated in degree $\left(
a_{1}+a_{2}+\cdots+a_{J}\right)  +a_{J+1}$. Since $\left(  f_{1}\ast f_{2}%
\ast\cdots\ast f_{J}\right)  \ast f_{J+1}=f_{1}\ast f_{2}\ast\cdots\ast
f_{J+1}$ and $\left(  a_{1}+a_{2}+\cdots+a_{J}\right)  +a_{J+1}=a_{1}%
+a_{2}+\cdots+a_{J+1}$, this rewrites as follows: The map $f_{1}\ast f_{2}%
\ast\cdots\ast f_{J+1}$ is graded and concentrated in degree $a_{1}%
+a_{2}+\cdots+a_{J+1}$. In other words, (\ref{236.pf.c}) holds for $j=J+1$.
This completes the induction step.

Thus, the induction proof of (\ref{236.pf.c}) is complete.

Now we can apply (\ref{236.pf.c}) to $j=\ell$, and conclude that the map
$f_{1}\ast f_{2}\ast\cdots\ast f_{\ell}$ is graded and concentrated in degree
$a_{1}+a_{2}+\cdots+a_{\ell}$. This proves Proposition~\ref{Proposition236}
\textbf{(c)}.

\textbf{(d)} Let $n\in\mathbb{N}$. Since $p_{n}$ was defined as $p_{n,H}$, we
can rewrite the identity $p_{n,H}\circ p_{n,H}=p_{n,H}$ (which follows from
(\ref{221.idp}), applied to $V=H$) as $p_{n}\circ p_{n}=p_{n}$.

Also, $p_{n}=p_{n,H}$ is graded (by (\ref{221.gr}), applied to $V=H$). Thus,
$p_{n}$ is a graded map satisfying $p_{n}=p_{n}\circ p_{n}$. By
Definition~\ref{Definition235}, this yields that $p_{n}$ is concentrated in
degree $n$. This proves Proposition~\ref{Proposition236} \textbf{(d)}.

\textbf{(e)} Assume that the graded $k$-bialgebra $H$ is connected. Then,
$\varepsilon_{H}\mid_{H_{0}}:H_{0}\rightarrow k$ is a $k$-vector space
isomorphism (because this is how a connected graded $k$-bialgebra was defined
in Definition~\ref{Definition212}).

Let us give $k$ the usual grading (the one where $k_{0}=k$ and $k_{n}=0$ for
all positive $n\in\mathbb{N}$). Since $H$ is a graded $k$-coalgebra, its
counity map $\varepsilon_{H}:H\rightarrow k$ is graded. Thus,
Proposition~\ref{Proposition223} (applied to $V=H$, $W=k$, $f=\varepsilon_{H}$
and $n=0$) yields $\varepsilon_{H}\circ p_{0,H}=p_{0,k}\circ\varepsilon
_{H}=p_{0,k}\circ\varepsilon_{H}\circ p_{0,H}$. But since $k_{0}=k$, we have
\begin{align*}
p_{0,k}  &  =p_{0,k}\mid_{k_{0}}=\operatorname{id}_{k}\mid_{k_{0}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{221.id}), applied to }V=k\text{
and }n=0\right) \\
&  =\operatorname{id}_{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
k_{0}=k\right)  ,
\end{align*}
so that $\varepsilon_{H}\circ p_{0,H}=p_{0,k}\circ\varepsilon_{H}$ simplifies
to $\varepsilon_{H}\circ p_{0,H}=\varepsilon_{H}$. Since $p_{0,H}=p_{0}$, this
further rewrites as $\varepsilon_{H}\circ p_{0}=\varepsilon_{H}$.

Since $H$ is a bialgebra, we have $\varepsilon_{H}\circ\eta_{H}%
=\operatorname{id}_{k}$ (by the axioms of a bialgebra). But $e_{H,H}=\eta
_{H}\circ\varepsilon_{H}$ (by the definition of $e_{H,H}$). Thus,
$\varepsilon_{H}\circ e_{H,H}=\underbrace{\varepsilon_{H}\circ\eta_{H}%
}_{=\operatorname{id}_{k}}\circ\varepsilon_{H}=\varepsilon_{H}$.

Now let $x\in H$. Then,
\[
\underbrace{p_{0}}_{=p_{0,H}}\left(  x\right)  =p_{0,H}\left(  x\right)  \in
p_{0,H}\left(  H\right)  =H_{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{221.im}), applied to }V=H\text{ and }n=0\right)  ,
\]
so that $\left(  \varepsilon_{H}\mid_{H_{0}}\right)  \left(  p_{0}\left(
x\right)  \right)  $ is well-defined. Also,%
\begin{align*}
\underbrace{e_{H,H}}_{=\eta_{H}\circ\varepsilon_{H}}\left(  x\right)   &
=\left(  \eta_{H}\circ\varepsilon_{H}\right)  \left(  x\right)  =\eta
_{H}\left(  \varepsilon_{H}\left(  x\right)  \right)  =\varepsilon_{H}\left(
x\right)  \cdot1_{H}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\eta_{H}\right) \\
&  \in H_{0}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }1_{H}\in H_{0}\text{ (because }H\text{ is a graded }%
k\text{-algebra),}\\
\text{and since }H_{0}\text{ is a }k\text{-vector space}%
\end{array}
\right)  ,
\end{align*}
so that $\left(  \varepsilon_{H}\mid_{H_{0}}\right)  \left(  e_{H,H}\left(
x\right)  \right)  $ is well-defined. Since%
\begin{align*}
\left(  \varepsilon_{H}\mid_{H_{0}}\right)  \left(  p_{0}\left(  x\right)
\right)   &  =\varepsilon_{H}\left(  p_{0}\left(  x\right)  \right)
=\underbrace{\left(  \varepsilon_{H}\circ p_{0}\right)  }_{=\varepsilon
_{H}=\varepsilon_{H}\circ e_{H,H}}\left(  x\right)  =\left(  \varepsilon
_{H}\circ e_{H,H}\right)  \left(  x\right) \\
&  =\varepsilon_{H}\left(  e_{H,H}\left(  x\right)  \right)  =\left(
\varepsilon_{H}\mid_{H_{0}}\right)  \left(  e_{H,H}\left(  x\right)  \right)
,
\end{align*}
we conclude that $p_{0}\left(  x\right)  =e_{H,H}\left(  x\right)  $ (since
$\varepsilon_{H}\mid_{H_{0}}$ is an isomorphism). Since we have proven this
for every $x\in H$, we conclude that $p_{0}=e_{H,H}$. This proves
Proposition~\ref{Proposition236} \textbf{(e)}.
\end{proof}

Proposition~\ref{Proposition236} has a very easy corollary:

\begin{corollary}
\label{Corollary237} Let $k$ be a field. Let $H$ be a graded $k$-bialgebra.
For every $n\in\mathbb{N}$, let $p_{n}$ denote the map $p_{n,H}$ (defined
according to Definition~\ref{Definition221}).

Let $n\in\mathbb{N}$ and $\ell\in\mathbb{N}$. Let $a_{i}$ be a nonnegative
integer for every $i\in\left\{  1,2,\ldots,\ell\right\}  $.

\textbf{(a)} We have $p_{n}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  =0$ if $n\neq a_{1}+a_{2}+\cdots+a_{\ell}$.

\textbf{(b)} We have $p_{n}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  =p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}$ if
$n=a_{1}+a_{2}+\cdots+a_{\ell}$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary237}.]For every $i\in\left\{  1,2,\ldots
,\ell\right\}  $, the map $p_{a_{i}}$ is graded and concentrated in degree
$a_{i}$ (by Proposition~\ref{Proposition236} \textbf{(d)}, applied to $a_{i}$
instead of $n$). Thus, Proposition~\ref{Proposition236} \textbf{(c)} (applied
to $f_{i}=p_{a_{i}}$) yields that $p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}$ is graded and concentrated in degree $a_{1}+a_{2}+\cdots
+a_{\ell}$. According to Definition~\ref{Definition235}, this means that%
\begin{equation}
p_{a_{1}+a_{2}+\cdots+a_{\ell}}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast
\cdots\ast p_{a_{\ell}}\right)  =p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}. \label{237.pf.1}%
\end{equation}


\textbf{(a)} Assume that $n\neq a_{1}+a_{2}+\cdots+a_{\ell}$. Then,
$p_{n,H}\mid_{H_{a_{1}+a_{2}+\cdots+a_{\ell}}}=0$ (by (\ref{221.0}), applied
to $m=a_{1}+a_{2}+\cdots+a_{\ell}$ and $V=H$).

But%
\begin{align*}
&  \underbrace{\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}%
}\right)  }_{\substack{=p_{a_{1}+a_{2}+\cdots+a_{\ell}}\circ\left(  p_{a_{1}%
}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \\\text{(by (\ref{237.pf.1}%
))}}}\left(  H\right) \\
&  =\left(  p_{a_{1}+a_{2}+\cdots+a_{\ell}}\circ\left(  p_{a_{1}}\ast
p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \right)  \left(  H\right) \\
&  =\underbrace{p_{a_{1}+a_{2}+\cdots+a_{\ell}}}_{\substack{=p_{a_{1}%
+a_{2}+\cdots+a_{\ell},H}\\\text{(by the definition of }p_{a_{1}+a_{2}%
+\cdots+a_{\ell}}\text{)}}}\left(  \underbrace{\left(  p_{a_{1}}\ast p_{a_{2}%
}\ast\cdots\ast p_{a_{\ell}}\right)  \left(  H\right)  }_{\subseteq H}\right)
\\
&  \subseteq p_{a_{1}+a_{2}+\cdots+a_{\ell},H}\left(  H\right)  =H_{a_{1}%
+a_{2}+\cdots+a_{\ell}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{221.im}), applied to }H\text{ and }a_{1}+a_{2}+\cdots+a_{\ell
}\\
\text{instead of }V\text{ and }n
\end{array}
\right)
\end{align*}
and thus%
\begin{align*}
p_{n}\left(  \left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}%
}\right)  \left(  H\right)  \right)   &  \subseteq\underbrace{p_{n}%
}_{\substack{=p_{n,H}\\\text{(by the definition of }p_{n}\text{)}}}\left(
H_{a_{1}+a_{2}+\cdots+a_{\ell}}\right)  =p_{n,H}\left(  H_{a_{1}+a_{2}%
+\cdots+a_{\ell}}\right) \\
&  =\underbrace{\left(  p_{n,H}\mid_{H_{a_{1}+a_{2}+\cdots+a_{\ell}}}\right)
}_{=0}\left(  H_{a_{1}+a_{2}+\cdots+a_{\ell}}\right)  =0\left(  H_{a_{1}%
+a_{2}+\cdots+a_{\ell}}\right)  =0.
\end{align*}
Since $p_{n}\left(  \left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}%
}\right)  \left(  H\right)  \right)  =\left(  p_{n}\circ\left(  p_{a_{1}}\ast
p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \right)  \left(  H\right)  $,
this rewrites as $\left(  p_{n}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast
\cdots\ast p_{a_{\ell}}\right)  \right)  \left(  H\right)  =0$. Thus,
$p_{n}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)
=0$. This proves Corollary~\ref{Corollary237} \textbf{(a)}.

\textbf{(b)} Now assume that $n=a_{1}+a_{2}+\cdots+a_{\ell}$. Then,%
\[
p_{n}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)
=p_{a_{1}+a_{2}+\cdots+a_{\ell}}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast
\cdots\ast p_{a_{\ell}}\right)  =p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}%
\]
(by (\ref{237.pf.1})). This proves Corollary~\ref{Corollary237} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition232}.]Let $n\in\mathbb{N}$.

Let $h$ denote the map $\operatorname*{id}-e_{H,H}$ (where $\operatorname*{id}%
$ means $\operatorname{id}_{H}$). Then, since $\operatorname*{id}\in G\left(
H,H\right)  =e_{H,H}+\mathfrak{g}\left(  H,H\right)  $, we have
$\operatorname*{id}-e_{H,H}\in\mathfrak{g}\left(  H,H\right)  $, so that
$h=\operatorname*{id}-e_{H,H}\in\mathfrak{g}\left(  H,H\right)  $.

We will now prove Proposition~\ref{Proposition232} in several steps:

\textbf{a)} Every $x\in H_{\leq n}$ satisfies%
\begin{equation}
h\left(  x\right)  =\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  \left(  x\right)
. \label{232.pf.1}%
\end{equation}


\textit{Proof of (\ref{232.pf.1}).} Let $x\in H_{\leq n}$. By (\ref{221.vsum})
(applied to $V=H$ and $v=x$), we have $x=\sum_{\ell\in\mathbb{N}}p_{\ell
,H}\left(  x\right)  $. Since $p_{\ell}=p_{\ell,H}$ for all $\ell\in
\mathbb{N}$ (by the definition of $p_{\ell}$), this becomes%
\begin{equation}
x=\sum_{\ell\in\mathbb{N}}\underbrace{p_{\ell,H}}_{=p_{\ell}}\left(  x\right)
=\sum_{\ell\in\mathbb{N}}p_{\ell}\left(  x\right)  =\sum_{\substack{\ell
\in\mathbb{N};\\\ell\leq n}}p_{\ell}\left(  x\right)  +\sum_{\substack{\ell
\in\mathbb{N};\\\ell>n}}p_{\ell}\left(  x\right)  . \label{232.pf.2}%
\end{equation}


But for every $\ell\in\mathbb{N}$ satisfying $\ell>n$, we have $p_{\ell
}\left(  H_{\leq n}\right)  =0$.\ \ \ \ \footnote{\textit{Proof.} By the
definition of $H_{\leq n}$, we have
\begin{align*}
H_{\leq n}  &  =\bigoplus_{\ell=0}^{n}H_{\ell}=\bigoplus_{m=0}^{n}%
H_{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\ell\text{ as }m\right) \\
&  =\sum_{m=0}^{n}H_{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are
sums}\right)  .
\end{align*}
Hence, for every $\ell\in\mathbb{N}$ satisfying $\ell>n$, we have%
\begin{align*}
p_{\ell}\left(  H_{\leq n}\right)   &  =p_{\ell}\left(  \sum_{m=0}^{n}%
H_{m}\right)  =\sum_{m=0}^{n}\underbrace{p_{\ell}\left(  H_{m}\right)
}_{=\left(  p_{\ell}\mid_{H_{m}}\right)  \left(  H_{m}\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{\ell}\text{ is }k\text{-linear}%
\right) \\
&  =\sum_{m=0}^{n}\left(  \underbrace{p_{\ell}}_{\substack{=p_{\ell
,H}\\\text{(by the definition of }H\text{)}}}\mid_{H_{m}}\right)  \left(
H_{m}\right)  =\sum_{m=0}^{n}\underbrace{\left(  p_{\ell,H}\mid_{H_{m}%
}\right)  }_{\substack{=0\text{ (by (\ref{221.0}) (applied}\\\text{to }%
\ell\text{ and }H\text{ instead of }n\text{ and }V\text{),}\\\text{since }%
\ell\neq m\text{ (since }\ell>n\geq m\text{))}}}\left(  H_{m}\right) \\
&  =\sum_{m=0}^{n}0\left(  H_{m}\right)  =0,
\end{align*}
qed.} Thus, for every $\ell\in\mathbb{N}$ satisfying $\ell>n$, we have
$p_{\ell}\left(  x\right)  =0$ (since $x\in H_{\leq n}$, and thus, for every
$\ell\in\mathbb{N}$ satisfying $\ell>n$, we obtain $p_{\ell}\left(  x\right)
\in p_{\ell}\left(  H_{\leq n}\right)  =0$, so that $p_{\ell}\left(  x\right)
=0$). Thus, (\ref{232.pf.2}) becomes%
\begin{align}
x  &  =\sum_{\substack{\ell\in\mathbb{N};\\\ell\leq n}}p_{\ell}\left(
x\right)  +\sum_{\substack{\ell\in\mathbb{N};\\\ell>n}}\underbrace{p_{\ell
}\left(  x\right)  }_{=0}=\sum_{\substack{\ell\in\mathbb{N};\\\ell\leq
n}}p_{\ell}\left(  x\right)  +\underbrace{\sum_{\substack{\ell\in
\mathbb{N};\\\ell>n}}0}_{=0}=\sum_{\substack{\ell\in\mathbb{N};\\\ell\leq
n}}p_{\ell}\left(  x\right)  =p_{0}\left(  x\right)  +p_{1}\left(  x\right)
+\cdots+p_{n}\left(  x\right) \label{232.pf.23}\\
&  =\underbrace{p_{0}}_{\substack{=e_{H,H}\\\text{(by
Proposition~\ref{Proposition236} \textbf{(e)},}\\\text{since }H\text{ is
connected)}}}\left(  x\right)  +\underbrace{\left(  p_{1}\left(  x\right)
+p_{2}\left(  x\right)  +\cdots+p_{n}\left(  x\right)  \right)  }_{=\left(
p_{1}+p_{2}+\cdots+p_{n}\right)  \left(  x\right)  }=e_{H,H}\left(  x\right)
+\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  \left(  x\right)  . \label{232.pf.3}%
\end{align}
Thus,%
\[
\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  \left(  x\right)  =\underbrace{x}%
_{=\operatorname*{id}\left(  x\right)  }-e_{H,H}\left(  x\right)
=\operatorname*{id}\left(  x\right)  -e_{H,H}\left(  x\right)
=\underbrace{\left(  \operatorname*{id}-e_{H,H}\right)  }_{=h}\left(
x\right)  =h\left(  x\right)  .
\]
This proves (\ref{232.pf.1}) for every $x\in H_{\leq n}$. Step \textbf{a)} is
thus done.

\textbf{b)} Every $x\in H_{\leq n}$ and every $\ell\in\mathbb{N}$ satisfy%
\begin{equation}
h^{\ast\ell}\left(  x\right)  =\left(  p_{1}+p_{2}+\cdots+p_{n}\right)
^{\ast\ell}\left(  x\right)  . \label{232.pf.4}%
\end{equation}


\textit{Proof of (\ref{232.pf.4}).} There are several ways to derive
(\ref{232.pf.4}) from (\ref{232.pf.1}). Here is one of them:

Let $\ell\in\mathbb{N}$.

We are going to use the notation $\mathcal{L}^{n}\left(  H,A\right)  $
introduced in Definition~\ref{Definition8} \textbf{(b)}.

By the definition of $\mathcal{L}^{n+1}\left(  H,H\right)  $, we have%
\[
\mathcal{L}^{n+1}\left(  H,H\right)  =\left\{  f\in\mathcal{L}\left(
H,H\right)  \ \mid\ \underbrace{f\mid_{H_{\leq n+1-1}}}_{=f\mid_{H_{\leq n}}%
}=0\right\}  =\left\{  f\in\mathcal{L}\left(  H,H\right)  \ \mid
\ f\mid_{H_{\leq n}}=0\right\}  .
\]
By Proposition~\ref{Proposition81} (applied to $n+1$ and $H$ instead of $n$
and $A$), the set $\mathcal{L}^{n+1}\left(  H,H\right)  $ is an ideal of the
$k$-algebra $\mathcal{L}\left(  H,H\right)  $.

Every $x\in H_{\leq n}$ satisfies%
\begin{align*}
\left(  h-\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  \right)  \left(  x\right)
&  =\underbrace{h\left(  x\right)  }_{\substack{=\left(  p_{1}+p_{2}%
+\cdots+p_{n}\right)  \left(  x\right)  \\\text{(by (\ref{232.pf.1}))}%
}}-\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  \left(  x\right) \\
&  =\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  \left(  x\right)  -\left(
p_{1}+p_{2}+\cdots+p_{n}\right)  \left(  x\right)  =0.
\end{align*}
In other words, $\left(  h-\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  \right)
\mid_{H_{\leq n}}=0$. Thus,%
\[
h-\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  \in\left\{  f\in\mathcal{L}\left(
H,H\right)  \ \mid\ f\mid_{H_{\leq n}}=0\right\}  =\mathcal{L}^{n+1}\left(
H,H\right)  .
\]
In other words, $h\equiv p_{1}+p_{2}+\cdots+p_{n}\operatorname{mod}%
\mathcal{L}^{n+1}\left(  H,H\right)  $. Thus,
\[
h^{\ast\ell}\equiv\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  ^{\ast\ell
}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,H\right)
\]
(because $\mathcal{L}^{n+1}\left(  H,H\right)  $ is an ideal of the
$k$-algebra $\mathcal{L}\left(  H,H\right)  $, and hence we can multiply
congruences modulo $\mathcal{L}^{n+1}\left(  H,H\right)  $). In other words,%
\[
h^{\ast\ell}-\left(  p_{1}+p_{2}+\cdots+p_{n}\right)  ^{\ast\ell}%
\in\mathcal{L}^{n+1}\left(  H,H\right)  =\left\{  f\in\mathcal{L}\left(
H,H\right)  \ \mid\ f\mid_{H_{\leq n}}=0\right\}  ,
\]
so that $\left(  h^{\ast\ell}-\left(  p_{1}+p_{2}+\cdots+p_{n}\right)
^{\ast\ell}\right)  \mid_{H_{\leq n}}=0$. Hence, every $x\in H_{\leq n}$
satisfies \newline$\left(  h^{\ast\ell}-\left(  p_{1}+p_{2}+\cdots
+p_{n}\right)  ^{\ast\ell}\right)  \left(  x\right)  =0$. As a consequence,
every $x\in H_{\leq n}$ satisfies $h^{\ast\ell}\left(  x\right)  =\left(
p_{1}+p_{2}+\cdots+p_{n}\right)  ^{\ast\ell}\left(  x\right)  $ (because
\[
h^{\ast\ell}\left(  x\right)  -\left(  p_{1}+p_{2}+\cdots+p_{n}\right)
^{\ast\ell}\left(  x\right)  =\left(  h^{\ast\ell}-\left(  p_{1}+p_{2}%
+\cdots+p_{n}\right)  ^{\ast\ell}\right)  \left(  x\right)  =0
\]
). In other words, (\ref{232.pf.4}) is proven for every $x\in H_{\leq n}$ and
every $\ell\in\mathbb{N}$. This completes step \textbf{b)}.

\textbf{c)} Every $x\in H_{\leq n}$ satisfies%
\begin{equation}
\zeta\left(  x\right)  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}%
}{\ell}h^{\ast\ell}\left(  x\right)  . \label{232.pf.5}%
\end{equation}


\textit{Proof of (\ref{232.pf.5}).} Since
\begin{align*}
\zeta &  =\operatorname*{Log}\operatorname*{id}=\operatorname{Log}%
_{1}\underbrace{\left(  \operatorname*{id}-e_{H,H}\right)  }_{=h}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Log}%
\right) \\
&  =\operatorname{Log}_{1}h,
\end{align*}
every $x\in H_{\leq n}$ satisfies%
\begin{align}
\zeta\left(  x\right)   &  =\left(  \operatorname{Log}_{1}h\right)  \left(
x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{log}), applied to
}f=h\right) \nonumber\\
&  =\sum_{\substack{i\geq1;\\i\leq n}}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}h^{\ast i}\left(  x\right)  +\sum_{\substack{i\geq1;\\i>n}}\dfrac{\left(
-1\right)  ^{i-1}}{i}\underbrace{h^{\ast i}\left(  x\right)  }%
_{\substack{=0\text{ (since }x\in H_{\leq n}\text{ and thus}\\h^{\ast
i}\left(  x\right)  \in h^{\ast i}\left(  H_{\leq n}\right)  =0\text{ (by
Remark~\ref{Remark9}}\\\text{(applied to }f=h\text{), since }i>n\text{),}%
\\\text{so that }h^{\ast i}\left(  x\right)  =0\text{)}}}\nonumber\\
&  =\underbrace{\sum_{\substack{i\geq1;\\i\leq n}}}_{=\sum_{i=1}^{n}}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}\left(  x\right)
+\underbrace{\sum_{\substack{i\geq1;\\i>n}}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}0}_{=0}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast
i}\left(  x\right) \nonumber\\
&  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}h^{\ast\ell
}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }i\text{ as }\ell\right)  .\nonumber
\end{align}
This proves (\ref{232.pf.5}), and thus our step \textbf{c)} is complete.

\textbf{d)} Every $x\in H_{\leq n}$ satisfies%
\begin{equation}
\zeta\left(  x\right)  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}%
}{\ell}\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast
\cdots\ast p_{a_{\ell}}\right)  \left(  x\right)  . \label{232.pf.6}%
\end{equation}


\textit{Proof of (\ref{232.pf.6}).} Let $x\in H^{n}$. Then, for every $\ell
\in\mathbb{N}$, we have%
\begin{align*}
h^{\ast\ell}\left(  x\right)   &  =\underbrace{\left(  p_{1}+p_{2}%
+\cdots+p_{n}\right)  ^{\ast\ell}}_{\substack{=\sum_{\left(  a_{1}%
,a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell
}}p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\\\text{(by the product
rule)}}}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{232.pf.4}%
)}\right) \\
&  =\left(  \sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell}}p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  \left(  x\right)  =\sum_{\left(  a_{1},a_{2}%
,\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}%
}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \left(
x\right)  .
\end{align*}
Now, (\ref{232.pf.5}) becomes%
\begin{align*}
\zeta\left(  x\right)   &  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)
^{\ell-1}}{\ell}\underbrace{h^{\ast\ell}\left(  x\right)  }_{=\sum_{\left(
a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}%
}\right)  \left(  x\right)  }\\
&  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}\sum_{\left(
a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}%
}\right)  \left(  x\right)  .
\end{align*}
This proves (\ref{232.pf.6}).

\textbf{e)} Every $x\in H_{n}$ satisfies\footnote{Note that here we require
$x\in H_{n}$ rather than $x\in H_{\leq n}$.}%
\begin{equation}
\zeta\left(  x\right)  =\left(  \sum_{\ell=1}^{n}\dfrac{\left(  -1\right)
^{\ell-1}}{\ell}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots
+a_{\ell}}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)
\right)  \left(  x\right)  . \label{232.pf.7}%
\end{equation}


\textit{Proof of (\ref{232.pf.7}).} Let $x\in H_{n}$. Then, $\zeta\left(
x\right)  \in\zeta\left(  H_{n}\right)  \subseteq H_{n}$ (since $\zeta$ is
graded (by Theorem~\ref{Theorem231} \textbf{(a)})). Thus,%
\begin{align}
p_{n}\left(  \zeta\left(  x\right)  \right)   &  =\left(  \underbrace{p_{n}%
}_{\substack{=p_{n,H}\\\text{(by the definition of }p_{n}\text{)}}}\mid
_{H_{n}}\right)  \left(  \zeta\left(  x\right)  \right)  =\underbrace{\left(
p_{n,H}\mid_{H_{n}}\right)  }_{\substack{=\operatorname{id}_{H}\mid_{H_{n}%
}\\\text{(by (\ref{221.id}), applied to }V=H\text{)}}}\left(  \zeta\left(
x\right)  \right) \nonumber\\
&  =\left(  \operatorname{id}_{H}\mid_{H_{n}}\right)  \left(  \zeta\left(
x\right)  \right)  =\operatorname{id}_{H}\left(  \zeta\left(  x\right)
\right)  =\zeta\left(  x\right)  . \label{232.pf.8}%
\end{align}


Since $H_{\leq n}=\bigoplus_{\ell=0}^{n}H_{\ell}$ by the definition of
$H_{\leq n}$, we have $H_{n}\subseteq H_{\leq n}$ (since $H_{n}$ is one addend
of the direct sum $\bigoplus_{\ell=0}^{n}H_{\ell}$, and thus $H_{n}%
\subseteq\bigoplus_{\ell=0}^{n}H_{\ell}=H_{\leq n}$).

Now, (\ref{232.pf.8}) yields%
\begin{align*}
\zeta\left(  x\right)   &  =p_{n}\left(  \zeta\left(  x\right)  \right)
=p_{n}\left(  \sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}%
\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{\times\ell}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  \left(  x\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{232.pf.6}), since }x\in
H_{n}\subseteq H_{\leq n}\right) \\
&  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}\sum_{\left(
a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell}}p_{n}\left(  \left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  \left(  x\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }p_{n}\text{ is }k\text{-linear}\right)  .
\end{align*}
Since every $\ell\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell}}\underbrace{p_{n}\left(  \left(  p_{a_{1}%
}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \left(  x\right)  \right)
}_{=\left(  p_{n}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  \right)  \left(  x\right)  }\\
&  =\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell}}\left(  p_{n}\circ\left(  p_{a_{1}}\ast
p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \right)  \left(  x\right) \\
&  =\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}%
}}\underbrace{\left(  p_{n}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  \right)  }_{\substack{=p_{a_{1}}\ast p_{a_{2}}\ast
\cdots\ast p_{a_{\ell}}\\\text{(by Corollary~\ref{Corollary237} \textbf{(b)},
since}\\n=a_{1}+a_{2}+\cdots+a_{\ell}\text{)}}}\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n\neq a_{1}%
+a_{2}+\cdots+a_{\ell}}}\underbrace{\left(  p_{n}\circ\left(  p_{a_{1}}\ast
p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \right)  }%
_{\substack{=0\\\text{(by Corollary~\ref{Corollary237} \textbf{(a)},
since}\\n\neq a_{1}+a_{2}+\cdots+a_{\ell}\text{)}}}\left(  x\right) \\
&  =\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \left(  x\right)
+\underbrace{\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n\neq a_{1}+a_{2}%
+\cdots+a_{\ell}}}0\left(  x\right)  }_{=0}\\
&  =\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \left(  x\right)  ,
\end{align*}
this rewrites as%
\begin{align*}
\zeta\left(  x\right)   &  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)
^{\ell-1}}{\ell}\underbrace{\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}}p_{n}\left(  \left(  p_{a_{1}%
}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \left(  x\right)  \right)
}_{=\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \left(  x\right)
}\\
&  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}\sum
_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \left(  x\right) \\
&  =\left(  \sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}%
\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \right)  \left(
x\right)  .
\end{align*}
This proves (\ref{232.pf.7}).

\textbf{f)} Now an auxiliary result:
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{Whenever }f:H\rightarrow H\text{ and }g:H\rightarrow H\text{ are two
graded maps}\\
\text{satisfying }\left(  f\left(  x\right)  =g\left(  x\right)  \text{ for
all }x\in H_{n}\right)  \text{, then }p_{n}\circ f=p_{n}\circ g
\end{array}
\right)  . \label{232.pf.10}%
\end{equation}


\textit{Proof of (\ref{232.pf.10}).} Let $f:H\rightarrow H$ and
$g:H\rightarrow H$ be two graded maps satisfying $\left(  f\left(  x\right)
=g\left(  x\right)  \text{ for all }x\in H_{n}\right)  $.

Since $f$ is graded, Proposition~\ref{Proposition223} (applied to $V=H$ and
$W=H$) yields that $p_{n,H}\circ f=f\circ p_{n,H}=p_{n,H}\circ f\circ p_{n,H}%
$. Since $p_{n,H}=p_{n}$, this rewrites as $p_{n}\circ f=f\circ p_{n}%
=p_{n}\circ f\circ p_{n}$. The same argument done for $g$ instead of $f$ shows
that $p_{n}\circ g=g\circ p_{n}=p_{n}\circ g\circ p_{n}$.

Since $p_{n}=p_{n,H}$, we have $p_{n}\left(  H\right)  =p_{n,H}\left(
H\right)  =H_{n}$ (by (\ref{221.im}), applied to $V=H$).

For every $y\in H$, we have
\begin{align*}
\left(  f\circ p_{n}\right)  \left(  y\right)   &  =f\left(  p_{n}\left(
y\right)  \right)  =g\left(  p_{n}\left(  y\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{this follows from the assumption that }\left(  f\left(  x\right)
=g\left(  x\right)  \text{ for all }x\in H_{n}\right)  \text{,}\\
\text{applied to }x=p_{n}\left(  y\right)  \text{ (because }p_{n}\left(
y\right)  \in p_{n}\left(  H\right)  =H_{n}\text{)}%
\end{array}
\right) \\
&  =\left(  g\circ p_{n}\right)  \left(  y\right)  .
\end{align*}
Thus, $f\circ p_{n}=g\circ p_{n}$, so that $p_{n}\circ f=f\circ p_{n}=g\circ
p_{n}=p_{n}\circ g$. This proves (\ref{232.pf.10}).

\textbf{g)} We have%
\begin{equation}
\zeta_{n}=\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}%
\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  . \label{232.pf.15}%
\end{equation}


\textit{Proof of (\ref{232.pf.15}).} For every $\ell\in\left\{  1,2,\ldots
,n\right\}  $ and every $\left(  a_{1},a_{2},\ldots,a_{\ell}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}$, the map $p_{a_{1}}\ast
p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}$ is graded\footnote{\textit{Proof.} Let
$\ell\in\left\{  1,2,\ldots,n\right\}  $ and $\left(  a_{1},a_{2}%
,\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}$.
Then, for every $i\in\left\{  1,2,\ldots,\ell\right\}  $, the map $p_{a_{i}}$
is graded and concentrated in degree $a_{i}$ (by
Proposition~\ref{Proposition236} \textbf{(d)}, applied to $a_{i}$ instead of
$n$). Hence, by Proposition~\ref{Proposition236} \textbf{(c)} (applied to
$f_{i}=p_{a_{i}}$), the map $p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell
}}$ is graded and concentrated in degree $a_{1}+a_{2}+\cdots+a_{\ell}$, qed.}.
Hence, the map
\[
\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}\sum
_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)
\]
is graded (since a $k$-linear combination of graded maps is always graded).
Also, $\zeta$ is graded (by Theorem~\ref{Theorem231} \textbf{(a)}). The two
latter facts, along with the fact that
\[
\zeta\left(  x\right)  =\left(  \sum_{\ell=1}^{n}\dfrac{\left(  -1\right)
^{\ell-1}}{\ell}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots
+a_{\ell}}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)
\right)  \left(  x\right)
\]
for all $x\in H_{n}$ (by (\ref{232.pf.7})), show that we can apply
(\ref{232.pf.10}) to $f=\zeta$ and \newline$g=\sum_{\ell=1}^{n}\dfrac{\left(
-1\right)  ^{\ell-1}}{\ell}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}%
+a_{2}+\cdots+a_{\ell}}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  $. As a result, we conclude that%
\[
p_{n}\circ\zeta=p_{n}\circ\left(  \sum_{\ell=1}^{n}\dfrac{\left(  -1\right)
^{\ell-1}}{\ell}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots
+a_{\ell}}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)
\right)  .
\]
Using this identity, we have%
\begin{align*}
\zeta_{n}  &  =p_{n}\circ\zeta\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{231.comm})}\right) \\
&  =p_{n}\circ\left(  \sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}%
}{\ell}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  \right) \\
&  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}\sum
_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}%
}}\underbrace{\left(  p_{n}\circ\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast
p_{a_{\ell}}\right)  \right)  }_{\substack{=p_{a_{1}}\ast p_{a_{2}}\ast
\cdots\ast p_{a_{\ell}}\\\text{(by Corollary~\ref{Corollary237} \textbf{(b)},
since}\\n=a_{1}+a_{2}+\cdots+a_{\ell}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is }k\text{-bilinear}\right) \\
&  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}\sum
_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  .
\end{align*}
This proves (\ref{232.pf.15}).

\textbf{h)} Now let us finally prove the claim of
Proposition~\ref{Proposition232}; this claims states that $p_{n}-\zeta_{n}%
\in\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}  $.

In fact, first we assume that $n\neq0$, because in the case $n=0$ this is very
easy\footnote{\textit{Proof.} Assume that $n=0$. Then, $p_{n}=p_{0}=e_{H,H}$
(by Proposition~\ref{Proposition236} \textbf{(e)}, since $H$ is connected).
Also, by (\ref{232.pf.15}), we have%
\begin{align*}
\zeta_{n}  &  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}%
\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  =\left(
\text{empty sum}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }n=0\right)
\\
&  =0.
\end{align*}
Thus,
\[
\underbrace{p_{n}}_{=e_{H,H}}-\underbrace{\zeta_{n}}_{=0}=e_{H,H}=\left(
\text{unity of the }k\text{-algebra }\mathcal{L}\left(  H,H\right)  \right)
\in\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}  .
\]
Thus, the claim of Proposition~\ref{Proposition232} is proven in the case
$n=0$.}.

Now, it is very easy to see that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for every }\ell\in\left\{  2,3,\ldots,n\right\}  \text{ and every
}\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell}\\
\text{satisfying }n=a_{1}+a_{2}+\cdots+a_{\ell}\text{, we have}\\
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\in\operatorname{AlgGen}%
_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}
\end{array}
\right)  . \label{232.pf.20}%
\end{equation}
\footnote{\textit{Proof of (\ref{232.pf.20}).} Let $\ell\in\left\{
2,3,\ldots,n\right\}  $ and $\left(  a_{1},a_{2},\ldots,a_{\ell}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}$ satisfy $n=a_{1}+a_{2}%
+\cdots+a_{\ell}$. Then, $a_{1}$, $a_{2}$, $\ldots$, $a_{\ell}$ are elements
of the set $\left\{  1,2,\ldots,n\right\}  $ and therefore positive integers.
In other words, $a_{i}$ is a positive integer for every $i\in\left\{
1,2,\ldots,\ell\right\}  $. Also, $\ell\in\left\{  2,3,\ldots,n\right\}  $, so
that $\ell\geq2$. Hence, for every $i\in\left\{  1,2,\ldots,\ell\right\}  $,
the set $\left\{  j\in\left\{  1,2,\ldots,\ell\right\}  ;\ j\neq i\right\}  $
is nonempty. Thus, for every $i\in\left\{  1,2,\ldots,\ell\right\}  $, the sum
$\sum_{\substack{j\in\left\{  1,2,\ldots,\ell\right\}  ;\\j\neq i}}a_{j}$ is
nonempty. Since all addends of this sum are positive (because $a_{1}$, $a_{2}%
$, $\ldots$, $a_{\ell}$ are positive), this sum $\sum_{\substack{j\in\left\{
1,2,\ldots,\ell\right\}  ;\\j\neq i}}a_{j}$ is therefore $>0$. Now, for every
$i\in\left\{  1,2,\ldots,\ell\right\}  $, we have%
\[
n=a_{1}+a_{2}+\cdots+a_{\ell}=\sum_{j\in\left\{  1,2,\ldots,\ell\right\}
}a_{j}=\underbrace{\sum_{\substack{j\in\left\{  1,2,\ldots,\ell\right\}
;\\j\neq i}}a_{j}}_{>0}+a_{i}>a_{i}.
\]
Thus, for every $i\in\left\{  1,2,\ldots,\ell\right\}  $, we have $a_{i}%
\in\left\{  1,2,\ldots,n-1\right\}  $ (since $a_{i}$ is a positive integer
satisfying $a_{i}<n$) and thus $p_{a_{i}}\in\left\{  p_{1},p_{2}%
,\ldots,p_{n-1}\right\}  \subseteq\operatorname{AlgGen}_{k}\left\{
p_{1},p_{2},\ldots,p_{n-1}\right\}  $. Hence, $p_{a_{1}}\ast p_{a_{2}}%
\ast\cdots\ast p_{a_{\ell}}\in\operatorname{AlgGen}_{k}\left\{  p_{1}%
,p_{2},\ldots,p_{n-1}\right\}  $ (because the product of elements of
$\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}  $
always lies in $\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots
,p_{n-1}\right\}  $). This proves (\ref{232.pf.20}).}

Now that we have assumed that $n\neq0$, we have%
\begin{align*}
\zeta_{n}  &  =\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}%
\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{232.pf.15})}\right) \\
&  =\underbrace{\dfrac{\left(  -1\right)  ^{1-1}}{1}}_{=\dfrac{1}{1}%
=1}\underbrace{\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{1}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times1};\\n=a_{1}+a_{2}+\cdots+a_{1}%
}}\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{1}}\right)
}_{\substack{=\sum_{\substack{\left(  a_{1}\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{\times1};\\n=a_{1}}}p_{a_{1}}=\sum_{\substack{a_{1}\in\left\{
1,2,\ldots,n\right\}  ;\\n=a_{1}}}p_{a_{1}}=p_{n}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\ell=2}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}%
}{\ell}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}%
}}\underbrace{\left(  p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}%
}\right)  }_{\substack{\in\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2}%
,\ldots,p_{n-1}\right\}  \\\text{(by (\ref{232.pf.20}))}}}\\
&  \in p_{n}+\underbrace{\sum_{\ell=2}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}%
}{\ell}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}%
}}\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}
}_{\substack{\subseteq\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2}%
,\ldots,p_{n-1}\right\}  \\\text{(since }\operatorname{AlgGen}_{k}\left\{
p_{1},p_{2},\ldots,p_{n-1}\right\}  \text{ is a }k\text{-vector space)}}}\\
&  \subseteq p_{n}+\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2}%
,\ldots,p_{n-1}\right\}  ,
\end{align*}
so that
\[
\zeta_{n}-p_{n}\in\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots
,p_{n-1}\right\}  .
\]
Hence,%
\[
p_{n}-\zeta_{n}=-\underbrace{\left(  \zeta_{n}-p_{n}\right)  }_{\in
\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}  }%
\in-\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}
\subseteq\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}
\]
(since $\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots,p_{n-1}\right\}
$ is a $k$-vector space). This proves Proposition~\ref{Proposition232}.
\end{proof}

The step from Proposition~\ref{Proposition232} to
Proposition~\ref{Proposition233} and Theorem~\ref{Theorem231} is a purely
formal one, and formalized in the following lemma:

\begin{lemma}
\label{Lemma240} Let $k$ be a field. Let $A$ be a $k$-algebra. Let $\left(
x_{1},x_{2},x_{3},\ldots\right)  $ and $\left(  y_{1},y_{2},y_{3}%
,\ldots\right)  $ be two sequences of elements of $A$. Assume that%
\begin{equation}
x_{n}-y_{n}\in\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{n-1}\right\}  \label{240.ass}%
\end{equation}
for every positive integer $n$.

\textbf{(a)} Then, every $n\in\mathbb{N}$ satisfies%
\[
\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{n}\right\}
=\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{n}\right\}  .
\]


\textbf{(b)} Besides,%
\[
\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},x_{3},\ldots\right\}
=\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3},\ldots\right\}  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma240}.]\textbf{(a)} Let us prove Lemma~\ref{Lemma240}
\textbf{(a)} by induction over $n$:

\textit{Induction base:} For $n=0$, both sets $\left\{  x_{1},x_{2}%
,\ldots,x_{n}\right\}  $ and $\left\{  y_{1},y_{2},\ldots,y_{n}\right\}  $ are
empty, and thus $\left\{  x_{1},x_{2},\ldots,x_{n}\right\}  =\left\{
y_{1},y_{2},\ldots,y_{n}\right\}  $, so that $\operatorname{AlgGen}%
_{k}\left\{  x_{1},x_{2},\ldots,x_{n}\right\}  =\operatorname{AlgGen}%
_{k}\left\{  y_{1},y_{2},\ldots,y_{n}\right\}  $. Thus, Lemma~\ref{Lemma240}
\textbf{(a)} is true if $n=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$ be positive. Assume that
Lemma~\ref{Lemma240} \textbf{(a)} holds for $n=N-1$. We now must prove that
Lemma~\ref{Lemma240} \textbf{(a)} also holds for $n=N$.

The assumption (\ref{240.ass}) (applied to $n=N$) yields
\begin{equation}
x_{N}-y_{N}\in\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{N-1}\right\}  . \label{240.pf.1}%
\end{equation}


Since Lemma~\ref{Lemma240} \textbf{(a)} holds for $n=N-1$, we have
$\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{N-1}\right\}
=\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{N-1}\right\}  $.
Thus, (\ref{240.pf.1}) becomes $x_{N}-y_{N}\in\operatorname{AlgGen}%
_{k}\left\{  y_{1},y_{2},\ldots,y_{N-1}\right\}  $. Hence,%
\begin{align}
y_{N}-x_{N}  &  =-\underbrace{\left(  x_{N}-y_{N}\right)  }_{\in
\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{N-1}\right\}  }%
\in-\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{N-1}\right\}
\nonumber\\
&  =\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{N-1}\right\}
\label{240.pf.2}%
\end{align}
(since $\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{N-1}\right\}
$ is a $k$-vector space).

From (\ref{240.pf.1}), we have
\begin{align*}
y_{N}  &  \in\underbrace{x_{N}}_{\in\operatorname{AlgGen}_{k}\left\{
x_{1},x_{2},\ldots,x_{N}\right\}  }-\operatorname{AlgGen}_{k}%
\underbrace{\left\{  x_{1},x_{2},\ldots,x_{N-1}\right\}  }_{\subseteq\left\{
x_{1},x_{2},\ldots,x_{N}\right\}  }\\
&  \subseteq\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{N}\right\}  -\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{N}\right\}  \subseteq\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2}%
,\ldots,x_{N}\right\}
\end{align*}
(since $\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $
is a $k$-vector space).

Now, it is easy to see that every $i\in\left\{  1,2,\ldots,N\right\}  $
satisfies $y_{i}\in\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{N}\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{
1,2,\ldots,N\right\}  $. Then, only two cases are possible:
\par
\textit{Case 1:} We have $i\neq N$.
\par
\textit{Case 2:} We have $i=N$.
\par
First let us consider Case 1. In this case, $i\in\left\{  1,2,\ldots
,N\right\}  $ but $i\neq N$; thus, $i\in\left\{  1,2,\ldots,N-1\right\}  $, so
that%
\begin{align*}
y_{i}  &  \in\left\{  y_{1},y_{2},\ldots,y_{N-1}\right\}  \subseteq
\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{N-1}\right\}
=\operatorname{AlgGen}_{k}\underbrace{\left\{  x_{1},x_{2},\ldots
,x_{N-1}\right\}  }_{\subseteq\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  }\\
&  \subseteq\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{N}\right\}  .
\end{align*}
Thus, $y_{i}\in\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{N}\right\}  $ is proven in Case 1.
\par
In Case 2, we have $i=N$ and thus $y_{i}=y_{N}\in\operatorname{AlgGen}%
_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $. Thus, $y_{i}\in
\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $ is also
proven in Case 2.
\par
Hence, $y_{i}\in\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{N}\right\}  $ is proven in both cases 1 and 2. Since these two cases are
the only ones possible, this yields that $y_{i}\in\operatorname{AlgGen}%
_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $ always holds, qed.}. Thus,
$\left\{  y_{1},y_{2},\ldots,y_{N}\right\}  \subseteq\operatorname{AlgGen}%
_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $. In other words,
$\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $
contains $\left\{  y_{1},y_{2},\ldots,y_{N}\right\}  $ as a subset. Hence,
$\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $ is a
$k$-subalgebra of $A$ containing $\left\{  y_{1},y_{2},\ldots,y_{N}\right\}  $
as a subset\footnote{because $\operatorname{AlgGen}_{k}\left\{  x_{1}%
,x_{2},\ldots,x_{N}\right\}  $ is clearly a $k$-subalgebra of $A$}.

But we know that $\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots
,y_{N}\right\}  $ is the smallest $k$-subalgebra of $A$ containing $\left\{
y_{1},y_{2},\ldots,y_{N}\right\}  $ as a subset. This means that whenever $U$
is a $k$-subalgebra of $A$ containing $\left\{  y_{1},y_{2},\ldots
,y_{N}\right\}  $ as a subset, we must necessarily have $\operatorname{AlgGen}%
_{k}\left\{  y_{1},y_{2},\ldots,y_{N}\right\}  \subseteq U$. Applied to
$U=\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $, this
yields that \newline$\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2}%
,\ldots,y_{N}\right\}  \subseteq\operatorname{AlgGen}_{k}\left\{  x_{1}%
,x_{2},\ldots,x_{N}\right\}  $.

So we have proven $\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots
,y_{N}\right\}  \subseteq\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2}%
,\ldots,x_{N}\right\}  $. But the same argument, with the sequences $\left(
x_{1},x_{2},x_{3},\ldots\right)  $ and $\left(  y_{1},y_{2},y_{3}%
,\ldots\right)  $ interchanged (and using the equality (\ref{240.pf.2})
instead of (\ref{240.pf.1})), shows that \newline$\operatorname{AlgGen}%
_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  \subseteq\operatorname{AlgGen}%
_{k}\left\{  y_{1},y_{2},\ldots,y_{N}\right\}  $.

Combining $\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{N}\right\}
\subseteq\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}  $
and $\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots,x_{N}\right\}
\subseteq\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots,y_{N}\right\}
$, we obtain $\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{N}\right\}  =\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},\ldots
,y_{N}\right\}  $. In other words, Lemma~\ref{Lemma240} \textbf{(a)} holds for
$n=N$. This completes the induction step.

Thus, the induction proof of Lemma~\ref{Lemma240} \textbf{(a)} is complete.

\textbf{(b)} For every positive integer $n$, we have%
\begin{align*}
x_{n}  &  \in\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},\ldots
,x_{n}\right\}  =\operatorname{AlgGen}_{k}\underbrace{\left\{  y_{1}%
,y_{2},\ldots,y_{n}\right\}  }_{\subseteq\left\{  y_{1},y_{2},y_{3}%
,\ldots\right\}  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma~\ref{Lemma240}
\textbf{(a)}}\right) \\
&  \subseteq\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3},\ldots
\right\}  .
\end{align*}
In other words, $\left\{  x_{1},x_{2},x_{3},\ldots\right\}  \subseteq
\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3},\ldots\right\}  $. In
other words, $\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3}%
,\ldots\right\}  $ contains $\left\{  x_{1},x_{2},x_{3},\ldots\right\}  $ as a
subset. Thus, $\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3}%
,\ldots\right\}  $ is a $k$-subalgebra of $A$ containing $\left\{  x_{1}%
,x_{2},x_{3},\ldots\right\}  $ as a subset\footnote{since
$\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3},\ldots\right\}  $ clearly
is a $k$-subalgebra of $A$}.

But we know that $\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},x_{3}%
,\ldots\right\}  $ is the smallest $k$-subalgebra of $A$ containing $\left\{
x_{1},x_{2},x_{3},\ldots\right\}  $ as a subset. This means that whenever $U$
is a $k$-subalgebra of $A$ containing $\left\{  x_{1},x_{2},x_{3}%
,\ldots\right\}  $ as a subset, we must necessarily have
$\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},x_{3},\ldots\right\}  \subseteq
U$. Applied to $U=\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3}%
,\ldots\right\}  $, this yields that $\operatorname{AlgGen}_{k}\left\{
x_{1},x_{2},x_{3},\ldots\right\}  \subseteq\operatorname{AlgGen}_{k}\left\{
y_{1},y_{2},y_{3},\ldots\right\}  $. The same argument, but with the sequences
$\left(  x_{1},x_{2},x_{3},\ldots\right)  $ and $\left(  y_{1},y_{2}%
,y_{3},\ldots\right)  $ interchanged, shows that $\operatorname{AlgGen}%
_{k}\left\{  y_{1},y_{2},y_{3},\ldots\right\}  \subseteq\operatorname{AlgGen}%
_{k}\left\{  x_{1},x_{2},x_{3},\ldots\right\}  $.

Combining $\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},x_{3},\ldots\right\}
\subseteq\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3},\ldots\right\}  $
with $\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3},\ldots\right\}
\subseteq\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},x_{3},\ldots\right\}
$, we obtain $\operatorname{AlgGen}_{k}\left\{  x_{1},x_{2},x_{3}%
,\ldots\right\}  =\operatorname{AlgGen}_{k}\left\{  y_{1},y_{2},y_{3}%
,\ldots\right\}  $. This proves Lemma~\ref{Lemma240} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition233}.]We have%
\[
p_{n}-\zeta_{n}\in\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots
,p_{n-1}\right\}
\]
for every $n\in\mathbb{N}$ (by Proposition~\ref{Proposition232}). Thus,
Lemma~\ref{Lemma240} \textbf{(a)} (applied to $A=\mathcal{L}\left(
H,H\right)  $, $\left(  x_{1},x_{2},x_{3},\ldots\right)  =\left(  p_{1}%
,p_{2},p_{3},\ldots\right)  $ and $\left(  y_{1},y_{2},y_{3},\ldots\right)
=\left(  \zeta_{1},\zeta_{2},\zeta_{3},\ldots\right)  $) yields that every
$n\in\mathbb{N}$ satisfies $\operatorname{AlgGen}_{k}\left\{  p_{1}%
,p_{2},\ldots,p_{n}\right\}  =\operatorname{AlgGen}_{k}\left\{  \zeta
_{1},\zeta_{2},\ldots,\zeta_{n}\right\}  $. This proves
Proposition~\ref{Proposition233}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem231}, second part.]We have already proven
Theorem~\ref{Theorem231} \textbf{(a)} and (\ref{231.comm}). Thus, the only
thing we still need to do is proving (\ref{231.gen}).

We have%
\[
p_{n}-\zeta_{n}\in\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},\ldots
,p_{n-1}\right\}
\]
for every $n\in\mathbb{N}$ (by Proposition~\ref{Proposition232}). Thus,
Lemma~\ref{Lemma240} \textbf{(b)} (applied to $A=\mathcal{L}\left(
H,H\right)  $, $\left(  x_{1},x_{2},x_{3},\ldots\right)  =\left(  p_{1}%
,p_{2},p_{3},\ldots\right)  $ and $\left(  y_{1},y_{2},y_{3},\ldots\right)
=\left(  \zeta_{1},\zeta_{2},\zeta_{3},\ldots\right)  $) yields that
$\operatorname{AlgGen}_{k}\left\{  p_{1},p_{2},p_{3},\ldots\right\}
=\operatorname{AlgGen}_{k}\left\{  \zeta_{1},\zeta_{2},\zeta_{3}%
,\ldots\right\}  $. This proves (\ref{231.gen}). This completes the proof of
Theorem~\ref{Theorem231}.
\end{proof}

\section{\label{S14}The surjectivity part of Cartier-Milnor-Moore}

In this Section \S \ref{S14}, we are going to continue what we began this
paper with: namely, studying cocommutative connected filtered bialgebras and
the properties of $\operatorname*{Log}\operatorname*{id}$ therein. Our first
goal in \S \ref{S14} is to derive the following corollary from
Theorem~\ref{Theorem1}:

\begin{theorem}
\label{Theorem270} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected filtered cocommutative bialgebra over $k$. Then,
$H=\operatorname{AlgGen}_{k}\left(  \operatorname*{Prim}H\right)  $, where
$\operatorname*{Prim}H$ denotes the set of all primitive elements of $H$.
\end{theorem}

Note that this theorem is often worded as follows: A connected filtered
cocommutative bialgebra over a field of characteristic $0$ is always
primitively generated.

Note also that Theorem~\ref{Theorem270} can be seen as one part of the
so-called \textit{Cartier-Milnor-Moore theorem}, which we are going to prove
later (in \S \ref{S31}):

\begin{theorem}
[the Cartier-Milnor-Moore theorem]\label{Theorem271} Let $k$ be a field of
characteristic $0$, and let $H$ be a connected filtered cocommutative
bialgebra over $k$. Let $\operatorname*{Prim}H$ denote the set of all
primitive elements of $H$. For every Lie algebra $\mathfrak{g}$, let $U\left(
\mathfrak{g}\right)  $ denote the universal enveloping algebra of
$\mathfrak{g}$. Then, the canonical $k$-algebra homomorphism $U\left(
\operatorname*{Prim}H\right)  \rightarrow H$ (which is obtained from the
inclusion map $\operatorname*{Prim}H\rightarrow H$ via the universal property
of the universal enveloping algebra) is an isomorphism of $k$-bialgebras.
\end{theorem}

Theorem~\ref{Theorem270} shows that the canonical $k$-algebra homomorphism
$U\left(  \operatorname*{Prim}H\right)  \rightarrow H$ in
Theorem~\ref{Theorem271} is surjective. Injectivity (and being a homomorphism
of $k$-bialgebras) will be proven in \S \ref{S31}.

We are going to show something stronger than Theorem~\ref{Theorem270}:

\begin{proposition}
\label{Proposition272} Let $k$ be a field of characteristic $0$, and let $H$
be a connected filtered cocommutative bialgebra over $k$. Then,
\[
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\left(  \operatorname*{Prim}H\right)
^{i}\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{N}.
\]
\footnote{Recall that $\left(  \operatorname*{Prim}H\right)  ^{i}$ means the
$i$-th power of the subspace $\operatorname*{Prim}H$ of the $k$-algebra $H$ as
explained in Convention~\ref{Convention100}.}
\end{proposition}

Actually, we shall show a somewhat more general fact:

\begin{proposition}
\label{Proposition272a} Let $k$ be a field of characteristic $0$, and let $H$
be a connected filtered bialgebra over $k$. Consider the convolution algebra
$\mathcal{L}\left(  H,H\right)  $. Let $\mathfrak{e}$ denote the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $. Let
$\mathfrak{E}$ be the $k$-vector subspace $\mathfrak{e}\left(  H\right)  $ of
$H$. Then,
\[
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\mathfrak{E}^{i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{N}.
\]
\footnote{Recall that $\mathfrak{E}^{i}$ means the $i$-th power of the
subspace $\mathfrak{E}$ of the $k$-algebra $H$ as explained in
Convention~\ref{Convention100}.}
\end{proposition}

This will be proven by induction over $\ell$, but before we start this proof,
we need a very basic concept:

\begin{definition}
\label{Definition273} Let $k$ be a field. Let $V$ and $W$ be two filtered
$k$-vector spaces. Let $f:V\rightarrow W$ be a $k$-linear map. We say that the
map $f$ \textit{respects the filtration} if and only if $\left(  f\left(
V_{\leq n}\right)  \subseteq W_{\leq n}\text{ for every }n\in\mathbb{N}%
\right)  $.
\end{definition}

First some very basic properties of this notion:

\begin{remark}
\label{Remark274} Let $k$ be a field. Let $V$ and $W$ be two filtered
$k$-vector spaces.

\textbf{(a)} The map $0:V\rightarrow W$ (which sends every $v\in V$ to $0$)
respects the filtration.

\textbf{(b)} If a $k$-linear map $f:V\rightarrow W$ respects the filtration,
and if $\lambda$ is an element of $k$, then $\lambda f$ respects the
filtration as well.

\textbf{(c)} If $f:V\rightarrow W$ and $g:V\rightarrow W$ are two $k$-linear
maps respecting the filtration, then $f+g$ also respects the filtration.

\textbf{(d)} The set of all $k$-linear maps $V\rightarrow W$ respecting the
filtration is a $k$-vector subspace of $\mathcal{L}\left(  V,W\right)  $.
\end{remark}

This remark is a well-known fact from basic algebra and has not much to do
with Hopf algebras. Purely for the sake of completeness, we prove part of it:

\begin{proof}
[Proof of Remark~\ref{Remark274}.]\textbf{(a)} For every $n\in\mathbb{N}$, the
map $0:V\rightarrow W$ satisfies $0\left(  V_{\leq n}\right)  =0\subseteq
W_{\leq n}$. Thus, the map $0:V\rightarrow W$ respects the filtration. This
proves Remark~\ref{Remark274} \textbf{(a)}.

\textbf{(b)} Let $f:V\rightarrow W$ be a $k$-linear map respecting the
filtration. Let $\lambda\in k$. Then, $f\left(  V_{\leq n}\right)  \subseteq
W_{\leq n}$ for every $n\in\mathbb{N}$ (since $f$ respects the filtration).
Now, $\left(  \lambda f\right)  \left(  V_{\leq n}\right)  =\lambda
\underbrace{f\left(  V_{\leq n}\right)  }_{\subseteq W_{\leq n}}%
\subseteq\lambda W_{\leq n}\subseteq W_{\leq n}$ (since $W_{\leq n}$ is a
$k$-vector space) for every $n\in\mathbb{N}$. In other words, $\lambda f$
respects the filtration. This proves Remark~\ref{Remark274} \textbf{(b)}.

\textbf{(c)} Let $f:V\rightarrow W$ and $g:V\rightarrow W$ be two $k$-linear
maps respecting the filtration. Then, $f\left(  V_{\leq n}\right)  \subseteq
W_{\leq n}$ for every $n\in\mathbb{N}$ (since $f$ respects the filtration).
Also, $g\left(  V_{\leq n}\right)  \subseteq W_{\leq n}$ for every
$n\in\mathbb{N}$ (since $g$ respects the filtration). Now,%
\[
\left(  f+g\right)  \left(  V_{\leq n}\right)  \subseteq\underbrace{f\left(
V_{\leq n}\right)  }_{\subseteq W_{\leq n}}+\underbrace{g\left(  V_{\leq
n}\right)  }_{\subseteq W_{\leq n}}\subseteq W_{\leq n}+W_{\leq n}\subseteq
W_{\leq n}%
\]
(since $W_{\leq n}$ is a $k$-vector space) for every $n\in\mathbb{N}$. In
other words, $f+g$ respects the filtration. This proves Remark~\ref{Remark274}
\textbf{(c)}.

\textbf{(d)} The set of all $k$-linear maps $V\rightarrow W$ respecting the
filtration is obviously a subset of $\mathcal{L}\left(  V,W\right)  $. But we
also know that this subset contains $0$ (by Remark~\ref{Remark274}
\textbf{(a)}), is closed under multiplication with every $\lambda\in k$ (by
Remark~\ref{Remark274} \textbf{(b)}), and is closed under addition (by
Remark~\ref{Remark274} \textbf{(c)}). Hence, this subset is a $k$-vector
subspace of $\mathcal{L}\left(  V,W\right)  $. This proves
Remark~\ref{Remark274} \textbf{(d)}.
\end{proof}

\begin{proposition}
\label{Proposition274a} Let $k$ be a field. Let $C$ be a connected filtered
$k$-coalgebra. Let $A$ be a $k$-algebra. If $f\in\mathfrak{g}\left(
C,A\right)  $ and $g\in\mathfrak{g}\left(  C,A\right)  $ are two $k$-linear
maps, then%
\[
\left(  f\ast g\right)  \left(  C_{\leq\ell}\right)  \subseteq\sum_{u=1}%
^{\ell-1}f\left(  C_{\leq u}\right)  g\left(  C_{\leq\ell-u}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every positive }\ell\in\mathbb{N}.
\]

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition274a}.]Since $C$ is filtered, we have
$C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}\subseteq\cdots$.

By the definition of $\mathfrak{g}\left(  C,A\right)  $, we have%
\[
\mathfrak{g}\left(  C,A\right)  =\left\{  f\in\mathcal{L}\left(  C,A\right)
\ \mid\ f\left(  1_{C}\right)  =0\right\}  =\left\{  h\in\mathcal{L}\left(
C,A\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}
\]
(here we renamed $f$ as $h$).

Let $f\in\mathfrak{g}\left(  C,A\right)  $ and $g\in\mathfrak{g}\left(
C,A\right)  $ be two $k$-linear maps. Let $\ell\in\mathbb{N}$ be positive.

Since $f\in\mathfrak{g}\left(  C,A\right)  =\left\{  h\in\mathcal{L}\left(
C,A\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}  $, we have $f\left(
1_{C}\right)  =0$. But Remark~\ref{Remark7l} gives $C_{\leq0}=k\cdot1_{C}$, so
that%
\begin{align*}
f\left(  C_{\leq0}\right)   &  =f\left(  k\cdot1_{C}\right)  =k\cdot
\underbrace{f\left(  1_{C}\right)  }_{=0}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }f\text{ is }k\text{-linear}\right) \\
&  =k\cdot0=0.
\end{align*}
The same argument (applied to $g$ instead of $f$) shows that $g\left(
C_{\leq0}\right)  =0$.

Now,%
\begin{align*}
&  \Delta_{C}\left(  C_{\leq\ell}\right) \\
&  \subseteq\sum_{u=0}^{\ell}C_{\leq u}\otimes C_{\leq\ell-u}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }C\text{ is a filtered coalgebra}%
\right) \\
&  =C_{\leq0}\otimes\underbrace{C_{\leq\ell-0}}_{=C_{\leq\ell}}+\sum
_{u=1}^{\ell-1}C_{\leq u}\otimes C_{\leq\ell-u}+C_{\leq\ell}\otimes
\underbrace{C_{\leq\ell-\ell}}_{=C_{\leq0}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }u=0\text{ and the}\\
\text{addend for }u=\ell\text{ from the sum (since }\ell>0\text{)}%
\end{array}
\right) \\
&  =C_{\leq0}\otimes C_{\leq\ell}+\sum_{u=1}^{\ell-1}C_{\leq u}\otimes
C_{\leq\ell-u}+C_{\leq\ell}\otimes C_{\leq0}.
\end{align*}
Applying the map $f\otimes g$ to both sides of this relation, we obtain%
\begin{align*}
&  \left(  f\otimes g\right)  \left(  \Delta_{C}\left(  C_{\leq\ell}\right)
\right) \\
&  \subseteq\left(  f\otimes g\right)  \left(  C_{\leq0}\otimes C_{\leq\ell
}+\sum_{u=1}^{\ell-1}C_{\leq u}\otimes C_{\leq\ell-u}+C_{\leq\ell}\otimes
C_{\leq0}\right) \\
&  \subseteq\underbrace{\left(  f\otimes g\right)  \left(  C_{\leq0}\otimes
C_{\leq\ell}\right)  }_{\subseteq f\left(  C_{\leq0}\right)  \otimes g\left(
C_{\leq\ell}\right)  }+\sum_{u=1}^{\ell-1}\underbrace{\left(  f\otimes
g\right)  \left(  C_{\leq u}\otimes C_{\leq\ell-u}\right)  }_{\subseteq
f\left(  C_{\leq u}\right)  \otimes g\left(  C_{\leq\ell-u}\right)
}+\underbrace{\left(  f\otimes g\right)  \left(  C_{\leq\ell}\otimes C_{\leq
0}\right)  }_{\subseteq f\left(  C_{\leq\ell}\right)  \otimes g\left(
C_{\leq0}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\otimes g\text{ is }%
k\text{-linear}\right) \\
&  \subseteq\underbrace{f\left(  C_{\leq0}\right)  }_{=0}\otimes g\left(
C_{\leq\ell}\right)  +\sum_{u=1}^{\ell-1}f\left(  C_{\leq u}\right)  \otimes
g\left(  C_{\leq\ell-u}\right)  +f\left(  C_{\leq\ell}\right)  \otimes
\underbrace{g\left(  C_{\leq0}\right)  }_{=0}\\
&  \subseteq\underbrace{0\otimes g\left(  C_{\leq\ell}\right)  }_{=0}%
+\sum_{u=1}^{\ell-1}f\left(  C_{\leq u}\right)  \otimes g\left(  C_{\leq
\ell-u}\right)  +\underbrace{f\left(  C_{\leq\ell}\right)  \otimes0}_{=0}%
=\sum_{u=1}^{\ell-1}f\left(  C_{\leq u}\right)  \otimes g\left(  C_{\leq
\ell-u}\right)  .
\end{align*}
Now, $f\ast g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}$, so that%
\begin{align*}
\left(  f\ast g\right)  \left(  C_{\leq\ell}\right)   &  =\left(  \mu_{A}%
\circ\left(  f\otimes g\right)  \circ\Delta_{C}\right)  \left(  C_{\leq\ell
}\right)  =\mu_{A}\left(  \underbrace{\left(  f\otimes g\right)  \left(
\Delta_{C}\left(  C_{\leq\ell}\right)  \right)  }_{\subseteq\sum_{u=1}%
^{\ell-1}f\left(  C_{\leq u}\right)  \otimes g\left(  C_{\leq\ell-u}\right)
}\right) \\
&  \subseteq\mu_{A}\left(  \sum_{u=1}^{\ell-1}f\left(  C_{\leq u}\right)
\otimes g\left(  C_{\leq\ell-u}\right)  \right) \\
&  \subseteq\sum_{u=1}^{\ell-1}\underbrace{\mu_{A}\left(  f\left(  C_{\leq
u}\right)  \otimes g\left(  C_{\leq\ell-u}\right)  \right)  }%
_{\substack{=f\left(  C_{\leq u}\right)  g\left(  C_{\leq\ell-u}\right)
\\\text{(by (\ref{104.UV}), applied to}\\U=f\left(  C_{\leq u}\right)  \text{
and }V=g\left(  C_{\leq\ell-u}\right)  \text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mu_{A}\text{ is }k\text{-linear}\right) \\
&  =\sum_{u=1}^{\ell-1}f\left(  C_{\leq u}\right)  g\left(  C_{\leq\ell
-u}\right)  .
\end{align*}
This proves Proposition~\ref{Proposition274a}.
\end{proof}

Now we shall show some facts about maps respecting the filtration:

\begin{proposition}
\label{Proposition275} Let $k$ be a field. Let $C$ be a filtered
$k$-coalgebra. Let $A$ be a filtered $k$-algebra.

\textbf{(a)} The map $e_{C,A}:C\rightarrow A$ respects the filtration.

\textbf{(b)} If $f:C\rightarrow A$ and $g:C\rightarrow A$ are two $k$-linear
maps respecting the filtration, then $f\ast g$ also respects the filtration.

\textbf{(c)} If $f:C\rightarrow A$ is a $k$-linear map respecting the
filtration, and if $n$ is a nonnegative integer, then $f^{\ast n}$ also
respects the filtration.

\textbf{(d)} If the filtered $k$-coalgebra $C$ is connected, and if
$f\in\mathfrak{g}\left(  C,A\right)  $ and $g\in\mathfrak{g}\left(
C,A\right)  $ are two $k$-linear maps respecting the filtration, then%
\begin{equation}
\left(  f\ast g\right)  \left(  C_{\leq\ell}\right)  \subseteq\sum_{u=1}%
^{\ell-1}A_{\leq u}A_{\leq\ell-u}\ \ \ \ \ \ \ \ \ \ \text{for every positive
}\ell\in\mathbb{N}. \label{275.c}%
\end{equation}

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition275}.]Since $C$ is filtered, we have
$C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}\subseteq\cdots$.

Since $A$ is filtered, we have $A_{\leq0}\subseteq A_{\leq1}\subseteq
A_{\leq2}\subseteq\cdots$. Since $A$ is a filtered $k$-algebra, we also have
$1_{A}\in A_{\leq0}$ and
\begin{equation}
A_{\leq a}A_{\leq b}\subseteq A_{\leq a+b}\ \ \ \ \ \ \ \ \ \ \text{for any
}a\in\mathbb{N}\text{ and }b\in\mathbb{N}. \label{275.pf.0}%
\end{equation}


\textbf{(a)} Now, for every $x\in C$, we have%
\begin{align*}
\underbrace{e_{C,A}}_{=\eta_{A}\circ\varepsilon_{C}}\left(  x\right)   &
=\left(  \eta_{A}\circ\varepsilon_{C}\right)  \left(  x\right)  =\eta
_{A}\left(  \varepsilon_{C}\left(  x\right)  \right)  =\varepsilon_{C}\left(
x\right)  \cdot\underbrace{1_{A}}_{\in A_{\leq0}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\eta_{A}\right) \\
&  \in\varepsilon_{C}\left(  x\right)  \cdot A_{\leq0}\subseteq A_{\leq
0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A_{\leq0}\text{ is a }%
k\text{-vector space}\right)  .
\end{align*}
In other words, $e_{C,A}\left(  C\right)  \subseteq A_{\leq0}$. Hence, for
every $n\in\mathbb{N}$, we have $e_{C,A}\left(  \underbrace{C_{\leq n}%
}_{\subseteq C}\right)  \subseteq e_{C,A}\left(  C\right)  \subseteq A_{\leq
0}\subseteq A_{\leq n}$ (since $0\leq n$ and $A_{\leq0}\subseteq A_{\leq
1}\subseteq A_{\leq2}\subseteq\cdots$). In other words, $e_{C,A}$ respects the
filtration. This proves Proposition~\ref{Proposition275} \textbf{(a)}.

\textbf{(b)} Let $f:C\rightarrow A$ and $g:C\rightarrow A$ be two $k$-linear
maps respecting the filtration. Then,
\begin{equation}
f\left(  C_{\leq n}\right)  \subseteq A_{\leq n}\ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N} \label{275.pf.1}%
\end{equation}
(since $f$ respects the filtration), and%
\begin{equation}
g\left(  C_{\leq n}\right)  \subseteq A_{\leq n}\ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N} \label{275.pf.2}%
\end{equation}
(since $g$ respects the filtration).

Let $n\in\mathbb{N}$. Let $x\in C_{\leq n}$ be arbitrary. Then,
\[
\Delta\left(  x\right)  \in\Delta\left(  C_{\leq n}\right)  \subseteq
\sum_{u=0}^{n}C_{\leq u}\otimes C_{\leq n-u}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }C\text{ is a filtered coalgebra}\right)  ,
\]
so that%
\begin{align*}
\left(  f\otimes g\right)  \left(  \Delta\left(  x\right)  \right)   &
\in\left(  f\otimes g\right)  \left(  \sum_{u=0}^{n}C_{\leq u}\otimes C_{\leq
n-u}\right)  \subseteq\sum_{u=0}^{n}\underbrace{\left(  f\otimes g\right)
\left(  C_{\leq u}\otimes C_{\leq n-u}\right)  }_{\subseteq f\left(  C_{\leq
u}\right)  \otimes g\left(  C_{\leq n-u}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\otimes g\text{ is }%
k\text{-linear}\right) \\
&  \subseteq\sum_{u=0}^{n}\underbrace{f\left(  C_{\leq u}\right)
}_{\substack{\subseteq A_{\leq u}\\\text{(by (\ref{275.pf.1}), applied to
}u\\\text{instead of }n\text{)}}}\otimes\underbrace{g\left(  C_{\leq
n-u}\right)  }_{\substack{\subseteq A_{\leq n-u}\\\text{(by (\ref{275.pf.2}),
applied to }n-u\\\text{instead of }n\text{)}}}\subseteq\sum_{u=0}^{n}A_{\leq
u}\otimes A_{\leq n-u}.
\end{align*}
Now, $f\ast g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}$ (by the
definition of convolution), so that%
\begin{align*}
\left(  f\ast g\right)  \left(  x\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{C}\right)  \left(  x\right)  =\mu_{A}\left(
\left(  f\otimes g\right)  \underbrace{\left(  \Delta_{C}\left(  x\right)
\right)  }_{=\Delta\left(  x\right)  }\right) \\
&  =\mu_{A}\left(  \underbrace{\left(  f\otimes g\right)  \left(
\Delta\left(  x\right)  \right)  }_{\in\sum_{u=0}^{n}A_{\leq u}\otimes A_{\leq
n-u}}\right)  \in\mu_{A}\left(  \sum_{u=0}^{n}A_{\leq u}\otimes A_{\leq
n-u}\right)  \subseteq\sum_{u=0}^{n}\underbrace{\mu_{A}\left(  A_{\leq
u}\otimes A_{\leq n-u}\right)  }_{\substack{=A_{\leq u}A_{\leq n-u}\\\text{(by
(\ref{104.UV}), applied to}\\U=A_{\leq u}\text{ and }V=A_{\leq n-u}\text{)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{A}\text{ is }k\text{-linear}%
\right) \\
&  =\sum_{u=0}^{n}\underbrace{A_{\leq u}A_{\leq n-u}}_{\substack{\subseteq
A_{\leq u+\left(  n-u\right)  }\\\text{(by (\ref{275.pf.0}), applied to
}a=u\text{ and }b=n-u\text{)}}}\subseteq\sum_{u=0}^{n}\underbrace{A_{\leq
u+\left(  n-u\right)  }}_{=A_{\leq n}}=\sum_{u=0}^{n}A_{\leq n}\subseteq
A_{\leq n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }A_{\leq n}\text{ is a
}k\text{-vector space}\right)  .
\end{align*}


Now forget that we fixed $x$. We thus have proven that $\left(  f\ast
g\right)  \left(  x\right)  \in A_{\leq n}$ for every $x\in C_{\leq n}$. In
other words, $\left(  f\ast g\right)  \left(  C_{\leq n}\right)  \subseteq
A_{\leq n}$.

Now forget that we fixed $n$. We thus have proven that $\left(  f\ast
g\right)  \left(  C_{\leq n}\right)  \subseteq A_{\leq n}$ for every
$n\in\mathbb{N}$. In other words, $f\ast g$ respects the filtration. This
proves Proposition~\ref{Proposition275} \textbf{(b)}.

\textbf{(c)} We are going to prove Proposition~\ref{Proposition275}
\textbf{(c)} by induction over $n$:

\textit{Induction base:} For any $k$-linear map $f:C\rightarrow A$, the map
$f^{\ast0}$ respects the filtration (since $f^{\ast0}=e_{C,A}$, and since
$e_{C,A}$ respects the filtration by Proposition~\ref{Proposition275}
\textbf{(a)}). In other words, Proposition~\ref{Proposition275} \textbf{(c)}
holds for $n=0$. This completes the induction base.

\textit{Induction base:} Let $N\in\mathbb{N}$. Assume that
Proposition~\ref{Proposition275} \textbf{(c)} holds for $n=N$. We now must
prove that Proposition~\ref{Proposition275} \textbf{(c)} holds for $n=N+1$.

Let $f:C\rightarrow A$ be a $k$-linear map respecting the filtration. Then,
$f^{\ast N}$ respects the filtration (by Proposition~\ref{Proposition275}
\textbf{(c)}). Hence, $f\ast f^{\ast N}$ respects the filtration (by
Proposition~\ref{Proposition275} \textbf{(b)}, applied to $g=f^{\ast N}$).
Since $f\ast f^{\ast N}=f^{\ast\left(  N+1\right)  }$, this rewrites as
follows: The map $f^{\ast\left(  N+1\right)  }$ respects the filtration. In
other words, Proposition~\ref{Proposition275} \textbf{(c)} holds for $n=N+1$.
This completes the induction step.

The induction proof of Proposition~\ref{Proposition275} \textbf{(c)} is thus complete.

\textbf{(d)} Assume that the filtered $k$-coalgebra $C$ is connected. Let
$f\in\mathfrak{g}\left(  C,A\right)  $ and $g\in\mathfrak{g}\left(
C,A\right)  $ be two $k$-linear maps respecting the filtration. Let $\ell
\in\mathbb{N}$ be positive.

Note that we have (\ref{275.pf.1}) (since $f$ respects the filtration) and
(\ref{275.pf.2}) (since $g$ respects the filtration).

Proposition~\ref{Proposition274a} yields%
\[
\left(  f\ast g\right)  \left(  C_{\leq\ell}\right)  \subseteq\sum_{u=1}%
^{\ell-1}\underbrace{f\left(  C_{\leq u}\right)  }_{\substack{\subseteq
A_{\leq u}\\\text{(by (\ref{275.pf.1}), applied to }u\\\text{instead of
}n\text{)}}}\underbrace{g\left(  C_{\leq\ell-u}\right)  }_{\substack{\subseteq
A_{\leq\ell-u}\\\text{(by (\ref{275.pf.2}), applied to }\ell-u\\\text{instead
of }n\text{)}}}\subseteq\sum_{u=1}^{\ell-1}A_{\leq u}A_{\leq\ell-u}.
\]
This proves Proposition~\ref{Proposition275} \textbf{(d)}.
\end{proof}

On a sidenote, in the above proof of Proposition~\ref{Proposition274a}, we saw
that
\[
\Delta_{C}\left(  C_{\leq\ell}\right)  \subseteq C_{\leq0}\otimes C_{\leq\ell
}+\sum_{u=1}^{\ell-1}C_{\leq u}\otimes C_{\leq\ell-u}+C_{\leq\ell}\otimes
C_{\leq0}\ \ \ \ \ \ \ \ \ \ \text{for every positive }\ell\in\mathbb{N}%
\]
when $C$ is a connected filtered $k$-coalgebra. This is a useful fact, but
more useful is its following strengthening:

\begin{proposition}
\label{Proposition276} Let $k$ be a field. Let $C$ be a connected filtered
$k$-coalgebra. Let $\ell\in\mathbb{N}$ be positive. Then,%
\[
\Delta_{C}\left(  x\right)  \in x\otimes1_{C}+1_{C}\otimes x+\sum_{u=1}%
^{\ell-1}C_{\leq u}\otimes C_{\leq\ell-u}\ \ \ \ \ \ \ \ \ \ \text{for every
}x\in C_{\leq\ell}\cap\operatorname*{Ker}\varepsilon\text{.}%
\]

\end{proposition}

This proposition is often used (e. g., it is Proposition II.2.1 in \cite{1};
however, the statement given in \cite{1} is slightly wrong, and the proof is
not a ``Straightforward adaptation of proof of proposition II.1.1'' as
asserted in \cite{1}). Let us give a proof of this proposition here, although
we are not going to use it until much further in this paper.

\begin{proof}
[Proof of Proposition~\ref{Proposition276}.]Since $C$ is filtered, we have
$C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}\subseteq\cdots$.

By Remark~\ref{Remark7l}, we have $C_{\leq0}=k\cdot1_{C}$. In particular,
$1_{C}\in k\cdot1_{C}=C_{\leq0}$. Also, since $1_{C}$ was defined as $\left(
\varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  $, we have
$1=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  \left(  1_{C}\right)
=\varepsilon_{C}\left(  1_{C}\right)  =\varepsilon\left(  1_{C}\right)  $.

For every $n\in\mathbb{N}$, let $C_{\leq n+}$ denote the $k$-vector subspace
$C_{\leq n}\cap\operatorname*{Ker}\varepsilon$ of $C_{\leq n}$. Then, every
$n\in\mathbb{N}$ satisfies%
\begin{equation}
C_{\leq n}=C_{\leq n+}+C_{\leq0} \label{276.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{276.pf.1}).} Let $n\in\mathbb{N}$. Then,
$0\leq n$, so that $C_{\leq0}\subseteq C_{\leq n}$ (since $C_{\leq0}\subseteq
C_{\leq1}\subseteq C_{\leq2}\subseteq\cdots$). Hence, $\underbrace{C_{\leq
n+}}_{\subseteq C_{\leq n}}+\underbrace{C_{\leq0}}_{\subseteq C_{\leq n}%
}\subseteq C_{\leq n}+C_{\leq n}\subseteq C_{\leq n}$ (since $C_{\leq n}$ is a
$k$-vector space).
\par
On the other hand, every $x\in C_{\leq n}$ satisfies $\underbrace{x}_{\in
C_{\leq n}}-\varepsilon\left(  x\right)  \cdot\underbrace{1_{C}}_{\in
k\cdot1_{C}=C_{\leq0}\subseteq C_{\leq n}}\in C_{\leq n}-\varepsilon\left(
x\right)  \cdot C_{\leq n}\subseteq C_{\leq n}$ (since $C_{\leq n}$ is a
$k$-vector space) and $x-\varepsilon\left(  x\right)  \cdot1_{C}%
\in\operatorname*{Ker}\varepsilon$ (since
\begin{align*}
\varepsilon\left(  x-\varepsilon\left(  x\right)  \cdot1_{C}\right)   &
=\varepsilon\left(  x\right)  -\varepsilon\left(  x\right)  \cdot
\underbrace{\varepsilon\left(  1_{C}\right)  }_{=1}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\varepsilon\text{ is }k\text{-linear}\right) \\
&  =\varepsilon\left(  x\right)  -\varepsilon\left(  x\right)  =0
\end{align*}
). Hence, every $x\in C_{\leq n}$ satisfies%
\begin{align*}
x-\varepsilon\left(  x\right)  \cdot1_{C}  &  \in C_{\leq n}\cap
\operatorname*{Ker}\varepsilon\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}x-\varepsilon\left(  x\right)  \cdot1_{C}\in C_{\leq n}\text{ and
}x-\varepsilon\left(  x\right)  \cdot1_{C}\in\operatorname*{Ker}%
\varepsilon\right) \\
&  =C_{\leq n+}.
\end{align*}
Thus, every $x\in C_{\leq n}$ satisfies%
\[
x=\underbrace{\varepsilon\left(  x\right)  \cdot1_{C}}_{\in k\cdot
1_{C}=C_{\leq0}}+\underbrace{\left(  x-\varepsilon\left(  x\right)  \cdot
1_{C}\right)  }_{\in C_{\leq n+}}\in C_{\leq0}+C_{\leq n+}=C_{\leq n+}%
+C_{\leq0}.
\]
In other words, $C_{\leq n}\subseteq C_{\leq n+}+C_{\leq0}$. Combined with
$C_{\leq n+}+C_{\leq0}\subseteq C_{\leq n}$, this yields $C_{\leq n}=C_{\leq
n+}+C_{\leq0}$, qed.}.

For each $u\in\left\{  1,2,\ldots,\ell-1\right\}  $, we have
\begin{align}
&  \underbrace{C_{\leq u}}_{\substack{=C_{\leq u+}+C_{\leq0}\\\text{(by
(\ref{276.pf.1}), applied to }n=u\text{)}}}\otimes\underbrace{C_{\leq\ell-u}%
}_{\substack{=C_{\leq\left(  \ell-u\right)  +}+C_{\leq0}\\\text{(by
(\ref{276.pf.1}), applied to }n=\ell-u\text{)}}}\nonumber\\
&  =\left(  C_{\leq u+}+C_{\leq0}\right)  \otimes\left(  C_{\leq\left(
\ell-u\right)  +}+C_{\leq0}\right) \nonumber\\
&  =C_{\leq u+}\otimes\left(  \underbrace{C_{\leq\left(  \ell-u\right)
+}+C_{\leq0}}_{=C_{\leq0}+C_{\leq\left(  \ell-u\right)  +}}\right)  +C_{\leq
0}\otimes\left(  C_{\leq\left(  \ell-u\right)  +}+C_{\leq0}\right) \nonumber\\
&  =\underbrace{C_{\leq u+}\otimes\left(  C_{\leq0}+C_{\leq\left(
\ell-u\right)  +}\right)  }_{\substack{=C_{\leq u+}\otimes C_{\leq0}+C_{\leq
u+}\otimes C_{\leq\left(  \ell-u\right)  +}\\\text{(since the tensor product
is distributive)}}}+\underbrace{C_{\leq0}\otimes\left(  C_{\leq\left(
\ell-u\right)  +}+C_{\leq0}\right)  }_{\substack{=C_{\leq0}\otimes
C_{\leq\left(  \ell-u\right)  +}+C_{\leq0}\otimes C_{\leq0}\\\text{(since the
tensor product is distributive)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the tensor product is
distributive}\right) \nonumber\\
&  =\underbrace{C_{\leq u+}}_{\substack{\subseteq C_{\leq u}\subseteq
C_{\leq\ell}\\\text{(since }u\leq\ell\\\text{and }C_{\leq0}\subseteq C_{\leq
1}\subseteq C_{\leq2}\subseteq\cdots\text{)}}}\otimes C_{\leq0}+C_{\leq
u+}\otimes C_{\leq\left(  \ell-u\right)  +}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +C_{\leq0}\otimes\underbrace{C_{\leq\left(
\ell-u\right)  +}}_{\substack{\subseteq C_{\leq\ell-u}\subseteq C_{\leq\ell
}\\\text{(since }\ell-u\leq\ell\\\text{and }C_{\leq0}\subseteq C_{\leq
1}\subseteq C_{\leq2}\subseteq\cdots\text{)}}}+C_{\leq0}\otimes
\underbrace{C_{\leq0}}_{\substack{\subseteq C_{\leq\ell}\\\text{(since }%
0\leq\ell\\\text{and }C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}%
\subseteq\cdots\text{)}}}\nonumber\\
&  \subseteq C_{\leq\ell}\otimes C_{\leq0}+C_{\leq u+}\otimes C_{\leq\left(
\ell-u\right)  +}+\underbrace{C_{\leq0}\otimes C_{\leq\ell}+C_{\leq0}\otimes
C_{\leq\ell}}_{\substack{\subseteq C_{\leq0}\otimes C_{\leq\ell}\\\text{(since
}C_{\leq0}\otimes C_{\leq\ell}\text{ is a }k\text{-vector space)}}}\nonumber\\
&  \subseteq C_{\leq\ell}\otimes C_{\leq0}+C_{\leq u+}\otimes C_{\leq\left(
\ell-u\right)  +}+C_{\leq0}\otimes C_{\leq\ell}. \label{276.pf.inc-for-each-u}%
\end{align}


Now, since $C$ is a filtered coalgebra, we have%
\begin{align*}
&  \Delta\left(  C_{\leq\ell}\right) \\
&  \subseteq\sum_{u=0}^{\ell}C_{\leq u}\otimes C_{\leq\ell-u}=C_{\leq0}\otimes
C_{\leq\ell-0}+\sum_{u=1}^{\ell-1}C_{\leq u}\otimes C_{\leq\ell-u}+C_{\leq
\ell}\otimes C_{\leq\ell-\ell}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell>0\right) \\
&  =C_{\leq0}\otimes\underbrace{C_{\leq\ell-0}}_{=C_{\leq\ell}}+C_{\leq\ell
}\otimes\underbrace{C_{\leq\ell-\ell}}_{=C_{\leq0}} +\sum_{u=1}^{\ell-1}
\underbrace{C_{\leq u}\otimes C_{\leq\ell-u} }_{\substack{\subseteq
C_{\leq\ell}\otimes C_{\leq0}+C_{\leq u+}\otimes C_{\leq\left(  \ell-u\right)
+}+C_{\leq0}\otimes C_{\leq\ell}\\\text{(by (\ref{276.pf.inc-for-each-u}))}%
}}\\
&  \subseteq C_{\leq0}\otimes C_{\leq\ell}+C_{\leq\ell}\otimes C_{\leq
0}+\underbrace{\sum_{u=1}^{\ell-1}\left(  C_{\leq\ell}\otimes C_{\leq
0}+C_{\leq u+}\otimes C_{\leq\left(  \ell-u\right)  +}+C_{\leq0}\otimes
C_{\leq\ell}\right)  }_{=\sum_{u=1}^{\ell-1}C_{\leq\ell}\otimes C_{\leq0}%
+\sum_{u=1}^{\ell-1}C_{\leq u+}\otimes C_{\leq\left(  \ell-u\right)  +}%
+\sum_{u=1}^{\ell-1}C_{\leq0}\otimes C_{\leq\ell}}\\
&  =C_{\leq0}\otimes C_{\leq\ell}+C_{\leq\ell}\otimes C_{\leq0}+\sum
_{u=1}^{\ell-1}C_{\leq\ell}\otimes C_{\leq0}+\sum_{u=1}^{\ell-1}C_{\leq
u+}\otimes C_{\leq\left(  \ell-u\right)  +}+\sum_{u=1}^{\ell-1}C_{\leq
0}\otimes C_{\leq\ell}\\
&  =\underbrace{C_{\leq0}\otimes C_{\leq\ell}+\sum_{u=1}^{\ell-1}C_{\leq
0}\otimes C_{\leq\ell}}_{\substack{\subseteq C_{\leq0}\otimes C_{\leq\ell
}\\\text{(since }C_{\leq0}\otimes C_{\leq\ell}\text{ is a }k\text{-vector
space)}}}+\underbrace{C_{\leq\ell}\otimes C_{\leq0}+\sum_{u=1}^{\ell-1}%
C_{\leq\ell}\otimes C_{\leq0}}_{\substack{\subseteq C_{\leq\ell}\otimes
C_{\leq0}\\\text{(since }C_{\leq\ell}\otimes C_{\leq0}\text{ is a
}k\text{-vector space)}}}+\sum_{u=1}^{\ell-1}C_{\leq u+}\otimes C_{\leq\left(
\ell-u\right)  +}\\
&  \subseteq C_{\leq0}\otimes C_{\leq\ell}+C_{\leq\ell}\otimes C_{\leq0}%
+\sum_{u=1}^{\ell-1}C_{\leq u+}\otimes C_{\leq\left(  \ell-u\right)  +}.
\end{align*}


It is easy to see that%
\begin{equation}
\text{for every }U\in C_{\leq0}\otimes C_{\leq\ell}\text{, there exists some
}u\in C_{\leq\ell}\text{ such that }U=1_{C}\otimes u\text{.} \label{276.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{276.pf.2}).} Let $\xi:C_{\leq\ell}\rightarrow
C_{\leq0}\otimes C_{\leq\ell}$ be the map which sends every $t\in C_{\leq\ell
}$ to $1_{C}\otimes t\in C_{\leq0}\otimes C_{\leq\ell}$ (this is well-defined
since every $t\in C_{\leq\ell}$ satisfies $\underbrace{1_{C}}_{\in C_{\leq0}%
}\otimes\underbrace{t}_{\in C_{\leq\ell}}\in C_{\leq0}\otimes C_{\leq\ell}$).
Let $\operatorname*{kan}:C_{\leq\ell}\rightarrow k\otimes C_{\leq\ell}$ be the
canonical isomorphism which sends every $t\in C_{\leq\ell}$ to $1\otimes t$.
\par
Since $C$ is a connected coalgebra, $\varepsilon_{C}\mid_{C_{\leq0}}:C_{\leq
0}\rightarrow k$ is an isomorphism. Hence, $\left(  \varepsilon_{C}%
\mid_{C_{\leq0}}\right)  \otimes\operatorname{id}_{C_{\leq\ell}}:C_{\leq
0}\otimes C_{\leq\ell}\rightarrow k\otimes C_{\leq\ell}$ is an isomorphism as
well.
\par
Now, every $t\in C_{\leq\ell}$ satisfies%
\begin{align*}
\left(  \left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  \otimes
\operatorname{id}_{C_{\leq\ell}}\right)  \circ\xi\right)  \left(  t\right)
&  =\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  \otimes
\operatorname{id}_{C_{\leq\ell}}\right)  \underbrace{\left(  \xi\left(
t\right)  \right)  }_{\substack{=1_{C}\otimes t\\\text{(by the definition of
}\xi\text{)}}}=\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
\otimes\operatorname{id}_{C_{\leq\ell}}\right)  \left(  1_{C}\otimes t\right)
\\
&  =\underbrace{\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  \left(
1_{C}\right)  }_{=1}\otimes\underbrace{\operatorname{id}_{C_{\leq\ell}}\left(
t\right)  }_{=t}=1\otimes t=\operatorname*{kan}t\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{kan}t=1\otimes
t\text{ by the definition of }\operatorname*{kan}\right)  .
\end{align*}
Hence, $\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  \otimes
\operatorname{id}_{C_{\leq\ell}}\right)  \circ\xi=\operatorname*{kan}$. Since
$\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  \otimes\operatorname{id}%
_{C_{\leq\ell}}$ and $\operatorname*{kan}$ are isomorphisms, this yields that
$\xi$ is an isomorphism. Thus, the map $\xi^{-1}$ is well-defined.
\par
Now, for every $U\in C_{\leq0}\otimes C_{\leq\ell}$, we have%
\[
U=\xi\left(  \xi^{-1}\left(  U\right)  \right)  =1_{C}\otimes\xi^{-1}\left(
U\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\xi\right)
.
\]
Hence, for every $U\in C_{\leq0}\otimes C_{\leq\ell}$, there exists some $u\in
C_{\leq\ell}$ such that $U=1_{C}\otimes u$ (namely, $u=\xi^{-1}\left(
U\right)  $). This proves (\ref{276.pf.2}).}

The same argument, but with the two tensorands transposed, shows that%
\begin{equation}
\text{for every }V\in C_{\leq\ell}\otimes C_{\leq0}\text{, there exists some
}v\in C_{\leq\ell}\text{ such that }V=v\otimes1_{C}\text{.} \label{276.pf.3}%
\end{equation}


Now, let $x\in C_{\leq\ell}\cap\operatorname*{Ker}\varepsilon$. Then, $x\in
C_{\leq\ell}\cap\operatorname*{Ker}\varepsilon\subseteq C_{\leq\ell}$ and
$x\in C_{\leq\ell}\cap\operatorname*{Ker}\varepsilon\subseteq
\operatorname*{Ker}\varepsilon$. From $x\in\operatorname*{Ker}\varepsilon$, we
conclude that $\varepsilon\left(  x\right)  =0$.

Since $x\in C_{\leq\ell}$, we have%
\[
\Delta\left(  x\right)  \in\Delta\left(  C_{\leq\ell}\right)  \subseteq
C_{\leq0}\otimes C_{\leq\ell}+C_{\leq\ell}\otimes C_{\leq0}+\sum_{u=1}%
^{\ell-1}C_{\leq u+}\otimes C_{\leq\left(  \ell-u\right)  +}.
\]
Thus, there exist some $U\in C_{\leq0}\otimes C_{\leq\ell}$, some $V\in
C_{\leq\ell}\otimes C_{\leq0}$ and some $W\in\sum_{u=1}^{\ell-1}C_{\leq
u+}\otimes C_{\leq\left(  \ell-u\right)  +}$ such that $\Delta\left(
x\right)  =U+V+W$. Consider these $U$, $V$ and $W$.

According to (\ref{276.pf.2}), there exists some $u\in C_{\leq\ell}$ such that
$U=1_{C}\otimes u$. Consider this $u$.

According to (\ref{276.pf.3}), there exists some $v\in C_{\leq\ell}$ such that
$V=v\otimes1_{C}$. Consider this $v$.

Now,%
\begin{equation}
\Delta\left(  x\right)  =\underbrace{U}_{=1_{C}\otimes u}+\underbrace{V}%
_{=v\otimes1_{C}}+W=1_{C}\otimes u+v\otimes1_{C}+W. \label{276.pf.5}%
\end{equation}
But
\begin{equation}
\text{every }n\in\mathbb{N}\text{ satisfies }\varepsilon\left(  C_{\leq
n+}\right)  =0 \label{276.pf.6}%
\end{equation}
(since $C_{\leq n+}=C_{\leq n}\cap\operatorname*{Ker}\varepsilon
\subseteq\operatorname*{Ker}\varepsilon$).

Let $\operatorname{kan}_{1}$ be the canonical isomorphism $C\otimes
k\rightarrow C$ which maps $c\otimes\lambda$ to $\lambda c$ for every $\left(
c,\lambda\right)  \in C\times k$. By the axioms of a coalgebra,
$\operatorname{kan}_{1}\circ\left(  \operatorname*{id}\otimes\varepsilon
\right)  \circ\Delta=\operatorname{id}_{C}$.

Let $\operatorname{kan}_{2}$ be the canonical isomorphism $k\otimes
C\rightarrow C$ which maps $\lambda\otimes c$ to $\lambda c$ for every
$\left(  c,\lambda\right)  \in C\times k$. By the axioms of a coalgebra,
$\operatorname{kan}_{2}\circ\left(  \varepsilon\otimes\operatorname*{id}%
\right)  \circ\Delta=\operatorname{id}_{C}$.

Now,%
\begin{align*}
\left(  \operatorname*{id}\otimes\varepsilon\right)  \left(  W\right)   &
\in\left(  \operatorname*{id}\otimes\varepsilon\right)  \left(  \sum
_{u=1}^{\ell-1}C_{\leq u+}\otimes C_{\leq\left(  \ell-u\right)  +}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }W\in\sum_{u=1}^{\ell-1}C_{\leq
u+}\otimes C_{\leq\left(  \ell-u\right)  +}\right) \\
&  \in\sum_{u=1}^{\ell-1}\underbrace{\left(  \operatorname*{id}\otimes
\varepsilon\right)  \left(  C_{\leq u+}\otimes C_{\leq\left(  \ell-u\right)
+}\right)  }_{\subseteq\operatorname*{id}\left(  C_{\leq u+}\right)
\otimes\varepsilon\left(  C_{\leq\left(  \ell-u\right)  +}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{id}\otimes
\varepsilon\text{ is }k\text{-linear}\right) \\
&  \subseteq\sum_{u=1}^{\ell-1}\operatorname*{id}\left(  C_{\leq u+}\right)
\otimes\underbrace{\varepsilon\left(  C_{\leq\left(  \ell-u\right)  +}\right)
}_{\substack{=0\\\text{(by (\ref{276.pf.6}), applied to }n=\ell-u\text{)}%
}}=\sum_{u=1}^{\ell-1}\underbrace{\operatorname*{id}\left(  C_{\leq
u+}\right)  \otimes0}_{=0}=\sum_{u=1}^{\ell-1}0=0,
\end{align*}
so that $\left(  \operatorname*{id}\otimes\varepsilon\right)  \left(
W\right)  =0$. Now,%
\begin{align*}
\left(  \operatorname*{id}\otimes\varepsilon\right)  \left(  \Delta\left(
x\right)  \right)   &  =\left(  \operatorname*{id}\otimes\varepsilon\right)
\left(  1_{C}\otimes u+v\otimes1_{C}+W\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{276.pf.5})}\right) \\
&  =\underbrace{\left(  \operatorname*{id}\otimes\varepsilon\right)  \left(
1_{C}\otimes u\right)  }_{=\operatorname*{id}\left(  1_{C}\right)
\otimes\varepsilon\left(  u\right)  }+\underbrace{\left(  \operatorname*{id}%
\otimes\varepsilon\right)  \left(  v\otimes1_{C}\right)  }%
_{=\operatorname*{id}\left(  v\right)  \otimes\varepsilon\left(  1_{C}\right)
}+\underbrace{\left(  \operatorname*{id}\otimes\varepsilon\right)  \left(
W\right)  }_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{id}\otimes
\varepsilon\text{ is }k\text{-linear}\right) \\
&  =\underbrace{\operatorname*{id}\left(  1_{C}\right)  }_{=1_{C}}%
\otimes\underbrace{\varepsilon\left(  u\right)  }_{=\varepsilon\left(
u\right)  1}+\underbrace{\operatorname*{id}\left(  v\right)  }_{=v}%
\otimes\underbrace{\varepsilon\left(  1_{C}\right)  }_{=1}=\underbrace{1_{C}%
\otimes\varepsilon\left(  u\right)  1}_{=\varepsilon\left(  u\right)
1_{C}\otimes1}+v\otimes1\\
&  =\varepsilon\left(  u\right)  1_{C}\otimes1+v\otimes1=\left(
\varepsilon\left(  u\right)  1_{C}+v\right)  \otimes1.
\end{align*}
Now,%
\begin{align*}
x  &  =\underbrace{\operatorname{id}_{C}}_{=\operatorname{kan}_{1}\circ\left(
\operatorname*{id}\otimes\varepsilon\right)  \circ\Delta}\left(  x\right)
=\left(  \operatorname{kan}_{1}\circ\left(  \operatorname*{id}\otimes
\varepsilon\right)  \circ\Delta\right)  \left(  x\right)  =\operatorname{kan}%
_{1}\underbrace{\left(  \left(  \operatorname*{id}\otimes\varepsilon\right)
\left(  \Delta\left(  x\right)  \right)  \right)  }_{=\left(  \varepsilon
\left(  u\right)  1_{C}+v\right)  \otimes1}\\
&  =\operatorname{kan}_{1}\left(  \left(  \varepsilon\left(  u\right)
1_{C}+v\right)  \otimes1\right)  =1\cdot\left(  \varepsilon\left(  u\right)
1_{C}+v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname{kan}_{1}\right) \\
&  =\varepsilon\left(  u\right)  1_{C}+v.
\end{align*}
Thus,%
\begin{equation}
v=x-\varepsilon\left(  u\right)  1_{C}. \label{276.pf.7}%
\end{equation}


On the other hand,%
\begin{align*}
\left(  \varepsilon\otimes\operatorname*{id}\right)  \left(  W\right)   &
\in\left(  \varepsilon\otimes\operatorname*{id}\right)  \left(  \sum
_{u=1}^{\ell-1}C_{\leq u+}\otimes C_{\leq\left(  \ell-u\right)  +}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }W\in\sum_{u=1}^{\ell-1}C_{\leq
u+}\otimes C_{\leq\left(  \ell-u\right)  +}\right) \\
&  \in\sum_{u=1}^{\ell-1}\underbrace{\left(  \varepsilon\otimes
\operatorname*{id}\right)  \left(  C_{\leq u+}\otimes C_{\leq\left(
\ell-u\right)  +}\right)  }_{\subseteq\varepsilon\left(  C_{\leq u+}\right)
\otimes\operatorname*{id}\left(  C_{\leq\left(  \ell-u\right)  +}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon\otimes\operatorname*{id}%
\text{ is }k\text{-linear}\right) \\
&  \subseteq\sum_{u=1}^{\ell-1}\underbrace{\varepsilon\left(  C_{\leq
u+}\right)  }_{\substack{=0\\\text{(by (\ref{276.pf.6}), applied to
}n=u\text{)}}}\otimes\operatorname*{id}\left(  C_{\leq\left(  \ell-u\right)
+}\right)  =\sum_{u=1}^{\ell-1}\underbrace{0\otimes\operatorname*{id}\left(
C_{\leq\left(  \ell-u\right)  +}\right)  }_{=0}=\sum_{u=1}^{\ell-1}0=0,
\end{align*}
so that $\left(  \varepsilon\otimes\operatorname*{id}\right)  \left(
W\right)  =0$. Now,%
\begin{align*}
\left(  \varepsilon\otimes\operatorname*{id}\right)  \left(  \Delta\left(
x\right)  \right)   &  =\left(  \varepsilon\otimes\operatorname*{id}\right)
\left(  1_{C}\otimes u+v\otimes1_{C}+W\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{276.pf.5})}\right) \\
&  =\underbrace{\left(  \varepsilon\otimes\operatorname*{id}\right)  \left(
1_{C}\otimes u\right)  }_{=\varepsilon\left(  1_{C}\right)  \otimes
\operatorname*{id}\left(  u\right)  }+\underbrace{\left(  \varepsilon
\otimes\operatorname*{id}\right)  \left(  v\otimes1_{C}\right)  }%
_{=\varepsilon\left(  v\right)  \otimes\operatorname*{id}\left(  1_{C}\right)
}+\underbrace{\left(  \varepsilon\otimes\operatorname*{id}\right)  \left(
W\right)  }_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon\otimes
\operatorname*{id}\text{ is }k\text{-linear}\right) \\
&  =\underbrace{\varepsilon\left(  1_{C}\right)  }_{=1}\otimes
\underbrace{\operatorname*{id}\left(  u\right)  }_{=u}+\underbrace{\varepsilon
\left(  v\right)  }_{=\varepsilon\left(  v\right)  1}\otimes
\underbrace{\operatorname*{id}\left(  1_{C}\right)  }_{=1_{C}}=1\otimes
u+\underbrace{\varepsilon\left(  v\right)  1\otimes1_{C}}_{=1\otimes
\varepsilon\left(  v\right)  1_{C}}\\
&  =1\otimes u+1\otimes\varepsilon\left(  v\right)  1_{C}=1\otimes\left(
u+\varepsilon\left(  v\right)  1_{C}\right)  .
\end{align*}
Now,%
\begin{align*}
x  &  =\underbrace{\operatorname{id}_{C}}_{=\operatorname{kan}_{2}\circ\left(
\varepsilon\otimes\operatorname*{id}\right)  \circ\Delta}\left(  x\right)
=\left(  \operatorname{kan}_{2}\circ\left(  \varepsilon\otimes
\operatorname*{id}\right)  \circ\Delta\right)  \left(  x\right)
=\operatorname{kan}_{2}\underbrace{\left(  \left(  \varepsilon\otimes
\operatorname*{id}\right)  \left(  \Delta\left(  x\right)  \right)  \right)
}_{=1\otimes\left(  u+\varepsilon\left(  v\right)  1_{C}\right)  }\\
&  =\operatorname{kan}_{2}\left(  1\otimes\left(  u+\varepsilon\left(
v\right)  1_{C}\right)  \right) \\
&  =1\cdot\left(  u+\varepsilon\left(  v\right)  1_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname{kan}%
_{2}\right) \\
&  =u+\varepsilon\left(  v\right)  1_{C}.
\end{align*}
Thus,%
\begin{equation}
u=x-\varepsilon\left(  v\right)  1_{C}. \label{276.pf.9}%
\end{equation}
Hence,%
\begin{align*}
\varepsilon\left(  u\right)   &  =\varepsilon\left(  x-\varepsilon\left(
v\right)  1_{C}\right)  =\underbrace{\varepsilon\left(  x\right)  }%
_{=0}-\varepsilon\left(  v\right)  \underbrace{\varepsilon\left(
1_{C}\right)  }_{=1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon\text{
is }k\text{-linear}\right) \\
&  =-\varepsilon\left(  v\right)  ,
\end{align*}
so that%
\begin{equation}
\varepsilon\left(  u\right)  +\varepsilon\left(  v\right)  =0.
\label{276.pf.10}%
\end{equation}


Now, (\ref{276.pf.5}) becomes%
\begin{align*}
\Delta\left(  x\right)   &  =1_{C}\otimes\underbrace{u}%
_{\substack{=x-\varepsilon\left(  v\right)  1_{C}\\\text{(by (\ref{276.pf.9}%
))}}}+\underbrace{v}_{\substack{=x-\varepsilon\left(  u\right)  1_{C}%
\\\text{(by (\ref{276.pf.7}))}}}\otimes1_{C}+W\\
&  =\underbrace{1_{C}\otimes\left(  x-\varepsilon\left(  v\right)
1_{C}\right)  }_{=1_{C}\otimes x-\varepsilon\left(  v\right)  1_{C}%
\otimes1_{C}}+\underbrace{\left(  x-\varepsilon\left(  u\right)  1_{C}\right)
\otimes1_{C}}_{=x\otimes1_{C}-\varepsilon\left(  u\right)  1_{C}\otimes1_{C}%
}+W\\
&  =1_{C}\otimes x-\varepsilon\left(  v\right)  1_{C}\otimes1_{C}%
+x\otimes1_{C}-\varepsilon\left(  u\right)  1_{C}\otimes1_{C}+W\\
&  =x\otimes1_{C}+1_{C}\otimes x-\underbrace{\left(  \varepsilon\left(
u\right)  1_{C}\otimes1_{C}+\varepsilon\left(  v\right)  1_{C}\otimes
1_{C}\right)  }_{\substack{=\left(  \varepsilon\left(  u\right)
+\varepsilon\left(  v\right)  \right)  1_{C}\otimes1_{C}=0\\\text{(since
}\varepsilon\left(  u\right)  +\varepsilon\left(  v\right)  =0\text{ by
(\ref{276.pf.10}))}}}+W\\
&  =x\otimes1_{C}+1_{C}\otimes x+W\in x\otimes1_{C}+1_{C}\otimes x+\sum
_{u=1}^{\ell-1}\underbrace{C_{\leq u+}}_{\subseteq C_{\leq u}}\otimes
\underbrace{C_{\leq\left(  \ell-u\right)  +}}_{\subseteq C_{\leq\ell-u}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }W\in\sum_{u=1}^{\ell-1}C_{\leq
u+}\otimes C_{\leq\left(  \ell-u\right)  +}\right) \\
&  \subseteq x\otimes1_{C}+1_{C}\otimes x+\sum_{u=1}^{\ell-1}C_{\leq u}\otimes
C_{\leq\ell-u}.
\end{align*}
This proves Proposition~\ref{Proposition276}.
\end{proof}

The next lemma is essentially obvious:

\begin{lemma}
\label{Lemma277} Let $k$ be a field, let $A$ be a $k$-algebra, and let $H$ be
a connected filtered $k$-coalgebra. Every positive $i\in\mathbb{N}$ and every
$h\in\mathfrak{g}\left(  H,A\right)  $ satisfy $h^{\ast i}\in\mathfrak{g}%
\left(  H,A\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma277}.]Let $i\in\mathbb{N}$ be positive. Let
$h\in\mathfrak{g}\left(  H,A\right)  $. Since $i>0$, we can apply
Remark~\ref{Remark9} to $n=0$ and $f=h$, and obtain $h^{\ast i}\left(
H_{\leq0}\right)  =0$. Since $1_{H}\in H_{\leq0}$ (because $H$ is a filtered
$k$-algebra), this yields $h^{\ast i}\left(  1_{H}\right)  =0$. Thus, $h^{\ast
i}\in\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(
1_{H}\right)  =0\right\}  =\mathfrak{g}\left(  H,A\right)  $. This proves
Lemma~\ref{Lemma277}.
\end{proof}

We can now prove Proposition~\ref{Proposition272a}:

\begin{proof}
[Proof of Proposition~\ref{Proposition272a}.]We will prove
Proposition~\ref{Proposition272a} by strong induction over $\ell$:

\textit{Induction step:} Let $L\in\mathbb{N}$ be a nonnegative integer. Assume
that Proposition~\ref{Proposition272a} has already been proven whenever
$\ell<L$. Now our aim is to prove Proposition~\ref{Proposition272a} for
$\ell=L$.

First of all, Proposition~\ref{Proposition272a} obviously holds for $\ell=L$
if $L=0$ (in fact,%
\begin{align*}
H_{\leq0}  &  =k\cdot1_{H}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Remark~\ref{Remark7l}, applied to }C=H\right) \\
&  =\mathfrak{E}^{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{E}%
^{0}=k\cdot1_{H}\right) \\
&  =\sum_{i=0}^{0}\mathfrak{E}^{i},
\end{align*}
so that Proposition~\ref{Proposition272a} holds for $\ell=0$; in other words,
Proposition~\ref{Proposition272a} holds for $\ell=L$ if $L=0$). Hence, in the
following, we can WLOG assume that $L\neq0$. Thus, $L$ is positive, so that
$L\geq1$.

Since Proposition~\ref{Proposition272a} has already been proven whenever
$\ell<L$, we have%
\begin{equation}
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\mathfrak{E}^{i}%
\ \ \ \ \ \ \ \ \ \ \text{for every nonnegative integer }\ell<L.
\label{272a.pf.indass}%
\end{equation}


Let $f=\operatorname*{id}-e_{H,H}$, where $\operatorname*{id}$ denotes the
identity map $\operatorname{id}_{H}:H\rightarrow H$. Then, $f\in
\mathfrak{g}\left(  H,H\right)  $\ \ \ \ \footnote{\textit{Proof.} We have
$\operatorname*{id}\in G\left(  H,H\right)  =e_{H,H}+\mathfrak{g}\left(
H,H\right)  $, and thus $\operatorname*{id}-e_{H,H}\in\mathfrak{g}\left(
H,H\right)  $, so that $f=\operatorname*{id}-e_{H,H}\in\mathfrak{g}\left(
H,H\right)  $.}.

Let $g:H\rightarrow H$ be the map $\sum_{i=2}^{L}\dfrac{\left(  -1\right)
^{i-1}}{i}f^{\ast\left(  i-1\right)  }$.

For every $i\in\left\{  2,3,\ldots,L\right\}  $, we have $f^{\ast\left(
i-1\right)  }\in\mathfrak{g}\left(  H,H\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  2,3,\ldots,L\right\}  $.
Then, $i-1\in\left\{  1,2,\ldots,L-1\right\}  $, so that $i-1$ is a positive
element of $\mathbb{N}$. Hence, Lemma~\ref{Lemma277} (applied to $H$, $f$ and
$i-1$ instead of $A$, $h$ and $i$) shows that $f^{\ast\left(  i-1\right)  }%
\in\mathfrak{g}\left(  H,H\right)  $. Qed.}. Thus,%
\[
g=\sum_{i=2}^{L}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{f^{\ast\left(
i-1\right)  }}_{\in\mathfrak{g}\left(  H,H\right)  }\in\sum_{i=2}^{L}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\mathfrak{g}\left(  H,H\right)
\subseteq\mathfrak{g}\left(  H,H\right)
\]
(since $\mathfrak{g}\left(  H,H\right)  $ is a $k$-vector space).

By Remark~\ref{Remark274} \textbf{(d)} (applied to $V=H$ and $W=H$), the set
of all $k$-linear maps $H\rightarrow H$ respecting the filtration is a
$k$-vector subspace of $\mathcal{L}\left(  H,H\right)  $.

Since the map $\operatorname{id}$ respects the filtration (obviously) and the
map $e_{H,H}$ respects the filtration (by Proposition~\ref{Proposition275}
\textbf{(a)}, applied to $C=H$ and $A=H$), the map $\operatorname{id}-e_{H,H}$
also respects the filtration (since the set of all $k$-linear maps
$H\rightarrow H$ respecting the filtration is a $k$-vector subspace of
$\mathcal{L}\left(  H,H\right)  $). Since $\operatorname{id}-e_{H,H}=f$, this
means that the map $f$ respects the filtration.

Thus, for every $i\in\left\{  2,3,\ldots,L\right\}  $, the map $f^{\ast\left(
i-1\right)  }$ respects the filtration (by Proposition~\ref{Proposition275}
\textbf{(c)}, applied to $C=H$, $A=H$ and $n=i-1$). Hence, the map $\sum
_{i=2}^{L}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast\left(  i-1\right)  }$
also respects the filtration (since the set of all $k$-linear maps
$H\rightarrow H$ respecting the filtration is a $k$-vector subspace of
$\mathcal{L}\left(  H,H\right)  $). Since $\sum_{i=2}^{L}\dfrac{\left(
-1\right)  ^{i-1}}{i}f^{\ast\left(  i-1\right)  }=g$, this means that the map
$g$ respects the filtration.

By the definition of $\operatorname*{Log}$, we have $\operatorname*{Log}%
\operatorname*{id}=\operatorname{Log}_{1}\underbrace{\left(
\operatorname*{id}-e_{H,H}\right)  }_{=f}=\operatorname{Log}_{1}f$. Thus,
$\mathfrak{e}=\operatorname*{Log}\operatorname*{id}=\operatorname{Log}_{1}f$.

Now let $x\in H_{\leq L}$ be arbitrary. Then,%
\begin{equation}
f^{\ast i}\left(  x\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every integer }i>L
\label{272a.pf.fix=0}%
\end{equation}
\footnote{\textit{Proof of (\ref{272a.pf.fix=0}):} Let $i>L$ be an integer.
Then, $i>L\geq0$, so that $i\in\mathbb{N}$. Now, Remark~\ref{Remark9} (applied
to $n=L$ and $A=H$) yields $f^{\ast i}\left(  H_{\leq L}\right)  =0$ (since
$i>L$). But $x\in H_{\leq L}$ yields $f^{\ast i}\left(  x\right)  \in f^{\ast
i}\left(  H_{\leq L}\right)  =0$, thus $f^{\ast i}\left(  x\right)  =0$. This
proves (\ref{272a.pf.fix=0}).}. Now, $\mathfrak{e}=\operatorname{Log}_{1}f$
leads to
\begin{align*}
\mathfrak{e}\left(  x\right)   &  =\left(  \operatorname{Log}_{1}f\right)
\left(  x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{log})}\right)
\\
&  =\sum_{\substack{i\geq1;\\i\leq L}}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}f^{\ast i}\left(  x\right)  +\sum_{\substack{i\geq1;\\i>L}}\dfrac{\left(
-1\right)  ^{i-1}}{i}\underbrace{f^{\ast i}\left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{272a.pf.fix=0}))}}}\\
&  =\underbrace{\sum_{\substack{i\geq1;\\i\leq L}}}_{=\sum_{i=1}^{L}}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)
+\underbrace{\sum_{\substack{i\geq1;\\i>L}}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}0}_{=0}=\sum_{i=1}^{L}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right) \\
&  =\underbrace{\dfrac{\left(  -1\right)  ^{1-1}}{1}}_{=\dfrac{1}{1}%
=1}\underbrace{f^{\ast1}}_{=f}\left(  x\right)  +\sum_{i=2}^{L}\dfrac{\left(
-1\right)  ^{i-1}}{i}\underbrace{f^{\ast i}}_{=f\ast f^{\ast\left(
i-1\right)  }}\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=1\\
\text{from the sum, because }L\geq1
\end{array}
\right) \\
&  =f\left(  x\right)  +\sum_{i=2}^{L}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  f\ast f^{\ast\left(  i-1\right)  }\right)  \left(  x\right)  ,
\end{align*}
so that%
\begin{align*}
\mathfrak{e}\left(  x\right)  -f\left(  x\right)   &  =\sum_{i=2}^{L}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f\ast f^{\ast\left(  i-1\right)
}\right)  \left(  x\right)  =\underbrace{\left(  \sum_{i=2}^{L}\dfrac{\left(
-1\right)  ^{i-1}}{i}f\ast f^{\ast\left(  i-1\right)  }\right)  }%
_{\substack{=f\ast\left(  \sum_{i=2}^{L}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}f^{\ast\left(  i-1\right)  }\right)  \\\text{(since convolution of
}k\text{-linear maps}\\\text{is }k\text{-bilinear)}}}\left(  x\right) \\
&  =\left(  f\ast\underbrace{\left(  \sum_{i=2}^{L}\dfrac{\left(  -1\right)
^{i-1}}{i}f^{\ast\left(  i-1\right)  }\right)  }_{=g}\right)  \left(
x\right)  =\left(  f\ast g\right)  \left(  x\right)  .
\end{align*}
This rewrites as%
\[
f\left(  x\right)  =\mathfrak{e}\left(  x\right)  -\left(  f\ast g\right)
\left(  x\right)  .
\]
Since
\begin{align*}
\underbrace{f}_{=\operatorname*{id}-e_{H,H}}\left(  x\right)   &  =\left(
\operatorname*{id}-e_{H,H}\right)  \left(  x\right)
=\underbrace{\operatorname*{id}\left(  x\right)  }_{=x}-\underbrace{e_{H,H}%
}_{=\eta_{H}\circ\varepsilon_{H}}\left(  x\right) \\
&  =x-\underbrace{\left(  \eta_{H}\circ\varepsilon_{H}\right)  \left(
x\right)  }_{\substack{=\eta_{H}\left(  \varepsilon_{H}\left(  x\right)
\right)  =\varepsilon_{H}\left(  x\right)  \cdot1_{H}\\\text{(by the
definition of }\eta_{H}\text{)}}}=x-\varepsilon_{H}\left(  x\right)
\cdot1_{H},
\end{align*}
this becomes $x-\varepsilon_{H}\left(  x\right)  \cdot1_{H}=\mathfrak{e}%
\left(  x\right)  -\left(  f\ast g\right)  \left(  x\right)  $. In other
words,%
\begin{equation}
x=\varepsilon_{H}\left(  x\right)  \cdot1_{H}+\mathfrak{e}\left(  x\right)
-\left(  f\ast g\right)  \left(  x\right)  . \label{272a.pf.3}%
\end{equation}


But since $f\in\mathfrak{g}\left(  H,H\right)  $ and $g\in\mathfrak{g}\left(
H,H\right)  $, and since the maps $f$ and $g$ respect the filtration,
Proposition~\ref{Proposition275} \textbf{(d)} (applied to $\ell=L$, $C=H$ and
$A=H$) yields%
\begin{equation}
\left(  f\ast g\right)  \left(  H_{\leq L}\right)  \subseteq\sum_{u=1}%
^{L-1}H_{\leq u}H_{\leq L-u}. \label{272a.pf.4}%
\end{equation}
But using (\ref{272a.pf.indass}), it is easy to see that every $u\in\left\{
1,2,\ldots,L-1\right\}  $ satisfies $H_{\leq u}H_{\leq L-u}\subseteq\sum
_{i=0}^{L}\mathfrak{E}^{i}$\ \ \ \ \footnote{\textit{Proof.} Let $u\in\left\{
1,2,\ldots,L-1\right\}  $. Then, $u\geq1$ and $u\leq L-1$.
\par
Since $u\leq L-1<L$, we have $H_{\leq u}\subseteq\sum_{i=0}^{u}\mathfrak{E}%
^{i}$ (by (\ref{272a.pf.indass}) (applied to $\ell=u$)).
\par
Since $L-\underbrace{u}_{\geq1}\leq L-1<L$, we have
\begin{align*}
H_{\leq L-u}  &  \subseteq\sum_{i=0}^{L-u}\mathfrak{E}^{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{272a.pf.indass}) (applied to }%
\ell=L-u\text{)}\right) \\
&  =\sum_{j=0}^{L-u}\mathfrak{E}^{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we
renamed the index }i\text{ as }j\text{ in the sum}\right)  .
\end{align*}
\par
Multiplying $H_{\leq u}\subseteq\sum_{i=0}^{u}\mathfrak{E}^{i}$ and $H_{\leq
L-u}\subseteq\sum_{j=0}^{L-u}\mathfrak{E}^{j}$, we obtain%
\begin{align*}
H_{\leq u}H_{\leq L-u}  &  \subseteq\left(  \sum_{i=0}^{u}\mathfrak{E}%
^{i}\right)  \left(  \sum_{j=0}^{L-u}\mathfrak{E}^{j}\right)  =\sum_{i=0}%
^{u}\sum_{j=0}^{L-u}\underbrace{\mathfrak{E}^{i}\mathfrak{E}^{j}%
}_{\substack{=\mathfrak{E}^{i+j}\subseteq\sum_{\nu=0}^{L}\mathfrak{E}^{\nu
}\\\text{(because }i\leq u\text{ and }j\leq L-u\text{, so that}\\i+j\leq
u+\left(  L-u\right)  =L\text{, so that}\\i+j\in\left\{  0,1,\ldots,L\right\}
\text{, so that }\mathfrak{E}^{i+j}\\\text{is an addend in the sum }\sum
_{\nu=0}^{L}\mathfrak{E}^{\nu}\text{)}}}\\
&  \subseteq\sum_{i=0}^{u}\sum_{j=0}^{L-u}\sum_{\nu=0}^{L}\mathfrak{E}^{\nu
}\subseteq\sum_{\nu=0}^{L}\mathfrak{E}^{v}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sum_{\nu=0}^{L}\mathfrak{E}^{\nu}\text{ is a }k\text{-vector
space}\right) \\
&  =\sum_{i=0}^{L}\mathfrak{E}^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we
renamed the summation index }\nu\text{ as }i\right)  ,
\end{align*}
qed.}. Hence, (\ref{272a.pf.4}) becomes%
\[
\left(  f\ast g\right)  \left(  H_{\leq L}\right)  \subseteq\sum_{u=1}%
^{L-1}\underbrace{H_{\leq u}H_{\leq L-u}}_{\subseteq\sum_{i=0}^{L}%
\mathfrak{E}^{i}}\subseteq\sum_{u=1}^{L-1}\sum_{i=0}^{L}\mathfrak{E}%
^{i}\subseteq\sum_{i=0}^{L}\mathfrak{E}^{i}%
\]
(since $\sum_{i=0}^{L}\mathfrak{E}^{i}$ is a $k$-vector space). Hence, $x\in
H_{\leq L}$ yields%
\[
\left(  f\ast g\right)  \left(  x\right)  \in\left(  f\ast g\right)  \left(
H_{\leq L}\right)  \subseteq\sum_{i=0}^{L}\mathfrak{E}^{i}.
\]
On the other hand,
\begin{align*}
\mathfrak{e}\left(  \underbrace{x}_{\in H}\right)   &  \in\mathfrak{e}\left(
H\right)  =\mathfrak{E}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mathfrak{E}\text{ was defined to be }\mathfrak{e}\left(  H\right)  \right) \\
&  =\mathfrak{E}^{1}\subseteq\sum_{i=0}^{L}\mathfrak{E}^{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }L\geq1\text{, so that }1\in\left\{  0,1,\ldots,L\right\}  \text{,
and thus}\\
\mathfrak{E}^{1}\text{ is an addend of the sum }\sum_{i=0}^{L}\mathfrak{E}^{i}%
\end{array}
\right)  .
\end{align*}
Also,%
\[
1_{H}\in\mathfrak{E}^{0}\subseteq\sum_{i=0}^{L}\mathfrak{E}^{i}%
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }0\in\left\{  0,1,\ldots,L\right\}  \text{, and thus}\\
\mathfrak{E}^{0}\text{ is an addend of the sum }\sum_{i=0}^{L}\mathfrak{E}^{i}%
\end{array}
\right)  .
\]
Using these all facts, (\ref{272a.pf.3}) becomes%
\begin{align*}
x  &  =\varepsilon_{H}\left(  x\right)  \cdot\underbrace{1_{H}}_{\in\sum
_{i=0}^{L}\mathfrak{E}^{i}}+\underbrace{\mathfrak{e}\left(  x\right)  }%
_{\in\sum_{i=0}^{L}\mathfrak{E}^{i}}-\underbrace{\left(  f\ast g\right)
\left(  x\right)  }_{\in\sum_{i=0}^{L}\mathfrak{E}^{i}}\\
&  \in\varepsilon_{H}\left(  x\right)  \cdot\left(  \sum_{i=0}^{L}%
\mathfrak{E}^{i}\right)  +\sum_{i=0}^{L}\mathfrak{E}^{i}-\sum_{i=0}%
^{L}\mathfrak{E}^{i}\\
&  \subseteq\sum_{i=0}^{L}\mathfrak{E}^{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sum_{i=0}^{L}\mathfrak{E}^{i}\text{ is a }k\text{-vector
space}\right)  .
\end{align*}


Now forget that we fixed $x$. We thus have shown that every $x\in H_{\leq L}$
satisfies $x\in\sum_{i=0}^{L}\mathfrak{E}^{i}$. In other words, $H_{\leq
L}\subseteq\sum_{i=0}^{L}\mathfrak{E}^{i}$. In other words, we have proven
Proposition~\ref{Proposition272a} for $\ell=L$. This completes the induction step.

Thus, the induction proof of Proposition~\ref{Proposition272a} is complete.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition272}.]Consider the convolution algebra
$\mathcal{L}\left(  H,H\right)  $. Let $\mathfrak{e}$ denote the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $. Let
$\mathfrak{E}$ be the $k$-vector subspace $\mathfrak{e}\left(  H\right)  $ of
$H$.

Theorem~\ref{Theorem1} yields that $\operatorname*{Log}\operatorname*{id}$ is
a projection to $\operatorname*{Prim}H$. In other words, $\mathfrak{e}$ is a
projection to $\operatorname*{Prim}H$ (since $\mathfrak{e}=\operatorname*{Log}%
\operatorname*{id}$). Hence, $\mathfrak{e}\left(  H\right)
=\operatorname*{Prim}H$. Thus, $\mathfrak{E}=\mathfrak{e}\left(  H\right)
=\operatorname*{Prim}H$.

Now, fix $\ell\in\mathbb{N}$. Proposition~\ref{Proposition272a} yields%
\[
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\underbrace{\mathfrak{E}^{i}%
}_{\substack{=\left(  \operatorname*{Prim}H\right)  ^{i}\\\text{(since
}\mathfrak{E}=\operatorname*{Prim}H\text{)}}}=\sum_{i=0}^{\ell}\left(
\operatorname*{Prim}H\right)  ^{i}.
\]
This proves Proposition~\ref{Proposition272}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem270}.]For every $\ell\in\mathbb{N}$,
Proposition~\ref{Proposition272} yields%
\[
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\underbrace{\left(  \operatorname*{Prim}%
H\right)  ^{i}}_{\subseteq\operatorname{AlgGen}_{k}\left(
\operatorname*{Prim}H\right)  }\subseteq\sum_{i=0}^{\ell}\operatorname{AlgGen}%
_{k}\left(  \operatorname*{Prim}H\right)  \subseteq\operatorname{AlgGen}%
_{k}\left(  \operatorname*{Prim}H\right)
\]
(since $\operatorname{AlgGen}_{k}\left(  \operatorname*{Prim}H\right)  $ is a
$k$-vector space). Since $H$ is filtered, we have%
\[
H=\bigcup_{\ell\in\mathbb{N}}\underbrace{H_{\leq\ell}}_{\subseteq
\operatorname{AlgGen}_{k}\left(  \operatorname*{Prim}H\right)  }%
\subseteq\bigcup_{\ell\in\mathbb{N}}\operatorname{AlgGen}_{k}\left(
\operatorname*{Prim}H\right)  =\operatorname{AlgGen}_{k}\left(
\operatorname*{Prim}H\right)  .
\]
Combined with $\operatorname{AlgGen}_{k}\left(  \operatorname*{Prim}H\right)
\subseteq H$ (which is trivial), this yields $H=\operatorname{AlgGen}%
_{k}\left(  \operatorname*{Prim}H\right)  $. This proves
Theorem~\ref{Theorem270}.
\end{proof}

We will next prove a kind of strengthening of Theorem~\ref{Theorem270} and
Proposition~\ref{Proposition272}. First a notation:

\begin{definition}
\label{Definition279} Let $k$ be a field.

\textbf{(a)} For every $k$-vector space $M$ and every subset $S$ of $M$, let
us denote by $\left\langle S\right\rangle $ the $k$-vector subspace of $M$
generated by the elements of $S$. (This is precisely
Convention~\ref{Convention99}; we just have repeated it for the sake of convenience.)

\textbf{(b)} For every $k$-vector space $M$, every set $\Phi$ and every map
$P:\Phi\rightarrow M$, let us denote by $\left\langle P\left(  v\right)
\ \mid\ v\in\Phi\right\rangle $ the subspace $\left\langle \left\{  P\left(
v\right)  \ \mid\ v\in\Phi\right\}  \right\rangle $ of $M$ (this is the
$k$-vector subspace of $M$ generated by all the elements $P\left(  v\right)  $
with $v\in\Phi$).

\textbf{(c)} Assume that the field $k$ has characteristic $0$. For every
$k$-algebra $A$, every $k$-vector subspace $V$ of $A$, and every
$n\in\mathbb{N}$, we denote by $\operatorname{symp}_{n}V$ the $k$-vector
subspace%
\[
\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}%
\right\rangle
\]
of $A$ (where $S_{n}$ denotes the $n$-th symmetric group). Note that this
definition yields that $\operatorname{symp}_{0}V=k\cdot1_{A}$%
\ \ \ \ \footnote{\textit{Proof.} Let $\operatorname*{pt}$ denote the
$0$-tuple of elements of $V$ (i. e., the only element of $V^{\times0}$), and
let $\operatorname{id}_{0}$ denote the identity map $\varnothing
\rightarrow\varnothing$. Then, $V^{\times0}=\left\{  \operatorname*{pt}%
\right\}  $ and $S_{0}=\left\{  \operatorname{id}_{0}\right\}  $.
\par
Since $S_{0}=\left\{  \operatorname{id}_{0}\right\}  $, we have $\sum
_{\sigma\in S_{0}}1_{A}=1_{A}$.
\par
But by the definition of $\operatorname{symp}_{0}V$, we have%
\begin{align*}
\operatorname{symp}_{0}V  &  =\left\langle \underbrace{\dfrac{1}{0!}}%
_{=\dfrac{1}{1}=1}\sum_{\sigma\in S_{0}}\underbrace{v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  0\right)  }}_{=\left(
\text{empty product}\right)  =1_{A}}\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{0}\right)  \in\underbrace{V^{\times0}}_{=\left\{  \operatorname*{pt}%
\right\}  }\right\rangle \\
&  =\left\langle \underbrace{\sum_{\sigma\in S_{0}}1_{A}}_{=1_{A}}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{0}\right)  \in\left\{
\operatorname*{pt}\right\}  \right\rangle \\
&  =\left\langle 1_{A}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{0}\right)
\in\left\{  \operatorname*{pt}\right\}  \right\rangle =k\cdot1_{A},
\end{align*}
qed.} and $\operatorname{symp}_{1}V=V$ (this is also immediate to see).
\end{definition}

Now, the strengthening of Theorem~\ref{Theorem270}:

\begin{theorem}
\label{Theorem280} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected filtered cocommutative bialgebra over $k$. Then,%
\[
H=\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}\left(  \operatorname*{Prim}%
H\right)  .
\]

\end{theorem}

And here the (more concrete) strengthening of Proposition~\ref{Proposition272}:

\begin{proposition}
\label{Proposition282} Let $k$ be a field of characteristic $0$, and let $H$
be a connected filtered cocommutative bialgebra over $k$. Then,
\[
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\operatorname{symp}_{i}\left(
\operatorname*{Prim}H\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\ell
\in\mathbb{N}.
\]

\end{proposition}

The path from Proposition~\ref{Proposition272} to
Proposition~\ref{Proposition282} consists of two parts: one (very easy)
Hopf-algebraic computation, and one elementary algebraic (i. e., using no
coalgebra structure) argument. First, the Hopf-algebraic computation:

\begin{proposition}
\label{Proposition283} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Then,
$xy-yx\in\operatorname*{Prim}H$ for any $x\in\operatorname*{Prim}H$ and
$y\in\operatorname*{Prim}H$.
\end{proposition}

This Proposition~\ref{Proposition283} is very easy (a straightforward
computation, which we will do in a moment) but immensely important: It shows
that $\operatorname*{Prim}H$ is a Lie subalgebra of the Lie algebra $H$ (with
the Lie bracket being given by the commutator).

Next, we shall need a basic fact about sums, known as the \textit{telescope
principle}:

\begin{lemma}
\label{Lemma283a} Let $k$ be a field. Let $A$ be a $k$-vector space. Let
$m\in\mathbb{N}$. Let $a_{0},a_{1},\ldots,a_{m}$ be $m+1$ elements of $A$.

\textbf{(a)} Then,%
\[
\sum_{j=1}^{m}\left(  a_{j-1}-a_{j}\right)  =a_{0}-a_{m}.
\]


\textbf{(b)} Let $M$ be a $k$-vector subspace of $A$. Assume that
$a_{j-1}-a_{j}\in M$ for each $j\in\left\{  1,2,\ldots,m\right\}  $. Then,
$a_{0}\equiv a_{m}\operatorname{mod}M$.
\end{lemma}

Less easy (but completely elementary -- it could be an exercise in basic
algebra) is the following proposition:

\begin{proposition}
\label{Proposition284} Let $k$ be a field. Let $A$ be a $k$-algebra. Let $V$
be a $k$-vector subspace of $A$. Assume that%
\begin{equation}
xy-yx\in V\ \ \ \ \ \ \ \ \ \ \text{for any }x\in V\text{ and }y\in V.
\label{284.cond}%
\end{equation}


\textbf{(a)} For any positive $n\in\mathbb{N}$, any $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ and any $\sigma\in S_{n}$, we
have%
\[
v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\equiv v_{1}v_{2}\cdots v_{n}\operatorname{mod}V^{n-1}.
\]


\textbf{(b)} Assume that the field $k$ has characteristic $0$. For any
$\ell\in\mathbb{N}$, we have
\[
\sum_{i=0}^{\ell}V^{i}=\sum_{i=0}^{\ell}\operatorname{symp}_{i}V.
\]
\footnote{Recall that $V^{i}$ is to be understood as in
Convention~\ref{Convention100}. Hence, $V^{i}$ means the $i$-th power of the
subspace $V$ of the $k$-algebra $A$.}
\end{proposition}

Let us now step to the proofs of these facts. We begin with the
straightforward proof of Proposition~\ref{Proposition283}:

\begin{proof}
[Proof of Proposition~\ref{Proposition283}.]Let $x\in\operatorname*{Prim}H$
and $y\in\operatorname*{Prim}H$.

Since $x\in\operatorname*{Prim}H=\left(  \text{the set of all primitive
elements of }H\right)  $, the element $x$ of $H$ is primitive. In other words,
$\Delta\left(  x\right)  =x\otimes1_{H}+1_{H}\otimes x$ (by the definition of
``primitive''). Similarly, $\Delta\left(  y\right)  =y\otimes1_{H}%
+1_{H}\otimes y$.

Since $H$ is a bialgebra,%
\begin{align*}
\Delta\left(  xy\right)   &  =\underbrace{\Delta\left(  x\right)  }%
_{=x\otimes1_{H}+1_{H}\otimes x}\cdot\underbrace{\Delta\left(  y\right)
}_{=y\otimes1_{H}+1_{H}\otimes y}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
axioms of a bialgebra}\right) \\
&  =\left(  x\otimes1_{H}+1_{H}\otimes x\right)  \cdot\left(  y\otimes
1_{H}+1_{H}\otimes y\right) \\
&  =\underbrace{\left(  x\otimes1_{H}\right)  \cdot\left(  y\otimes
1_{H}\right)  }_{=xy\otimes1_{H}1_{H}}+\underbrace{\left(  x\otimes
1_{H}\right)  \cdot\left(  1_{H}\otimes y\right)  }_{=x1_{H}\otimes1_{H}%
y}+\underbrace{\left(  1_{H}\otimes x\right)  \cdot\left(  y\otimes
1_{H}\right)  }_{=1_{H}y\otimes x1_{H}}+\underbrace{\left(  1_{H}\otimes
x\right)  \cdot\left(  1_{H}\otimes y\right)  }_{=1_{H}1_{H}\otimes xy}\\
&  =xy\otimes\underbrace{1_{H}1_{H}}_{=1_{H}}+\underbrace{x1_{H}}_{=x}%
\otimes\underbrace{1_{H}y}_{=y}+\underbrace{1_{H}y}_{=y}\otimes
\underbrace{x1_{H}}_{=x}+\underbrace{1_{H}1_{H}}_{=1_{H}}\otimes xy\\
&  =xy\otimes1_{H}+x\otimes y+y\otimes x+1_{H}\otimes xy.
\end{align*}
The same argument, but with $x$ and $y$ transposed, yields%
\[
\Delta\left(  yx\right)  =yx\otimes1_{H}+y\otimes x+x\otimes y+1_{H}\otimes
yx.
\]
Now, since $\Delta$ is $k$-linear, we have%
\begin{align*}
\Delta\left(  xy-yx\right)   &  =\underbrace{\Delta\left(  xy\right)
}_{=xy\otimes1_{H}+x\otimes y+y\otimes x+1_{H}\otimes xy}-\underbrace{\Delta
\left(  yx\right)  }_{=yx\otimes1_{H}+y\otimes x+x\otimes y+1_{H}\otimes yx}\\
&  =\left(  xy\otimes1_{H}+x\otimes y+y\otimes x+1_{H}\otimes xy\right)
-\left(  yx\otimes1_{H}+y\otimes x+x\otimes y+1_{H}\otimes yx\right) \\
&  =\underbrace{xy\otimes1_{H}-yx\otimes1_{H}}_{=\left(  xy-yx\right)
\otimes1_{H}}+\underbrace{1_{H}\otimes xy-1_{H}\otimes yx}_{=1_{H}%
\otimes\left(  xy-yx\right)  }\\
&  =\left(  xy-yx\right)  \otimes1_{H}+1_{H}\otimes\left(  xy-yx\right)  .
\end{align*}
In other words, $xy-yx$ is primitive (by the definition of \textquotedblleft
primitive\textquotedblright). Thus, $xy-yx\in\left(  \text{the set of all
primitive elements of }H\right)  =\operatorname*{Prim}H$. This proves
Proposition~\ref{Proposition283}.
\end{proof}

Next, let us prove Lemma~\ref{Lemma283a}.

\begin{proof}
[Proof of Lemma~\ref{Lemma283a}.]\textbf{(a)} If $m=0$, then
Lemma~\ref{Lemma283a} \textbf{(a)} holds\footnote{\textit{Proof.} Assume that
$m=0$. We must show that Lemma~\ref{Lemma283a} \textbf{(a)} holds.
\par
We have $m=0$ and thus $\sum_{j=1}^{m}\left(  a_{j-1}-a_{j}\right)
=\sum_{j=1}^{0}\left(  a_{j-1}-a_{j}\right)  =\left(  \text{empty sum}\right)
=0$. Comparing this with $a_{0}-\underbrace{a_{m}}_{\substack{=a_{0}%
\\\text{(since }m=0\text{)}}}=a_{0}-a_{0}=0$, we obtain $\sum_{j=1}^{m}\left(
a_{j-1}-a_{j}\right)  =a_{0}-a_{m}$. Thus, Lemma~\ref{Lemma283a} \textbf{(a)}
holds. Qed.}. Hence, for the rest of our proof of Lemma~\ref{Lemma283a}
\textbf{(a)}, we can WLOG assume that we don't have $m=0$. Assume this.

We have $m\neq0$ (since we don't have $m=0$). Thus, $m\geq1$ (since
$m\in\mathbb{N}$). Now,%
\begin{align*}
&  \sum_{j=1}^{m}\left(  a_{j-1}-a_{j}\right) \\
&  =\sum_{j=1}^{m}a_{j-1}-\sum_{j=1}^{m}a_{j}\\
&  =\underbrace{\sum_{j=0}^{m-1}a_{j}}_{\substack{=a_{0}+\sum_{j=1}^{m-1}%
a_{j}\\\text{(here, we have split off the addend for }j=0\\\text{from the sum,
because }m\geq1\text{)}}}-\underbrace{\sum_{j=1}^{m}a_{j}}_{\substack{=\sum
_{j=1}^{m-1}a_{j}+a_{m}\\\text{(here, we have split off the addend for
}j=m\\\text{from the sum, because }m\geq1\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for
}j-1\text{ in the first sum}\right) \\
&  =\left(  a_{0}+\sum_{j=1}^{m-1}a_{j}\right)  -\left(  \sum_{j=1}^{m-1}%
a_{j}+a_{m}\right)  =a_{0}-a_{m}.
\end{align*}
This proves Lemma~\ref{Lemma283a} \textbf{(a)}.

\textbf{(b)} We assumed that $a_{j-1}-a_{j}\in M$ for each $j\in\left\{
1,2,\ldots,m\right\}  $. Thus, $\sum_{j=1}^{m}\underbrace{\left(
a_{j-1}-a_{j}\right)  }_{\in M}\in\sum_{j=1}^{m}M\subseteq M$ (since $M$ is a
$k$-vector space). Now, Lemma~\ref{Lemma283a} \textbf{(a)} yields%
\[
a_{0}-a_{m}=\sum_{j=1}^{m}\left(  a_{j-1}-a_{j}\right)  \in M.
\]
In other words, $a_{0}\equiv a_{m}\operatorname{mod}M$. This proves
Lemma~\ref{Lemma283a} \textbf{(b)}.
\end{proof}

Before we step to the proof of Proposition~\ref{Proposition284}, let us recall
a known fact from linear algebra:%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }M\text{ is a }k\text{-vector space, if }S\text{ is a subset of
}M\text{, and if }Q\text{ is a}\\
k\text{-vector subspace of }M\text{ such that }S\subseteq Q\text{, then
}\left\langle S\right\rangle \subseteq Q
\end{array}
\right)  . \label{284.pf.b.2}%
\end{equation}


Now, let us show Proposition~\ref{Proposition284}:

\begin{proof}
[Proof of Proposition~\ref{Proposition284}.]\textbf{(a)} Let $n\in\mathbb{N}$,
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ and $\sigma\in
S_{n}$ be arbitrary.

For every $i\in\left\{  1,2,\ldots,n-1\right\}  $, let $\tau_{i}$ denote the
transposition $\left(  i,i+1\right)  \in S_{n}$.

Now, it is known that every element of the symmetric group $S_{n}$ can be
written as a product of some transpositions from the set $\left\{  \tau
_{1},\tau_{2},\ldots,\tau_{n-1}\right\}  $\ \ \ \ \footnote{Indeed, this is
precisely the claim of \cite[Exercise 5.1 \textbf{(b)}]{detnotes} (since the
transpositions that we denote by $\tau_{1},\tau_{2},\ldots,\tau_{n-1}$ have
been denoted by $s_{1},s_{2},\ldots,s_{n-1}$ in \cite{detnotes}).}. Applying
this to the element $\sigma\in S_{n}$, we conclude that $\sigma$ can be
written as a product of some transpositions from the set $\left\{  \tau
_{1},\tau_{2},\ldots,\tau_{n-1}\right\}  $. In other words, there exists a
natural number $m\in\mathbb{N}$ and a sequence $\left(  i_{1},i_{2}%
,\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,n-1\right\}  ^{\times m}$ such
that $\sigma=\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{m}}$. Consider this $m$
and this $\left(  i_{1},i_{2},\ldots,i_{m}\right)  $. For every $j\in\left\{
0,1,\ldots,m\right\}  $, let $\sigma_{j}$ denote the permutation $\tau_{i_{1}%
}\tau_{i_{2}}\cdots\tau_{i_{j}}\in S_{n}$. Then, $\sigma_{0}=\tau_{i_{1}}%
\tau_{i_{2}}\cdots\tau_{i_{0}}=\left(  \text{empty product}\right)
=\operatorname{id}$ and $\sigma_{m}=\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{m}%
}=\sigma$. Moreover, every $j\in\left\{  1,2,\ldots,m\right\}  $ satisfies
$v_{\sigma_{j-1}\left(  1\right)  }v_{\sigma_{j-1}\left(  2\right)  }\cdots
v_{\sigma_{j-1}\left(  n\right)  }-v_{\sigma_{j}\left(  1\right)  }%
v_{\sigma_{j}\left(  2\right)  }\cdots v_{\sigma_{j}\left(  n\right)  }\in
V^{n-1}$.\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,\ldots
,m\right\}  $ be arbitrary. Then,
\[
\underbrace{\sigma_{j-1}}_{\substack{=\tau_{i_{1}}\tau_{i_{2}}\cdots
\tau_{i_{j-1}}\\\text{(by the formula }\sigma_{j}=\tau_{i_{1}}\tau_{i_{2}%
}\cdots\tau_{i_{j}}\text{,}\\\text{applied to }j-1\text{ instead of }%
j\text{)}}}\tau_{i_{j}}=\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{j-1}}%
\tau_{i_{j}}=\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{j}}=\sigma_{j},
\]
so that $\sigma_{j}=\sigma_{j-1}\tau_{i_{j}}$. Denote $i_{j}$ by $\mathbf{I}$.
Then, $\sigma_{j}=\sigma_{j-1}\tau_{i_{j}}$ rewrites as $\sigma_{j}%
=\sigma_{j-1}\tau_{\mathbf{I}}$.
\par
Since $\tau_{\mathbf{I}}=\left(  \mathbf{I},\mathbf{I}+1\right)  $, every
$\ell\in\left\{  1,2,\ldots,\mathbf{I}-1\right\}  $ satisfies%
\[
\tau_{\mathbf{I}}\left(  \ell\right)  =\left(  \mathbf{I},\mathbf{I}+1\right)
\left(  \ell\right)  =\ell\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell
\in\left\{  1,2,\ldots,\mathbf{I}-1\right\}  \text{, so that }\ell
\notin\left\{  \mathbf{I},\mathbf{I}+1\right\}  \right)
\]
and thus%
\begin{align*}
v_{\sigma_{j}\left(  \ell\right)  }  &  =v_{\sigma_{j-1}\left(  \tau
_{\mathbf{I}}\left(  \ell\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sigma_{j}=\sigma_{j-1}\tau_{\mathbf{I}}\text{ and thus }%
\sigma_{j}\left(  \ell\right)  =\left(  \sigma_{j-1}\tau_{\mathbf{I}}\right)
\left(  \ell\right)  =\sigma_{j-1}\left(  \tau_{\mathbf{I}}\left(
\ell\right)  \right)  \right) \\
&  =v_{\sigma_{j-1}\left(  \ell\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\tau_{\mathbf{I}}\left(  \ell\right)  =\ell\right)  .
\end{align*}
In other words, $v_{\sigma_{j}\left(  1\right)  }=v_{\sigma_{j-1}\left(
1\right)  }$, $v_{\sigma_{j}\left(  2\right)  }=v_{\sigma_{j-1}\left(
2\right)  }$, $\ldots$, $v_{\sigma_{j}\left(  \mathbf{I}-1\right)  }%
=v_{\sigma_{j-1}\left(  \mathbf{I}-1\right)  }$. Multiplying these
$\mathbf{I}-1$ equations, we obtain $v_{\sigma_{j}\left(  1\right)  }%
v_{\sigma_{j}\left(  2\right)  }\cdots v_{\sigma_{j}\left(  \mathbf{I}%
-1\right)  }=v_{\sigma_{j-1}\left(  1\right)  }v_{\sigma_{j-1}\left(
2\right)  }\cdots v_{\sigma_{j-1}\left(  \mathbf{I}-1\right)  }$.
\par
Since $\tau_{\mathbf{I}}=\left(  \mathbf{I},\mathbf{I}+1\right)  $, every
$\ell\in\left\{  \mathbf{I}+2,\mathbf{I}+3,\ldots,n\right\}  $ satisfies%
\[
\tau_{\mathbf{I}}\left(  \ell\right)  =\left(  \mathbf{I},\mathbf{I}+1\right)
\left(  \ell\right)  =\ell\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell
\in\left\{  \mathbf{I}+2,\mathbf{I}+3,\ldots,n\right\}  \text{, so that }%
\ell\notin\left\{  \mathbf{I},\mathbf{I}+1\right\}  \right)
\]
and thus%
\begin{align*}
v_{\sigma_{j}\left(  \ell\right)  }  &  =v_{\sigma_{j-1}\left(  \tau
_{\mathbf{I}}\left(  \ell\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sigma_{j}=\sigma_{j-1}\tau_{\mathbf{I}}\text{ and thus }%
\sigma_{j}\left(  \ell\right)  =\left(  \sigma_{j-1}\tau_{\mathbf{I}}\right)
\left(  \ell\right)  =\sigma_{j-1}\left(  \tau_{\mathbf{I}}\left(
\ell\right)  \right)  \right) \\
&  =v_{\sigma_{j-1}\left(  \ell\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\tau_{\mathbf{I}}\left(  \ell\right)  =\ell\right)  .
\end{align*}
In other words, $v_{\sigma_{j}\left(  \mathbf{I}+2\right)  }=v_{\sigma
_{j-1}\left(  \mathbf{I}+2\right)  }$, $v_{\sigma_{j}\left(  \mathbf{I}%
+3\right)  }=v_{\sigma_{j-1}\left(  \mathbf{I}+3\right)  }$, $\ldots$,
$v_{\sigma_{j}\left(  n\right)  }=v_{\sigma_{j-1}\left(  n\right)  }$.
Multiplying these $n-\left(  \mathbf{I}+1\right)  $ equations, we obtain
$v_{\sigma_{j}\left(  \mathbf{I}+2\right)  }v_{\sigma_{j}\left(
\mathbf{I}+3\right)  }\cdots v_{\sigma_{j}\left(  n\right)  }=v_{\sigma
_{j-1}\left(  \mathbf{I}+2\right)  }v_{\sigma_{j-1}\left(  \mathbf{I}%
+3\right)  }\cdots v_{\sigma_{j-1}\left(  n\right)  }$.
\par
Since $\tau_{\mathbf{I}}=\left(  \mathbf{I},\mathbf{I}+1\right)  $, we have
$\tau_{\mathbf{I}}\left(  \mathbf{I}\right)  =\mathbf{I}+1$ and
\begin{align*}
v_{\sigma_{j}\left(  \mathbf{I}\right)  }  &  =v_{\sigma_{j-1}\left(
\tau_{\mathbf{I}}\left(  \mathbf{I}\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma_{j}=\sigma_{j-1}%
\tau_{\mathbf{I}}\text{ and thus }\sigma_{j}\left(  \mathbf{I}\right)
=\left(  \sigma_{j-1}\tau_{\mathbf{I}}\right)  \left(  \mathbf{I}\right)
=\sigma_{j-1}\left(  \tau_{\mathbf{I}}\left(  \mathbf{I}\right)  \right)
\right) \\
&  =v_{\sigma_{j-1}\left(  \mathbf{I}+1\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\tau_{\mathbf{I}}\left(  \mathbf{I}\right)  =\mathbf{I}+1\right)
.
\end{align*}
\par
Since $\tau_{\mathbf{I}}=\left(  \mathbf{I},\mathbf{I}+1\right)  $, we have
$\tau_{\mathbf{I}}\left(  \mathbf{I}+1\right)  =\mathbf{I}$ and
\begin{align*}
v_{\sigma_{j}\left(  \mathbf{I}+1\right)  }  &  =v_{\sigma_{j-1}\left(
\tau_{\mathbf{I}}\left(  \mathbf{I}+1\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma_{j}=\sigma_{j-1}%
\tau_{\mathbf{I}}\text{ and thus }\sigma_{j}\left(  \mathbf{I}+1\right)
=\left(  \sigma_{j-1}\tau_{\mathbf{I}}\right)  \left(  \mathbf{I}+1\right)
=\sigma_{j-1}\left(  \tau_{\mathbf{I}}\left(  \mathbf{I}+1\right)  \right)
\right) \\
&  =v_{\sigma_{j-1}\left(  \mathbf{I}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\tau_{\mathbf{I}}\left(  \mathbf{I}+1\right)  =\mathbf{I}\right)
.
\end{align*}
\par
Since $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$, we have
$v_{\sigma_{j-1}\left(  \mathbf{I}\right)  }\in V$ and $v_{\sigma_{j-1}\left(
\mathbf{I}+1\right)  }\in V$. Thus, (\ref{284.cond}) (applied to
$x=v_{\sigma_{j-1}\left(  \mathbf{I}\right)  }$ and $y=v_{\sigma_{j-1}\left(
\mathbf{I}+1\right)  }$) yields that $v_{\sigma_{j-1}\left(  \mathbf{I}%
\right)  }v_{\sigma_{j-1}\left(  \mathbf{I}+1\right)  }-v_{\sigma_{j-1}\left(
\mathbf{I}+1\right)  }v_{\sigma_{j-1}\left(  \mathbf{I}\right)  }\in V$.
\par
Now,%
\begin{align*}
&  \underbrace{v_{\sigma_{j-1}\left(  1\right)  }v_{\sigma_{j-1}\left(
2\right)  }\cdots v_{\sigma_{j-1}\left(  n\right)  }}_{=\left(  v_{\sigma
_{j-1}\left(  1\right)  }v_{\sigma_{j-1}\left(  2\right)  }\cdots
v_{\sigma_{j-1}\left(  \mathbf{I}-1\right)  }\right)  v_{\sigma_{j-1}\left(
\mathbf{I}\right)  }v_{\sigma_{j-1}\left(  \mathbf{I}+1\right)  }\left(
v_{\sigma_{j-1}\left(  \mathbf{I}+2\right)  }v_{\sigma_{j-1}\left(
\mathbf{I}+3\right)  }\cdots v_{\sigma_{j-1}\left(  n\right)  }\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{v_{\sigma_{j}\left(  1\right)  }%
v_{\sigma_{j}\left(  2\right)  }\cdots v_{\sigma_{j}\left(  n\right)  }%
}_{=\left(  v_{\sigma_{j}\left(  1\right)  }v_{\sigma_{j}\left(  2\right)
}\cdots v_{\sigma_{j}\left(  \mathbf{I}-1\right)  }\right)  v_{\sigma
_{j}\left(  \mathbf{I}\right)  }v_{\sigma_{j}\left(  \mathbf{I}+1\right)
}\left(  v_{\sigma_{j}\left(  \mathbf{I}+2\right)  }v_{\sigma_{j}\left(
\mathbf{I}+3\right)  }\cdots v_{\sigma_{j}\left(  n\right)  }\right)  }\\
&  =\left(  v_{\sigma_{j-1}\left(  1\right)  }v_{\sigma_{j-1}\left(  2\right)
}\cdots v_{\sigma_{j-1}\left(  \mathbf{I}-1\right)  }\right)  v_{\sigma
_{j-1}\left(  \mathbf{I}\right)  }v_{\sigma_{j-1}\left(  \mathbf{I}+1\right)
}\left(  v_{\sigma_{j-1}\left(  \mathbf{I}+2\right)  }v_{\sigma_{j-1}\left(
\mathbf{I}+3\right)  }\cdots v_{\sigma_{j-1}\left(  n\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\left(  v_{\sigma_{j}\left(  1\right)
}v_{\sigma_{j}\left(  2\right)  }\cdots v_{\sigma_{j}\left(  \mathbf{I}%
-1\right)  }\right)  }_{=v_{\sigma_{j-1}\left(  1\right)  }v_{\sigma
_{j-1}\left(  2\right)  }\cdots v_{\sigma_{j-1}\left(  \mathbf{I}-1\right)  }%
}\underbrace{v_{\sigma_{j}\left(  \mathbf{I}\right)  }}_{=v_{\sigma
_{j-1}\left(  \mathbf{I}+1\right)  }}\underbrace{v_{\sigma_{j}\left(
\mathbf{I}+1\right)  }}_{=v_{\sigma_{j-1}\left(  \mathbf{I}\right)  }%
}\underbrace{\left(  v_{\sigma_{j}\left(  \mathbf{I}+2\right)  }v_{\sigma
_{j}\left(  \mathbf{I}+3\right)  }\cdots v_{\sigma_{j}\left(  n\right)
}\right)  }_{=v_{\sigma_{j-1}\left(  \mathbf{I}+2\right)  }v_{\sigma
_{j-1}\left(  \mathbf{I}+3\right)  }\cdots v_{\sigma_{j-1}\left(  n\right)  }%
}\\
&  =\left(  v_{\sigma_{j-1}\left(  1\right)  }v_{\sigma_{j-1}\left(  2\right)
}\cdots v_{\sigma_{j-1}\left(  \mathbf{I}-1\right)  }\right)  v_{\sigma
_{j-1}\left(  \mathbf{I}\right)  }v_{\sigma_{j-1}\left(  \mathbf{I}+1\right)
}\left(  v_{\sigma_{j-1}\left(  \mathbf{I}+2\right)  }v_{\sigma_{j-1}\left(
\mathbf{I}+3\right)  }\cdots v_{\sigma_{j-1}\left(  n\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  v_{\sigma_{j-1}\left(  1\right)  }%
v_{\sigma_{j-1}\left(  2\right)  }\cdots v_{\sigma_{j-1}\left(  \mathbf{I}%
-1\right)  }\right)  v_{\sigma_{j-1}\left(  \mathbf{I}+1\right)  }%
v_{\sigma_{j-1}\left(  \mathbf{I}\right)  }\left(  v_{\sigma_{j-1}\left(
\mathbf{I}+2\right)  }v_{\sigma_{j-1}\left(  \mathbf{I}+3\right)  }\cdots
v_{\sigma_{j-1}\left(  n\right)  }\right) \\
&  =\underbrace{\left(  v_{\sigma_{j-1}\left(  1\right)  }v_{\sigma
_{j-1}\left(  2\right)  }\cdots v_{\sigma_{j-1}\left(  \mathbf{I}-1\right)
}\right)  }_{\substack{\in V^{\mathbf{I}-1}\\\text{(since }\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{\times n}\text{, so that}\\v_{\sigma
_{j-1}\left(  1\right)  }\text{, }v_{\sigma_{j-1}\left(  2\right)  }\text{,
}\ldots\text{, }v_{\sigma_{j-1}\left(  \mathbf{I}-1\right)  }\\\text{all lie
in }V\text{)}}}\underbrace{\left(  v_{\sigma_{j-1}\left(  \mathbf{I}\right)
}v_{\sigma_{j-1}\left(  \mathbf{I}+1\right)  }-v_{\sigma_{j-1}\left(
\mathbf{I}+1\right)  }v_{\sigma_{j-1}\left(  \mathbf{I}\right)  }\right)
}_{\in V=V^{1}}\underbrace{\left(  v_{\sigma_{j-1}\left(  \mathbf{I}+2\right)
}v_{\sigma_{j-1}\left(  \mathbf{I}+3\right)  }\cdots v_{\sigma_{j-1}\left(
n\right)  }\right)  }_{\substack{\in V^{n-\left(  \mathbf{I}+1\right)
}\\\text{(since }\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\text{, so that}\\v_{\sigma_{j-1}\left(  \mathbf{I}+2\right)  }\text{,
}v_{\sigma_{j-1}\left(  \mathbf{I}+3\right)  }\text{, }\ldots\text{,
}v_{\sigma_{j-1}\left(  n\right)  }\\\text{all lie in }V\text{)}}}\\
&  \in V^{\mathbf{I}-1}\cdot V^{1}\cdot V^{n-\left(  \mathbf{I}+1\right)
}=V^{\left(  \mathbf{I}-1\right)  +1+\left(  n-\left(  \mathbf{I}+1\right)
\right)  }=V^{n-1},
\end{align*}
qed.} Thus, Lemma~\ref{Lemma283a} \textbf{(b)} (applied to $M=V^{n-1}$ and
$a_{j}=v_{\sigma_{j}\left(  1\right)  }v_{\sigma_{j}\left(  2\right)  }\cdots
v_{\sigma_{j}\left(  n\right)  }$) yields%
\[
v_{\sigma_{0}\left(  1\right)  }v_{\sigma_{0}\left(  2\right)  }\cdots
v_{\sigma_{0}\left(  n\right)  }\equiv v_{\sigma_{m}\left(  1\right)
}v_{\sigma_{m}\left(  2\right)  }\cdots v_{\sigma_{m}\left(  n\right)
}\operatorname{mod}V^{n-1}%
\]
(since $V^{n-1}$ is a $k$-vector subspace of $A$). But from $\sigma
_{0}=\operatorname*{id}$, we obtain%
\[
v_{\sigma_{0}\left(  1\right)  }v_{\sigma_{0}\left(  2\right)  }\cdots
v_{\sigma_{0}\left(  n\right)  }=v_{\operatorname*{id}\left(  1\right)
}v_{\operatorname*{id}\left(  2\right)  }\cdots v_{\operatorname*{id}\left(
n\right)  }=v_{1}v_{2}\cdots v_{n},
\]
so that%
\begin{align*}
v_{1}v_{2}\cdots v_{n}  &  =v_{\sigma_{0}\left(  1\right)  }v_{\sigma
_{0}\left(  2\right)  }\cdots v_{\sigma_{0}\left(  n\right)  }\equiv
v_{\sigma_{m}\left(  1\right)  }v_{\sigma_{m}\left(  2\right)  }\cdots
v_{\sigma_{m}\left(  n\right)  }\\
&  =v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots
v_{\sigma\left(  n\right)  }\operatorname{mod}V^{n-1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma_{m}=\sigma\right)  .
\end{align*}
This proves Proposition~\ref{Proposition284} \textbf{(a)}.

\textbf{(b)} Let us prove Proposition~\ref{Proposition284} \textbf{(b)} by
induction over $\ell$:

\textit{Induction base:} We have%
\[
\sum_{i=0}^{0}V^{i}=V^{0}=k\cdot1_{A}%
\]
and%
\[
\sum_{i=0}^{0}\operatorname{symp}_{i}V=\operatorname{symp}_{0}V=k\cdot1_{A},
\]
so that $\sum_{i=0}^{0}V^{i}=\sum_{i=0}^{0}\operatorname{symp}_{i}V$. In other
words, Proposition~\ref{Proposition284} \textbf{(b)} holds for $\ell=0$. This
completes the induction base.

\textit{Induction step:} Let $n\in\mathbb{N}$ be positive. Assume that
Proposition~\ref{Proposition284} \textbf{(b)} holds for $\ell=n-1$. We now
must prove that Proposition~\ref{Proposition284} \textbf{(b)} holds for
$\ell=n$ as well.

Since Proposition~\ref{Proposition284} \textbf{(b)} holds for $\ell=n-1$, we
have%
\begin{equation}
\sum_{i=0}^{n-1}V^{i}=\sum_{i=0}^{n-1}\operatorname{symp}_{i}V.
\label{284.pf.b.1}%
\end{equation}


Next, we notice that%
\begin{align*}
&  \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\} \\
&  \subseteq\left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}%
v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\}  \right\rangle \\
&  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle
=\operatorname{symp}_{n}V.
\end{align*}
Thus,%
\begin{equation}
\text{every }\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\text{
satisfies }\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\in\operatorname{symp}_{n}V. \label{284.pf.b.3}%
\end{equation}


On the other hand, every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}$ satisfies%
\begin{align*}
\dfrac{1}{n!}\sum_{\sigma\in S_{n}}\underbrace{v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
}_{\substack{\equiv v_{1}v_{2}\cdots v_{n}\operatorname{mod}V^{n-1}\\\text{(by
Proposition~\ref{Proposition284} \textbf{(a)})}}}  &  \equiv\dfrac{1}%
{n!}\underbrace{\sum_{\sigma\in S_{n}}v_{1}v_{2}\cdots v_{n}}_{=\left\vert
S_{n}\right\vert \cdot v_{1}v_{2}\cdots v_{n}}=\dfrac{1}{n!}%
\underbrace{\left\vert S_{n}\right\vert }_{=n!}\cdot v_{1}v_{2}\cdots v_{n}\\
&  =\dfrac{1}{n!}n!\cdot v_{1}v_{2}\cdots v_{n}=v_{1}v_{2}\cdots
v_{n}\operatorname{mod}V^{n-1},
\end{align*}
so that $v_{1}v_{2}\cdots v_{n}\equiv\dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\operatorname{mod}V^{n-1}$, and thus%
\begin{align*}
v_{1}v_{2}\cdots v_{n}  &  \in\underbrace{\dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }}_{\substack{\in\operatorname{symp}_{n}V\\\text{(by
(\ref{284.pf.b.3}))}}}+\underbrace{V^{n-1}}_{\substack{\subseteq\sum
_{i=0}^{n-2}V^{i}+V^{n-1}=\sum_{i=0}^{n-1}V^{i}\\=\sum_{i=0}^{n-1}%
\operatorname{symp}_{i}V\\\text{(by (\ref{284.pf.b.1}))}}}\\
&  \subseteq\operatorname{symp}_{n}V+\sum_{i=0}^{n-1}\operatorname{symp}%
_{i}V=\sum_{i=0}^{n}\operatorname{symp}_{i}V.
\end{align*}
In other words,%
\[
\left\{  v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}\right\}  \subseteq\sum_{i=0}^{n}%
\operatorname{symp}_{i}V.
\]
Hence, (\ref{284.pf.b.2}) (applied to $M=A$, $S=\left\{  v_{1}v_{2}\cdots
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\right\}  $ and $Q=\sum_{i=0}^{n}\operatorname{symp}_{i}V$) yields
\[
\left\langle \left\{  v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\}  \right\rangle \subseteq
\sum_{i=0}^{n}\operatorname{symp}_{i}V.
\]
Since%
\begin{align*}
\left\langle \left\{  v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\}  \right\rangle  &
=\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}\right\rangle \\
&  =V^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{100.Vn})}\right)  ,
\end{align*}
this becomes $V^{n}\subseteq\sum_{i=0}^{n}\operatorname{symp}_{i}V$. Thus,%
\begin{align}
\sum_{i=0}^{n}V^{i}  &  =\underbrace{\sum_{i=0}^{n-1}V^{i}}_{\substack{=\sum
_{i=0}^{n-1}\operatorname{symp}_{i}V\\\text{(by (\ref{284.pf.b.1}))}%
}}+\underbrace{V^{n}}_{\subseteq\sum_{i=0}^{n}\operatorname{symp}_{i}%
V}\subseteq\underbrace{\sum_{i=0}^{n-1}\operatorname{symp}_{i}V}%
_{\substack{\subseteq\sum_{i=0}^{n}\operatorname{symp}_{i}V\\\text{(since
}n-1\leq n\text{)}}}+\sum_{i=0}^{n}\operatorname{symp}_{i}V\nonumber\\
&  \subseteq\sum_{i=0}^{n}\operatorname{symp}_{i}V+\sum_{i=0}^{n}%
\operatorname{symp}_{i}V\subseteq\sum_{i=0}^{n}\operatorname{symp}_{i}V
\label{284.pf.b.6}%
\end{align}
(since $\sum_{i=0}^{n}\operatorname{symp}_{i}V$ is a $k$-vector space).

On the other hand, every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}$ satisfies%
\[
\dfrac{1}{n!}\sum_{\sigma\in S_{n}}\underbrace{v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
}_{\substack{\in V^{n}\\\text{(since }v_{\sigma\left(  1\right)  }\text{,
}v_{\sigma\left(  2\right)  }\text{, }\ldots\text{, }v_{\sigma\left(
n\right)  }\text{ all lie in }V\text{)}}}\in\dfrac{1}{n!}\sum_{\sigma\in
S_{n}}V^{n}\subseteq V^{n}%
\]
(since $V^{n}$ is a $k$-vector space). In other words,%
\[
\left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\subseteq V^{n}.
\]
Hence, (\ref{284.pf.b.2}) (applied to $M=A$, $S=\left\{  \dfrac{1}{n!}%
\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}\right\}  $ and $Q=V^{n}$) yields%
\[
\left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle \subseteq V^{n}.
\]
Since
\begin{align*}
&  \left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle \\
&  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle
=\operatorname{symp}_{n}V,
\end{align*}
this rewrites as $\operatorname{symp}_{n}V\subseteq V^{n}$. Thus,%
\[
\sum_{i=0}^{n}\operatorname{symp}_{i}V=\underbrace{\sum_{i=0}^{n-1}%
\operatorname{symp}_{i}V}_{\substack{=\sum_{i=0}^{n-1}V^{i}\\\text{(by
(\ref{284.pf.b.1}))}}}+\underbrace{\operatorname{symp}_{n}V}_{\subseteq V^{n}%
}\subseteq\sum_{i=0}^{n-1}V^{i}+V^{n}=\sum_{i=0}^{n}V^{i}.
\]
Combined with (\ref{284.pf.b.6}), this yields
\[
\sum_{i=0}^{n}V^{i}=\sum_{i=0}^{n}\operatorname{symp}_{i}V.
\]
In other words, Proposition~\ref{Proposition284} \textbf{(b)} is proven for
$\ell=n$. This completes the induction step. Thus, the induction proof of
Proposition~\ref{Proposition284} \textbf{(b)} is complete.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition282}.]Let $\ell\in\mathbb{N}$.

Let $A=H$ and $V=\operatorname*{Prim}H$. Then, the condition (\ref{284.cond})
of Proposition~\ref{Proposition284} is satisfied (because
Proposition~\ref{Proposition283} tells us that $xy-yx\in\operatorname*{Prim}H$
for any $x\in\operatorname*{Prim}H$ and $y\in\operatorname*{Prim}H$). Thus, we
can apply Proposition~\ref{Proposition284} \textbf{(b)} and conclude that
$\sum_{i=0}^{\ell}V^{i}=\sum_{i=0}^{\ell}\operatorname{symp}_{i}V$. Since
$V=\operatorname*{Prim}H$, this becomes $\sum_{i=0}^{\ell}\left(
\operatorname*{Prim}H\right)  ^{i}=\sum_{i=0}^{\ell}\operatorname{symp}%
_{i}\left(  \operatorname*{Prim}H\right)  $. But
Proposition~\ref{Proposition272} yields
\[
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\left(  \operatorname*{Prim}H\right)
^{i}=\sum_{i=0}^{\ell}\operatorname{symp}_{i}\left(  \operatorname*{Prim}%
H\right)  .
\]
This proves Proposition~\ref{Proposition282}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem280}.]For every $\ell\in\mathbb{N}$,
Proposition~\ref{Proposition282} yields%
\[
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\operatorname{symp}_{i}\left(
\operatorname*{Prim}H\right)  \subseteq\sum_{i\in\mathbb{N}}%
\operatorname{symp}_{i}\left(  \operatorname*{Prim}H\right)  .
\]
Since $H$ is filtered, we have%
\[
H=\bigcup_{\ell\in\mathbb{N}}\underbrace{H_{\leq\ell}}_{\subseteq\sum
_{i\in\mathbb{N}}\operatorname{symp}_{i}\left(  \operatorname*{Prim}H\right)
}\subseteq\bigcup_{\ell\in\mathbb{N}}\sum_{i\in\mathbb{N}}\operatorname{symp}%
_{i}\left(  \operatorname*{Prim}H\right)  =\sum_{i\in\mathbb{N}}%
\operatorname{symp}_{i}\left(  \operatorname*{Prim}H\right)  .
\]
Combined with $\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}\left(
\operatorname*{Prim}H\right)  \subseteq H$ (which is trivial), this yields
$H=\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}\left(  \operatorname*{Prim}%
H\right)  $. This proves Theorem~\ref{Theorem280}.
\end{proof}

\section{\label{S15}Intermezzo on homogeneous subspaces}

As often happens with properties of filtered bialgebras, the particular case
of Theorem~\ref{Theorem280} for a \textit{graded} bialgebra $H$ can be
somewhat extended using the grading. Before we formulate this extension, we
introduce an auxiliary notion. This is the notion of \textit{homogeneous
subspaces} of graded vector spaces. There are several equivalent definitions
of this notion; here, we will use the following one:

\begin{definition}
\label{Definition300} Let $k$ be a field. Let $V$ be a graded $k$-vector
space. Let $W$ be a $k$-vector subspace of $V$.

We say that $W$ is a \textit{homogeneous subspace} of $V$ if $W=\bigoplus
_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  $.

Note that the internal direct sum $\bigoplus_{n\in\mathbb{N}}\left(  W\cap
V_{n}\right)  $ is always well-defined\footnote{\textit{Proof.} Since $V$ is a
graded $k$-vector space, we have $V=\bigoplus_{n\in\mathbb{N}}V_{n}$. Thus,
$\sum_{n\in\mathbb{N}}V_{n}$ is a direct sum. Hence,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{every family }\left(  v_{n}\right)  _{n\in\mathbb{N}}\in\prod
_{n\in\mathbb{N}}V_{n}\text{ satisfying }\sum_{n\in\mathbb{N}}v_{n}=0\text{
and}\\
\left(  v_{n}=0\text{ for all but finitely many }n\in\mathbb{N}\right)  \text{
must satisfy }\left(  v_{n}\right)  _{n\in\mathbb{N}}=\left(  0\right)
_{n\in\mathbb{N}}%
\end{array}
\right)  . \label{300.pf.1}%
\end{equation}
\par
Now, it follows that every family $\left(  v_{n}\right)  _{n\in\mathbb{N}}%
\in\prod_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  $ satisfying $\sum
_{n\in\mathbb{N}}v_{n}=0$ and $\left(  v_{n}=0\text{ for all but finitely many
}n\in\mathbb{N}\right)  $ must satisfy $\left(  v_{n}\right)  _{n\in
\mathbb{N}}=\left(  0\right)  _{n\in\mathbb{N}}$ (by (\ref{300.pf.1}), since
$\left(  v_{n}\right)  _{n\in\mathbb{N}}\in\prod_{n\in\mathbb{N}}\left(  W\cap
V_{n}\right)  $ yields $\left(  v_{n}\right)  _{n\in\mathbb{N}}\in\prod
_{n\in\mathbb{N}}V_{n}$). In other words, $\sum_{n\in\mathbb{N}}\left(  W\cap
V_{n}\right)  $ is a direct sum. Hence, the internal direct sum $\bigoplus
_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  $ is well-defined, qed.}, and
thus always is a $k$-vector subspace of $W$\ \ \ \ \footnote{\textit{Proof.}
Since direct sums are sums, we have $\bigoplus_{n\in\mathbb{N}}\left(  W\cap
V_{n}\right)  =\sum_{n\in\mathbb{N}}\underbrace{\left(  W\cap V_{n}\right)
}_{\subseteq W}\subseteq\sum_{n\in\mathbb{N}}W\subseteq W$ (since $W$ is a
$k$-vector space), qed.}. But it is not always the whole $W$.
\end{definition}

A basic fact from linear algebra:

\begin{remark}
\label{Remark301} Let $k$ be a field. Let $U$ and $V$ be two graded $k$-vector
spaces. Let $f:V\rightarrow U$ be a graded $k$-linear map. Then,
$\operatorname*{Ker}f$ is a homogeneous subspace of $V$.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark301}.]Let $W=\operatorname*{Ker}f$. By
Definition~\ref{Definition300}, we know that the internal direct sum
$\bigoplus_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  $ is well-defined and a
$k$-vector subspace of $W$. We are now going to show that $W=\bigoplus
_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  $.

Let $w\in W$. Then, $w\in W=\operatorname*{Ker}f\subseteq V$ and $f\left(
w\right)  =0$ (since $w\in\operatorname*{Ker}f$). Now, for every
$n\in\mathbb{N}$, we have%
\[
f\left(  p_{n,V}\left(  w\right)  \right)  =\underbrace{\left(  f\circ
p_{n,V}\right)  }_{\substack{=p_{n,U}\circ f\\\text{(by
Proposition~\ref{Proposition223},} \\\text{applied to }U\text{ instead of
}W\text{)}}}\left(  w\right)  =\left(  p_{n,U}\circ f\right)  \left(
w\right)  =p_{n,U}\left(  \underbrace{f\left(  w\right)  }_{=0}\right)
=p_{n,U}\left(  0\right)  =0
\]
and thus $p_{n,V}\left(  w\right)  \in\operatorname*{Ker}f$. But for every
$n\in\mathbb{N}$, we also have $p_{n,V}\left(  \underbrace{w}_{\in V}\right)
\in p_{n,V}\left(  V\right)  =V_{n}$ (by (\ref{221.im})). Thus, for every
$n\in\mathbb{N}$, we have $p_{n,V}\left(  w\right)  \in W\cap V_{n}$ (this
results from combining $p_{n,V}\left(  w\right)  \in\operatorname*{Ker}f=W$
and $p_{n,V}\left(  w\right)  \in V_{n}$). But (\ref{221.vsum}) (applied to
$v=w$) yields%
\begin{align*}
w  &  =\sum_{\ell\in\mathbb{N}}p_{\ell,V}\left(  w\right)  =\sum
_{n\in\mathbb{N}}\underbrace{p_{n,V}\left(  w\right)  }_{\in W\cap V_{n}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }\ell\text{ as
}n\text{ in the sum}\right) \\
&  \in\sum_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  =\bigoplus
_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  .
\end{align*}


Now forget that we fixed $w$. We thus have proven that every $w\in W$
satisfies $w\in\bigoplus_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  $. Thus,
$W\subseteq\bigoplus_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  $. Combined
with the (already known) relation $\bigoplus_{n\in\mathbb{N}}\left(  W\cap
V_{n}\right)  \subseteq W$, this yields $W=\bigoplus_{n\in\mathbb{N}}\left(
W\cap V_{n}\right)  $. Thus, $W$ is a homogeneous subspace of $V$ (by the
definition of ``homogeneous subspace''). Since $W=\operatorname*{Ker}f$, this
yields that $\operatorname*{Ker}f$ is a homogeneous subspace of $V$. This
proves Remark~\ref{Remark301}.
\end{proof}

Now we can see the following fundamental fact:

\begin{proposition}
\label{Proposition302} Let $k$ be a field. Let $H$ be a graded $k$-bialgebra.
Then, $\operatorname*{Prim}H$ (this is the set of all primitive elements of
$H$) is a homogeneous subspace of $H$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition302}.]Define a map $f:H\rightarrow H$ by%
\begin{equation}
\left(  f\left(  x\right)  =\Delta\left(  x\right)  -x\otimes1_{H}%
-1_{H}\otimes x\ \ \ \ \ \ \ \ \ \ \text{for every }x\in H\right)  .
\label{302.pf.1}%
\end{equation}
Then, this map $f$ is $k$-linear (because $\Delta$ is $k$-linear, so that the
term $\Delta\left(  x\right)  -x\otimes1_{H}-1_{H}\otimes x$ depends
$k$-linearly on $x$). We are now going to prove that $\operatorname*{Prim}%
H=\operatorname*{Ker}f$ and that $f$ is a graded map. Then, an application of
Remark~\ref{Remark301} will do the rest. Here are the details:

\textbf{a)} We have $\operatorname*{Prim}H=\operatorname*{Ker}f$.

\textit{Proof.} We have%
\begin{align*}
\operatorname*{Prim}H  &  =\left(  \text{the set of all primitive elements of
}H\right) \\
&  =\left\{  x\in H\ \mid\ \underbrace{x\text{ is primitive}}%
_{\substack{\text{this is equivalent to }\Delta\left(  x\right)
=x\otimes1_{H}+1_{H}\otimes x\\\text{(by the definition of ``primitive'')}%
}}\right\} \\
&  =\left\{  x\in H\ \mid\ \underbrace{\Delta\left(  x\right)  =x\otimes
1_{H}+1_{H}\otimes x}_{\substack{\text{this is equivalent to}\\\Delta\left(
x\right)  -x\otimes1_{H}-1_{H}\otimes x=0}}\right\} \\
&  =\left\{  x\in H\ \mid\ \underbrace{\Delta\left(  x\right)  -x\otimes
1_{H}-1_{H}\otimes x}_{=f\left(  x\right)  \text{ (by (\ref{302.pf.1}))}%
}=0\right\} \\
&  =\left\{  x\in H\ \mid\ f\left(  x\right)  =0\right\}  =\operatorname*{Ker}%
f.
\end{align*}
This proves part \textbf{a)}.

\textbf{b)} The map $f$ is graded.

\textit{Proof.} Let $n\in\mathbb{N}$. Since $\Delta$ is graded (because $H$ is
a graded $k$-coalgebra), we have $\Delta\left(  H_{n}\right)  \subseteq\left(
H\otimes H\right)  _{n}$. But by the definition of the tensor product of two
graded $k$-vector spaces, we have $\left(  H\otimes H\right)  _{n}=\sum
_{\ell=0}^{n}H_{\ell}\otimes H_{n-\ell}$. Hence,%
\[
\left(  H\otimes H\right)  _{n}=\sum_{\ell=0}^{n}H_{\ell}\otimes H_{n-\ell
}=\sum_{\ell=0}^{n-1}H_{\ell}\otimes H_{n-\ell}+H_{n}\otimes H_{n-n}\supseteq
H_{n}\otimes\underbrace{H_{n-n}}_{=H_{0}}=H_{n}\otimes H_{0}%
\]
and%
\[
\left(  H\otimes H\right)  _{n}=\sum_{\ell=0}^{n}H_{\ell}\otimes H_{n-\ell
}=H_{0}\otimes H_{n-0}+\sum_{\ell=1}^{n}H_{\ell}\otimes H_{n-\ell}\supseteq
H_{0}\otimes\underbrace{H_{n-0}}_{=H_{n}}=H_{0}\otimes H_{n}.
\]
Also, $1_{H}\in H_{0}$ (since $H$ is a graded $k$-algebra).

Now, every $x\in H_{n}$ satisfies%
\begin{align*}
f\left(  x\right)   &  =\Delta\left(  \underbrace{x}_{\in H_{n}}\right)
-\underbrace{x}_{\in H_{n}}\otimes\underbrace{1_{H}}_{\in H_{0}}%
-\underbrace{1_{H}}_{\in H_{0}}\otimes\underbrace{x}_{\in H_{n}}\\
&  \in\underbrace{\Delta\left(  H_{n}\right)  }_{\subseteq\left(  H\otimes
H\right)  _{n}}-\underbrace{H_{n}\otimes H_{0}}_{\subseteq\left(  H\otimes
H\right)  _{n}}-\underbrace{H_{0}\otimes H_{n}}_{\subseteq\left(  H\otimes
H\right)  _{n}}\\
&  \subseteq\left(  H\otimes H\right)  _{n}-\left(  H\otimes H\right)
_{n}-\left(  H\otimes H\right)  _{n}\subseteq\left(  H\otimes H\right)  _{n}%
\end{align*}
(since $\left(  H\otimes H\right)  _{n}$ is a $k$-vector space). In other
words, $f\left(  H_{n}\right)  \subseteq\left(  H\otimes H\right)  _{n}$.

Now forget that we fixed $n$. Thus, we have proven that every $n\in\mathbb{N}$
satisfies $f\left(  H_{n}\right)  \subseteq\left(  H\otimes H\right)  _{n}$.
In other words, the map $f$ is graded. This proves part \textbf{b)}.

\textbf{c)} Since $f$ is graded (by part \textbf{b)}), we can apply
Remark~\ref{Remark301} to $U=H$ and $V=H$, and conclude that
$\operatorname*{Ker}f$ is a homogeneous subspace of $H$. Since
$\operatorname*{Ker}f=\operatorname*{Prim}H$ (by part \textbf{a)}), this
yields that $\operatorname*{Prim}H$ is a homogeneous subspace of $H$. This
proves Proposition~\ref{Proposition302}.
\end{proof}

\section{\label{S16}A graded Theorem~\ref{Theorem280}}

Now, to formulate the graded strengthening of Theorem~\ref{Theorem280}, we
define a notion:

\begin{definition}
\label{Definition310} Let $k$ be a field of characteristic $0$. Let $A$ be a
graded $k$-algebra. Let $V$ be a $k$-vector subspace of $A$. Let
$n\in\mathbb{N}$. Let $\mu$ be an $n$-tuple of nonnegative integers. Then, we
define a $k$-vector subspace $\operatorname{symp}^{\mu}V$ of $A$ as follows:
Let $V_{i}$ denote $V\cap A_{i}$ for every $i\in\mathbb{N}$. Write the
$n$-tuple $\mu$ in the form $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu
_{n}\right)  $. Then, we define $\operatorname{symp}^{\mu}V$ as the $k$-vector
subspace%
\[
\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times
V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\rangle
\]
of $A$ (where $S_{n}$ denotes the $n$-th symmetric group).
\end{definition}

\begin{definition}
\label{Definition311} A \textit{partition} will mean a nonincreasing finite
sequence\footnote{A \textit{finite sequence}, of course, means the same as a
\textit{tuple} (i. e., an object which, for a suitable $n$, is an $n$-tuple).
The word ``nonincreasing'' means the same as ``weakly decreasing'': i.e., a
sequence $\left(  a_{1}, a_{2}, \ldots, a_{n}\right)  $ is said to be
\textit{nonincreasing} if and only if $a_{1} \geq a_{2} \geq\cdots\geq a_{n}$.
(The concept of ``nonincreasing'' is defined similarly for infinite
sequences.)} of positive integers.
\end{definition}

Note that Definition~\ref{Definition311} is not the only definition of a
partition. There is a different definition, which defines a partition as a
nonincreasing \textit{infinite but essentially finite}\footnote{A sequence of
integers is said to be \textit{essentially finite} if all but finitely many of
its entries are $0$.} sequence of nonnegative integers. This definition of a
partition is, of course, not directly equivalent to
Definition~\ref{Definition311}, but it is essentially the same, since there
exists a canonical bijection%
\begin{align*}
&  \left(  \text{the set of nonincreasing finite sequences of positive
integers}\right) \\
&  \rightarrow\left(  \text{the set of nonincreasing infinite but essentially
finite sequences of nonnegative integers}\right)  .
\end{align*}
\footnote{This bijection sends every nonincreasing finite sequence $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  $ of positive integers to
the infinite sequence $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{n},0,0,\ldots\right)  $. (The inverse of this bijection trims an infinite
sequence after its last nonzero entry.)}. We will not care about this here (we
are not really studying partitions here) and just define a partition by
Definition~\ref{Definition311}.

We now claim:

\begin{theorem}
\label{Theorem312} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected graded cocommutative bialgebra over $k$. Then,%
\[
H=\sum_{\lambda\text{ is a partition}}\operatorname{symp}^{\lambda}\left(
\operatorname*{Prim}H\right)  .
\]
Note that for every partition $\lambda$, the $k$-vector subspace
$\operatorname{symp}^{\lambda}\left(  \operatorname*{Prim}H\right)  $ of $H$
is well-defined (due to Definition~\ref{Definition310}, since $\lambda$ is an
$n$-tuple for an appropriate $n$).
\end{theorem}

Note that Theorem~\ref{Theorem312} can be strengthened to:

\begin{theorem}
\label{Theorem313} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected graded cocommutative bialgebra over $k$. Then,%
\[
H=\bigoplus_{\lambda\text{ is a partition}}\operatorname*{symp}%
\nolimits^{\lambda}\left(  \operatorname*{Prim}H\right)  .
\]
Note that for every partition $\lambda$, the $k$-vector subspace
$\operatorname{symp}^{\lambda}\left(  \operatorname*{Prim}H\right)  $ of $H$
is well-defined (due to Definition~\ref{Definition310}, since $\lambda$ is an
$n$-tuple for an appropriate $n$).
\end{theorem}

This is, however, a significantly harder result, requiring the
Poincar\'{e}-Birkhoff-Witt and Cartier-Milnor-Moore theorems to prove; we will
not prove this here.

We notice that \cite{2} makes use of Theorem~\ref{Theorem313} in the proof of
Theorem 4.3 in \cite{2}.\footnote{Namely, if we work with the notations of
Section 4 of \cite{2}, then applying our Theorem~\ref{Theorem313} to $H=A$
yields $A=\bigoplus_{\lambda\text{ is a partition}}\operatorname*{symp}%
\nolimits^{\lambda}\left(  \operatorname*{Prim}A\right)  $. But
$\operatorname{symp}^{\lambda}\left(  \operatorname*{Prim}A\right)  $ is
exactly what was denoted by $A^{\lambda}$ in \cite{2}. Thus, we get
$A=\bigoplus_{\lambda\text{ is a partition}}A^{\lambda}$ (still working with
the notations of \cite{2}). This fact is used in the proof of Theorem 4.3 in
\cite{2} (where it is proven using the Cartier-Milnor-Moore theorem and the
Poincar\'{e}-Birkhoff-Witt theorem).} \textit{However}, the proof of Theorem
4.3 in \cite{2} does not need the whole strength of Theorem~\ref{Theorem313};
in fact it only uses Theorem~\ref{Theorem312}.\footnote{In fact, if we work
with the notations of Section 4 of \cite{2}, then applying our
Theorem~\ref{Theorem312} to $H=A$ yields $A=\sum_{\lambda\text{ is a
partition}}\operatorname{symp}^{\lambda}\left(  \operatorname*{Prim}A\right)
$. But $\operatorname{symp}^{\lambda}\left(  \operatorname*{Prim}A\right)  $
is exactly what was denoted by $A^{\lambda}$ in \cite{2}. Thus, we get
$A=\sum_{\lambda\text{ is a partition}}A^{\lambda}$ (still working with the
notations of \cite{2}). This is already enough to prove that, if the
restriction of $E_{\lambda}^{\iota}$ to $A^{\lambda}$ is the identity for
every partition $\lambda$, then $\sum_{\left\vert \mu\right\vert
=n}\operatorname*{Im}\left(  E_{\mu}^{\iota}\right)  =A_{n}$.} Hence, our
proof of Theorem~\ref{Theorem312} further below will relieve the proof of
Theorem 4.3 in \cite{2} from having to use the Cartier-Milnor-Moore and
Poincar\'{e}-Birkhoff-Witt theorems.\footnote{Note that the proof of Corollary
4.4 in \cite{2} may still require these two theorems - I don't know (I don't
understand how Corollary 4.4 in \cite{2} is proven).}

We will derive Theorem~\ref{Theorem312} from a rather elementary fact once again:

\begin{proposition}
\label{Proposition314} Let $k$ be a field of characteristic $0$. Let $A$ be a
graded $k$-algebra. Let $V$ be a homogeneous subspace of $A$ such that $V\cap
A_{0}=0$. Assume that%
\begin{equation}
xy-yx\in V\ \ \ \ \ \ \ \ \ \ \text{for any }x\in V\text{ and }y\in V.
\label{314.cond}%
\end{equation}


Then,%
\[
\sum_{i\in\mathbb{N}}V^{i}=\sum_{\lambda\text{ is a partition}}%
\operatorname{symp}^{\lambda}V.
\]
\footnote{Recall that $V^{i}$ is to be understood according to
Convention~\ref{Convention100}. Hence, $V^{i}$ means the $i$-th power of the
subspace $V$ of the $k$-algebra $A$.}
\end{proposition}

To prove this, we begin with a trivial consequence of
Proposition~\ref{Proposition284}:

\begin{proposition}
\label{Proposition315} Let $k$ be a field of characteristic $0$. Let $A$ be a
$k$-algebra. Let $V$ be a $k$-vector subspace of $A$. Assume that%
\[
xy-yx\in V\ \ \ \ \ \ \ \ \ \ \text{for any }x\in V\text{ and }y\in V.
\]


Then,
\[
\sum_{i\in\mathbb{N}}V^{i}=\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}V.
\]


(Here, $V^{i}$ is to be understood as in Proposition~\ref{Proposition314}.)
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition315}.]Every $\ell\in\mathbb{N}$
satisfies%
\begin{align}
V^{\ell}  &  \subseteq V^{\ell}+\sum_{i=0}^{\ell-1}V^{i}=\sum_{i=0}^{\ell
}V^{i}=\sum_{i=0}^{\ell}\operatorname{symp}_{i}V\ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition~\ref{Proposition284} \textbf{(b)}}\right) \nonumber\\
&  \subseteq\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}V \label{315.pf.1}%
\end{align}
and%
\begin{align}
\operatorname{symp}_{\ell}V  &  \subseteq\operatorname{symp}_{\ell}%
V+\sum_{i=0}^{\ell-1}\operatorname{symp}_{i}V=\sum_{i=0}^{\ell}%
\operatorname{symp}_{i}V\nonumber\\
&  =\sum_{i=0}^{\ell}V^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{Proposition284} \textbf{(b)}}\right) \nonumber\\
&  \subseteq\sum_{i\in\mathbb{N}}V^{i}. \label{315.pf.2}%
\end{align}
Now,
\begin{align*}
\sum_{i\in\mathbb{N}}V^{i}  &  =\sum_{\ell\in\mathbb{N}}\underbrace{V^{\ell}%
}_{\subseteq\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}V\text{ (by
(\ref{315.pf.1}))}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
index }i\text{ as }\ell\text{ in the sum}\right) \\
&  \subseteq\sum_{\ell\in\mathbb{N}}\sum_{i\in\mathbb{N}}\operatorname{symp}%
_{i}V\subseteq\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}%
V\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{i\in\mathbb{N}}%
\operatorname{symp}_{i}V\text{ is a }k\text{-vector space}\right)  ,
\end{align*}
combined with%
\begin{align*}
\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}V  &  =\sum_{\ell\in\mathbb{N}%
}\underbrace{\operatorname{symp}_{\ell}V}_{\subseteq\sum_{i\in\mathbb{N}}%
V^{i}\text{ (by (\ref{315.pf.2}))}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we
renamed the index }i\text{ as }\ell\text{ in the sum}\right) \\
&  \subseteq\sum_{\ell\in\mathbb{N}}\sum_{i\in\mathbb{N}}V^{i}\subseteq
\sum_{i\in\mathbb{N}}V^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum
_{i\in\mathbb{N}}V^{i}\text{ is a }k\text{-vector space}\right)  ,
\end{align*}
yields $\sum_{i\in\mathbb{N}}V^{i}=\sum_{i\in\mathbb{N}}\operatorname{symp}%
_{i}V$. This proves Proposition~\ref{Proposition315}.
\end{proof}

The next step is less obvious, but still not a large step:

\begin{proposition}
\label{Proposition316} Let $k$ be a field of characteristic $0$. Let $A$ be a
graded $k$-algebra. Let $V$ be a homogeneous subspace of $A$ such that $V\cap
A_{0}=0$. Then,%
\[
\operatorname{symp}_{n}V=\sum_{\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times
n}}\operatorname{symp}^{\mu}V\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\mathbb{N}.
\]

\end{proposition}

Note that we are not summing over all partitions $\mu$, but over all
$n$-tuples of positive integers.

To prove this, we need the following purely linear-algebraic fact (which is an
analogue of the product rule for tensor products, and is proven in the same
way as the product rule):

\begin{remark}
\label{Remark317} Let $k$ be a field. Let $\left(  W_{i}\right)  _{i\in I}$ be
a family of $k$-vector spaces (where $I$ is some index set). Let
$m\in\mathbb{N}$. For every $m$-tuple $\mu\in I^{\times m}$, let us write the
$m$-tuple $\mu$ in the form $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu
_{m}\right)  $.\ \ \ \ \footnote{This means that, for every $s\in\left\{
1,2,\ldots,m\right\}  $, we denote by $\mu_{s}$ the $s$-th element of the
$m$-tuple $\mu$.}. Then,%
\[
\left(  \bigoplus_{i\in I}W_{i}\right)  ^{\otimes m}=\bigoplus_{\mu\in
I^{\times m}}\left(  W_{\mu_{1}}\otimes W_{\mu_{2}}\otimes\cdots\otimes
W_{\mu_{m}}\right)
\]
(where $W_{j}$ is considered a subspace of $\bigoplus_{i\in I}W_{i}$ for every
$j\in I$, and thus $W_{\mu_{1}}\otimes W_{\mu_{2}}\otimes\cdots\otimes
W_{\mu_{m}}$ is considered a subspace of $\left(  \bigoplus_{i\in I}%
W_{i}\right)  ^{\otimes m}$ for every $\mu\in I^{\times m}$).
\end{remark}

Remark~\ref{Remark317} is a particular case of the following, even more
general fact:

\begin{remark}
\label{Remark318} Let $k$ be a field. Let $m\in\mathbb{N}$. Let $I_{1}$,
$I_{2}$, $\ldots$, $I_{m}$ be $m$ sets. For every $j\in\left\{  1,2,\ldots
,m\right\}  $, let $\left(  W_{j,i}\right)  _{i\in I_{j}}$ be a family of
$k$-vector spaces. For every $m$-tuple $\mu\in I_{1}\times I_{2}\times
\cdots\times I_{m}$, let us write the $m$-tuple $\mu$ in the form $\mu=\left(
\mu_{1},\mu_{2},\ldots,\mu_{m}\right)  $.\ \ \ \ \footnote{This means that,
for every $s\in\left\{  1,2,\ldots,m\right\}  $, we denote by $\mu_{s}$ the
$s$-th element of the $m$-tuple $\mu$.}. Then,%
\begin{align*}
&  \left(  \bigoplus_{i\in I_{1}}W_{1,i}\right)  \otimes\left(  \bigoplus
_{i\in I_{2}}W_{2,i}\right)  \otimes\cdots\otimes\left(  \bigoplus_{i\in
I_{m}}W_{m,i}\right) \\
&  =\bigoplus_{\mu\in I_{1}\times I_{2}\times\cdots\times I_{m}}\left(
W_{1,\mu_{1}}\otimes W_{2,\mu_{2}}\otimes\cdots\otimes W_{m,\mu_{m}}\right)
\end{align*}
(where $W_{j,\ell}$ is considered a subspace of $\bigoplus_{i\in I_{j}}%
W_{j,i}$ for every $j\in\left\{  1,2,\ldots,m\right\}  $ and $\ell\in I_{j}$,
and thus $W_{1,\mu_{1}}\otimes W_{2,\mu_{2}}\otimes\cdots\otimes W_{m,\mu_{m}%
}$ is considered a subspace of $\left(  \bigoplus_{i\in I_{1}}W_{1,i}\right)
\otimes\left(  \bigoplus_{i\in I_{2}}W_{2,i}\right)  \otimes\cdots
\otimes\left(  \bigoplus_{i\in I_{m}}W_{m,i}\right)  $ for every $\mu\in
I_{1}\times I_{2}\times\cdots\times I_{m}$).
\end{remark}

Remark~\ref{Remark318} can be proven by induction on $m$ (just as the product
rule). Remark~\ref{Remark317} follows from Remark~\ref{Remark318} (applied to
$I_{1}=I$, $I_{2}=I$, $\ldots$, $I_{m}=I$ and $W_{j,i}=W_{i}$).

\begin{proof}
[Proof of Proposition~\ref{Proposition316}.]For every $n$-tuple $\mu\in
I^{\times n}$, let us write the $n$-tuple $\mu$ in the form $\mu=\left(
\mu_{1},\mu_{2},\ldots,\mu_{n}\right)  $.\ \ \ \ \footnote{This means that,
for every $s\in\left\{  1,2,\ldots,n\right\}  $, we denote by $\mu_{s}$ the
$s$-th element of the $n$-tuple $\mu$.}

Let $V_{i}$ denote $V\cap A_{i}$ for every $i\in\mathbb{N}$. Since $V$ is a
homogeneous subspace of $A$, we have
\begin{align*}
V  &  =\bigoplus_{n\in\mathbb{N}}\left(  V\cap A_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of ``homogeneous
subspace''}\right) \\
&  =\underbrace{V\cap A_{0}}_{=0}\oplus\underbrace{\bigoplus_{\substack{n\in
\mathbb{N};\\n\geq1}}}_{=\bigoplus_{n\in\left\{  1,2,3,\ldots\right\}  }%
}\underbrace{\left(  V\cap A_{n}\right)  }_{\substack{=V_{n}\\\text{(since
}V_{n}\text{ was defined as }V\cap A_{n}\text{)}}}=\bigoplus_{n\in\left\{
1,2,3,\ldots\right\}  }V_{n}\\
&  =\bigoplus_{i\in\left\{  1,2,3,\ldots\right\}  }V_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }n\text{ as
}i\right)  .
\end{align*}


Now, let $n\in\mathbb{N}$ be arbitrary. Then, $V=\bigoplus_{i\in\left\{
1,2,3,\ldots\right\}  }V_{i}$ leads to%
\begin{align}
V^{\otimes n}  &  =\left(  \bigoplus_{i\in\left\{  1,2,3,\ldots\right\}
}V_{i}\right)  ^{\otimes n}=\bigoplus_{\mu\in\left\{  1,2,3,\ldots\right\}
^{\times n}}\left(  V_{\mu_{1}}\otimes V_{\mu_{2}}\otimes\cdots\otimes
V_{\mu_{n}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Remark~\ref{Remark317}, applied to
}I=\left\{  1,2,3,\ldots\right\}  \text{, }m=n\text{ and }W_{i}=V_{i}\right)
\nonumber\\
&  =\sum_{\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times n}}\left(  V_{\mu_{1}%
}\otimes V_{\mu_{2}}\otimes\cdots\otimes V_{\mu_{n}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are sums}\right)  .
\label{316.pf.1}%
\end{align}


Let $\operatorname*{tensor}\nolimits_{n}:V^{\times n}\rightarrow V^{\otimes
n}$ be the map defined by%
\[
\left(  \operatorname*{tensor}\nolimits_{n}\left(  v_{1},v_{2},\ldots
,v_{n}\right)  =v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{\times n}\right)  .
\]
Then, by the universal property of the $n$-th tensor power, every
$n$-multilinear map from $V^{\times n}$ factors through
$\operatorname*{tensor}\nolimits_{n}$.

Now let us define a map $\varphi_{n}:V^{\times n}\rightarrow A$ by%
\[
\left(  \varphi_{n}\left(  v_{1},v_{2},\ldots,v_{n}\right)  =\dfrac{1}{n!}%
\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right)  .
\]
Then, this map $\varphi_{n}$ is clearly $n$-multilinear. Thus, $\varphi_{n}$
factors through $\operatorname*{tensor}\nolimits_{n}$ (since every
$n$-multilinear map from $V^{\times n}$ factors through
$\operatorname*{tensor}\nolimits_{n}$). This means that there exists a
$k$-linear map $\psi_{n}:V^{\otimes n}\rightarrow A$ such that $\varphi
_{n}=\psi_{n}\circ\operatorname*{tensor}\nolimits_{n}$. Consider this
$\psi_{n}$. Every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$
satisfies%
\begin{align}
\psi_{n}\underbrace{\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  }_{=\operatorname*{tensor}\nolimits_{n}\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  }  &  =\psi_{n}\left(  \operatorname*{tensor}%
\nolimits_{n}\left(  v_{1},v_{2},\ldots,v_{n}\right)  \right)
=\underbrace{\left(  \psi_{n}\circ\operatorname*{tensor}\nolimits_{n}\right)
}_{=\varphi_{n}}\left(  v_{1},v_{2},\ldots,v_{n}\right) \nonumber\\
&  =\varphi_{n}\left(  v_{1},v_{2},\ldots,v_{n}\right) \nonumber\\
&  =\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }%
v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }.
\label{316.pf.2}%
\end{align}


Now, we recall the basic linear-algebraic fact that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for any two }k\text{-vector spaces }M\text{ and }R\text{, any
}k\text{-linear map }\phi:M\rightarrow R\\
\text{and every subset }S\text{ of }M\text{ satisfy }\phi\left(  \left\langle
S\right\rangle \right)  =\left\langle \phi\left(  S\right)  \right\rangle
\end{array}
\right)  . \label{316.pf.3}%
\end{equation}


Since the tensor power $V^{\otimes n}$ is generated by pure tensors, we have%
\begin{align*}
V^{\otimes n}  &  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\right\rangle \\
&  =\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle ,
\end{align*}
so that%
\begin{align}
\psi_{n}\left(  V^{\otimes n}\right)   &  =\psi_{n}\left(  \left\langle
\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}  \right\rangle
\right) \nonumber\\
&  =\left\langle \underbrace{\psi_{n}\left(  \left\{  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}\right\}  \right)  }_{=\left\{  \psi_{n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}  }\right\rangle
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{316.pf.3}), applied to }M=V^{\otimes n}\text{, }R=A\text{,
}\phi=\psi_{n}\text{ and}\\
S=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\end{array}
\right) \nonumber\\
&  =\left\langle \left\{  \psi_{n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{\times n}\right\}  \right\rangle \nonumber\\
&  =\left\langle \underbrace{\psi_{n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  }_{\substack{=\dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\\\text{(by (\ref{316.pf.2}))}}}\ \mid\ \left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle \nonumber\\
&  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle
=\operatorname{symp}_{n}V. \label{316.pf.6}%
\end{align}


On the other hand, for every $\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times
n}$, the tensor product $V_{\mu_{1}}\otimes V_{\mu_{2}}\otimes\cdots\otimes
V_{\mu_{n}}$ is generated by pure tensors, so that%
\begin{align*}
V_{\mu_{1}}\otimes V_{\mu_{2}}\otimes\cdots\otimes V_{\mu_{n}}  &
=\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times V_{\mu_{2}}\times
\cdots\times V_{\mu_{n}}\right\rangle \\
&  =\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times
V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\}  \right\rangle ,
\end{align*}
so that%
\begin{align}
&  \psi_{n}\left(  V_{\mu_{1}}\otimes V_{\mu_{2}}\otimes\cdots\otimes
V_{\mu_{n}}\right) \nonumber\\
&  =\psi_{n}\left(  \left\langle \left\{  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V_{\mu_{1}}\times V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\}
\right\rangle \right) \nonumber\\
&  =\left\langle \underbrace{\psi_{n}\left(  \left\{  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V_{\mu_{1}}\times V_{\mu_{2}}\times\cdots\times V_{\mu_{n}%
}\right\}  \right)  }_{=\left\{  \psi_{n}\left(  v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V_{\mu_{1}}\times V_{\mu_{2}}\times\cdots\times V_{\mu_{n}%
}\right\}  }\right\rangle \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{316.pf.3}), applied to }M=V_{\mu_{1}}\otimes V_{\mu_{2}}%
\otimes\cdots\otimes V_{\mu_{n}}\text{, }R=A\text{, }\phi=\psi_{n}\text{
and}\\
S=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times V_{\mu_{2}}\times
\cdots\times V_{\mu_{n}}\right\}
\end{array}
\right) \nonumber\\
&  =\left\langle \left\{  \psi_{n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V_{\mu_{1}}\times V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\}
\right\rangle \nonumber\\
&  =\left\langle \underbrace{\psi_{n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  }_{\substack{=\dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\\\text{(by (\ref{316.pf.2}))}}}\ \mid\ \left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times V_{\mu_{2}}\times
\cdots\times V_{\mu_{n}}\right\rangle \nonumber\\
&  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times
V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\rangle =\operatorname*{symp}%
\nolimits^{\mu}V. \label{316.pf.7}%
\end{align}


Now, (\ref{316.pf.6}) yields%
\begin{align*}
\operatorname{symp}_{n}V  &  =\psi_{n}\left(  V^{\otimes n}\right)  =\psi
_{n}\left(  \sum_{\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times n}}\left(
V_{\mu_{1}}\otimes V_{\mu_{2}}\otimes\cdots\otimes V_{\mu_{n}}\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{316.pf.1})}\right) \\
&  =\sum_{\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times n}}\underbrace{\psi
_{n}\left(  V_{\mu_{1}}\otimes V_{\mu_{2}}\otimes\cdots\otimes V_{\mu_{n}%
}\right)  }_{\substack{=\operatorname{symp}^{\mu}V\\\text{(by (\ref{316.pf.7}%
))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\psi_{n}\text{ is }%
k\text{-linear}\right) \\
&  =\sum_{\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times n}}\operatorname{symp}%
^{\mu}V.
\end{align*}
This proves Proposition~\ref{Proposition316}.
\end{proof}

The next easy step is to prove the following fact:

\begin{proposition}
\label{Proposition319} Let $k$ be a field of characteristic $0$. Let $A$ be a
graded $k$-algebra. Let $n\in\mathbb{N}$. Let $\mu\in\left\{  1,2,3,\ldots
\right\}  ^{\times n}$ be arbitrary. Then,%
\[
\operatorname{symp}^{\mu}V\subseteq\sum_{\lambda\text{ is a partition}%
}\operatorname{symp}^{\lambda}V.
\]

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition319}.]Let us write the $n$-tuple $\mu$
in the form $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{n}\right)  $%
.\ \ \ \ \footnote{This means that, for every $s\in\left\{  1,2,\ldots
,n\right\}  $, we denote by $\mu_{s}$ the $s$-th element of the $n$-tuple
$\mu$.}

Every finite sequence of integers can be sorted into nonincreasing order by a
permutation. Applying this to the sequence $\left(  \mu_{1},\mu_{2},\ldots
,\mu_{n}\right)  $, we see that there exists a permutation $\tau\in S_{n}$
such that $\mu_{\tau\left(  1\right)  }\geq\mu_{\tau\left(  2\right)  }%
\geq\cdots\geq\mu_{\tau\left(  n\right)  }$. Consider such a $\tau$.

The numbers $\mu_{\tau\left(  1\right)  }$, $\mu_{\tau\left(  2\right)  }$,
$\ldots$, $\mu_{\tau\left(  n\right)  }$ are positive integers (since $\mu
\in\left\{  1,2,3,\ldots\right\}  ^{\times n}$) and satisfy $\mu_{\tau\left(
1\right)  }\geq\mu_{\tau\left(  2\right)  }\geq\cdots\geq\mu_{\tau\left(
n\right)  }$. Hence, $\left(  \mu_{\tau\left(  1\right)  },\mu_{\tau\left(
2\right)  },\ldots,\mu_{\tau\left(  n\right)  }\right)  $ is a nonincreasing
finite sequence of positive integers. In other words, $\left(  \mu
_{\tau\left(  1\right)  },\mu_{\tau\left(  2\right)  },\ldots,\mu_{\tau\left(
n\right)  }\right)  $ is a partition (by Definition~\ref{Definition311}).
Denote this partition by $\rho$. Thus, $\rho=\left(  \mu_{\tau\left(
1\right)  },\mu_{\tau\left(  2\right)  },\ldots,\mu_{\tau\left(  n\right)
}\right)  $.

Let us write the $n$-tuple $\rho$ in the form $\rho=\left(  \rho_{1},\rho
_{2},\ldots,\rho_{n}\right)  $. Since we already know that $\rho=\left(
\mu_{\tau\left(  1\right)  },\mu_{\tau\left(  2\right)  },\ldots,\mu
_{\tau\left(  n\right)  }\right)  $, we obtain $\left(  \rho_{1},\rho
_{2},\ldots,\rho_{n}\right)  =\rho=\left(  \mu_{\tau\left(  1\right)  }%
,\mu_{\tau\left(  2\right)  },\ldots,\mu_{\tau\left(  n\right)  }\right)  $.
Hence,
\begin{equation}
\rho_{j}=\mu_{\tau\left(  j\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }%
j\in\left\{  1,2,\ldots,n\right\}  . \label{319.pf.rho-vs-mu}%
\end{equation}


Now, by the definition of $\operatorname{symp}^{\rho}V$, we have%
\[
\operatorname{symp}^{\rho}V=\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V_{\rho_{1}}\times V_{\rho_{2}}\times\cdots\times V_{\rho_{n}}\right\rangle ,
\]
so that%
\begin{align*}
&  \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\rho_{1}}\times
V_{\rho_{2}}\times\cdots\times V_{\rho_{n}}\right\} \\
&  \subseteq\left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}%
v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V_{\rho_{1}}\times V_{\rho_{2}}\times\cdots\times V_{\rho_{n}}\right\}
\right\rangle \\
&  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\rho_{1}}\times
V_{\rho_{2}}\times\cdots\times V_{\rho_{n}}\right\rangle \\
&  =\operatorname{symp}^{\rho}V.
\end{align*}
In other words,%
\begin{equation}
\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma
\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }\in\operatorname*{symp}%
\nolimits^{\rho}V\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V_{\rho_{1}}\times V_{\rho_{2}}\times
\cdots\times V_{\rho_{n}}. \label{319.pf.1}%
\end{equation}


Now, let us prove that%
\begin{equation}
\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma
\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }\in\operatorname*{symp}%
\nolimits^{\rho}V\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times V_{\mu_{2}}\times
\cdots\times V_{\mu_{n}}. \label{319.pf.2}%
\end{equation}


\textit{Proof of (\ref{319.pf.2}).} Since $S_{n}$ is a group, the map%
\[
S_{n}\rightarrow S_{n},\ \ \ \ \ \ \ \ \ \ \sigma\mapsto\tau\circ\sigma
\]
is a bijection.

Let $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times V_{\mu
_{2}}\times\cdots\times V_{\mu_{n}}$ be arbitrary. Then, for every
$i\in\left\{  1,2,\ldots,n\right\}  $, we have $v_{i}\in V_{\mu_{i}}$. Hence,
for every $j\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
v_{\tau\left(  j\right)  }  &  \in V_{\mu_{\tau\left(  j\right)  }%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the formula }v_{i}\in V_{\mu_{i}}\text{,
applied to }i=\tau\left(  j\right)  \right) \\
&  =V_{\rho_{j}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{\tau\left(
j\right)  }=\rho_{j}\text{ (by (\ref{319.pf.rho-vs-mu}))}\right)  .
\end{align*}
In other words, $\left(  v_{\tau\left(  1\right)  },v_{\tau\left(  2\right)
},\ldots,v_{\tau\left(  n\right)  }\right)  \in V_{\rho_{1}}\times V_{\rho
_{2}}\times\cdots\times V_{\rho_{n}}$. Thus, (\ref{319.pf.1}) (applied to
$\left(  v_{\tau\left(  1\right)  },v_{\tau\left(  2\right)  },\ldots
,v_{\tau\left(  n\right)  }\right)  $ instead of $\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  $) yields%
\[
\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\tau\left(  \sigma\left(  1\right)
\right)  }v_{\tau\left(  \sigma\left(  2\right)  \right)  }\cdots
v_{\tau\left(  \sigma\left(  n\right)  \right)  }\in\operatorname*{symp}%
\nolimits^{\rho}V.
\]
But since%
\begin{align*}
&  \sum_{\sigma\in S_{n}}\underbrace{v_{\tau\left(  \sigma\left(  1\right)
\right)  }v_{\tau\left(  \sigma\left(  2\right)  \right)  }\cdots
v_{\tau\left(  \sigma\left(  n\right)  \right)  }}_{\substack{=v_{\left(
\tau\circ\sigma\right)  \left(  1\right)  }v_{\left(  \tau\circ\sigma\right)
\left(  2\right)  }\cdots v_{\left(  \tau\circ\sigma\right)  \left(  n\right)
}\\\text{(since }\tau\left(  \sigma\left(  1\right)  \right)  =\left(
\tau\circ\sigma\right)  \left(  1\right)  \text{, }\tau\left(  \sigma\left(
2\right)  \right)  =\left(  \tau\circ\sigma\right)  \left(  2\right)  \text{,
}\ldots\text{, }\tau\left(  \sigma\left(  n\right)  \right)  =\left(
\tau\circ\sigma\right)  \left(  n\right)  \text{)}}}\\
&  =\sum_{\sigma\in S_{n}}v_{\left(  \tau\circ\sigma\right)  \left(  1\right)
}v_{\left(  \tau\circ\sigma\right)  \left(  2\right)  }\cdots v_{\left(
\tau\circ\sigma\right)  \left(  n\right)  }\\
&  =\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(
2\right)  }\cdots v_{\sigma\left(  n\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\sigma\text{ for }\tau\circ\sigma\text{ in the
sum, since the map}\\
S_{n}\rightarrow S_{n},\ \ \ \ \ \ \ \ \ \ \sigma\mapsto\tau\circ\sigma\\
\text{is a bijection}%
\end{array}
\right)  ,
\end{align*}
this rewrites as $\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\in\operatorname{symp}^{\rho}V$. This proves (\ref{319.pf.2}).

Now that (\ref{319.pf.2}) is proven, we can rewrite (\ref{319.pf.2}) as
follows:%
\begin{align*}
&  \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times
V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\} \\
&  \subseteq\operatorname{symp}^{\rho}V.
\end{align*}
Thus, (\ref{284.pf.b.2}) (applied to $M=A$, \newline$S=\left\{  \dfrac{1}%
{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(
2\right)  }\cdots v_{\sigma\left(  n\right)  }\ \mid\ \left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times V_{\mu_{2}}\times
\cdots\times V_{\mu_{n}}\right\}  $ and $Q=\operatorname{symp}^{\rho}V$)
yields%
\begin{align*}
&  \left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times
V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\}  \right\rangle \\
&  \subseteq\operatorname{symp}^{\rho}V.
\end{align*}
Thus,%
\begin{align*}
&  \operatorname{symp}^{\mu}V\\
&  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times
V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\rangle \\
&  =\left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{\mu_{1}}\times
V_{\mu_{2}}\times\cdots\times V_{\mu_{n}}\right\}  \right\rangle \\
&  \subseteq\operatorname{symp}^{\rho}V\subseteq\sum_{\lambda\text{ is a
partition}}\operatorname{symp}^{\lambda}V
\end{align*}
(since $\rho$ is a partition, and thus $\operatorname{symp}^{\rho}V$ is an
addend of the sum $\sum_{\lambda\text{ is a partition}}\operatorname{symp}%
^{\lambda}V$). This proves Proposition~\ref{Proposition319}.
\end{proof}

Now, finally, we prove Proposition~\ref{Proposition314} and
Theorem~\ref{Theorem312}:

\begin{proof}
[Proof of Proposition~\ref{Proposition314}.]Proposition~\ref{Proposition315}
yields%
\begin{align}
\sum_{i\in\mathbb{N}}V^{i}  &  =\sum_{i\in\mathbb{N}}%
\underbrace{\operatorname{symp}_{i}V}_{\substack{=\sum_{\mu\in\left\{
1,2,3,\ldots\right\}  ^{\times i}}\operatorname*{symp}\nolimits^{\mu
}V\\\text{(by Proposition~\ref{Proposition316}, applied to }n=i\text{)}}%
}=\sum_{i\in\mathbb{N}}\sum_{\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times i}%
}\underbrace{\operatorname{symp}^{\mu}V}_{\substack{\subseteq\sum
_{\lambda\text{ is a partition}}\operatorname*{symp}\nolimits^{\lambda
}V\\\text{(by Proposition~\ref{Proposition319}, applied to }n=i\text{)}%
}}\nonumber\\
&  \subseteq\sum_{i\in\mathbb{N}}\sum_{\mu\in\left\{  1,2,3,\ldots\right\}
^{\times i}}\sum_{\lambda\text{ is a partition}}\operatorname*{symp}%
\nolimits^{\lambda}V\subseteq\sum_{\lambda\text{ is a partition}%
}\operatorname{symp}^{\lambda}V\label{314.pf.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{\lambda\text{ is a
partition}}\operatorname{symp}^{\lambda}V\text{ is a }k\text{-vector
space}\right)  .\nonumber
\end{align}


On the other hand, for every partition $\lambda$, we have $\operatorname{symp}%
^{\lambda}V\subseteq\sum_{n\in\mathbb{N}}\sum_{\mu\in\left\{  1,2,3,\ldots
\right\}  ^{\times n}}\operatorname*{symp}\nolimits^{\mu}V$%
.\ \ \ \ \footnote{\textit{Proof.} Let $\lambda$ be a partition. Then,
$\lambda$ is a tuple of positive integers. In other words, there exists an
$m\in\mathbb{N}$ such that $\lambda$ is an $m$-tuple of positive integers.
Consider this $m$. Then, $\lambda$ is an $m$-tuple of positive integers, so
that $\lambda\in\left\{  1,2,3,\ldots\right\}  ^{\times m}$. Hence,
$\operatorname{symp}^{\lambda}V$ is an addend of the sum $\sum_{\mu\in\left\{
1,2,3,\ldots\right\}  ^{\times m}}\operatorname*{symp}\nolimits^{\mu}V$. Thus,
$\operatorname{symp}^{\lambda}V\subseteq\sum_{\mu\in\left\{  1,2,3,\ldots
\right\}  ^{\times m}}\operatorname{symp}^{\mu}V$. But $\sum_{\mu\in\left\{
1,2,3,\ldots\right\}  ^{\times m}}\operatorname{symp}^{\mu}V$ is an addend of
the sum $\sum_{n\in\mathbb{N}}\sum_{\mu\in\left\{  1,2,3,\ldots\right\}
^{\times n}}\operatorname{symp}^{\mu}V$. Thus, $\sum_{\mu\in\left\{
1,2,3,\ldots\right\}  ^{\times m}}\operatorname*{symp}\nolimits^{\mu
}V\subseteq\sum_{n\in\mathbb{N}}\sum_{\mu\in\left\{  1,2,3,\ldots\right\}
^{\times n}}\operatorname{symp}^{\mu}V$. Hence,%
\[
\operatorname{symp}^{\lambda}V\subseteq\sum_{\mu\in\left\{  1,2,3,\ldots
\right\}  ^{\times m}}\operatorname{symp}^{\mu}V\subseteq\sum_{n\in\mathbb{N}%
}\sum_{\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times n}}\operatorname{symp}%
^{\mu}V,
\]
qed.} Thus,%
\begin{align*}
&  \sum_{\lambda\text{ is a partition}}\underbrace{\operatorname*{symp}%
\nolimits^{\lambda}V}_{\subseteq\sum_{n\in\mathbb{N}}\sum_{\mu\in\left\{
1,2,3,\ldots\right\}  ^{\times n}}\operatorname{symp}^{\mu}V}\\
&  \subseteq\sum_{\lambda\text{ is a partition}}\sum_{n\in\mathbb{N}}\sum
_{\mu\in\left\{  1,2,3,\ldots\right\}  ^{\times n}}\operatorname*{symp}%
\nolimits^{\mu}V\\
&  \subseteq\sum_{n\in\mathbb{N}}\underbrace{\sum_{\mu\in\left\{
1,2,3,\ldots\right\}  ^{\times n}}\operatorname{symp}^{\mu}V}%
_{=\operatorname{symp}_{n}V\text{ (by Proposition~\ref{Proposition316})}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{n\in\mathbb{N}}\sum_{\mu
\in\left\{  1,2,3,\ldots\right\}  ^{\times n}}\operatorname{symp}^{\mu}V\text{
is a }k\text{-vector space}\right) \\
&  =\sum_{n\in\mathbb{N}}\operatorname{symp}_{n}V=\sum_{i\in\mathbb{N}%
}\operatorname{symp}_{i}V\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed
the index }n\text{ as }i\text{ in the sum}\right) \\
&  =\sum_{i\in\mathbb{N}}V^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{Proposition315}}\right)  .
\end{align*}
Combined with (\ref{314.pf.1}), this yields $\sum_{i\in\mathbb{N}}V^{i}%
=\sum_{\lambda\text{ is a partition}}\operatorname{symp}^{\lambda}V$. This
proves Proposition~\ref{Proposition314}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem312}.]First, $\operatorname*{Prim}H$ is a
homogeneous subspace of $H$ (by Proposition~\ref{Proposition302}).

Second, it is easy to see $\left(  \operatorname*{Prim}H\right)  \cap H_{0}%
=0$\ \ \ \ \footnote{\textit{Proof.} Let $x\in\left(  \operatorname*{Prim}%
H\right)  \cap H_{0}$. Then, $x\in\left(  \operatorname*{Prim}H\right)  \cap
H_{0}\subseteq H_{0}=H_{\leq0}$ (because the definition of $H_{\leq0}$ yields
$H_{\leq0}=\bigoplus_{\ell=0}^{0}H_{\ell}=H_{0}$). Since $H_{\leq0}%
=k\cdot1_{H}$ (by Remark~\ref{Remark7l}, applied to $C=H$), this rewrites as
$x\in k\cdot1_{H}$. Hence, there exists some $\lambda\in k$ such that
$x=\lambda\cdot1_{H}$. Consider such a $\lambda$.
\par
On the other hand, $x\in\left(  \operatorname*{Prim}H\right)  \cap
H_{0}\subseteq\operatorname*{Prim}H$. Since $H$ is a unital coalgebra (by
Proposition~\ref{Proposition7d}, applied to $C=H$), this yields $\varepsilon
\left(  x\right)  =0$ (by Remark~\ref{Remark21}). Since $x=\lambda\cdot1_{H}$,
this becomes $\varepsilon\left(  \lambda\cdot1_{H}\right)  =0$. But since
\begin{align*}
\varepsilon\left(  \lambda\cdot1_{H}\right)   &  =\lambda\cdot
\underbrace{\varepsilon\left(  1_{H}\right)  }_{\substack{=1\\\text{(by the
axioms of a bialgebra,}\\\text{since }H\text{ is a bialgebra)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon\text{ is }%
k\text{-linear}\right) \\
&  =\lambda,
\end{align*}
this means that $\lambda=0$, so that $x=\underbrace{\lambda}_{=0}\cdot1_{H}%
=0$.
\par
Now forget that we fixed $x$. We thus have shown that every $x\in\left(
\operatorname*{Prim}H\right)  \cap H_{0}$ satisfies $x=0$. In other words,
$\left(  \operatorname*{Prim}H\right)  \cap H_{0}=0$, qed.}.

Third, we know from Proposition~\ref{Proposition283} that $xy-yx\in
\operatorname*{Prim}H$ for any $x\in\operatorname*{Prim}H$ and $y\in
\operatorname*{Prim}H$.

Thus, we can apply Proposition~\ref{Proposition314} to $A=H$ and
$V=\operatorname*{Prim}H$. We conclude that%
\[
\sum_{i\in\mathbb{N}}\left(  \operatorname*{Prim}H\right)  ^{i}=\sum
_{\lambda\text{ is a partition}}\operatorname{symp}^{\lambda}\left(
\operatorname*{Prim}H\right)  .
\]
But since $H$ is filtered, we know that%
\begin{align*}
H  &  =\bigcup_{\ell\in\mathbb{N}}\underbrace{H_{\leq\ell}}%
_{\substack{\subseteq\sum_{i=0}^{\ell}\left(  \operatorname*{Prim}H\right)
^{i}\\\text{(by Proposition~\ref{Proposition272})}}}\subseteq\bigcup_{\ell
\in\mathbb{N}}\left(  \underbrace{\sum_{i=0}^{\ell}\left(
\operatorname*{Prim}H\right)  ^{i}}_{\subseteq\sum_{i\in\mathbb{N}}\left(
\operatorname*{Prim}H\right)  ^{i}}\right)  \subseteq\bigcup_{\ell
\in\mathbb{N}}\left(  \sum_{i\in\mathbb{N}}\left(  \operatorname*{Prim}%
H\right)  ^{i}\right) \\
&  =\sum_{i\in\mathbb{N}}\left(  \operatorname*{Prim}H\right)  ^{i}%
=\sum_{\lambda\text{ is a partition}}\operatorname{symp}^{\lambda}\left(
\operatorname*{Prim}H\right)  .
\end{align*}
Combined with the obvious relation $\sum_{\lambda\text{ is a partition}%
}\operatorname{symp}^{\lambda}\left(  \operatorname*{Prim}H\right)  \subseteq
H$, this yields $H=\sum_{\lambda\text{ is a partition}}\operatorname{symp}%
^{\lambda}\left(  \operatorname*{Prim}H\right)  $. This proves
Theorem~\ref{Theorem312}.
\end{proof}

\section{\label{S17}Writing $p_{n}$ as a sum of convolutions of $\zeta_{m}$'s}

We continue to study connected graded bialgebras. First let us make explicit
one result that we proved during the proof of Proposition~\ref{Proposition232}:

\begin{theorem}
\label{Theorem340} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected graded $k$-bialgebra. Let $\zeta$ be the map $\operatorname*{Log}%
\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $. For every
$n\in\mathbb{N}$, let $p_{n}$ denote the map $p_{n,H}$ (defined according to
Definition~\ref{Definition221}), and let $\zeta_{n}$ denote the map
$\zeta\circ p_{n}$. Then,%
\[
\zeta_{n}=\sum_{\ell=1}^{n}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}%
\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)
\]
for every $n\in\mathbb{N}$.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem340}.]Theorem~\ref{Theorem340} directly follows
from (\ref{232.pf.15}).
\end{proof}

Theorem~\ref{Theorem340} gives us a formula for writing $\zeta_{n}$ in terms
of convolutions of $p_{m}$'s. We can also get a formula for writing $p_{n}$ in
terms of convolutions of $\zeta_{m}$'s:

\begin{theorem}
\label{Theorem341} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected graded $k$-bialgebra. Let $\zeta$ be the map $\operatorname*{Log}%
\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $. For every
$n\in\mathbb{N}$, let $p_{n}$ denote the map $p_{n,H}$ (defined according to
Definition~\ref{Definition221}), and let $\zeta_{n}$ denote the map
$\zeta\circ p_{n}$. Then,%
\[
p_{n}=\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(  a_{1}%
,a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell
};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)
\]
for every $n\in\mathbb{N}$.
\end{theorem}

We are going to prove Theorem~\ref{Theorem341} now. As we could expect, the
proof will be similar to the proof of Theorem~\ref{Theorem340} which we did in
\S \ref{S13}. We begin with a few lemmata:

\begin{lemma}
\label{Lemma342} Let $k$ be a field of characteristic $0$, and let $H$ be a
connected graded $k$-bialgebra. Let $\zeta$ be the map $\operatorname*{Log}%
\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $. For every
$n\in\mathbb{N}$, let $p_{n}$ denote the map $p_{n,H}$ (defined according to
Definition~\ref{Definition221}), and let $\zeta_{n}$ denote the map
$\zeta\circ p_{n}$. Then, $\zeta_{0}=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma342}.]Theorem~\ref{Theorem340} (applied to $n=0$)
yields%
\[
\zeta_{0}=\sum_{\ell=1}^{0}\dfrac{\left(  -1\right)  ^{\ell-1}}{\ell}%
\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,0\right\}  ^{\times\ell};\\0=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
p_{a_{1}}\ast p_{a_{2}}\ast\cdots\ast p_{a_{\ell}}\right)  =\left(
\text{empty sum}\right)  =0.
\]
This proves Lemma~\ref{Lemma342}.
\end{proof}

(Of course, we could have proven Lemma~\ref{Lemma342} much more easily.)

The following proposition is reminiscent of Proposition~\ref{Proposition236}
\textbf{(d)}:

\begin{proposition}
\label{Proposition343} Let $k$ be a field of characteristic $0$, and let $H$
be a connected graded $k$-bialgebra. Then, for every $n\in\mathbb{N}$, the map
$\zeta_{n}$ (defined in Theorem~\ref{Theorem341}) is graded and concentrated
in degree $n$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition343}.]Let $n\in\mathbb{N}$. Then,
$\zeta_{n}=\zeta\circ p_{n}$ is the composition of two graded maps (since
$\zeta$ is graded (by Theorem~\ref{Theorem231}) and since $p_{n}$ is graded
(by Proposition~\ref{Proposition236} \textbf{(d)})). Since the composition of
two graded maps must always be graded, this yields that $\zeta_{n}$ is graded.

We have $\zeta_{n}=\zeta\circ p_{n}=p_{n}\circ\zeta\circ p_{n}$ (by
(\ref{231.comm})), thus $\zeta_{n}=p_{n}\circ\underbrace{\zeta\circ p_{n}%
}_{=\zeta_{n}}=p_{n}\circ\zeta_{n}$. Since $\zeta_{n}$ is graded, this yields
that $\zeta_{n}$ is concentrated in degree $n$. This proves
Proposition~\ref{Proposition343}.
\end{proof}

Next, here is an analogue of Corollary~\ref{Corollary237}:

\begin{corollary}
\label{Corollary344} Let $k$ be a field. Let $H$ be a graded $k$-bialgebra.
Let $\zeta$ be the map $\operatorname*{Log}\operatorname*{id}\in
\mathcal{L}\left(  H,H\right)  $. For every $n\in\mathbb{N}$, let $p_{n}$
denote the map $p_{n,H}$ (defined according to Definition~\ref{Definition221}%
), and let $\zeta_{n}$ denote the map $\zeta\circ p_{n}$.

Let $n\in\mathbb{N}$ and $\ell\in\mathbb{N}$. Let $a_{i}$ be a nonnegative
integer for every $i\in\left\{  1,2,\ldots,\ell\right\}  $.

\textbf{(a)} We have $p_{n}\circ\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)  =0$ if $n\neq a_{1}+a_{2}%
+\cdots+a_{\ell}$.

\textbf{(b)} We have $p_{n}\circ\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)  =\zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}$ if $n=a_{1}+a_{2}+\cdots+a_{\ell}$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary344}.]For every $i\in\left\{  1,2,\ldots
,\ell\right\}  $, the map $\zeta_{a_{i}}$ is graded and concentrated in degree
$a_{i}$ (by Proposition~\ref{Proposition343}, applied to $a_{i}$ instead of
$n$). Thus, Proposition~\ref{Proposition236} \textbf{(c)} (applied to
$f_{i}=\zeta_{a_{i}}$) yields that $\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast
\cdots\ast\zeta_{a_{\ell}}$ is graded and concentrated in degree $a_{1}%
+a_{2}+\cdots+a_{\ell}$. According to Definition~\ref{Definition235}, this
means that%
\begin{equation}
p_{a_{1}+a_{2}+\cdots+a_{\ell}}\circ\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)  =\zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}. \label{344.pf.1}%
\end{equation}


\textbf{(a)} Assume that $n\neq a_{1}+a_{2}+\cdots+a_{\ell}$. Then,
$p_{n,H}\mid_{H_{a_{1}+a_{2}+\cdots+a_{\ell}}}=0$ (by (\ref{221.0}), applied
to $m=a_{1}+a_{2}+\cdots+a_{\ell}$ and $V=H$).

But%
\begin{align*}
&  \underbrace{\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast
\zeta_{a_{\ell}}\right)  }_{\substack{=p_{a_{1}+a_{2}+\cdots+a_{\ell}}%
\circ\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}%
}\right)  \\\text{(by (\ref{344.pf.1}))}}}\left(  H\right) \\
&  =\left(  p_{a_{1}+a_{2}+\cdots+a_{\ell}}\circ\left(  \zeta_{a_{1}}\ast
\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)  \left(  H\right)
=\underbrace{p_{a_{1}+a_{2}+\cdots+a_{\ell}}}_{\substack{=p_{a_{1}%
+a_{2}+\cdots+a_{\ell},H}\\\text{(by the definition of }p_{a_{1}+a_{2}%
+\cdots+a_{\ell}}\text{)}}}\left(  \underbrace{\left(  \zeta_{a_{1}}\ast
\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(  H\right)
}_{\subseteq H}\right) \\
&  \subseteq p_{a_{1}+a_{2}+\cdots+a_{\ell},H}\left(  H\right)  =H_{a_{1}%
+a_{2}+\cdots+a_{\ell}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{221.im}), applied to }H\text{ and }a_{1}+a_{2}+\cdots+a_{\ell
}\\
\text{instead of }V\text{ and }n
\end{array}
\right)
\end{align*}
and thus%
\begin{align*}
p_{n}\left(  \left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast
\zeta_{a_{\ell}}\right)  \left(  H\right)  \right)   &  \subseteq
\underbrace{p_{n}}_{\substack{=p_{n,H}\\\text{(by the definition of }%
p_{n}\text{)}}}\left(  H_{a_{1}+a_{2}+\cdots+a_{\ell}}\right)  =p_{n,H}\left(
H_{a_{1}+a_{2}+\cdots+a_{\ell}}\right) \\
&  =\underbrace{\left(  p_{n,H}\mid_{H_{a_{1}+a_{2}+\cdots+a_{\ell}}}\right)
}_{=0}\left(  H_{a_{1}+a_{2}+\cdots+a_{\ell}}\right)  =0\left(  H_{a_{1}%
+a_{2}+\cdots+a_{\ell}}\right)  =0.
\end{align*}
Since $p_{n}\left(  \left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast
\zeta_{a_{\ell}}\right)  \left(  H\right)  \right)  =\left(  p_{n}\circ\left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)
\left(  H\right)  $, this rewrites as \newline$\left(  p_{n}\circ\left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)
\left(  H\right)  =0$. Thus, $p_{n}\circ\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}%
}\ast\cdots\ast\zeta_{a_{\ell}}\right)  =0$. This proves
Corollary~\ref{Corollary344} \textbf{(a)}.

\textbf{(b)} Now assume that $n=a_{1}+a_{2}+\cdots+a_{\ell}$. Then,%
\[
p_{n}\circ\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}%
}\right)  =p_{a_{1}+a_{2}+\cdots+a_{\ell}}\circ\left(  \zeta_{a_{1}}\ast
\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  =\zeta_{a_{1}}\ast
\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}%
\]
(by (\ref{344.pf.1})). This proves Corollary~\ref{Corollary344} \textbf{(b)}.
\end{proof}

The next proof of Theorem~\ref{Theorem341} proceeds in analogy to (and
occasional use of) the proof of Proposition~\ref{Proposition232} above.

\begin{proof}
[Proof of Theorem~\ref{Theorem341}.]Let $n\in\mathbb{N}$.

Notice that $\zeta=\operatorname*{Log}\operatorname*{id}\in\mathfrak{g}\left(
H,H\right)  $ (by the definition of $\operatorname*{Log}$).

We will now prove Theorem~\ref{Theorem341} in several steps:

\textbf{a)} Every $x\in H_{\leq n}$ satisfies%
\begin{equation}
\zeta\left(  x\right)  =\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)
\left(  x\right)  . \label{341.pf.1}%
\end{equation}


\textit{Proof of (\ref{341.pf.1}).} Let $x\in H_{\leq n}$. We know that
$x=p_{0}\left(  x\right)  +p_{1}\left(  x\right)  +\cdots+p_{n}\left(
x\right)  $ (this is the equality (\ref{232.pf.23}) which we proved during the
proof of (\ref{232.pf.1}), while we were proving
Proposition~\ref{Proposition232}). Thus,%
\begin{align*}
\zeta\left(  x\right)   &  =\zeta\underbrace{\left(  p_{0}\left(  x\right)
+p_{1}\left(  x\right)  +\cdots+p_{n}\left(  x\right)  \right)  }_{=\sum
_{i=0}^{n}p_{i}\left(  x\right)  }=\zeta\left(  \sum_{i=0}^{n}p_{i}\left(
x\right)  \right)  =\sum_{i=0}^{n}\underbrace{\zeta\left(  p_{i}\left(
x\right)  \right)  }_{=\left(  \zeta\circ p_{i}\right)  \left(  x\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\zeta\text{ is }k\text{-linear}%
\right) \\
&  =\sum_{i=0}^{n}\underbrace{\left(  \zeta\circ p_{i}\right)  }
_{\substack{=\zeta_{i} \\\text{(since }\zeta_{i}\text{ was defined as }%
\zeta\circ p_{i}\text{)}}}\left(  x\right)  =\sum_{i=0}^{n}\zeta_{i}\left(
x\right)  =\underbrace{\zeta_{0}}_{\substack{=0 \\\text{(by
Lemma~\ref{Lemma342})}}} \left(  x\right)  +\sum_{i=1}^{n}\zeta_{i}\left(
x\right) \\
&  =\underbrace{0\left(  x\right)  }_{=0}+\underbrace{\sum_{i=1}^{n}\zeta
_{i}\left(  x\right)  }_{=\zeta_{1}\left(  x\right)  +\zeta_{2}\left(
x\right)  +\cdots+\zeta_{n}\left(  x\right)  }=\zeta_{1}\left(  x\right)
+\zeta_{2}\left(  x\right)  +\cdots+\zeta_{n}\left(  x\right)  =\left(
\zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)  \left(  x\right)  .
\end{align*}
This proves (\ref{341.pf.1}). Step \textbf{a)} is thus done.

\textbf{b)} Every $x\in H_{\leq n}$ and every $\ell\in\mathbb{N}$ satisfy%
\begin{equation}
\zeta^{\ast\ell}\left(  x\right)  =\left(  \zeta_{1}+\zeta_{2}+\cdots
+\zeta_{n}\right)  ^{\ast\ell}\left(  x\right)  . \label{341.pf.4}%
\end{equation}


\textit{Proof of (\ref{341.pf.4}).} There are several ways to derive
(\ref{341.pf.4}) from (\ref{341.pf.1}) (just as there were several ways to
derive (\ref{232.pf.4}) from (\ref{232.pf.1})). Here is one of them:

Let $\ell\in\mathbb{N}$.

We are going to use the notation $\mathcal{L}^{n}\left(  H,A\right)  $
introduced in Definition~\ref{Definition8} \textbf{(b)}.

By the definition of $\mathcal{L}^{n+1}\left(  H,H\right)  $, we have%
\[
\mathcal{L}^{n+1}\left(  H,H\right)  =\left\{  f\in\mathcal{L}\left(
H,H\right)  \ \mid\ \underbrace{f\mid_{H_{\leq n+1-1}}}_{=f\mid_{H_{\leq n}}%
}=0\right\}  =\left\{  f\in\mathcal{L}\left(  H,H\right)  \ \mid
\ f\mid_{H_{\leq n}}=0\right\}  .
\]
By Proposition~\ref{Proposition81} (applied to $n+1$ and $H$ instead of $n$
and $A$), the set $\mathcal{L}^{n+1}\left(  H,H\right)  $ is an ideal of the
$k$-algebra $\mathcal{L}\left(  H,H\right)  $.

Every $x\in H_{\leq n}$ satisfies%
\begin{align*}
\left(  \zeta-\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)  \right)
\left(  x\right)   &  =\underbrace{\zeta\left(  x\right)  }%
_{\substack{=\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)  \left(
x\right)  \\\text{(by (\ref{341.pf.1}))}}}-\left(  \zeta_{1}+\zeta_{2}%
+\cdots+\zeta_{n}\right)  \left(  x\right) \\
&  =\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)  \left(  x\right)
-\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)  \left(  x\right)  =0.
\end{align*}
In other words, $\left(  \zeta-\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta
_{n}\right)  \right)  \mid_{H_{\leq n}}=0$. Thus,%
\[
\zeta-\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)  \in\left\{
f\in\mathcal{L}\left(  H,H\right)  \ \mid\ f\mid_{H_{\leq n}}=0\right\}
=\mathcal{L}^{n+1}\left(  H,H\right)  .
\]
In other words, $\zeta\equiv\zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}%
\operatorname{mod}\mathcal{L}^{n+1}\left(  H,H\right)  $. Thus,
\[
\zeta^{\ast\ell}\equiv\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)
^{\ast\ell}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,H\right)
\]
(because $\mathcal{L}^{n+1}\left(  H,H\right)  $ is an ideal of the
$k$-algebra $\mathcal{L}\left(  H,H\right)  $, and hence we can multiply
congruences modulo $\mathcal{L}^{n+1}\left(  H,H\right)  $). In other words,%
\[
\zeta^{\ast\ell}-\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)
^{\ast\ell}\in\mathcal{L}^{n+1}\left(  H,H\right)  =\left\{  f\in
\mathcal{L}\left(  H,H\right)  \ \mid\ f\mid_{H_{\leq n}}=0\right\}  ,
\]
so that $\left(  \zeta^{\ast\ell}-\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta
_{n}\right)  ^{\ast\ell}\right)  \mid_{H_{\leq n}}=0$. Hence, every $x\in
H_{\leq n}$ satisfies \newline$\left(  \zeta^{\ast\ell}-\left(  \zeta
_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)  ^{\ast\ell}\right)  \left(  x\right)
=0$. As a consequence, every $x\in H_{\leq n}$ satisfies $\zeta^{\ast\ell
}\left(  x\right)  =\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)
^{\ast\ell}\left(  x\right)  $ (because
\[
\zeta^{\ast\ell}\left(  x\right)  -\left(  \zeta_{1}+\zeta_{2}+\cdots
+\zeta_{n}\right)  ^{\ast\ell}\left(  x\right)  =\left(  \zeta^{\ast\ell
}-\left(  \zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}\right)  ^{\ast\ell}\right)
\left(  x\right)  =0
\]
). In other words, (\ref{341.pf.4}) is proven for every $x\in H_{\leq n}$ and
every $\ell\in\mathbb{N}$. This completes step \textbf{b)}.

\textbf{c)} Every $x\in H_{\leq n}$ satisfies%
\begin{equation}
x=\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\zeta^{\ast\ell}\left(  x\right)  .
\label{341.pf.5}%
\end{equation}


\textit{Proof of (\ref{341.pf.5}).} Since $\zeta=\operatorname*{Log}%
\operatorname*{id}$, we have $e^{\ast\zeta}=e^{\ast\left(  \operatorname*{Log}%
\operatorname*{id}\right)  }=\operatorname*{id}$ (by
Proposition~\ref{Proposition13} \textbf{(b)}, applied to $A=H$ and
$F=\operatorname*{id}$). Thus, every $x\in H_{\leq n}$ satisfies%
\begin{align*}
x  &  =\underbrace{\operatorname*{id}}_{=e^{\ast\zeta}}\left(  x\right)
=e^{\ast\zeta}\left(  x\right)  =\sum_{i\geq0}\underbrace{\dfrac{\zeta^{\ast
i}\left(  x\right)  }{i!}}_{=\dfrac{1}{i!}\zeta^{\ast i}\left(  x\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{exp}), applied to }f=\zeta\right)
\\
&  =\sum_{i\geq0}\dfrac{1}{i!}\zeta^{\ast i}\left(  x\right)  =\sum
_{\substack{i\geq0;\\i\leq n}}\dfrac{1}{i!}\zeta^{\ast i}\left(  x\right)
+\sum_{\substack{i\geq0;\\i>n}}\dfrac{1}{i!}\underbrace{\zeta^{\ast i}\left(
x\right)  }_{\substack{=0\text{ (since }x\in H_{\leq n}\text{ and thus}%
\\\zeta^{\ast i}\left(  x\right)  \in\zeta^{\ast i}\left(  H_{\leq n}\right)
=0\text{ (by Remark~\ref{Remark9}}\\\text{(applied to }f=\zeta\text{), since
}i>n\text{),}\\\text{so that }\zeta^{\ast i}\left(  x\right)  =0\text{)}}}\\
&  =\underbrace{\sum_{\substack{i\geq0;\\i\leq n}}}_{=\sum_{i=0}^{n}}\dfrac
{1}{i!}\zeta^{\ast i}\left(  x\right)  +\underbrace{\sum_{\substack{i\geq
0;\\i>n}}\dfrac{1}{i!}0}_{=0}=\sum_{i=0}^{n}\dfrac{1}{i!}\zeta^{\ast i}\left(
x\right)  =\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\zeta^{\ast\ell}\left(  x\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}i\text{ as }\ell\right)  .
\end{align*}
This proves (\ref{341.pf.5}), and thus our step \textbf{c)} is complete.

\textbf{d)} Every $x\in H_{\leq n}$ satisfies%
\begin{equation}
x=\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}}\left(  \zeta
_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(
x\right)  . \label{341.pf.6}%
\end{equation}


\textit{Proof of (\ref{341.pf.6}).} Let $x\in H_{\leq n}$. Then, for every
$\ell\in\mathbb{N}$, we have%
\begin{align*}
\zeta^{\ast\ell}\left(  x\right)   &  =\underbrace{\left(  \zeta_{1}+\zeta
_{2}+\cdots+\zeta_{n}\right)  ^{\ast\ell}}_{\substack{=\sum_{\left(
a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell}}\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}
\\\text{(by the product rule)}}}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{341.pf.4})}\right) \\
&  =\left(  \sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell}}\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast
\cdots\ast\zeta_{a_{\ell}}\right)  \left(  x\right)  =\sum_{\left(
a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast
\zeta_{a_{\ell}}\right)  \left(  x\right)  .
\end{align*}
Now, (\ref{341.pf.5}) becomes%
\begin{align*}
x  &  =\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\underbrace{\zeta^{\ast\ell}\left(
x\right)  }_{=\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(  x\right)  }\\
&  =\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}}\left(  \zeta
_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(
x\right)  .
\end{align*}
This proves (\ref{341.pf.6}).

\textbf{e)} Every $x\in H_{n}$ satisfies\footnote{Note that here we require
$x\in H_{n}$ rather than $x\in H_{\leq n}$.}%
\begin{equation}
x=\left(  \sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(  a_{1}%
,a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell
}; \\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)  \left(  x\right)  .
\label{341.pf.7}%
\end{equation}


\textit{Proof of (\ref{341.pf.7}).} Let $x\in H_{n}$.

Since $H_{\leq n}=\bigoplus_{\ell=0}^{n}H_{\ell}$ by the definition of
$H_{\leq n}$, we have $H_{n}\subseteq H_{\leq n}$ (since $H_{n}$ is one addend
of the direct sum $\bigoplus_{\ell=0}^{n}H_{\ell}$, and thus $H_{n}%
\subseteq\bigoplus_{\ell=0}^{n}H_{\ell}=H_{\leq n}$).

Now, $p_{n,H}\mid_{H_{n}}=\operatorname{id}_{H}\mid_{H_{n}}$ (by
(\ref{221.id}), applied to $V=H$). Since $p_{n}=p_{n,H}$ (by the definition of
$p_{n}$), this rewrites as $p_{n}\mid_{H_{n}}=\operatorname{id}_{H}\mid
_{H_{n}}$. Since $x\in H_{n}$, we have%
\[
p_{n}\left(  x\right)  =\underbrace{\left(  p_{n}\mid_{H_{n}}\right)
}_{=\operatorname{id}_{H}\mid_{H_{n}}}\left(  x\right)  =\left(
\operatorname{id}_{H}\mid_{H_{n}}\right)  \left(  x\right)  =\operatorname{id}%
_{H}\left(  x\right)  =x.
\]
Thus,%
\begin{align*}
x  &  =p_{n}\left(  x\right)  =p_{n}\left(  \sum_{\ell=0}^{n}\dfrac{1}{\ell
!}\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{\times\ell}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots
\ast\zeta_{a_{\ell}}\right)  \left(  x\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{341.pf.6}), since }x\in
H_{n}\subseteq H_{\leq n}\right) \\
&  =\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}}p_{n}\left(  \left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(
x\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{n}\text{ is
}k\text{-linear}\right)  .
\end{align*}
Since every $\ell\in\left\{  0,1,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell}}\underbrace{p_{n}\left(  \left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(
x\right)  \right)  }_{=\left(  p_{n}\circ\left(  \zeta_{a_{1}}\ast\zeta
_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)  \left(  x\right)  }\\
&  =\sum_{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell}}\left(  p_{n}\circ\left(  \zeta_{a_{1}%
}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)  \left(
x\right) \\
&  =\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}%
}}\underbrace{\left(  p_{n}\circ\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)  }_{\substack{=\zeta_{a_{1}%
}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\\\text{(by
Corollary~\ref{Corollary344} \textbf{(b)}, since}\\n=a_{1}+a_{2}%
+\cdots+a_{\ell}\text{)}}}\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n\neq a_{1}%
+a_{2}+\cdots+a_{\ell}}}\underbrace{\left(  p_{n}\circ\left(  \zeta_{a_{1}%
}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)
}_{\substack{=0\\\text{(by Corollary~\ref{Corollary344} \textbf{(a)},
since}\\n\neq a_{1}+a_{2}+\cdots+a_{\ell}\text{)}}}\left(  x\right) \\
&  =\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(
x\right)  +\underbrace{\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n\neq a_{1}%
+a_{2}+\cdots+a_{\ell}}}0\left(  x\right)  }_{=0}\\
&  =\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(
x\right)  ,
\end{align*}
this rewrites as%
\begin{align*}
x  &  =\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\underbrace{\sum_{\left(  a_{1}%
,a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell
}}p_{n}\left(  \left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast
\zeta_{a_{\ell}}\right)  \left(  x\right)  \right)  }_{=\sum
_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(
x\right)  }\\
&  =\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(  a_{1}%
,a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell
};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)  \left(  x\right) \\
&  =\left(  \sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(
a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(  \zeta_{a_{1}}\ast
\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)  \left(  x\right)
.
\end{align*}
This proves (\ref{341.pf.7}).

\textbf{f)} We are going to use the result (\ref{232.pf.10}) once again (but
for obvious reasons, we are not going to prove it once again).

\textbf{g)} Now let us finally prove Theorem~\ref{Theorem341}:

For every $\ell\in\left\{  1,2,\ldots,n\right\}  $ and every $\left(
a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell}$, the map $\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast
\zeta_{a_{\ell}}$ is graded\footnote{\textit{Proof.} Let $\ell\in\left\{
1,2,\ldots,n\right\}  $ and $\left(  a_{1},a_{2},\ldots,a_{\ell}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\times\ell}$. Then, for every
$i\in\left\{  1,2,\ldots,\ell\right\}  $, the map $\zeta_{a_{i}}$ is graded
and concentrated in degree $a_{i}$ (by Proposition~\ref{Proposition343},
applied to $a_{i}$ instead of $n$). Hence, by Proposition~\ref{Proposition236}
\textbf{(c)} (applied to $f_{i}=\zeta_{a_{i}}$), the map $\zeta_{a_{1}}%
\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}$ is graded and concentrated in
degree $a_{1}+a_{2}+\cdots+a_{\ell}$, qed.}. Hence, the map $\sum_{\ell=0}%
^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}%
+a_{2}+\cdots+a_{\ell}}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots
\ast\zeta_{a_{\ell}}\right)  $ is graded (since a $k$-linear combination of
graded maps is always graded). Also, $\operatorname*{id}$ is graded. The two
latter facts, along with the fact that
\[
\operatorname*{id}\left(  x\right)  =x=\left(  \sum_{\ell=0}^{n}\dfrac{1}%
{\ell!}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)
\left(  x\right)
\]
for all $x\in H_{n}$ (by (\ref{341.pf.7})), show that we can apply
(\ref{232.pf.10}) to $f=\operatorname*{id}$ and \newline$g=\sum_{\ell=0}%
^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}%
+a_{2}+\cdots+a_{\ell}}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots
\ast\zeta_{a_{\ell}}\right)  $. As a result, we conclude that%
\[
p_{n}\circ\operatorname*{id}=p_{n}\circ\left(  \sum_{\ell=0}^{n}\dfrac{1}%
{\ell!}\sum_{\substack{\left(  a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)
.
\]
Using this identity, we have%
\begin{align*}
p_{n}  &  =p_{n}\circ\operatorname*{id}\\
&  =p_{n}\circ\left(  \sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(
a_{1},a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}
^{\times\ell};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(  \zeta_{a_{1}}\ast
\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right) \\
&  =\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(  a_{1}%
,a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell
};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\underbrace{\left(  p_{n}\circ\left(
\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}}\right)  \right)
}_{\substack{=\zeta_{a_{1}}\ast\zeta_{a_{2}}\ast\cdots\ast\zeta_{a_{\ell}%
}\\\text{(by Corollary~\ref{Corollary344} \textbf{(b)}, since}\\n=a_{1}%
+a_{2}+\cdots+a_{\ell}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is }k\text{-bilinear}\right) \\
&  =\sum_{\ell=0}^{n}\dfrac{1}{\ell!}\sum_{\substack{\left(  a_{1}%
,a_{2},\ldots,a_{\ell}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\times\ell
};\\n=a_{1}+a_{2}+\cdots+a_{\ell}}}\left(  \zeta_{a_{1}}\ast\zeta_{a_{2}}%
\ast\cdots\ast\zeta_{a_{\ell}}\right)  .
\end{align*}
This proves Theorem~\ref{Theorem341}.
\end{proof}

\section{\label{S18}Logarithms of commutative convolutions}

In this section, we are going to study the logarithm (as defined in
Definition~\ref{Definition12}) further. We will prove its following property:

\begin{theorem}
\label{Theorem400} Let $k$ be a field of characteristic $0$. Let $A$ be a
$k$-algebra, and let $C$ be a connected filtered $k$-coalgebra. Let $F\in
G\left(  C,A\right)  $ and $H\in G\left(  C,A\right)  $ be maps satisfying
$F\ast H=H\ast F$. Then, $F\ast H\in G\left(  C,A\right)  $ and
$\operatorname*{Log}\left(  F\ast H\right)  =\operatorname*{Log}%
F+\operatorname*{Log}H$.
\end{theorem}

Note that this Theorem~\ref{Theorem400} is a kind of logarithmic ``sibling''
of Proposition~\ref{Proposition47}. It will be harder to prove, though. The
proof will require a fact about power series:

\begin{theorem}
\label{Theorem401} Let $k$ be a field of characteristic $0$. Consider the ring
of formal power series $k\left[  \left[  X,Y\right]  \right]  $ in two
(commuting) indeterminates $X$ and $Y$. For every power series $Q\in k\left[
\left[  X,Y\right]  \right]  $ whose coefficient before $X^{0}Y^{0}$ is $1$,
let $\log Q$ denote the power series in $k\left[  \left[  X,Y\right]  \right]
$ defined by $\log Q=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
Q-1\right)  ^{i}$.

Let $Q$ and $R$ be two power series in $k\left[  \left[  X,Y\right]  \right]
$ whose coefficients before $X^{0}Y^{0}$ are both equal to $1$. Then, the
coefficient of $QR$ before $X^{0}Y^{0}$ is also equal to $1$, and we have
$\log\left(  QR\right)  =\log Q+\log R$.
\end{theorem}

We are not going to prove this fact, as it is rather fundamental and
well-known. But we derive a ``finite version'' from it (just as we derived
Corollary~\ref{Corollary15} from Theorem~\ref{Theorem14}):

\begin{corollary}
\label{Corollary402} Let $k$ be a field of characteristic $0$. Let
$n\in\mathbb{N}$.

Let $\mathfrak{A}$ be a commutative $k$-algebra. Let $\mathfrak{I}$ be an
ideal of $\mathfrak{A}$ such that $\mathfrak{I}^{n+1}=0$. Let $a\in
\mathfrak{I}$ and $b\in\mathfrak{I}$. Then,%
\[
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}+\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}b^{i}=\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  a+b+ab\right)  ^{i}.
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary402}.]\textbf{a)} First let us notice that%
\begin{equation}
\text{every }\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}\text{ such
that }\alpha+\beta>n\text{ satisfies }a^{\alpha}b^{\beta}=0. \label{402.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{402.pf.1}).} Let $\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2}$ satisfy $\alpha+\beta>n$. Then,
$\alpha+\beta\geq n+1$ (since $\alpha+\beta$ and $n+1$ are integers), so that
$\mathfrak{I}^{\alpha+\beta}\subseteq\mathfrak{I}^{n+1}$ (since $\mathfrak{I}$
is an ideal of $\mathfrak{A}$).
\par
On the other hand, $a\in\mathfrak{I}$, so that $a^{\alpha}\in\mathfrak{I}%
^{\alpha}$. Also, $b\in\mathfrak{I}$, so that $b^{\beta}\in\mathfrak{I}%
^{\beta}$. Thus, $\underbrace{a^{\alpha}}_{\in\mathfrak{I}^{\alpha}%
}\underbrace{b^{\beta}}_{\in\mathfrak{I}^{\beta}}\in\mathfrak{I}^{\alpha
}\mathfrak{I}^{\beta}=\mathfrak{I}^{\alpha+\beta}\subseteq\mathfrak{I}%
^{n+1}=0$. Hence, $a^{\alpha}b^{\beta}=0$. This proves (\ref{402.pf.1}).}
Also, $a^{n+1}=0$ (since $a\in\mathfrak{I}$, so that $a^{n+1}\in
\mathfrak{I}^{n+1}=0$), and $b^{n+1}=0$ (for the same reason). Finally,
$ab\in\mathfrak{I}$ (since $a\in\mathfrak{I}$ and since $\mathfrak{I}$ is an
ideal), so that $\underbrace{a}_{\in\mathfrak{I}}+\underbrace{b}%
_{\in\mathfrak{I}}+\underbrace{ab}_{\in\mathfrak{I}}\in\mathfrak{I}%
+\mathfrak{I}+\mathfrak{I}\subseteq\mathfrak{I}$ (since $\mathfrak{I}$ is an
ideal), so that $\left(  a+b+ab\right)  ^{n+1}=0$ (since $a+b+ab\in
\mathfrak{I}$, so that $\left(  a+b+ab\right)  ^{n+1}\in\mathfrak{I}^{n+1}=0$).

\textbf{b)} Consider the ring of formal power series $k\left[  \left[
X,Y\right]  \right]  $ in two (commuting) indeterminates $X$ and $Y$. For
every power series $P\in k\left[  \left[  X,Y\right]  \right]  $ and every
$\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}$, we let
$\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)  $ denote the
coefficient of the power series $P$ before $X^{\alpha}Y^{\beta}$. Then,
clearly,%
\begin{equation}
\text{every power series }P\in k\left[  \left[  X,Y\right]  \right]  \text{
satisfies }P=\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}%
}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)  \cdot
X^{\alpha}Y^{\beta}. \label{402.pf.2}%
\end{equation}


By the definition of the sum of two power series, we have%
\begin{align}
\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P+Q\right)   &
=\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)
+\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  Q\right)
\label{402.pf.sum}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for any }\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2}\text{, any }P\in k\left[  \left[  X,Y\right]  \right]
\text{ and any }Q\in k\left[  \left[  X,Y\right]  \right]  .\nonumber
\end{align}
By the definition of the product of a power series with a scalar, we have%
\begin{equation}
\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  \lambda P\right)
=\lambda\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }\left(  \alpha,\beta\right)  \in
\mathbb{N}^{\times2}\text{, any }P\in k\left[  \left[  X,Y\right]  \right]
\text{ and any }\lambda\in k. \label{402.pf.scal}%
\end{equation}
By the definition of the product of two power series, we have%
\begin{align}
\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  PQ\right)   &
=\sum_{\substack{\left(  \gamma,\delta\right)  \in\mathbb{N}^{\times
2};\\\gamma\leq\alpha;\ \delta\leq\beta}}\operatorname*{coeff}%
\nolimits_{\gamma,\delta}\left(  P\right)  \cdot\operatorname*{coeff}%
\nolimits_{\alpha-\gamma,\beta-\delta}\left(  Q\right) \label{402.pf.prod}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for any }\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2}\text{, any }P\in k\left[  \left[  X,Y\right]  \right]
\text{ and any }Q\in k\left[  \left[  X,Y\right]  \right]  .\nonumber
\end{align}


\textbf{c)} Let us now define a map $\rho:k\left[  \left[  X,Y\right]
\right]  \rightarrow\mathfrak{A}$ by%
\begin{equation}
\left(  \rho\left(  P\right)  =\sum_{\substack{\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2};\\\alpha+\beta\leq n}}\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  P\right)  \cdot a^{\alpha}b^{\beta
}\ \ \ \ \ \ \ \ \ \ \text{for every }P\in k\left[  \left[  X,Y\right]
\right]  \right)  . \label{402.pf.3}%
\end{equation}
Note that the sum $\sum_{\substack{\left(  \alpha,\beta\right)  \in
\mathbb{N}^{\times2};\\\alpha+\beta\leq n}}\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  P\right)  \cdot a^{\alpha}b^{\beta}$ is a
finite sum for every $P\in k\left[  \left[  X,Y\right]  \right]  $ (since
there are only finitely many pairs $\left(  \alpha,\beta\right)  \in
\mathbb{N}^{\times2}$ satisfying $\alpha+\beta\leq n$), so this map $\rho$ is well-defined.

\textbf{d)} Let us show that%
\begin{equation}
\rho\left(  P\right)  =\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}%
^{\times2}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)
\cdot a^{\alpha}b^{\beta}\ \ \ \ \ \ \ \ \ \ \text{for every }P\in k\left[
\left[  X,Y\right]  \right]  . \label{402.pf.4}%
\end{equation}
Here, the infinite sum $\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}%
^{\times2}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)
\cdot a^{\alpha}b^{\beta}$ has a well-defined value because all but finitely
many addends of this sum are zero.\footnote{\textit{Proof.} Only finitely many
pairs $\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}$ satisfy
$\alpha+\beta\leq n$. Hence, all but finitely many pairs $\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2}$ don't satisfy $\alpha+\beta\leq n$. In
other words, all but finitely many pairs $\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2}$ satisfy $\alpha+\beta>n$. But since every pair
$\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}$ which satisfies
$\alpha+\beta>n$ must satisfy $\operatorname*{coeff}\nolimits_{\alpha,\beta
}\left(  P\right)  \cdot\underbrace{a^{\alpha}b^{\beta}}_{=0\text{ (by
(\ref{402.pf.1}))}}=0$, this yields that all but finitely many pairs $\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2}$ satisfy $\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  P\right)  \cdot a^{\alpha}b^{\beta}=0$. In
other words, all but finitely many addends of the sum $\sum_{\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2}}\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  P\right)  \cdot a^{\alpha}b^{\beta}$ are zero.
Qed.}

\textit{Proof of (\ref{402.pf.4}).} We already know that the sum
$\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}}%
\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)  \cdot
a^{\alpha}b^{\beta}$ is well-defined. Thus, we have%
\begin{align*}
\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}}%
\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)  \cdot
a^{\alpha}b^{\beta}  &  =\underbrace{\sum_{\substack{\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2};\\\alpha+\beta\leq n}%
}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)  \cdot
a^{\alpha}b^{\beta}}_{\substack{=\rho\left(  P\right)  \\\text{(by
(\ref{402.pf.3}))}}}+\sum_{\substack{\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2};\\\alpha+\beta>n}}\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  P\right)  \cdot\underbrace{a^{\alpha}b^{\beta
}}_{\substack{=0\\\text{(by (\ref{402.pf.1}),}\\\text{since }\alpha
+\beta>n\text{)}}}\\
&  =\rho\left(  P\right)  +\underbrace{\sum_{\substack{\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2};\\\alpha+\beta>n}}\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  P\right)  \cdot0}_{=0}=\rho\left(  P\right)  .
\end{align*}
This proves (\ref{402.pf.4}).

\textbf{e)} The map $\rho$ is $k$-linear.

\textit{Proof.} Every $P\in k\left[  \left[  X,Y\right]  \right]  $ and $Q\in
k\left[  \left[  X,Y\right]  \right]  $ satisfy%
\begin{align*}
\rho\left(  P+Q\right)   &  =\sum_{\substack{\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2};\\\alpha+\beta\leq n}%
}\underbrace{\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P+Q\right)
}_{\substack{=\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)
+\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  Q\right)  \\\text{(by
(\ref{402.pf.sum}))}}}\cdot a^{\alpha}b^{\beta}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{402.pf.3}), applied to }P+Q\text{ instead of }P\right) \\
&  =\sum_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times
2};\\\alpha+\beta\leq n}}\left(  \operatorname*{coeff}\nolimits_{\alpha,\beta
}\left(  P\right)  +\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
Q\right)  \right)  \cdot a^{\alpha}b^{\beta}%
\end{align*}
and%
\begin{align*}
&  \underbrace{\rho\left(  P\right)  }_{\substack{=\sum_{\substack{\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2};\\\alpha+\beta\leq n}%
}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  P\right)  \cdot
a^{\alpha}b^{\beta}\\\text{(by (\ref{402.pf.3}))}}}+\underbrace{\rho\left(
Q\right)  }_{\substack{=\sum_{\substack{\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2};\\\alpha+\beta\leq n}}\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  Q\right)  \cdot a^{\alpha}b^{\beta}\\\text{(by
(\ref{402.pf.3}), applied to }Q\text{ instead of }P\text{)}}}\\
&  =\sum_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times
2};\\\alpha+\beta\leq n}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
P\right)  \cdot a^{\alpha}b^{\beta}+\sum_{\substack{\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2};\\\alpha+\beta\leq n}%
}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  Q\right)  \cdot
a^{\alpha}b^{\beta}\\
&  =\sum_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times
2};\\\alpha+\beta\leq n}}\left(  \operatorname*{coeff}\nolimits_{\alpha,\beta
}\left(  P\right)  +\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
Q\right)  \right)  \cdot a^{\alpha}b^{\beta}.
\end{align*}
Hence, every $P\in k\left[  \left[  X,Y\right]  \right]  $ and $Q\in k\left[
\left[  X,Y\right]  \right]  $ satisfy%
\begin{equation}
\rho\left(  P+Q\right)  =\sum_{\substack{\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2};\\\alpha+\beta\leq n}}\left(  \operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  P\right)  +\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  Q\right)  \right)  \cdot a^{\alpha}b^{\beta
}=\rho\left(  P\right)  +\rho\left(  Q\right)  . \label{402.pf.10}%
\end{equation}


Also, every $P\in k\left[  \left[  X,Y\right]  \right]  $ and $\lambda\in k$
satisfy%
\begin{align}
\rho\left(  \lambda P\right)   &  =\sum_{\substack{\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2};\\\alpha+\beta\leq n}%
}\underbrace{\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  \lambda
P\right)  }_{\substack{=\lambda\operatorname*{coeff}\nolimits_{\alpha,\beta
}\left(  P\right)  \\\text{(by (\ref{402.pf.scal}))}}}\cdot a^{\alpha}%
b^{\beta}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{402.pf.3}), applied to
}\lambda P\text{ instead of }P\right) \nonumber\\
&  =\sum_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times
2};\\\alpha+\beta\leq n}}\lambda\operatorname*{coeff}\nolimits_{\alpha,\beta
}\left(  P\right)  \cdot a^{\alpha}b^{\beta}=\lambda\underbrace{\sum
_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}%
;\\\alpha+\beta\leq n}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
P\right)  \cdot a^{\alpha}b^{\beta}}_{\substack{=\rho\left(  P\right)
\\\text{(by (\ref{402.pf.3}))}}}=\lambda\rho\left(  P\right)  .
\label{402.pf.12}%
\end{align}


So we have proven that every $P\in k\left[  \left[  X,Y\right]  \right]  $ and
$Q\in k\left[  \left[  X,Y\right]  \right]  $ satisfy (\ref{402.pf.10}), and
that every $P\in k\left[  \left[  X,Y\right]  \right]  $ and $\lambda\in k$
satisfy (\ref{402.pf.12}). In other words, we have proven that $\rho$ is $k$-linear.

\textbf{f)} The map $\rho$ is a $k$-algebra homomorphism.

\textit{Proof.} First of all, it is clear the power series $1$ has coefficient
$1$ before $X^{0}Y^{0}$, while all its other coefficients are zero. In other
words, $\operatorname*{coeff}\nolimits_{0,0}\left(  1\right)  =1$, while
$\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  1\right)  =0$ for every
$\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}$ satisfying $\left(
\alpha,\beta\right)  \neq\left(  0,0\right)  $.

Now, (\ref{402.pf.4}) (applied to $P=1$) yields
\begin{align*}
\rho\left(  1\right)   &  =\sum_{\left(  \alpha,\beta\right)  \in
\mathbb{N}^{\times2}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
1\right)  \cdot a^{\alpha}b^{\beta}=\underbrace{\sum_{\substack{\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2};\\\left(  \alpha,\beta\right)
=\left(  0,0\right)  }}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
1\right)  \cdot a^{\alpha}b^{\beta}}_{=\operatorname*{coeff}\nolimits_{0,0}%
\left(  1\right)  \cdot a^{0}b^{0}}+\sum_{\substack{\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2};\\\left(  \alpha,\beta\right)
\neq\left(  0,0\right)  }}\underbrace{\operatorname*{coeff}\nolimits_{\alpha
,\beta}\left(  1\right)  }_{\substack{=0\\\text{(since }\left(  \alpha
,\beta\right)  \neq\left(  0,0\right)  \text{)}}}\cdot a^{\alpha}b^{\beta}\\
&  =\underbrace{\operatorname*{coeff}\nolimits_{0,0}\left(  1\right)  }%
_{=1}\cdot\underbrace{a^{0}}_{=1}\underbrace{b^{0}}_{=1}+\underbrace{\sum
_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2};\\\left(
\alpha,\beta\right)  \neq\left(  0,0\right)  }}0\cdot a^{\alpha}b^{\beta}%
}_{=0}=1.
\end{align*}


Now let $P\in k\left[  \left[  X,Y\right]  \right]  $ and $Q\in k\left[
\left[  X,Y\right]  \right]  $ be arbitrary. Then, multiplying the equalities%
\begin{align*}
\rho\left(  P\right)   &  =\sum_{\left(  \alpha,\beta\right)  \in
\mathbb{N}^{\times2}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
P\right)  \cdot a^{\alpha}b^{\beta}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{402.pf.4})}\right) \\
&  =\sum_{\left(  \gamma,\delta\right)  \in\mathbb{N}^{\times2}}%
\operatorname*{coeff}\nolimits_{\gamma,\delta}\left(  P\right)  \cdot
a^{\gamma}b^{\delta}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
index }\left(  \alpha,\beta\right)  \text{ as }\left(  \gamma,\delta\right)
\text{ in the sum}\right)
\end{align*}
and%
\[
\rho\left(  Q\right)  =\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}%
^{\times2}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  Q\right)
\cdot a^{\alpha}b^{\beta}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{402.pf.4}%
), applied to }Q\text{ instead of }P\right)  ,
\]
we obtain%
\begin{align*}
\rho\left(  P\right)  \cdot\rho\left(  Q\right)   &  =\left(  \sum_{\left(
\gamma,\delta\right)  \in\mathbb{N}^{\times2}}\operatorname*{coeff}%
\nolimits_{\gamma,\delta}\left(  P\right)  \cdot a^{\gamma}b^{\delta}\right)
\cdot\left(  \sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}%
}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  Q\right)  \cdot
a^{\alpha}b^{\beta}\right) \\
&  =\sum_{\left(  \gamma,\delta\right)  \in\mathbb{N}^{\times2}}\sum_{\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2}}\underbrace{\operatorname*{coeff}%
\nolimits_{\gamma,\delta}\left(  P\right)  \cdot a^{\gamma}b^{\delta}%
\cdot\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  Q\right)  \cdot
a^{\alpha}b^{\beta}}_{=\operatorname*{coeff}\nolimits_{\gamma,\delta}\left(
P\right)  \cdot\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  Q\right)
\cdot a^{\gamma}a^{\alpha}\cdot b^{\delta}b^{\beta}}\\
&  =\sum_{\left(  \gamma,\delta\right)  \in\mathbb{N}^{\times2}}\sum_{\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2}}\operatorname*{coeff}%
\nolimits_{\gamma,\delta}\left(  P\right)  \cdot\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  Q\right)  \cdot\underbrace{a^{\gamma}%
a^{\alpha}}_{=a^{\gamma+\alpha}}\cdot\underbrace{b^{\delta}b^{\beta}%
}_{=b^{\delta+\beta}}\\
&  =\sum_{\left(  \gamma,\delta\right)  \in\mathbb{N}^{\times2}}\sum_{\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2}}\operatorname*{coeff}%
\nolimits_{\gamma,\delta}\left(  P\right)  \cdot\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  Q\right)  \cdot a^{\gamma+\alpha}\cdot
b^{\delta+\beta}\\
&  =\underbrace{\sum_{\left(  \gamma,\delta\right)  \in\mathbb{N}^{\times2}%
}\sum_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}%
;\\\alpha\geq\gamma;\ \beta\geq\delta}}}_{=\sum_{\left(  \alpha,\beta\right)
\in\mathbb{N}^{\times2}}\sum_{\substack{\left(  \gamma,\delta\right)
\in\mathbb{N}^{\times2};\\\alpha\geq\gamma;\ \beta\geq\delta}}}%
\operatorname*{coeff}\nolimits_{\gamma,\delta}\left(  P\right)  \cdot
\operatorname*{coeff}\nolimits_{\alpha-\gamma,\beta-\delta}\left(  Q\right)
\cdot\underbrace{a^{\gamma+\left(  \alpha-\gamma\right)  }}_{=a^{\alpha}}%
\cdot\underbrace{b^{\delta+\left(  \beta-\delta\right)  }}_{=b^{\beta}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }\left(
\alpha,\beta\right)  \text{ for }\left(  \gamma+\alpha,\delta+\beta\right)
\text{ in the second sum}\right) \\
&  =\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}}%
\underbrace{\sum_{\substack{\left(  \gamma,\delta\right)  \in\mathbb{N}%
^{\times2};\\\alpha\geq\gamma;\ \beta\geq\delta}}}_{=\sum_{\substack{\left(
\gamma,\delta\right)  \in\mathbb{N}^{\times2};\\\gamma\leq\alpha;\ \delta
\leq\beta\\\text{(since the assertion }\left(  \alpha\geq\gamma\text{ and
}\beta\geq\delta\right)  \\\text{is equivalent to }\left(  \gamma\leq
\alpha\text{ and }\delta\leq\beta\right)  \text{)}}}}\operatorname*{coeff}%
\nolimits_{\gamma,\delta}\left(  P\right)  \cdot\operatorname*{coeff}%
\nolimits_{\alpha-\gamma,\beta-\delta}\left(  Q\right)  \cdot a^{\alpha
}b^{\beta}\\
&  =\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}}\sum
_{\substack{\left(  \gamma,\delta\right)  \in\mathbb{N}^{\times2};\\\gamma
\leq\alpha;\ \delta\leq\beta}}\operatorname*{coeff}\nolimits_{\gamma,\delta
}\left(  P\right)  \cdot\operatorname*{coeff}\nolimits_{\alpha-\gamma
,\beta-\delta}\left(  Q\right)  \cdot a^{\alpha}b^{\beta}.
\end{align*}
Compared to
\begin{align*}
\rho\left(  PQ\right)   &  =\sum_{\left(  \alpha,\beta\right)  \in
\mathbb{N}^{\times2}}\underbrace{\operatorname*{coeff}\nolimits_{\alpha,\beta
}\left(  PQ\right)  }_{\substack{=\sum_{\substack{\left(  \gamma
,\delta\right)  \in\mathbb{N}^{\times2};\\\gamma\leq\alpha;\ \delta\leq\beta
}}\operatorname*{coeff}\nolimits_{\gamma,\delta}\left(  P\right)
\cdot\operatorname*{coeff}\nolimits_{\alpha-\gamma,\beta-\delta}\left(
Q\right)  \\\left(  \text{by (\ref{402.pf.prod})}\right)  }}\cdot a^{\alpha
}b^{\beta}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{402.pf.4}), applied to
}PQ\text{ instead of }P\right) \\
&  =\sum_{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}}\sum
_{\substack{\left(  \gamma,\delta\right)  \in\mathbb{N}^{\times2};\\\gamma
\leq\alpha;\ \delta\leq\beta}}\operatorname*{coeff}\nolimits_{\gamma,\delta
}\left(  P\right)  \cdot\operatorname*{coeff}\nolimits_{\alpha-\gamma
,\beta-\delta}\left(  Q\right)  \cdot a^{\alpha}b^{\beta},
\end{align*}
this yields $\rho\left(  P\right)  \cdot\rho\left(  Q\right)  =\rho\left(
PQ\right)  $.

Now forget that we fixed $P$ and $Q$. We thus have shown that any $P\in
k\left[  \left[  X,Y\right]  \right]  $ and $Q\in k\left[  \left[  X,Y\right]
\right]  $ satisfy $\rho\left(  P\right)  \cdot\rho\left(  Q\right)
=\rho\left(  PQ\right)  $. Combined with $\rho\left(  1\right)  =1$, and with
the fact that $\rho$ is $k$-linear, this yields that $\rho$ is a $k$-algebra
homomorphism. This proves part \textbf{f)}.

\textbf{g)} Both $1+X$ and $1+Y$ are power series in $k\left[  \left[
X,Y\right]  \right]  $ whose coefficients before $X^{0}Y^{0}$ are equal to
$1$. Hence, we can apply Theorem~\ref{Theorem401} to $Q=1+X$ and $R=1+Y$. As a
result, we obtain that the coefficient of $\left(  1+X\right)  \left(
1+Y\right)  $ before $X^{0}Y^{0}$ is also equal to $1$ (of course, this is
obvious for simpler reasons...) and that%
\begin{equation}
\log\left(  \left(  1+X\right)  \left(  1+Y\right)  \right)  =\log\left(
1+X\right)  +\log\left(  1+Y\right)  . \label{402.pf.crux}%
\end{equation}


\textbf{h)} The power series $X$ has the coefficient $1$ before $X^{1}Y^{0}$,
while all its other coefficients are zero. In other words,
$\operatorname*{coeff}\nolimits_{1,0}\left(  X\right)  =1$, while
$\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(  X\right)  =0$ for all
$\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2}$ satisfying $\left(
\alpha,\beta\right)  \neq\left(  1,0\right)  $. Now, (\ref{402.pf.4}) (applied
to $P=X$) yields
\begin{align*}
\rho\left(  X\right)   &  =\sum_{\left(  \alpha,\beta\right)  \in
\mathbb{N}^{\times2}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
X\right)  \cdot a^{\alpha}b^{\beta}=\underbrace{\sum_{\substack{\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2};\\\left(  \alpha,\beta\right)
=\left(  1,0\right)  }}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
X\right)  \cdot a^{\alpha}b^{\beta}}_{=\operatorname*{coeff}\nolimits_{1,0}%
\left(  X\right)  \cdot a^{1}b^{0}}+\sum_{\substack{\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2};\\\left(  \alpha,\beta\right)
\neq\left(  1,0\right)  }}\underbrace{\operatorname*{coeff}\nolimits_{\alpha
,\beta}\left(  X\right)  }_{\substack{=0\\\text{(since }\left(  \alpha
,\beta\right)  \neq\left(  1,0\right)  \text{)}}}\cdot a^{\alpha}b^{\beta}\\
&  =\underbrace{\operatorname*{coeff}\nolimits_{1,0}\left(  X\right)  }%
_{=1}\cdot\underbrace{a^{1}}_{=a}\underbrace{b^{0}}_{=1}+\underbrace{\sum
_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2};\\\left(
\alpha,\beta\right)  \neq\left(  1,0\right)  }}0\cdot a^{\alpha}b^{\beta}%
}_{=0}=a.
\end{align*}


The power series $Y$ has the coefficient $1$ before $X^{0}Y^{1}$, while all
its other coefficients are zero. In other words, $\operatorname*{coeff}%
\nolimits_{0,1}\left(  Y\right)  =1$, while $\operatorname*{coeff}%
\nolimits_{\alpha,\beta}\left(  Y\right)  =0$ for all $\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2}$ satisfying $\left(  \alpha
,\beta\right)  \neq\left(  0,1\right)  $. Now, (\ref{402.pf.4}) (applied to
$P=Y$) yields
\begin{align*}
\rho\left(  Y\right)   &  =\sum_{\left(  \alpha,\beta\right)  \in
\mathbb{N}^{\times2}}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
Y\right)  \cdot a^{\alpha}b^{\beta}=\underbrace{\sum_{\substack{\left(
\alpha,\beta\right)  \in\mathbb{N}^{\times2};\\\left(  \alpha,\beta\right)
=\left(  0,1\right)  }}\operatorname*{coeff}\nolimits_{\alpha,\beta}\left(
Y\right)  \cdot a^{\alpha}b^{\beta}}_{=\operatorname*{coeff}\nolimits_{0,1}%
\left(  Y\right)  \cdot a^{0}b^{1}}+\sum_{\substack{\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2};\\\left(  \alpha,\beta\right)
\neq\left(  0,1\right)  }}\underbrace{\operatorname*{coeff}\nolimits_{\alpha
,\beta}\left(  Y\right)  }_{\substack{=0\\\text{(since }\left(  \alpha
,\beta\right)  \neq\left(  0,1\right)  \text{)}}}\cdot a^{\alpha}b^{\beta}\\
&  =\underbrace{\operatorname*{coeff}\nolimits_{0,1}\left(  Y\right)  }%
_{=1}\cdot\underbrace{a^{0}}_{=1}\underbrace{b^{1}}_{=b}+\underbrace{\sum
_{\substack{\left(  \alpha,\beta\right)  \in\mathbb{N}^{\times2};\\\left(
\alpha,\beta\right)  \neq\left(  0,1\right)  }}0\cdot a^{\alpha}b^{\beta}%
}_{=0}=b.
\end{align*}


Since $\rho$ is a $k$-algebra homomorphism, we have $\rho\left(
X+Y+XY\right)  =\underbrace{\rho\left(  X\right)  }_{=a}+\underbrace{\rho
\left(  Y\right)  }_{=b}+\underbrace{\rho\left(  X\right)  }_{=a}%
\cdot\underbrace{\rho\left(  Y\right)  }_{=b}=a+b+ab$.

\textbf{i)} By the definition of $\log$, we have%
\begin{align*}
\log\left(  1+X\right)   &  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  \underbrace{\left(  1+X\right)  -1}_{=X}\right)  ^{i}=\sum_{i\geq
1}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\\
&  =\underbrace{\sum_{\substack{i\geq1;\\i<n+1}}}_{=\sum_{i=1}^{n}}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}+\sum_{\substack{i\geq1;\\i\geq
n+1}}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{X^{i}}%
_{\substack{=X^{n+1}\cdot X^{i-\left(  n+1\right)  }\\\text{(since }i\geq
n+1\text{)}}}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}+\underbrace{\sum
_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}{i}%
X^{n+1}\cdot X^{i-\left(  n+1\right)  }}_{=X^{n+1}\sum_{\substack{i\geq
1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i-\left(  n+1\right)  }%
}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}+X^{n+1}%
\sum_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}X^{i-\left(  n+1\right)  }.
\end{align*}
Hence,%
\begin{align}
\rho\left(  \log\left(  1+X\right)  \right)   &  =\rho\left(  \sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}+X^{n+1}\sum_{\substack{i\geq
1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i-\left(  n+1\right)
}\right) \nonumber\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \rho\left(
X\right)  \right)  ^{i}+\left(  \rho\left(  X\right)  \right)  ^{n+1}%
\sum_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  \rho\left(  X\right)  \right)  ^{i-\left(  n+1\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho\text{ is a }k\text{-algebra
homomorphism}\right) \nonumber\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}%
+\underbrace{a^{n+1}}_{=0}\sum_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(
-1\right)  ^{i-1}}{i}a^{i-\left(  n+1\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\rho\left(  X\right)  =a\right) \nonumber\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}+\underbrace{0\sum
_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}a^{i-\left(  n+1\right)  }}_{=0}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}a^{i}. \label{402.pf.X}%
\end{align}
The same argument, but with $X$ and $a$ replaced by $Y$ and $b$, yields%
\begin{equation}
\rho\left(  \log\left(  1+Y\right)  \right)  =\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}b^{i}. \label{402.pf.Y}%
\end{equation}


On the other hand, the definition of $\log$ yields%
\begin{align*}
&  \log\left(  \left(  1+X\right)  \left(  1+Y\right)  \right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
\underbrace{\left(  1+X\right)  \left(  1+Y\right)  -1}_{=X+Y+XY}\right)
^{i}=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  X+Y+XY\right)
^{i}\\
&  =\underbrace{\sum_{\substack{i\geq1;\\i<n+1}}}_{=\sum_{i=1}^{n}}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  X+Y+XY\right)  ^{i}+\sum
_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{\left(  X+Y+XY\right)  ^{i}}_{\substack{=\left(  X+Y+XY\right)
^{n+1}\cdot\left(  X+Y+XY\right)  ^{i-\left(  n+1\right)  }\\\text{(since
}i\geq n+1\text{)}}}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  X+Y+XY\right)
^{i}+\underbrace{\sum_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(  -1\right)
^{i-1}}{i}\left(  X+Y+XY\right)  ^{n+1}\cdot\left(  X+Y+XY\right)  ^{i-\left(
n+1\right)  }}_{=\left(  X+Y+XY\right)  ^{n+1}\sum_{\substack{i\geq1;\\i\geq
n+1}}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  X+Y+XY\right)  ^{i-\left(
n+1\right)  }}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  X+Y+XY\right)
^{i}+\left(  X+Y+XY\right)  ^{n+1}\sum_{\substack{i\geq1;\\i\geq n+1}%
}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  X+Y+XY\right)  ^{i-\left(
n+1\right)  }.
\end{align*}
Hence,%
\begin{align*}
&  \rho\left(  \log\left(  \left(  1+X\right)  \left(  1+Y\right)  \right)
\right) \\
&  =\rho\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
X+Y+XY\right)  ^{i}+\left(  X+Y+XY\right)  ^{n+1}\sum_{\substack{i\geq
1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  X+Y+XY\right)
^{i-\left(  n+1\right)  }\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \rho\left(
X+Y+XY\right)  \right)  ^{i}+\left(  \rho\left(  X+Y+XY\right)  \right)
^{n+1}\sum_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  \rho\left(  X+Y+XY\right)  \right)  ^{i-\left(  n+1\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  a+b+ab\right)
^{i}+\underbrace{\left(  a+b+ab\right)  ^{n+1}}_{=0}\sum_{\substack{i\geq
1;\\i\geq n+1}}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  a+b+ab\right)
^{i-\left(  n+1\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho\left(  X+Y+XY\right)
=a+b+ab\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  a+b+ab\right)
^{i}+\underbrace{0\sum_{\substack{i\geq1;\\i\geq n+1}}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  a+b+ab\right)  ^{i-\left(  n+1\right)  }}%
_{=0}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  a+b+ab\right)
^{i}.
\end{align*}
Compared with%
\begin{align*}
&  \rho\left(  \underbrace{\log\left(  \left(  1+X\right)  \left(  1+Y\right)
\right)  }_{\substack{=\log\left(  1+X\right)  +\log\left(  1+Y\right)
\\\text{(by (\ref{402.pf.crux}))}}}\right) \\
&  =\rho\left(  \log\left(  1+X\right)  +\log\left(  1+Y\right)  \right)
=\underbrace{\rho\left(  \log\left(  1+X\right)  \right)  }_{\substack{=\sum
_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}\\\text{(by (\ref{402.pf.X}%
))}}}+\underbrace{\rho\left(  \log\left(  1+Y\right)  \right)  }%
_{\substack{=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}b^{i}%
\\\text{(by (\ref{402.pf.Y}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\rho\text{ is }k\text{-linear}\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}+\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}b^{i},
\end{align*}
this yields%
\[
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}+\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}b^{i}=\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  a+b+ab\right)  ^{i}.
\]
This proves Corollary~\ref{Corollary402}.
\end{proof}

We further prepare for proving Theorem~\ref{Theorem400} by showing a useful
result that we ought to have proven long ago:

\begin{proposition}
\label{Proposition403} Let $k$ be a field. Let $A$ be a $k$-algebra, and let
$C$ be a filtered $k$-coalgebra. In this proposition, we shall denote the
convolution on $\mathcal{L}\left(  C,A\right)  $ as ordinary multiplication
(i.e., we write $fg$ for the convolution $f\ast g$ of two maps $f,g\in
\mathcal{L}\left(  C,A\right)  $).

\textbf{(a)} Any $a\in\mathbb{N}$ and $b\in\mathbb{N}$ satisfy $\left(
\mathcal{L}^{a}\left(  C,A\right)  \right)  \cdot\left(  \mathcal{L}%
^{b}\left(  C,A\right)  \right)  \subseteq\mathcal{L}^{a+b}\left(  C,A\right)
$.

\textbf{(b)} Any $n\in\mathbb{N}$ satisfies $\left(  \mathcal{L}^{1}\left(
C,A\right)  \right)  ^{n}\subseteq\mathcal{L}^{n}\left(  C,A\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition403}.]\textbf{(a)} Let $a\in\mathbb{N}$
and $b\in\mathbb{N}$.

Due to how we defined $\mathcal{L}^{n}\left(  C,A\right)  $ (in
Definition~\ref{Definition8} \textbf{(b)}), we have%
\begin{align}
\mathcal{L}^{n}\left(  C,A\right)   &  =\left\{  f\in\mathcal{L}\left(
C,A\right)  \ \mid\ f\mid_{C_{\leq n-1}}=0\right\} \nonumber\\
&  =\left\{  h\in\mathcal{L}\left(  C,A\right)  \ \mid\ h\mid_{C_{\leq n-1}%
}=0\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }f\text{ as
}h\right)  \label{403.pf.0}%
\end{align}
for every $n\in\mathbb{N}$.

It is easy to see that every $\left(  f,g\right)  \in\left(  \mathcal{L}%
^{a}\left(  C,A\right)  \right)  \times\left(  \mathcal{L}^{b}\left(
C,A\right)  \right)  $ satisfies $f\ast g\in\mathcal{L}^{a+b}\left(
C,A\right)  $.\ \ \ \ \footnote{\textit{Proof.} Let $\left(  f,g\right)
\in\left(  \mathcal{L}^{a}\left(  C,A\right)  \right)  \times\left(
\mathcal{L}^{b}\left(  C,A\right)  \right)  $ be arbitrary. Then,
$f\in\mathcal{L}^{a}\left(  C,A\right)  $ and $g\in\mathcal{L}^{b}\left(
C,A\right)  $. Now, $f\in\mathcal{L}^{a}\left(  C,A\right)  =\left\{
h\in\mathcal{L}\left(  C,A\right)  \ \mid\ h\mid_{C_{\leq a-1}}=0\right\}  $
(by (\ref{403.pf.0}), applied to $n=a$), so that $f\mid_{C_{\leq a-1}}=0$.
Hence, $f\left(  C_{\leq a-1}\right)  =\underbrace{\left(  f\mid_{C_{\leq
a-1}}\right)  }_{=0}\left(  C_{\leq a-1}\right)  =0\left(  C_{\leq
a-1}\right)  =0$. Similarly, $g\left(  C_{\leq b-1}\right)  =0$.
\par
Since $C$ is a filtered $k$-coalgebra, we have $\Delta_{C}\left(  C_{\leq
a+b-1}\right)  \subseteq\sum_{u=0}^{a+b-1}C_{\leq u}\otimes C_{\leq a+b-1-u}$.
\par
Since $C$ is filtered, we have $C_{\leq0}\subseteq C_{\leq1}\subseteq
C_{\leq2}\subseteq\cdots$. This yields%
\begin{equation}
C_{\leq u}\subseteq C_{\leq a-1}\text{ for every }u\in\mathbb{N}\text{
satisfying }u\leq a-1, \label{403.pf.1}%
\end{equation}
and%
\begin{equation}
C_{\leq v}\subseteq C_{\leq b-1}\text{ for every }v\in\mathbb{N}\text{
satisfying }v\leq b-1. \label{403.pf.2}%
\end{equation}
\par
Thus,%
\begin{equation}
C_{\leq a+b-1-u}\subseteq C_{\leq b-1}\text{ for every }u\in\mathbb{N}\text{
satisfying }u\geq a\text{ and }u\leq a+b-1 \label{403.pf.3}%
\end{equation}
(because for every $u\in\mathbb{N}$ satisfying $u\geq a$ and $u\leq a+b-1$, we
have $a+b-1-u\in\mathbb{N}$ (since $\underbrace{a+b-1}_{\geq u}-u\geq u-u=0$)
and $a+b-1-\underbrace{u}_{\geq a}\leq a+b-1-a=b-1$, and thus $C_{\leq
a+b-1-u}\subseteq C_{\leq b-1}$ (by (\ref{403.pf.2}), applied to
$v=a+b-1-u$)).
\par
By the definition of convolution, $f\ast g=\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$, so that%
\begin{align*}
&  \left(  f\ast g\right)  \left(  C_{\leq a+b-1}\right) \\
&  =\left(  \mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}\right)
\left(  C_{\leq a+b-1}\right)  =\mu_{A}\left(  \left(  f\otimes g\right)
\underbrace{\left(  \Delta_{C}\left(  C_{\leq a+b-1}\right)  \right)
}_{\subseteq\sum_{u=0}^{a+b-1}C_{\leq u}\otimes C_{\leq a+b-1-u}}\right) \\
&  \subseteq\mu_{A}\left(  \underbrace{\left(  f\otimes g\right)  \left(
\sum_{u=0}^{a+b-1}C_{\leq u}\otimes C_{\leq a+b-1-u}\right)  }%
_{\substack{\subseteq\sum_{u=0}^{a+b-1}\left(  f\otimes g\right)  \left(
C_{\leq u}\otimes C_{\leq a+b-1-u}\right)  \\\text{(since }f\otimes g\text{ is
}k\text{-linear)}}}\right)  \subseteq\mu_{A}\left(  \sum_{u=0}^{a+b-1}%
\underbrace{\left(  f\otimes g\right)  \left(  C_{\leq u}\otimes C_{\leq
a+b-1-u}\right)  }_{\subseteq f\left(  C_{\leq u}\right)  \otimes g\left(
C_{\leq a+b-1-u}\right)  }\right) \\
&  \subseteq\mu_{A}\left(  \underbrace{\sum_{u=0}^{a+b-1}f\left(  C_{\leq
u}\right)  \otimes g\left(  C_{\leq a+b-1-u}\right)  }_{=\sum_{u=0}%
^{a-1}f\left(  C_{\leq u}\right)  \otimes g\left(  C_{\leq a+b-1-u}\right)
+\sum_{u=a}^{a+b-1}f\left(  C_{\leq u}\right)  \otimes g\left(  C_{\leq
a+b-1-u}\right)  }\right) \\
&  =\mu_{A}\left(  \sum_{u=0}^{a-1}f\left(  \underbrace{C_{\leq u}%
}_{\substack{\subseteq C_{\leq a-1}\\\text{(by (\ref{403.pf.1}), since }u\leq
a-1\text{)}}}\right)  \otimes g\left(  C_{\leq a+b-1-u}\right)  +\sum
_{u=a}^{a+b-1}f\left(  C_{\leq u}\right)  \otimes g\left(  \underbrace{C_{\leq
a+b-1-u}}_{\substack{\subseteq C_{\leq b-1}\\\text{(by (\ref{403.pf.3}), since
}u\geq a\text{ and }u\leq a+b-1\text{)}}}\right)  \right) \\
&  =\mu_{A}\left(  \sum_{u=0}^{a-1}\underbrace{f\left(  C_{\leq a-1}\right)
}_{=0}\otimes g\left(  C_{\leq a+b-1-u}\right)  +\sum_{u=a}^{a+b-1}f\left(
C_{\leq u}\right)  \otimes\underbrace{g\left(  C_{\leq b-1}\right)  }%
_{=0}\right) \\
&  =\mu_{A}\left(  \sum_{u=0}^{a-1}\underbrace{0\otimes g\left(  C_{\leq
a+b-1-u}\right)  }_{=0}+\sum_{u=a}^{a+b-1}\underbrace{f\left(  C_{\leq
u}\right)  \otimes0}_{=0}\right)  =\mu_{A}\left(  \underbrace{\sum_{u=0}%
^{a-1}0+\sum_{u=a}^{a+b-1}0}_{=0}\right)  =\mu_{A}\left(  0\right)  =0.
\end{align*}
Hence, $f\ast g\in\left\{  h\in\mathcal{L}\left(  C,A\right)  \ \mid
\ h\mid_{C_{\leq a+b-1}}=0\right\}  $. Since $\mathcal{L}^{a+b}\left(
C,A\right)  =\left\{  h\in\mathcal{L}\left(  C,A\right)  \ \mid\ h\mid
_{C_{\leq a+b-1}}=0\right\}  $ (by (\ref{403.pf.0}), applied to $n=a+b$), this
becomes%
\[
f\ast g\in\left\{  h\in\mathcal{L}\left(  C,A\right)  \ \mid\ h\mid_{C_{\leq
a+b-1}}=0\right\}  =\mathcal{L}^{a+b}\left(  C,A\right)  ,
\]
qed.} In other words,%
\[
\left\{  f\ast g\ \mid\ \left(  f,g\right)  \in\left(  \mathcal{L}^{a}\left(
C,A\right)  \right)  \times\left(  \mathcal{L}^{b}\left(  C,A\right)  \right)
\right\}  \subseteq\mathcal{L}^{a+b}\left(  C,A\right)  .
\]
Thus, (\ref{284.pf.b.2}) (applied to $M=\mathcal{L}\left(  C,A\right)  $,
$S=\left\{  f\ast g\ \mid\ \left(  f,g\right)  \in\left(  \mathcal{L}%
^{a}\left(  C,A\right)  \right)  \times\left(  \mathcal{L}^{b}\left(
C,A\right)  \right)  \right\}  $ and $Q=\mathcal{L}^{a+b}\left(  C,A\right)
$) yields%
\[
\left\langle \left\{  f\ast g\ \mid\ \left(  f,g\right)  \in\left(
\mathcal{L}^{a}\left(  C,A\right)  \right)  \times\left(  \mathcal{L}%
^{b}\left(  C,A\right)  \right)  \right\}  \right\rangle \subseteq
\mathcal{L}^{a+b}\left(  C,A\right)  .
\]
Since%
\begin{align*}
&  \left\langle \left\{  f\ast g\ \mid\ \left(  f,g\right)  \in\left(
\mathcal{L}^{a}\left(  C,A\right)  \right)  \times\left(  \mathcal{L}%
^{b}\left(  C,A\right)  \right)  \right\}  \right\rangle \\
&  =\left\langle \underbrace{f\ast g}_{=fg}\ \mid\ \left(  f,g\right)
\in\left(  \mathcal{L}^{a}\left(  C,A\right)  \right)  \times\left(
\mathcal{L}^{b}\left(  C,A\right)  \right)  \right\rangle \\
&  =\left\langle fg\ \mid\ \left(  f,g\right)  \in\left(  \mathcal{L}%
^{a}\left(  C,A\right)  \right)  \times\left(  \mathcal{L}^{b}\left(
C,A\right)  \right)  \right\rangle =\left(  \mathcal{L}^{a}\left(  C,A\right)
\right)  \cdot\left(  \mathcal{L}^{b}\left(  C,A\right)  \right)  ,
\end{align*}
this becomes $\left(  \mathcal{L}^{a}\left(  C,A\right)  \right)  \cdot\left(
\mathcal{L}^{b}\left(  C,A\right)  \right)  \subseteq\mathcal{L}^{a+b}\left(
C,A\right)  $. This proves Proposition~\ref{Proposition403} \textbf{(a)}.

\textbf{(b)} We are going to prove Proposition~\ref{Proposition403}
\textbf{(b)} by induction over $n$:

\textit{Induction base:} From Definition~\ref{Definition8} \textbf{(b)}
(applied to $C$ instead of $H$), we know that $\mathcal{L}^{0}\left(
C,A\right)  =\mathcal{L}\left(  C,A\right)  $. Now, clearly, $\left(
\mathcal{L}^{1}\left(  C,A\right)  \right)  ^{0}\subseteq\mathcal{L}\left(
C,A\right)  =\mathcal{L}^{0}\left(  C,A\right)  $. In other words,
Proposition~\ref{Proposition403} \textbf{(b)} holds for $n=0$. This completes
the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
Proposition~\ref{Proposition403} \textbf{(b)} holds for $n=N$. We must now
prove that Proposition~\ref{Proposition403} \textbf{(b)} also holds for
$n=N+1$.

Since Proposition~\ref{Proposition403} \textbf{(b)} holds for $n=N$, we have
$\left(  \mathcal{L}^{1}\left(  C,A\right)  \right)  ^{N}\subseteq
\mathcal{L}^{N}\left(  C,A\right)  $. Now,%
\begin{align*}
\left(  \mathcal{L}^{1}\left(  C,A\right)  \right)  ^{N+1}  &
=\underbrace{\left(  \mathcal{L}^{1}\left(  C,A\right)  \right)  ^{N}%
}_{\subseteq\mathcal{L}^{N}\left(  C,A\right)  }\cdot\left(  \mathcal{L}%
^{1}\left(  C,A\right)  \right)  \subseteq\left(  \mathcal{L}^{N}\left(
C,A\right)  \right)  \cdot\left(  \mathcal{L}^{1}\left(  C,A\right)  \right)
\\
&  \subseteq\mathcal{L}^{N+1}\left(  C,A\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition~\ref{Proposition403} \textbf{(a)}, applied to }a=N\text{
and }b=1\right)  .
\end{align*}
In other words, Proposition~\ref{Proposition403} \textbf{(b)} holds for
$n=N+1$. This completes the induction step. Thus, the induction proof of
Proposition~\ref{Proposition403} \textbf{(b)} is complete.
\end{proof}

Here is, finally, an equivalent version of Corollary~\ref{Corollary402} for
use in our proof of Theorem~\ref{Theorem400}:

\begin{corollary}
\label{Corollary404} Let $k$ be a field of characteristic $0$. Let
$n\in\mathbb{N}$.

Let $\mathfrak{B}$ be a commutative $k$-algebra. Let $\mathfrak{K}$ be an
ideal of $\mathfrak{B}$. Let $a\in\mathfrak{K}$ and $b\in\mathfrak{K}$. Then,%
\[
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}+\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}b^{i}-\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  a+b+ab\right)  ^{i}\in\mathfrak{K}^{n+1}.
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary404}.]Since $\mathfrak{K}$ is an ideal of
$\mathfrak{B}$, it is clear that $\mathfrak{K}^{n+1}$ is an ideal of
$\mathfrak{B}$.

Let $\pi$ be the canonical projection $\mathfrak{B}\rightarrow\mathfrak{B}%
\diagup\mathfrak{K}^{n+1}$. Then, $\pi$ is a $k$-algebra homomorphism (since
$\mathfrak{K}^{n+1}$ is an ideal of $\mathfrak{B}$, so that $\mathfrak{B}%
\diagup\mathfrak{K}^{n+1}$ is a factor algebra), so that
\[
\left(  \pi\left(  \mathfrak{K}\right)  \right)  ^{n+1}=\pi\left(
\mathfrak{K}^{n+1}\right)  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\pi\text{ is the canonical projection }\mathfrak{B}\rightarrow\mathfrak{B}%
\diagup\mathfrak{K}^{n+1}\right)  .
\]


Let $\mathfrak{A}$ be the $k$-algebra $\mathfrak{B}\diagup\mathfrak{K}^{n+1}$,
and let $\mathfrak{I}$ be the subset $\pi\left(  \mathfrak{K}\right)  $ of
$\mathfrak{B}\diagup\mathfrak{K}^{n+1}=\mathfrak{A}$. The $k$-algebra
$\mathfrak{A}$ is commutative (because $\mathfrak{A}$ is a quotient of the
commutative $k$-algebra $\mathfrak{B}$).

Since $\pi$ is the canonical projection $\mathfrak{B}\rightarrow
\mathfrak{B}\diagup\mathfrak{K}^{n+1}$, this map $\pi$ is surjective. Thus,
$\pi\left(  \mathfrak{B}\right)  =\mathfrak{B}\diagup\mathfrak{K}%
^{n+1}=\mathfrak{A}$.

Since $\mathfrak{K}$ is an ideal of $\mathfrak{B}$, we have $\mathfrak{BK}%
\subseteq\mathfrak{K}$. Now,%
\begin{align*}
\underbrace{\mathfrak{A}}_{=\pi\left(  \mathfrak{B}\right)  }%
\underbrace{\mathfrak{I}}_{=\pi\left(  \mathfrak{K}\right)  }  &  =\pi\left(
\mathfrak{B}\right)  \pi\left(  \mathfrak{K}\right)  =\pi\left(
\underbrace{\mathfrak{BK}}_{\subseteq\mathfrak{K}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  \subseteq\pi\left(  \mathfrak{K}\right)  =\mathfrak{I}.
\end{align*}
In other words, $\mathfrak{I}$ is a right ideal of $\mathfrak{A}$. Since
$\mathfrak{A}$ is commutative, this yields that $\mathfrak{I}$ is an ideal of
$\mathfrak{A}$. Also, $\mathfrak{I}=\pi\left(  \mathfrak{K}\right)  $ leads to
$\mathfrak{I}^{n+1}=\left(  \pi\left(  \mathfrak{K}\right)  \right)  ^{n+1}=0$.

Since $a\in\mathfrak{K}$, we have $\pi\left(  a\right)  \in\pi\left(
\mathfrak{K}\right)  =\mathfrak{I}$. Similarly, $\pi\left(  b\right)
\in\mathfrak{I}$. Therefore, we can apply Corollary~\ref{Corollary402} to
$\pi\left(  a\right)  $ and $\pi\left(  b\right)  $ instead of $a$ and $b$. We
obtain%
\[
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \pi\left(  a\right)
\right)  ^{i}+\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
\pi\left(  b\right)  \right)  ^{i}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}\left(  \pi\left(  a\right)  +\pi\left(  b\right)  +\pi\left(
a\right)  \pi\left(  b\right)  \right)  ^{i}.
\]
Hence,%
\begin{align*}
0  &  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \pi\left(
a\right)  \right)  ^{i}+\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  \pi\left(  b\right)  \right)  ^{i}-\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  \pi\left(  a\right)  +\pi\left(  b\right)
+\pi\left(  a\right)  \pi\left(  b\right)  \right)  ^{i}\\
&  =\pi\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}%
+\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}b^{i}-\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  a+b+ab\right)  ^{i}\right)
\end{align*}
(since $\pi$ is a $k$-algebra homomorphism). In other words,%
\[
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}a^{i}+\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}b^{i}-\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}\left(  a+b+ab\right)  ^{i}\in\operatorname*{Ker}%
\pi=\mathfrak{K}^{n+1}%
\]
(since $\pi$ is the canonical projection $\mathfrak{B}\rightarrow
\mathfrak{B}\diagup\mathfrak{K}^{n+1}$). This proves
Corollary~\ref{Corollary404}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem400}.]\textbf{a)} First, we notice that $F\ast
H\in G\left(  C,A\right)  $.

\textit{Proof.} By the definition of $G\left(  C,A\right)  $, we have%
\[
G\left(  C,A\right)  =\left\{  f\in\mathcal{L}\left(  C,A\right)
\ \mid\ f\left(  1_{C}\right)  =1_{A}\right\}  .
\]
Now, $F\in G\left(  C,A\right)  =\left\{  f\in\mathcal{L}\left(  C,A\right)
\ \mid\ f\left(  1_{C}\right)  =1_{A}\right\}  $, so that $F\left(
1_{C}\right)  =1_{A}$. Similarly, $G\left(  1_{C}\right)  =1_{A}$. But since
$1_{C}$ is the unity of the unital coalgebra $C$, we have $\Delta_{C}\left(
1_{C}\right)  =1_{C}\otimes1_{C}$ and $\varepsilon_{C}\left(  1_{C}\right)
=1$ (by the definition of a unital coalgebra).

By the definition of convolution, $F\ast H=\mu_{A}\circ\left(  F\otimes
H\right)  \circ\Delta_{C}$. Hence,%
\begin{align*}
\left(  F\ast H\right)  \left(  1_{C}\right)   &  =\left(  \mu_{A}\circ\left(
F\otimes H\right)  \circ\Delta_{C}\right)  \left(  1_{C}\right)  =\mu
_{A}\left(  \left(  F\otimes H\right)  \underbrace{\left(  \Delta_{C}\left(
1_{C}\right)  \right)  }_{=1_{C}\otimes1_{C}}\right)  =\mu_{A}\left(
\underbrace{\left(  F\otimes H\right)  \left(  1_{C}\otimes1_{C}\right)
}_{=F\left(  1_{C}\right)  \otimes H\left(  1_{C}\right)  }\right) \\
&  =\mu_{A}\left(  \underbrace{F\left(  1_{C}\right)  }_{=1_{A}}%
\otimes\underbrace{H\left(  1_{C}\right)  }_{=1_{A}}\right)  =\mu_{A}\left(
1_{A}\otimes1_{A}\right) \\
&  =1_{A}1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{A}\text{ is the
multiplication map}\right) \\
&  =1_{A}.
\end{align*}
In other words,%
\[
F\ast H\in\left\{  f\in\mathcal{L}\left(  C,A\right)  \ \mid\ f\left(
1_{C}\right)  =1_{A}\right\}  =G\left(  C,A\right)  .
\]
Thus we have proven that $F\ast H\in G\left(  C,A\right)  $. This completes
the proof of \textbf{a)}.

\textbf{b)} Now let us prove that $\operatorname*{Log}\left(  F\ast H\right)
=\operatorname*{Log}F+\operatorname*{Log}H$.

\textit{Proof.} Let $x\in C$. We are now going to prove that $\left(
\operatorname*{Log}F+\operatorname*{Log}H-\operatorname*{Log}\left(  F\ast
H\right)  \right)  \left(  x\right)  =0$.

Since $C$ is filtered, there exists some $n\in\mathbb{N}$ such that $x\in
C_{\leq n}$. Consider this $n$.

Due to how we defined $\mathcal{L}^{n+1}\left(  C,A\right)  $ (in
Definition~\ref{Definition8} \textbf{(b)}), we have%
\begin{align}
\mathcal{L}^{n+1}\left(  C,A\right)   &  =\left\{  f\in\mathcal{L}\left(
C,A\right)  \ \mid\ \underbrace{f\mid_{C_{\leq n+1-1}}}_{=f\mid_{C_{\leq n}}%
}=0\right\}  =\left\{  f\in\mathcal{L}\left(  C,A\right)  \ \mid
\ f\mid_{C_{\leq n}}=0\right\} \nonumber\\
&  =\left\{  g\in\mathcal{L}\left(  C,A\right)  \ \mid\ g\mid_{C_{\leq n}%
}=0\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }f\text{ as
}g\right)  . \label{400.pf.0}%
\end{align}


Let $f=F-e_{C,A}$ and $h=H-e_{C,A}$. Then, $\underbrace{f}_{=F-e_{C,A}%
}+e_{C,A}=F-e_{C,A}+e_{C,A}=F$ and similarly $h+e_{C,A}=H$. Hence,%
\begin{align}
\underbrace{F}_{=f+e_{C,A}}\ast\underbrace{H}_{=h+e_{C,A}}-e_{C,A}  &
=\underbrace{\left(  f+e_{C,A}\right)  \ast\left(  h+e_{C,A}\right)  }_{=f\ast
h+f\ast e_{C,A}+e_{C,A}\ast h+e_{C,A}\ast e_{C,A}}-e_{C,A}\nonumber\\
&  =f\ast h+\underbrace{f\ast e_{C,A}}_{=f}+\underbrace{e_{C,A}\ast h}%
_{=h}+\underbrace{e_{C,A}\ast e_{C,A}}_{=e_{C,A}}-e_{C,A}\nonumber\\
&  =f\ast h+f+h+e_{C,A}-e_{C,A}=f+h+f\ast h. \label{400.pf.FH}%
\end{align}


Note that $f=F-e_{C,A}$ and $h=H-e_{C,A}$ lead to%
\begin{align*}
\underbrace{f}_{=F-e_{C,A}}\ast\underbrace{h}_{=H-e_{C,A}}  &  =\left(
F-e_{C,A}\right)  \ast\left(  H-e_{C,A}\right)  =\underbrace{F\ast H}_{=H\ast
F}-\underbrace{F\ast e_{C,A}}_{=F=e_{C,A}\ast F}-\underbrace{e_{C,A}\ast
H}_{=H=H\ast e_{C,A}}+e_{C,A}\ast e_{C,A}\\
&  =H\ast F-e_{C,A}\ast F-H\ast e_{C,A}+e_{C,A}\ast e_{C,A}\\
&  =\underbrace{\left(  H-e_{C,A}\right)  }_{=h}\ast\underbrace{\left(
F-e_{C,A}\right)  }_{=f}=h\ast f.
\end{align*}


Since $F\in G\left(  C,A\right)  =e_{C,A}+\mathfrak{g}\left(  C,A\right)  $,
we have $F-e_{C,A}\in\mathfrak{g}\left(  C,A\right)  $. Since $F-e_{C,A}=f$,
this rewrites as $f\in\mathfrak{g}\left(  C,A\right)  $. Similarly,
$h\in\mathfrak{g}\left(  C,A\right)  $. Moreover, since $F\ast H\in G\left(
C,A\right)  =e_{C,A}+\mathfrak{g}\left(  C,A\right)  $, we have $F\ast
H-e_{C,A}\in\mathfrak{g}\left(  C,A\right)  $. By (\ref{400.pf.FH}), this
becomes $f+h+f\ast h\in\mathfrak{g}\left(  C,A\right)  $.

We recall that $\mathcal{L}^{1}\left(  C,A\right)  =\mathfrak{g}\left(
C,A\right)  $ (by Definition~\ref{Definition8} \textbf{(b)}, applied to $C$
instead of $H$). Since $\mathcal{L}^{1}\left(  C,A\right)  $ is an ideal of
$\mathcal{L}\left(  C,A\right)  $ (by Proposition~\ref{Proposition81}, applied
to $1$ and $C$ instead of $n$ and $H$), this yields that $\mathfrak{g}\left(
C,A\right)  $ is an ideal of $\mathcal{L}\left(  C,A\right)  $.

We know that $f\ast h=h\ast f$. In other words, $f$ and $h$ commute (as
elements of $\mathcal{L}\left(  C,A\right)  $). Let $\mathfrak{B}$ be the
$k$-subalgebra of $\mathcal{L}\left(  C,A\right)  $ generated by $f$ and $h$.
Then, the $k$-algebra $\mathfrak{B}$ is commutative (by
Corollary~\ref{Corollary47b}, applied to $\mathcal{L}\left(  C,A\right)  $,
$h$ and $\mathfrak{B}$ instead of $A$, $g$ and $\mathfrak{H}$).

Let $\mathfrak{K}=\mathfrak{B}\cap\mathfrak{g}\left(  C,A\right)  $. Then,
$\mathfrak{K}=\mathfrak{B}\cap\mathfrak{g}\left(  C,A\right)  \subseteq
\mathfrak{B}$ and $\mathfrak{K}=\mathfrak{B}\cap\mathfrak{g}\left(
C,A\right)  \subseteq\mathfrak{g}\left(  C,A\right)  $.

Since $\mathfrak{g}\left(  C,A\right)  $ is an ideal of $\mathcal{L}\left(
C,A\right)  $, we have $\left(  \mathcal{L}\left(  C,A\right)  \right)
\cdot\left(  \mathfrak{g}\left(  C,A\right)  \right)  \subseteq\mathfrak{g}%
\left(  C,A\right)  $. Now, combining%
\[
\underbrace{\mathfrak{B}}_{\subseteq\mathcal{L}\left(  C,A\right)  }%
\cdot\underbrace{\mathfrak{K}}_{\subseteq\mathfrak{g}\left(  C,A\right)
}\subseteq\left(  \mathcal{L}\left(  C,A\right)  \right)  \cdot\left(
\mathfrak{g}\left(  C,A\right)  \right)  \subseteq\mathfrak{g}\left(
C,A\right)
\]
with%
\[
\mathfrak{B}\cdot\underbrace{\mathfrak{K}}_{\subseteq\mathfrak{B}}%
\subseteq\mathfrak{B}\cdot\mathfrak{B}\subseteq\mathfrak{B}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{B}\text{ is a
}k\text{-algebra}\right)  ,
\]
we get $\mathfrak{B}\cdot\mathfrak{K}\subseteq\mathfrak{B}\cap\mathfrak{g}%
\left(  C,A\right)  =\mathfrak{K}$. Thus, $\mathfrak{K}$ is a left ideal of
$\mathfrak{B}$. Since $\mathfrak{B}$ is commutative, this means that
$\mathfrak{K}$ is an ideal of $\mathfrak{B}$.

Since $f\in\mathfrak{B}$ (by the definition of $\mathfrak{B}$) and
$f\in\mathfrak{g}\left(  C,A\right)  $, we have $f\in\mathfrak{B}%
\cap\mathfrak{g}\left(  C,A\right)  =\mathfrak{K}$. Similarly, $h\in
\mathfrak{K}$. Hence, Corollary~\ref{Corollary404} (applied to $a=f$ and
$b=h$) yields%
\begin{equation}
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}+\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}-\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast h\right)  ^{\ast i}%
\in\mathfrak{K}^{n+1}. \label{400.pf.1}%
\end{equation}
Since $\mathfrak{K}\subseteq\mathfrak{g}\left(  C,A\right)  =\mathcal{L}%
^{1}\left(  C,A\right)  $, we have $\mathfrak{K}^{n+1}\subseteq\left(
\mathcal{L}^{1}\left(  C,A\right)  \right)  ^{n+1}\subseteq\mathcal{L}%
^{n+1}\left(  C,A\right)  $ (by Proposition~\ref{Proposition403} \textbf{(b)},
applied to $n+1$ instead of $n$). Hence, (\ref{400.pf.1}) becomes%
\begin{align*}
&  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}+\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}-\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast h\right)  ^{\ast i}\\
&  \in\mathfrak{K}^{n+1}\subseteq\mathcal{L}^{n+1}\left(  C,A\right)
=\left\{  g\in\mathcal{L}\left(  C,A\right)  \ \mid\ g\mid_{C_{\leq n}%
}=0\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{400.pf.0})}\right)  .
\end{align*}
In other words,%
\[
\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}%
+\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}-\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast h\right)  ^{\ast
i}\right)  \mid_{C_{\leq n}}=0.
\]
Hence,%
\begin{align}
&  \left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}%
+\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}-\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast h\right)  ^{\ast
i}\right)  \left(  x\right) \nonumber\\
&  =\underbrace{\left(  \left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}f^{\ast i}+\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast
i}-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast
h\right)  ^{\ast i}\right)  \mid_{C_{\leq n}}\right)  }_{=0}\left(  x\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }x\in C_{\leq n}\right) \nonumber\\
&  =0\left(  x\right)  =0. \label{400.pf.2}%
\end{align}


Now, by the definition of $\operatorname*{Log}$, we have $\operatorname*{Log}%
F=\operatorname{Log}_{1}\underbrace{\left(  F-e_{C,A}\right)  }_{=f}%
=\operatorname{Log}_{1}f$, so that%
\begin{align}
\left(  \operatorname*{Log}F\right)  \left(  x\right)   &  =\left(
\operatorname{Log}_{1}f\right)  \left(  x\right) \nonumber\\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname{Log}_{1}\right) \nonumber\\
&  =\underbrace{\sum_{\substack{i\geq1;\\i\leq n}}}_{=\sum_{i=1}^{n}}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)  +\sum
_{\substack{i\geq1;\\i>n}}\dfrac{\left(  -1\right)  ^{i-1}}{i}%
\underbrace{f^{\ast i}\left(  x\right)  }_{\substack{=0\text{ (since }\\x\in
C_{\leq n}\text{ and thus }f^{\ast i}\left(  x\right)  \in f^{\ast i}\left(
C_{\leq n}\right)  =0\\\text{(by Remark~\ref{Remark9} (applied to
}C\\\text{instead of }H\text{), since }i>n\text{))}}}\nonumber\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
x\right)  +\underbrace{\sum_{\substack{i\geq1;\\i>n}}\dfrac{\left(  -1\right)
^{i-1}}{i}0}_{=0}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  . \label{400.pf.3}%
\end{align}
The same argument, applied to $H$ and $h$ instead of $F$ and $f$, shows that%
\begin{equation}
\left(  \operatorname*{Log}H\right)  \left(  x\right)  =\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}\left(  x\right)  .
\label{400.pf.4}%
\end{equation}


But for every $i\in\mathbb{N}$ satisfying $i>n$, we have%
\begin{equation}
\left(  f+h+f\ast h\right)  ^{\ast i}\left(  x\right)  =0 \label{400.pf.0ity}%
\end{equation}
\footnote{\textit{Proof of (\ref{400.pf.0ity}):} Let $i\in\mathbb{N}$ be such
that $i>n$. Then, Remark~\ref{Remark9} (applied to $C$ instead of $H$) yields
$\left(  f+h+f\ast h\right)  ^{\ast i}\left(  C_{\leq n}\right)  =0$ (since
$f+h+f\ast h\in\mathfrak{g}\left(  C,A\right)  $ and $i>n$). Now,%
\[
\left(  f+h+f\ast h\right)  ^{\ast i}\left(  \underbrace{x}_{\in C_{\leq n}%
}\right)  \in\left(  f+h+f\ast h\right)  ^{\ast i}\left(  C_{\leq n}\right)
=0,
\]
so that $\left(  f+h+f\ast h\right)  ^{\ast i}\left(  x\right)  =0$. This
proves (\ref{400.pf.0ity}).}.

Finally, by the definition of $\operatorname*{Log}$, we have
$\operatorname*{Log}\left(  F\ast H\right)  =\operatorname{Log}_{1}%
\underbrace{\left(  F\ast H-e_{C,A}\right)  }_{\substack{=f+h+f\ast
h\\\text{(by (\ref{400.pf.FH}))}}}=\operatorname{Log}_{1}\left(  f+h+f\ast
h\right)  $. Thus,%
\begin{align}
&  \left(  \operatorname*{Log}\left(  F\ast H\right)  \right)  \left(
x\right) \nonumber\\
&  =\left(  f+h+f\ast h\right)  \left(  x\right) \nonumber\\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast
h\right)  ^{\ast i}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\operatorname{Log}_{1}\right) \nonumber\\
&  =\underbrace{\sum_{\substack{i\geq1;\\i\leq n}}}_{=\sum_{i=1}^{n}}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast h\right)  ^{\ast
i}\left(  x\right)  +\sum_{\substack{i\geq1;\\i>n}}\dfrac{\left(  -1\right)
^{i-1}}{i}\underbrace{\left(  f+h+f\ast h\right)  ^{\ast i}\left(  x\right)
}_{\substack{=0\\\text{(by (\ref{400.pf.0ity}))}}}\nonumber\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast
h\right)  ^{\ast i}\left(  x\right)  +\underbrace{\sum_{\substack{i\geq
1;\\i>n}}\dfrac{\left(  -1\right)  ^{i-1}}{i}0}_{=0}\nonumber\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast
h\right)  ^{\ast i}\left(  x\right)  . \label{400.pf.5}%
\end{align}


Now,%
\begin{align*}
&  \left(  \operatorname*{Log}F+\operatorname*{Log}H-\operatorname*{Log}%
\left(  F\ast H\right)  \right)  \left(  x\right) \\
&  =\underbrace{\left(  \operatorname*{Log}F\right)  \left(  x\right)
}_{\substack{=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  \\\text{(by (\ref{400.pf.3}))}}}+\underbrace{\left(
\operatorname*{Log}H\right)  \left(  x\right)  }_{\substack{=\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}\left(  x\right)
\\\text{(by (\ref{400.pf.4}))}}}-\underbrace{\left(  \operatorname*{Log}%
\left(  F\ast H\right)  \right)  \left(  x\right)  }_{\substack{=\sum
_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast h\right)
^{\ast i}\left(  x\right)  \\\text{(by (\ref{400.pf.5}))}}}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
x\right)  +\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}\left(
x\right)  -\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast
h\right)  ^{\ast i}\left(  x\right) \\
&  =\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}%
+\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}h^{\ast i}-\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f+h+f\ast h\right)  ^{\ast
i}\right)  \left(  x\right)  =0
\end{align*}
(by (\ref{400.pf.2})).

Now forget that we fixed $x$. We thus have shown that every $x\in C$ satisfies
$\left(  \operatorname*{Log}F+\operatorname*{Log}H-\operatorname*{Log}\left(
F\ast H\right)  \right)  \left(  x\right)  =0$. In other words,
$\operatorname*{Log}F+\operatorname*{Log}H-\operatorname*{Log}\left(  F\ast
H\right)  =0$. Thus, $\operatorname*{Log}F+\operatorname*{Log}%
H=\operatorname*{Log}\left(  F\ast H\right)  $. This completes the proof of
\textbf{b)}.

The proof of Theorem~\ref{Theorem400} is thus complete.
\end{proof}

\section{\label{S19}Logarithms of tensor products}

In this Section \S \ref{S19}, we will apply Theorem~\ref{Theorem400} to
compute the logarithm of the \textit{tensor} product of two maps (using the
logarithms of these maps). First, to make sure that we can talk about the
logarithm of a tensor product, we need a proposition:

\begin{proposition}
\label{Proposition410} Let $k$ be a field. Let $C$ and $D$ be two filtered
$k$-coalgebras. For every $\ell\in\mathbb{N}$, define a subspace $\left(
C\otimes D\right)  _{\leq\ell}$ of $C\otimes D$ by
\[
\left(  C\otimes D\right)  _{\leq\ell}=\sum_{\substack{\left(  a,b\right)
\in\mathbb{N}^{\times2};\\a+b=\ell}}C_{\leq a}\otimes D_{\leq b}.
\]


\textbf{(a)} Then, $\left(  C\otimes D,\left(  \left(  C\otimes D\right)
_{\leq\ell}\right)  _{\ell\geq0}\right)  $ is a filtered $k$-coalgebra.

We are going to denote this filtered $k$-coalgebra $\left(  C\otimes D,\left(
\left(  C\otimes D\right)  _{\leq\ell}\right)  _{\ell\geq0}\right)  $ by
$C\otimes D$. Thus, when we speak of ``the filtered $k$-coalgebra $C\otimes
D$'', we actually mean the filtered $k$-coalgebra $\left(  C\otimes D,\left(
\left(  C\otimes D\right)  _{\leq\ell}\right)  _{\ell\geq0}\right)  $.

\textbf{(b)} If the filtered $k$-coalgebras $C$ and $D$ are connected, then
the filtered $k$-coalgebra $C\otimes D$ is connected, too.

\textbf{(c)} Assume that the filtered $k$-coalgebras $C$ and $D$ are
connected. Then, all three filtered $k$-coalgebras $C$, $D$ and $C\otimes D$
are connected\footnote{\textit{Proof.} For $C$ and $D$, this is clear. For
$C\otimes D$, this follows from Proposition~\ref{Proposition410}
\textbf{(b)}.}, and hence induce unital coalgebras $C$, $D$ and $C\otimes D$
according to Definition~\ref{Definition7h}. These unital coalgebras satisfy
$1_{C\otimes D}=1_{C}\otimes1_{D}$.
\end{proposition}

A boring linear-algebraic lemma ahead:

\begin{lemma}
\label{Lemma411} Let $k$ be a field.

\textbf{(a)} Let $U$ and $V$ be two $k$-vector spaces. Let $a$ be a
nonnegative integer. Let $U_{c}$ be a subspace of $U$ for every $c\in\left\{
0,1,\ldots,a\right\}  $. Then,
\begin{equation}
\left(  \sum_{c=0}^{a}U_{c}\right)  \otimes V=\sum_{c=0}^{a}U_{c}\otimes V,
\label{411.a}%
\end{equation}
where both sides $\left(  \sum_{c=0}^{a}U_{c}\right)  \otimes V$ and
$\sum_{c=0}^{a}U_{c}\otimes V$ of this equality are to be understood as
subspaces of $U\otimes V$.

\textbf{(b)} Let $U$ and $V$ be two $k$-vector spaces. Let $b$ be a
nonnegative integer. Let $V_{d}$ be a subspace of $V$ for every $d\in\left\{
0,1,\ldots,b\right\}  $. Then,%
\begin{equation}
U\otimes\left(  \sum_{d=0}^{b}V_{d}\right)  =\sum_{d=0}^{b}U\otimes V_{d},
\label{411.b}%
\end{equation}
where both sides $U\otimes\left(  \sum_{d=0}^{b}V_{d}\right)  $ and
$\sum_{d=0}^{b}U\otimes V_{d}$ of this equality are to be understood as
subspaces of $U\otimes V$.

\textbf{(c)} Let $U$ and $V$ be two $k$-vector spaces. Let $a$ and $b$ be two
nonnegative integers. Let $U_{c}$ be a subspace of $U$ for every $c\in\left\{
0,1,\ldots,a\right\}  $. Let $V_{d}$ be a subspace of $V$ for every
$d\in\left\{  0,1,\ldots,b\right\}  $. Then,%
\begin{equation}
\left(  \sum_{c=0}^{a}U_{c}\right)  \otimes\left(  \sum_{d=0}^{b}V_{d}\right)
=\sum_{c=0}^{a}\sum_{d=0}^{b}U_{c}\otimes V_{d}, \label{411.c}%
\end{equation}
where both sides $\left(  \sum_{c=0}^{a}U_{c}\right)  \otimes\left(
\sum_{d=0}^{b}V_{d}\right)  $ and $\sum_{c=0}^{a}\sum_{d=0}^{b}U_{c}\otimes
V_{d}$ of this equality are to be understood as subspaces of $U\otimes V$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma411}.]\textbf{(a)} First, let us introduce some conventions:

\begin{itemize}
\item Whenever $T$ is a $k$-vector space, $W_{c}$ is a $k$-vector space for
every $c\in\left\{  0,1,\ldots,a\right\}  $, and $f_{c}$ is a $k$-linear map
$W_{c}\rightarrow T$ for every $c\in\left\{  0,1,\ldots,a\right\}  $, we are
going to denote by $\sum_{c=0}^{a}f_{c}$ the canonical $k$-linear map
$\bigoplus_{c=0}^{a}W_{c}\rightarrow T$ induced by the family $\left(
f_{c}\right)  _{c\in\left\{  0,1,\ldots,a\right\}  }$ of maps by the universal
property of the direct sum. This map $\sum_{c=0}^{a}f_{c}$ sends every
$\left(  w_{0},w_{1},\ldots,w_{a}\right)  \in\bigoplus_{c=0}^{a}W_{c}$ to
$f_{0}\left(  w_{0}\right)  +f_{1}\left(  w_{1}\right)  +\cdots+f_{a}\left(
w_{a}\right)  \in T$.

\item Whenever $T$ is a $k$-vector space and $U$ is a subspace of $T$, we
denote by $\operatorname*{inc}\nolimits_{U,T}$ the inclusion map $U\rightarrow
T$.
\end{itemize}

It is clear that, whenever $T$ is a $k$-vector space and $T_{c}$ is a subspace
of $T$ for every $c\in\left\{  0,1,\ldots,a\right\}  $, then%
\begin{equation}
\left(  \sum_{c=0}^{a}\operatorname*{inc}\nolimits_{T_{c},T}\right)  \left(
\bigoplus_{c=0}^{a}T_{c}\right)  =\sum_{c=0}^{a}T_{c} \label{411.pf.1}%
\end{equation}
where the $\bigoplus_{c=0}^{a}T_{c}$ on the left hand side denotes an
\textbf{external} direct sum.\footnote{\textit{Proof of (\ref{411.pf.1}):} Let
$T$ be a $k$-vector space. Let $T_{c}$ be a $k$-vector subspace of $T$ for
every $c\in\left\{  0,1,\ldots,a\right\}  $. We have%
\begin{align}
&  \left(  \sum_{c=0}^{a}\operatorname*{inc}\nolimits_{T_{c},T}\right)
\left(  \bigoplus_{c=0}^{a}T_{c}\right) \nonumber\\
&  =\left\{  \left(  \sum_{c=0}^{a}\operatorname*{inc}\nolimits_{T_{c}%
,T}\right)  \left(  w_{0},w_{1},\ldots,w_{a}\right)  \ \mid\ \left(
w_{0},w_{1},\ldots,w_{a}\right)  \in\bigoplus_{c=0}^{a}T_{c}\right\}  .
\label{411.pf.1.pf.1}%
\end{align}
But every $\left(  w_{0},w_{1},\ldots,w_{a}\right)  \in\bigoplus_{c=0}%
^{a}T_{c}$ satisfies%
\begin{align*}
&  \left(  \sum_{c=0}^{a}\operatorname*{inc}\nolimits_{T_{c},T}\right)
\left(  w_{0},w_{1},\ldots,w_{a}\right) \\
&  =\operatorname*{inc}\nolimits_{T_{0},T}\left(  w_{0}\right)
+\operatorname*{inc}\nolimits_{T_{1},T}\left(  w_{1}\right)  +\cdots
+\operatorname*{inc}\nolimits_{T_{a},T}\left(  w_{a}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\sum_{c=0}%
^{a}\operatorname*{inc}\nolimits_{T_{c},T}\right) \\
&  =\sum_{c=0}^{a}\underbrace{\left(  \operatorname*{inc}\nolimits_{T_{c}%
,T}\left(  w_{c}\right)  \right)  }_{\substack{=w_{c}\\\text{(since
}\operatorname*{inc}\nolimits_{T_{c},T}\text{ is an inclusion map)}}%
}=\sum_{c=0}^{a}w_{c}.
\end{align*}
Hence, (\ref{411.pf.1.pf.1}) becomes%
\begin{align*}
&  \left(  \sum_{c=0}^{a}\operatorname*{inc}\nolimits_{T_{c},T}\right)
\left(  \bigoplus_{c=0}^{a}T_{c}\right) \\
&  =\left\{  \underbrace{\left(  \sum_{c=0}^{a}\operatorname*{inc}%
\nolimits_{T_{c},T}\right)  \left(  w_{0},w_{1},\ldots,w_{a}\right)  }%
_{=\sum_{c=0}^{a}w_{c}}\ \mid\ \left(  w_{0},w_{1},\ldots,w_{a}\right)
\in\bigoplus_{c=0}^{a}T_{c}\right\} \\
&  =\left\{  \sum_{c=0}^{a}w_{c}\ \mid\ \underbrace{\left(  w_{0},w_{1}%
,\ldots,w_{a}\right)  \in\bigoplus_{c=0}^{a}T_{c}}_{\substack{\text{this is
equivalent to}\\\left(
\begin{array}
[c]{c}%
\left(  w_{0},w_{1},\ldots,w_{a}\right)  \in\prod_{c=0}^{a}T_{c}\text{, and}\\
\text{all but finitely many }i\in\left\{  0,1,\ldots,a\right\}  \text{ satisfy
}w_{i}=0
\end{array}
\right)  \\\text{(since }\bigoplus_{c=0}^{a}T_{c}\text{ is the set of all
}\left(  u_{0},u_{1},\ldots,u_{a}\right)  \in\prod_{c=0}^{a}T_{c}\text{ such
that}\\\text{all but finitely many }i\in\left\{  0,1,\ldots,a\right\}  \text{
satisfy }u_{i}=0\text{)}}}\right\} \\
&  =\left\{  \sum_{c=0}^{a}w_{c}\ \mid\ \left(
\begin{array}
[c]{c}%
\left(  w_{0},w_{1},\ldots,w_{a}\right)  \in\prod_{c=0}^{a}T_{c}\text{, and}\\
\text{all but finitely many }i\in\left\{  0,1,\ldots,a\right\}  \text{ satisfy
}w_{i}=0
\end{array}
\right)  \right\}  .
\end{align*}
Compared with%
\[
\sum_{c=0}^{a}T_{c}=\left\{  \sum_{c=0}^{a}w_{c}\ \mid\ \left(
\begin{array}
[c]{c}%
\left(  w_{0},w_{1},\ldots,w_{a}\right)  \in\prod_{c=0}^{a}T_{c}\text{, and}\\
\text{all but finitely many }i\in\left\{  0,1,\ldots,a\right\}  \text{ satisfy
}w_{i}=0
\end{array}
\right)  \right\}
\]
(by the definition of $\sum_{c=0}^{a}T_{c}$), this yields $\left(  \sum
_{c=0}^{a}\operatorname*{inc}\nolimits_{T_{c},T}\right)  \left(
\bigoplus_{c=0}^{a}T_{c}\right)  =\sum_{c=0}^{a}T_{c}$. This proves
(\ref{411.pf.1}).}

Now let us actually prove Lemma~\ref{Lemma411} \textbf{(a)}. In the following,
let $\operatorname*{id}$ denote the identity map $\operatorname{id}_{V}$. For
every $c\in\left\{  0,1,\ldots,a\right\}  $, let $i_{c}$ denote the canonical
inclusion map $\operatorname*{inc}\nolimits_{U_{c},U}:U_{c}\rightarrow U$.

For every $c\in\left\{  0,1,\ldots,a\right\}  $, we are identifying
$U_{c}\otimes V$ with a subspace of $U\otimes V$. We are doing this
identification by means of the map $\underbrace{\operatorname*{inc}%
\nolimits_{U_{c},U}}_{=i_{c}}\otimes\underbrace{\operatorname*{inc}%
\nolimits_{V,V}}_{=\operatorname*{id}}=i_{c}\otimes\operatorname*{id}$. Hence,
for every $c\in\left\{  0,1,\ldots,a\right\}  $, we consider $i_{c}%
\otimes\operatorname*{id}$ to be the canonical inclusion map $U_{c}\otimes
V\rightarrow U\otimes V$. Thus, for every $c\in\left\{  0,1,\ldots,a\right\}
$, we have%
\[
i_{c}\otimes\operatorname*{id}=\left(  \text{the canonical inclusion map
}U_{c}\otimes V\rightarrow U\otimes V\right)  =\operatorname*{inc}%
\nolimits_{U_{c}\otimes V,U\otimes V}.
\]
Hence,%
\begin{align*}
\left(  \sum_{c=0}^{a}\underbrace{\left(  i_{c}\otimes\operatorname*{id}%
\right)  }_{=\operatorname*{inc}\nolimits_{U_{c}\otimes V,U\otimes V}}\right)
\left(  \bigoplus_{c=0}^{a}\left(  U_{c}\otimes V\right)  \right)   &
=\left(  \sum_{c=0}^{a}\operatorname*{inc}\nolimits_{U_{c}\otimes V,U\otimes
V}\right)  \left(  \bigoplus_{c=0}^{a}\left(  U_{c}\otimes V\right)  \right)
\\
&  =\sum_{c=0}^{a}U_{c}\otimes V
\end{align*}
(by (\ref{411.pf.1}), applied to $T=U\otimes V$ and $T_{c}=U_{c}\otimes V$).

On the other hand, the map $\sum_{c=0}^{a}i_{c}:\bigoplus_{c=0}^{a}%
U_{c}\rightarrow U$ satisfies
\[
\left(  \sum_{c=0}^{a}\underbrace{i_{c}}_{=\operatorname*{inc}\nolimits_{U_{c}%
,U}}\right)  \left(  \bigoplus_{c=0}^{a}U_{c}\right)  =\left(  \sum_{c=0}%
^{a}\operatorname*{inc}\nolimits_{U_{c},U}\right)  \left(  \bigoplus_{c=0}%
^{a}U_{c}\right)  =\sum_{c=0}^{a}U_{c}%
\]
(by (\ref{411.pf.1}), applied to $T=U$ and $T_{c}=U_{c}$).

Since the tensor product is known to commute with direct sums, there is a
canonical $k$-vector space isomorphism $\left(  \bigoplus_{c=0}^{a}%
U_{c}\right)  \otimes V\rightarrow\bigoplus_{c=0}^{a}\left(  U_{c}\otimes
V\right)  $. Denote this isomorphism by $I$. By the universal property of $I$,
we know that whenever $W$ is a $k$-vector space, and $f_{c}$ is a $k$-vector
space homomorphism $U_{c}\rightarrow W$ for every $c\in\left\{  0,1,\ldots
,a\right\}  $, the diagram%
\[%
%TCIMACRO{\TeXButton{consequence of universality}{\xymatrixcolsep{5pc}
%\xymatrixrowsep{4pc}
%\xymatrix{
%\left( \bigoplus_{c=0}^a U_c \right) \otimes V \ar[r]^I \ar[dr]_{\left
%(\sum_{c=0}^a f_c\right) \otimes\id} & \bigoplus_{c=0}^a \left(U_c\otimes
%V\right) \ar[d]^{\sum_{c=0}^a \left(f_c \otimes\id\right)}\\
%& W \otimes V
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrixrowsep{4pc}
\xymatrix{
\left( \bigoplus_{c=0}^a U_c \right) \otimes V \ar[r]^I \ar[dr]_{\left
(\sum_{c=0}^a f_c\right) \otimes\id} & \bigoplus_{c=0}^a \left(U_c\otimes
V\right) \ar[d]^{\sum_{c=0}^a \left(f_c \otimes\id\right)}\\
& W \otimes V
}%
%EndExpansion
\]
commutes (where $\operatorname*{id}$ denotes the identity map
$\operatorname{id}_{V}$). Applying this to $W=U$ and $f_{c}=i_{c}$, we obtain
the following result: The diagram%
\[%
%TCIMACRO{\TeXButton{consequence of universality for V}{\xymatrixcolsep{5pc}
%\xymatrixrowsep{4pc}
%\xymatrix{
%\left( \bigoplus_{c=0}^a U_c \right) \otimes V \ar[r]^I \ar[dr]_{\left
%(\sum_{c=0}^a i_c\right) \otimes\id} & \bigoplus_{c=0}^a \left(U_c\otimes
%V\right) \ar[d]^{\sum_{c=0}^a \left(i_c \otimes\id\right)}\\
%& U \otimes V
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrixrowsep{4pc}
\xymatrix{
\left( \bigoplus_{c=0}^a U_c \right) \otimes V \ar[r]^I \ar[dr]_{\left
(\sum_{c=0}^a i_c\right) \otimes\id} & \bigoplus_{c=0}^a \left(U_c\otimes
V\right) \ar[d]^{\sum_{c=0}^a \left(i_c \otimes\id\right)}\\
& U \otimes V
}%
%EndExpansion
\]
commutes (where $\operatorname*{id}$ denotes the identity map
$\operatorname{id}_{V}$). In other words,%
\[
\left(  \sum_{c=0}^{a}i_{c}\right)  \otimes\operatorname*{id}=\left(
\sum_{c=0}^{a}\left(  i_{c}\otimes\operatorname*{id}\right)  \right)  \circ
I.
\]
Thus,%
\begin{align*}
&  \left(  \left(  \sum_{c=0}^{a}i_{c}\right)  \otimes\operatorname*{id}%
\right)  \left(  \left(  \bigoplus_{c=0}^{a}U_{c}\right)  \otimes V\right) \\
&  =\left(  \left(  \sum_{c=0}^{a}\left(  i_{c}\otimes\operatorname*{id}%
\right)  \right)  \circ I\right)  \left(  \left(  \bigoplus_{c=0}^{a}%
U_{c}\right)  \otimes V\right) \\
&  =\left(  \sum_{c=0}^{a}\left(  i_{c}\otimes\operatorname*{id}\right)
\right)  \underbrace{\left(  I\left(  \left(  \bigoplus_{c=0}^{a}U_{c}\right)
\otimes V\right)  \right)  }_{\substack{=\bigoplus_{c=0}^{a}\left(
U_{c}\otimes V\right)  \\\text{(since }I\text{ is an isomorphism)}}}\\
&  =\left(  \sum_{c=0}^{a}\left(  i_{c}\otimes\operatorname*{id}\right)
\right)  \left(  \bigoplus_{c=0}^{a}\left(  U_{c}\otimes V\right)  \right)
=\sum_{c=0}^{a}U_{c}\otimes V.
\end{align*}
Compared with%
\begin{align*}
\left(  \left(  \sum_{c=0}^{a}i_{c}\right)  \otimes\operatorname*{id}\right)
\left(  \left(  \bigoplus_{c=0}^{a}U_{c}\right)  \otimes V\right)   &
=\underbrace{\left(  \left(  \sum_{c=0}^{a}i_{c}\right)  \left(
\bigoplus_{c=0}^{a}U_{c}\right)  \right)  }_{=\sum_{c=0}^{a}U_{c}}%
\otimes\underbrace{\operatorname*{id}\left(  V\right)  }_{=V}\\
&  =\left(  \sum_{c=0}^{a}U_{c}\right)  \otimes V,
\end{align*}
this yields $\left(  \sum_{c=0}^{a}U_{c}\right)  \otimes V=\sum_{c=0}^{a}%
U_{c}\otimes V$. This proves Lemma~\ref{Lemma411} \textbf{(a)}.

\textbf{(b)} The proof of Lemma~\ref{Lemma411} \textbf{(b)} proceeds by the
same arguments as the proof of Lemma~\ref{Lemma411} \textbf{(a)} that we gave
above (the only real difference is that this time, the sum stands in the
second tensorand rather than in the first one).

\textbf{(c)} We have%
\begin{align*}
\left(  \sum_{c=0}^{a}U_{c}\right)  \otimes\left(  \sum_{d=0}^{b}V_{d}\right)
&  =\sum_{c=0}^{a}\left(  \underbrace{U_{c}\otimes\left(  \sum_{d=0}^{b}%
V_{d}\right)  }_{\substack{=\sum_{d=0}^{b}U_{c}\otimes V_{d}\\\text{(by
Lemma~\ref{Lemma411} \textbf{(b)}, applied to }U_{c}\text{ instead of
}U\text{)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma~\ref{Lemma411} \textbf{(a)},
applied to }\sum_{d=0}^{b}V_{d}\text{ instead of }V\right) \\
&  =\sum_{c=0}^{a}\sum_{d=0}^{b}U_{c}\otimes V_{d}.
\end{align*}
This proves Lemma~\ref{Lemma411} \textbf{(c)}.
\end{proof}

One more linear-algebraic triviality will be useful:

\begin{lemma}
\label{Lemma412} Let $k$ be a field. Let $V$ and $W$ be two $k$-vector spaces.
Let $V^{\prime}$ be a vector subspace of $V$. Let $W^{\prime}$ be a vector
subspace of $W$. Then, $\tau_{V,W}\left(  V^{\prime}\otimes W^{\prime}\right)
=W^{\prime}\otimes V^{\prime}$, where $\tau_{V,W}$ is the $\left(  V,W\right)
$-flip defined in Definition~\ref{Definition34}.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma412}.]Since a tensor product of $k$-vector spaces is
always generated by pure tensors, we have%
\[
V^{\prime}\otimes W^{\prime}=\left\langle v\otimes w\ \mid\ \left(
v,w\right)  \in V^{\prime}\times W^{\prime}\right\rangle =\left\langle
\left\{  v\otimes w\ \mid\ \left(  v,w\right)  \in V^{\prime}\times W^{\prime
}\right\}  \right\rangle .
\]
Hence,%
\begin{align*}
\tau_{V,W}\left(  V^{\prime}\otimes W^{\prime}\right)   &  =\tau_{V,W}\left(
\left\langle \left\{  v\otimes w\ \mid\ \left(  v,w\right)  \in V^{\prime
}\times W^{\prime}\right\}  \right\rangle \right) \\
&  =\left\langle \underbrace{\tau_{V,W}\left(  \left\{  v\otimes
w\ \mid\ \left(  v,w\right)  \in V^{\prime}\times W^{\prime}\right\}  \right)
}_{=\left\{  \tau_{V,W}\left(  v\otimes w\right)  \ \mid\ \left(  v,w\right)
\in V^{\prime}\times W^{\prime}\right\}  }\right\rangle
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{316.pf.3})}\right) \\
&  =\left\langle \left\{  \underbrace{\tau_{V,W}\left(  v\otimes w\right)
}_{\substack{=w\otimes v\\\text{(by the definition of }\tau_{V,W}\text{)}%
}}\ \mid\ \left(  v,w\right)  \in V^{\prime}\times W^{\prime}\right\}
\right\rangle \\
&  =\left\langle \left\{  w\otimes v\ \mid\ \left(  v,w\right)  \in V^{\prime
}\times W^{\prime}\right\}  \right\rangle =\left\langle w\otimes
v\ \mid\ \left(  v,w\right)  \in V^{\prime}\times W^{\prime}\right\rangle \\
&  =\left\langle x\otimes y\ \mid\ \left(  x,y\right)  \in W^{\prime}\times
V^{\prime}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\left(  w,v\right)  \text{ by }\left(  x,y\right)
\text{,}\\
\text{since the map }V^{\prime}\times W^{\prime}\rightarrow W^{\prime}\times
V^{\prime},\ \left(  v,w\right)  \mapsto\left(  w,v\right)  \text{ is a
bijection}%
\end{array}
\right)  .
\end{align*}
Compared with%
\begin{align*}
W^{\prime}\otimes V^{\prime}  &  =\left\langle x\otimes y\ \mid\ \left(
x,y\right)  \in W^{\prime}\times V^{\prime}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since a tensor product is always
generated by pure tensors}\right)  ,
\end{align*}
this yields $\tau_{V,W}\left(  V^{\prime}\otimes W^{\prime}\right)
=W^{\prime}\otimes V^{\prime}$. This proves Lemma~\ref{Lemma412}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition410}.]\textbf{(a)} For every
$a\in\mathbb{N}$, we have $\Delta_{C}\left(  C_{\leq a}\right)  \subseteq
\sum_{c=0}^{a}C_{\leq c}\otimes C_{\leq a-c}$ (since $C$ is a filtered
$k$-coalgebra). For every $b\in\mathbb{N}$, we have $\Delta_{D}\left(  D_{\leq
b}\right)  \subseteq\sum_{d=0}^{b}D_{\leq d}\otimes D_{\leq b-d}$ (since $D$
is a filtered $k$-coalgebra).

We notice that
\begin{equation}
C_{\leq\alpha}\otimes D_{\leq\beta}\subseteq\left(  C\otimes D\right)
_{\leq\alpha+\beta}\ \ \ \ \ \ \ \ \ \ \text{for any }\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2}. \label{410.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{410.pf.1}).} Let $\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2}$. Then, this pair $\left(  \alpha
,\beta\right)  \in\mathbb{N}^{\times2}$ satisfies $\alpha+\beta=\alpha+\beta$.
Hence, $C_{\leq\alpha}\otimes D_{\leq\beta}$ is an addend of the sum
$\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\alpha
+\beta}}C_{\leq a}\otimes D_{\leq b}$ (namely, the addend for $\left(
a,b\right)  =\left(  \alpha,\beta\right)  $). Thus, $C_{\leq\alpha}\otimes
D_{\leq\beta}\subseteq\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}%
^{\times2};\\a+b=\alpha+\beta}}C_{\leq a}\otimes D_{\leq b}$. Since $\left(
C\otimes D\right)  _{\leq\alpha+\beta}=\sum_{\substack{\left(  a,b\right)
\in\mathbb{N}^{\times2};\\a+b=\alpha+\beta}}C_{\leq a}\otimes D_{\leq b}$ (by
the definition of $\left(  C\otimes D\right)  _{\leq\alpha+\beta}$), we thus
have%
\[
C_{\leq\alpha}\otimes D_{\leq\beta}\subseteq\sum_{\substack{\left(
a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\alpha+\beta}}C_{\leq a}\otimes
D_{\leq b}=\left(  C\otimes D\right)  _{\leq\alpha+\beta}.
\]
This proves (\ref{410.pf.1}).}

Now, let $\ell\in\mathbb{N}$.

Let $\tau_{C,D}:C\otimes D\rightarrow D\otimes C$ be the $\left(  C,D\right)
$-flip, defined as in Definition~\ref{Definition34}. Then, by the definition
of the $k$-coalgebra $C\otimes D$, we have $\Delta_{C\otimes D}=\left(
\operatorname{id}_{C}\otimes\tau_{C,D}\otimes\operatorname{id}_{D}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  $. Hence,%
\begin{align*}
\Delta_{C\otimes D}\left(  \left(  C\otimes D\right)  _{\leq\ell}\right)   &
=\left(  \left(  \operatorname{id}_{C}\otimes\tau_{C,D}\otimes
\operatorname{id}_{D}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)
\right)  \left(  \left(  C\otimes D\right)  _{\leq\ell}\right) \\
&  =\left(  \operatorname{id}_{C}\otimes\tau_{C,D}\otimes\operatorname{id}%
_{D}\right)  \left(  \left(  \Delta_{C}\otimes\Delta_{D}\right)
\underbrace{\left(  \left(  C\otimes D\right)  _{\leq\ell}\right)  }%
_{=\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell
}}C_{\leq a}\otimes D_{\leq b}}\right) \\
&  =\left(  \operatorname{id}_{C}\otimes\tau_{C,D}\otimes\operatorname{id}%
_{D}\right)  \left(  \left(  \Delta_{C}\otimes\Delta_{D}\right)  \left(
\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell
}}C_{\leq a}\otimes D_{\leq b}\right)  \right)  .
\end{align*}
But since%
\begin{align*}
&  \left(  \Delta_{C}\otimes\Delta_{D}\right)  \left(  \sum_{\substack{\left(
a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell}}C_{\leq a}\otimes D_{\leq
b}\right) \\
&  =\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell
}}\underbrace{\left(  \Delta_{C}\otimes\Delta_{D}\right)  \left(  C_{\leq
a}\otimes D_{\leq b}\right)  }_{\subseteq\left(  \Delta_{C}\left(  C_{\leq
a}\right)  \right)  \otimes\left(  \Delta_{D}\left(  D_{\leq b}\right)
\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{C}\otimes\Delta
_{D}\text{ is }k\text{-linear}\right) \\
&  \subseteq\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times
2};\\a+b=\ell}}\underbrace{\left(  \Delta_{C}\left(  C_{\leq a}\right)
\right)  }_{\subseteq\sum_{c=0}^{a}C_{\leq c}\otimes C_{\leq a-c}}%
\otimes\underbrace{\left(  \Delta_{D}\left(  D_{\leq b}\right)  \right)
}_{\subseteq\sum_{d=0}^{b}D_{\leq d}\otimes D_{\leq b-d}}\\
&  \subseteq\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times
2};\\a+b=\ell}}\underbrace{\left(  \sum_{c=0}^{a}C_{\leq c}\otimes C_{\leq
a-c}\right)  \otimes\left(  \sum_{d=0}^{b}D_{\leq d}\otimes D_{\leq
b-d}\right)  }_{\substack{=\sum_{c=0}^{a}\sum_{d=0}^{b}\left(  C_{\leq
c}\otimes C_{\leq a-c}\right)  \otimes\left(  D_{\leq d}\otimes D_{\leq
b-d}\right)  \\\text{(by Lemma~\ref{Lemma411} \textbf{(c)}, applied to
}U=C\otimes C\text{, }V=D\otimes D\text{,}\\U_{c}=C_{\leq c}\otimes C_{\leq
a-c}\text{ and }V_{d}=D_{\leq d}\otimes D_{\leq b-d}\text{)}}}\\
&  =\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell
}}\sum_{c=0}^{a}\sum_{d=0}^{b}\underbrace{\left(  C_{\leq c}\otimes C_{\leq
a-c}\right)  \otimes\left(  D_{\leq d}\otimes D_{\leq b-d}\right)  }_{=C_{\leq
c}\otimes C_{\leq a-c}\otimes D_{\leq d}\otimes D_{\leq b-d}}\\
&  =\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell
}}\sum_{c=0}^{a}\sum_{d=0}^{b}C_{\leq c}\otimes C_{\leq a-c}\otimes D_{\leq
d}\otimes D_{\leq b-d},
\end{align*}
this becomes%
\begin{align}
&  \Delta_{C\otimes D}\left(  \left(  C\otimes D\right)  _{\leq\ell}\right)
\nonumber\\
&  \subseteq\left(  \operatorname{id}_{C}\otimes\tau_{C,D}\otimes
\operatorname{id}_{D}\right)  \left(  \underbrace{\left(  \Delta_{C}%
\otimes\Delta_{D}\right)  \left(  \sum_{\substack{\left(  a,b\right)
\in\mathbb{N}^{\times2};\\a+b=\ell}}C_{\leq a}\otimes D_{\leq b}\right)
}_{\subseteq\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times
2};\\a+b=\ell}}\sum_{c=0}^{a}\sum_{d=0}^{b}C_{\leq c}\otimes C_{\leq
a-c}\otimes D_{\leq d}\otimes D_{\leq b-d}}\right) \nonumber\\
&  \subseteq\left(  \operatorname{id}_{C}\otimes\tau_{C,D}\otimes
\operatorname{id}_{D}\right)  \left(  \sum_{\substack{\left(  a,b\right)
\in\mathbb{N}^{\times2};\\a+b=\ell}}\sum_{c=0}^{a}\sum_{d=0}^{b}C_{\leq
c}\otimes C_{\leq a-c}\otimes D_{\leq d}\otimes D_{\leq b-d}\right)
\nonumber\\
&  =\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell
}}\sum_{c=0}^{a}\sum_{d=0}^{b}\underbrace{\left(  \operatorname{id}_{C}%
\otimes\tau_{C,D}\otimes\operatorname{id}_{D}\right)  \left(  C_{\leq
c}\otimes C_{\leq a-c}\otimes D_{\leq d}\otimes D_{\leq b-d}\right)
}_{=\operatorname{id}_{C}\left(  C_{\leq c}\right)  \otimes\tau_{C,D}\left(
C_{\leq a-c}\otimes D_{\leq d}\right)  \otimes\operatorname{id}_{D}\left(
D_{\leq b-d}\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname{id}_{C}\otimes
\tau_{C,D}\otimes\operatorname{id}_{D}\text{ is }k\text{-linear}\right)
\nonumber\\
&  =\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell
}}\sum_{c=0}^{a}\sum_{d=0}^{b}\underbrace{\operatorname{id}_{C}\left(  C_{\leq
c}\right)  }_{=C_{\leq c}}\otimes\underbrace{\tau_{C,D}\left(  C_{\leq
a-c}\otimes D_{\leq d}\right)  }_{\substack{=D_{\leq d}\otimes C_{\leq
a-c}\\\text{(by Lemma~\ref{Lemma412}, applied to}\\V=C\text{, }W=D\text{,
}V^{\prime}=C_{\leq a-c}\text{ and }W^{\prime}=D_{\leq d}\text{)}}%
}\otimes\underbrace{\operatorname{id}_{D}\left(  D_{\leq b-d}\right)
}_{=D_{\leq b-d}}\nonumber\\
&  =\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times2};\\a+b=\ell
}}\sum_{c=0}^{a}\sum_{d=0}^{b}C_{\leq c}\otimes D_{\leq d}\otimes C_{\leq
a-c}\otimes D_{\leq b-d}. \label{410.pf.2}%
\end{align}
But we can easily see that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for every }\left(  a,b\right)  \in\mathbb{N}^{\times2}\text{ satisfying
}a+b=\ell\text{, for every}\\
c\in\left\{  0,1,\ldots,a\right\}  \text{ and for every }d\in\left\{
0,1,\ldots,b\right\}  \text{, we have}\\
C_{\leq c}\otimes D_{\leq d}\otimes C_{\leq a-c}\otimes D_{\leq b-d}%
\subseteq\sum_{m=0}^{\ell}\left(  C\otimes D\right)  _{\leq m}\otimes\left(
C\otimes D\right)  _{\leq\ell-m}%
\end{array}
\right)  . \label{410.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{410.pf.3}).} Let $\left(  a,b\right)
\in\mathbb{N}^{\times2}$ satisfy $a+b=\ell$. Let $c\in\left\{  0,1,\ldots
,a\right\}  $ and $d\in\left\{  0,1,\ldots,b\right\}  $ be arbitrary. Then,
$0\leq c\leq a$ (since $c\in\left\{  0,1,\ldots,a\right\}  $) and $0\leq d\leq
b$ (since $d\in\left\{  0,1,\ldots,b\right\}  $). Hence, $\underbrace{c}_{\leq
a}+\underbrace{d}_{\leq b}\leq a+b=\ell$ and $\underbrace{c}_{\geq
0}+\underbrace{d}_{\geq0}\geq0+0=0$. Thus, $c+d\in\left\{  0,1,\ldots
,\ell\right\}  $. Hence, $\left(  C\otimes D\right)  _{\leq c+d}\otimes\left(
C\otimes D\right)  _{\leq\ell-\left(  c+d\right)  }$ is an addend of the sum
$\sum_{m=0}^{\ell}\left(  C\otimes D\right)  _{\leq m}\otimes\left(  C\otimes
D\right)  _{\leq\ell-m}$ (namely, the addend for $m=c+d$). Thus,%
\[
\left(  C\otimes D\right)  _{\leq c+d}\otimes\left(  C\otimes D\right)
_{\leq\ell-\left(  c+d\right)  }\subseteq\sum_{m=0}^{\ell}\left(  C\otimes
D\right)  _{\leq m}\otimes\left(  C\otimes D\right)  _{\leq\ell-m}.
\]
But now,%
\begin{align*}
&  \underbrace{C_{\leq c}\otimes D_{\leq d}}_{\substack{\subseteq\left(
C\otimes D\right)  _{\leq c+d}\\\text{(by (\ref{410.pf.1}), applied to
}\left(  \alpha,\beta\right)  =\left(  c,d\right)  \text{)}}}\otimes
\underbrace{C_{\leq a-c}\otimes D_{\leq b-d}}_{\substack{\subseteq\left(
C\otimes D\right)  _{\leq\left(  a-c\right)  +\left(  b-d\right)  }\\\text{(by
(\ref{410.pf.1}), applied to }\left(  \alpha,\beta\right)  =\left(
a-c,b-d\right)  \text{)}}}\\
&  \subseteq\left(  C\otimes D\right)  _{\leq c+d}\otimes\left(  C\otimes
D\right)  _{\leq\left(  a-c\right)  +\left(  b-d\right)  }=\left(  C\otimes
D\right)  _{\leq c+d}\otimes\left(  C\otimes D\right)  _{\leq\ell-\left(
c+d\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  a-c\right)  +\left(
b-d\right)  =\underbrace{\left(  a+b\right)  }_{=\ell}-\left(  c+d\right)
=\ell-\left(  c+d\right)  \right) \\
&  \subseteq\sum_{m=0}^{\ell}\left(  C\otimes D\right)  _{\leq m}%
\otimes\left(  C\otimes D\right)  _{\leq\ell-m}.
\end{align*}
This proves (\ref{410.pf.3}).} Hence, (\ref{410.pf.2}) becomes%
\begin{align*}
\Delta_{C\otimes D}\left(  \left(  C\otimes D\right)  _{\leq\ell}\right)   &
\subseteq\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times
2};\\a+b=\ell}}\sum_{c=0}^{a}\sum_{d=0}^{b}\underbrace{C_{\leq c}\otimes
D_{\leq d}\otimes C_{\leq a-c}\otimes D_{\leq b-d}}_{\substack{\subseteq
\sum_{m=0}^{\ell}\left(  C\otimes D\right)  _{\leq m}\otimes\left(  C\otimes
D\right)  _{\leq\ell-m}\\\text{(by (\ref{410.pf.3}) (since }\left(
a,b\right)  \in\mathbb{N}^{\times2}\text{, }a+b=\ell\text{,}\\c\in\left\{
0,1,\ldots,a\right\}  \text{ and }d\in\left\{  0,1,\ldots,b\right\}
\text{))}}}\\
&  \subseteq\sum_{\substack{\left(  a,b\right)  \in\mathbb{N}^{\times
2};\\a+b=\ell}}\sum_{c=0}^{a}\sum_{d=0}^{b}\sum_{m=0}^{\ell}\left(  C\otimes
D\right)  _{\leq m}\otimes\left(  C\otimes D\right)  _{\leq\ell-m}\\
&  \subseteq\sum_{m=0}^{\ell}\left(  C\otimes D\right)  _{\leq m}%
\otimes\left(  C\otimes D\right)  _{\leq\ell-m}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{m=0}^{\ell}\left(  C\otimes
D\right)  _{\leq m}\otimes\left(  C\otimes D\right)  _{\leq\ell-m}\text{ is a
}k\text{-vector space}\right)  .
\end{align*}


Now forget that we fixed $\ell$. We thus have proved that every $\ell
\in\mathbb{N}$ satisfies $\Delta_{C\otimes D}\left(  \left(  C\otimes
D\right)  _{\leq\ell}\right)  \subseteq\sum_{m=0}^{\ell}\left(  C\otimes
D\right)  _{\leq m}\otimes\left(  C\otimes D\right)  _{\leq\ell-m}$. In other
words, $\left(  C\otimes D,\left(  \left(  C\otimes D\right)  _{\leq\ell
}\right)  _{\ell\geq0}\right)  $ is a filtered $k$-coalgebra. This proves
Proposition~\ref{Proposition410} \textbf{(a)}.

\textbf{(b)} Assume that the filtered $k$-coalgebras $C$ and $D$ are connected.

By the definition of $\left(  C\otimes D\right)  _{\leq0}$, we have $\left(
C\otimes D\right)  _{\leq0}=\sum_{\substack{\left(  a,b\right)  \in
\mathbb{N}^{\times2};\\a+b=0}}C_{\leq a}\otimes D_{\leq b}$. But the only pair
$\left(  a,b\right)  \in\mathbb{N}^{\times2}$ satisfying $a+b=0$ is $\left(
0,0\right)  $. Hence, the sum $\sum_{\substack{\left(  a,b\right)
\in\mathbb{N}^{\times2};\\a+b=0}}C_{\leq a}\otimes D_{\leq b}$ has only one
addend, namely $C_{\leq0}\otimes D_{\leq0}$. Thus, $\sum_{\substack{\left(
a,b\right)  \in\mathbb{N}^{\times2};\\a+b=0}}C_{\leq a}\otimes D_{\leq
b}=C_{\leq0}\otimes D_{\leq0}$, so we conclude that%
\[
\left(  C\otimes D\right)  _{\leq0}=\sum_{\substack{\left(  a,b\right)
\in\mathbb{N}^{\times2};\\a+b=0}}C_{\leq a}\otimes D_{\leq b}=C_{\leq0}\otimes
D_{\leq0}.
\]


Since $C$ is connected, the map $\varepsilon_{C}\mid_{C_{\leq0}}:C_{\leq
0}\rightarrow k$ is a $k$-vector space isomorphism (by
Definition~\ref{Definition7a}). Similarly, the map $\varepsilon_{D}%
\mid_{D_{\leq0}}:D_{\leq0}\rightarrow k$ is a $k$-vector space isomorphism.

Since the maps $\varepsilon_{C}\mid_{C_{\leq0}}:C_{\leq0}\rightarrow k$ and
$\varepsilon_{D}\mid_{D_{\leq0}}:D_{\leq0}\rightarrow k$ are $k$-vector space
isomorphisms, their tensor product $\left(  \varepsilon_{C}\mid_{C_{\leq0}%
}\right)  \otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}}\right)  $ must also
be a $k$-vector space isomorphism (since the tensor product of two $k$-vector
space isomorphisms always is a $k$-vector space isomorphism).

Let $\mu_{k}$ be the canonical $k$-vector space isomorphism $k\otimes
k\rightarrow k$. Now, we have $\mu_{k}\circ\left(  \left(  \varepsilon_{C}%
\mid_{C_{\leq0}}\right)  \otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}%
}\right)  \right)  =\varepsilon_{C\otimes D}\mid_{C_{\leq0}\otimes D_{\leq0}}%
$\ \ \ \ \footnote{\textit{Proof.} For every $\left(  x,y\right)  \in
C_{\leq0}\times D_{\leq0}$, we have%
\begin{align*}
&  \left(  \mu_{k}\circ\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
\otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}}\right)  \right)  \right)
\left(  x\otimes y\right) \\
&  =\mu_{k}\left(  \underbrace{\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}%
}\right)  \otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}}\right)  \right)
\left(  x\otimes y\right)  }_{=\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}%
}\right)  \left(  x\right)  \right)  \otimes\left(  \left(  \varepsilon
_{D}\mid_{D_{\leq0}}\right)  \left(  y\right)  \right)  }\right)  =\mu
_{k}\left(  \underbrace{\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}%
}\right)  \left(  x\right)  \right)  }_{=\varepsilon_{C}\left(  x\right)
}\otimes\underbrace{\left(  \left(  \varepsilon_{D}\mid_{D_{\leq0}}\right)
\left(  y\right)  \right)  }_{=\varepsilon_{D}\left(  y\right)  }\right) \\
&  =\mu_{k}\left(  \underbrace{\left(  \varepsilon_{C}\left(  x\right)
\right)  \otimes\left(  \varepsilon_{D}\left(  y\right)  \right)  }_{=\left(
\varepsilon_{C}\otimes\varepsilon_{D}\right)  \left(  x\otimes y\right)
}\right)  =\mu_{k}\left(  \left(  \varepsilon_{C}\otimes\varepsilon
_{D}\right)  \left(  x\otimes y\right)  \right)  =\left(  \mu_{k}\circ\left(
\varepsilon_{C}\otimes\varepsilon_{D}\right)  \right)  \left(  x\otimes
y\right)
\end{align*}
and%
\[
\left(  \varepsilon_{C\otimes D}\mid_{C_{\leq0}\otimes D_{\leq0}}\right)
\left(  x\otimes y\right)  =\varepsilon_{C\otimes D}\left(  x\otimes y\right)
=\left(  \mu_{k}\circ\left(  \varepsilon_{C}\otimes\varepsilon_{D}\right)
\right)  \left(  x\otimes y\right)
\]
(since, by the definition of the $k$-coalgebra $C\otimes D$, we have
$\varepsilon_{C\otimes D}=\mu_{k}\circ\left(  \varepsilon_{C}\otimes
\varepsilon_{D}\right)  $). Thus, for every $\left(  x,y\right)  \in C_{\leq
0}\times D_{\leq0}$, we have%
\[
\left(  \mu_{k}\circ\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
\otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}}\right)  \right)  \right)
\left(  x\otimes y\right)  =\left(  \mu_{k}\circ\left(  \varepsilon_{C}%
\otimes\varepsilon_{D}\right)  \right)  \left(  x\otimes y\right)  =\left(
\varepsilon_{C\otimes D}\mid_{C_{\leq0}\otimes D_{\leq0}}\right)  \left(
x\otimes y\right)  .
\]
In other words, the two maps $\mu_{k}\circ\left(  \left(  \varepsilon_{C}%
\mid_{C_{\leq0}}\right)  \otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}%
}\right)  \right)  $ and $\varepsilon_{C\otimes D}\mid_{C_{\leq0}\otimes
D_{\leq0}}$ are equal to each other on each pure tensor in $C_{\leq0}\otimes
D_{\leq0}$. Since these two maps are $k$-linear, this yields that these two
maps must be identic (because any two $k$-linear maps from a tensor product
which are equal to each other on each pure tensor must be identic). In other
words, $\mu_{k}\circ\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
\otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}}\right)  \right)  =\varepsilon
_{C\otimes D}\mid_{C_{\leq0}\otimes D_{\leq0}}$, qed.}. Hence,%
\[
\mu_{k}\circ\left(  \left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)
\otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}}\right)  \right)  =\varepsilon
_{C\otimes D}\mid_{C_{\leq0}\otimes D_{\leq0}}=\varepsilon_{C\otimes D}%
\mid_{\left(  C\otimes D\right)  _{\leq0}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }C_{\leq0}\otimes D_{\leq0}=\left(  C\otimes D\right)  _{\leq
0}\right)  .
\]
Since we know that $\mu_{k}\circ\left(  \left(  \varepsilon_{C}\mid_{C_{\leq
0}}\right)  \otimes\left(  \varepsilon_{D}\mid_{D_{\leq0}}\right)  \right)  $
is a $k$-vector space isomorphism (because $\mu_{k}$ and $\left(
\varepsilon_{C}\mid_{C_{\leq0}}\right)  \otimes\left(  \varepsilon_{D}%
\mid_{D_{\leq0}}\right)  $ are $k$-vector space isomorphisms, and because the
composition of two $k$-vector space isomorphisms is always a $k$-vector space
isomorphism), we can thus conclude that $\varepsilon_{C\otimes D}\mid_{\left(
C\otimes D\right)  _{\leq0}}$ is a $k$-vector space isomorphism.

Now, by Definition~\ref{Definition7a} (applied to $C\otimes D$ instead of
$C$), the filtered $k$-coalgebra $C\otimes D$ is connected if and only if the
map $\varepsilon_{C\otimes D}\mid_{\left(  C\otimes D\right)  _{\leq0}%
}:\left(  C\otimes D\right)  _{\leq0}\rightarrow k$ is a $k$-vector space
isomorphism. Since we already know that the map $\varepsilon_{C\otimes D}%
\mid_{\left(  C\otimes D\right)  _{\leq0}}$ is a $k$-vector space isomorphism,
we can thus conclude that the filtered $k$-coalgebra $C\otimes D$ is
connected. This proves Proposition~\ref{Proposition410} \textbf{(b)}.

\textbf{(c)} Assume that the filtered $k$-coalgebras $C$ and $D$ are connected.

Since $1_{C}$ is the unity of the unital coalgebra $C$, we see that $\left(
C,1_{C}\right)  $ is a unital coalgebra, so that $\Delta_{C}\left(
1_{C}\right)  =1_{C}\otimes1_{C}$ and $\varepsilon_{C}\left(  1_{C}\right)
=1$ (by the definition of a unital coalgebra). Similarly to $\varepsilon
_{C}\left(  1_{C}\right)  =1$, we can prove that $\varepsilon_{D}\left(
1_{D}\right)  =1$.

Let $\mu_{k}$ be the canonical $k$-vector space isomorphism $k\otimes
k\rightarrow k$. Now, by the definition of the $k$-coalgebra $C\otimes D$, we
have $\varepsilon_{C\otimes D}=\mu_{k}\circ\left(  \varepsilon_{C}%
\otimes\varepsilon_{D}\right)  $, so that%
\begin{align*}
\varepsilon_{C\otimes D}\left(  1_{C}\otimes1_{D}\right)   &  =\left(  \mu
_{k}\circ\left(  \varepsilon_{C}\otimes\varepsilon_{D}\right)  \right)
\left(  1_{C}\otimes1_{D}\right)  =\mu_{k}\left(  \underbrace{\left(
\varepsilon_{C}\otimes\varepsilon_{D}\right)  \left(  1_{C}\otimes
1_{D}\right)  }_{=\varepsilon_{C}\left(  1_{C}\right)  \otimes\varepsilon
_{D}\left(  1_{D}\right)  }\right) \\
&  =\mu_{k}\left(  \underbrace{\varepsilon_{C}\left(  1_{C}\right)  }%
_{=1}\otimes\underbrace{\varepsilon_{D}\left(  1_{D}\right)  }_{=1}\right)
=\mu_{k}\left(  1\otimes1\right)  =1\cdot1\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\mu_{k}\right) \\
&  =1.
\end{align*}


By Remark~\ref{Remark7k}, we have $1_{C}=\left(  \varepsilon_{C}\mid
_{C_{\leq0}}\right)  ^{-1}\left(  1\right)  \in C_{\leq0}$. Similarly,
$1_{D}\in D_{\leq0}$. Thus, $\underbrace{1_{C}}_{\in C_{\leq0}}\otimes
\underbrace{1_{D}}_{\in D_{\leq0}}\in C_{\leq0}\otimes D_{\leq0}=\left(
C\otimes D\right)  _{\leq0}$ (by the above proof of
Proposition~\ref{Proposition410} \textbf{(b)}). Hence, $\left(  \varepsilon
_{C\otimes D}\mid_{\left(  C\otimes D\right)  _{\leq0}}\right)  \left(
1_{C}\otimes1_{D}\right)  $ is well-defined, and we have%
\[
\left(  \varepsilon_{C\otimes D}\mid_{\left(  C\otimes D\right)  _{\leq0}%
}\right)  \left(  1_{C}\otimes1_{D}\right)  =\varepsilon_{C\otimes D}\left(
1_{C}\otimes1_{D}\right)  =1.
\]


Since the filtered $k$-coalgebra $C\otimes D$ is connected (by
Proposition~\ref{Proposition410} \textbf{(b)}), the map $\varepsilon_{C\otimes
D}\mid_{\left(  C\otimes D\right)  _{\leq0}}$ is a $k$-vector space
isomorphism (by the definition of ``connected''). Thus, the identity $\left(
\varepsilon_{C\otimes D}\mid_{\left(  C\otimes D\right)  _{\leq0}}\right)
\left(  1_{C}\otimes1_{D}\right)  =1$ (which was proven above) rewrites as
$1_{C}\otimes1_{D}=\left(  \varepsilon_{C\otimes D}\mid_{\left(  C\otimes
D\right)  _{\leq0}}\right)  ^{-1}\left(  1\right)  $.

Now,%
\begin{align*}
1_{C\otimes D}  &  =\left(  \varepsilon_{C\otimes D}\mid_{\left(  C\otimes
D\right)  _{\leq0}}\right)  ^{-1}\left(  1\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by Remark~\ref{Remark7k}, applied to }C\otimes D\text{ instead of
}C\right) \\
&  =1_{C}\otimes1_{D}.
\end{align*}
This proves Proposition~\ref{Proposition410} \textbf{(c)}.
\end{proof}

Now to the question of how the logarithm acts on tensor products:

\begin{proposition}
\label{Proposition420} Let $k$ be a field of characteristic $0$. Let $C$ and
$D$ be two connected filtered $k$-coalgebras. Let $A$ and $B$ be $k$-algebras.
Let a map $e_{D,B}:D\rightarrow B$ be defined as in Definition~\ref{def.eHA}.
Recall that $C\otimes D$ is a $k$-coalgebra and $A\otimes B$ is a $k$-algebra;
hence, $\mathcal{L}\left(  C\otimes D,A\otimes B\right)  $ becomes a
$k$-algebra with respect to convolution. We notice that the filtered
$k$-coalgebra $C\otimes D$ is connected (by Proposition~\ref{Proposition410}
\textbf{(b)}), so that there is a well-defined notion of $\operatorname*{Log}%
T$ for maps $T\in G\left(  C\otimes D,A\otimes B\right)  $.

\textbf{(a)} For any $F\in G\left(  C,A\right)  $, we have $F\otimes
e_{D,B}\in G\left(  C\otimes D,A\otimes B\right)  $ and $\operatorname*{Log}%
\left(  F\otimes e_{D,B}\right)  =\left(  \operatorname*{Log}F\right)  \otimes
e_{D,B}$.

\textbf{(b)} For any $H\in G\left(  D,B\right)  $, we have $e_{C,A}\otimes
H\in G\left(  C\otimes D,A\otimes B\right)  $ and $\operatorname*{Log}\left(
e_{C,A}\otimes H\right)  =e_{C,A}\otimes\left(  \operatorname*{Log}H\right)  $.

\textbf{(c)} For any $F\in G\left(  C,A\right)  $ and $H\in G\left(
D,B\right)  $, we have $F\otimes H\in G\left(  C\otimes D,A\otimes B\right)  $
and $\operatorname*{Log}\left(  F\otimes H\right)  =\left(
\operatorname*{Log}F\right)  \otimes e_{D,B}+e_{C,A}\otimes\left(
\operatorname*{Log}H\right)  $.
\end{proposition}

First, again, a harmless lemma:

\begin{lemma}
\label{Lemma421} Let $k$ be a field. Let $C$ and $D$ be two connected filtered
$k$-coalgebras. Let $A$ and $B$ be two $k$-algebras. Let $F\in G\left(
C,A\right)  $ and $H\in G\left(  D,B\right)  $ be arbitrary. Then, $F\otimes
H\in G\left(  C\otimes D,A\otimes B\right)  $. (Here, we are using the
notation $G\left(  C\otimes D,A\otimes B\right)  $; this notation makes sense
since $C\otimes D$ is a connected filtered $k$-coalgebra (by
Proposition~\ref{Proposition410} \textbf{(b)}).)
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma421}.]We have%
\[
F\in G\left(  C,A\right)  =\left\{  f\in\mathcal{L}\left(  C,A\right)
\ \mid\ f\left(  1_{C}\right)  =1_{A}\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }G\left(  C,A\right)  \right)  ,
\]
so that $F\left(  1_{C}\right)  =1_{A}$. Similarly, $H\left(  1_{D}\right)
=1_{B}$. Now, $1_{C\otimes D}=1_{C}\otimes1_{D}$, and thus%
\[
\left(  F\otimes H\right)  \left(  1_{C\otimes D}\right)  =\left(  F\otimes
H\right)  \left(  1_{C}\otimes1_{D}\right)  =\underbrace{F\left(
1_{C}\right)  }_{=1_{A}}\otimes\underbrace{H\left(  1_{D}\right)  }_{=1_{B}%
}=1_{A}\otimes1_{B}=1_{A\otimes B},
\]
so that%
\begin{align*}
F\otimes H  &  \in\left\{  f\in\mathcal{L}\left(  C\otimes D,A\otimes
B\right)  \ \mid\ f\left(  1_{C\otimes D}\right)  =1_{A\otimes B}\right\}
=G\left(  C\otimes D,A\otimes B\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }G\left(  C\otimes D,A\otimes B\right)  =\left\{  f\in
\mathcal{L}\left(  C\otimes D,A\otimes B\right)  \ \mid\ f\left(  1_{C\otimes
D}\right)  =1_{A\otimes B}\right\} \\
\text{by the definition of }G\left(  C\otimes D,A\otimes B\right)
\end{array}
\right)  .
\end{align*}
This proves Lemma~\ref{Lemma421}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition420}.]\textbf{(a)} Let $F\in G\left(
C,A\right)  $. By Lemma~\ref{Lemma421} (applied to $H=e_{D,B}$), we obtain
$F\otimes e_{D,B}\in G\left(  C\otimes D,A\otimes B\right)  $. In order to
prove Proposition~\ref{Proposition420} \textbf{(a)}, it thus remains to show
that $\operatorname*{Log}\left(  F\otimes e_{D,B}\right)  =\left(
\operatorname*{Log}F\right)  \otimes e_{D,B}$.

Let $f=F-e_{C,A}$. Then,%
\begin{align}
\underbrace{f}_{=F-e_{C,A}}\otimes e_{D,B}  &  =\left(  F-e_{C,A}\right)
\otimes e_{D,B}=F\otimes e_{D,B}-\underbrace{e_{C,A}\otimes e_{D,B}%
}_{\substack{=e_{C\otimes D,A\otimes B}\\\text{(by Corollary~\ref{Corollary42}
\textbf{(b)})}}}\nonumber\\
&  =F\otimes e_{D,B}-e_{C\otimes D,A\otimes B}. \label{420.pf.1}%
\end{align}


Now let $\left(  c,d\right)  \in C\times D$. Then, $c\in C$ and $d\in D$.

By the definition of $\operatorname*{Log}$, we have $\operatorname*{Log}%
F=\operatorname{Log}_{1}\underbrace{\left(  F-e_{C,A}\right)  }_{=f}%
=\operatorname{Log}_{1}f$, so that%
\[
\left(  \operatorname*{Log}F\right)  \left(  c\right)  =\left(
\operatorname{Log}_{1}f\right)  \left(  c\right)  =\sum_{i\geq1}\dfrac{\left(
-1\right)  ^{i-1}}{i}f^{\ast i}\left(  c\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{log}), applied to }C\text{ and }c\text{ instead of }H\text{ and
}x\right)  .
\]


Now,%
\begin{align*}
&  \left(  \left(  \operatorname*{Log}F\right)  \otimes e_{D,B}\right)
\left(  c\otimes d\right) \\
&  =\underbrace{\left(  \operatorname*{Log}F\right)  \left(  c\right)
}_{=\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
c\right)  }\otimes e_{D,B}\left(  d\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\left(  \operatorname*{Log}F\right)  \otimes
e_{D,B}\right) \\
&  =\left(  \sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
c\right)  \right)  \otimes e_{D,B}\left(  d\right) \\
&  =\sum_{i\geq1}\left(  \dfrac{\left(  -1\right)  ^{i-1}}{i}%
\underbrace{f^{\ast i}\left(  c\right)  \otimes e_{D,B}\left(  d\right)
}_{=\left(  f^{\ast i}\otimes e_{D,B}\right)  \left(  c\otimes d\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the tensor product is
}k\text{-bilinear}\right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f^{\ast i}\otimes
e_{D,B}\right)  \left(  c\otimes d\right)  .
\end{align*}


On the other hand, by the definition of $\operatorname*{Log}$, we have
\[
\operatorname*{Log}\left(  F\otimes e_{D,B}\right)  =\operatorname{Log}%
_{1}\underbrace{\left(  F\otimes e_{D,B}-e_{C\otimes D,A\otimes B}\right)
}_{\substack{=f\otimes e_{D,B}\\\text{(by (\ref{420.pf.1}))}}%
}=\operatorname{Log}_{1}\left(  f\otimes e_{D,B}\right)  ,
\]
so that%
\begin{align*}
\left(  \operatorname*{Log}\left(  F\otimes e_{D,B}\right)  \right)  \left(
c\otimes d\right)   &  =\left(  \operatorname{Log}_{1}\left(  f\otimes
e_{D,B}\right)  \right)  \left(  c\otimes d\right)  =\sum_{i\geq1}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{\left(  f\otimes
e_{D,B}\right)  ^{\ast i}}_{\substack{=f^{\ast i}\otimes e_{D,B}\\\text{(by
Corollary~\ref{Corollary42} \textbf{(c)})}}}\left(  c\otimes d\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{log}), applied to }f\otimes e_{D,B}\text{, }C\otimes D\text{,
}A\otimes B\text{ and }c\otimes d\\
\text{instead of }f\text{, }H\text{, }A\text{ and }x
\end{array}
\right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  f^{\ast i}\otimes
e_{D,B}\right)  \left(  c\otimes d\right)  .
\end{align*}
Hence,
\[
\left(  \left(  \operatorname*{Log}F\right)  \otimes e_{D,B}\right)  \left(
c\otimes d\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
f^{\ast i}\otimes e_{D,B}\right)  \left(  c\otimes d\right)  =\left(
\operatorname*{Log}\left(  F\otimes e_{D,B}\right)  \right)  \left(  c\otimes
d\right)  .
\]


Now forget that we fixed $\left(  c,d\right)  $. We have proven that every
$\left(  c,d\right)  \in C\times D$ satisfies $\left(  \left(
\operatorname*{Log}F\right)  \otimes e_{D,B}\right)  \left(  c\otimes
d\right)  =\left(  \operatorname*{Log}\left(  F\otimes e_{D,B}\right)
\right)  \left(  c\otimes d\right)  $. In other words, we have shown that the
two maps $\left(  \operatorname*{Log}F\right)  \otimes e_{D,B}$ and
$\operatorname*{Log}\left(  F\otimes e_{D,B}\right)  $ are equal to each other
on each pure tensor in $C\otimes D$. Since these two maps are $k$-linear, this
yields that these two maps are identic (because if two $k$-linear maps from a
tensor product are equal to each other on each pure tensor, then these two
maps must be identic). In other words, $\left(  \operatorname*{Log}F\right)
\otimes e_{D,B}=\operatorname*{Log}\left(  F\otimes e_{D,B}\right)  $. This
proves Proposition~\ref{Proposition420} \textbf{(a)}.

\textbf{(b)} The proof of Proposition~\ref{Proposition420} \textbf{(b)}
differs from the proof of Proposition~\ref{Proposition420} \textbf{(a)} only
in the order of the tensorands, so there is no need to write down this proof here.

\textbf{(c)} Let $F\in G\left(  C,A\right)  $ and $H\in G\left(  D,B\right)
$. Lemma~\ref{Lemma421} yields $F\otimes H\in G\left(  C\otimes D,A\otimes
B\right)  $. In order to prove Proposition~\ref{Proposition420} \textbf{(c)},
it thus remains only to prove that $\operatorname*{Log}\left(  F\otimes
H\right)  =\left(  \operatorname*{Log}F\right)  \otimes e_{D,B}+e_{C,A}%
\otimes\left(  \operatorname*{Log}H\right)  $.

Applying Corollary~\ref{Corollary41} to $f=F$ and $g=H$, we obtain%
\[
\left(  F\otimes e_{D,B}\right)  \ast\left(  e_{C,A}\otimes H\right)
=F\otimes H=\left(  e_{C,A}\otimes H\right)  \ast\left(  F\otimes
e_{D,B}\right)  .
\]
In particular, this yields $\left(  F\otimes e_{D,B}\right)  \ast\left(
e_{C,A}\otimes H\right)  =\left(  e_{C,A}\otimes H\right)  \ast\left(
F\otimes e_{D,B}\right)  $. Due to this equality, and also due to the facts
that $F\otimes e_{D,B}\in G\left(  C\otimes D,A\otimes B\right)  $ (by
Proposition~\ref{Proposition420} \textbf{(a)}) and $e_{C,A}\otimes H\in
G\left(  C\otimes D,A\otimes B\right)  $ (by Proposition~\ref{Proposition420}
\textbf{(b)}), we can apply Theorem~\ref{Theorem400} to $C\otimes D$,
$A\otimes B$, $F\otimes e_{D,B}$ and $e_{C,A}\otimes H$ instead of $C$, $A$,
$F$ and $H$. This gives us $\left(  F\otimes e_{D,B}\right)  \ast\left(
e_{C,A}\otimes H\right)  \in G\left(  C\otimes D,A\otimes B\right)  $ and%
\[
\operatorname*{Log}\left(  \left(  F\otimes e_{D,B}\right)  \ast\left(
e_{C,A}\otimes H\right)  \right)  =\operatorname*{Log}\left(  F\otimes
e_{D,B}\right)  +\operatorname*{Log}\left(  e_{C,A}\otimes H\right)  .
\]
Thus,%
\begin{align*}
\operatorname*{Log}\left(  \underbrace{F\otimes H}_{=\left(  F\otimes
e_{D,B}\right)  \ast\left(  e_{C,A}\otimes H\right)  }\right)   &
=\operatorname*{Log}\left(  \left(  F\otimes e_{D,B}\right)  \ast\left(
e_{C,A}\otimes H\right)  \right) \\
&  =\underbrace{\operatorname*{Log}\left(  F\otimes e_{D,B}\right)
}_{\substack{=\left(  \operatorname*{Log}F\right)  \otimes e_{D,B}\\\text{(by
Proposition~\ref{Proposition420} \textbf{(a)})}}%
}+\underbrace{\operatorname*{Log}\left(  e_{C,A}\otimes H\right)
}_{\substack{=e_{C,A}\otimes\left(  \operatorname*{Log}H\right)  \\\text{(by
Proposition~\ref{Proposition420} \textbf{(b)})}}}\\
&  =\left(  \operatorname*{Log}F\right)  \otimes e_{D,B}+e_{C,A}\otimes\left(
\operatorname*{Log}H\right)  .
\end{align*}
This completes the proof of Proposition~\ref{Proposition420} \textbf{(c)}.
\end{proof}

\section{\label{S20}When graded bialgebras are Hopf}

The following section (\S \ref{S20}) is only tangentially related to the
above. The only reason I am putting this section here is that it extends
Proposition~\ref{Proposition224} \textbf{(c)} to the case of $n\in\mathbb{Z}$
for $\ast$-invertible $f$. More precisely, this section will (among other
things) prove the following fact:

\begin{proposition}
\label{Proposition501} Let $k$ be a field. Let $C$ be a graded $k$-coalgebra.
Let $A$ be a graded $k$-algebra. Let $f:C\rightarrow A$ be a graded $\ast
$-invertible linear map (i. e., a graded linear map which has an inverse with
respect to the operation $\ast$ in $\mathcal{L}\left(  C,A\right)  $).

\textbf{(a)} The map $f^{\ast\left(  -1\right)  }$ (that is, the inverse of
$f$ with respect to the operation $\ast$ in $\mathcal{L}\left(  C,A\right)  $)
is graded.

\textbf{(b)} For every $n\in\mathbb{Z}$, the map $f^{\ast n}$ is graded.
\end{proposition}

A consequence of this proposition is a curious property of graded bialgebras
which are Hopf algebras:

\begin{theorem}
\label{Theorem502} Let $k$ be a field. Let $H$ be a graded $k$-bialgebra.
Assume that the $k$-bialgebra $H$ is a Hopf algebra. Then, the antipode of the
Hopf algebra $H$ is a graded map.
\end{theorem}

Note that there is a notion of a ``graded $k$-Hopf algebra''. There are two
possible ways to define this notion. One way is to define a graded $k$-Hopf
algebra as a graded $k$-bialgebra whose underlying $k$-bialgebra (without
grading) is a Hopf algebra. The other way is to define a graded $k$-Hopf
algebra as a graded $k$-bialgebra whose underlying $k$-bialgebra (without
grading) is a Hopf algebra with its antipode being a graded map.
Theorem~\ref{Theorem502} shows that these two ways are equivalent.

Here is how to prove Theorem~\ref{Theorem502} using
Proposition~\ref{Proposition501}:

\begin{proof}
[Proof of Theorem~\ref{Theorem502}.]Let $S$ be the antipode of the Hopf
algebra $H$. Then, $S$ is the $\ast$-inverse of the identity map
$\operatorname{id}_{H}$ (because the antipode of a Hopf algebra is defined as
the $\ast$-inverse of its identity map). In other words, $S=\operatorname{id}%
_{H}^{\ast\left(  -1\right)  }$. As a consequence, the map $\operatorname{id}%
_{H}$ is $\ast$-invertible. Since the map $\operatorname{id}_{H}$ is also
graded, we can thus apply Proposition~\ref{Proposition501} \textbf{(a)} to
$f=\operatorname{id}_{H}$, and conclude that the map $\operatorname{id}%
_{H}^{\ast\left(  -1\right)  }$ is graded. Since $\operatorname{id}_{H}%
^{\ast\left(  -1\right)  }=S$ is the antipode of $H$, this rewrites as
follows: The antipode of $H$ is graded. This proves Theorem~\ref{Theorem502}.
\end{proof}

What remains to be done now is proving Proposition~\ref{Proposition501}. For
this, we will construct a way to assign to every map $f\in\mathcal{L}\left(
C,A\right)  $ a map $T_{f}\in\operatorname*{End}\left(  A\otimes C\right)  $
in such a way that the convolution of maps in $\mathcal{L}\left(  C,A\right)
$ corresponds to the composition of maps in $\operatorname*{End}\left(
A\otimes C\right)  $ (in reverse order). (In other words, we will make
$A\otimes C$ into a representation of the $k$-algebra $\left(  \mathcal{L}%
\left(  C,A\right)  \right)  ^{\operatorname*{op}}$.) This construction does
not require $C$ or $A$ to be graded; it works in general.

\begin{definition}
\label{Definition503} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $A$
be a $k$-algebra. For every $k$-linear map $f:C\rightarrow A$, we define a
$k$-linear map $T_{f}:A\otimes C\rightarrow A\otimes C$ by%
\[
T_{f}=\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  .
\]

\end{definition}

\begin{proposition}
\label{Proposition504} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $A$
be a $k$-algebra.

\textbf{(a)} For any two $k$-linear maps $f:C\rightarrow A$ and
$g:C\rightarrow A$, we have $T_{f+g}=T_{f}+T_{g}$.

\textbf{(b)} We have $T_{0}=0$.

\textbf{(c)} For any $k$-linear map $f:C\rightarrow A$ and any $\lambda\in k$,
we have $T_{\lambda f}=\lambda T_{f}$.

\textbf{(d)} For any two $k$-linear maps $f:C\rightarrow A$ and
$g:C\rightarrow A$, we have $T_{g\ast f}=T_{f}\circ T_{g}$.

\textbf{(e)} We have $T_{e_{C,A}}=\operatorname{id}_{A\otimes C}$.

\textbf{(f)} If $C$ is a graded $k$-coalgebra, $A$ is a graded $k$-algebra,
and $f$ is a graded map, then $T_{f}$ also is a graded map.

\textbf{(g)} For any $\ast$-invertible $k$-linear map $f:C\rightarrow A$, the
map $T_{f}$ is invertible and satisfies $T_{f^{\ast\left(  -1\right)  }%
}=\left(  T_{f}\right)  ^{-1}$.

\textbf{(h)} Let $\operatorname{kan}_{C,k\otimes C}:C\rightarrow k\otimes C$
be the canonical isomorphism which sends $c$ to $1\otimes c$ for every $c\in
C$. Let $\operatorname{kan}_{A\otimes k,A}:A\otimes k\rightarrow A$ be the
canonical isomorphism which sends $a\otimes\lambda$ to $\lambda a$ for every
$\left(  a,\lambda\right)  \in A\times k$. For any $k$-linear map
$f:C\rightarrow A$, we have%
\[
\operatorname{kan}_{A\otimes k,A}\circ\left(  \operatorname{id}_{A}%
\otimes\varepsilon_{C}\right)  \circ T_{f}\circ\left(  \eta_{A}\otimes
\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes C}=f.
\]


\textbf{(i)} If $C$ is a graded $k$-coalgebra, $A$ is a graded $k$-algebra,
and $f$ is a $k$-linear map such that $T_{f}$ is graded, then $f$ is graded.
\end{proposition}

Proposition~\ref{Proposition504} is again an easy exercise to prove with the
use of the Sweedler notation, but we are going to stay pure and prove it by
manipulation of maps. First, let us make two general observations about vector
spaces and tensor products:

\begin{itemize}
\item \textit{Observation 1:}%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{Any five }k\text{-vector spaces }P\text{, }Q\text{, }R_{1}\text{, }%
R_{2}\text{, }R_{3}\text{ and any }k\text{-linear maps}\\
\varphi:R_{1}\rightarrow R_{2}\text{ and }\psi:R_{2}\rightarrow R_{3}\text{
satisfy}\\
\left(  \operatorname{id}_{P}\otimes\psi\otimes\operatorname{id}_{Q}\right)
\circ\left(  \operatorname{id}_{P}\otimes\varphi\otimes\operatorname{id}%
_{Q}\right)  =\operatorname{id}_{P}\otimes\left(  \psi\circ\varphi\right)
\otimes\operatorname{id}_{Q}%
\end{array}
\right)  . \label{504.pf.idid}%
\end{equation}
\footnote{\textit{Proof of (\ref{504.pf.idid}).} Let $P$, $Q$, $R_{1}$,
$R_{2}$, $R_{3}$ be five $k$-vector space, and let $\varphi:R_{1}\rightarrow
R_{2}$ and $\psi:R_{2}\rightarrow R_{3}$ be two $k$-linear maps. By
(\ref{(X)(X)}) (applied to $U=P$, $V=P$, $W=P$, $U^{\prime}=R_{1}$,
$V^{\prime}=R_{2}$, $W^{\prime}=R_{3}$, $\alpha=\operatorname{id}_{P}$,
$\beta=\operatorname{id}_{P}$, $\alpha^{\prime}=\varphi$ and $\beta^{\prime
}=\psi$), we have $\left(  \operatorname{id}_{P}\circ\operatorname{id}%
_{P}\right)  \otimes\left(  \psi\circ\varphi\right)  =\left(
\operatorname{id}_{P}\otimes\psi\right)  \circ\left(  \operatorname{id}%
_{P}\otimes\varphi\right)  $. By (\ref{(X)(X)}) (applied to $U=P\otimes R_{1}%
$, $V=P\otimes R_{2}$, $W=P\otimes R_{3}$, $U^{\prime}=Q$, $V^{\prime}=Q$,
$W^{\prime}=Q$, $\alpha=\operatorname{id}_{P}\otimes\varphi$, $\beta
=\operatorname{id}_{P}\otimes\psi$, $\alpha^{\prime}=\operatorname{id}_{Q}$
and $\beta^{\prime}=\operatorname{id}_{Q}$), we have%
\[
\left(  \left(  \operatorname{id}_{P}\otimes\psi\right)  \circ\left(
\operatorname{id}_{P}\otimes\varphi\right)  \right)  \otimes\left(
\operatorname{id}_{Q}\circ\operatorname{id}_{Q}\right)  =\left(
\operatorname{id}_{P}\otimes\psi\otimes\operatorname{id}_{Q}\right)
\circ\left(  \operatorname{id}_{P}\otimes\varphi\otimes\operatorname{id}%
_{Q}\right)  .
\]
Thus,%
\begin{align*}
\left(  \operatorname{id}_{P}\otimes\psi\otimes\operatorname{id}_{Q}\right)
\circ\left(  \operatorname{id}_{P}\otimes\varphi\otimes\operatorname{id}%
_{Q}\right)   &  =\underbrace{\left(  \left(  \operatorname{id}_{P}\otimes
\psi\right)  \circ\left(  \operatorname{id}_{P}\otimes\varphi\right)  \right)
}_{=\left(  \operatorname{id}_{P}\circ\operatorname{id}_{P}\right)
\otimes\left(  \psi\circ\varphi\right)  }\otimes\underbrace{\left(
\operatorname{id}_{Q}\circ\operatorname{id}_{Q}\right)  }_{=\operatorname{id}%
_{Q}}\\
&  =\underbrace{\left(  \operatorname{id}_{P}\circ\operatorname{id}%
_{P}\right)  }_{=\operatorname{id}_{P}}\otimes\left(  \psi\circ\varphi\right)
\otimes\operatorname{id}_{Q}=\operatorname{id}_{P}\otimes\left(  \psi
\circ\varphi\right)  \otimes\operatorname{id}_{Q}.
\end{align*}
This proves (\ref{504.pf.idid}).}

\item \textit{Observation 2:}%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{Any four }k\text{-vector spaces }P\text{, }Q\text{, }R\text{ and
}S\text{ and any two}\\
k\text{-linear maps }\gamma:P\rightarrow Q\text{ and }\delta:R\rightarrow
S\text{ satisfy}\\
\left(  \operatorname{id}_{Q}\otimes\delta\right)  \circ\left(  \gamma
\otimes\operatorname{id}_{R}\right)  =\left(  \gamma\otimes\operatorname{id}%
_{S}\right)  \circ\left(  \operatorname{id}_{P}\otimes\delta\right)
\end{array}
\right)  . \label{504.pf.comm}%
\end{equation}
\footnote{\textit{Proof of (\ref{504.pf.comm}).} Let $P$, $Q$, $R$ and $S$ be
four $k$-vector spaces. Let $\gamma:P\rightarrow Q$ and $\delta:R\rightarrow
S$ be two $k$-linear maps.
\par
Applying (\ref{(X)(X)}) to $U=P$, $V=Q$, $W=Q$, $U^{\prime}=R$, $V^{\prime}%
=R$, $W^{\prime}=S$, $\alpha=\gamma$, $\beta=\operatorname{id}_{Q}$,
$\alpha^{\prime}=\operatorname{id}_{R}$ and $\beta^{\prime}=\delta$, we obtain
$\left(  \operatorname{id}_{Q}\circ\gamma\right)  \otimes\left(  \delta
\circ\operatorname{id}_{R}\right)  =\left(  \operatorname{id}_{Q}\otimes
\delta\right)  \circ\left(  \gamma\otimes\operatorname{id}_{R}\right)  $, so
that%
\[
\left(  \operatorname{id}_{Q}\otimes\delta\right)  \circ\left(  \gamma
\otimes\operatorname{id}_{R}\right)  =\underbrace{\left(  \operatorname{id}%
_{Q}\circ\gamma\right)  }_{=\gamma}\otimes\underbrace{\left(  \delta
\circ\operatorname{id}_{R}\right)  }_{=\delta}=\gamma\otimes\delta.
\]
\par
Applying (\ref{(X)(X)}) to $U=P$, $V=P$, $W=Q$, $U^{\prime}=R$, $V^{\prime}%
=S$, $W^{\prime}=S$, $\alpha=\operatorname{id}_{P}$, $\beta=\gamma$,
$\alpha^{\prime}=\delta$ and $\beta^{\prime}=\operatorname{id}_{S}$, we obtain
$\left(  \gamma\circ\operatorname{id}_{P}\right)  \otimes\left(
\operatorname{id}_{S}\circ\delta\right)  =\left(  \gamma\otimes
\operatorname{id}_{S}\right)  \circ\left(  \operatorname{id}_{P}\otimes
\delta\right)  $, so that%
\[
\left(  \gamma\otimes\operatorname{id}_{S}\right)  \circ\left(
\operatorname{id}_{P}\otimes\delta\right)  =\underbrace{\left(  \gamma
\circ\operatorname{id}_{P}\right)  }_{=\gamma}\circ\underbrace{\left(
\operatorname{id}_{S}\circ\delta\right)  }_{=\delta}=\gamma\otimes
\delta=\left(  \operatorname{id}_{Q}\otimes\delta\right)  \circ\left(
\gamma\otimes\operatorname{id}_{R}\right)  .
\]
This proves (\ref{504.pf.comm}).}
\end{itemize}

\begin{proof}
[Proof of Proposition~\ref{Proposition504}.]\textbf{(a)} Let $f:C\rightarrow
A$ and $g:C\rightarrow A$ be two $k$-linear maps. Since tensoring of
$k$-linear maps is distributive, we have $\left(  f+g\right)  \otimes
\operatorname{id}_{C}=f\otimes\operatorname{id}_{C}+g\otimes\operatorname{id}%
_{C}$, so that%
\[
\operatorname{id}_{A}\otimes\left(  f+g\right)  \otimes\operatorname{id}%
_{C}=\operatorname{id}_{A}\otimes\left(  f\otimes\operatorname{id}%
_{C}+g\otimes\operatorname{id}_{C}\right)  =\operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{C}+\operatorname{id}_{A}\otimes g\otimes
\operatorname{id}_{C}%
\]
(since tensoring of $k$-linear maps is distributive), so that%
\begin{align*}
\left(  \operatorname{id}_{A}\otimes\left(  f+g\right)  \otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\Delta_{C}\right)   &  =\left(  \operatorname{id}_{A}\otimes f\otimes
\operatorname{id}_{C}+\operatorname{id}_{A}\otimes g\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right) \\
&  =\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  +\left(
\operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
\end{align*}
(since composition of $k$-linear maps is distributive), so that%
\begin{align*}
&  \left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\left(  f+g\right)  \otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right) \\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  +\left(
\operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  \right) \\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  +\left(  \mu
_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes g\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right)
\end{align*}
(since composition of $k$-linear maps is distributive). Now, by the definition
of $T_{f+g}$, we have%
\begin{align*}
T_{f+g}  &  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\underbrace{\left(  \operatorname{id}_{A}\otimes\left(  f+g\right)
\otimes\operatorname{id}_{C}\right)  }_{\substack{=\operatorname{id}%
_{A}\otimes f\otimes\operatorname{id}_{C}+\operatorname{id}_{A}\otimes
g\otimes\operatorname{id}_{C}\\\text{(since the tensor product of
}k\text{-linear maps is distributive)}}}\circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right) \\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}+\operatorname{id}%
_{A}\otimes g\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right) \\
&  =\underbrace{\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
}_{\substack{=T_{f}\\\text{(since }T_{f}\text{ was defined as }\left(  \mu
_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes f\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \mu_{A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes g\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\Delta_{C}\right)  }_{\substack{=T_{g}\\\text{(since }T_{g}\text{ was defined
as }\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the composition of }k\text{-linear
maps is distributive}\right) \\
&  =T_{f}+T_{g}.
\end{align*}
This proves Proposition~\ref{Proposition504} \textbf{(a)}.

\textbf{(c)} Let $f:C\rightarrow A$ be a $k$-linear map, and let $\lambda\in
k$. Since tensoring of $k$-linear maps is $k$-bilinear, we have $\left(
\lambda f\right)  \otimes\operatorname{id}_{C}=\lambda\left(  f\otimes
\operatorname{id}_{C}\right)  $ and $\operatorname{id}_{A}\otimes\left(
\lambda\left(  f\otimes\operatorname{id}_{C}\right)  \right)  =\lambda\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)  $. Thus,%
\[
\operatorname{id}_{A}\otimes\underbrace{\left(  \lambda f\right)
\otimes\operatorname{id}_{C}}_{=\lambda\left(  f\otimes\operatorname{id}%
_{C}\right)  }=\operatorname{id}_{A}\otimes\left(  \lambda\left(
f\otimes\operatorname{id}_{C}\right)  \right)  =\lambda\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)  .
\]
Since composition of $k$-linear maps is $k$-bilinear, we have%
\[
\left(  \lambda\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}%
_{C}\right)  \right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta
_{C}\right)  =\lambda\left(  \left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right)  \right)  .
\]


By the definition of $T_{\lambda f}$, we have%
\begin{align*}
T_{f}  &  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ
\underbrace{\left(  \operatorname{id}_{A}\otimes\left(  \lambda f\right)
\otimes\operatorname{id}_{C}\right)  }_{=\lambda\left(  \operatorname{id}%
_{A}\otimes f\otimes\operatorname{id}_{C}\right)  }\circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right) \\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ
\underbrace{\left(  \lambda\left(  \operatorname{id}_{A}\otimes f\otimes
\operatorname{id}_{C}\right)  \right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right)  }_{=\lambda\left(  \left(  \operatorname{id}%
_{A}\otimes f\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right)  \right)  }\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\lambda\left(  \left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
\right)  \right) \\
&  =\lambda\cdot\underbrace{\left(  \mu_{A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes f\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\Delta_{C}\right)  }_{=T_{f}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is }k\text{-bilinear}\right) \\
&  =\lambda T_{f}.
\end{align*}
This proves Proposition~\ref{Proposition504} \textbf{(c)}.

\textbf{(b)} Applying Proposition~\ref{Proposition504} \textbf{(b)} to
$\lambda=0$ and $f=0$, we obtain $T_{0\cdot0}=0T_{0}=0$. In other words,
$T_{0}=0$. This proves Proposition~\ref{Proposition504} \textbf{(c)}.

\textbf{(d)} Let $f:C\rightarrow A$ and $g:C\rightarrow A$ be two $k$-linear maps.

Applying (\ref{504.pf.comm}) to $P=A\otimes A$, $Q=A$, $R=C$, $S=C\otimes C$,
$\gamma=\mu_{A}$ and $\delta=\Delta_{C}$, we obtain%
\begin{equation}
\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  \circ\left(  \mu
_{A}\otimes\operatorname{id}_{C}\right)  =\left(  \mu_{A}\otimes
\operatorname{id}_{C\otimes C}\right)  \circ\left(  \operatorname{id}%
_{A\otimes A}\otimes\Delta_{C}\right)  . \label{504.pf.1}%
\end{equation}


Next, we notice that%
\begin{equation}
\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)  =\left(
\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(
\operatorname{id}_{A\otimes A}\otimes f\right)  \right)  \otimes
\operatorname{id}_{C} \label{504.pf.2}%
\end{equation}
\footnote{\textit{Proof.} Applying (\ref{504.pf.comm}) to $P=A\otimes A$,
$Q=A$, $R=C$, $S=A$, $\gamma=\mu_{A}$ and $\delta=f$, we obtain%
\[
\left(  \operatorname{id}_{A}\otimes f\right)  \circ\left(  \mu_{A}%
\otimes\operatorname{id}_{C}\right)  =\left(  \mu_{A}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)  .
\]
\par
Applying (\ref{(X)(X)}) to $U=A\otimes A\otimes C$, $V=A\otimes C$,
$W=A\otimes A$, $U^{\prime}=C$, $V^{\prime}=C$, $W^{\prime}=C$, $\alpha
=\mu_{A}\otimes\operatorname{id}_{C}$, $\beta=\operatorname{id}_{A}\otimes f$,
$\alpha^{\prime}=\operatorname{id}_{C}$ and $\beta^{\prime}=\operatorname{id}%
_{C}$, we obtain%
\begin{align*}
\left(  \left(  \operatorname{id}_{A}\otimes f\right)  \circ\left(  \mu
_{A}\otimes\operatorname{id}_{C}\right)  \right)  \otimes\left(
\operatorname{id}_{C}\circ\operatorname{id}_{C}\right)   &  =\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \mu_{A}\otimes\underbrace{\operatorname{id}_{C}\otimes
\operatorname{id}_{C}}_{=\operatorname{id}_{C\otimes C}}\right) \\
&  =\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)  ,
\end{align*}
so that%
\begin{align*}
\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)   &
=\underbrace{\left(  \left(  \operatorname{id}_{A}\otimes f\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \right)  }_{=\left(
\mu_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{A\otimes A}\otimes f\right)  }\otimes\underbrace{\left(  \operatorname{id}%
_{C}\circ\operatorname{id}_{C}\right)  }_{=\operatorname{id}_{C}}\\
&  =\left(  \left(  \mu_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(
\operatorname{id}_{A\otimes A}\otimes f\right)  \right)  \otimes
\operatorname{id}_{C}.
\end{align*}
This proves (\ref{504.pf.2}).}. Using this equality, we can easily get%
\begin{equation}
\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)  =\left(
\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(
\operatorname{id}_{A\otimes A}\otimes f\right)  \right)  \otimes
\operatorname{id}_{C} \label{504.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{504.pf.3}).} Applying (\ref{(X)(X)}) to
$U=A\otimes A\otimes C$, $V=A\otimes A$, $W=A$, $U^{\prime}=C$, $V^{\prime}%
=C$, $W^{\prime}=C$, $\alpha=\left(  \mu_{A}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)  $,
$\beta=\mu_{A}$, $\alpha^{\prime}=\operatorname{id}_{C}$ and $\beta^{\prime
}=\operatorname{id}_{C}$, we obtain%
\[
\left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)  \right)
\otimes\left(  \operatorname{id}_{C}\circ\operatorname{id}_{C}\right)
=\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \left(
\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(
\operatorname{id}_{A\otimes A}\otimes f\right)  \right)  \otimes
\operatorname{id}_{C}\right)  ,
\]
so that%
\begin{align*}
&  \left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \left(
\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(
\operatorname{id}_{A\otimes A}\otimes f\right)  \right)  \otimes
\operatorname{id}_{C}\right) \\
&  =\left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)  \right)
\otimes\underbrace{\left(  \operatorname{id}_{C}\circ\operatorname{id}%
_{C}\right)  }_{=\operatorname{id}_{C}}\\
&  =\left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)  \right)
\otimes\operatorname{id}_{C}.
\end{align*}
\par
Now,%
\begin{align*}
\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\underbrace{\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)
}_{\substack{=\left(  \left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)  \right)
\otimes\operatorname{id}_{C}\\\text{(by (\ref{504.pf.2}))}}}  &  =\left(
\mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \left(  \left(
\mu_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{A\otimes A}\otimes f\right)  \right)  \otimes\operatorname{id}_{C}\right) \\
&  =\left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)  \right)
\otimes\operatorname{id}_{C}.
\end{align*}
This proves (\ref{504.pf.3}).}. Since $\mu_{A}\circ\left(  \mu_{A}%
\otimes\operatorname{id}_{A}\right)  =\mu_{A}\circ\left(  \operatorname{id}%
_{A}\otimes\mu_{A}\right)  $ (by the axioms of a $k$-algebra, since $A$ is a
$k$-algebra), this becomes%
\begin{align}
&  \left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right) \nonumber\\
&  =\left(  \underbrace{\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}%
_{A}\right)  }_{=\mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\mu
_{A}\right)  }\circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)
\right)  \otimes\underbrace{\operatorname{id}_{C}}_{=\operatorname{id}%
_{C}\circ\operatorname{id}_{C}}\nonumber\\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)
\circ\left(  \operatorname{id}_{A\otimes A}\otimes f\right)  \right)
\otimes\left(  \operatorname{id}_{C}\circ\operatorname{id}_{C}\right)
\nonumber\\
&  =\left(  \left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\mu
_{A}\right)  \right)  \otimes\underbrace{\operatorname{id}_{C}}%
_{=\operatorname{id}_{C}\circ\operatorname{id}_{C}}\right)  \circ\left(
\underbrace{\operatorname{id}_{A\otimes A}}_{=\operatorname{id}_{A}%
\otimes\operatorname{id}_{A}}\otimes f\otimes\operatorname{id}_{C}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}), applied to }U=A\otimes A\otimes C\text{, }V=A\otimes
A\otimes A\text{, }W=A\text{, }U^{\prime}=C\text{, }V^{\prime}=C\text{,}\\
W^{\prime}=C\text{, }\alpha=\operatorname{id}_{A\otimes A}\otimes f\text{,
}\beta=\mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \text{,
}\alpha^{\prime}=\operatorname{id}_{C}\text{ and }\beta^{\prime}%
=\operatorname{id}_{C}\text{)}%
\end{array}
\right) \nonumber\\
&  =\underbrace{\left(  \left(  \mu_{A}\circ\left(  \operatorname{id}%
_{A}\otimes\mu_{A}\right)  \right)  \otimes\left(  \operatorname{id}_{C}%
\circ\operatorname{id}_{C}\right)  \right)  }_{\substack{=\left(  \mu
_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\mu_{A}\otimes\operatorname{id}_{C}\right)  \\\text{(by
(\ref{(X)(X)}), applied to }U=A\otimes A\otimes A\text{, }V=A\otimes A\text{,
}W=A\text{,}\\U^{\prime}=C\text{, }V^{\prime}=C\text{, }W^{\prime}=C\text{,
}\alpha=\operatorname{id}_{A}\otimes\mu_{A}\text{, }\beta=\mu_{A}\text{,
}\alpha^{\prime}=\operatorname{id}_{C}\text{ and }\beta^{\prime}%
=\operatorname{id}_{C}\text{)}}}\circ\left(  \operatorname{id}_{A}%
\otimes\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{C}\right)  . \label{504.pf.4}%
\end{align}


On the other hand,%
\begin{equation}
\left(  \operatorname{id}_{A\otimes A}\otimes\Delta_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}\right)
=\operatorname{id}_{A}\otimes\left(  \left(  g\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right)  \right)  \label{504.pf.-2}%
\end{equation}
\footnote{\textit{Proof.} Applying (\ref{504.pf.comm}) to $P=C$, $Q=A$, $R=C$,
$S=C\otimes C$, $\gamma=g$ and $\delta=\Delta_{C}$, we obtain%
\[
\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  \circ\left(
g\otimes\operatorname{id}_{C}\right)  =\left(  g\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right)  .
\]
\par
Applying (\ref{(X)(X)}) to $U=A$, $V=A$, $W=A$, $U^{\prime}=C\otimes C$,
$V^{\prime}=A\otimes C$, $W^{\prime}=A\otimes C\otimes C$, $\alpha
=\operatorname{id}_{A}$, $\beta=\operatorname{id}_{A}$, $\alpha^{\prime
}=g\otimes\operatorname{id}_{C}$ and $\beta^{\prime}=\operatorname{id}%
_{A}\otimes\Delta_{C}$, we obtain%
\begin{align*}
\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)  \otimes\left(
\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  \circ\left(
g\otimes\operatorname{id}_{C}\right)  \right)   &  =\left(
\underbrace{\operatorname{id}_{A}\otimes\operatorname{id}_{A}}%
_{=\operatorname{id}_{A\otimes A}}\otimes\Delta_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}\right) \\
&  =\left(  \operatorname{id}_{A\otimes A}\otimes\Delta_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes g\otimes\operatorname{id}%
_{C}\right)  ,
\end{align*}
so that%
\begin{align*}
\left(  \operatorname{id}_{A\otimes A}\otimes\Delta_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}\right)   &
=\underbrace{\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)
}_{=\operatorname{id}_{A}}\otimes\underbrace{\left(  \left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right)  \circ\left(  g\otimes\operatorname{id}%
_{C}\right)  \right)  }_{=\left(  g\otimes\operatorname{id}_{C\otimes
C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)  }\\
&  =\operatorname{id}_{A}\otimes\left(  \left(  g\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right)  \right)  .
\end{align*}
This proves (\ref{504.pf.-2}).}. Using this equality, we can easily get%
\begin{equation}
\left(  \operatorname{id}_{A\otimes A}\otimes\Delta_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  =\operatorname{id}%
_{A}\otimes\left(  \left(  g\otimes\operatorname{id}_{C\otimes C}\right)
\circ\left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)  \circ\Delta
_{C}\right)  \label{504.pf.-3}%
\end{equation}
\footnote{\textit{Proof of (\ref{504.pf.-3}).} Applying (\ref{(X)(X)}) to
$U=A$, $V=A$, $W=A$, $U^{\prime}=C$, $V^{\prime}=C\otimes C$, $W^{\prime
}=A\otimes C\otimes C$, $\alpha=\operatorname{id}_{A}$, $\beta
=\operatorname{id}_{A}$, $\alpha^{\prime}=\Delta_{C}$ and $\beta^{\prime
}=\left(  g\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right)  $, we obtain%
\[
\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)  \otimes\left(
\left(  g\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\right)
=\left(  \operatorname{id}_{A}\otimes\left(  \left(  g\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right)  \right)  \right)  \circ\left(  \operatorname{id}_{A}\otimes
\Delta_{C}\right)  ,
\]
so that%
\begin{align*}
&  \left(  \operatorname{id}_{A}\otimes\left(  \left(  g\otimes
\operatorname{id}_{C\otimes C}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\Delta_{C}\right)  \right)  \right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right) \\
&  =\underbrace{\left(  \operatorname{id}_{A}\circ\operatorname{id}%
_{A}\right)  }_{=\operatorname{id}_{A}}\otimes\left(  \left(  g\otimes
\operatorname{id}_{C\otimes C}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\right) \\
&  =\operatorname{id}_{A}\otimes\left(  \left(  g\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right)  \circ\Delta_{C}\right)  .
\end{align*}
\par
Now,%
\begin{align*}
&  \underbrace{\left(  \operatorname{id}_{A\otimes A}\otimes\Delta_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes g\otimes\operatorname{id}%
_{C}\right)  }_{\substack{=\operatorname{id}_{A}\otimes\left(  \left(
g\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\Delta_{C}\right)  \right)  \\\text{(by (\ref{504.pf.-2}))}}%
}\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right) \\
&  =\left(  \operatorname{id}_{A}\otimes\left(  \left(  g\otimes
\operatorname{id}_{C\otimes C}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\Delta_{C}\right)  \right)  \right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right) \\
&  =\operatorname{id}_{A}\otimes\left(  \left(  g\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right)  \circ\Delta_{C}\right)  .
\end{align*}
This proves (\ref{504.pf.-3}).}. Since $\left(  \operatorname{id}_{C}%
\otimes\Delta_{C}\right)  \circ\Delta_{C}=\left(  \Delta_{C}\otimes
\operatorname{id}_{C}\right)  \circ\Delta_{C}$ (by the axioms of a
$k$-coalgebra, since $C$ is a $k$-coalgebra), this becomes%
\begin{align}
&  \left(  \operatorname{id}_{A\otimes A}\otimes\Delta_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes g\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
\nonumber\\
&  =\underbrace{\operatorname{id}_{A}}_{=\operatorname{id}_{A}\circ
\operatorname{id}_{A}}\otimes\left(  \left(  g\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\underbrace{\left(  \operatorname{id}_{C}%
\otimes\Delta_{C}\right)  \circ\Delta_{C}}_{=\left(  \Delta_{C}\otimes
\operatorname{id}_{C}\right)  \circ\Delta_{C}}\right) \nonumber\\
&  =\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)
\otimes\left(  \left(  g\otimes\operatorname{id}_{C\otimes C}\right)
\circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta
_{C}\right) \nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes g\otimes\underbrace{\operatorname{id}%
_{C\otimes C}}_{=\operatorname{id}_{C}\otimes\operatorname{id}_{C}}\right)
\circ\left(  \underbrace{\operatorname{id}_{A}}_{=\operatorname{id}_{A}%
\circ\operatorname{id}_{A}}\otimes\left(  \left(  \Delta_{C}\otimes
\operatorname{id}_{C}\right)  \circ\Delta_{C}\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}), applied to }U=A\text{, }V=A\text{, }W=A\text{,
}U^{\prime}=C\text{, }V^{\prime}=C\otimes C\otimes C\text{,}\\
W^{\prime}=A\otimes C\otimes C\text{, }\alpha=\operatorname{id}_{A}\text{,
}\beta=\operatorname{id}_{A}\text{, }\alpha^{\prime}=\left(  \Delta_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}\text{ and }\beta^{\prime
}=g\otimes\operatorname{id}_{C\otimes C}\text{)}%
\end{array}
\right) \nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\underbrace{\left(  \left(
\operatorname{id}_{A}\circ\operatorname{id}_{A}\right)  \otimes\left(  \left(
\Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}\right)
\right)  }_{\substack{=\left(  \operatorname{id}_{A}\otimes\Delta_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}%
\otimes\Delta_{C}\right)  \\\text{(by (\ref{(X)(X)}), applied to }U=A\text{,
}V=A\text{, }W=A\text{, }U^{\prime}=C\text{, }V^{\prime}=C\otimes C\text{,
}W^{\prime}=C\otimes C\otimes C\text{,}\\\alpha=\operatorname{id}_{A}\text{,
}\beta=\operatorname{id}_{A}\text{, }\alpha^{\prime}=\Delta_{C}\text{ and
}\beta^{\prime}=\Delta_{C}\otimes\operatorname{id}_{C}\text{)}}}\nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}%
\otimes\Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right)  . \label{504.pf.-4}%
\end{align}


Applying (\ref{(X)(X)}) to $U=C$, $V=A$, $W=A$, $U^{\prime}=C$, $V^{\prime}%
=C$, $W^{\prime}=A$, $\alpha=g$, $\beta=\operatorname{id}_{A}$, $\alpha
^{\prime}=\operatorname{id}_{C}$ and $\beta^{\prime}=f$, we obtain $\left(
\operatorname{id}_{A}\circ g\right)  \otimes\left(  f\circ\operatorname{id}%
_{C}\right)  =\left(  \operatorname{id}_{A}\otimes f\right)  \circ\left(
g\otimes\operatorname{id}_{C}\right)  $, so that%
\begin{equation}
\left(  \operatorname{id}_{A}\otimes f\right)  \circ\left(  g\otimes
\operatorname{id}_{C}\right)  =\underbrace{\left(  \operatorname{id}_{A}\circ
g\right)  }_{=g}\otimes\underbrace{\left(  f\circ\operatorname{id}_{C}\right)
}_{=f}=g\otimes f. \label{504.pf.9}%
\end{equation}


Now, $T_{f}=\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  $ (by the
definition of $T_{f}$) and $T_{g}=\left(  \mu_{A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes g\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\Delta_{C}\right)  $ (by the definition of $T_{g}$), so that%
\begin{align}
&  T_{f}\circ T_{g}\nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\underbrace{\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  }_{\substack{=\left(
\mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{A\otimes A}\otimes\Delta_{C}\right)  \\\text{(by
(\ref{504.pf.1}))}}}\circ\left(  \operatorname{id}_{A}\otimes g\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\Delta_{C}\right) \nonumber\\
&  =\underbrace{\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)
}_{\substack{=\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{C}\right)  \\\text{(by (\ref{504.pf.4}))}}%
}\circ\underbrace{\left(  \operatorname{id}_{A\otimes A}\otimes\Delta
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes g\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\Delta_{C}\right)  }_{\substack{=\left(  \operatorname{id}_{A}\otimes
g\otimes\operatorname{id}_{C}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  \\\text{(by
(\ref{504.pf.-4}))}}}\nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\underbrace{\left(  \operatorname{id}_{A}\otimes\operatorname{id}%
_{A}\otimes f\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes\operatorname{id}_{C}\otimes
\operatorname{id}_{C}\right)  }_{\substack{=\operatorname{id}_{A}%
\otimes\left(  \left(  \operatorname{id}_{A}\otimes f\right)  \circ\left(
g\otimes\operatorname{id}_{C}\right)  \right)  \otimes\operatorname{id}%
_{C}\\\text{(by (\ref{504.pf.idid}), applied to }P=A\text{, }Q=C\text{, }%
R_{1}=C\otimes C\text{,}\\R_{2}=A\otimes C\text{, }R_{3}=A\otimes A\text{,
}\varphi=g\otimes\operatorname{id}_{C}\text{ and }\psi=\operatorname{id}%
_{A}\otimes f\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \operatorname{id}_{A}\otimes\Delta
_{C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right) \nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\underbrace{\left(  \left(
\operatorname{id}_{A}\otimes f\right)  \circ\left(  g\otimes\operatorname{id}%
_{C}\right)  \right)  }_{\substack{=g\otimes f\\\text{(by (\ref{504.pf.9}))}%
}}\otimes\operatorname{id}_{C}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \operatorname{id}_{A}\otimes\Delta
_{C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right) \nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ
\underbrace{\left(  \operatorname{id}_{A}\otimes\mu_{A}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
g\otimes f\otimes\operatorname{id}_{C}\right)  }_{\substack{=\operatorname{id}%
_{A}\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)
\otimes\operatorname{id}_{C}\\\text{(by (\ref{504.pf.idid}), applied to
}P=A\text{, }Q=C\text{, }R_{1}=C\otimes C\text{,}\\R_{2}=A\otimes A\text{,
}R_{3}=A\text{, }\varphi=g\otimes f\text{ and }\psi=\mu_{A}\text{)}}%
}\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
\nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ
\underbrace{\left(  \operatorname{id}_{A}\otimes\left(  \mu_{A}\circ\left(
g\otimes f\right)  \right)  \otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\otimes\operatorname{id}_{C}\right)
}_{\substack{=\operatorname{id}_{A}\otimes\left(  \mu_{A}\circ\left(  g\otimes
f\right)  \circ\Delta_{C}\right)  \otimes\operatorname{id}_{C}\\\text{(by
(\ref{504.pf.idid}), applied to }P=A\text{, }Q=C\text{, }R_{1}=C\text{,}%
\\R_{2}=C\otimes C\text{, }R_{3}=A\text{, }\varphi=\Delta_{C}\text{ and }%
\psi=\mu_{A}\circ\left(  g\otimes f\right)  \text{)}}}\circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right) \nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)
\circ\Delta_{C}\right)  \otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right)  . \label{504.pf.15}%
\end{align}


But by the definition of $T_{g\ast f}$, we have%
\begin{align*}
T_{g\ast f}  &  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\underbrace{\left(  g\ast f\right)
}_{\substack{=\mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta
_{C}\\\text{(by the definition of convolution)}}}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right) \\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)
\circ\Delta_{C}\right)  \otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right)  =T_{f}\circ T_{g}%
\end{align*}
(by (\ref{504.pf.15})). This proves Proposition~\ref{Proposition504}
\textbf{(d)}.

\textbf{(e)} By (\ref{504.pf.idid}) (applied to $P=A$, $Q=C$, $R_{1}=C$,
$R_{2}=k$, $R_{3}=A$, $\varphi=\varepsilon_{C}$ and $\psi=\eta_{A}$), we have%
\[
\left(  \operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}%
\otimes\operatorname{id}_{C}\right)  =\operatorname{id}_{A}\otimes
\underbrace{\left(  \eta_{A}\circ\varepsilon_{C}\right)  }_{=e_{C,A}}%
\otimes\operatorname{id}_{C}=\operatorname{id}_{A}\otimes e_{C,A}%
\otimes\operatorname{id}_{C}.
\]
Now, by the definition of $T_{e_{C,A}}$, we have%
\begin{align}
T_{e_{C,A}}  &  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\underbrace{\left(  \operatorname{id}_{A}\otimes e_{C,A}\otimes
\operatorname{id}_{C}\right)  }_{=\left(  \operatorname{id}_{A}\otimes\eta
_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\varepsilon_{C}\otimes\operatorname{id}_{C}\right)  }\circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right) \nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\Delta_{C}\right)  . \label{504.pf.b0}%
\end{align}


Now, consider the isomorphisms $\operatorname{kan}_{A\otimes k,A}$ and
$\operatorname{kan}_{C,k\otimes C}$ defined in
Proposition~\ref{Proposition504} \textbf{(h)}. Then, by the axioms of a
$k$-coalgebra, we have $\left(  \varepsilon_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\Delta_{C}=\operatorname{kan}_{C,k\otimes C}$ (since $C$ is
a $k$-coalgebra). Also, by the axioms of a $k$-algebra, we have $\mu_{A}%
\circ\left(  \operatorname{id}_{A}\otimes\eta_{A}\right)  =\operatorname{kan}%
_{A\otimes k,A}$ (since $A$ is a $k$-algebra).

By (\ref{(X)(X)}) (applied to $U=A\otimes k$, $V=A\otimes A$, $W=A$,
$U^{\prime}=C$, $V^{\prime}=C$, $W^{\prime}=C$, $\alpha=\operatorname{id}%
_{A}\otimes\eta_{A}$, $\beta=\mu_{A}$, $\alpha^{\prime}=\operatorname{id}_{C}$
and $\beta^{\prime}=\operatorname{id}_{C}$), we have%
\[
\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\eta_{A}\right)
\right)  \otimes\left(  \operatorname{id}_{C}\circ\operatorname{id}%
_{C}\right)  =\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  ,
\]
so that%
\begin{equation}
\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)
=\underbrace{\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\eta
_{A}\right)  \right)  }_{=\operatorname{kan}_{A\otimes k,A}}\otimes
\underbrace{\left(  \operatorname{id}_{C}\circ\operatorname{id}_{C}\right)
}_{=\operatorname{id}_{C}}=\operatorname{kan}_{A\otimes k,A}\otimes
\operatorname{id}_{C}. \label{504.pf.b1}%
\end{equation}


By (\ref{(X)(X)}) (applied to $U=A$, $V=A$, $W=A$, $U^{\prime}=C$, $V^{\prime
}=C\otimes C$, $W^{\prime}=k\otimes C$, $\alpha=\operatorname{id}_{A}$,
$\beta=\operatorname{id}_{A}$, $\alpha^{\prime}=\Delta_{C}$ and $\beta
^{\prime}=\varepsilon_{C}\otimes\operatorname{id}_{C}$), we have%
\[
\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)  \otimes\left(
\left(  \varepsilon_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta
_{C}\right)  =\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}%
\otimes\Delta_{C}\right)  ,
\]
so that%
\begin{equation}
\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
=\underbrace{\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)
}_{=\operatorname{id}_{A}}\otimes\underbrace{\left(  \left(  \varepsilon
_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}\right)
}_{=\operatorname{kan}_{C,k\otimes C}}=\operatorname{id}_{A}\otimes
\operatorname{kan}_{C,k\otimes C}. \label{504.pf.b2}%
\end{equation}


Now, (\ref{504.pf.b0}) becomes%
\begin{align*}
T_{e_{C,A}}  &  =\underbrace{\left(  \mu_{A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\eta_{A}\otimes
\operatorname{id}_{C}\right)  }_{=\operatorname{kan}_{A\otimes k,A}%
\otimes\operatorname{id}_{C}}\circ\underbrace{\left(  \operatorname{id}%
_{A}\otimes\varepsilon_{C}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right)  }_{=\operatorname{id}%
_{A}\otimes\operatorname{kan}_{C,k\otimes C}}\\
&  =\left(  \operatorname{kan}_{A\otimes k,A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{C,k\otimes C}\right)  =\operatorname{id}_{A\otimes C}%
\end{align*}
(because a very easy linear-algebraic fact says that $\left(
\operatorname{kan}_{A\otimes k,A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}_{C,k\otimes
C}\right)  =\operatorname{id}_{A\otimes C}$\ \ \ \ \footnote{\textit{Proof.}
For any $\left(  a,c\right)  \in A\times C$, we have%
\begin{align*}
\left(  \left(  \operatorname{kan}_{A\otimes k,A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{C,k\otimes C}\right)  \right)  \left(  a\otimes c\right)   &  =\left(
\operatorname{kan}_{A\otimes k,A}\otimes\operatorname{id}_{C}\right)
\underbrace{\left(  \left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{C,k\otimes C}\right)  \left(  a\otimes c\right)  \right)  }%
_{=\operatorname{id}_{A}\left(  a\right)  \otimes\operatorname{kan}%
_{C,k\otimes C}\left(  c\right)  }\\
&  =\left(  \operatorname{kan}_{A\otimes k,A}\otimes\operatorname{id}%
_{C}\right)  \left(  \underbrace{\operatorname{id}_{A}\left(  a\right)  }%
_{=a}\otimes\underbrace{\operatorname{kan}_{C,k\otimes C}\left(  c\right)
}_{\substack{=1\otimes c\\\text{(by the definition of }\operatorname{kan}%
_{C,k\otimes C}\text{)}}}\right) \\
&  =\left(  \operatorname{kan}_{A\otimes k,A}\otimes\operatorname{id}%
_{C}\right)  \left(  a\otimes1\otimes c\right)
=\underbrace{\operatorname{kan}_{A\otimes k,A}\left(  a\otimes1\right)
}_{\substack{=1a\\\text{(by the definition of }\operatorname{kan}_{A\otimes
k,A}\text{)}}}\otimes\underbrace{\operatorname{id}_{C}\left(  c\right)  }%
_{=c}\\
&  =1a\otimes c=a\otimes c=\operatorname{id}_{A\otimes C}\left(  a\otimes
c\right)  .
\end{align*}
In other words, the two $k$-linear maps $\left(  \operatorname{kan}_{A\otimes
k,A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\operatorname{kan}_{C,k\otimes C}\right)  $ and $\operatorname{id}%
_{A\otimes C}$ are equal to each other on each pure tensor in $A\otimes C$.
Thus, these two maps must be identical (since two $k$-linear maps from a
tensor product which are equal to each other on each pure tensor must
necessarily be identical). In other words, $\left(  \operatorname{kan}%
_{A\otimes k,A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\operatorname{kan}_{C,k\otimes C}\right)
=\operatorname{id}_{A\otimes C}$, qed.}). This proves
Proposition~\ref{Proposition504} \textbf{(e)}.

\textbf{(f)} Assume that $C$ is a graded $k$-coalgebra, and that $A$ is a
graded $k$-algebra, and that $f$ is a graded map.

The map $\mu_{A}$ is graded (since $A$ is a graded $k$-algebra), and so is the
map $\Delta_{C}$ (since $C$ is a graded $k$-coalgebra). Thus, the maps
$\mu_{A}$, $\operatorname{id}_{C}$, $\operatorname{id}_{A}$, $f$ and
$\Delta_{C}$ are all graded. Hence, the maps $\mu_{A}\otimes\operatorname{id}%
_{C}$, $\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}$ and
$\operatorname{id}_{A}\otimes\Delta_{C}$ are graded (since tensor products of
graded maps are always graded). Hence, the map $\left(  \mu_{A}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right)  $ is graded (since compositions of graded maps
are always graded). Since $\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
=T_{f}$, this rewrites as follows: The map $T_{f}$ is graded.
Proposition~\ref{Proposition504} \textbf{(f)} is thus proven.

\textbf{(g)} Let $f:C\rightarrow A$ be a $\ast$-invertible $k$-linear map.

Applying Proposition~\ref{Proposition504} \textbf{(d)} to $g=f^{\ast\left(
-1\right)  }$, we obtain $T_{f^{\ast\left(  -1\right)  }\ast f}=T_{f}\circ
T_{f^{\ast\left(  -1\right)  }}$. Since $f^{\ast\left(  -1\right)  }\ast
f=e_{C,A}$, this rewrites as $T_{e_{C,A}}=T_{f}\circ T_{f^{\ast\left(
-1\right)  }}$. Since $T_{e_{C,A}}=\operatorname{id}_{A\otimes C}$ (by
Proposition~\ref{Proposition504} \textbf{(e)}), this rewrites as
$\operatorname{id}_{A\otimes C}=T_{f}\circ T_{f^{\ast\left(  -1\right)  }}$.

Applying Proposition~\ref{Proposition504} \textbf{(d)} to $f^{\ast\left(
-1\right)  }$ and $f$ instead of $f$ and $g$, we obtain $T_{f\ast
f^{\ast\left(  -1\right)  }}=T_{f^{\ast\left(  -1\right)  }}\circ T_{f}$.
Since $f\ast f^{\ast\left(  -1\right)  }=e_{C,A}$, this rewrites as
$T_{e_{C,A}}=T_{f^{\ast\left(  -1\right)  }}\circ T_{f}$. Since $T_{e_{C,A}%
}=\operatorname{id}_{A\otimes C}$, this rewrites as $\operatorname{id}%
_{A\otimes C}=T_{f^{\ast\left(  -1\right)  }}\circ T_{f}$.

From $T_{f^{\ast\left(  -1\right)  }}\circ T_{f}=\operatorname{id}_{A\otimes
C}$ and $T_{f}\circ T_{f^{\ast\left(  -1\right)  }}=\operatorname{id}%
_{A\otimes C}$, we conclude that the map $T_{f}$ is invertible and satisfies
$T_{f^{\ast\left(  -1\right)  }}=\left(  T_{f}\right)  ^{-1}$.
Proposition~\ref{Proposition504} \textbf{(g)} is thus proven.

\textbf{(h)} Let $f:C\rightarrow A$ be a $k$-linear map.

By the axioms of a $k$-coalgebra, we have $\left(  \varepsilon_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}=\operatorname{kan}%
_{C,k\otimes C}$ (since $C$ is a $k$-coalgebra).

By the axioms of a $k$-algebra, we have $\mu_{A}\circ\left(  \operatorname{id}%
_{A}\otimes\eta_{A}\right)  =\operatorname{kan}_{A\otimes k,A}$ (since $A$ is
a $k$-algebra).

Let $\operatorname{kan}_{C,C\otimes k}:C\rightarrow C\otimes k$ be the
canonical isomorphism which sends every $c\in C$ to $c\otimes1\in C\otimes k$.

Every $x\in C$ satisfies%
\begin{equation}
\operatorname{kan}_{C,k\otimes C}\left(  x\right)  =1\otimes x
\label{504.pf.kan1}%
\end{equation}
(by the definition of $\operatorname{kan}_{C,k\otimes C}$).

We have $\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  =\left(  \mu
_{A}\otimes\operatorname{id}_{k}\right)  \circ\left(  \operatorname{id}%
_{A\otimes A}\otimes\varepsilon_{C}\right)  $ (by (\ref{504.pf.comm}), applied
to $P=A\otimes A$, $Q=A$, $R=C$, $S=k$, $\gamma=\mu_{A}$ and $\delta
=\varepsilon_{C}$).

We have $\left(  \operatorname{id}_{A\otimes A}\otimes\varepsilon_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}%
_{C}\right)  =\left(  \operatorname{id}_{A}\otimes f\otimes\operatorname{id}%
_{k}\right)  \circ\left(  \operatorname{id}_{A\otimes C}\otimes\varepsilon
_{C}\right)  $ (by (\ref{504.pf.comm}), applied to $P=A\otimes C$, $Q=A\otimes
A$, $R=C$, $S=k$, $\gamma=\operatorname{id}_{A}\otimes f$ and $\delta
=\varepsilon_{C}$).

By the definition of $\eta_{A}$, we have $\eta_{A}\left(  1\right)
=1\cdot1_{A}=1_{A}$.

By (\ref{(X)(X)}) (applied to $U=A$, $V=A$, $W=A$, $U^{\prime}=C$, $V^{\prime
}=C\otimes C$, $W^{\prime}=C\otimes k$, $\alpha=\operatorname{id}_{A}$,
$\beta=\operatorname{id}_{A}$, $\alpha^{\prime}=\Delta_{C}$ and $\beta
^{\prime}=\operatorname{id}_{C}\otimes\varepsilon_{C}$), we have%
\begin{align*}
\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)  \otimes\left(
\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \circ\Delta
_{C}\right)   &  =\left(  \underbrace{\operatorname{id}_{A}\otimes
\operatorname{id}_{C}}_{=\operatorname{id}_{A\otimes C}}\otimes\varepsilon
_{C}\right)  \circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right) \\
&  =\left(  \operatorname{id}_{A\otimes C}\otimes\varepsilon_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  ,
\end{align*}
so that%
\[
\left(  \operatorname{id}_{A\otimes C}\otimes\varepsilon_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)
=\underbrace{\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)
}_{=\operatorname{id}_{A}}\otimes\underbrace{\left(  \left(  \operatorname{id}%
_{C}\otimes\varepsilon_{C}\right)  \circ\Delta_{C}\right)  }%
_{\substack{=\operatorname{kan}_{C,C\otimes k}\\\text{(by the axioms of a
}k\text{-coalgebra,}\\\text{since }C\text{ is a }k\text{-coalgebra)}%
}}=\operatorname{id}_{A}\otimes\operatorname{kan}_{C,C\otimes k}.
\]


Since $T_{f}=\left(  \mu_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  $, we have%
\begin{align*}
&  \operatorname{kan}_{A\otimes k,A}\circ\left(  \operatorname{id}_{A}%
\otimes\varepsilon_{C}\right)  \circ T_{f}\circ\left(  \eta_{A}\otimes
\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes C}\\
&  =\operatorname{kan}_{A\otimes k,A}\circ\underbrace{\left(
\operatorname{id}_{A}\otimes\varepsilon_{C}\right)  \circ\left(  \mu
_{A}\otimes\operatorname{id}_{C}\right)  }_{=\left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \circ\left(  \operatorname{id}_{A\otimes
A}\otimes\varepsilon_{C}\right)  }\circ\left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{C}\right)  \circ\left(  \eta_{A}\otimes\operatorname{id}%
_{C}\right)  \circ\operatorname{kan}_{C,k\otimes C}\\
&  =\operatorname{kan}_{A\otimes k,A}\circ\left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \circ\underbrace{\left(  \operatorname{id}%
_{A\otimes A}\otimes\varepsilon_{C}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes f\otimes\operatorname{id}_{C}\right)  }_{=\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{k}\right)
\circ\left(  \operatorname{id}_{A\otimes C}\otimes\varepsilon_{C}\right)
}\circ\left(  \operatorname{id}_{A}\otimes\Delta_{C}\right)  \circ\left(
\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}%
_{C,k\otimes C}\\
&  =\operatorname{kan}_{A\otimes k,A}\circ\left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \circ\left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{k}\right)  \circ\underbrace{\left(
\operatorname{id}_{A\otimes C}\otimes\varepsilon_{C}\right)  \circ\left(
\operatorname{id}_{A}\otimes\Delta_{C}\right)  }_{=\operatorname{id}%
_{A}\otimes\operatorname{kan}_{C,C\otimes k}}\circ\left(  \eta_{A}%
\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes C}\\
&  =\operatorname{kan}_{A\otimes k,A}\circ\left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \circ\left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{k}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\operatorname{kan}_{C,C\otimes k}\right)  \circ\left(  \eta
_{A}\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes
C}.
\end{align*}
Thus, every $x\in C$ satisfies%
\begin{align*}
&  \left(  \operatorname{kan}_{A\otimes k,A}\circ\left(  \operatorname{id}%
_{A}\otimes\varepsilon_{C}\right)  \circ T_{f}\circ\left(  \eta_{A}%
\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes
C}\right)  \left(  x\right) \\
&  =\left(  \operatorname{kan}_{A\otimes k,A}\circ\left(  \mu_{A}%
\otimes\operatorname{id}_{k}\right)  \circ\left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{k}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\operatorname{kan}_{C,C\otimes k}\right)  \circ\left(  \eta
_{A}\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes
C}\right)  \left(  x\right) \\
&  =\operatorname{kan}_{A\otimes k,A}\left(  \left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \left(  \left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{k}\right)  \left(  \left(  \operatorname{id}%
_{A}\otimes\operatorname{kan}_{C,C\otimes k}\right)  \left(  \left(  \eta
_{A}\otimes\operatorname{id}_{C}\right)  \underbrace{\left(
\operatorname{kan}_{C,k\otimes C}\left(  x\right)  \right)  }%
_{\substack{=1\otimes x\\\text{(by (\ref{504.pf.kan1}))}}}\right)  \right)
\right)  \right) \\
&  =\operatorname{kan}_{A\otimes k,A}\left(  \left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \left(  \left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{k}\right)  \left(  \left(  \operatorname{id}%
_{A}\otimes\operatorname{kan}_{C,C\otimes k}\right)  \underbrace{\left(
\left(  \eta_{A}\otimes\operatorname{id}_{C}\right)  \left(  1\otimes
x\right)  \right)  }_{=\eta_{A}\left(  1\right)  \otimes\operatorname{id}%
_{C}\left(  x\right)  }\right)  \right)  \right) \\
&  =\operatorname{kan}_{A\otimes k,A}\left(  \left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \left(  \left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{k}\right)  \left(  \left(  \operatorname{id}%
_{A}\otimes\operatorname{kan}_{C,C\otimes k}\right)  \left(  \underbrace{\eta
_{A}\left(  1\right)  }_{=1_{A}}\otimes\underbrace{\operatorname{id}%
_{C}\left(  x\right)  }_{=x}\right)  \right)  \right)  \right) \\
&  =\operatorname{kan}_{A\otimes k,A}\left(  \left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \left(  \left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{k}\right)  \underbrace{\left(  \left(
\operatorname{id}_{A}\otimes\operatorname{kan}_{C,C\otimes k}\right)  \left(
1_{A}\otimes x\right)  \right)  }_{=\operatorname{id}_{A}\left(  1_{A}\right)
\otimes\operatorname{kan}_{C,C\otimes k}\left(  x\right)  }\right)  \right) \\
&  =\operatorname{kan}_{A\otimes k,A}\left(  \left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \left(  \left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{k}\right)  \left(  \underbrace{\operatorname{id}%
_{A}\left(  1_{A}\right)  }_{=1_{A}}\otimes\underbrace{\operatorname{kan}%
_{C,C\otimes k}\left(  x\right)  }_{\substack{=x\otimes1\\\text{(by the
definition of }\operatorname{kan}_{C,C\otimes k}\text{)}}}\right)  \right)
\right) \\
&  =\operatorname{kan}_{A\otimes k,A}\left(  \left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \underbrace{\left(  \left(  \operatorname{id}%
_{A}\otimes f\otimes\operatorname{id}_{k}\right)  \left(  1_{A}\otimes
x\otimes1\right)  \right)  }_{=\operatorname{id}_{A}\left(  1_{A}\right)
\otimes f\left(  x\right)  \otimes\operatorname{id}_{k}\left(  1\right)
}\right) \\
&  =\operatorname{kan}_{A\otimes k,A}\left(  \left(  \mu_{A}\otimes
\operatorname{id}_{k}\right)  \left(  \underbrace{\operatorname{id}_{A}\left(
1_{A}\right)  }_{=1_{A}}\otimes f\left(  x\right)  \otimes
\underbrace{\operatorname{id}_{k}\left(  1\right)  }_{=1}\right)  \right) \\
&  =\operatorname{kan}_{A\otimes k,A}\underbrace{\left(  \left(  \mu
_{A}\otimes\operatorname{id}_{k}\right)  \left(  1_{A}\otimes f\left(
x\right)  \otimes1\right)  \right)  }_{=\mu_{A}\left(  1_{A}\otimes f\left(
x\right)  \right)  \otimes\operatorname{id}_{k}\left(  1\right)  }\\
&  =\operatorname{kan}_{A\otimes k,A}\left(  \underbrace{\mu_{A}\left(
1_{A}\otimes f\left(  x\right)  \right)  }_{\substack{=1_{A}f\left(  x\right)
\\\text{(since }\mu_{A}\text{ is the multiplication map of }A\text{)}}%
}\otimes\underbrace{\operatorname{id}_{k}\left(  1\right)  }_{=1}\right) \\
&  =\operatorname{kan}_{A\otimes k,A}\left(  1_{A}f\left(  x\right)
,1\right)  =1\cdot1_{A}f\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\operatorname{kan}_{A\otimes k,A}\right) \\
&  =f\left(  x\right)  .
\end{align*}
Hence, $\operatorname{kan}_{A\otimes k,A}\circ\left(  \operatorname{id}%
_{A}\otimes\varepsilon_{C}\right)  \circ T_{f}\circ\left(  \eta_{A}%
\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes C}%
=f$. This proves Proposition~\ref{Proposition504} \textbf{(h)}.

\textbf{(i)} Assume that $C$ is a graded $k$-coalgebra, that $A$ is a graded
$k$-algebra, and that $f$ is a $k$-linear map such that $T_{f}$ is graded.

Consider the isomorphisms $\operatorname{kan}_{A\otimes k,A}$ and
$\operatorname{kan}_{C,k\otimes C}$ defined in
Proposition~\ref{Proposition504} \textbf{(h)}. These isomorphisms are clearly graded.

Since $C$ is a graded $k$-coalgebra, the map $\varepsilon_{C}$ is graded.
Since $A$ is a graded $k$-algebra, the map $\eta_{A}$ is graded.

Thus, the maps $\operatorname{kan}_{A\otimes k,A}$, $\operatorname{id}_{A}$,
$\varepsilon_{C}$, $T_{f}$, $\eta_{A}$, $\operatorname{id}_{C}$ and
$\operatorname{kan}_{C,k\otimes C}$ all are graded. Hence, the maps
$\operatorname{kan}_{A\otimes k,A}$, $\operatorname{id}_{A}\otimes
\varepsilon_{C}$, $T_{f}$, $\eta_{A}\otimes\operatorname{id}_{C}$,
$\operatorname{kan}_{C,k\otimes C}$ all are graded (since tensor products of
graded maps are always graded). Hence, the map $\operatorname{kan}_{A\otimes
k,A}\circ\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}\right)  \circ
T_{f}\circ\left(  \eta_{A}\otimes\operatorname{id}_{C}\right)  \circ
\operatorname{kan}_{C,k\otimes C}$ is graded (since compositions of graded
maps are always graded). Since $\operatorname{kan}_{A\otimes k,A}\circ\left(
\operatorname{id}_{A}\otimes\varepsilon_{C}\right)  \circ T_{f}\circ\left(
\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}%
_{C,k\otimes C}=f$ (by Proposition~\ref{Proposition504} \textbf{(h)}), this
rewrites as follows: The map$\ f$ is graded. Thus, we have verified
Proposition~\ref{Proposition504} \textbf{(i)}.
\end{proof}

Now that Proposition~\ref{Proposition504} is completely proven, we can fulfill
our debt of proving Proposition~\ref{Proposition501}:

\begin{proof}
[Proof of Proposition~\ref{Proposition501}.]\textbf{(a)} Let us use
Definition~\ref{Definition503}. By Proposition~\ref{Proposition504}
\textbf{(f)}, the map $T_{f}$ is graded. By Proposition~\ref{Proposition504}
\textbf{(g)}, this map $T_{f}$ is invertible. Thus, $T_{f}$ is a graded
invertible $k$-linear map. Hence, $\left(  T_{f}\right)  ^{-1}$ is also a
graded map (because it is well-known that the inverse of a graded invertible
$k$-linear map must always be graded). In other words, $T_{f^{\ast\left(
-1\right)  }}$ is a graded map (because Proposition~\ref{Proposition504}
\textbf{(g)} yields $T_{f^{\ast\left(  -1\right)  }}=\left(  T_{f}\right)
^{-1}$). Now, Proposition~\ref{Proposition504} \textbf{(i)} (applied to
$f^{\ast\left(  -1\right)  }$ instead of $f$) yields that $f^{\ast\left(
-1\right)  }$ is graded. This proves Proposition~\ref{Proposition501}
\textbf{(a)}.

\textbf{(b)} Let $n\in\mathbb{Z}$. If $n\in\mathbb{N}$, then
Proposition~\ref{Proposition501} \textbf{(b)} immediately follows from
Proposition~\ref{Proposition224} \textbf{(c)}. Hence, for the rest of the
proof of Proposition~\ref{Proposition501} \textbf{(b)}, we can WLOG assume
that $n\in\mathbb{N}$ does not hold. Assume this. Then, $n\notin\mathbb{N}$,
so that $n$ is negative, and thus $-n\in\mathbb{N}$. Hence,
Proposition~\ref{Proposition224} \textbf{(c)} (applied to $f^{\ast\left(
-1\right)  }$ and $-n$ instead of $f$ and $n$) yields that $\left(
f^{\ast\left(  -1\right)  }\right)  ^{\ast\left(  -n\right)  }$ is graded
(because Proposition~\ref{Proposition501} \textbf{(a)} shows that
$f^{\ast\left(  -1\right)  }$ is a graded map). Since $\left(  f^{\ast\left(
-1\right)  }\right)  ^{\ast\left(  -n\right)  }=f^{\ast\left(  \left(
-1\right)  \cdot\left(  -n\right)  \right)  }=f^{\ast n}$, this rewrites as
follows: The map $f^{\ast n}$ is graded. This proves
Proposition~\ref{Proposition501} \textbf{(b)}.

The proof of Proposition~\ref{Proposition501} is thus complete, and with it
Theorem~\ref{Theorem502} is proven.
\end{proof}

We have used Proposition~\ref{Proposition504} to prove
Proposition~\ref{Proposition501}. This, however, is not its only application.
For a different application, let us show an alternative proof of
Proposition~\ref{Proposition45} with its help. First, a lemma:

\begin{lemma}
\label{Lemma511} Let $k$ be a field. Let $C$ be a cocommutative $k$-coalgebra.
Then, $\Delta_{C}$ is a $k$-coalgebra homomorphism from $C$ to $C\otimes C$,
and $\varepsilon_{C}$ is a $k$-coalgebra homomorphism from $C$ to $k$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma511}.]Lemma~\ref{lem.coalg.epsi-hom} \textbf{(b)}
shows that $\varepsilon_{C}$ is a $k$-coalgebra homomorphism from $C$ to $k$.
Hence, the only thing that remains to be done in order to prove
Lemma~\ref{Lemma511} is to show that $\Delta_{C}$ is a $k$-coalgebra
homomorphism from $C$ to $C\otimes C$.

By the definition of the $k$-coalgebra $C\otimes C$, we have $\Delta_{C\otimes
C}=\left(  \operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  $. Thus,%
\[
\Delta_{C\otimes C}\circ\Delta_{C}=\left(  \operatorname{id}_{C}\otimes
\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \circ\Delta_{C}=\left(  \Delta_{C}\otimes\Delta
_{C}\right)  \circ\Delta_{C}%
\]
(by Lemma~\ref{Lemma36}). Also, $\varepsilon_{C\otimes C}\circ\Delta
_{C}=\varepsilon_{C}$\ \ \ \ \footnote{\textit{Proof.}
Lemma~\ref{lem.coalg.epsi-hom} \textbf{(a)} shows that every $x\in C$
satisfies $\left(  \left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)
\circ\Delta_{C}\right)  \left(  x\right)  =\varepsilon_{C}\left(  x\right)
1\otimes1$.
\par
By the definition of the $k$-coalgebra $C\otimes C$, we have $\varepsilon
_{C\otimes C}=\operatorname{kan}_{k\otimes k,k}\circ\left(  \varepsilon
_{C}\otimes\varepsilon_{C}\right)  $, where $\operatorname{kan}_{k\otimes
k,k}:k\otimes k\rightarrow k$ is the canonical isomorphism which sends every
$\lambda\otimes\lambda^{\prime}\in k\otimes k$ to $\lambda\lambda^{\prime}\in
k$. Now, for every $x\in C$, we have
\begin{align*}
\left(  \underbrace{\varepsilon_{C\otimes C}}_{=\operatorname{kan}_{k\otimes
k,k}\circ\left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)  }\circ
\Delta_{C}\right)  \left(  x\right)   &  =\left(  \operatorname{kan}_{k\otimes
k,k}\circ\left(  \varepsilon_{C}\otimes\varepsilon_{C}\right)  \circ\Delta
_{C}\left(  x\right)  \right)  =\operatorname{kan}_{k\otimes k,k}%
\underbrace{\left(  \left(  \left(  \varepsilon_{C}\otimes\varepsilon
_{C}\right)  \circ\Delta_{C}\right)  \left(  x\right)  \right)  }%
_{=\varepsilon_{C}\left(  x\right)  1\otimes1}\\
&  =\operatorname{kan}_{k\otimes k,k}\left(  \varepsilon_{C}\left(  x\right)
1\otimes1\right)  =\varepsilon_{C}\left(  x\right)  1\cdot
1\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname{kan}%
_{k\otimes k,k}\right) \\
&  =\varepsilon_{C}\left(  x\right)  .
\end{align*}
In other words, $\varepsilon_{C\otimes C}\circ\Delta_{C}=\varepsilon_{C}$,
qed.}.

We have thus proven that $\Delta_{C\otimes C}\circ\Delta_{C}=\left(
\Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}$ and $\varepsilon
_{C\otimes C}\circ\Delta_{C}=\varepsilon_{C}$. These two equations combined
yield that $\Delta_{C}$ is a $k$-coalgebra homomorphism from $C$ to $C\otimes
C$. This completes the proof of Lemma~\ref{Lemma511}.
\end{proof}

\begin{proof}
[Alternative proof of Proposition~\ref{Proposition45}.]Applying
Definition~\ref{Definition503} to $A=H$, we obtain a map $T_{f}:H\otimes
C\rightarrow H\otimes C$. Similarly, we obtain maps $T_{g}:H\otimes
C\rightarrow H\otimes C$ and $T_{f\ast g}:H\otimes C\rightarrow H\otimes C$.
Proposition~\ref{Proposition504} \textbf{(d)} (applied to $H$, $g$ and $f$
instead of $A$, $f$ and $g$) yields $T_{f\ast g}=T_{g}\circ T_{f}$.

By Lemma~\ref{Lemma511}, we know that $\Delta_{C}$ and $\varepsilon_{C}$ are
$k$-coalgebra homomorphisms.

Also, $H$ is a $k$-bialgebra, so that $\mu_{H}$ and $\eta_{H}$ are
$k$-coalgebra homomorphisms (by the axioms of a bialgebra).

Since $\mu_{H}$ and $\operatorname{id}_{C}$ are $k$-coalgebra homomorphisms,
the map $\mu_{H}\otimes\operatorname{id}_{C}$ is also a $k$-coalgebra
homomorphism (because a tensor product of $k$-coalgebra homomorphisms must
always be a $k$-coalgebra homomorphism).

Since $\operatorname{id}_{H}$, $f$ and $\operatorname{id}_{C}$ are
$k$-coalgebra homomorphisms, the map $\operatorname{id}_{H}\otimes
f\otimes\operatorname{id}_{C}$ is also a $k$-coalgebra homomorphism (because a
tensor product of $k$-coalgebra homomorphisms must always be a $k$-coalgebra homomorphism).

Since $\operatorname{id}_{H}$ and $\Delta_{C}$ are $k$-coalgebra
homomorphisms, the map $\operatorname{id}_{H}\otimes\Delta_{C}$ is also a
$k$-coalgebra homomorphism (because a tensor product of $k$-coalgebra
homomorphisms must always be a $k$-coalgebra homomorphism).

Since $\mu_{H}\otimes\operatorname{id}_{C}$, $\operatorname{id}_{H}\otimes
f\otimes\operatorname{id}_{C}$ and $\operatorname{id}_{H}\otimes\Delta_{C}$
are $k$-coalgebra homomorphisms, the map $\left(  \mu_{H}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{H}\otimes
f\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{H}\otimes\Delta_{C}\right)  $ is also a $k$-coalgebra homomorphism (because
a composition of $k$-coalgebra homomorphisms must always be a $k$-coalgebra homomorphism).

By the definition of $T_{f}$, we have%
\[
T_{f}=\left(  \mu_{H}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{H}\otimes f\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{H}\otimes\Delta_{C}\right)  .
\]
Thus, $T_{f}$ is a $k$-coalgebra homomorphism (because $\left(  \mu_{H}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{H}\otimes
f\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{H}\otimes\Delta_{C}\right)  $ is a $k$-coalgebra homomorphism). Similarly,
$T_{g}$ is a $k$-coalgebra homomorphism.

Since $T_{g}$ and $T_{f}$ are $k$-coalgebra homomorphisms, the map $T_{g}\circ
T_{f}$ is also a $k$-coalgebra homomorphism (because a composition of
$k$-coalgebra homomorphisms must always be a $k$-coalgebra homomorphism).
Since $T_{g}\circ T_{f}=T_{f\ast g}$, this rewrites as follows: The map
$T_{f\ast g}$ is a $k$-coalgebra homomorphism.

Let $\operatorname{kan}_{C,k\otimes C}:C\rightarrow k\otimes C$ be the
canonical isomorphism which sends $c$ to $1\otimes c$ for every $c\in C$. Let
$\operatorname{kan}_{H\otimes k,H}:H\otimes k\rightarrow H$ be the canonical
isomorphism which sends $a\otimes\lambda$ to $\lambda a$ for every $\left(
a,\lambda\right)  \in H\times k$. Both $\operatorname{kan}_{C,k\otimes C}$ and
$\operatorname{kan}_{H\otimes k,H}$ are $k$-coalgebra homomorphisms.

Since $\operatorname{id}_{H}$ and $\varepsilon_{C}$ are $k$-coalgebra
homomorphisms, the map $\operatorname{id}_{H}\otimes\varepsilon_{C}$ is also a
$k$-coalgebra homomorphism (because a tensor product of $k$-coalgebra
homomorphisms must always be a $k$-coalgebra homomorphism).

Since $\eta_{H}$ and $\operatorname{id}_{C}$ are $k$-coalgebra homomorphisms,
the map $\eta_{H}\otimes\operatorname{id}_{C}$ is also a $k$-coalgebra
homomorphism (because a tensor product of $k$-coalgebra homomorphisms must
always be a $k$-coalgebra homomorphism).

Since $\operatorname{kan}_{H\otimes k,H}$, $\operatorname{id}_{H}%
\otimes\varepsilon_{C}$, $T_{f\ast g}$, $\eta_{H}\otimes\operatorname{id}_{C}$
and $\operatorname{kan}_{C,k\otimes C}$ are $k$-coalgebra homomorphisms, the
map $\operatorname{kan}_{H\otimes k,H}\circ\left(  \operatorname{id}%
_{H}\otimes\varepsilon_{C}\right)  \circ T_{f\ast g}\circ\left(  \eta
_{H}\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes
C}$ is also a $k$-coalgebra homomorphism (because a composition of
$k$-coalgebra homomorphisms must always be a $k$-coalgebra homomorphism).

Proposition~\ref{Proposition504} \textbf{(h)} (applied to $H$ and $f\ast g$
instead of $A$ and $f$) yields%
\[
\operatorname{kan}_{H\otimes k,H}\circ\left(  \operatorname{id}_{H}%
\otimes\varepsilon_{C}\right)  \circ T_{f\ast g}\circ\left(  \eta_{H}%
\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes
C}=f\ast g.
\]
Thus, $f\ast g$ is a $k$-coalgebra homomorphism (since we know that
$\operatorname{kan}_{H\otimes k,H}\circ\left(  \operatorname{id}_{H}%
\otimes\varepsilon_{C}\right)  \circ T_{f\ast g}\circ\left(  \eta_{H}%
\otimes\operatorname{id}_{C}\right)  \circ\operatorname{kan}_{C,k\otimes C}$
is a $k$-coalgebra homomorphism). Proposition~\ref{Proposition45} is thus proven.
\end{proof}

A similar argument allows us to construct a new proof of
Proposition~\ref{Proposition135}. The analogue of Lemma~\ref{Lemma511} now
takes the following form:

\begin{lemma}
\label{Lemma512} Let $k$ be a field. Let $A$ be a commutative $k$-algebra.
Then, $\mu_{A}$ is a $k$-algebra homomorphism from $A\otimes A$ to $A$, and
$\eta_{A}$ is a $k$-algebra homomorphism from $A$ to $k$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma512}.]Every $a\in A$, $b\in A$, $c\in A$ and $d\in
A$ satisfy%
\begin{align*}
&  \left(  \mu_{A}\circ\mu_{A\otimes A}\right)  \left(  a\otimes b\otimes
c\otimes d\right) \\
&  =\mu_{A}\underbrace{\left(  \mu_{A\otimes A}\left(  a\otimes b\otimes
c\otimes d\right)  \right)  }_{\substack{=\left(  a\otimes b\right)
\cdot\left(  c\otimes d\right)  \\\text{(by the definition of }\mu_{A\otimes
A}\text{)}}}=\mu_{A}\underbrace{\left(  \left(  a\otimes b\right)
\cdot\left(  c\otimes d\right)  \right)  }_{\substack{=ac\otimes bd\\\text{(by
the definition of the }k\text{-algebra }A\otimes A\text{)}}}\\
&  =\mu_{A}\left(  ac\otimes bd\right)  =\left(  ac\right)  \left(  bd\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu_{A}\right) \\
&  =acbd=abcd\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A\text{ is
commutative}\right)
\end{align*}
and%
\begin{align*}
&  \left(  \mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)  \right)  \left(
a\otimes b\otimes c\otimes d\right) \\
&  =\mu_{A}\underbrace{\left(  \left(  \mu_{A}\otimes\mu_{A}\right)  \left(
a\otimes b\otimes c\otimes d\right)  \right)  }_{=\mu_{A}\left(  a\otimes
b\right)  \otimes\mu_{A}\left(  c\otimes d\right)  }=\mu_{A}\left(
\underbrace{\mu_{A}\left(  a\otimes b\right)  }_{\substack{=ab\\\text{(by the
definition of }\mu_{A}\text{)}}}\otimes\underbrace{\mu_{A}\left(  c\otimes
d\right)  }_{\substack{=cd\\\text{(by the definition of }\mu_{A}\text{)}%
}}\right) \\
&  =\mu_{A}\left(  ab\otimes cd\right)  =\left(  ab\right)  \left(  cd\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu_{A}\right)  .
\end{align*}
Thus, every $a\in A$, $b\in A$, $c\in A$ and $d\in A$ satisfy%
\begin{align*}
&  \left(  \mu_{A}\circ\mu_{A\otimes A}\right)  \left(  a\otimes b\otimes
c\otimes d\right) \\
&  =abcd=\left(  ab\right)  \left(  cd\right)  =\left(  \mu_{A}\circ\left(
\mu_{A}\otimes\mu_{A}\right)  \right)  \left(  a\otimes b\otimes c\otimes
d\right)  .
\end{align*}
In other words, the two maps $\mu_{A}\circ\mu_{A\otimes A}$ and $\mu_{A}%
\circ\left(  \mu_{A}\otimes\mu_{A}\right)  $ are equal on every pure tensor.
But since these two maps are $k$-linear, this yields that these maps must be
identic (because whenever two $k$-linear maps from a tensor product are equal
on every pure tensor, they must be identic). In other words, $\mu_{A}\circ
\mu_{A\otimes A}=\mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)  $.

Besides, by the definition of the $k$-algebra $A\otimes A$, we have
$1_{A\otimes A}=1_{A}\otimes1_{A}$, so that
\begin{align*}
\mu_{A}\left(  1_{A\otimes A}\right)   &  =\mu_{A}\left(  1_{A}\otimes
1_{A}\right)  =1_{A}\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mu_{A}\right) \\
&  =1_{A}.
\end{align*}
Now, every $\lambda\in k$ satisfies%
\begin{align*}
\left(  \mu_{A}\circ\eta_{A\otimes A}\right)  \left(  \lambda\right)   &
=\mu_{A}\underbrace{\left(  \eta_{A\otimes A}\left(  \lambda\right)  \right)
}_{\substack{=\lambda\cdot1_{A\otimes A}\\\text{(by the definition of }%
\eta_{A\otimes A}\text{)}}}=\mu_{A}\left(  \lambda\cdot1_{A\otimes A}\right)
\\
&  =\lambda\cdot\underbrace{\mu_{A}\left(  1_{A\otimes A}\right)  }_{=1_{A}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{A}\text{ is }k\text{-linear}%
\right) \\
&  =\lambda\cdot1_{A}=\eta_{A}\left(  \lambda\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{A}\left(  \lambda\right)
=\lambda\cdot1_{A}\text{ by the definition of }\eta_{A}\right)  .
\end{align*}
In other words, $\mu_{A}\circ\eta_{A\otimes A}=\eta_{A}$. Combined with
$\mu_{A}\circ\mu_{A\otimes A}=\mu_{A}\circ\left(  \mu_{A}\otimes\mu
_{A}\right)  $ (which we proved above), this yields that $\mu_{A}$ is a
$k$-algebra homomorphism (because $\mu_{A}$ is a $k$-algebra homomorphism if
and only if it satisfies $\mu_{A}\circ\eta_{A\otimes A}=\eta_{A}$ and $\mu
_{A}\circ\mu_{A\otimes A}=\mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)  $
(due to the definition of $k$-algebra homomorphisms using arrows)).

Now, it remains to prove that $\eta_{A}$ is a $k$-algebra homomorphism.

Any $x\in k$ and $y\in k$ satisfy%
\[
\underbrace{\eta_{A}\left(  x\right)  }_{\substack{=x\cdot1_{A}\\\text{(by the
definition}\\\text{of }\eta_{A}\text{)}}}\cdot\underbrace{\eta_{A}\left(
y\right)  }_{\substack{=y\cdot1_{A}\\\text{(by the definition}\\\text{of }%
\eta_{A}\text{)}}}=\left(  x\cdot1_{A}\right)  \cdot\left(  y\cdot
1_{A}\right)  =xy\cdot1_{A}=\eta_{A}\left(  xy\right)
\]
(since $\eta_{A}\left(  xy\right)  =xy\cdot1_{A}$ (by the definition of
$\eta_{A}$)). Combined with the fact that $\eta_{A}\left(  1\right)  =1_{A}$
(because the definition of $\eta_{A}$ yields $\eta_{A}\left(  1\right)
=1\cdot1_{A}=1_{A}$), this yields that $\eta_{A}$ is a $k$-algebra
homomorphism. This completes the proof of Lemma~\ref{Lemma512}.
\end{proof}

\begin{proof}
[Alternative proof of Proposition~\ref{Proposition135}.]Applying
Definition~\ref{Definition503} to $C=H$, we obtain a map $T_{f}:A\otimes
H\rightarrow A\otimes H$. Similarly, we obtain maps $T_{g}:A\otimes
H\rightarrow A\otimes H$ and $T_{f\ast g}:A\otimes H\rightarrow A\otimes H$.
Proposition~\ref{Proposition504} \textbf{(d)} (applied to $H$, $g$ and $f$
instead of $C$, $f$ and $g$) yields $T_{f\ast g}=T_{g}\circ T_{f}$.

Since $H$ is a $k$-bialgebra, the maps $\Delta_{H}$ and $\varepsilon_{H}$ are
$k$-algebra homomorphisms (by the axioms of a bialgebra).

By Lemma~\ref{Lemma512}, the maps $\mu_{A}$ and $\eta_{A}$ are $k$-algebra homomorphisms.

Since $\mu_{A}$ and $\operatorname{id}_{H}$ are $k$-algebra homomorphisms, the
map $\mu_{A}\otimes\operatorname{id}_{H}$ is also a $k$-algebra homomorphism
(because a tensor product of $k$-algebra homomorphisms must always be a
$k$-algebra homomorphism).

Since $\operatorname{id}_{A}$, $f$ and $\operatorname{id}_{H}$ are $k$-algebra
homomorphisms, the map $\operatorname{id}_{A}\otimes f\otimes\operatorname{id}%
_{H}$ is also a $k$-algebra homomorphism (because a tensor product of
$k$-algebra homomorphisms must always be a $k$-algebra homomorphism).

Since $\operatorname{id}_{A}$ and $\Delta_{H}$ are $k$-algebra homomorphisms,
the map $\operatorname{id}_{A}\otimes\Delta_{H}$ is also a $k$-algebra
homomorphism (because a tensor product of $k$-algebra homomorphisms must
always be a $k$-algebra homomorphism).

Since $\mu_{A}\otimes\operatorname{id}_{H}$, $\operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{H}$ and $\operatorname{id}_{A}\otimes\Delta_{H}$
are $k$-algebra homomorphisms, the map $\left(  \mu_{A}\otimes
\operatorname{id}_{H}\right)  \circ\left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{H}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{H}\right)  $ is also a $k$-algebra homomorphism (because a
composition of $k$-algebra homomorphisms must always be a $k$-algebra homomorphism).

By the definition of $T_{f}$, we have%
\[
T_{f}=\left(  \mu_{A}\otimes\operatorname{id}_{H}\right)  \circ\left(
\operatorname{id}_{A}\otimes f\otimes\operatorname{id}_{H}\right)
\circ\left(  \operatorname{id}_{A}\otimes\Delta_{H}\right)  .
\]
Thus, $T_{f}$ is a $k$-algebra homomorphism (because $\left(  \mu_{A}%
\otimes\operatorname{id}_{H}\right)  \circ\left(  \operatorname{id}_{A}\otimes
f\otimes\operatorname{id}_{H}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\Delta_{H}\right)  $ is a $k$-algebra homomorphism). Similarly,
$T_{g}$ is a $k$-algebra homomorphism.

Since $T_{g}$ and $T_{f}$ are $k$-algebra homomorphisms, the map $T_{g}\circ
T_{f}$ is also a $k$-algebra homomorphism (because a composition of
$k$-algebra homomorphisms must always be a $k$-algebra homomorphism). Since
$T_{g}\circ T_{f}=T_{f\ast g}$, this rewrites as follows: The map $T_{f\ast
g}$ is a $k$-algebra homomorphism.

Let $\operatorname{kan}_{H,k\otimes H}:H\rightarrow k\otimes H$ be the
canonical isomorphism which sends $c$ to $1\otimes c$ for every $c\in H$. Let
$\operatorname{kan}_{A\otimes k,A}:A\otimes k\rightarrow A$ be the canonical
isomorphism which sends $a\otimes\lambda$ to $\lambda a$ for every $\left(
a,\lambda\right)  \in A\times k$. Both $\operatorname{kan}_{H,k\otimes H}$ and
$\operatorname{kan}_{A\otimes k,A}$ are $k$-algebra homomorphisms.

Since $\operatorname{id}_{A}$ and $\varepsilon_{H}$ are $k$-algebra
homomorphisms, the map $\operatorname{id}_{A}\otimes\varepsilon_{H}$ is also a
$k$-algebra homomorphism (because a tensor product of $k$-algebra
homomorphisms must always be a $k$-algebra homomorphism).

Since $\eta_{A}$ and $\operatorname{id}_{H}$ are $k$-algebra homomorphisms,
the map $\eta_{A}\otimes\operatorname{id}_{H}$ is also a $k$-algebra
homomorphism (because a tensor product of $k$-algebra homomorphisms must
always be a $k$-algebra homomorphism).

Since $\operatorname{kan}_{A\otimes k,A}$, $\operatorname{id}_{A}%
\otimes\varepsilon_{H}$, $T_{f\ast g}$, $\eta_{A}\otimes\operatorname{id}_{H}$
and $\operatorname{kan}_{H,k\otimes H}$ are $k$-algebra homomorphisms, the map
$\operatorname{kan}_{A\otimes k,A}\circ\left(  \operatorname{id}_{A}%
\otimes\varepsilon_{H}\right)  \circ T_{f\ast g}\circ\left(  \eta_{A}%
\otimes\operatorname{id}_{H}\right)  \circ\operatorname{kan}_{H,k\otimes H}$
is also a $k$-algebra homomorphism (because a composition of $k$-algebra
homomorphisms must always be a $k$-algebra homomorphism).

Proposition~\ref{Proposition504} \textbf{(h)} (applied to $H$ and $f\ast g$
instead of $C$ and $f$) yields%
\[
\operatorname{kan}_{A\otimes k,A}\circ\left(  \operatorname{id}_{A}%
\otimes\varepsilon_{H}\right)  \circ T_{f\ast g}\circ\left(  \eta_{A}%
\otimes\operatorname{id}_{H}\right)  \circ\operatorname{kan}_{H,k\otimes
H}=f\ast g.
\]
Thus, $f\ast g$ is a $k$-algebra homomorphism (since we know that
$\operatorname{kan}_{A\otimes k,A}\circ\left(  \operatorname{id}_{A}%
\otimes\varepsilon_{H}\right)  \circ T_{f\ast g}\circ\left(  \eta_{A}%
\otimes\operatorname{id}_{H}\right)  \circ\operatorname{kan}_{H,k\otimes H}$
is a $k$-algebra homomorphism). Proposition~\ref{Proposition135} is thus proven.
\end{proof}

\section{\label{S21}A graded comultiplication makes a coalgebra graded}

Next we discuss another elementary property of coalgebras which has not much
to do with what we did above. We will prove the following:

\begin{theorem}
\label{Theorem531} Let $k$ be a field. Let $C$ be a $k$-coalgebra which is, at
the same time, a graded $k$-vector space such that the underlying vector space
of the $k$-coalgebra $C$ is identical with the underlying vector space of the
graded $k$-vector space $C$. Assume that the map $\Delta_{C}:C\rightarrow
C\otimes C$ is graded. Then, $C$ is a graded $k$-coalgebra (i. e., the map
$\varepsilon_{C}:C\rightarrow k$ is also graded).
\end{theorem}

This is an analogue of the following well-known result about algebras:

\begin{theorem}
\label{Theorem532} Let $k$ be a field. Let $A$ be a $k$-algebra which is, at
the same time, a graded $k$-vector space such that the underlying vector space
of the $k$-algebra $A$ is identical with the underlying vector space of the
graded $k$-vector space $A$. Assume that $A_{i}A_{j}\subseteq A_{i+j}$ for all
$i\in\mathbb{N}$ and $j\in\mathbb{N}$. Then, $A$ is a graded $k$-algebra (i.
e., we have $1_{A}\in A_{0}$).
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem531}.]Consider the $k$-algebra $k$. Clearly,
$\eta_{k}=\operatorname{id}_{k}$, and the map $\mu_{k}$ is the canonical
isomorphism $k\otimes k\rightarrow k$.

By applying Definition~\ref{Definition6} to $A=k$, we obtain a convolution
algebra $\left(  \mathcal{L}\left(  C,k\right)  ,\ast\right)  $. The unity of
this $k$-algebra is $e_{k,C}=\underbrace{\eta_{k}}_{=\operatorname{id}_{k}%
}\circ\varepsilon_{C}=\varepsilon_{C}$.

Let us use the notations of Definition~\ref{Definition221}. Clearly,
$\varepsilon_{C}\circ p_{0,C}:C\rightarrow k$ is also an element of the
convolution algebra $\left(  \mathcal{L}\left(  C,k\right)  ,\ast\right)  $.
Thus, $\varepsilon_{C}\ast\left(  \varepsilon_{C}\circ p_{0,C}\right)
=\varepsilon_{C}\circ p_{0,C}$ (since $\varepsilon_{C}$ is the unity of the
$k$-algebra $\left(  \mathcal{L}\left(  C,k\right)  ,\ast\right)  $).

We will now prove that $\varepsilon_{C}\circ p_{0,C}=\varepsilon_{C}$. This
will very easily yield that $\varepsilon_{C}$ is graded.

Let $\ell\in\mathbb{N}$ be arbitrary. Let $x\in C_{\ell}$. As in the proof of
Proposition~\ref{Proposition206}, we can show that (\ref{206.pf.2}) holds, i.
e., that $\Delta_{C}\left(  C_{\ell}\right)  \subseteq\sum_{i=0}^{\ell}%
C_{i}\otimes C_{\ell-i}$. Since $x\in C_{\ell}$, we now have $\Delta
_{C}\left(  x\right)  \in\Delta_{C}\left(  C_{\ell}\right)  \subseteq
\sum_{i=0}^{\ell}C_{i}\otimes C_{\ell-i}$. Hence, there exists an $\left(
\ell+1\right)  $-tuple of $\left(  t_{0},t_{1},\ldots,t_{\ell}\right)  $ of
elements of $C\otimes C$ such that $\Delta_{C}\left(  x\right)  =\sum
_{i=0}^{\ell}t_{i}$ and such that every $i\in\left\{  0,1,\ldots,\ell\right\}
$ satisfies $t_{i}\in C_{i}\otimes C_{\ell-i}$. Consider this $\left(
\ell+1\right)  $-tuple.

For every $i\in\left\{  0,1,\ldots,\ell-1\right\}  $, we have $p_{0,C}\left(
C_{\ell-i}\right)  =0$\ \ \ \ \footnote{\textit{Proof.} For every
$i\in\left\{  0,1,\ldots,\ell-1\right\}  $, we have $i\neq\ell$, thus
$0\neq\ell-i$, thus $p_{0,C}\left(  C_{\ell-i}\right)  =\left(  p_{0,C}%
\mid_{C_{\ell-i}}\right)  \left(  C_{\ell-i}\right)  =0$ (since (\ref{221.0})
(applied to $V=C$, $n=0$ and $m=\ell-i$) yields $p_{0,C}\mid_{C_{\ell-i}}=0$),
qed.} and thus%
\begin{align*}
\left(  \operatorname{id}_{C}\otimes p_{0,C}\right)  \left(  t_{i}\right)   &
\in\left(  \operatorname{id}_{C}\otimes p_{0,C}\right)  \left(  C_{i}\otimes
C_{\ell-i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }t_{i}\in
C_{i}\otimes C_{\ell-i}\right) \\
&  =\operatorname{id}_{C}\left(  C_{i}\right)  \otimes\underbrace{p_{0,C}%
\left(  C_{\ell-i}\right)  }_{=0}=0,
\end{align*}
so that $\left(  \operatorname{id}_{C}\otimes p_{0,C}\right)  \left(
t_{i}\right)  =0$.

On the other hand, it is easy to see that $\left(  p_{0,C}-\operatorname{id}%
_{C}\right)  \left(  C_{0}\right)  =0$\ \ \ \ \footnote{\textit{Proof.} Since
$p_{0,C}\mid_{C_{0}}=\operatorname{id}_{C}\mid_{C_{0}}$ (by (\ref{221.id}),
applied to $V=C$ and $n=0$), we have%
\[
\left(  p_{0,C}-\operatorname{id}_{C}\right)  \mid_{C_{0}}=\underbrace{p_{0,C}%
\mid_{C_{0}}}_{=\operatorname{id}_{C}\mid_{C_{0}}}-\operatorname{id}_{C}%
\mid_{C_{0}}=\operatorname{id}_{C}\mid_{C_{0}}-\operatorname{id}_{C}%
\mid_{C_{0}}=0,
\]
so that $\left(  p_{0,C}-\operatorname{id}_{C}\right)  \left(  C_{0}\right)
=\underbrace{\left(  \left(  p_{0,C}-\operatorname{id}_{C}\right)  \mid
_{C_{0}}\right)  }_{=0}\left(  C_{0}\right)  =0$, qed.}, so that%
\begin{align*}
\left(  \operatorname{id}_{C}\otimes\left(  p_{0,C}-\operatorname{id}%
_{C}\right)  \right)  \left(  t_{\ell}\right)   &  \in\left(
\operatorname{id}_{C}\otimes\left(  p_{0,C}-\operatorname{id}_{C}\right)
\right)  \left(  C_{\ell}\otimes\underbrace{C_{\ell-\ell}}_{=C_{0}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }t_{\ell}\in C_{\ell}\otimes C_{\ell-\ell}\text{ (because every}\\
i\in\left\{  0,1,\ldots,\ell-1\right\}  \text{ satisfies }t_{i}\in
C_{i}\otimes C_{\ell-i}\text{)}%
\end{array}
\right) \\
&  =\left(  \operatorname{id}_{C}\otimes\left(  p_{0,C}-\operatorname{id}%
_{C}\right)  \right)  \left(  C_{\ell}\otimes C_{0}\right)  =\operatorname{id}%
_{C}\left(  C_{\ell}\right)  \otimes\underbrace{\left(  p_{0,C}%
-\operatorname{id}_{C}\right)  \left(  C_{0}\right)  }_{=0}=0,
\end{align*}
so that $\left(  \operatorname{id}_{C}\otimes\left(  p_{0,C}-\operatorname{id}%
_{C}\right)  \right)  \left(  t_{\ell}\right)  =0$.

From $\Delta_{C}\left(  x\right)  =\sum_{i=0}^{\ell}t_{i}$, we conclude that
\begin{align}
\left(  \operatorname{id}_{C}\otimes p_{0,C}\right)  \left(  \Delta_{C}\left(
x\right)  \right)   &  =\left(  \operatorname{id}_{C}\otimes p_{0,C}\right)
\left(  \sum_{i=0}^{\ell}t_{i}\right)  =\sum_{i=0}^{\ell}\left(
\operatorname{id}_{C}\otimes p_{0,C}\right)  \left(  t_{i}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname{id}_{C}\otimes
p_{0,C}\text{ is }k\text{-linear}\right) \nonumber\\
&  =\sum_{i=0}^{\ell-1}\underbrace{\left(  \operatorname{id}_{C}\otimes
p_{0,C}\right)  \left(  t_{i}\right)  }_{\substack{=0\\\text{(since we know
that }\left(  \operatorname{id}_{C}\otimes p_{0,C}\right)  \left(
t_{i}\right)  =0\\\text{for every }i\in\left\{  0,1,\ldots,\ell-1\right\}
\text{)}}}+\left(  \operatorname{id}_{C}\otimes\underbrace{p_{0,C}%
}_{=\operatorname{id}_{C}+\left(  p_{0,C}-\operatorname{id}_{C}\right)
}\right)  \left(  t_{\ell}\right) \nonumber\\
&  =\underbrace{\sum_{i=0}^{\ell-1}0}_{=0}+\left(  \operatorname{id}%
_{C}\otimes\left(  \operatorname{id}_{C}+\left(  p_{0,C}-\operatorname{id}%
_{C}\right)  \right)  \right)  \left(  t_{\ell}\right) \nonumber\\
&  =\underbrace{\left(  \operatorname{id}_{C}\otimes\left(  \operatorname{id}%
_{C}+\left(  p_{0,C}-\operatorname{id}_{C}\right)  \right)  \right)
}_{\substack{=\operatorname{id}_{C}\otimes\operatorname{id}_{C}%
+\operatorname{id}_{C}\otimes\left(  p_{0,C}-\operatorname{id}_{C}\right)
\\\text{(since tensoring of }k\text{-linear maps is }k\text{-bilinear)}%
}}\left(  t_{\ell}\right) \nonumber\\
&  =\left(  \underbrace{\operatorname{id}_{C}\otimes\operatorname{id}_{C}%
}_{=\operatorname{id}_{C\otimes C}}+\operatorname{id}_{C}\otimes\left(
p_{0,C}-\operatorname{id}_{C}\right)  \right)  \left(  t_{\ell}\right)
=\left(  \operatorname{id}_{C\otimes C}+\operatorname{id}_{C}\otimes\left(
p_{0,C}-\operatorname{id}_{C}\right)  \right)  \left(  t_{\ell}\right)
\nonumber\\
&  =\underbrace{\operatorname{id}_{C\otimes C}\left(  t_{\ell}\right)
}_{=t_{\ell}}+\underbrace{\left(  \operatorname{id}_{C}\otimes\left(
p_{0,C}-\operatorname{id}_{C}\right)  \right)  \left(  t_{\ell}\right)  }%
_{=0}=t_{\ell}+0=t_{\ell}. \label{531.pf.0}%
\end{align}


On the other hand, let $\operatorname*{kan}$ denote the canonical isomorphism
$C\otimes k\rightarrow C$ which sends $c\otimes\lambda$ to $\lambda c$ for
every $\left(  c,\lambda\right)  \in C\times k$. Then, clearly,
$\operatorname*{kan}\nolimits^{-1}$ is the isomorphism $C\rightarrow C\otimes
k$ which sends every $\xi\in C$ to $\xi\otimes1$.

Every $i\in\left\{  0,1,\ldots,\ell\right\}  $ satisfies%
\begin{align*}
\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \left(
t_{i}\right)   &  \in\left(  \operatorname{id}_{C}\otimes\varepsilon
_{C}\right)  \left(  C_{i}\otimes C_{\ell-i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }t_{i}\in C_{i}\otimes C_{\ell
-i}\right) \\
&  =\underbrace{\operatorname{id}_{C}\left(  C_{i}\right)  }_{=C_{i}}%
\otimes\underbrace{\varepsilon_{C}\left(  C_{\ell-i}\right)  }_{\subseteq
k}\subseteq C_{i}\otimes k=\left(  C\otimes k\right)  _{i},
\end{align*}
and thus%
\begin{equation}
\operatorname*{kan}\left(  \left(  \operatorname{id}_{C}\otimes\varepsilon
_{C}\right)  \left(  t_{i}\right)  \right)  \in\operatorname*{kan}\left(
\left(  C\otimes k\right)  _{i}\right)  \subseteq C_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{kan}\text{ is a graded
map}\right)  . \label{531.pf.1}%
\end{equation}


But every $i\in\left\{  0,1,\ldots,\ell-1\right\}  $ satisfies $p_{\ell
,C}\left(  C_{i}\right)  =0$\ \ \ \ \footnote{\textit{Proof.} For every
$i\in\left\{  0,1,\ldots,\ell-1\right\}  $, we have $\ell\neq i$, and thus
$p_{\ell,C}\left(  C_{i}\right)  =\left(  p_{\ell,C}\mid_{C_{i}}\right)
\left(  C_{i}\right)  =0$ (since (\ref{221.0}) (applied to $V=C$, $n=\ell$ and
$m=i$) yields $p_{\ell,C}\mid_{C_{i}}=0$), qed.} and thus $p_{\ell,C}\left(
\underbrace{\operatorname*{kan}\left(  \left(  \operatorname{id}_{C}%
\otimes\varepsilon_{C}\right)  \left(  t_{i}\right)  \right)  }_{\substack{\in
C_{i}\\\text{(by (\ref{531.pf.1}))}}}\right)  \in p_{\ell,C}\left(
C_{i}\right)  =0$, so that
\begin{equation}
p_{\ell,C}\left(  \operatorname*{kan}\left(  \left(  \operatorname{id}%
_{C}\otimes\varepsilon_{C}\right)  \left(  t_{i}\right)  \right)  \right)  =0.
\label{531.pf.2}%
\end{equation}


On the other hand, (\ref{531.pf.1}) (applied to $i=\ell$) yields
$\operatorname*{kan}\left(  \left(  \operatorname{id}_{C}\otimes
\varepsilon_{C}\right)  \left(  t_{\ell}\right)  \right)  \in C_{\ell}$, so
that%
\begin{align}
p_{\ell,C}\left(  \operatorname*{kan}\left(  \left(  \operatorname{id}%
_{C}\otimes\varepsilon_{C}\right)  \left(  t_{\ell}\right)  \right)  \right)
&  =\underbrace{\left(  p_{\ell,C}\mid_{C_{\ell}}\right)  }%
_{\substack{=\operatorname{id}_{C}\mid_{C_{\ell}}\\\text{(by (\ref{221.id}),
applied to }V=C\text{ and }n=\ell\text{)}}}\left(  \operatorname*{kan}\left(
\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \left(  t_{\ell
}\right)  \right)  \right) \nonumber\\
&  =\left(  \operatorname{id}_{C}\mid_{C_{\ell}}\right)  \left(
\operatorname*{kan}\left(  \left(  \operatorname{id}_{C}\otimes\varepsilon
_{C}\right)  \left(  t_{\ell}\right)  \right)  \right) \nonumber\\
&  =\operatorname*{kan}\left(  \left(  \operatorname{id}_{C}\otimes
\varepsilon_{C}\right)  \left(  t_{\ell}\right)  \right)  . \label{531.pf.3}%
\end{align}


Now, since $C$ is a $k$-coalgebra, we have $\operatorname*{kan}\circ\left(
\operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \circ\Delta_{C}%
=\operatorname{id}_{C}$ (by the axioms of a $k$-coalgebra), so that $\left(
\operatorname*{kan}\circ\left(  \operatorname{id}_{C}\otimes\varepsilon
_{C}\right)  \circ\Delta_{C}\right)  \left(  x\right)  =\operatorname{id}%
_{C}\left(  x\right)  =x$. Thus,%
\begin{align*}
x  &  =\left(  \operatorname*{kan}\circ\left(  \operatorname{id}_{C}%
\otimes\varepsilon_{C}\right)  \circ\Delta_{C}\right)  \left(  x\right)
=\left(  \operatorname*{kan}\circ\left(  \operatorname{id}_{C}\otimes
\varepsilon_{C}\right)  \right)  \underbrace{\left(  \Delta_{C}\left(
x\right)  \right)  }_{=\sum_{i=0}^{\ell}t_{i}}=\left(  \operatorname*{kan}%
\circ\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \right)
\left(  \sum_{i=0}^{\ell}t_{i}\right) \\
&  =\sum_{i=0}^{\ell}\underbrace{\left(  \operatorname*{kan}\circ\left(
\operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \right)  \left(
t_{i}\right)  }_{=\operatorname*{kan}\left(  \left(  \operatorname{id}%
_{C}\otimes\varepsilon_{C}\right)  \left(  t_{i}\right)  \right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{kan}\circ\left(
\operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \text{ is }k\text{-linear}%
\right) \\
&  =\sum_{i=0}^{\ell}\operatorname*{kan}\left(  \left(  \operatorname{id}%
_{C}\otimes\varepsilon_{C}\right)  \left(  t_{i}\right)  \right)  .
\end{align*}
Hence,%
\begin{align}
p_{\ell,C}\left(  x\right)   &  =p_{\ell,C}\left(  \sum_{i=0}^{\ell
}\operatorname*{kan}\left(  \left(  \operatorname{id}_{C}\otimes
\varepsilon_{C}\right)  \left(  t_{i}\right)  \right)  \right) \nonumber\\
&  =\sum_{i=0}^{\ell}p_{\ell,C}\left(  \operatorname*{kan}\left(  \left(
\operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \left(  t_{i}\right)
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{\ell,C}\text{ is
}k\text{-linear}\right) \nonumber\\
&  =\sum_{i=0}^{\ell-1}\underbrace{p_{\ell,C}\left(  \operatorname*{kan}%
\left(  \left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \left(
t_{i}\right)  \right)  \right)  }_{\substack{=0\\\text{(by (\ref{531.pf.2}))}%
}}+\underbrace{p_{\ell,C}\left(  \operatorname*{kan}\left(  \left(
\operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \left(  t_{\ell}\right)
\right)  \right)  }_{\substack{=\operatorname*{kan}\left(  \left(
\operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \left(  t_{\ell}\right)
\right)  \\\text{(by (\ref{531.pf.3}))}}}\nonumber\\
&  =\underbrace{\sum_{i=0}^{\ell-1}0}_{=0}+\operatorname*{kan}\left(  \left(
\operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \left(  t_{\ell}\right)
\right)  =\operatorname*{kan}\left(  \left(  \operatorname{id}_{C}%
\otimes\varepsilon_{C}\right)  \left(  t_{\ell}\right)  \right)  .
\label{531.pf.5}%
\end{align}
Since $p_{\ell,C}\mid_{C_{\ell}}=\operatorname{id}_{C}\mid_{C_{\ell}}$ (by
(\ref{221.id}), applied to $V=C$ and $n=\ell$), we have $\left(  p_{\ell
,C}\mid_{C_{\ell}}\right)  \left(  x\right)  =\left(  \operatorname{id}%
_{C}\mid_{C_{\ell}}\right)  \left(  x\right)  =x$, so that $p_{\ell,C}\left(
x\right)  =\left(  p_{\ell,C}\mid_{C_{\ell}}\right)  \left(  x\right)  =x$.
Thus, (\ref{531.pf.5}) rewrites as%
\begin{align*}
x  &  =\operatorname*{kan}\left(  \left(  \operatorname{id}_{C}\otimes
\varepsilon_{C}\right)  \left(  t_{\ell}\right)  \right)  =\operatorname*{kan}%
\left(  \underbrace{\left(  \operatorname{id}_{C}\otimes\varepsilon
_{C}\right)  \left(  \left(  \operatorname{id}_{C}\otimes p_{0,C}\right)
\left(  \Delta_{C}\left(  x\right)  \right)  \right)  }_{=\left(
\operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes p_{0,C}\right)  \circ\Delta_{C}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }t_{\ell}=\left(  \operatorname{id}%
_{C}\otimes p_{0,C}\right)  \left(  \Delta_{C}\left(  x\right)  \right)
\text{ by (\ref{531.pf.0})}\right) \\
&  =\operatorname*{kan}\left(  \left(  \left(  \operatorname{id}_{C}%
\otimes\varepsilon_{C}\right)  \circ\left(  \operatorname{id}_{C}\otimes
p_{0,C}\right)  \circ\Delta_{C}\right)  \left(  x\right)  \right)  .
\end{align*}
Since $\operatorname*{kan}$ is an isomorphism, this rewrites as%
\begin{equation}
\left(  \left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)
\circ\left(  \operatorname{id}_{C}\otimes p_{0,C}\right)  \circ\Delta
_{C}\right)  \left(  x\right)  =\operatorname*{kan}\nolimits^{-1}\left(
x\right)  =x\otimes1 \label{531.pf.6}%
\end{equation}
(because $\operatorname*{kan}\nolimits^{-1}$ is the isomorphism $C\rightarrow
C\otimes k$ which sends every $\xi\in C$ to $\xi\otimes1$). Now it is easy to
see that $\operatorname{id}_{C}\otimes\left(  \varepsilon_{C}\circ
p_{0,C}\right)  =\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)
\circ\left(  \operatorname{id}_{C}\otimes p_{0,C}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} By (\ref{(X)(X)}) (applied to $U=C$, $V=C$,
$W=C$, $U^{\prime}=C$, $V^{\prime}=C$, $W^{\prime}=k$, $\alpha
=\operatorname{id}_{C}$, $\beta=\operatorname{id}_{C}$, $\alpha^{\prime
}=p_{0,C}$ and $\beta^{\prime}=\varepsilon_{C}$), we have $\left(
\operatorname{id}_{C}\circ\operatorname{id}_{C}\right)  \otimes\left(
\varepsilon_{C}\circ p_{0,C}\right)  =\left(  \operatorname{id}_{C}%
\otimes\varepsilon_{C}\right)  \circ\left(  \operatorname{id}_{C}\otimes
p_{0,C}\right)  $, so that%
\[
\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes p_{0,C}\right)  =\underbrace{\left(
\operatorname{id}_{C}\circ\operatorname{id}_{C}\right)  }_{=\operatorname{id}%
_{C}}\otimes\left(  \varepsilon_{C}\circ p_{0,C}\right)  =\operatorname{id}%
_{C}\otimes\left(  \varepsilon_{C}\circ p_{0,C}\right)  ,
\]
qed.}. Hence,%
\begin{align*}
\left(  \underbrace{\left(  \operatorname{id}_{C}\otimes\left(  \varepsilon
_{C}\circ p_{0,C}\right)  \right)  }_{=\left(  \operatorname{id}_{C}%
\otimes\varepsilon_{C}\right)  \circ\left(  \operatorname{id}_{C}\otimes
p_{0,C}\right)  }\circ\Delta_{C}\right)  \left(  x\right)   &  =\left(
\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes p_{0,C}\right)  \circ\Delta_{C}\right)  \left(
x\right) \\
&  =x\otimes1\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{531.pf.6})}\right)  ,
\end{align*}
so that%
\[
\left(  \varepsilon_{C}\otimes\operatorname{id}_{k}\right)  \left(  \left(
\left(  \operatorname{id}_{C}\otimes\left(  \varepsilon_{C}\circ
p_{0,C}\right)  \right)  \circ\Delta_{C}\right)  \left(  x\right)  \right)
=\left(  \varepsilon_{C}\otimes\operatorname{id}_{k}\right)  \left(
x\otimes1\right)  =\varepsilon_{C}\left(  x\right)  \otimes
\underbrace{\operatorname{id}_{k}\left(  1\right)  }_{=1}=\varepsilon
_{C}\left(  x\right)  \otimes1.
\]
This rewrites as%
\begin{align}
\varepsilon_{C}\left(  x\right)  \otimes1  &  =\left(  \varepsilon_{C}%
\otimes\operatorname{id}_{k}\right)  \left(  \left(  \left(  \operatorname{id}%
_{C}\otimes\left(  \varepsilon_{C}\circ p_{0,C}\right)  \right)  \circ
\Delta_{C}\right)  \left(  x\right)  \right) \nonumber\\
&  =\left(  \underbrace{\left(  \varepsilon_{C}\otimes\operatorname{id}%
_{k}\right)  \circ\left(  \operatorname{id}_{C}\otimes\left(  \varepsilon
_{C}\circ p_{0,C}\right)  \right)  }_{\substack{=\left(  \varepsilon_{C}%
\circ\operatorname{id}_{C}\right)  \otimes\left(  \operatorname{id}_{k}%
\circ\varepsilon_{C}\circ p_{0,C}\right)  \\\text{(because (\ref{(X)(X)})
(applied to }U=C\text{, }V=C\text{, }W=k\text{, }U^{\prime}=C\text{,
}V^{\prime}=k\text{, }W^{\prime}=k\text{,}\\\alpha=\operatorname{id}%
_{C}\text{, }\beta=\varepsilon_{C}\text{, }\alpha^{\prime}=\varepsilon
_{C}\circ p_{0,C}\text{ and }\beta^{\prime}=\operatorname{id}_{k}\text{)
yields}\\\left(  \varepsilon_{C}\circ\operatorname{id}_{C}\right)
\otimes\left(  \operatorname{id}_{k}\circ\varepsilon_{C}\circ p_{0,C}\right)
=\left(  \varepsilon_{C}\otimes\operatorname{id}_{k}\right)  \circ\left(
\operatorname{id}_{C}\otimes\left(  \varepsilon_{C}\circ p_{0,C}\right)
\right)  }}\circ\Delta_{C}\right)  \left(  x\right) \nonumber\\
&  =\left(  \left(  \underbrace{\left(  \varepsilon_{C}\circ\operatorname{id}%
_{C}\right)  }_{=\varepsilon_{C}}\otimes\underbrace{\left(  \operatorname{id}%
_{k}\circ\varepsilon_{C}\circ p_{0,C}\right)  }_{=\varepsilon_{C}\circ
p_{0,C}}\right)  \circ\Delta_{C}\right)  \left(  x\right) \nonumber\\
&  =\left(  \left(  \varepsilon_{C}\otimes\left(  \varepsilon_{C}\circ
p_{0,C}\right)  \right)  \circ\Delta_{C}\right)  \left(  x\right)  .
\label{531.pf.11}%
\end{align}


But by the definition of convolution, we have $\varepsilon_{C}\ast\left(
\varepsilon_{C}\circ p_{0,C}\right)  =\mu_{k}\circ\left(  \varepsilon
_{C}\otimes\left(  \varepsilon_{C}\circ p_{0,C}\right)  \right)  \circ
\Delta_{C}$, so that%
\begin{align*}
\left(  \varepsilon_{C}\ast\left(  \varepsilon_{C}\circ p_{0,C}\right)
\right)  \left(  x\right)   &  =\left(  \mu_{k}\circ\left(  \varepsilon
_{C}\otimes\left(  \varepsilon_{C}\circ p_{0,C}\right)  \right)  \circ
\Delta_{C}\right)  \left(  x\right) \\
&  =\mu_{k}\underbrace{\left(  \left(  \left(  \varepsilon_{C}\otimes\left(
\varepsilon_{C}\circ p_{0,C}\right)  \right)  \circ\Delta_{C}\right)  \left(
x\right)  \right)  }_{\substack{=\varepsilon_{C}\left(  x\right)
\otimes1\\\text{(by (\ref{531.pf.11}))}}}\\
&  =\mu_{k}\left(  \varepsilon_{C}\left(  x\right)  \otimes1\right)
=\varepsilon_{C}\left(  x\right)  \cdot1\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\mu_{k}\right) \\
&  =\varepsilon_{C}\left(  x\right)  .
\end{align*}
Thus,
\begin{align*}
\varepsilon_{C}\left(  x\right)   &  =\left(  \varepsilon_{C}\ast\left(
\varepsilon_{C}\circ p_{0,C}\right)  \right)  \left(  x\right)  =\left(
\varepsilon_{C}\circ p_{0,C}\right)  \left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\varepsilon_{C}\text{ is the unity of the }k\text{-algebra
}\left(  \mathcal{L}\left(  C,k\right)  ,\ast\right)  \text{, and thus
satisfies}\\
\varepsilon_{C}\ast\left(  \varepsilon_{C}\circ p_{0,C}\right)  =\varepsilon
_{C}\circ p_{0,C}%
\end{array}
\right)  .
\end{align*}


Now forget that we fixed $x$. We thus have shown that
\begin{equation}
\varepsilon_{C}\left(  x\right)  =\left(  \varepsilon_{C}\circ p_{0,C}\right)
\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }x\in C_{\ell}.
\label{531.pf.20}%
\end{equation}
Now, it is very easy to see that $\varepsilon_{C}\left(  C_{\ell}\right)
\subseteq k_{\ell}$\ \ \ \ \footnote{\textit{Proof.} We distinguish between
two cases:
\par
\textit{Case 1:} We have $\ell=0$.
\par
\textit{Case 2:} We have $\ell\neq0$.
\par
Let us consider Case 1 first. In this case, $\ell=0$, so that $k_{\ell}%
=k_{0}=k$ (because this is how the grading on $k$ is defined), so that
$\varepsilon_{C}\left(  C_{\ell}\right)  \subseteq k=k_{\ell}$. Thus,
$\varepsilon_{C}\left(  C_{\ell}\right)  \subseteq k_{\ell}$ is proven in Case
1.
\par
Next, let us consider Case 2. In this case, $\ell\neq0$, thus $0\neq\ell$, so
that $p_{0,C}\mid_{C_{\ell}}=0$ (by (\ref{221.0}) (applied to $V=C$, $n=0$ and
$m=\ell$)), so that every $x\in C_{\ell}$ satisfies $p_{0,C}\left(  x\right)
=\underbrace{\left(  p_{0,C}\mid_{C_{\ell}}\right)  }_{=0}\left(  x\right)
=0$. Thus, every $x\in C_{\ell}$ satisfies%
\begin{align*}
\varepsilon_{C}\left(  x\right)   &  =\left(  \varepsilon_{C}\circ
p_{0,C}\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{531.pf.20})}\right) \\
&  =\varepsilon_{C}\underbrace{\left(  p_{0,C}\left(  x\right)  \right)
}_{=0}=\varepsilon_{C}\left(  0\right)  =0\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\varepsilon_{C}\text{ is }k\text{-linear}\right) \\
&  \in k_{\ell}.
\end{align*}
In other words, $\varepsilon_{C}\left(  C_{\ell}\right)  \subseteq k_{\ell}$.
We thus have proven $\varepsilon_{C}\left(  C_{\ell}\right)  \subseteq
k_{\ell}$ in Case 2.
\par
Hence, $\varepsilon_{C}\left(  C_{\ell}\right)  \subseteq k_{\ell}$ is proven
in both possible cases. Thus, $\varepsilon_{C}\left(  C_{\ell}\right)
\subseteq k_{\ell}$ always holds, qed.}.

Now forget that we fixed $\ell$. We thus have shown that $\varepsilon
_{C}\left(  C_{\ell}\right)  \subseteq k_{\ell}$ for every $\ell\in\mathbb{N}%
$. In other words, $\varepsilon_{C}$ is a graded map. Combined with the fact
that $\Delta_{C}$ is a graded map, this yields that $C$ is a graded
$k$-coalgebra. Theorem~\ref{Theorem531} is thus proven.
\end{proof}

\section{\label{S22}$\ast$-inverses of coalgebra homomorphisms}

In \S \ref{S20}, we extended Proposition~\ref{Proposition224} \textbf{(c)} to
the case of negative $n$ whenever $f$ is $\ast$-invertible. This extension
(which we formulated as Proposition~\ref{Proposition504}) was proven using
Proposition~\ref{Proposition504} (particularly, parts \textbf{(g)} and
\textbf{(h)}) as the main ingredient. Let us now, in a similar manner, extend
Corollary~\ref{Corollary46} to the case of negative $n$ whenever $f$ is $\ast$-invertible:

\begin{proposition}
\label{Proposition550} Let $k$ be a field. Let $C$ be a cocommutative
$k$-coalgebra. Let $H$ be a $k$-bialgebra. Let $f:C\rightarrow H$ be a $\ast
$-invertible $k$-coalgebra homomorphism.

\textbf{(a)} Then, $f^{\ast\left(  -1\right)  }:C\rightarrow H$ is also a
$k$-coalgebra homomorphism.

\textbf{(b)} Let $n\in\mathbb{Z}$. Then, $f^{\ast n}:C\rightarrow H$ is also a
$k$-coalgebra homomorphism.
\end{proposition}

Our goal here is not just to prove Proposition~\ref{Proposition550}; in fact,
that would be pretty easy: Proposition~\ref{Proposition550} \textbf{(a)} can
be proven with the help of Proposition~\ref{Proposition504} in the same way as
we proved Proposition~\ref{Proposition45} in \S \ref{S21} with the help of
Proposition~\ref{Proposition504}. And Proposition~\ref{Proposition550}
\textbf{(b)} is a quick corollary of Proposition~\ref{Proposition550}
\textbf{(a)}. But instead of doing this proof, we will show something more
general than Proposition~\ref{Proposition550}:

\begin{proposition}
\label{Proposition551} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $A$
be a $k$-bialgebra. Let $f:C\rightarrow A$ be a $\ast$-invertible
$k$-coalgebra homomorphism. Then, $f^{\ast\left(  -1\right)  }$ is a
$k$-coalgebra homomorphism from $C^{\operatorname*{cop}}$ to $A$.
\end{proposition}

Here, we are using the following definition:

\begin{definition}
\label{Definition552} Let $k$ be a field. Let $C$ be a $k$-coalgebra. The
\textit{coopposite coalgebra} of $C$ is defined to be the $k$-coalgebra
$\left(  C,\tau_{C,C}\circ\Delta_{C},\varepsilon_{C}\right)  $ (this is easily
seen to be a $k$-coalgebra), and denoted by $C^{\operatorname*{cop}}$.
\end{definition}

Note that Proposition~\ref{Proposition551} cannot be easily derived from
Proposition~\ref{Proposition504} anymore (mainly because there is no
generalization of Lemma~\ref{Lemma511} to non-cocommutative coalgebras $C$).

Before we prove Proposition~\ref{Proposition551}, let us show how
Proposition~\ref{Proposition550} can be derived from it:

\begin{proof}
[Proof of Proposition~\ref{Proposition550}.]\textbf{(a)} Since $C$ is
cocommutative, we have $\tau_{C,C}\circ\Delta_{C}=\Delta_{C}$. By the
definition of $C^{\operatorname*{cop}}$, we have $C^{\operatorname*{cop}%
}=\left(  C,\underbrace{\tau_{C,C}\circ\Delta_{C}}_{=\Delta_{C}}%
,\varepsilon_{C}\right)  =\left(  C,\Delta_{C},\varepsilon_{C}\right)  =C$. By
Proposition~\ref{Proposition551} (applied to $A=H$), we know that
$f^{\ast\left(  -1\right)  }$ is a $k$-coalgebra homomorphism from
$C^{\operatorname*{cop}}$ to $H$. Since $C^{\operatorname*{cop}}=C$, this
rewrites as follows: $f^{\ast\left(  -1\right)  }$ is a $k$-coalgebra
homomorphism from $C$ to $H$. This proves Proposition~\ref{Proposition550}
\textbf{(a)}.

\textbf{(b)} Let us first check that, for every $m\in\mathbb{N}$,
\begin{equation}
\text{the map }f^{\ast\left(  -m\right)  }:C\rightarrow H\text{ is a
}k\text{-coalgebra homomorphism.} \label{550.pf.1}%
\end{equation}


\textit{Proof of (\ref{550.pf.1}):} We will prove (\ref{550.pf.1}) by
induction over $m$:

\textit{Induction base:} From the proof of Corollary~\ref{Corollary46}, we
know that $f^{\ast0}$ is a $k$-coalgebra homomorphism. In other words,
$f^{\ast\left(  -0\right)  }$ is a $k$-coalgebra homomorphism (since $0=-0$).
In other words, (\ref{550.pf.1}) holds for $m=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that (\ref{550.pf.1})
holds for $m=N$. We must now prove that (\ref{550.pf.1}) also holds for
$m=N+1$.

Since (\ref{550.pf.1}) holds for $m=N$, we know that $f^{\ast\left(
-N\right)  }$ is a $k$-coalgebra homomorphism. Proposition~\ref{Proposition45}
(applied to $f^{\ast\left(  -1\right)  }$ and $f^{\ast\left(  -N\right)  }$
instead of $f$ and $g$) now yields that $f^{\ast\left(  -1\right)  }\ast
f^{\ast\left(  -N\right)  }$ is a $k$-coalgebra homomorphism (because we know
from Proposition~\ref{Proposition550} \textbf{(a)} that $f^{\ast\left(
-1\right)  }$ is a $k$-coalgebra homomorphism). Since $f^{\ast\left(
-1\right)  }\ast f^{\ast\left(  -N\right)  }=f^{\ast\left(  -\left(
N+1\right)  \right)  }$, this yields that $f^{\ast\left(  -\left(  N+1\right)
\right)  }$ is a $k$-coalgebra homomorphism. In other words, (\ref{550.pf.1})
holds for $m=N+1$. This completes the induction step. The induction proof of
(\ref{550.pf.1}) is thus complete.

Now, let us distinguish between two cases:

\textit{Case 1:} We have $n\geq0$.

\textit{Case 2:} We have $n<0$.

Let us consider Case 1 first. In this case, $n\geq0$, so that $n\in\mathbb{N}%
$, and thus Corollary~\ref{Corollary46} shows that $f^{\ast n}:C\rightarrow H$
is a $k$-coalgebra homomorphism. Hence, Proposition~\ref{Proposition550}
\textbf{(b)} is proven in Case 1.

Now, let us consider Case 2. In this case, $n<0$, so that $-n>0$ and thus
$-n\in\mathbb{N}$. Hence, (\ref{550.pf.1}) (applied to $m=-n$) yields that the
map $f^{\ast\left(  -\left(  -n\right)  \right)  }:C\rightarrow H$ is a
$k$-coalgebra homomorphism. Since $-\left(  -n\right)  =n$, this rewrites as
follows: The map $f^{\ast n}:C\rightarrow H$ is a $k$-coalgebra homomorphism.
Thus, Proposition~\ref{Proposition550} \textbf{(b)} is proven in Case 2.

Hence, Proposition~\ref{Proposition550} \textbf{(b)} is proven in each of the
cases 1 and 2. Since these two cases cover all possibilities, this yields that
Proposition~\ref{Proposition550} \textbf{(b)} always holds. The proof of
Proposition~\ref{Proposition550} \textbf{(b)} is thus complete (up to proving
Proposition~\ref{Proposition551}).
\end{proof}

Let us show another consequence of Proposition~\ref{Proposition551}:

\begin{proposition}
\label{Proposition553} Let $k$ be a field. Let $H$ be a $k$-Hopf algebra.
Then, the antipode of $H$ is a $k$-coalgebra homomorphism from
$H^{\operatorname*{cop}}$ to $H$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition553}.]The antipode of $H$ is the $\ast
$-inverse of the identity map $\operatorname{id}_{H}:H\rightarrow H$ (by the
definition of the antipode of a Hopf algebra). In other words, the antipode of
$H$ is the map $\operatorname{id}_{H}^{\ast\left(  -1\right)  }$. In
particular, this yields that the map $\operatorname{id}_{H}$ is $\ast
$-invertible. Thus, Proposition~\ref{Proposition551} (applied to $C=H$, $A=H$
and $f=\operatorname{id}_{H}$) yields that $\operatorname{id}_{H}^{\ast\left(
-1\right)  }$ is a $k$-coalgebra homomorphism from $H^{\operatorname*{cop}}$
to $H$. Since $\operatorname{id}_{H}^{\ast\left(  -1\right)  }$ is the
antipode of $H$, this rewrites as follows: The antipode of $H$ is a
$k$-coalgebra homomorphism from $H^{\operatorname*{cop}}$ to $H$. This proves
Proposition~\ref{Proposition553}.
\end{proof}

Let us now prepare for proving Proposition~\ref{Proposition551}. Our proof
will involve lengthy computations, but we can simplify them by showing some
lemmas first:

\begin{lemma}
\label{Lemma555} Let $k$ be a field. Let $U$, $V$ and $W$ be three $k$-vector
spaces. Then,%
\[
\left(  \operatorname{id}_{V}\otimes\tau_{U,W}\right)  \circ\left(  \tau
_{U,V}\otimes\operatorname{id}_{W}\right)  =\tau_{U,V\otimes W}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma555}.]Every $u\in U$, $v\in V$ and $w\in W$ satisfy%
\begin{align*}
&  \left(  \left(  \operatorname{id}_{V}\otimes\tau_{U,W}\right)  \circ\left(
\tau_{U,V}\otimes\operatorname{id}_{W}\right)  \right)  \left(  u\otimes
v\otimes w\right) \\
&  =\left(  \operatorname{id}_{V}\otimes\tau_{U,W}\right)  \underbrace{\left(
\left(  \tau_{U,V}\otimes\operatorname{id}_{W}\right)  \left(  u\otimes
v\otimes w\right)  \right)  }_{=\tau_{U,V}\left(  u\otimes v\right)
\otimes\operatorname{id}_{W}\left(  w\right)  }\\
&  =\left(  \operatorname{id}_{V}\otimes\tau_{U,W}\right)  \left(
\underbrace{\tau_{U,V}\left(  u\otimes v\right)  }_{\substack{=v\otimes
u\\\text{(by the definition of }\tau_{U,V}\text{)}}}\otimes
\underbrace{\operatorname{id}_{W}\left(  w\right)  }_{=w}\right) \\
&  =\left(  \operatorname{id}_{V}\otimes\tau_{U,W}\right)  \left(  v\otimes
u\otimes w\right)  =\underbrace{\operatorname{id}_{V}\left(  v\right)  }%
_{=v}\otimes\underbrace{\tau_{U,W}\left(  u\otimes w\right)  }%
_{\substack{=w\otimes u\\\text{(by the definition of }\tau_{U,W}\text{)}}}\\
&  =v\otimes w\otimes u=\tau_{U,V\otimes W}\left(  u\otimes v\otimes w\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau_{U,V\otimes W}\left(
u\otimes v\otimes w\right)  =v\otimes w\otimes u\text{ by the definition of
}\tau_{U,V\otimes W}\right)  .
\end{align*}
In other words, the two maps $\left(  \operatorname{id}_{V}\otimes\tau
_{U,W}\right)  \circ\left(  \tau_{U,V}\otimes\operatorname{id}_{W}\right)  $
and $\tau_{U,V\otimes W}$ are equal on every pure tensor. But since these two
maps are $k$-linear, this yields that these maps must be identic (because
whenever two $k$-linear maps from a tensor product are equal on every pure
tensor, they must be identic). In other words, $\left(  \operatorname{id}%
_{V}\otimes\tau_{U,W}\right)  \circ\left(  \tau_{U,V}\otimes\operatorname{id}%
_{W}\right)  =\tau_{U,V\otimes W}$. This proves Lemma~\ref{Lemma555}.
\end{proof}

\begin{lemma}
\label{Lemma556} Let $k$ be a field. Let $U$, $V$, $W$, $T$ and $Q$ be five
$k$-vector spaces. Let $r:Q\rightarrow V\otimes W$ be a $k$-linear map. Then,%
\begin{align*}
&  \left(  \operatorname{id}_{V}\otimes\tau_{U,W}\otimes\operatorname{id}%
_{T}\right)  \circ\left(  \tau_{U,V}\otimes\operatorname{id}_{W}%
\otimes\operatorname{id}_{T}\right)  \circ\left(  \operatorname{id}_{U}\otimes
r\otimes\operatorname{id}_{T}\right) \\
&  =\left(  r\otimes\operatorname{id}_{U}\otimes\operatorname{id}_{T}\right)
\circ\left(  \tau_{U,Q}\otimes\operatorname{id}_{T}\right)  .
\end{align*}

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma556}.]By (\ref{(X)(X)}) (applied to $U\otimes
V\otimes W$, $V\otimes U\otimes W$, $V\otimes W\otimes U$, $T$, $T$, $T$,
$\tau_{U,V}\otimes\operatorname{id}_{W}$, $\operatorname{id}_{V}\otimes
\tau_{U,W}$, $\operatorname{id}_{T}$, $\operatorname{id}_{T}$ instead of $U$,
$V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$,
$\alpha^{\prime}$, $\beta^{\prime}$), we have%
\[
\left(  \left(  \operatorname{id}_{V}\otimes\tau_{U,W}\right)  \circ\left(
\tau_{U,V}\otimes\operatorname{id}_{W}\right)  \right)  \otimes\left(
\operatorname{id}_{T}\circ\operatorname{id}_{T}\right)  =\left(
\operatorname{id}_{V}\otimes\tau_{U,W}\otimes\operatorname{id}_{T}\right)
\circ\left(  \tau_{U,V}\otimes\operatorname{id}_{W}\otimes\operatorname{id}%
_{T}\right)  .
\]
Thus,%
\begin{align}
&  \left(  \operatorname{id}_{V}\otimes\tau_{U,W}\otimes\operatorname{id}%
_{T}\right)  \circ\left(  \tau_{U,V}\otimes\operatorname{id}_{W}%
\otimes\operatorname{id}_{T}\right) \nonumber\\
&  =\underbrace{\left(  \left(  \operatorname{id}_{V}\otimes\tau_{U,W}\right)
\circ\left(  \tau_{U,V}\otimes\operatorname{id}_{W}\right)  \right)
}_{\substack{=\tau_{U,V\otimes W}\\\text{(by Lemma~\ref{Lemma555})}}%
}\otimes\underbrace{\left(  \operatorname{id}_{T}\circ\operatorname{id}%
_{T}\right)  }_{=\operatorname{id}_{T}}=\tau_{U,V\otimes W}\otimes
\operatorname{id}_{T}. \label{556.pf.0}%
\end{align}


On the other hand, by (\ref{(X)(X)}) (applied to $U\otimes Q$, $U\otimes
V\otimes W$, $V\otimes W\otimes U$, $T$, $T$, $T$, $\operatorname{id}%
_{U}\otimes r$, $\tau_{U,V\otimes W}$, $\operatorname{id}_{T}$,
$\operatorname{id}_{T}$ instead of $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$,
$W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$, $\beta^{\prime}$), we have%
\begin{equation}
\left(  \tau_{U,V\otimes W}\circ\left(  \operatorname{id}_{U}\otimes r\right)
\right)  \otimes\left(  \operatorname{id}_{T}\circ\operatorname{id}%
_{T}\right)  =\left(  \tau_{U,V\otimes W}\otimes\operatorname{id}_{T}\right)
\circ\left(  \operatorname{id}_{U}\otimes r\otimes\operatorname{id}%
_{T}\right)  . \label{556.pf.2}%
\end{equation}


By Proposition~\ref{Proposition35} \textbf{(a)} (applied to $U$, $Q$, $U$,
$V\otimes W$, $\operatorname{id}_{U}$ and $r$ instead of $V$, $W$, $V^{\prime
}$, $W^{\prime}$, $f$ and $g$), we have%
\begin{equation}
\left(  r\otimes\operatorname{id}_{U}\right)  \circ\tau_{U,Q}=\tau_{U,V\otimes
W}\circ\left(  \operatorname{id}_{U}\otimes r\right)  . \label{556.pf.3}%
\end{equation}


Besides, by (\ref{(X)(X)}) (applied to $U\otimes Q$, $Q\otimes U$, $V\otimes
W\otimes U$, $T$, $T$, $T$, $\tau_{U,Q}$, $r\otimes\operatorname{id}_{U}$,
$\operatorname{id}_{T}$, $\operatorname{id}_{T}$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}%
$, $\beta^{\prime}$), we have%
\begin{equation}
\left(  \left(  r\otimes\operatorname{id}_{U}\right)  \circ\tau_{U,Q}\right)
\otimes\left(  \operatorname{id}_{T}\circ\operatorname{id}_{T}\right)
=\left(  r\otimes\operatorname{id}_{U}\otimes\operatorname{id}_{T}\right)
\circ\left(  \tau_{U,Q}\otimes\operatorname{id}_{T}\right)  . \label{556.pf.4}%
\end{equation}


Now,%
\begin{align*}
&  \underbrace{\left(  \operatorname{id}_{V}\otimes\tau_{U,W}\otimes
\operatorname{id}_{T}\right)  \circ\left(  \tau_{U,V}\otimes\operatorname{id}%
_{W}\otimes\operatorname{id}_{T}\right)  }_{\substack{=\tau_{U,V\otimes
W}\otimes\operatorname{id}_{T}\\\text{(by (\ref{556.pf.0}))}}}\circ\left(
\operatorname{id}_{U}\otimes r\otimes\operatorname{id}_{T}\right) \\
&  =\left(  \tau_{U,V\otimes W}\otimes\operatorname{id}_{T}\right)
\circ\left(  \operatorname{id}_{U}\otimes r\otimes\operatorname{id}_{T}\right)
\\
&  =\underbrace{\left(  \tau_{U,V\otimes W}\circ\left(  \operatorname{id}%
_{U}\otimes r\right)  \right)  }_{\substack{=\left(  r\otimes\operatorname{id}%
_{U}\right)  \circ\tau_{U,Q}\\\text{(by (\ref{556.pf.3}))}}}\otimes\left(
\operatorname{id}_{T}\circ\operatorname{id}_{T}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{556.pf.2})}\right) \\
&  =\left(  \left(  r\otimes\operatorname{id}_{U}\right)  \circ\tau
_{U,Q}\right)  \otimes\left(  \operatorname{id}_{T}\circ\operatorname{id}%
_{T}\right)  =\left(  r\otimes\operatorname{id}_{U}\otimes\operatorname{id}%
_{T}\right)  \circ\left(  \tau_{U,Q}\otimes\operatorname{id}_{T}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{556.pf.4})}\right)  .
\end{align*}
This proves Lemma~\ref{Lemma556}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition551}.]Let $g=f^{\ast\left(  -1\right)
}$. Then, $f\ast g=f\ast f^{\ast\left(  -1\right)  }=e_{C,A}$ and $g\ast
f=f^{\ast\left(  -1\right)  }\ast f=e_{C,A}$.

We are now going to show that%
\begin{equation}
\left(  \left(  g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}%
}\right)  \ast\left(  \Delta_{A}\circ f\right)  =e_{C,A\otimes A}
\label{551.pf.left}%
\end{equation}
and%
\begin{equation}
\left(  \Delta_{A}\circ f\right)  \ast\left(  \Delta_{A}\circ g\right)
=e_{C,A\otimes A} \label{551.pf.right}%
\end{equation}
(where $\ast$ stands for convolution of maps from $C$ to $A\otimes A$). Once
these formulas are shown, it will be easy to conclude that $\left(  g\otimes
g\right)  \circ\Delta_{C^{\operatorname*{cop}}}=\Delta_{A}\circ g$, which will
settle the hardest part of Proposition~\ref{Proposition551}.

But let us prove (\ref{551.pf.left}) and (\ref{551.pf.right}) now:

\textit{Proof of (\ref{551.pf.left}):} Since $f$ is a $k$-coalgebra
homomorphism, we have $\Delta_{A}\circ f=\left(  f\otimes f\right)
\circ\Delta_{C}$. Thus,%
\begin{align}
&  \left(  \left(  g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}%
}\right)  \ast\underbrace{\left(  \Delta_{A}\circ f\right)  }_{=\left(
f\otimes f\right)  \circ\Delta_{C}}\nonumber\\
&  =\left(  \left(  g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}%
}\right)  \ast\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)
\nonumber\\
&  =\underbrace{\mu_{A\otimes A}}_{\substack{=\left(  \mu_{A}\otimes\mu
_{A}\right)  \circ\left(  \operatorname{id}_{A}\otimes\tau_{A,A}%
\otimes\operatorname{id}_{A}\right)  \\\text{(by the definition of the
}k\text{-algebra }A\otimes A\text{)}}}\circ\underbrace{\left(  \left(  \left(
g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}}\right)
\otimes\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)  \right)
}_{\substack{=\left(  g\otimes g\otimes f\otimes f\right)  \circ\left(
\Delta_{C^{\operatorname*{cop}}}\otimes\Delta_{C}\right)  \\\text{(by
(\ref{(X)(X)}), applied to }C\text{, }C\otimes C\text{, }A\otimes A\text{,
}C\text{, }C\otimes C\text{, }A\otimes A\text{,}\\\Delta
_{C^{\operatorname*{cop}}}\text{, }g\otimes g\text{, }\Delta_{C}\text{,
}f\otimes f\text{ instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{,
}V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }%
\alpha^{\prime}\text{, }\beta^{\prime}\text{)}}}\circ\Delta_{C}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of convolution}\right)
\nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\underbrace{\left(
\operatorname{id}_{A}\otimes\tau_{A,A}\otimes\operatorname{id}_{A}\right)
\circ\left(  g\otimes g\otimes f\otimes f\right)  }_{\substack{=\left(
g\otimes f\otimes g\otimes f\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \\\text{(by
Proposition~\ref{Proposition35} \textbf{(b)}, applied to }C\text{, }C\text{,
}C\text{, }C\text{,}\\A\text{, }A\text{, }A\text{, }A\text{, }g\text{,
}g\text{, }f\text{, }f\text{ instead of}\\U\text{, }V\text{, }W\text{,
}T\text{, }U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }T^{\prime
}\text{, }e\text{, }f\text{, }g\text{, }h\text{)}}}\circ\left(
\underbrace{\Delta_{C^{\operatorname*{cop}}}}_{\substack{=\tau_{C,C}%
\circ\Delta_{C}\\\text{(by the definition}\\\text{of the}\\k\text{-coalgebra
}C^{\operatorname*{cop}}\text{)}}}\otimes\underbrace{\Delta_{C}}%
_{=\operatorname{id}_{C\otimes C}\circ\Delta_{C}}\right)  \circ\Delta
_{C}\nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\left(  g\otimes f\otimes
g\otimes f\right)  \circ\left(  \operatorname{id}_{C}\otimes\tau_{C,C}%
\otimes\operatorname{id}_{C}\right)  \circ\underbrace{\left(  \left(
\tau_{C,C}\circ\Delta_{C}\right)  \otimes\left(  \operatorname{id}_{C\otimes
C}\circ\Delta_{C}\right)  \right)  }_{\substack{=\left(  \tau_{C,C}%
\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \\\text{(by (\ref{(X)(X)}), applied to }C\text{,
}C\otimes C\text{, }C\otimes C\text{,}\\C\text{, }C\otimes C\text{, }C\otimes
C\text{, }\Delta_{C}\text{, }\tau_{C,C}\text{, }\Delta_{C}\text{,
}\operatorname{id}_{C\otimes C}\\\text{instead of }U\text{, }V\text{,
}W\text{, }U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }%
\alpha\text{, }\beta\text{, }\alpha^{\prime}\text{, }\beta^{\prime}\text{)}%
}}\circ\Delta_{C}\nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\left(  g\otimes f\otimes
g\otimes f\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \operatorname{id}_{C}\otimes\tau
_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \tau_{C,C}%
\otimes\underbrace{\operatorname{id}_{C\otimes C}}_{=\operatorname{id}%
_{C}\otimes\operatorname{id}_{C}}\right)  \circ\underbrace{\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}}_{\substack{=\left(
\operatorname{id}_{C}\otimes\Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta
_{C}\\\text{(by Lemma~\ref{Lemma554})}}}\nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\left(  g\otimes f\otimes
g\otimes f\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\underbrace{\left(  \operatorname{id}_{C}%
\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \tau
_{C,C}\otimes\operatorname{id}_{C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{C}\otimes\Delta_{C}\otimes\operatorname{id}%
_{C}\right)  }_{\substack{=\left(  \Delta_{C}\otimes\operatorname{id}%
_{C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \tau_{C,C}%
\otimes\operatorname{id}_{C}\right)  \\\text{(by Lemma~\ref{Lemma556}, applied
to }U=C\text{, }V=C\text{, }W=C\text{, }T=C\text{, }Q=C\text{ and }%
r=\Delta_{C}\text{)}}}\circ\left(  \Delta_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\Delta_{C}\nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\left(  g\otimes f\otimes
g\otimes f\right)  \circ\left(  \Delta_{C}\otimes\underbrace{\operatorname{id}%
_{C}\otimes\operatorname{id}_{C}}_{=\operatorname{id}_{C\otimes C}}\right)
\circ\left(  \tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(
\Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}\nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\left(  g\otimes f\otimes
g\otimes f\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C\otimes
C}\right)  \circ\left(  \tau_{C,C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}.
\label{551.pf.2}%
\end{align}


Applying (\ref{(X)(X)}) to $A\otimes A$, $A$, $A$, $A\otimes A$, $A\otimes A$,
$A$, $\mu_{A}$, $\operatorname{id}_{A}$, $\operatorname{id}_{A\otimes A}$,
$\mu_{A}$ instead of $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$,
$\alpha$, $\beta$, $\alpha^{\prime}$, $\beta^{\prime}$, we obtain%
\[
\left(  \operatorname{id}_{A}\circ\mu_{A}\right)  \otimes\left(  \mu_{A}%
\circ\operatorname{id}_{A\otimes A}\right)  =\left(  \operatorname{id}%
_{A}\otimes\mu_{A}\right)  \circ\left(  \mu_{A}\otimes\operatorname{id}%
_{A\otimes A}\right)  .
\]
Since $\operatorname{id}_{A}\circ\mu_{A}=\mu_{A}$ and $\mu_{A}\circ
\operatorname{id}_{A\otimes A}=\mu_{A}$, this rewrites as
\begin{equation}
\mu_{A}\otimes\mu_{A}=\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{A\otimes A}\right)  .
\label{551.pf.4}%
\end{equation}


Applying (\ref{(X)(X)}) to $C\otimes C$, $A\otimes A$, $A\otimes A$, $C\otimes
C$, $C\otimes C$, $A\otimes A$, $g\otimes f$, $\operatorname{id}_{A\otimes A}%
$, $\operatorname{id}_{C\otimes C}$, $g\otimes f$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}%
$, $\beta^{\prime}$, we obtain%
\[
\left(  \operatorname{id}_{A\otimes A}\circ\left(  g\otimes f\right)  \right)
\otimes\left(  \left(  g\otimes f\right)  \circ\operatorname{id}_{C\otimes
C}\right)  =\left(  \operatorname{id}_{A\otimes A}\otimes g\otimes f\right)
\circ\left(  g\otimes f\otimes\operatorname{id}_{C\otimes C}\right)  .
\]
Since $\operatorname{id}_{A\otimes A}\circ\left(  g\otimes f\right)  =g\otimes
f$ and $\left(  g\otimes f\right)  \circ\operatorname{id}_{C\otimes
C}=g\otimes f$, this rewrites as
\begin{equation}
g\otimes f\otimes g\otimes f=\left(  \operatorname{id}_{A\otimes A}\otimes
g\otimes f\right)  \circ\left(  g\otimes f\otimes\operatorname{id}_{C\otimes
C}\right)  . \label{551.pf.5}%
\end{equation}


By (\ref{504.pf.comm}) (applied to $P=A\otimes A$, $Q=A$, $R=C\otimes C$,
$S=A\otimes A$, $\gamma=\mu_{A}$ and $\delta=g\otimes f$), we have%
\begin{equation}
\left(  \operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \mu
_{A}\otimes\operatorname{id}_{C\otimes C}\right)  =\left(  \mu_{A}%
\otimes\operatorname{id}_{A\otimes A}\right)  \circ\left(  \operatorname{id}%
_{A\otimes A}\otimes g\otimes f\right)  . \label{551.pf.6}%
\end{equation}


By the definition of convolution, $g\ast f=\mu_{A}\circ\left(  g\otimes
f\right)  \circ\Delta_{C}$, so that%
\begin{equation}
\mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta_{C}=g\ast f=e_{C,A}.
\label{551.pf.7}%
\end{equation}


Applying (\ref{(X)(X)}) to $C$, $C\otimes C$, $A\otimes A$, $C\otimes C$,
$C\otimes C$, $C\otimes C$, $\Delta_{C}$, $g\otimes f$, $\operatorname{id}%
_{C\otimes C}$, $\operatorname{id}_{C\otimes C}$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}%
$, $\beta^{\prime}$, we obtain%
\[
\left(  \left(  g\otimes f\right)  \circ\Delta_{C}\right)  \otimes\left(
\operatorname{id}_{C\otimes C}\circ\operatorname{id}_{C\otimes C}\right)
=\left(  g\otimes f\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\Delta_{C}\otimes\operatorname{id}_{C\otimes C}\right)  .
\]
Since $\operatorname{id}_{C\otimes C}\circ\operatorname{id}_{C\otimes
C}=\operatorname{id}_{C\otimes C}$, this rewrites as
\begin{equation}
\left(  \left(  g\otimes f\right)  \circ\Delta_{C}\right)  \otimes
\operatorname{id}_{C\otimes C}=\left(  g\otimes f\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}%
_{C\otimes C}\right)  . \label{551.pf.9}%
\end{equation}


Applying (\ref{(X)(X)}) to $C$, $A\otimes A$, $A$, $C\otimes C$, $C\otimes C$,
$C\otimes C$, $\left(  g\otimes f\right)  \circ\Delta_{C}$, $\mu_{A}$,
$\operatorname{id}_{C\otimes C}$, $\operatorname{id}_{C\otimes C}$ instead of
$U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$,
$\alpha^{\prime}$, $\beta^{\prime}$, we obtain%
\begin{equation}
\left(  \mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta_{C}\right)
\otimes\left(  \operatorname{id}_{C\otimes C}\circ\operatorname{id}_{C\otimes
C}\right)  =\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)
\circ\left(  \left(  \left(  g\otimes f\right)  \circ\Delta_{C}\right)
\otimes\operatorname{id}_{C\otimes C}\right)  . \label{551.pf.10}%
\end{equation}
Thus,%
\begin{align}
&  \left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)  \circ
\underbrace{\left(  g\otimes f\otimes\operatorname{id}_{C\otimes C}\right)
\circ\left(  \Delta_{C}\otimes\operatorname{id}_{C\otimes C}\right)
}_{\substack{=\left(  \left(  g\otimes f\right)  \circ\Delta_{C}\right)
\otimes\operatorname{id}_{C\otimes C}\\\text{(by (\ref{551.pf.9}))}%
}}\nonumber\\
&  =\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\left(  \left(  g\otimes f\right)  \circ\Delta_{C}\right)  \otimes
\operatorname{id}_{C\otimes C}\right) \nonumber\\
&  =\underbrace{\left(  \mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta
_{C}\right)  }_{\substack{=e_{C,A}\\\text{(by (\ref{551.pf.7}))}}%
}\otimes\underbrace{\left(  \operatorname{id}_{C\otimes C}\circ
\operatorname{id}_{C\otimes C}\right)  }_{=\operatorname{id}_{C\otimes C}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{551.pf.10})}\right) \nonumber\\
&  =e_{C,A}\otimes\operatorname{id}_{C\otimes C}. \label{551.pf.12}%
\end{align}


Now, (\ref{551.pf.2}) becomes%
\begin{align}
&  \left(  \left(  g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}%
}\right)  \ast\left(  \Delta_{A}\circ f\right) \nonumber\\
&  =\underbrace{\left(  \mu_{A}\otimes\mu_{A}\right)  }_{\substack{=\left(
\operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(  \mu_{A}%
\otimes\operatorname{id}_{A\otimes A}\right)  \\\text{(by (\ref{551.pf.4}))}%
}}\circ\underbrace{\left(  g\otimes f\otimes g\otimes f\right)  }%
_{\substack{=\left(  \operatorname{id}_{A\otimes A}\otimes g\otimes f\right)
\circ\left(  g\otimes f\otimes\operatorname{id}_{C\otimes C}\right)
\\\text{(by (\ref{551.pf.5}))}}}\circ\left(  \Delta_{C}\otimes
\operatorname{id}_{C\otimes C}\right)  \circ\left(  \tau_{C,C}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\Delta_{C}\nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ
\underbrace{\left(  \mu_{A}\otimes\operatorname{id}_{A\otimes A}\right)
\circ\left(  \operatorname{id}_{A\otimes A}\otimes g\otimes f\right)
}_{\substack{=\left(  \operatorname{id}_{A}\otimes g\otimes f\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)  \\\text{(by
(\ref{551.pf.6}))}}}\circ\left(  g\otimes f\otimes\operatorname{id}_{C\otimes
C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C\otimes C}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\Delta_{C}\nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\underbrace{\left(
\mu_{A}\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(  g\otimes
f\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(  \Delta_{C}%
\otimes\operatorname{id}_{C\otimes C}\right)  }_{\substack{=e_{C,A}%
\otimes\operatorname{id}_{C\otimes C}\\\text{(by (\ref{551.pf.12}))}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C}\right)
\circ\Delta_{C}\nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  e_{C,A}%
\otimes\underbrace{\operatorname{id}_{C\otimes C}}_{=\operatorname{id}%
_{C}\otimes\operatorname{id}_{C}}\right)  \circ\left(  \tau_{C,C}%
\otimes\operatorname{id}_{C}\right)  \circ\underbrace{\left(  \Delta
_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}}_{\substack{=\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\\\text{(by the
axioms of a}\\\text{coalgebra, since }C\text{ is}\\\text{a coalgebra)}%
}}\nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  e_{C,A}%
\otimes\operatorname{id}_{C}\otimes\operatorname{id}_{C}\right)  \circ\left(
\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}. \label{551.pf.20}%
\end{align}


On the other hand, Proposition~\ref{Proposition35} \textbf{(a)} (applied to
$C$, $C$, $C$, $A$, $\operatorname{id}_{C}$, $e_{C,A}$ instead of $V$, $W$,
$V^{\prime}$, $W^{\prime}$, $f$, $g$) yields%
\begin{equation}
\left(  e_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\tau_{C,C}=\tau
_{C,A}\circ\left(  \operatorname{id}_{C}\otimes e_{C,A}\right)  .
\label{551.pf.22}%
\end{equation}
But applying (\ref{(X)(X)}) to $C\otimes C$, $C\otimes C$, $A\otimes C$, $C$,
$C$, $C$, $\tau_{C,C}$, $e_{C,A}\otimes\operatorname{id}_{C}$,
$\operatorname{id}_{C}$, $\operatorname{id}_{C}$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}%
$, $\beta^{\prime}$, we obtain%
\[
\left(  \left(  e_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\tau
_{C,C}\right)  \otimes\left(  \operatorname{id}_{C}\circ\operatorname{id}%
_{C}\right)  =\left(  e_{C,A}\otimes\operatorname{id}_{C}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  .
\]
Thus,%
\begin{align}
&  \left(  e_{C,A}\otimes\operatorname{id}_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \tau_{C,C}\otimes\operatorname{id}_{C}\right)
\nonumber\\
&  =\underbrace{\left(  \left(  e_{C,A}\otimes\operatorname{id}_{C}\right)
\circ\tau_{C,C}\right)  }_{\substack{=\tau_{C,A}\circ\left(  \operatorname{id}%
_{C}\otimes e_{C,A}\right)  \\\text{(by (\ref{551.pf.22}))}}}\otimes\left(
\operatorname{id}_{C}\circ\operatorname{id}_{C}\right)  =\left(  \tau
_{C,A}\circ\left(  \operatorname{id}_{C}\otimes e_{C,A}\right)  \right)
\otimes\left(  \operatorname{id}_{C}\circ\operatorname{id}_{C}\right)
\nonumber\\
&  =\left(  \tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\underbrace{e_{C,A}}_{\substack{=\eta_{A}%
\circ\varepsilon_{C}\\\text{(by the definition of }e_{C,A}\text{)}}%
}\otimes\operatorname{id}_{C}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}), applied to }C\otimes C\text{, }C\otimes A\text{,
}A\otimes C\text{, }C\text{, }C\text{, }C\text{, }\operatorname{id}_{C}\otimes
e_{C,A}\text{, }\tau_{C,A}\text{, }\operatorname{id}_{C}\text{, }%
\operatorname{id}_{C}\\
\text{instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{, }V^{\prime
}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }\alpha^{\prime
}\text{, }\beta^{\prime}%
\end{array}
\right) \nonumber\\
&  =\left(  \tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\left(  \eta_{A}\circ\varepsilon_{C}\right)
\otimes\operatorname{id}_{C}\right)  . \label{551.pf.23}%
\end{align}


But (\ref{504.pf.idid}) (applied to $P=C$, $Q=C$, $R_{1}=C$, $R_{2}=k$,
$R_{3}=A$, $\varphi=\varepsilon_{C}$ and $\psi=\eta_{A}$) yields
\begin{equation}
\left(  \operatorname{id}_{C}\otimes\eta_{A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}%
\otimes\operatorname{id}_{C}\right)  =\operatorname{id}_{C}\otimes\left(
\eta_{A}\circ\varepsilon_{C}\right)  \otimes\operatorname{id}_{C}.
\label{551.pf.24}%
\end{equation}
Hence, (\ref{551.pf.23}) becomes%
\begin{align}
&  \left(  e_{C,A}\otimes\operatorname{id}_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \tau_{C,C}\otimes\operatorname{id}_{C}\right)
\nonumber\\
&  =\left(  \tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ
\underbrace{\left(  \operatorname{id}_{C}\otimes\left(  \eta_{A}%
\circ\varepsilon_{C}\right)  \otimes\operatorname{id}_{C}\right)
}_{\substack{=\left(  \operatorname{id}_{C}\otimes\eta_{A}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\varepsilon_{C}\otimes\operatorname{id}_{C}\right)  \\\text{(by
(\ref{551.pf.24}))}}}\nonumber\\
&  =\left(  \tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\otimes
\operatorname{id}_{C}\right)  . \label{551.pf.26}%
\end{align}


Now, let $\operatorname{kan}_{C,k\otimes C}$ be the canonical isomorphism
$C\rightarrow k\otimes C$ which sends every $c\in C$ to $1\otimes c\in
k\otimes C$. Then, by the axioms of a coalgebra, we have $\left(
\varepsilon_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}%
=\operatorname{kan}_{C,k\otimes C}$ (since $C$ is a $k$-coalgebra).

Applying (\ref{(X)(X)}) to $C$, $C$, $C$, $C$, $C\otimes C$, $k\otimes C$,
$\operatorname{id}_{C}$, $\operatorname{id}_{C}$, $\Delta_{C}$, $\varepsilon
_{C}\otimes\operatorname{id}_{C}$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain%
\[
\left(  \operatorname{id}_{C}\circ\operatorname{id}_{C}\right)  \otimes\left(
\left(  \varepsilon_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta
_{C}\right)  =\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\Delta_{C}\right)  .
\]
Thus,%
\begin{align}
\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)   &
=\underbrace{\left(  \operatorname{id}_{C}\circ\operatorname{id}_{C}\right)
}_{=\operatorname{id}_{C}}\otimes\underbrace{\left(  \left(  \varepsilon
_{C}\otimes\operatorname{id}_{C}\right)  \circ\Delta_{C}\right)
}_{=\operatorname{kan}_{C,k\otimes C}}\nonumber\\
&  =\operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes C}.
\label{551.pf.27}%
\end{align}


Now,%
\begin{align}
&  \underbrace{\left(  e_{C,A}\otimes\operatorname{id}_{C}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \tau_{C,C}\otimes\operatorname{id}%
_{C}\right)  }_{\substack{=\left(  \tau_{C,A}\otimes\operatorname{id}%
_{C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\eta_{A}\otimes
\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\varepsilon_{C}\otimes\operatorname{id}_{C}\right)  \\\text{(by
(\ref{551.pf.26}))}}}\circ\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right) \nonumber\\
&  =\left(  \tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)
\circ\underbrace{\left(  \operatorname{id}_{C}\otimes\varepsilon_{C}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\Delta_{C}\right)  }_{\substack{=\operatorname{id}_{C}\otimes
\operatorname{kan}_{C,k\otimes C}\\\text{(by (\ref{551.pf.27}))}}}\nonumber\\
&  =\left(  \tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes
C}\right)  . \label{551.pf.29}%
\end{align}
Thus, (\ref{551.pf.20}) becomes%
\begin{align}
&  \left(  \left(  g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}%
}\right)  \ast\left(  \Delta_{A}\circ f\right) \nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\underbrace{\left(
e_{C,A}\otimes\operatorname{id}_{C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right)  }_{\substack{=\left(
\tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes C}\right)
\\\text{(by (\ref{551.pf.29}))}}}\circ\Delta_{C}\nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes C}\right)
\circ\Delta_{C}. \label{551.pf.32}%
\end{align}


Now, let $\operatorname{kan}_{A,k\otimes A}$ be the canonical isomorphism
$A\rightarrow k\otimes A$ which sends every $a\in A$ to $1\otimes a\in
k\otimes A$. Then, it is easy to see that%
\begin{align}
&  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes C}\right)
\nonumber\\
&  =\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \circ
\operatorname{kan}_{A,k\otimes A}\circ\mu_{A}\circ\left(  g\otimes f\right)  .
\label{551.pf.39}%
\end{align}
\footnote{\textit{Proof of (\ref{551.pf.39}):} By the definition of $\eta_{A}%
$, we have $\eta_{A}\left(  1\right)  =1\cdot1_{A}=1_{A}$. Every $c\in C$ and
$d\in C$ satisfy%
\begin{align*}
&  \left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes C}\right)  \right)
\left(  c\otimes d\right) \\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \right)
\underbrace{\left(  \left(  \operatorname{id}_{C}\otimes\operatorname{kan}%
_{C,k\otimes C}\right)  \left(  c\otimes d\right)  \right)  }%
_{=\operatorname{id}_{C}\left(  c\right)  \otimes\operatorname{kan}%
_{C,k\otimes C}\left(  d\right)  }\\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \right)  \left(
\underbrace{\operatorname{id}_{C}\left(  c\right)  }_{=c}\otimes
\underbrace{\operatorname{kan}_{C,k\otimes C}\left(  d\right)  }%
_{\substack{=1\otimes d\\\text{(by the definition}\\\text{of }%
\operatorname{kan}_{C,k\otimes C}\text{)}}}\right) \\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \right)  \left(
c\otimes1\otimes d\right) \\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \right)  \underbrace{\left(  \left(
\operatorname{id}_{C}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)
\left(  c\otimes1\otimes d\right)  \right)  }_{=\operatorname{id}_{C}\left(
c\right)  \otimes\eta_{A}\left(  1\right)  \otimes\operatorname{id}_{C}\left(
d\right)  }\\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \right)  \left(
\underbrace{\operatorname{id}_{C}\left(  c\right)  }_{=c}\otimes
\underbrace{\eta_{A}\left(  1\right)  }_{=1_{A}}\otimes
\underbrace{\operatorname{id}_{C}\left(  d\right)  }_{=d}\right) \\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \right)  \left(  c\otimes1_{A}\otimes
d\right) \\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \right)  \underbrace{\left(
\left(  \tau_{C,A}\otimes\operatorname{id}_{C}\right)  \left(  c\otimes
1_{A}\otimes d\right)  \right)  }_{=\tau_{C,A}\left(  c\otimes1_{A}\right)
\otimes\operatorname{id}_{C}\left(  d\right)  }\\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \right)  \left(
\underbrace{\tau_{C,A}\left(  c\otimes1_{A}\right)  }_{\substack{=1_{A}\otimes
c\\\text{(by the definition}\\\text{of }\tau_{C,A}\text{)}}}\otimes
\underbrace{\operatorname{id}_{C}\left(  d\right)  }_{=d}\right) \\
&  =\left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \right)  \left(  1_{A}\otimes
c\otimes d\right)  =\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)
\underbrace{\left(  \left(  \operatorname{id}_{A}\otimes g\otimes f\right)
\left(  1_{A}\otimes c\otimes d\right)  \right)  }_{=\operatorname{id}%
_{A}\left(  1_{A}\right)  \otimes g\left(  c\right)  \otimes f\left(
d\right)  }\\
&  =\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \left(
\underbrace{\operatorname{id}_{A}\left(  1_{A}\right)  }_{=1_{A}}\otimes
g\left(  c\right)  \otimes f\left(  d\right)  \right)  =\left(
\operatorname{id}_{A}\otimes\mu_{A}\right)  \left(  1_{A}\otimes g\left(
c\right)  \otimes f\left(  d\right)  \right) \\
&  =\underbrace{\operatorname{id}_{A}\left(  1_{A}\right)  }_{=1_{A}}%
\otimes\underbrace{\mu_{A}\left(  g\left(  c\right)  \otimes f\left(
d\right)  \right)  }_{\substack{=g\left(  c\right)  f\left(  d\right)
\\\text{(by the definition of }\mu_{A}\text{)}}}=1_{A}\otimes g\left(
c\right)  f\left(  d\right)
\end{align*}
and%
\begin{align*}
&  \left(  \left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \circ
\operatorname{kan}_{A,k\otimes A}\circ\mu_{A}\circ\left(  g\otimes f\right)
\right)  \left(  c\otimes d\right) \\
&  =\left(  \left(  \eta_{A}\otimes\operatorname{id}_{A}\right)
\circ\operatorname{kan}_{A,k\otimes A}\circ\mu_{A}\right)  \underbrace{\left(
\left(  g\otimes f\right)  \left(  c\otimes d\right)  \right)  }_{=g\left(
c\right)  \otimes f\left(  d\right)  }=\left(  \left(  \eta_{A}\otimes
\operatorname{id}_{A}\right)  \circ\operatorname{kan}_{A,k\otimes A}\circ
\mu_{A}\right)  \left(  g\left(  c\right)  \otimes f\left(  d\right)  \right)
\\
&  =\left(  \left(  \eta_{A}\otimes\operatorname{id}_{A}\right)
\circ\operatorname{kan}_{A,k\otimes A}\right)  \underbrace{\left(  \mu
_{A}\left(  g\left(  c\right)  \otimes f\left(  d\right)  \right)  \right)
}_{\substack{=g\left(  c\right)  f\left(  d\right)  \\\text{(by the definition
of }\mu_{A}\text{)}}}=\left(  \left(  \eta_{A}\otimes\operatorname{id}%
_{A}\right)  \circ\operatorname{kan}_{A,k\otimes A}\right)  \left(  g\left(
c\right)  f\left(  d\right)  \right) \\
&  =\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \underbrace{\left(
\operatorname{kan}_{A,k\otimes A}\left(  g\left(  c\right)  f\left(  d\right)
\right)  \right)  }_{\substack{=1\otimes g\left(  c\right)  f\left(  d\right)
\\\text{(by the definition of }\operatorname{kan}_{A,k\otimes A}\text{)}%
}}=\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \left(  1\otimes
g\left(  c\right)  f\left(  d\right)  \right) \\
&  =\underbrace{\eta_{A}\left(  1\right)  }_{=1}\otimes
\underbrace{\operatorname{id}_{A}\left(  g\left(  c\right)  f\left(  d\right)
\right)  }_{=g\left(  c\right)  f\left(  d\right)  }=1\otimes g\left(
c\right)  f\left(  d\right)  .
\end{align*}
Thus, every $c\in C$ and $d\in C$ satisfy%
\begin{align*}
&  \left(  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes C}\right)  \right)
\left(  c\otimes d\right) \\
&  =1\otimes g\left(  c\right)  f\left(  d\right)  =\left(  \left(  \eta
_{A}\otimes\operatorname{id}_{A}\right)  \circ\operatorname{kan}_{A,k\otimes
A}\circ\mu_{A}\circ\left(  g\otimes f\right)  \right)  \left(  c\otimes
d\right)  .
\end{align*}
In other words, the two maps \newline$\left(  \operatorname{id}_{A}\otimes
\mu_{A}\right)  \circ\left(  \operatorname{id}_{A}\otimes g\otimes f\right)
\circ\left(  \tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)
\circ\left(  \operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes
C}\right)  $ and \newline$\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)
\circ\operatorname{kan}_{A,k\otimes A}\circ\mu_{A}\circ\left(  g\otimes
f\right)  $ are equal on every pure tensor. But since these two maps are
$k$-linear, this yields that these maps must be identic (because whenever two
$k$-linear maps from a tensor product are equal on every pure tensor, they
must be identic). In other words,
\begin{align*}
&  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(  \tau_{C,A}%
\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes C}\right) \\
&  =\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \circ
\operatorname{kan}_{A,k\otimes A}\circ\mu_{A}\circ\left(  g\otimes f\right)  .
\end{align*}
This proves (\ref{551.pf.39}).} Hence, (\ref{551.pf.32}) becomes%
\begin{align*}
&  \left(  \left(  g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}%
}\right)  \ast\left(  \Delta_{A}\circ f\right) \\
&  =\underbrace{\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)
\circ\left(  \operatorname{id}_{A}\otimes g\otimes f\right)  \circ\left(
\tau_{C,A}\otimes\operatorname{id}_{C}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\eta_{A}\otimes\operatorname{id}_{C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\operatorname{kan}_{C,k\otimes C}\right)
}_{\substack{=\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)
\circ\operatorname{kan}_{A,k\otimes A}\circ\mu_{A}\circ\left(  g\otimes
f\right)  \\\text{(by (\ref{551.pf.39}))}}}\circ\Delta_{C}\\
&  =\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \circ
\operatorname{kan}_{A,k\otimes A}\circ\underbrace{\mu_{A}\circ\left(  g\otimes
f\right)  \circ\Delta_{C}}_{\substack{=e_{C,A}\\\text{(by (\ref{551.pf.7}))}%
}}=\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \circ
\operatorname{kan}_{A,k\otimes A}\circ\underbrace{e_{C,A}}_{\substack{=\eta
_{A}\circ\varepsilon_{C}\\\text{(by the definition}\\\text{of }e_{C,A}%
\text{)}}}\\
&  =\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \circ
\operatorname{kan}_{A,k\otimes A}\circ\eta_{A}\circ\varepsilon_{C}.
\end{align*}
Thus, every $c\in C$ satisfies%
\begin{align*}
&  \left(  \left(  \left(  g\otimes g\right)  \circ\Delta
_{C^{\operatorname*{cop}}}\right)  \ast\left(  \Delta_{A}\circ f\right)
\right)  \left(  c\right) \\
&  =\left(  \left(  \eta_{A}\otimes\operatorname{id}_{A}\right)
\circ\operatorname{kan}_{A,k\otimes A}\circ\eta_{A}\circ\varepsilon
_{C}\right)  \left(  c\right)  =\left(  \left(  \eta_{A}\otimes
\operatorname{id}_{A}\right)  \circ\operatorname{kan}_{A,k\otimes A}\circ
\eta_{A}\right)  \left(  \varepsilon_{C}\left(  c\right)  \right) \\
&  =\left(  \left(  \eta_{A}\otimes\operatorname{id}_{A}\right)
\circ\operatorname{kan}_{A,k\otimes A}\right)  \underbrace{\left(  \eta
_{A}\left(  \varepsilon_{C}\left(  c\right)  \right)  \right)  }%
_{\substack{=\varepsilon_{C}\left(  c\right)  \cdot1_{A}\\\text{(by the
definition}\\\text{of }\eta_{A}\text{)}}}=\left(  \left(  \eta_{A}%
\otimes\operatorname{id}_{A}\right)  \circ\operatorname{kan}_{A,k\otimes
A}\right)  \left(  \varepsilon_{C}\left(  c\right)  \cdot1_{A}\right) \\
&  =\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \underbrace{\left(
\operatorname{kan}_{A,k\otimes A}\left(  \varepsilon_{C}\left(  c\right)
\cdot1_{A}\right)  \right)  }_{\substack{=1\otimes\varepsilon_{C}\left(
c\right)  \cdot1_{A}\\\text{(by the definition}\\\text{of }\operatorname{kan}%
_{A,k\otimes A}\text{)}}}=\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)
\left(  1\otimes\varepsilon_{C}\left(  c\right)  \cdot1_{A}\right) \\
&  =\underbrace{\eta_{A}\left(  1\right)  }_{\substack{=1\cdot1_{A}\\\text{(by
the definition of }\eta_{A}\text{)}}}\otimes\underbrace{\operatorname{id}%
_{A}\left(  \varepsilon_{C}\left(  c\right)  \cdot1_{A}\right)  }%
_{=\varepsilon_{C}\left(  c\right)  \cdot1_{A}}\\
&  =1\cdot1_{A}\otimes\varepsilon_{C}\left(  c\right)  \cdot1_{A}%
=\varepsilon_{C}\left(  c\right)  \cdot\underbrace{1_{A}\otimes1_{A}%
}_{=1_{A\otimes A}}=\varepsilon_{C}\left(  c\right)  \cdot1_{A\otimes A}%
=\eta_{A\otimes A}\left(  \varepsilon_{C}\left(  c\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{A\otimes A}\left(
\varepsilon_{C}\left(  c\right)  \right)  =\varepsilon_{C}\left(  c\right)
\cdot1_{A\otimes A}\text{ by the definition of }\eta_{A\otimes A}\right) \\
&  =\left(  \eta_{A\otimes A}\circ\varepsilon_{C}\right)  \left(  c\right)
=e_{C,A\otimes A}\left(  c\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }e_{C,A\otimes A}\text{ is defined
as }\eta_{A\otimes A}\circ\varepsilon_{C}\text{, so that }\eta_{A\otimes
A}\circ\varepsilon_{C}=e_{C,A\otimes A}\right)  .
\end{align*}
In other words, $\left(  \left(  g\otimes g\right)  \circ\Delta
_{C^{\operatorname*{cop}}}\right)  \ast\left(  \Delta_{A}\circ f\right)
=e_{C,A\otimes A}$. This proves (\ref{551.pf.left}).

\textit{Proof of (\ref{551.pf.right}):} Since $A$ is a $k$-bialgebra, the
comultiplication map $\Delta_{A}$ of $A$ is a $k$-algebra homomorphism (by the
axioms of a bialgebra). Thus, $\mu_{A\otimes A}\circ\left(  \Delta_{A}%
\otimes\Delta_{A}\right)  =\Delta_{A}\circ\mu_{A}$ and $\eta_{A\otimes
A}=\Delta_{A}\circ\eta_{A}$.

By the definition of convolution, $f\ast g=\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$. Thus,%
\begin{equation}
\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}=f\ast g=e_{C,A}%
=\eta_{A}\circ\varepsilon_{C} \label{551.pf.90}%
\end{equation}
(by the definition of $e_{C,A}$).

By the definition of convolution,%
\begin{align*}
\left(  \Delta_{A}\circ f\right)  \ast\left(  \Delta_{A}\circ g\right)   &
=\mu_{A\otimes A}\circ\underbrace{\left(  \left(  \Delta_{A}\circ f\right)
\otimes\left(  \Delta_{A}\circ g\right)  \right)  }_{\substack{=\left(
\Delta_{A}\otimes\Delta_{A}\right)  \circ\left(  f\otimes g\right)
\\\text{(by (\ref{(X)(X)}), applied to }C\text{, }A\text{, }A\otimes A\text{,
}C\text{, }A\text{, }A\otimes A\text{,}\\f\text{, }\Delta_{A}\text{, }g\text{,
}\Delta_{A}\text{ instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{,
}V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }%
\alpha^{\prime}\text{, }\beta^{\prime}\text{)}}}\circ\Delta_{C}\\
&  =\underbrace{\mu_{A\otimes A}\circ\left(  \Delta_{A}\otimes\Delta
_{A}\right)  }_{=\Delta_{A}\circ\mu_{A}}\circ\left(  f\otimes g\right)
\circ\Delta_{C}=\Delta_{A}\circ\underbrace{\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}}_{\substack{=\eta_{A}\circ\varepsilon_{C}\\\text{(by
(\ref{551.pf.90}))}}}\\
&  =\underbrace{\Delta_{A}\circ\eta_{A}}_{=\eta_{A\otimes A}}\circ
\varepsilon_{C}=\eta_{A\otimes A}\circ\varepsilon_{C}=e_{C,A\otimes A}%
\end{align*}
(since $e_{C,A\otimes A}$ is defined as $\eta_{A\otimes A}\circ\varepsilon
_{C}$). This proves (\ref{551.pf.right}).

Now, let us finish the proof of Proposition~\ref{Proposition551}: Comparing
the equalities%
\begin{align*}
&  \underbrace{\left(  \left(  g\otimes g\right)  \circ\Delta
_{C^{\operatorname*{cop}}}\right)  \ast\left(  \Delta_{A}\circ f\right)
}_{\substack{=e_{C,A\otimes A}\\\text{(by (\ref{551.pf.left}))}}}\ast\left(
\Delta_{A}\circ g\right) \\
&  =e_{C,A\otimes A}\ast\left(  \Delta_{A}\circ g\right)  =\Delta_{A}\circ g
\end{align*}
and%
\begin{align*}
&  \left(  \left(  g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}%
}\right)  \ast\underbrace{\left(  \Delta_{A}\circ f\right)  \ast\left(
\Delta_{A}\circ g\right)  }_{\substack{=e_{C,A\otimes A}\\\text{(by
(\ref{551.pf.right}))}}}\\
&  =\left(  \left(  g\otimes g\right)  \circ\Delta_{C^{\operatorname*{cop}}%
}\right)  \ast e_{C,A\otimes A}=\left(  g\otimes g\right)  \circ
\Delta_{C^{\operatorname*{cop}}},
\end{align*}
we obtain
\begin{equation}
\Delta_{A}\circ g=\left(  g\otimes g\right)  \circ\Delta
_{C^{\operatorname*{cop}}}. \label{551.pf.Delta}%
\end{equation}


We will now prove that $\varepsilon_{A}\circ g=\varepsilon
_{C^{\operatorname*{cop}}}$.

Since $A$ is a $k$-bialgebra, the counit map $\varepsilon_{A}$ of $A$ is a
$k$-algebra homomorphism (by the axioms of a bialgebra). Thus, $\varepsilon
_{A}\circ\mu_{A}=\mu_{k}\circ\left(  \varepsilon_{A}\otimes\varepsilon
_{A}\right)  $ and $\varepsilon_{A}\circ\eta_{A}=\eta_{k}$.

Since $f$ is a $k$-coalgebra homomorphism $C\rightarrow A$, we have
$\varepsilon_{A}\circ f=\varepsilon_{C}$.

Applying (\ref{(X)(X)}) to $C$, $A$, $k$, $C$, $A$, $k$, $f$, $\varepsilon
_{A}$, $g$, $\varepsilon_{A}$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain%
\[
\left(  \varepsilon_{A}\circ f\right)  \otimes\left(  \varepsilon_{A}\circ
g\right)  =\left(  \varepsilon_{A}\otimes\varepsilon_{A}\right)  \circ\left(
f\otimes g\right)  .
\]
Thus,%
\begin{equation}
\left(  \varepsilon_{A}\otimes\varepsilon_{A}\right)  \circ\left(  f\otimes
g\right)  =\underbrace{\left(  \varepsilon_{A}\circ f\right)  }_{=\varepsilon
_{C}}\otimes\left(  \varepsilon_{A}\circ g\right)  =\varepsilon_{C}%
\otimes\left(  \varepsilon_{A}\circ g\right)  . \label{551.pf.93}%
\end{equation}


The unity of the $k$-algebra $\left(  \mathcal{L}\left(  C,k\right)  ,
\ast\right)  $ is $\underbrace{\eta_{k}}_{=\operatorname{id}} \circ
\varepsilon_{C} = \varepsilon_{C}$.

But now,%
\[
\varepsilon_{A}\circ\underbrace{\mu_{A}\circ\left(  f\otimes g\right)
\circ\Delta_{C}}_{\substack{=\eta_{A}\circ\varepsilon_{C}\\\text{(by
(\ref{551.pf.90}))}}}=\underbrace{\varepsilon_{A}\circ\eta_{A}}_{=\eta
_{k}=\operatorname{id}_{k}}\circ\varepsilon_{C}=\varepsilon_{C}=\varepsilon
_{C^{\operatorname*{cop}}}%
\]
(since $\varepsilon_{C^{\operatorname*{cop}}}=\varepsilon_{C}$ by the
definition of $C^{\operatorname*{cop}}$). Thus,%
\begin{align*}
\varepsilon_{C^{\operatorname*{cop}}}  &  =\underbrace{\varepsilon_{A}\circ
\mu_{A}}_{=\mu_{k}\circ\left(  \varepsilon_{A}\otimes\varepsilon_{A}\right)
}\circ\left(  f\otimes g\right)  \circ\Delta_{C}=\mu_{k}\circ
\underbrace{\left(  \varepsilon_{A}\otimes\varepsilon_{A}\right)  \circ\left(
f\otimes g\right)  }_{\substack{=\varepsilon_{C}\otimes\left(  \varepsilon
_{A}\circ g\right)  \\\text{(by (\ref{551.pf.93}))}}}\circ\Delta_{C}\\
&  =\mu_{k}\circ\left(  \varepsilon_{C}\otimes\left(  \varepsilon_{A}\circ
g\right)  \right)  \circ\Delta_{C}=\varepsilon_{C}\ast\left(  \varepsilon
_{A}\circ g\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\varepsilon_{C}\ast\left(  \varepsilon_{A}\circ g\right)
=\mu_{k}\circ\left(  \varepsilon_{C}\otimes\left(  \varepsilon_{A}\circ
g\right)  \right)  \circ\Delta_{C}\\
\text{by the definition of convolution}%
\end{array}
\right) \\
&  =\varepsilon_{A}\circ g\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\varepsilon_{C}\text{ is the unity of the }k\text{-algebra }\left(
\mathcal{L}\left(  C,k\right)  , \ast\right)  \right)  .
\end{align*}
We thus have proven that $\varepsilon_{A}\circ g=\varepsilon
_{C^{\operatorname*{cop}}}$. Combined with (\ref{551.pf.Delta}), this yields
that $g$ is a $k$-coalgebra homomorphism from $C^{\operatorname*{cop}}$ to
$A$. Since $g=f^{\ast\left(  -1\right)  }$, this rewrites as follows:
$f^{\ast\left(  -1\right)  }$ is a $k$-coalgebra homomorphism from
$C^{\operatorname*{cop}}$ to $A$. This proves Proposition~\ref{Proposition551}.
\end{proof}

\section{\label{S23}$\ast$-inverses of algebra homomorphisms}

In this section, we are going to prove the dual versions of the results of
\S \ref{S22}, dual in the sense that coalgebras become algebras. The proofs
are mostly identical to the ones of \S \ref{S22} up to ``reversing arrows'',
except in some of the cases when we worked with elements in \S \ref{S22}.

Let us extend Corollary~\ref{Corollary136} to the case of negative $n$
whenever $f$ is $\ast$-invertible:

\begin{proposition}
\label{Proposition600} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a commutative $k$-algebra. Let $f:H\rightarrow A$ be a $\ast$-invertible
$k$-algebra homomorphism.

\textbf{(a)} Then, $f^{\ast\left(  -1\right)  }:H\rightarrow A$ is also a
$k$-algebra homomorphism.

\textbf{(b)} Let $n\in\mathbb{Z}$. Then, $f^{\ast n}:H\rightarrow A$ is also a
$k$-algebra homomorphism.
\end{proposition}

Again, this proposition (like Proposition~\ref{Proposition550}) can be easily
shown using Proposition~\ref{Proposition504} in the same way as we proved
Proposition~\ref{Proposition135} in \S \ref{S21} with the help of
Proposition~\ref{Proposition504}. Again, we are not going to present this
proof, but instead we will show something more general than
Proposition~\ref{Proposition600}:

\begin{proposition}
\label{Proposition601} Let $k$ be a field. Let $C$ be a $k$-bialgebra. Let $A$
be a $k$-algebra. Let $f:C\rightarrow A$ be a $\ast$-invertible $k$-algebra
homomorphism. Then, $f^{\ast\left(  -1\right)  }$ is a $k$-algebra
homomorphism from $C$ to $A^{\operatorname*{op}}$.
\end{proposition}

Here, we are using the following definition:

\begin{definition}
\label{Definition602} Let $k$ be a field. Let $A$ be a $k$-algebra. The
\textit{opposite algebra} of $A$ is defined to be the $k$-algebra $\left(
A,\mu_{A}\circ\tau_{A,A},\eta_{A}\right)  $ (this is easily seen to be a
$k$-algebra), and denoted by $A^{\operatorname*{op}}$.
\end{definition}

Before we prove Proposition~\ref{Proposition601}, let us show how
Proposition~\ref{Proposition600} can be derived from it:

\begin{proof}
[Proof of Proposition~\ref{Proposition600}.]\textbf{(a)} Since $A$ is
commutative, we have $\mu_{A}\circ\tau_{A,A}=\mu_{A}$%
\ \ \ \ \footnote{\textit{Proof.} Every $a\in A$ and $b\in A$ satisfy%
\begin{align*}
\left(  \mu_{A}\circ\tau_{A,A}\right)  \left(  a\otimes b\right)   &  =\mu
_{A}\underbrace{\left(  \tau_{A,A}\left(  a\otimes b\right)  \right)
}_{\substack{=b\otimes a\\\text{(by the definition}\\\text{of }\tau
_{A,A}\text{)}}}=\mu_{A}\left(  b\otimes a\right)
=ba\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu_{A}\right) \\
&  =ab\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A\text{ is commutative}\right)
\\
&  =\mu_{A}\left(  a\otimes b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mu_{A}\left(  a\otimes b\right)  =ab\text{ by the definition of }\mu
_{A}\right)  .
\end{align*}
In other words, the two maps $\mu_{A}\circ\tau_{A,A}$ and $\mu_{A}$ are equal
on every pure tensor. But since these two maps are $k$-linear, this yields
that these maps must be identic (because whenever two $k$-linear maps from a
tensor product are equal on every pure tensor, they must be identic). In other
words, $\mu_{A}\circ\tau_{A,A}=\mu_{A}$, qed.}. By the definition of
$A^{\operatorname*{op}}$, we have $A^{\operatorname*{op}}=\left(
A,\underbrace{\mu_{A}\circ\tau_{A,A}}_{=\mu_{A}},\eta_{A}\right)  =\left(
A,\mu_{A},\eta_{A}\right)  =A$. By Proposition~\ref{Proposition601} (applied
to $C=H$), we know that $f^{\ast\left(  -1\right)  }$ is a $k$-algebra
homomorphism from $H$ to $A^{\operatorname*{op}}$. Since
$A^{\operatorname*{op}}=A$, this rewrites as follows: $f^{\ast\left(
-1\right)  }$ is a $k$-algebra homomorphism from $H$ to $A$. This proves
Proposition~\ref{Proposition600} \textbf{(a)}.

\textbf{(b)} Let us first check that, for every $m\in\mathbb{N}$,
\begin{equation}
\text{the map }f^{\ast\left(  -m\right)  }:H\rightarrow A\text{ is a
}k\text{-algebra homomorphism.} \label{600.pf.1}%
\end{equation}


\textit{Proof of (\ref{600.pf.1}):} We will prove (\ref{600.pf.1}) by
induction over $m$:

\textit{Induction base:} From the proof of Corollary~\ref{Corollary136}, we
know that $f^{\ast0}$ is a $k$-algebra homomorphism. In other words,
$f^{\ast\left(  -0\right)  }$ is a $k$-algebra homomorphism (since $0=-0$). In
other words, (\ref{600.pf.1}) holds for $m=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that (\ref{600.pf.1})
holds for $m=N$. We must now prove that (\ref{600.pf.1}) also holds for
$m=N+1$.

Since (\ref{600.pf.1}) holds for $m=N$, we know that $f^{\ast\left(
-N\right)  }$ is a $k$-algebra homomorphism. Proposition~\ref{Proposition135}
(applied to $f^{\ast\left(  -1\right)  }$ and $f^{\ast\left(  -N\right)  }$
instead of $f$ and $g$) now yields that $f^{\ast\left(  -1\right)  }\ast
f^{\ast\left(  -N\right)  }$ is a $k$-algebra homomorphism (because we know
from Proposition~\ref{Proposition600} \textbf{(a)} that $f^{\ast\left(
-1\right)  }$ is a $k$-algebra homomorphism). Since $f^{\ast\left(  -1\right)
}\ast f^{\ast\left(  -N\right)  }=f^{\ast\left(  -\left(  N+1\right)  \right)
}$, this yields that $f^{\ast\left(  -\left(  N+1\right)  \right)  }$ is a
$k$-algebra homomorphism. In other words, (\ref{600.pf.1}) holds for $m=N+1$.
This completes the induction step. The induction proof of (\ref{600.pf.1}) is
thus complete.

Now, let us distinguish between two cases:

\textit{Case 1:} We have $n\geq0$.

\textit{Case 2:} We have $n<0$.

Let us consider Case 1 first. In this case, $n\geq0$, so that $n\in\mathbb{N}%
$, and thus Corollary~\ref{Corollary136} shows that $f^{\ast n}:H\rightarrow
A$ is a $k$-algebra homomorphism. Hence, Proposition~\ref{Proposition600}
\textbf{(b)} is proven in Case 1.

Now, let us consider Case 2. In this case, $n<0$, so that $-n>0$ and thus
$-n\in\mathbb{N}$. Hence, (\ref{600.pf.1}) (applied to $m=-n$) yields that the
map $f^{\ast\left(  -\left(  -n\right)  \right)  }:H\rightarrow A$ is a
$k$-algebra homomorphism. Since $-\left(  -n\right)  =n$, this rewrites as
follows: The map $f^{\ast n}:H\rightarrow A$ is a $k$-algebra homomorphism.
Thus, Proposition~\ref{Proposition600} \textbf{(b)} is proven in Case 2.

Hence, Proposition~\ref{Proposition600} \textbf{(b)} is proven in each of the
cases 1 and 2. Since these two cases cover all possibilities, this yields that
Proposition~\ref{Proposition600} \textbf{(b)} always holds. The proof of
Proposition~\ref{Proposition600} \textbf{(b)} is thus complete (up to proving
Proposition~\ref{Proposition601}).
\end{proof}

Let us show another consequence of Proposition~\ref{Proposition601}:

\begin{proposition}
\label{Proposition603} Let $k$ be a field. Let $H$ be a $k$-Hopf algebra.
Then, the antipode of $H$ is a $k$-algebra homomorphism from $H$ to
$H^{\operatorname*{op}}$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition603}.]The antipode of $H$ is the $\ast
$-inverse of the identity map $\operatorname{id}_{H}:H\rightarrow H$ (by the
definition of the antipode of a Hopf algebra). In other words, the antipode of
$H$ is the map $\operatorname{id}_{H}^{\ast\left(  -1\right)  }$. In
particular, this yields that the map $\operatorname{id}_{H}$ is $\ast
$-invertible. Thus, Proposition~\ref{Proposition601} (applied to $C=H$, $A=H$
and $f=\operatorname{id}_{H}$) yields that $\operatorname{id}_{H}^{\ast\left(
-1\right)  }$ is a $k$-algebra homomorphism from $H$ to $H^{\operatorname*{op}%
}$. Since $\operatorname{id}_{H}^{\ast\left(  -1\right)  }$ is the antipode of
$H$, this rewrites as follows: The antipode of $H$ is a $k$-algebra
homomorphism from $H$ to $H^{\operatorname*{op}}$. This proves
Proposition~\ref{Proposition603}.
\end{proof}

Let us now prepare for proving Proposition~\ref{Proposition601}. Our proof
will involve lengthy computations, but we can simplify them by showing some
lemmas first:

\begin{lemma}
\label{Lemma604} Let $k$ be a field. Let $A$ be a $k$-algebra. Then,%
\[
\mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)  =\mu_{A}\circ\left(  \mu
_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\mu_{A}\otimes\operatorname{id}_{A}\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma604}.]Every $a\in A$, $b\in A$, $c\in A$ and $d\in
A$ satisfy%
\begin{align*}
&  \left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{A}\otimes\mu_{A}\otimes\operatorname{id}%
_{A}\right)  \right)  \left(  a\otimes b\otimes c\otimes d\right) \\
&  =\left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\right)  \underbrace{\left(  \left(  \operatorname{id}_{A}\otimes\mu
_{A}\otimes\operatorname{id}_{A}\right)  \left(  a\otimes b\otimes c\otimes
d\right)  \right)  }_{=\operatorname{id}_{A}\left(  a\right)  \otimes\mu
_{A}\left(  b\otimes c\right)  \otimes\operatorname{id}_{A}\left(  d\right)
}\\
&  =\left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\right)  \left(  \underbrace{\operatorname{id}_{A}\left(  a\right)  }%
_{=a}\otimes\underbrace{\mu_{A}\left(  b\otimes c\right)  }%
_{\substack{=bc\\\text{(by the definition of }\mu_{A}\text{)}}}\otimes
\underbrace{\operatorname{id}_{A}\left(  d\right)  }_{=d}\right) \\
&  =\left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\right)  \left(  a\otimes bc\otimes d\right)  =\mu_{A}\underbrace{\left(
\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)  \left(  a\otimes bc\otimes
d\right)  \right)  }_{=\mu_{A}\left(  a\otimes bc\right)  \otimes
\operatorname{id}_{A}\left(  d\right)  }\\
&  =\mu_{A}\left(  \underbrace{\mu_{A}\left(  a\otimes bc\right)
}_{\substack{=a\left(  bc\right)  \\\text{(by the definition of }\mu
_{A}\text{)}}}\otimes\underbrace{\operatorname{id}_{A}\left(  d\right)  }%
_{=d}\right)  =\mu_{A}\left(  a\left(  bc\right)  \otimes d\right)  =\left(
a\left(  bc\right)  \right)  d\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu_{A}\right)
\end{align*}
and%
\begin{align*}
&  \left(  \mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)  \right)  \left(
a\otimes b\otimes c\otimes d\right) \\
&  =\mu_{A}\underbrace{\left(  \left(  \mu_{A}\otimes\mu_{A}\right)  \left(
a\otimes b\otimes c\otimes d\right)  \right)  }_{=\mu_{A}\left(  a\otimes
b\right)  \otimes\mu_{A}\left(  c\otimes d\right)  }=\mu_{A}\left(
\underbrace{\mu_{A}\left(  a\otimes b\right)  }_{\substack{=ab\\\text{(by the
definition of }\mu_{A}\text{)}}}\otimes\underbrace{\mu_{A}\left(  c\otimes
d\right)  }_{\substack{=cd\\\text{(by the definition of }\mu_{A}\text{)}%
}}\right) \\
&  =\mu_{A}\left(  ab\otimes cd\right)  =\left(  ab\right)  \left(  cd\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu_{A}\right)  .
\end{align*}
Thus, every $a\in A$, $b\in A$, $c\in A$ and $d\in A$ satisfy%
\begin{align*}
&  \left(  \mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)  \right)  \left(
a\otimes b\otimes c\otimes d\right) \\
&  =\left(  ab\right)  \left(  cd\right)  =abcd=\left(  a\left(  bc\right)
\right)  d=\left(  \mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  \operatorname{id}_{A}\otimes\mu_{A}\otimes
\operatorname{id}_{A}\right)  \right)  \left(  a\otimes b\otimes c\otimes
d\right)  .
\end{align*}
In other words, the two maps $\mu_{A}\circ\left(  \mu_{A}\otimes\mu
_{A}\right)  $ and $\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  \operatorname{id}_{A}\otimes\mu_{A}\otimes
\operatorname{id}_{A}\right)  $ are equal on every pure tensor. But since
these two maps are $k$-linear, this yields that these maps must be identic
(because whenever two $k$-linear maps from a tensor product are equal on every
pure tensor, they must be identic). In other words, $\mu_{A}\circ\left(
\mu_{A}\otimes\mu_{A}\right)  =\mu_{A}\circ\left(  \mu_{A}\otimes
\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}_{A}\otimes\mu
_{A}\otimes\operatorname{id}_{A}\right)  $, so that Lemma~\ref{Lemma604} is proven.
\end{proof}

\begin{lemma}
\label{Lemma605} Let $k$ be a field. Let $U$, $V$ and $W$ be three $k$-vector
spaces. Then,%
\[
\left(  \tau_{U,W}\otimes\operatorname{id}_{V}\right)  \circ\left(
\operatorname{id}_{U}\otimes\tau_{V,W}\right)  =\tau_{U\otimes V,W}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma605}.]Every $u\in U$, $v\in V$ and $w\in W$ satisfy%
\begin{align*}
&  \left(  \left(  \tau_{U,W}\otimes\operatorname{id}_{V}\right)  \circ\left(
\operatorname{id}_{U}\otimes\tau_{V,W}\right)  \right)  \left(  u\otimes
v\otimes w\right) \\
&  =\left(  \tau_{U,W}\otimes\operatorname{id}_{V}\right)  \underbrace{\left(
\left(  \operatorname{id}_{U}\otimes\tau_{V,W}\right)  \left(  u\otimes
v\otimes w\right)  \right)  }_{=\operatorname{id}_{U}\left(  u\right)
\otimes\tau_{V,W}\left(  v\otimes w\right)  }\\
&  =\left(  \tau_{U,W}\otimes\operatorname{id}_{V}\right)  \left(
\underbrace{\operatorname{id}_{U}\left(  u\right)  }_{=u}\otimes
\underbrace{\tau_{V,W}\left(  v\otimes w\right)  }_{\substack{=w\otimes
v\\\text{(by the definition of }\tau_{V,W}\text{)}}}\right) \\
&  =\left(  \tau_{U,W}\otimes\operatorname{id}_{V}\right)  \left(  u\otimes
w\otimes v\right)  =\underbrace{\tau_{U,W}\left(  u\otimes w\right)
}_{\substack{=w\otimes u\\\text{(by the definition of }\tau_{U,W}\text{)}%
}}\otimes\underbrace{\operatorname{id}_{V}\left(  v\right)  }_{=v}\\
&  =w\otimes u\otimes v=\tau_{U\otimes V,W}\left(  u\otimes v\otimes w\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau_{U\otimes V,W}\left(
u\otimes v\otimes w\right)  =w\otimes u\otimes v\text{ by the definition of
}\tau_{U\otimes V,W}\right)  .
\end{align*}
In other words, the two maps $\left(  \tau_{U,W}\otimes\operatorname{id}%
_{V}\right)  \circ\left(  \operatorname{id}_{U}\otimes\tau_{V,W}\right)  $ and
$\tau_{U\otimes V,W}$ are equal on every pure tensor. But since these two maps
are $k$-linear, this yields that these maps must be identic (because whenever
two $k$-linear maps from a tensor product are equal on every pure tensor, they
must be identic). In other words, $\left(  \tau_{U,W}\otimes\operatorname{id}%
_{V}\right)  \circ\left(  \operatorname{id}_{U}\otimes\tau_{V,W}\right)
=\tau_{U\otimes V,W}$. This proves Lemma~\ref{Lemma605}.
\end{proof}

\begin{lemma}
\label{Lemma606} Let $k$ be a field. Let $U$, $V$, $W$, $T$ and $Q$ be five
$k$-vector spaces. Let $r:U\otimes V\rightarrow Q$ be a $k$-linear map. Then,%
\begin{align*}
&  \left(  \operatorname{id}_{W}\otimes r\otimes\operatorname{id}_{T}\right)
\circ\left(  \tau_{U,W}\otimes\operatorname{id}_{V}\otimes\operatorname{id}%
_{T}\right)  \circ\left(  \operatorname{id}_{U}\otimes\tau_{V,W}%
\otimes\operatorname{id}_{T}\right) \\
&  =\left(  \tau_{Q,W}\otimes\operatorname{id}_{T}\right)  \circ\left(
r\otimes\operatorname{id}_{W}\otimes\operatorname{id}_{T}\right)  .
\end{align*}

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma606}.]By (\ref{(X)(X)}) (applied to $U\otimes
V\otimes W$, $U\otimes W\otimes V$, $W\otimes U\otimes V$, $T$, $T$, $T$,
$\operatorname{id}_{U}\otimes\tau_{V,W}$, $\tau_{U,W}\otimes\operatorname{id}%
_{V}$, $\operatorname{id}_{T}$, $\operatorname{id}_{T}$ instead of $U$, $V$,
$W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$,
$\alpha^{\prime}$, $\beta^{\prime}$), we have%
\[
\left(  \left(  \tau_{U,W}\otimes\operatorname{id}_{V}\right)  \circ\left(
\operatorname{id}_{U}\otimes\tau_{V,W}\right)  \right)  \otimes\left(
\operatorname{id}_{T}\circ\operatorname{id}_{T}\right)  =\left(  \tau
_{U,W}\otimes\operatorname{id}_{V}\otimes\operatorname{id}_{T}\right)
\circ\left(  \operatorname{id}_{U}\otimes\tau_{V,W}\otimes\operatorname{id}%
_{T}\right)  .
\]
Thus,%
\begin{align}
&  \left(  \tau_{U,W}\otimes\operatorname{id}_{V}\otimes\operatorname{id}%
_{T}\right)  \circ\left(  \operatorname{id}_{U}\otimes\tau_{V,W}%
\otimes\operatorname{id}_{T}\right) \nonumber\\
&  =\underbrace{\left(  \left(  \tau_{U,W}\otimes\operatorname{id}_{V}\right)
\circ\left(  \operatorname{id}_{U}\otimes\tau_{V,W}\right)  \right)
}_{\substack{=\tau_{U\otimes V,W}\\\text{(by Lemma~\ref{Lemma605})}}%
}\otimes\underbrace{\left(  \operatorname{id}_{T}\circ\operatorname{id}%
_{T}\right)  }_{=\operatorname{id}_{T}}=\tau_{U\otimes V,W}\otimes
\operatorname{id}_{T}. \label{606.pf.0}%
\end{align}


On the other hand, by (\ref{(X)(X)}) (applied to $U\otimes V\otimes W$,
$W\otimes U\otimes V$, $W\otimes Q$, $T$, $T$, $T$, $\tau_{U\otimes V,W}$,
$\operatorname{id}_{W}\otimes r$, $\operatorname{id}_{T}$, $\operatorname{id}%
_{T}$ instead of $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$,
$\alpha$, $\beta$, $\alpha^{\prime}$, $\beta^{\prime}$), we have%
\begin{equation}
\left(  \left(  \operatorname{id}_{W}\otimes r\right)  \circ\tau_{U\otimes
V,W}\right)  \otimes\left(  \operatorname{id}_{T}\circ\operatorname{id}%
_{T}\right)  =\left(  \operatorname{id}_{W}\otimes r\otimes\operatorname{id}%
_{T}\right)  \circ\left(  \tau_{U\otimes V,W}\otimes\operatorname{id}%
_{T}\right)  . \label{606.pf.2}%
\end{equation}


By Proposition~\ref{Proposition35} \textbf{(a)} (applied to $U\otimes V$, $W$,
$Q$, $W$, $r$ and $\operatorname{id}_{W}$ instead of $V$, $W$, $V^{\prime}$,
$W^{\prime}$, $f$ and $g$), we have%
\begin{equation}
\left(  \operatorname{id}_{W}\otimes r\right)  \circ\tau_{U\otimes V,W}%
=\tau_{Q,W}\circ\left(  r\otimes\operatorname{id}_{W}\right)  .
\label{606.pf.3}%
\end{equation}


Besides, by (\ref{(X)(X)}) (applied to $U\otimes V\otimes W$, $Q\otimes W$,
$W\otimes Q$, $T$, $T$, $T$, $r\otimes\operatorname{id}_{W}$, $\tau_{Q,W}$,
$\operatorname{id}_{T}$, $\operatorname{id}_{T}$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}%
$, $\beta^{\prime}$), we have%
\begin{equation}
\left(  \tau_{Q,W}\circ\left(  r\otimes\operatorname{id}_{W}\right)  \right)
\otimes\left(  \operatorname{id}_{T}\circ\operatorname{id}_{T}\right)
=\left(  \tau_{Q,W}\otimes\operatorname{id}_{T}\right)  \circ\left(
r\otimes\operatorname{id}_{W}\otimes\operatorname{id}_{T}\right)  .
\label{606.pf.4}%
\end{equation}


Now,%
\begin{align*}
&  \left(  \operatorname{id}_{W}\otimes r\otimes\operatorname{id}_{T}\right)
\circ\underbrace{\left(  \tau_{U,W}\otimes\operatorname{id}_{V}\otimes
\operatorname{id}_{T}\right)  \circ\left(  \operatorname{id}_{U}\otimes
\tau_{V,W}\otimes\operatorname{id}_{T}\right)  }_{\substack{=\tau_{U\otimes
V,W}\otimes\operatorname{id}_{T}\\\text{(by (\ref{606.pf.0}))}}}\\
&  =\left(  \operatorname{id}_{W}\otimes r\otimes\operatorname{id}_{T}\right)
\circ\left(  \tau_{U\otimes V,W}\otimes\operatorname{id}_{T}\right) \\
&  =\underbrace{\left(  \left(  \operatorname{id}_{W}\otimes r\right)
\circ\tau_{U\otimes V,W}\right)  }_{\substack{=\tau_{Q,W}\circ\left(
r\otimes\operatorname{id}_{W}\right)  \\\text{(by (\ref{606.pf.3}))}}%
}\otimes\left(  \operatorname{id}_{T}\circ\operatorname{id}_{T}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{606.pf.2})}\right) \\
&  =\left(  \tau_{Q,W}\circ\left(  r\otimes\operatorname{id}_{W}\right)
\right)  \otimes\left(  \operatorname{id}_{T}\circ\operatorname{id}%
_{T}\right)  =\left(  \tau_{Q,W}\otimes\operatorname{id}_{T}\right)
\circ\left(  r\otimes\operatorname{id}_{W}\otimes\operatorname{id}_{T}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{606.pf.4})}\right)  .
\end{align*}
This proves Lemma~\ref{Lemma606}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition601}.]Let $g=f^{\ast\left(  -1\right)
}$. Then, $f\ast g=f\ast f^{\ast\left(  -1\right)  }=e_{C,A}$ and $g\ast
f=f^{\ast\left(  -1\right)  }\ast f=e_{C,A}$.

We are now going to show that%
\begin{equation}
\left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)  \right)
\ast\left(  f\circ\mu_{C}\right)  =e_{C\otimes C,A} \label{601.pf.left}%
\end{equation}
and%
\begin{equation}
\left(  f\circ\mu_{C}\right)  \ast\left(  g\circ\mu_{C}\right)  =e_{C\otimes
C,A} \label{601.pf.right}%
\end{equation}
(where $\ast$ stands for convolution of maps from $C\otimes C$ to $A$). Once
these formulas are shown, it will be easy to conclude that $\mu
_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)  =g\circ\mu_{C}$,
which will settle the hardest part of Proposition~\ref{Proposition601}.

But let us prove (\ref{601.pf.left}) and (\ref{601.pf.right}) now:

\textit{Proof of (\ref{601.pf.left}):} Since $f$ is a $k$-algebra
homomorphism, we have $f\circ\mu_{C}=\mu_{A}\circ\left(  f\otimes f\right)  $.
Thus,%
\begin{align}
&  \left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast\underbrace{\left(  f\circ\mu_{C}\right)  }_{=\mu_{A}\circ\left(
f\otimes f\right)  }\nonumber\\
&  =\left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast\left(  \mu_{A}\circ\left(  f\otimes f\right)  \right)
\nonumber\\
&  =\mu_{A}\circ\underbrace{\left(  \left(  \mu_{A^{\operatorname*{op}}}%
\circ\left(  g\otimes g\right)  \right)  \otimes\left(  \mu_{A}\circ\left(
f\otimes f\right)  \right)  \right)  }_{\substack{=\left(  \mu
_{A^{\operatorname*{op}}}\otimes\mu_{A}\right)  \circ\left(  g\otimes g\otimes
f\otimes f\right)  \\\text{(by (\ref{(X)(X)}), applied to }C\otimes C\text{,
}A\otimes A\text{, }A\text{, }C\otimes C\text{, }A\otimes A\text{, }%
A\text{,}\\g\otimes g\text{, }\mu_{A^{\operatorname*{op}}}\text{, }f\otimes
f\text{, }\mu_{A}\text{ instead of }U\text{, }V\text{, }W\text{, }U^{\prime
}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{,
}\alpha^{\prime}\text{, }\beta^{\prime}\text{)}}}\circ\underbrace{\Delta
_{C\otimes C}}_{\substack{=\left(  \operatorname{id}_{C}\otimes\tau
_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \\\text{(by the definition of the }k\text{-coalgebra
}C\otimes C\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of convolution}\right)
\nonumber\\
&  =\mu_{A}\circ\left(  \underbrace{\mu_{A^{\operatorname*{op}}}%
}_{\substack{=\mu_{A}\circ\tau_{A,A}\\\text{(by the definition}\\\text{of
the}\\k\text{-algebra }A^{\operatorname*{op}}\text{)}}}\otimes\underbrace{\mu
_{A}}_{=\mu_{A}\circ\operatorname{id}_{A\otimes A}}\right)  \circ
\underbrace{\left(  g\otimes g\otimes f\otimes f\right)  \circ\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
}_{\substack{=\left(  \operatorname{id}_{A}\otimes\tau_{A,A}\otimes
\operatorname{id}_{A}\right)  \circ\left(  g\otimes f\otimes g\otimes
f\right)  \\\text{(since Proposition~\ref{Proposition35} \textbf{(b)} (applied
to }C\text{, }C\text{, }C\text{, }C\text{,}\\A\text{, }A\text{, }A\text{,
}A\text{, }g\text{, }f\text{, }g\text{, }f\text{ instead of}\\U\text{,
}V\text{, }W\text{, }T\text{, }U^{\prime}\text{, }V^{\prime}\text{, }%
W^{\prime}\text{, }T^{\prime}\text{, }e\text{, }f\text{, }g\text{, }h\text{)
yields}\\\left(  \operatorname{id}_{A}\otimes\tau_{A,A}\otimes
\operatorname{id}_{A}\right)  \circ\left(  g\otimes f\otimes g\otimes
f\right)  \\=\left(  g\otimes g\otimes f\otimes f\right)  \circ\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
\text{)}}}\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =\mu_{A}\circ\underbrace{\left(  \left(  \mu_{A}\circ\tau_{A,A}\right)
\otimes\left(  \mu_{A}\circ\operatorname{id}_{A\otimes A}\right)  \right)
}_{\substack{=\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\left(  \tau
_{A,A}\otimes\operatorname{id}_{A\otimes A}\right)  \\\text{(by (\ref{(X)(X)}%
), applied to }A\otimes A\text{, }A\otimes A\text{, }A\text{,}\\A\otimes
A\text{, }A\otimes A\text{, }A\text{, }\tau_{A,A}\text{, }\mu_{A}\text{,
}\operatorname{id}_{A\otimes A}\text{, }\mu_{A}\\\text{instead of }U\text{,
}V\text{, }W\text{, }U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{,
}\alpha\text{, }\beta\text{, }\alpha^{\prime}\text{, }\beta^{\prime}\text{)}%
}}\circ\left(  \operatorname{id}_{A}\otimes\tau_{A,A}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  g\otimes f\otimes g\otimes f\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =\underbrace{\mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)
}_{\substack{=\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{A}\otimes\mu_{A}\otimes\operatorname{id}%
_{A}\right)  \\\text{(by Lemma~\ref{Lemma604})}}}\circ\left(  \tau
_{A,A}\otimes\underbrace{\operatorname{id}_{A\otimes A}}_{=\operatorname{id}%
_{A}\otimes\operatorname{id}_{A}}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\tau_{A,A}\otimes\operatorname{id}_{A}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  g\otimes f\otimes g\otimes f\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\underbrace{\left(  \operatorname{id}_{A}\otimes\mu_{A}\otimes
\operatorname{id}_{A}\right)  \circ\left(  \tau_{A,A}\otimes\operatorname{id}%
_{A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\tau_{A,A}\otimes\operatorname{id}_{A}\right)  }%
_{\substack{=\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\otimes\operatorname{id}%
_{A}\right)  \\\text{(by Lemma~\ref{Lemma606}, applied to }U=A\text{,
}V=A\text{, }W=A\text{, }T=A\text{, }Q=A\text{ and }r=\mu_{A}\text{)}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  g\otimes f\otimes g\otimes f\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)  \circ\left(
\mu_{A}\otimes\underbrace{\operatorname{id}_{A}\otimes\operatorname{id}_{A}%
}_{=\operatorname{id}_{A\otimes A}}\right)  \circ\left(  g\otimes f\otimes
g\otimes f\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)  \circ\left(
\mu_{A}\otimes\operatorname{id}_{A\otimes A}\right)  \circ\left(  g\otimes
f\otimes g\otimes f\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{601.pf.2}%
\end{align}


Applying (\ref{(X)(X)}) to $C$, $C$, $C\otimes C$, $C$, $C\otimes C$,
$C\otimes C$, $\operatorname{id}_{C}$, $\Delta_{C}$, $\Delta_{C}$,
$\operatorname{id}_{C\otimes C}$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain%
\[
\left(  \Delta_{C}\circ\operatorname{id}_{C}\right)  \otimes\left(
\operatorname{id}_{C\otimes C}\circ\Delta_{C}\right)  =\left(  \Delta
_{C}\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right)  .
\]
Since $\Delta_{C}\circ\operatorname{id}_{C}=\Delta_{C}$ and $\operatorname{id}%
_{C\otimes C}\circ\Delta_{C}=\Delta_{C}$, this rewrites as
\begin{equation}
\Delta_{C}\otimes\Delta_{C}=\left(  \Delta_{C}\otimes\operatorname{id}%
_{C\otimes C}\right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta
_{C}\right)  . \label{601.pf.4}%
\end{equation}


Applying (\ref{(X)(X)}) to $C\otimes C$, $C\otimes C$, $A\otimes A$, $C\otimes
C$, $A\otimes A$, $A\otimes A$, $\operatorname{id}_{C\otimes C}$, $g\otimes
f$, $g\otimes f$, $\operatorname{id}_{A\otimes A}$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}%
$, $\beta^{\prime}$, we obtain%
\[
\left(  \left(  g\otimes f\right)  \circ\operatorname{id}_{C\otimes C}\right)
\otimes\left(  \operatorname{id}_{A\otimes A}\circ\left(  g\otimes f\right)
\right)  =\left(  g\otimes f\otimes\operatorname{id}_{A\otimes A}\right)
\circ\left(  \operatorname{id}_{C\otimes C}\otimes g\otimes f\right)  .
\]
Since $\left(  g\otimes f\right)  \circ\operatorname{id}_{C\otimes C}=g\otimes
f$ and $\operatorname{id}_{A\otimes A}\circ\left(  g\otimes f\right)
=g\otimes f$, this rewrites as
\begin{equation}
g\otimes f\otimes g\otimes f=\left(  g\otimes f\otimes\operatorname{id}%
_{A\otimes A}\right)  \circ\left(  \operatorname{id}_{C\otimes C}\otimes
g\otimes f\right)  . \label{601.pf.5}%
\end{equation}


By (\ref{504.pf.comm}) (applied to $P=C$, $Q=C\otimes C$, $R=C\otimes C$,
$S=A\otimes A$, $\gamma=\Delta_{C}$ and $\delta=g\otimes f$), we have%
\begin{equation}
\left(  \operatorname{id}_{C\otimes C}\otimes g\otimes f\right)  \circ\left(
\Delta_{C}\otimes\operatorname{id}_{C\otimes C}\right)  =\left(  \Delta
_{C}\otimes\operatorname{id}_{A\otimes A}\right)  \circ\left(
\operatorname{id}_{C}\otimes g\otimes f\right)  . \label{601.pf.6}%
\end{equation}


By the definition of convolution, $g\ast f=\mu_{A}\circ\left(  g\otimes
f\right)  \circ\Delta_{C}$, so that%
\begin{equation}
\mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta_{C}=g\ast f=e_{C,A}.
\label{601.pf.7}%
\end{equation}


Applying (\ref{(X)(X)}) to $C\otimes C$, $A\otimes A$, $A$, $A\otimes A$,
$A\otimes A$, $A\otimes A$, $g\otimes f$, $\mu_{A}$, $\operatorname{id}%
_{A\otimes A}$, $\operatorname{id}_{A\otimes A}$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}%
$, $\beta^{\prime}$, we obtain%
\[
\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)  \otimes\left(
\operatorname{id}_{A\otimes A}\circ\operatorname{id}_{A\otimes A}\right)
=\left(  \mu_{A}\otimes\operatorname{id}_{A\otimes A}\right)  \circ\left(
g\otimes f\otimes\operatorname{id}_{A\otimes A}\right)  .
\]
Since $\operatorname{id}_{A\otimes A}\circ\operatorname{id}_{A\otimes
A}=\operatorname{id}_{A\otimes A}$, this rewrites as
\begin{equation}
\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)  \otimes
\operatorname{id}_{A\otimes A}=\left(  \mu_{A}\otimes\operatorname{id}%
_{A\otimes A}\right)  \circ\left(  g\otimes f\otimes\operatorname{id}%
_{A\otimes A}\right)  . \label{601.pf.9}%
\end{equation}


Applying (\ref{(X)(X)}) to $C$, $C\otimes C$, $A$, $A\otimes A$, $A\otimes A$,
$A\otimes A$, $\Delta_{C}$, $\mu_{A}\circ\left(  g\otimes f\right)  $,
$\operatorname{id}_{A\otimes A}$, $\operatorname{id}_{A\otimes A}$ instead of
$U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$,
$\alpha^{\prime}$, $\beta^{\prime}$, we obtain%
\begin{equation}
\left(  \mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta_{C}\right)
\otimes\left(  \operatorname{id}_{A\otimes A}\circ\operatorname{id}_{A\otimes
A}\right)  =\left(  \left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)
\otimes\operatorname{id}_{A\otimes A}\right)  \circ\left(  \Delta_{C}%
\otimes\operatorname{id}_{A\otimes A}\right)  . \label{601.pf.10}%
\end{equation}
Thus,%
\begin{align}
&  \underbrace{\left(  \mu_{A}\otimes\operatorname{id}_{A\otimes A}\right)
\circ\left(  g\otimes f\otimes\operatorname{id}_{A\otimes A}\right)
}_{\substack{=\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)
\otimes\operatorname{id}_{A\otimes A}\\\text{(by (\ref{601.pf.9}))}}%
}\circ\left(  \Delta_{C}\otimes\operatorname{id}_{A\otimes A}\right)
\nonumber\\
&  =\left(  \left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)
\otimes\operatorname{id}_{A\otimes A}\right)  \circ\left(  \Delta_{C}%
\otimes\operatorname{id}_{A\otimes A}\right) \nonumber\\
&  =\underbrace{\left(  \mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta
_{C}\right)  }_{\substack{=e_{C,A}\\\text{(by (\ref{601.pf.7}))}}%
}\otimes\underbrace{\left(  \operatorname{id}_{A\otimes A}\circ
\operatorname{id}_{A\otimes A}\right)  }_{=\operatorname{id}_{C\otimes C}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{601.pf.10})}\right) \nonumber\\
&  =e_{C,A}\otimes\operatorname{id}_{A\otimes A}. \label{601.pf.12}%
\end{align}


Now, (\ref{601.pf.2}) becomes%
\begin{align}
&  \left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast\left(  f\circ\mu_{C}\right) \nonumber\\
&  =\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)  \circ\left(
\mu_{A}\otimes\operatorname{id}_{A\otimes A}\right)  \circ\underbrace{\left(
g\otimes f\otimes g\otimes f\right)  }_{\substack{=\left(  g\otimes
f\otimes\operatorname{id}_{A\otimes A}\right)  \circ\left(  \operatorname{id}%
_{C\otimes C}\otimes g\otimes f\right)  \\\text{(by (\ref{601.pf.5}))}}%
}\circ\underbrace{\left(  \Delta_{C}\otimes\Delta_{C}\right)  }%
_{\substack{=\left(  \Delta_{C}\otimes\operatorname{id}_{C\otimes C}\right)
\circ\left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)  \\\text{(by
(\ref{601.pf.4}))}}}\nonumber\\
&  =\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \mu_{A}\otimes\operatorname{id}_{A\otimes
A}\right)  \circ\left(  g\otimes f\otimes\operatorname{id}_{A\otimes
A}\right)  \circ\underbrace{\left(  \operatorname{id}_{C\otimes C}\otimes
g\otimes f\right)  \circ\left(  \Delta_{C}\otimes\operatorname{id}_{C\otimes
C}\right)  }_{\substack{=\left(  \Delta_{C}\otimes\operatorname{id}_{A\otimes
A}\right)  \circ\left(  \operatorname{id}_{C}\otimes g\otimes f\right)
\\\text{(by (\ref{601.pf.6}))}}}\circ\left(  \operatorname{id}_{C}%
\otimes\Delta_{C}\right) \nonumber\\
&  =\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \circ\underbrace{\left(  \mu_{A}\otimes
\operatorname{id}_{A\otimes A}\right)  \circ\left(  g\otimes f\otimes
\operatorname{id}_{A\otimes A}\right)  \circ\left(  \Delta_{C}\otimes
\operatorname{id}_{A\otimes A}\right)  }_{\substack{=e_{C,A}\otimes
\operatorname{id}_{A\otimes A}\\\text{(by (\ref{601.pf.12}))}}}\circ\left(
\operatorname{id}_{C}\otimes g\otimes f\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =\underbrace{\mu_{A}\circ\left(  \mu_{A}\otimes\operatorname{id}%
_{A}\right)  }_{\substack{=\mu_{A}\circ\left(  \operatorname{id}_{A}\otimes
\mu_{A}\right)  \\\text{(by the axioms of an}\\\text{algebra, since }A\text{
is an}\\\text{algebra)}}}\circ\left(  \tau_{A,A}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  e_{C,A}\otimes\underbrace{\operatorname{id}%
_{A\otimes A}}_{=\operatorname{id}_{A}\otimes\operatorname{id}_{A}}\right)
\circ\left(  \operatorname{id}_{C}\otimes g\otimes f\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =\mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)
\circ\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)  \circ\left(
e_{C,A}\otimes\operatorname{id}_{A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{C}\otimes g\otimes f\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right)  . \label{601.pf.20}%
\end{align}


On the other hand, Proposition~\ref{Proposition35} \textbf{(a)} (applied to
$C$, $A$, $A$, $A$, $e_{C,A}$, $\operatorname{id}_{A}$ instead of $V$, $W$,
$V^{\prime}$, $W^{\prime}$, $f$, $g$) yields%
\begin{equation}
\left(  \operatorname{id}_{A}\otimes e_{C,A}\right)  \circ\tau_{C,A}%
=\tau_{A,A}\circ\left(  e_{C,A}\otimes\operatorname{id}_{A}\right)  .
\label{601.pf.22}%
\end{equation}
But applying (\ref{(X)(X)}) to $C\otimes A$, $A\otimes A$, $A\otimes A$, $A$,
$A$, $A$, $e_{C,A}\otimes\operatorname{id}_{A}$, $\tau_{A,A}$,
$\operatorname{id}_{A}$, $\operatorname{id}_{A}$ instead of $U$, $V$, $W$,
$U^{\prime}$, $V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}%
$, $\beta^{\prime}$, we obtain%
\[
\left(  \tau_{A,A}\circ\left(  e_{C,A}\otimes\operatorname{id}_{A}\right)
\right)  \otimes\left(  \operatorname{id}_{A}\circ\operatorname{id}%
_{A}\right)  =\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)
\circ\left(  e_{C,A}\otimes\operatorname{id}_{A}\otimes\operatorname{id}%
_{A}\right)  .
\]
Thus,%
\begin{align}
&  \left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)  \circ\left(
e_{C,A}\otimes\operatorname{id}_{A}\otimes\operatorname{id}_{A}\right)
\nonumber\\
&  =\underbrace{\left(  \tau_{A,A}\circ\left(  e_{C,A}\otimes\operatorname{id}%
_{A}\right)  \right)  }_{\substack{=\left(  \operatorname{id}_{A}\otimes
e_{C,A}\right)  \circ\tau_{C,A}\\\text{(by (\ref{601.pf.22}))}}}\otimes\left(
\operatorname{id}_{A}\circ\operatorname{id}_{A}\right)  =\left(  \left(
\operatorname{id}_{A}\otimes e_{C,A}\right)  \circ\tau_{C,A}\right)
\otimes\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)
\nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\underbrace{e_{C,A}}_{\substack{=\eta
_{A}\circ\varepsilon_{C}\\\text{(by the definition of }e_{C,A}\text{)}%
}}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau_{C,A}\otimes
\operatorname{id}_{A}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}), applied to }C\otimes A\text{, }A\otimes C\text{,
}A\otimes A\text{, }A\text{, }A\text{, }A\text{, }\tau_{C,A}\text{,
}\operatorname{id}_{A}\otimes e_{C,A}\text{, }\operatorname{id}_{A}\text{,
}\operatorname{id}_{A}\\
\text{instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{, }V^{\prime
}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }\alpha^{\prime
}\text{, }\beta^{\prime}%
\end{array}
\right) \nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\left(  \eta_{A}\circ\varepsilon
_{C}\right)  \otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  . \label{601.pf.23}%
\end{align}


But (\ref{504.pf.idid}) (applied to $P=A$, $Q=A$, $R_{1}=C$, $R_{2}=k$,
$R_{3}=A$, $\varphi=\varepsilon_{C}$ and $\psi=\eta_{A}$) yields
\begin{equation}
\left(  \operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  =\operatorname{id}_{A}\otimes\left(
\eta_{A}\circ\varepsilon_{C}\right)  \otimes\operatorname{id}_{A}.
\label{601.pf.24}%
\end{equation}
Hence, (\ref{601.pf.23}) becomes%
\begin{align}
&  \left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)  \circ\left(
e_{C,A}\otimes\operatorname{id}_{A}\otimes\operatorname{id}_{A}\right)
\nonumber\\
&  =\underbrace{\left(  \operatorname{id}_{A}\otimes\left(  \eta_{A}%
\circ\varepsilon_{C}\right)  \otimes\operatorname{id}_{A}\right)
}_{\substack{=\left(  \operatorname{id}_{A}\otimes\eta_{A}\otimes
\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}_{A}%
\otimes\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \\\text{(by
(\ref{601.pf.24}))}}}\circ\left(  \tau_{C,A}\otimes\operatorname{id}%
_{A}\right) \nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau_{C,A}\otimes
\operatorname{id}_{A}\right)  . \label{601.pf.26}%
\end{align}


Now, let $\operatorname{kan}_{k\otimes A,A}$ be the canonical isomorphism
$k\otimes A\rightarrow A$ which sends every $\lambda\otimes a\in k\otimes A$
to $\lambda a\in A$. Then, by the axioms of an algebra, we have $\mu_{A}%
\circ\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  =\operatorname{kan}%
_{k\otimes A,A}$ (since $A$ is a $k$-algebra).

Applying (\ref{(X)(X)}) to $A$, $A$, $A$, $k\otimes A$, $A\otimes A$, $A$,
$\operatorname{id}_{A}$, $\operatorname{id}_{A}$, $\eta_{A}\otimes
\operatorname{id}_{A}$, $\mu_{A}$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain%
\[
\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)  \otimes\left(
\mu_{A}\circ\left(  \eta_{A}\otimes\operatorname{id}_{A}\right)  \right)
=\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}_{A}\right)  .
\]
Thus,%
\begin{align}
\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ\left(
\operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}_{A}\right)   &
=\underbrace{\left(  \operatorname{id}_{A}\circ\operatorname{id}_{A}\right)
}_{=\operatorname{id}_{A}}\otimes\underbrace{\left(  \mu_{A}\circ\left(
\eta_{A}\otimes\operatorname{id}_{A}\right)  \right)  }_{=\operatorname{kan}%
_{k\otimes A,A}}\nonumber\\
&  =\operatorname{id}_{A}\otimes\operatorname{kan}_{k\otimes A,A}.
\label{601.pf.27}%
\end{align}


Now,%
\begin{align}
&  \left(  \operatorname{id}_{A}\otimes\mu_{A}\right)  \circ
\underbrace{\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)
\circ\left(  e_{C,A}\otimes\operatorname{id}_{A}\otimes\operatorname{id}%
_{A}\right)  }_{\substack{=\left(  \operatorname{id}_{A}\otimes\eta_{A}%
\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}_{A}%
\otimes\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(
\tau_{C,A}\otimes\operatorname{id}_{A}\right)  \\\text{(by (\ref{601.pf.26}%
))}}}\nonumber\\
&  =\underbrace{\left(  \operatorname{id}_{A}\otimes\mu_{A}\right)
\circ\left(  \operatorname{id}_{A}\otimes\eta_{A}\otimes\operatorname{id}%
_{A}\right)  }_{\substack{=\operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\\\text{(by (\ref{601.pf.27}))}}}\circ\left(  \operatorname{id}%
_{A}\otimes\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(
\tau_{C,A}\otimes\operatorname{id}_{A}\right) \nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\operatorname{kan}_{k\otimes
A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes\varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau_{C,A}\otimes
\operatorname{id}_{A}\right)  . \label{601.pf.29}%
\end{align}
Thus, (\ref{601.pf.20}) becomes%
\begin{align}
&  \left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast\left(  f\circ\mu_{C}\right) \nonumber\\
&  =\mu_{A}\circ\underbrace{\left(  \operatorname{id}_{A}\otimes\mu
_{A}\right)  \circ\left(  \tau_{A,A}\otimes\operatorname{id}_{A}\right)
\circ\left(  e_{C,A}\otimes\operatorname{id}_{A}\otimes\operatorname{id}%
_{A}\right)  }_{\substack{=\left(  \operatorname{id}_{A}\otimes
\operatorname{kan}_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(
\tau_{C,A}\otimes\operatorname{id}_{A}\right)  \\\text{(by (\ref{601.pf.29}%
))}}}\circ\left(  \operatorname{id}_{C}\otimes g\otimes f\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =\mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes g\otimes f\right)  \circ\left(  \operatorname{id}_{C}\otimes
\Delta_{C}\right)  . \label{601.pf.32}%
\end{align}


From this point on, our proof will not be analogous to the proof of
Proposition~\ref{Proposition551} anymore. We can easily show that%
\begin{align}
&  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes g\otimes f\right) \nonumber\\
&  =\operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)  \right)  .
\label{601.pf.39}%
\end{align}
\footnote{\textit{Proof of (\ref{601.pf.39}):} Every $c\in C$, $d\in C$ and
$e\in C$ satisfy%
\begin{align*}
&  \left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes g\otimes f\right)  \right)  \left(  c\otimes d\otimes e\right) \\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  \right)  \underbrace{\left(
\left(  \operatorname{id}_{C}\otimes g\otimes f\right)  \left(  c\otimes
d\otimes e\right)  \right)  }_{=\operatorname{id}_{C}\left(  c\right)  \otimes
g\left(  d\right)  \otimes f\left(  e\right)  }\\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  \right)  \left(
\underbrace{\operatorname{id}_{C}\left(  c\right)  }_{=c}\otimes g\left(
d\right)  \otimes f\left(  e\right)  \right) \\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  \right)  \left(  c\otimes g\left(
d\right)  \otimes f\left(  e\right)  \right) \\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \right)
\underbrace{\left(  \left(  \tau_{C,A}\otimes\operatorname{id}_{A}\right)
\left(  c\otimes g\left(  d\right)  \otimes f\left(  e\right)  \right)
\right)  }_{=\tau_{C,A}\left(  c\otimes g\left(  d\right)  \right)
\otimes\operatorname{id}_{A}\left(  f\left(  e\right)  \right)  }\\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(
\underbrace{\tau_{C,A}\left(  c\otimes g\left(  d\right)  \right)
}_{\substack{=g\left(  d\right)  \otimes c\\\text{(by the definition}%
\\\text{of }\tau_{C,A}\text{)}}}\otimes\underbrace{\operatorname{id}%
_{A}\left(  f\left(  e\right)  \right)  }_{=f\left(  e\right)  }\right) \\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(  g\left(
d\right)  \otimes c\otimes f\left(  e\right)  \right) \\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \right)  \underbrace{\left(  \left(  \operatorname{id}%
_{A}\otimes\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \left(
g\left(  d\right)  \otimes c\otimes f\left(  e\right)  \right)  \right)
}_{=\operatorname{id}_{A}\left(  g\left(  d\right)  \right)  \otimes
\varepsilon_{C}\left(  c\right)  \otimes\operatorname{id}_{A}\left(  f\left(
e\right)  \right)  }\\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \right)  \left(  \underbrace{\operatorname{id}%
_{A}\left(  g\left(  d\right)  \right)  }_{=g\left(  d\right)  }%
\otimes\varepsilon_{C}\left(  c\right)  \otimes\underbrace{\operatorname{id}%
_{A}\left(  f\left(  e\right)  \right)  }_{=f\left(  e\right)  }\right) \\
&  =\left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \right)  \left(  g\left(  d\right)  \otimes
\varepsilon_{C}\left(  c\right)  \otimes f\left(  e\right)  \right) \\
&  =\mu_{A}\underbrace{\left(  \left(  \operatorname{id}_{A}\otimes
\operatorname{kan}_{k\otimes A,A}\right)  \left(  g\left(  d\right)
\otimes\varepsilon_{C}\left(  c\right)  \otimes f\left(  e\right)  \right)
\right)  }_{=\operatorname{id}_{A}\left(  g\left(  d\right)  \right)
\otimes\operatorname{kan}_{k\otimes A,A}\left(  \varepsilon_{C}\left(
c\right)  \otimes f\left(  e\right)  \right)  }\\
&  =\mu_{A}\left(  \operatorname{id}_{A}\left(  g\left(  d\right)  \right)
\otimes\operatorname{kan}_{k\otimes A,A}\left(  \varepsilon_{C}\left(
c\right)  \otimes f\left(  e\right)  \right)  \right) \\
&  =\underbrace{\operatorname{id}_{A}\left(  g\left(  d\right)  \right)
}_{=g\left(  d\right)  }\cdot\underbrace{\operatorname{kan}_{k\otimes
A,A}\left(  \varepsilon_{C}\left(  c\right)  \otimes f\left(  e\right)
\right)  }_{\substack{=\varepsilon_{C}\left(  c\right)  f\left(  e\right)
\\\text{(by the definition of }\operatorname{kan}_{k\otimes A,A}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu_{A}\right) \\
&  =g\left(  d\right)  \varepsilon_{C}\left(  c\right)  f\left(  e\right)
=\varepsilon_{C}\left(  c\right)  g\left(  d\right)  f\left(  e\right)
\end{align*}
and%
\begin{align*}
&  \left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)  \right)
\right)  \left(  c\otimes d\otimes e\right) \\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \underbrace{\left(  \left(
\operatorname{id}_{C}\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)
\right)  \right)  \left(  c\otimes d\otimes e\right)  \right)  }%
_{=\operatorname{id}_{C}\left(  c\right)  \otimes\left(  \mu_{A}\circ\left(
g\otimes f\right)  \right)  \left(  d\otimes e\right)  }\\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(
\underbrace{\operatorname{id}_{C}\left(  c\right)  }_{=c}\otimes
\underbrace{\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)  \left(
d\otimes e\right)  }_{=\mu_{A}\left(  \left(  g\otimes f\right)  \left(
d\otimes e\right)  \right)  }\right) \\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(  c\otimes\mu
_{A}\underbrace{\left(  \left(  g\otimes f\right)  \left(  d\otimes e\right)
\right)  }_{=g\left(  d\right)  \otimes f\left(  e\right)  }\right) \\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(  c\otimes
\underbrace{\mu_{A}\left(  g\left(  d\right)  \otimes f\left(  e\right)
\right)  }_{\substack{=g\left(  d\right)  f\left(  e\right)  \\\text{(by the
definition of }\mu_{A}\text{)}}}\right) \\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(  c\otimes g\left(
d\right)  f\left(  e\right)  \right)  =\left(  \mu_{A}\circ\operatorname{kan}%
_{k\otimes A,A}\right)  \underbrace{\left(  \left(  \varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \left(  c\otimes g\left(  d\right)
f\left(  e\right)  \right)  \right)  }_{=\varepsilon_{C}\left(  c\right)
\otimes\operatorname{id}_{A}\left(  g\left(  d\right)  f\left(  e\right)
\right)  }\\
&  =\operatorname{kan}_{k\otimes A,A}\left(  \varepsilon_{C}\left(  c\right)
\otimes\underbrace{\operatorname{id}_{A}\left(  g\left(  d\right)  f\left(
e\right)  \right)  }_{=g\left(  d\right)  f\left(  e\right)  }\right)
=\operatorname{kan}_{k\otimes A,A}\left(  \varepsilon_{C}\left(  c\right)
\otimes g\left(  d\right)  f\left(  e\right)  \right) \\
&  =\varepsilon_{C}\left(  c\right)  g\left(  d\right)  f\left(  e\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname{kan}%
_{k\otimes A,A}\right)  .
\end{align*}
Thus, every $c\in C$, $d\in C$ and $e\in C$ satisfy%
\begin{align*}
&  \left(  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes g\otimes f\right)  \right)  \left(  c\otimes d\otimes e\right) \\
&  =\varepsilon_{C}\left(  c\right)  g\left(  d\right)  f\left(  e\right)
=\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)  \right)
\right)  \left(  c\otimes d\otimes e\right)  .
\end{align*}
In other words, the two maps \newline$\mu_{A}\circ\left(  \operatorname{id}%
_{A}\otimes\operatorname{kan}_{k\otimes A,A}\right)  \circ\left(
\operatorname{id}_{A}\otimes\varepsilon_{C}\otimes\operatorname{id}%
_{A}\right)  \circ\left(  \tau_{C,A}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{C}\otimes g\otimes f\right)  $ and
\newline$\operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)  \right)  $ are
equal on every pure tensor. But since these two maps are $k$-linear, this
yields that these maps must be identic (because whenever two $k$-linear maps
from a tensor product are equal on every pure tensor, they must be identic).
In other words,
\begin{align*}
&  \mu_{A}\circ\left(  \operatorname{id}_{A}\otimes\operatorname{kan}%
_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}_{A}\otimes
\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \tau
_{C,A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes g\otimes f\right) \\
&  =\operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)  \right)  \right)  .
\end{align*}
This proves (\ref{601.pf.39}).}

Also, applying (\ref{(X)(X)}) to $C$, $C$, $C$, $C$, $C\otimes C$, $A$,
$\operatorname{id}_{C}$, $\operatorname{id}_{C}$, $\Delta_{C}$, $\mu_{A}%
\circ\left(  g\otimes f\right)  $ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain
\[
\left(  \operatorname{id}_{C}\circ\operatorname{id}_{C}\right)  \otimes\left(
\mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta_{C}\right)  =\left(
\operatorname{id}_{C}\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)
\right)  \right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)
,
\]
so that%
\begin{align}
\left(  \operatorname{id}_{C}\otimes\left(  \mu_{A}\circ\left(  g\otimes
f\right)  \right)  \right)  \circ\left(  \operatorname{id}_{C}\otimes
\Delta_{C}\right)   &  =\underbrace{\left(  \operatorname{id}_{C}%
\circ\operatorname{id}_{C}\right)  }_{=\operatorname{id}_{C}}\otimes
\underbrace{\left(  \mu_{A}\circ\left(  g\otimes f\right)  \circ\Delta
_{C}\right)  }_{\substack{=e_{C,A}\\\text{(by (\ref{601.pf.7}))}}}\nonumber\\
&  =\operatorname{id}_{C}\otimes e_{C,A}. \label{601.pf.40}%
\end{align}


Now, (\ref{601.pf.32}) becomes%
\begin{align*}
&  \left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast\left(  f\circ\mu_{C}\right) \\
&  =\underbrace{\mu_{A}\circ\left(  \operatorname{id}_{A}\otimes
\operatorname{kan}_{k\otimes A,A}\right)  \circ\left(  \operatorname{id}%
_{A}\otimes\varepsilon_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(
\tau_{C,A}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes g\otimes f\right)  }_{\substack{=\operatorname{kan}_{k\otimes
A,A}\circ\left(  \varepsilon_{C}\otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{C}\otimes\left(  \mu_{A}\circ\left(  g\otimes
f\right)  \right)  \right)  \\\text{(by (\ref{601.pf.39}))}}}\circ\left(
\operatorname{id}_{C}\otimes\Delta_{C}\right) \\
&  =\operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \circ\underbrace{\left(
\operatorname{id}_{C}\otimes\left(  \mu_{A}\circ\left(  g\otimes f\right)
\right)  \right)  \circ\left(  \operatorname{id}_{C}\otimes\Delta_{C}\right)
}_{\substack{=\operatorname{id}_{C}\otimes e_{C,A}\\\text{(by (\ref{601.pf.40}%
))}}}\\
&  =\operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon_{C}%
\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}_{C}\otimes
e_{C,A}\right)  .
\end{align*}
Thus, any $c\in C$ and $d\in C$ satisfy%
\begin{align}
&  \left(  \left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast\left(  f\circ\mu_{C}\right)  \right)  \left(  c\otimes d\right)
\nonumber\\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \circ\left(  \operatorname{id}%
_{C}\otimes e_{C,A}\right)  \right)  \left(  c\otimes d\right) \nonumber\\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \underbrace{\left(  \left(
\operatorname{id}_{C}\otimes e_{C,A}\right)  \left(  c\otimes d\right)
\right)  }_{=\operatorname{id}_{C}\left(  c\right)  \otimes e_{C,A}\left(
d\right)  }\nonumber\\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(
\underbrace{\operatorname{id}_{C}\left(  c\right)  }_{=c}\otimes
\underbrace{e_{C,A}}_{\substack{=\eta_{A}\circ\varepsilon_{C}\\\text{(by the
definition}\\\text{of }e_{C,A}\text{)}}}\left(  d\right)  \right) \nonumber\\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(  c\otimes
\underbrace{\left(  \eta_{A}\circ\varepsilon_{C}\right)  \left(  d\right)
}_{\substack{=\eta_{A}\left(  \varepsilon_{C}\left(  d\right)  \right)
}}\right) \nonumber\\
&  =\left(  \operatorname{kan}_{k\otimes A,A}\circ\left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \right)  \left(  c\otimes\eta
_{A}\left(  \varepsilon_{C}\left(  d\right)  \right)  \right)
=\operatorname{kan}_{k\otimes A,A}\underbrace{\left(  \left(  \varepsilon
_{C}\otimes\operatorname{id}_{A}\right)  \left(  c\otimes\eta_{A}\left(
\varepsilon_{C}\left(  d\right)  \right)  \right)  \right)  }_{=\varepsilon
_{C}\left(  c\right)  \otimes\operatorname{id}_{A}\left(  \eta_{A}\left(
\varepsilon_{C}\left(  d\right)  \right)  \right)  }\nonumber\\
&  =\operatorname{kan}_{k\otimes A,A}\left(  \varepsilon_{C}\left(  c\right)
\otimes\underbrace{\operatorname{id}_{A}\left(  \eta_{A}\left(  \varepsilon
_{C}\left(  d\right)  \right)  \right)  }_{=\eta_{A}\left(  \varepsilon
_{C}\left(  d\right)  \right)  }\right)  =\operatorname{kan}_{k\otimes
A,A}\left(  \varepsilon_{C}\left(  c\right)  \otimes\eta_{A}\left(
\varepsilon_{C}\left(  d\right)  \right)  \right) \nonumber\\
&  =\varepsilon_{C}\left(  c\right)  \eta_{A}\left(  \varepsilon_{C}\left(
d\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname{kan}_{k\otimes A,A}\right) \nonumber\\
&  =\eta_{A}\left(  \varepsilon_{C}\left(  c\right)  \varepsilon_{C}\left(
d\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{A}\text{ is
}k\text{-linear}\right)  . \label{601.pf.75}%
\end{align}
On the other hand, any $c\in C$ and $d\in C$ satisfy $\varepsilon_{C\otimes
C}\left(  c\otimes d\right)  =\varepsilon_{C}\left(  c\right)  \varepsilon
_{C}\left(  d\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $c\in C$ and
$d\in C$. By the definition of the $k$-coalgebra $C\otimes C$, we have
$\varepsilon_{C\otimes C}=\operatorname{kan}_{k\otimes k,k}\circ\left(
\varepsilon_{C}\otimes\varepsilon_{C}\right)  $, where $\operatorname{kan}%
_{k\otimes k,k}:k\otimes k\rightarrow k$ is the canonical isomorphism which
sends every $\lambda\otimes\lambda^{\prime}\in k\otimes k$ to $\lambda
\lambda^{\prime}\in k$. Thus,%
\begin{align*}
\varepsilon_{C\otimes C}\left(  c\otimes d\right)   &  =\left(
\operatorname{kan}_{k\otimes k,k}\circ\left(  \varepsilon_{C}\otimes
\varepsilon_{C}\right)  \right)  \left(  c\otimes d\right)
=\operatorname{kan}_{k\otimes k,k}\underbrace{\left(  \left(  \varepsilon
_{C}\otimes\varepsilon_{C}\right)  \left(  c\otimes d\right)  \right)
}_{=\varepsilon_{C}\left(  c\right)  \otimes\varepsilon_{C}\left(  d\right)
}\\
&  =\operatorname{kan}_{k\otimes k,k}\left(  \varepsilon_{C}\left(  c\right)
\otimes\varepsilon_{C}\left(  d\right)  \right)  =\varepsilon_{C}\left(
c\right)  \varepsilon_{C}\left(  d\right)
\end{align*}
(by the definition of $\operatorname{kan}_{k\otimes k,k}$), qed.}. Hence,
every $c\in C$ and $d\in C$ satisfy%
\begin{align*}
e_{C\otimes C,A}\left(  c\otimes d\right)   &  =\left(  \eta_{A}%
\circ\varepsilon_{C\otimes C}\right)  \left(  c\otimes d\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }e_{C\otimes C,A}=\eta_{A}%
\circ\varepsilon_{C\otimes C}\text{ by the definition of }e_{C\otimes
C,A}\right) \\
&  =\eta_{A}\underbrace{\left(  \varepsilon_{C\otimes C}\left(  c\otimes
d\right)  \right)  }_{=\varepsilon_{C}\left(  c\right)  \varepsilon_{C}\left(
d\right)  }=\eta_{A}\left(  \varepsilon_{C}\left(  c\right)  \varepsilon
_{C}\left(  d\right)  \right)  .
\end{align*}
Comparing this with (\ref{601.pf.75}), we see that any $c\in C$ and $d\in C$
satisfy%
\[
\left(  \left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast\left(  f\circ\mu_{C}\right)  \right)  \left(  c\otimes d\right)
=e_{C\otimes C,A}\left(  c\otimes d\right)  .
\]
In other words, the two maps $\left(  \mu_{A^{\operatorname*{op}}}\circ\left(
g\otimes g\right)  \right)  \ast\left(  f\circ\mu_{C}\right)  $ and
$e_{C\otimes C,A}$ are equal on every pure tensor. But since these two maps
are $k$-linear, this yields that these maps must be identic (because whenever
two $k$-linear maps from a tensor product are equal on every pure tensor, they
must be identic). In other words, $\left(  \mu_{A^{\operatorname*{op}}}%
\circ\left(  g\otimes g\right)  \right)  \ast\left(  f\circ\mu_{C}\right)
=e_{C\otimes C,A}$, so that (\ref{601.pf.left}) is proven.

\textit{Proof of (\ref{601.pf.right}):} Since $C$ is a $k$-bialgebra, the
multiplication map $\mu_{C}$ of $C$ is a $k$-coalgebra homomorphism (by the
axioms of a bialgebra). Thus, $\left(  \mu_{C}\otimes\mu_{C}\right)
\circ\Delta_{C\otimes C}=\Delta_{C}\circ\mu_{C}$ and $\varepsilon_{C\otimes
C}=\varepsilon_{C}\circ\mu_{C}$.

By the definition of convolution, $f\ast g=\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$. Thus,%
\begin{equation}
\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}=f\ast g=e_{C,A}%
=\eta_{A}\circ\varepsilon_{C} \label{601.pf.90}%
\end{equation}
(by the definition of $e_{C,A}$).

By the definition of convolution,%
\begin{align*}
\left(  f\circ\mu_{C}\right)  \ast\left(  g\circ\mu_{C}\right)   &  =\mu
_{A}\circ\underbrace{\left(  \left(  f\circ\mu_{C}\right)  \otimes\left(
g\circ\mu_{C}\right)  \right)  }_{\substack{=\left(  f\otimes g\right)
\circ\left(  \mu_{C}\otimes\mu_{C}\right)  \\\text{(by (\ref{(X)(X)}), applied
to }C\otimes C\text{, }C\text{, }A\text{, }C\otimes C\text{, }C\text{,
}A\text{,}\\\mu_{C}\text{, }f\text{, }\mu_{C}\text{, }g\text{ instead of
}U\text{, }V\text{, }W\text{, }U^{\prime}\text{, }V^{\prime}\text{, }%
W^{\prime}\text{, }\alpha\text{, }\beta\text{, }\alpha^{\prime}\text{, }%
\beta^{\prime}\text{)}}}\circ\Delta_{C\otimes C}\\
&  =\mu_{A}\circ\left(  f\otimes g\right)  \circ\underbrace{\left(  \mu
_{C}\otimes\mu_{C}\right)  \circ\Delta_{C\otimes C}}_{=\Delta_{C}\circ\mu_{C}%
}=\underbrace{\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
}_{\substack{=\eta_{A}\circ\varepsilon_{C}\\\text{(by (\ref{601.pf.90}))}%
}}\circ\mu_{C}\\
&  =\eta_{A}\circ\underbrace{\varepsilon_{C}\circ\mu_{C}}_{=\varepsilon
_{C\otimes C}}=\eta_{A}\circ\varepsilon_{C\otimes C}=e_{C\otimes C,A}%
\end{align*}
(since $e_{C\otimes C,A}$ is defined as $\eta_{A}\circ\varepsilon_{C\otimes
C}$). This proves (\ref{601.pf.right}).

Now, let us finish the proof of Proposition~\ref{Proposition601}: Comparing
the equalities%
\begin{align*}
&  \underbrace{\left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes
g\right)  \right)  \ast\left(  f\circ\mu_{C}\right)  }_{\substack{=e_{C\otimes
C,A}\\\text{(by (\ref{601.pf.left}))}}}\ast\left(  g\circ\mu_{C}\right) \\
&  =e_{C\otimes C,A}\ast\left(  g\circ\mu_{C}\right)  =g\circ\mu_{C}%
\end{align*}
and%
\begin{align*}
&  \left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast\underbrace{\left(  f\circ\mu_{C}\right)  \ast\left(  g\circ
\mu_{C}\right)  }_{\substack{=e_{C\otimes C,A}\\\text{(by (\ref{601.pf.right}%
))}}}\\
&  =\left(  \mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)
\right)  \ast e_{C\otimes C,A}=\mu_{A^{\operatorname*{op}}}\circ\left(
g\otimes g\right)  ,
\end{align*}
we obtain
\begin{equation}
g\circ\mu_{C}=\mu_{A^{\operatorname*{op}}}\circ\left(  g\otimes g\right)  .
\label{601.pf.Delta}%
\end{equation}


We will now prove that $g\circ\eta_{C}=\eta_{A^{\operatorname*{op}}}$.

We know that $C$ is a $k$-bialgebra. By the axioms of a bialgebra, this yields
$\Delta_{C}\left(  1_{C}\right)  =1_{C}\otimes1_{C}$ and $\varepsilon
_{C}\left(  1_{C}\right)  =1$. Now,%
\begin{align*}
\underbrace{\left(  \eta_{A}\circ\varepsilon_{C}\right)  }_{\substack{=\mu
_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}\\\text{(by
(\ref{601.pf.90}))}}}\left(  1_{C}\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{C}\right)  \left(  1_{C}\right)  =\left(
\mu_{A}\circ\left(  f\otimes g\right)  \right)  \underbrace{\left(  \Delta
_{C}\left(  1_{C}\right)  \right)  }_{=1_{C}\otimes1_{C}}=\left(  \mu_{A}%
\circ\left(  f\otimes g\right)  \right)  \left(  1_{C}\otimes1_{C}\right) \\
&  =\mu_{A}\underbrace{\left(  \left(  f\otimes g\right)  \left(  1_{C}%
\otimes1_{C}\right)  \right)  }_{=f\left(  1_{C}\right)  \otimes g\left(
1_{C}\right)  }=\mu_{A}\left(  f\left(  1_{C}\right)  \otimes g\left(
1_{C}\right)  \right) \\
&  =\underbrace{f\left(  1_{C}\right)  }_{\substack{=1_{A}\\\text{(since
}f\text{ is a }k\text{-algebra}\\\text{homomorphism)}}}g\left(  1_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu_{A}\right) \\
&  =1_{A}g\left(  1_{C}\right)  =g\left(  1_{C}\right)  .
\end{align*}
Compared to%
\begin{align*}
\left(  \eta_{A}\circ\varepsilon_{C}\right)  \left(  1_{C}\right)   &
=\eta_{A}\underbrace{\left(  \varepsilon_{C}\left(  1_{C}\right)  \right)
}_{=1}=\eta_{A}\left(  1\right)  =1\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\eta_{A}\right) \\
&  =1_{A},
\end{align*}
this yields $g\left(  1_{C}\right)  =1_{A}$. Thus, every $\lambda\in k$
satisfies%
\begin{align*}
\left(  g\circ\eta_{C}\right)  \left(  \lambda\right)   &
=g\underbrace{\left(  \eta_{C}\left(  \lambda\right)  \right)  }%
_{\substack{=\lambda\cdot1_{C}\\\text{(by the definition of }\eta_{C}\text{)}%
}}=g\left(  \lambda\cdot1_{C}\right)  =\lambda\underbrace{g\left(
1_{C}\right)  }_{=1_{A}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }g\text{ is
}k\text{-linear}\right) \\
&  =\lambda\cdot1_{A}=\eta_{A}\left(  \lambda\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{A}\left(  \lambda\right)
=\lambda\cdot1_{A}\text{ by the definition of }\eta_{A}\right) \\
&  =\eta_{A^{\operatorname*{op}}}\left(  \lambda\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{A^{\operatorname*{op}}}%
=\eta_{A}\text{ (by the definition of }A^{\operatorname*{op}}\text{), so that
}\eta_{A^{\operatorname*{op}}}\left(  \lambda\right)  =\eta_{A}\left(
\lambda\right)  \right)
\end{align*}
In other words, $g\circ\eta_{C}=\eta_{A^{\operatorname*{op}}}$. Combined with
(\ref{601.pf.Delta}), this yields that $g$ is a $k$-algebra homomorphism from
$C$ to $A^{\operatorname*{op}}$. Since $g=f^{\ast\left(  -1\right)  }$, this
rewrites as follows: $f^{\ast\left(  -1\right)  }$ is a $k$-algebra
homomorphism from $C$ to $A^{\operatorname*{op}}$. This proves
Proposition~\ref{Proposition601}.
\end{proof}

\section{\label{S24}The Euler operator}

Recall what we did in Theorems~\ref{Theorem1} and~\ref{Theorem101}: We showed
that if $k$ is a field of characteristic $0$, and $H$ is a connected filtered
$k$-bialgebra, then the map $\operatorname*{Log}\operatorname*{id}%
\in\mathcal{L}\left(  H,H\right)  $ is a projection if $H$ is either
commutative or cocommutative. (This is not an \textquotedblleft if and only
if\textquotedblright\ assertion since $\operatorname*{Log}\operatorname*{id}$
can also be a projection for some connected filtered $k$-bialgebras $H$ which
are neither commutative nor cocommutative; but in general,
$\operatorname*{Log}\operatorname*{id}$ is not a projection\footnote{Not even
when $H$ is a connected graded involutive $k$-Hopf algebra.}.) We will now
study two maps that have similar properties to those of $\operatorname*{Log}%
\operatorname*{id}$, but are defined for \textbf{graded} Hopf algebras only.
Whereas $\operatorname*{Log}\operatorname*{id}$ is often referred to in
literature as the \textit{Eulerian idempotent}, these new maps are known as
the called \textit{Dynkin idempotents}. But let us first develop some theory
of graded vector spaces.

\begin{definition}
\label{Definition701} Let $k$ be a field. Let $V$ be a graded $k$-vector
space. Then, $V=\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$. The map $\bigoplus
_{\ell\in\mathbb{N}}\left(  \ell\cdot\operatorname{id}_{V_{\ell}}\right)
:\bigoplus_{\ell\in\mathbb{N}}V_{\ell}\rightarrow\bigoplus_{\ell\in\mathbb{N}%
}V_{\ell}$ is a $k$-linear map from $V$ to $V$ (because it is a $k$-linear map
from $\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ to $\bigoplus_{\ell\in\mathbb{N}%
}V_{\ell}$, but we have $\bigoplus_{\ell\in\mathbb{N}}V_{\ell}=V$). This map
will be denoted by $E_{V}$ and referred to as the \textit{Euler operator}%
\footnote{not to be confused with the Eulerian idempotent} of the graded
$k$-vector space $V$.
\end{definition}

Here is something pretty obvious:

\begin{proposition}
\label{Proposition702} Let $k$ be a field. Let $V$ be a graded $k$-vector space.

\textbf{(a)} For any $n\in\mathbb{N}$ and $v\in V_{n}$, we have $E_{V}\left(
v\right)  =nv$.

\textbf{(b)} The map $E_{V}$ is graded.

\textbf{(c)} Let $k$ be a field of characteristic $0$. Then,
$\operatorname*{Ker}\left(  E_{V}\right)  =V_{0}$.
\end{proposition}

We will give a detailed proof of this later.

We next state a fact about Euler operators on tensor products:

\begin{proposition}
\label{Proposition703} Let $k$ be a field. Let $V$ and $W$ be two graded
$k$-vector spaces. Then,%
\[
E_{V\otimes W}=E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes
E_{W}.
\]

\end{proposition}

Again, this will be shown later. Also:

\begin{proposition}
\label{Proposition704} Let $k$ be a field. Let $V$ and $W$ be graded
$k$-vector spaces. Let $f:V\rightarrow W$ be a graded $k$-linear map. Then,
$f\circ E_{V}=E_{W}\circ f$.
\end{proposition}

Before we show these facts, we make a definition that generalizes
Definition~\ref{Definition701}:

\begin{definition}
\label{Definition705} Let $k$ be a field. Let $\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}$ be a sequence of elements of $k$. Let $V$ be a graded
$k$-vector space. Then, $V=\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$. The map
$\bigoplus_{\ell\in\mathbb{N}}\left(  a_{\ell}\cdot\operatorname{id}_{V_{\ell
}}\right)  :\bigoplus_{\ell\in\mathbb{N}}V_{\ell}\rightarrow\bigoplus_{\ell
\in\mathbb{N}}V_{\ell}$ is a $k$-linear map from $V$ to $V$ (because it is a
$k$-linear map from $\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ to $\bigoplus
_{\ell\in\mathbb{N}}V_{\ell}$, but we have $\bigoplus_{\ell\in\mathbb{N}%
}V_{\ell}=V$). This map will be denoted by $E_{V}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}$ and referred to as the $\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}$\textit{-Euler operator}\footnote{not to be confused with
the Eulerian idempotent} of the graded $k$-vector space $V$.
\end{definition}

\begin{remark}
\label{Remark706} Let $k$ be a field. Let $V$ be a graded $k$-vector space.
Then, $E_{V}=E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}$.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark706}.]By the definition of $E_{V}$, we have
$E_{V}=\bigoplus_{\ell\in\mathbb{N}}\left(  \ell\cdot\operatorname{id}%
_{V_{\ell}}\right)  $. By the definition of $E_{V}^{\left(  \ell\right)
_{\ell\in\mathbb{N}}}$, we have $E_{V}^{\left(  \ell\right)  _{\ell
\in\mathbb{N}}}=\bigoplus_{\ell\in\mathbb{N}}\left(  \ell\cdot
\operatorname{id}_{V_{\ell}}\right)  $. Thus, $E_{V}=\bigoplus_{\ell
\in\mathbb{N}}\left(  \ell\cdot\operatorname{id}_{V_{\ell}}\right)
=E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}$. This proves
Remark~\ref{Remark706}.
\end{proof}

\begin{proposition}
\label{Proposition707} Let $k$ be a field. Let $\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}$ be a sequence of elements of $k$. Let $V$ be a graded
$k$-vector space.

\textbf{(a)} For any $n\in\mathbb{N}$ and $v\in V_{n}$, we have $E_{V}%
^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\left(  v\right)  =a_{n}v$.

\textbf{(b)} The map $E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}$
is graded.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition707}.]\textbf{(a)} Let $n\in\mathbb{N}$.
Let $\iota_{n}:V_{n}\rightarrow\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ be the
canonical injection of the $n$-th addend into the direct sum. Then, by the
universal property of the direct sum of linear maps, the diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrixcolsep{5pc}
%\xymatrix{
%V_n \arinj[r]^{\iota_n} \ar[d]_{a_n\cdot\id_{V_n}} & \bigoplus_{\ell\in\mathbb
%N}
%V_{\ell} \ar[d]^{\bigoplus_{\ell\in\mathbb N}
%\left(a_{\ell}\cdot\id_{V_{\ell}}\right)} \\
%V_n \arinj[r]_{\iota_n} & \bigoplus_{\ell\in\mathbb N} V_{\ell}
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
V_n \arinj[r]^{\iota_n} \ar[d]_{a_n\cdot\id_{V_n}} & \bigoplus_{\ell\in\mathbb
N}
V_{\ell} \ar[d]^{\bigoplus_{\ell\in\mathbb N}
\left(a_{\ell}\cdot\id_{V_{\ell}}\right)} \\
V_n \arinj[r]_{\iota_n} & \bigoplus_{\ell\in\mathbb N} V_{\ell}
}%
%EndExpansion
\]
is commutative. In other words, $\left(  \bigoplus_{\ell\in\mathbb{N}}\left(
a_{\ell}\cdot\operatorname{id}_{V_{\ell}}\right)  \right)  \circ\iota
_{n}=\iota_{n}\circ\left(  a_{n}\cdot\operatorname{id}_{V_{n}}\right)  $. Now,
let $v\in V_{n}$ be arbitrary. Then, $\iota_{n}\left(  v\right)  =v$ (because
we regard the canonical injection $\iota_{n}$ as an inclusion). Also, by the
definition of $E_{V}$, we have $E_{V}^{\left(  a_{\ell}\right)  _{\ell
\in\mathbb{N}}}=\bigoplus_{\ell\in\mathbb{N}}\left(  a_{\ell}\cdot
\operatorname{id}_{V_{\ell}}\right)  $ and thus%
\begin{align*}
\underbrace{E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}}%
_{=\bigoplus_{\ell\in\mathbb{N}}\left(  a_{\ell}\cdot\operatorname{id}%
_{V_{\ell}}\right)  }\underbrace{\left(  v\right)  }_{=\iota_{n}\left(
v\right)  }  &  =\left(  \bigoplus_{\ell\in\mathbb{N}}\left(  a_{\ell}%
\cdot\operatorname{id}_{V_{\ell}}\right)  \right)  \left(  \iota_{n}\left(
v\right)  \right)  =\underbrace{\left(  \left(  \bigoplus_{\ell\in\mathbb{N}%
}\left(  a_{\ell}\cdot\operatorname{id}_{V_{\ell}}\right)  \right)  \circ
\iota_{n}\right)  }_{=\iota_{n}\circ\left(  a_{n}\cdot\operatorname{id}%
_{V_{n}}\right)  }\left(  v\right) \\
&  =\left(  \iota_{n}\circ\left(  a_{n}\cdot\operatorname{id}_{V_{n}}\right)
\right)  \left(  v\right)  =\iota_{n}\underbrace{\left(  \left(  a_{n}%
\cdot\operatorname{id}_{V_{n}}\right)  \left(  v\right)  \right)  }_{=a_{n}%
v}=\iota_{n}\left(  a_{n}v\right)  =a_{n}v\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since we regard the canonical injection
}\iota_{n}\text{ as an inclusion}\right)  .
\end{align*}
This proves Proposition~\ref{Proposition707} \textbf{(a)}.

\textbf{(b)} Let $n\in\mathbb{N}$. Then, every $v\in V_{n}$ satisfies%
\begin{align*}
E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\left(  v\right)   &
=\underbrace{a_{n}}_{\in k}\underbrace{v}_{\in V_{n}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition~\ref{Proposition702}
\textbf{(a)}}\right) \\
&  \in kV_{n}\subseteq V_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
V_{n}\text{ is a }k\text{-vector space}\right)  .
\end{align*}
In other words, $E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\left(
V_{n}\right)  \subseteq V_{n}$.

Now, forget that we fixed $v$. We have thus shown that $E_{V}^{\left(
a_{\ell}\right)  _{\ell\in\mathbb{N}}}\left(  V_{n}\right)  \subseteq V_{n}$
for every $n\in\mathbb{N}$. In other words, $E_{V}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}$ is graded. This proves Proposition~\ref{Proposition707}
\textbf{(b)}.
\end{proof}

\begin{proposition}
\label{Proposition708} Let $k$ be a field. Let $\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}$ be a sequence of elements of $k$. Let $V$ and $W$ be
graded $k$-vector spaces. Let $f:V\rightarrow W$ be a graded $k$-linear map.
Then, $f\circ E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}%
=E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ f$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition708}.]Let $n\in\mathbb{N}$ be arbitrary.
Let $v\in V_{n}$. Then, $f\left(  v\right)  \in f\left(  V_{n}\right)
\subseteq W_{n}$ (since $f$ is graded), so that $E_{W}^{\left(  a_{\ell
}\right)  _{\ell\in\mathbb{N}}}\left(  f\left(  v\right)  \right)
=a_{n}f\left(  v\right)  $ (by Proposition~\ref{Proposition707} \textbf{(a)},
applied to $f\left(  v\right)  $ and $W$ instead of $v$ and $V$). On the other
hand, since $v\in V_{n}$, we have $E_{V}^{\left(  a_{\ell}\right)  _{\ell
\in\mathbb{N}}}\left(  v\right)  =a_{n}v$ (by Proposition~\ref{Proposition707}
\textbf{(a)}). Thus,%
\begin{align*}
\left(  f\circ E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\right)
\left(  v\right)   &  =f\underbrace{\left(  E_{V}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}\left(  v\right)  \right)  }_{=a_{n}v}=f\left(
a_{n}v\right)  =a_{n}f\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }f\text{ is }k\text{-linear}\right) \\
&  =E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\left(  f\left(
v\right)  \right)  =\left(  E_{W}^{\left(  a_{\ell}\right)  _{\ell
\in\mathbb{N}}}\circ f\right)  \left(  v\right)  ,
\end{align*}
so that
\begin{align*}
\left(  f\circ E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}%
-E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ f\right)  \left(
v\right)   &  =\underbrace{\left(  f\circ E_{V}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}\right)  \left(  v\right)  }_{=\left(  E_{W}^{\left(
a_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ f\right)  \left(  v\right)
}-\left(  E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ f\right)
\left(  v\right) \\
&  =\left(  E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ
f\right)  \left(  v\right)  -\left(  E_{W}^{\left(  a_{\ell}\right)  _{\ell
\in\mathbb{N}}}\circ f\right)  \left(  v\right)  =0.
\end{align*}
Thus, $v\in\operatorname*{Ker}\left(  f\circ E_{V}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}-E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\circ f\right)  $.

Now, forget that we fixed $v$. We thus have shown that every $v\in V_{n}$
satisfies $v\in\operatorname*{Ker}\left(  f\circ E_{V}^{\left(  a_{\ell
}\right)  _{\ell\in\mathbb{N}}}-E_{W}^{\left(  a_{\ell}\right)  _{\ell
\in\mathbb{N}}}\circ f\right)  $. In other words, $V_{n}\subseteq
\operatorname*{Ker}\left(  f\circ E_{V}^{\left(  a_{\ell}\right)  _{\ell
\in\mathbb{N}}}-E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ
f\right)  $.

Now forget that we fixed $n$. We thus have shown that every $n\in\mathbb{N}$
satisfies $V_{n}\subseteq\operatorname*{Ker}\left(  f\circ E_{V}^{\left(
a_{\ell}\right)  _{\ell\in\mathbb{N}}}-E_{W}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}\circ f\right)  $. But $V$ is a graded $k$-vector space,
so that $V=\bigoplus_{n\in\mathbb{N}}V_{n}=\sum_{n\in\mathbb{N}}V_{n}$ (since
direct sums are sums). Thus,%
\begin{align*}
V  &  =\sum_{n\in\mathbb{N}}\underbrace{V_{n}}_{\subseteq\operatorname*{Ker}%
\left(  f\circ E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}%
-E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ f\right)
}\subseteq\sum_{n\in\mathbb{N}}\operatorname*{Ker}\left(  f\circ
E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}-E_{W}^{\left(  a_{\ell
}\right)  _{\ell\in\mathbb{N}}}\circ f\right) \\
&  \subseteq\operatorname*{Ker}\left(  f\circ E_{V}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}-E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\circ f\right)
\end{align*}
(since $\operatorname*{Ker}\left(  f\circ E_{V}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}-E_{W}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\circ f\right)  $ is a $k$-vector space). In other words, $f\circ
E_{V}^{\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}}-E_{W}^{\left(  a_{\ell
}\right)  _{\ell\in\mathbb{N}}}\circ f=0$, so that $f\circ E_{V}^{\left(
a_{\ell}\right)  _{\ell\in\mathbb{N}}}=E_{W}^{\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}}\circ f$. This proves Proposition~\ref{Proposition708}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition702}.]\textbf{(a)} Let $n\in\mathbb{N}$
and $v\in V_{n}$. Applying Proposition~\ref{Proposition707} \textbf{(a)} to
$\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  \ell\right)  _{\ell
\in\mathbb{N}}$, we see that $E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}%
}}\left(  v\right)  =nv$. Since $E_{V}^{\left(  \ell\right)  _{\ell
\in\mathbb{N}}}=E_{V}$ (by Remark~\ref{Remark706}), this rewrites as
$E_{V}\left(  v\right)  =nv$. This proves Proposition~\ref{Proposition702}
\textbf{(a)}.

\textbf{(b)} Applying Proposition~\ref{Proposition707} \textbf{(b)} to
$\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  \ell\right)  _{\ell
\in\mathbb{N}}$, we see that $E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}%
}}$ is graded. Since $E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}=E_{V}$
(by Remark~\ref{Remark706}), this shows that $E_{V}$ is graded. This proves
Proposition~\ref{Proposition702} \textbf{(b)}.

\textbf{(c)} Let $v\in\operatorname*{Ker}\left(  E_{V}\right)  $. Then, $v\in
V$ and $E_{V}\left(  v\right)  =0$. Since $v\in V=\bigoplus_{\ell\in
\mathbb{N}}V_{\ell}$, we can write $v$ in the form $v=\sum_{\ell\in\mathbb{N}%
}v_{\ell}$ for some family $\left(  v_{\ell}\right)  _{\ell\in\mathbb{N}}$ of
vectors in $V$ satisfying the following properties:

\begin{itemize}
\item All but finitely many $\ell\in\mathbb{N}$ satisfy $v_{\ell}=0$.

\item We have $v_{\ell}\in V_{\ell}$ for every $\ell\in\mathbb{N}$.
\end{itemize}

Consider this family $\left(  v_{\ell}\right)  _{\ell\in\mathbb{N}}$. Every
$\ell\in\mathbb{N}$ satisfies $v_{\ell}\in V_{\ell}$ and thus $E_{V}\left(
v_{\ell}\right)  =\ell v_{\ell}$ (by Proposition~\ref{Proposition702}
\textbf{(a)}, applied to $\ell$ and $v_{\ell}$ instead of $n$ and $v$). Since
$E_{V}\left(  v\right)  =0$, we have%
\begin{align*}
0  &  =E_{V}\left(  v\right)  =E_{V}\left(  \sum_{\ell\in\mathbb{N}}v_{\ell
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }v=\sum_{\ell\in\mathbb{N}%
}v_{\ell}\right) \\
&  =\sum_{\ell\in\mathbb{N}}\underbrace{E_{V}\left(  v_{\ell}\right)  }_{=\ell
v_{\ell}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }E_{V}\text{ is
}k\text{-linear}\right) \\
&  =\sum_{\ell\in\mathbb{N}}\ell v_{\ell}.
\end{align*}
In other words, $\sum_{\ell\in\mathbb{N}}\ell v_{\ell}=0$.

Note that all but finitely many $\ell\in\mathbb{N}$ satisfy $\ell v_{\ell}=0$
(because all but finitely many $\ell\in\mathbb{N}$ satisfy $v_{\ell}=0$).
Also, $\ell v_{\ell}\in V_{\ell}$ for every $\ell\in\mathbb{N}$ (because
$v_{\ell}\in V_{\ell}$ for every $\ell\in\mathbb{N}$, and therefore
$\underbrace{\ell}_{\in k}\underbrace{v_{\ell}}_{\in V_{\ell}}\in kV_{\ell
}\subseteq V_{\ell}$ (since $V_{\ell}$ is a $k$-vector space) for every
$\ell\in\mathbb{N}$).

But $\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ is a direct sum. Therefore, the
vector spaces $V_{\ell}$ for $\ell\in\mathbb{N}$ are linearly disjoint. In
other words, every family $\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}$
satisfying the conditions
\begin{align*}
&  \left(  \text{all but finitely many }\ell\in\mathbb{N}\text{ satisfy
}a_{\ell}=0\right)  ,\\
&  \left(  a_{\ell}\in V_{\ell}\text{ for every }\ell\in\mathbb{N}\right)
,\ \ \ \ \ \ \ \ \ \ \text{and}\\
&  \sum_{\ell\in\mathbb{N}}a_{\ell}=0
\end{align*}
must satisfy $\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  0\right)
_{\ell\in\mathbb{N}}$. Applied to $\left(  a_{\ell}\right)  _{\ell
\in\mathbb{N}}=\left(  \ell v_{\ell}\right)  _{\ell\in\mathbb{N}}$, this
yields that $\left(  \ell v_{\ell}\right)  _{\ell\in\mathbb{N}}=0$ (because we
have shown that $\left(  \text{all but finitely many }\ell\in\mathbb{N}\text{
satisfy }\ell v_{\ell}=0\right)  $, that \newline$\left(  \ell v_{\ell}\in
V_{\ell}\text{ for every }\ell\in\mathbb{N}\right)  $ and that $\sum_{\ell
\in\mathbb{N}}\ell v_{\ell}=0$). In other words, every $\ell\in\mathbb{N}$
satisfies $\ell v_{\ell}=0$.

Now, for every positive integer $\ell$, we have $v_{\ell}=\dfrac{1}{\ell
}\underbrace{\ell v_{\ell}}_{=0}=\dfrac{1}{\ell}0=0$ (here, we used that $k$
is a field of characteristic $0$, so that $\dfrac{1}{\ell}$ is well-defined in
$k$). Thus,%
\[
v=\sum_{\ell\in\mathbb{N}}v_{\ell}=v_{0}+\sum_{\substack{\ell\in
\mathbb{N};\\\ell>0}}\underbrace{v_{\ell}}_{\substack{=0\\\text{(since }%
\ell\text{ is a}\\\text{positive integer)}}}=v_{0}+\underbrace{\sum
_{\substack{\ell\in\mathbb{N};\\\ell>0}}0}_{=0}=v_{0}\in V_{0}%
\]
(since $v_{\ell}\in V_{\ell}$ for every $\ell\in\mathbb{N}$).

Now, forget that we fixed $v$. We have thus shown that every $v\in
\operatorname*{Ker}\left(  E_{V}\right)  $ satisfies $v\in V_{0}$. In other
words, $\operatorname*{Ker}\left(  E_{V}\right)  \subseteq V_{0}$. Combined
with the fact that $V_{0}\subseteq\operatorname*{Ker}\left(  E_{V}\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $v\in V_{0}$ be arbitrary. Then,
Proposition~\ref{Proposition702} \textbf{(a)} (applied to $n=0$) yields that
$E_{V}\left(  v\right)  =0v=0$, so that $v\in\operatorname*{Ker}\left(
E_{V}\right)  $. Now forget that we fixed $v$. We have thus shown that every
$v\in V_{0}$ satisfies $v\in\operatorname*{Ker}\left(  E_{V}\right)  $. In
other words, $V_{0}\subseteq\operatorname*{Ker}\left(  E_{V}\right)  $, qed.},
this yields that $\operatorname*{Ker}\left(  E_{V}\right)  =V_{0}$. This
proves Proposition~\ref{Proposition702} \textbf{(c)}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition704}.]By Remark~\ref{Remark706}, we have
$E_{V}=E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}$. By
Remark~\ref{Remark706} (applied to $W$ instead of $V$), we have $E_{W}%
=E_{W}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}$. By
Proposition~\ref{Proposition708} (applied to $\left(  a_{\ell}\right)
_{\ell\in\mathbb{N}}=\left(  \ell\right)  _{\ell\in\mathbb{N}}$), we have
$f\circ E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}=E_{W}^{\left(
\ell\right)  _{\ell\in\mathbb{N}}}\circ f$. Since $E_{V}^{\left(  \ell\right)
_{\ell\in\mathbb{N}}}=E_{V}$ and $E_{W}^{\left(  \ell\right)  _{\ell
\in\mathbb{N}}}=E_{W}$, this rewrites as $f\circ E_{V}=E_{W}\circ f$. This
proves Proposition~\ref{Proposition704}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition703}.]Since $V$ is a graded $k$-vector
space, we have $V=\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$. Since $W$ is a
graded $k$-vector space, we have $W=\bigoplus_{m\in\mathbb{N}}W_{m}$. Since
$V=\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ and $W=\bigoplus_{m\in\mathbb{N}%
}W_{m}$, we have%
\[
V\otimes W=\left(  \bigoplus_{\ell\in\mathbb{N}}V_{\ell}\right)
\otimes\left(  \bigoplus_{m\in\mathbb{N}}W_{m}\right)  =\bigoplus_{\left(
\ell,m\right)  \in\mathbb{N}^{\times2}}V_{\ell}\otimes W_{m}%
\]
(by the distributivity of the tensor product).

Now, fix any $\left(  \ell,m\right)  \in\mathbb{N}^{\times2}$. We are going to
prove that $\left(  E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}%
_{V}\otimes E_{W}-E_{V\otimes W}\right)  \mid_{V_{\ell}\otimes W_{m}}=0$.

By the usual definition of the grading on the tensor product $V\otimes W$, we
have%
\begin{equation}
\left(  V\otimes W\right)  _{n}=\sum_{i=0}^{n}V_{i}\otimes W_{n-i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{.}
\label{703.pf.grading}%
\end{equation}


Now, let $v\in V_{\ell}$ and $w\in W_{m}$ be arbitrary. Let $n=\ell+m$. Then,
$0\leq\ell\leq n$. We have%
\begin{align*}
\underbrace{v}_{\in V_{\ell}}\otimes\underbrace{w}_{\in W_{m}}  &  \in
V_{\ell}\otimes W_{m}=V_{\ell}\otimes W_{n-\ell}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }m=\underbrace{\ell+m}_{=n}-\ell=n-\ell\right) \\
&  \subseteq\sum_{i=0}^{n}V_{i}\otimes W_{n-i}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }V_{\ell}\otimes W_{n-\ell}\text{ is an addend of the sum }%
\sum_{i=0}^{n}V_{i}\otimes W_{n-i}\\
\text{(namely, the addend for }i=\ell\text{)}%
\end{array}
\right) \\
&  =\left(  V\otimes W\right)  _{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{703.pf.grading})}\right)  ,
\end{align*}
so that $E_{V\otimes W}\left(  v\otimes w\right)  =n\left(  v\otimes w\right)
$ (by Proposition~\ref{Proposition702} \textbf{(a)}, applied to $v\otimes w$
and $V\otimes W$ instead of $v$ and $V$). On the other hand, due to $v\in
V_{\ell}$, we have $E_{V}\left(  v\right)  =\ell v$ (by
Proposition~\ref{Proposition702} \textbf{(a)}, applied to $\ell$ instead of
$n$). Due to $w\in W_{m}$, we have $E_{W}\left(  w\right)  =mw$ (by
Proposition~\ref{Proposition702} \textbf{(a)}, applied to $m$, $w$ and $W$
instead of $n$, $v$ and $V$). Thus,%
\begin{align*}
&  \left(  \left(  E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}%
_{V}\otimes E_{W}-E_{V\otimes W}\right)  \mid_{V_{\ell}\otimes W_{m}}\right)
\left(  v\otimes w\right) \\
&  =\left(  E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes
E_{W}-E_{V\otimes W}\right)  \left(  v\otimes w\right) \\
&  =\underbrace{\left(  E_{V}\otimes\operatorname{id}_{W}\right)  \left(
v\otimes w\right)  }_{=E_{V}\left(  v\right)  \otimes\operatorname{id}%
_{W}\left(  w\right)  }+\underbrace{\left(  \operatorname{id}_{V}\otimes
E_{W}\right)  \left(  v\otimes w\right)  }_{=\operatorname{id}_{V}\left(
v\right)  \otimes E_{W}\left(  w\right)  }-\underbrace{E_{V\otimes W}\left(
v\otimes w\right)  }_{=n\left(  v\otimes w\right)  }\\
&  =\underbrace{E_{V}\left(  v\right)  }_{=\ell v}\otimes
\underbrace{\operatorname{id}_{W}\left(  w\right)  }_{=w}%
+\underbrace{\operatorname{id}_{V}\left(  v\right)  }_{=v}\otimes
\underbrace{E_{W}\left(  w\right)  }_{=mw}-\underbrace{n}_{=\ell+m}\left(
v\otimes w\right) \\
&  =\underbrace{\ell v\otimes w}_{=\ell\left(  v\otimes w\right)
}+\underbrace{v\otimes mw}_{=m\left(  v\otimes w\right)  }-\left(
\ell+m\right)  \left(  v\otimes w\right)  =\ell\left(  v\otimes w\right)
+m\left(  v\otimes w\right)  -\left(  \ell+m\right)  \left(  v\otimes w\right)
\\
&  =\underbrace{\left(  \ell+m-\left(  \ell+m\right)  \right)  }_{=0}\left(
v\otimes w\right)  =0=0\left(  v\otimes w\right)  .
\end{align*}


Now, forget that we fixed $v$ and $w$. We thus have proven that \newline%
$\left(  \left(  E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}%
_{V}\otimes E_{W}-E_{V\otimes W}\right)  \mid_{V_{\ell}\otimes W_{m}}\right)
\left(  v\otimes w\right)  =0\left(  v\otimes w\right)  $ for any $v\in
V_{\ell}$ and $w\in W_{m}$. In other words, the two maps $\left(  E_{V}%
\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes E_{W}-E_{V\otimes
W}\right)  \mid_{V_{\ell}\otimes W_{m}}$ and $0$ are equal on every pure
tensor. But since these two maps are $k$-linear, this yields that these maps
must be identic (because whenever two $k$-linear maps from a tensor product
are equal on every pure tensor, they must be identic). In other words,
$\left(  E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes
E_{W}-E_{V\otimes W}\right)  \mid_{V_{\ell}\otimes W_{m}}=0$. Hence,
\begin{align*}
\left(  E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes
E_{W}-E_{V\otimes W}\right)  \left(  V_{\ell}\otimes W_{m}\right)   &
=\underbrace{\left(  \left(  E_{V}\otimes\operatorname{id}_{W}%
+\operatorname{id}_{V}\otimes E_{W}-E_{V\otimes W}\right)  \mid_{V_{\ell
}\otimes W_{m}}\right)  }_{=0}\left(  V_{\ell}\otimes W_{m}\right) \\
&  =0\left(  V_{\ell}\otimes W_{m}\right)  =0.
\end{align*}


Now, forget that we fixed $\ell$ and $m$. We thus have proven that every
$\ell\in\mathbb{N}$ and $m\in\mathbb{N}$ satisfy%
\begin{equation}
\left(  E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes
E_{W}-E_{V\otimes W}\right)  \left(  V_{\ell}\otimes W_{m}\right)  =0.
\label{703.pf.5}%
\end{equation}


But%
\[
V\otimes W=\bigoplus_{\left(  \ell,m\right)  \in\mathbb{N}^{\times2}}V_{\ell
}\otimes W_{m}=\sum_{\left(  \ell,m\right)  \in\mathbb{N}^{\times2}}V_{\ell
}\otimes W_{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are
sums}\right)  ,
\]
so that%
\begin{align*}
\left(  E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes
E_{W}-E_{V\otimes W}\right)  \left(  V\otimes W\right)   &  =\left(
E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes E_{W}%
-E_{V\otimes W}\right)  \left(  \sum_{\left(  \ell,m\right)  \in
\mathbb{N}^{\times2}}V_{\ell}\otimes W_{m}\right) \\
&  =\sum_{\left(  \ell,m\right)  \in\mathbb{N}^{\times2}}\underbrace{\left(
E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes E_{W}%
-E_{V\otimes W}\right)  \left(  V_{\ell}\otimes W_{m}\right)  }%
_{\substack{=0\\\text{(by (\ref{703.pf.5}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }E_{V}\otimes\operatorname{id}%
_{W}+\operatorname{id}_{V}\otimes E_{W}-E_{V\otimes W}\text{ is }%
k\text{-linear}\right) \\
&  =\sum_{\left(  \ell,m\right)  \in\mathbb{N}^{\times2}}0=0.
\end{align*}
Thus, $E_{V}\otimes\operatorname{id}_{W}+\operatorname{id}_{V}\otimes
E_{W}-E_{V\otimes W}=0$, so that $E_{V}\otimes\operatorname{id}_{W}%
+\operatorname{id}_{V}\otimes E_{W}=E_{V\otimes W}$. This proves
Proposition~\ref{Proposition703}.
\end{proof}

We will now construct a ``partial inverse'' to $E_{V}$ in characteristic $0$:

\begin{definition}
\label{Definition721} Let $k$ be a field of characteristic $0$. Let $\left(
b_{\ell}\right)  _{\ell\in\mathbb{N}}$ be the sequence of elements of $k$
defined by
\[
\left(  b_{\ell}=\left\{
\begin{array}
[c]{c}%
\dfrac{1}{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\ell>0;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }\ell=0
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{N}\right)  .
\]
Let $V$ be a graded $k$-vector space. Then, $V=\bigoplus_{\ell\in\mathbb{N}%
}V_{\ell}$. The map $\bigoplus_{\ell\in\mathbb{N}}\left(  b_{\ell}%
\cdot\operatorname{id}_{V_{\ell}}\right)  :\bigoplus_{\ell\in\mathbb{N}%
}V_{\ell}\rightarrow\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$ is a $k$-linear map
from $V$ to $V$ (because it is a $k$-linear map from $\bigoplus_{\ell
\in\mathbb{N}}V_{\ell}$ to $\bigoplus_{\ell\in\mathbb{N}}V_{\ell}$, but we
have $\bigoplus_{\ell\in\mathbb{N}}V_{\ell}=V$). This map will be denoted by
$E_{V}^{\operatorname*{inv}}$.
\end{definition}

\begin{remark}
\label{Remark722} Let $k$ be a field of characteristic $0$. Let $\left(
b_{\ell}\right)  _{\ell\in\mathbb{N}}$ be as defined in
Definition~\ref{Definition721}. Let $V$ be a graded $k$-vector space. Then,
$E_{V}^{\operatorname*{inv}}=E_{V}^{\left(  b_{\ell}\right)  _{\ell
\in\mathbb{N}}}$.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{Remark722}.]By the definition of $E_{V}%
^{\operatorname*{inv}}$, we have $E_{V}^{\operatorname*{inv}}=\bigoplus
_{\ell\in\mathbb{N}}\left(  b_{\ell}\cdot\operatorname{id}_{V_{\ell}}\right)
$. By the definition of $E_{V}^{\left(  b_{\ell}\right)  _{\ell\in\mathbb{N}}%
}$, we have $E_{V}^{\left(  b_{\ell}\right)  _{\ell\in\mathbb{N}}}%
=\bigoplus_{\ell\in\mathbb{N}}\left(  b_{\ell}\cdot\operatorname{id}_{V_{\ell
}}\right)  $. Thus, $E_{V}^{\operatorname*{inv}}=\bigoplus_{\ell\in\mathbb{N}%
}\left(  b_{\ell}\cdot\operatorname{id}_{V_{\ell}}\right)  =E_{V}^{\left(
b_{\ell}\right)  _{\ell\in\mathbb{N}}}$. This proves Remark~\ref{Remark722}.
\end{proof}

\begin{proposition}
\label{Proposition723} Let $k$ be a field. Let $\left(  c_{\ell}\right)
_{\ell\in\mathbb{N}}$ and $\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}$ be
two sequences of elements of $k$. Let $V$ be a graded $k$-vector space. Then,
\[
E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ E_{V}^{\left(
d_{\ell}\right)  _{\ell\in\mathbb{N}}}=E_{V}^{\left(  c_{\ell}d_{\ell}\right)
_{\ell\in\mathbb{N}}}=E_{V}^{\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\circ E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}.
\]

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition723}.]Let $n\in\mathbb{N}$. Every $v\in
V_{n}$ satisfies%
\begin{align*}
\left(  E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ
E_{V}^{\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}}\right)  \left(
v\right)   &  =E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\underbrace{\left(  E_{V}^{\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\left(  v\right)  \right)  }_{\substack{=d_{n}v\\\text{(by
Proposition~\ref{Proposition707} \textbf{(a)},}\\\text{applied to }\left(
a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  d_{\ell}\right)  _{\ell
\in\mathbb{N}}\text{)}}}=E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\left(  d_{n}v\right) \\
&  =d_{n}\underbrace{E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\left(  v\right)  }_{\substack{=c_{n}v\\\text{(by
Proposition~\ref{Proposition707} \textbf{(a)},}\\\text{applied to }\left(
a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  c_{\ell}\right)  _{\ell
\in\mathbb{N}}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}\text{ is }%
k\text{-linear}\right) \\
&  =d_{n}c_{n}v=c_{n}d_{n}v=E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell
\in\mathbb{N}}}\left(  v\right)
\end{align*}
(since $E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}}%
=c_{n}d_{n}v$ by Proposition~\ref{Proposition707} \textbf{(a)} (applied to
$\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  c_{\ell}d_{\ell
}\right)  _{\ell\in\mathbb{N}}$)) and thus%
\begin{align*}
\left(  E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ
E_{V}^{\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}}-E_{V}^{\left(  c_{\ell
}d_{\ell}\right)  _{\ell\in\mathbb{N}}}\right)  \left(  v\right)   &
=\underbrace{\left(  E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\circ E_{V}^{\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}}\right)  \left(
v\right)  }_{=E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\left(  v\right)  }-E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell
\in\mathbb{N}}}\left(  v\right) \\
&  =E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}}\left(
v\right)  -E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\left(  v\right)  =0,
\end{align*}
so that $v\in\operatorname*{Ker}\left(  E_{V}^{\left(  c_{\ell}\right)
_{\ell\in\mathbb{N}}}\circ E_{V}^{\left(  d_{\ell}\right)  _{\ell\in
\mathbb{N}}}-E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\right)  $. Thus, $V_{n}\subseteq\operatorname*{Ker}\left(  E_{V}^{\left(
c_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ E_{V}^{\left(  d_{\ell}\right)
_{\ell\in\mathbb{N}}}-E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell
\in\mathbb{N}}}\right)  $.

Now, forget that we fixed $n$. We have thus shown that every $n\in\mathbb{N}$
satisfies $V_{n}\subseteq\operatorname*{Ker}\left(  E_{V}^{\left(  c_{\ell
}\right)  _{\ell\in\mathbb{N}}}\circ E_{V}^{\left(  d_{\ell}\right)  _{\ell
\in\mathbb{N}}}-E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\right)  $. Since $V$ is a graded $k$-vector space, we have%
\begin{align*}
V  &  =\bigoplus_{n\in\mathbb{N}}V_{n}=\sum_{n\in\mathbb{N}}\underbrace{V_{n}%
}_{\subseteq\operatorname*{Ker}\left(  E_{V}^{\left(  c_{\ell}\right)
_{\ell\in\mathbb{N}}}\circ E_{V}^{\left(  d_{\ell}\right)  _{\ell\in
\mathbb{N}}}-E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\right)  }\subseteq\sum_{n\in\mathbb{N}}\operatorname*{Ker}\left(
E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ E_{V}^{\left(
d_{\ell}\right)  _{\ell\in\mathbb{N}}}-E_{V}^{\left(  c_{\ell}d_{\ell}\right)
_{\ell\in\mathbb{N}}}\right) \\
&  \subseteq\operatorname*{Ker}\left(  E_{V}^{\left(  c_{\ell}\right)
_{\ell\in\mathbb{N}}}\circ E_{V}^{\left(  d_{\ell}\right)  _{\ell\in
\mathbb{N}}}-E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}\right)
\end{align*}
(since $\operatorname*{Ker}\left(  E_{V}^{\left(  c_{\ell}\right)  _{\ell
\in\mathbb{N}}}\circ E_{V}^{\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}-E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}}\right)  $ is a
$k$-vector space), so that $E_{V}^{\left(  c_{\ell}\right)  _{\ell
\in\mathbb{N}}}\circ E_{V}^{\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}%
}-E_{V}^{\left(  c_{\ell}d_{\ell}\right)  _{\ell\in\mathbb{N}}}=0$, and thus
$E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ E_{V}^{\left(
d_{\ell}\right)  _{\ell\in\mathbb{N}}}=E_{V}^{\left(  c_{\ell}d_{\ell}\right)
_{\ell\in\mathbb{N}}}$. The same argument, but with the roles of $\left(
c_{\ell}\right)  _{\ell\in\mathbb{N}}$ and $\left(  d_{\ell}\right)  _{\ell
\in\mathbb{N}}$ interchanged, proves that $E_{V}^{\left(  d_{\ell}\right)
_{\ell\in\mathbb{N}}}\circ E_{V}^{\left(  c_{\ell}\right)  _{\ell\in
\mathbb{N}}}=E_{V}^{\left(  d_{\ell}c_{\ell}\right)  _{\ell\in\mathbb{N}}}$.
Thus,%
\[
E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ E_{V}^{\left(
d_{\ell}\right)  _{\ell\in\mathbb{N}}}=E_{V}^{\left(  c_{\ell}d_{\ell}\right)
_{\ell\in\mathbb{N}}}=E_{V}^{\left(  d_{\ell}c_{\ell}\right)  _{\ell
\in\mathbb{N}}}=E_{V}^{\left(  d_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ
E_{V}^{\left(  c_{\ell}\right)  _{\ell\in\mathbb{N}}}.
\]
This proves Proposition~\ref{Proposition723}.
\end{proof}

\begin{corollary}
\label{Corollary724} Let $k$ be a field of characteristic $0$. Let $V$ be a
graded $k$-vector space. Then:

\textbf{(a)} We have $E_{V}\circ E_{V}^{\operatorname*{inv}}=E_{V}%
^{\operatorname*{inv}}\circ E_{V}$.

\textbf{(b)} Every $v\in\bigoplus_{n\geq1}V_{n}$ satisfies $\left(
E_{V}^{\operatorname*{inv}}\circ E_{V}\right)  \left(  v\right)  =v$.

\textbf{(c)} The map $E_{V}^{\operatorname*{inv}}$ is graded.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary724}.]Define $\left(  b_{\ell}\right)
_{\ell\in\mathbb{N}}$ as in Definition~\ref{Definition721}.
Proposition~\ref{Proposition723} (applied to $\left(  c_{\ell}\right)
_{\ell\in\mathbb{N}}=\left(  \ell\right)  _{\ell\in\mathbb{N}}$ and $\left(
d_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  b_{\ell}\right)  _{\ell
\in\mathbb{N}}$) yields%
\[
E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}\circ E_{V}^{\left(  b_{\ell
}\right)  _{\ell\in\mathbb{N}}}=E_{V}^{\left(  \ell b_{\ell}\right)  _{\ell
\in\mathbb{N}}}=E_{V}^{\left(  b_{\ell}\right)  _{\ell\in\mathbb{N}}}\circ
E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}.
\]
Since $E_{V}^{\left(  \ell\right)  _{\ell\in\mathbb{N}}}=E_{V}$ (by
Remark~\ref{Remark706}) and $E_{V}^{\left(  b_{\ell}\right)  _{\ell
\in\mathbb{N}}}=E_{V}^{\operatorname*{inv}}$ (by Remark~\ref{Remark722}), this
rewrites as%
\[
E_{V}\circ E_{V}^{\operatorname*{inv}}=E_{V}^{\left(  \ell b_{\ell}\right)
_{\ell\in\mathbb{N}}}=E_{V}^{\operatorname*{inv}}\circ E_{V}.
\]
This proves Corollary~\ref{Corollary724} \textbf{(a)}.

\textbf{(b)} Since $V$ is a graded $k$-vector space, we have $V=\bigoplus
_{n\in\mathbb{N}}V_{n}$. Thus, the internal direct sum $\bigoplus_{n\geq
1}V_{n}$ is well-defined (as a subsum of the direct sum $\bigoplus
_{n\in\mathbb{N}}V_{n}$).

Now, let $n\in\mathbb{N}$ be positive. By the definition of $b_{n}$, we then
have
\[
b_{n}=\left\{
\begin{array}
[c]{c}%
\dfrac{1}{n},\ \ \ \ \ \ \ \ \ \ \text{if }n>0;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }n=0
\end{array}
\right.  =\dfrac{1}{n}%
\]
(since $n>0$), so that $nb_{n}=1$. Every $v\in V_{n}$ satisfies%
\begin{align*}
E_{V}^{\left(  \ell b_{\ell}\right)  _{\ell\in\mathbb{N}}}\left(  v\right)
&  =\underbrace{nb_{n}}_{=1}v \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{Proposition707} \textbf{(a)}, applied to } \left(  a_{\ell
}\right)  _{\ell\in\mathbb{N}}=\left(  \ell b_{\ell}\right)  _{\ell
\in\mathbb{N}}\right) \\
&  =v,
\end{align*}
so that $\left(  E_{V}^{\left(  \ell b_{\ell}\right)  _{\ell\in\mathbb{N}}%
}-\operatorname*{id}\right)  v=\underbrace{E_{V}^{\left(  \ell b_{\ell
}\right)  _{\ell\in\mathbb{N}}}\left(  v\right)  }_{=v}%
-\underbrace{\operatorname*{id}\left(  v\right)  }_{=v}=v-v=0$. Thus, $\left(
E_{V}^{\left(  \ell b_{\ell}\right)  _{\ell\in\mathbb{N}}}-\operatorname*{id}%
\right)  \left(  V_{n}\right)  =0$.

Now, forget that we fixed $n$. We thus have showed that every positive
$n\in\mathbb{N}$ satisfies $\left(  E_{V}^{\left(  \ell b_{\ell}\right)
_{\ell\in\mathbb{N}}}-\operatorname*{id}\right)  \left(  V_{n}\right)  =0$.
Thus, $\sum_{n\geq1}\underbrace{\left(  E_{V}^{\left(  \ell b_{\ell}\right)
_{\ell\in\mathbb{N}}}-\operatorname*{id}\right)  \left(  V_{n}\right)  }%
_{=0}=\sum_{n\geq1}0=0$.

Since direct sums are sums, we have $\bigoplus_{n\geq1}V_{n}=\sum_{n\geq
1}V_{n}$ and thus%
\begin{align*}
\left(  E_{V}^{\left(  \ell b_{\ell}\right)  _{\ell\in\mathbb{N}}%
}-\operatorname*{id}\right)  \left(  \bigoplus_{n\geq1}V_{n}\right)   &
=\left(  E_{V}^{\left(  \ell b_{\ell}\right)  _{\ell\in\mathbb{N}}%
}-\operatorname*{id}\right)  \left(  \sum_{n\geq1}V_{n}\right)  =\sum_{n\geq
1}\left(  E_{V}^{\left(  \ell b_{\ell}\right)  _{\ell\in\mathbb{N}}%
}-\operatorname*{id}\right)  \left(  V_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }E_{V}^{\left(  \ell
b_{\ell}\right)  _{\ell\in\mathbb{N}}}-\operatorname*{id}\text{ is
}k\text{-linear}\right) \\
&  =0.
\end{align*}
Hence, every $v\in\bigoplus_{n\geq1}V_{n}$ satisfies $\left(  E_{V}^{\left(
\ell b_{\ell}\right)  _{\ell\in\mathbb{N}}}-\operatorname*{id}\right)  \left(
v\right)  =0$ and thus%
\[
0=\left(  E_{V}^{\left(  \ell b_{\ell}\right)  _{\ell\in\mathbb{N}}%
}-\operatorname*{id}\right)  \left(  v\right)  =\underbrace{E_{V}^{\left(
\ell b_{\ell}\right)  _{\ell\in\mathbb{N}}}}_{=E_{V}^{\operatorname*{inv}%
}\circ E_{V}}\left(  v\right)  -\underbrace{\operatorname*{id}\left(
v\right)  }_{=v}=\left(  E_{V}^{\operatorname*{inv}}\circ E_{V}\right)
\left(  v\right)  -v,
\]
so that $\left(  E_{V}^{\operatorname*{inv}}\circ E_{V}\right)  \left(
v\right)  =v$. This proves Corollary~\ref{Corollary724} \textbf{(b)}.

\textbf{(c)} Proposition~\ref{Proposition707} \textbf{(b)} (applied to
$\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  b_{\ell}\right)
_{\ell\in\mathbb{N}}$) shows that the map $E_{V}^{\left(  b_{\ell}\right)
_{\ell\in\mathbb{N}}}$ is graded. Since $E_{V}^{\left(  b_{\ell}\right)
_{\ell\in\mathbb{N}}}=E_{V}^{\operatorname*{inv}}$ (by Remark~\ref{Remark722}%
), this rewrites as follows: The map $E_{V}^{\operatorname*{inv}}$ is graded.
This proves Corollary~\ref{Corollary724} \textbf{(c)}.
\end{proof}

\section{\label{S25}The Dynkin idempotents in cocommutative Hopf algebras}

We are now ready to state the main property of the Dynkin idempotents in
cocommutative Hopf algebras:

\begin{theorem}
\label{Theorem801} Let $k$ be a field of characteristic $0$. Let $H$ be a
cocommutative graded $k$-Hopf algebra. Let $S$ be the antipode of $H$. Define
a map $E_{H}:H\rightarrow H$ according to Definition~\ref{Definition701}.
Define a map $E_{H}^{\operatorname*{inv}}:H\rightarrow H$ according to
Definition~\ref{Definition721}. Let $\operatorname*{Prim}H$ denote the
subspace of $H$ consisting of all primitive elements of $H$. Let $\left(
\operatorname*{Prim}H\right)  ^{+}$ denote the intersection $\left(
\operatorname*{Prim}H\right)  \cap\left(  \bigoplus_{n\geq1}H_{n}\right)  $. Then:

\textbf{(a)} The map $E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast
S\right)  $ is a projection from $H$ to the subspace $\left(
\operatorname*{Prim}H\right)  ^{+}$.

\textbf{(b)} The map $E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\right)  $ is a projection from $H$ to the subspace $\left(
\operatorname*{Prim}H\right)  ^{+}$.
\end{theorem}

Note that in the case when $H$ is connected, it is easy to see that $\left(
\operatorname*{Prim}H\right)  ^{+}=\operatorname*{Prim}H$, and thus this yields:

\begin{theorem}
\label{Theorem802} Let $k$ be a field of characteristic $0$. Let $H$ be a
cocommutative connected graded $k$-Hopf algebra. Let $S$ be the antipode of
$H$. Define a map $E_{H}:H\rightarrow H$ according to
Definition~\ref{Definition701}. Define a map $E_{H}^{\operatorname*{inv}%
}:H\rightarrow H$ according to Definition~\ref{Definition721}. Let
$\operatorname*{Prim}H$ denote the subspace of $H$ consisting of all primitive
elements of $H$. Then:

\textbf{(a)} The map $E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast
S\right)  $ is a projection from $H$ to the subspace $\operatorname*{Prim}H$.

\textbf{(b)} The map $E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\right)  $ is a projection from $H$ to the subspace $\operatorname*{Prim}%
H$.
\end{theorem}

The maps $E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  $ and
$E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  $ are called the
\textit{Dynkin idempotents} of $H$.

We will prove Theorems~\ref{Theorem801} and~\ref{Theorem802}, but also more
general facts, in the rest of \S \ref{S25}. In \S \ref{S26}, we will dualize
them to commutative Hopf algebras (again obtaining projections, but not
necessarily on $\left(  \operatorname*{Prim}H\right)  ^{+}$ anymore). In
\S \ref{S27}, we will ``interpolate'' between $E_{H}\ast S$ and $S\ast E_{H}$,
obtaining infinitely many ``intermediate'' Dynkin idempotents.

But let us first define the notion of a \textit{coderivation}:

\begin{definition}
\label{Definition803} Let $k$ be a field. Let $H$ be a $k$-coalgebra. Let
$f:H\rightarrow H$ be a $k$-linear map. Then, $f$ is said to be a
\textit{coderivation} if and only if $\Delta_{H}\circ f=\left(  f\otimes
\operatorname{id}_{H}+\operatorname{id}_{H}\otimes f\right)  \circ\Delta_{H}$.
\end{definition}

Keep in mind that \textbf{a coderivation is not the same as an }$\left(
\varepsilon,\varepsilon\right)  $\textbf{-coderivation}. (The latter has been
defined in Definition~\ref{Definition22}.) We will, however, connect these two
notions in the following results.

Definition~\ref{Definition803} could be generalized to $k$-linear maps
$f:M\rightarrow H$ with $M$ being a $\left(  H,H\right)  $-bicomodule; but we
will not need this generalization and we will not even define the notion of a
$\left(  H,H\right)  $-bicomodule.

Before we prove anything about coderivations, let us show a technical lemma:

\begin{lemma}
\label{Lemma804} Let $k$ be a field. Let $C$ be a cocommutative $k$-coalgebra.
Let $A$ be a $k$-bialgebra. Let $\alpha:C\rightarrow A$ and $\beta
:C\rightarrow A$ be any $k$-linear maps. Let $g:C\rightarrow A$ be a
$k$-coalgebra homomorphism. Let $f:C\rightarrow A$ be a $k$-linear map
satisfying%
\[
\Delta_{A}\circ f=\left(  \alpha\otimes f+f\otimes\beta\right)  \circ
\Delta_{C}.
\]
Then:

\textbf{(a)} We have%
\[
\Delta_{A}\circ\left(  f\ast g\right)  =\left(  \left(  \alpha\ast g\right)
\otimes\left(  f\ast g\right)  +\left(  f\ast g\right)  \otimes\left(
\beta\ast g\right)  \right)  \circ\Delta_{C}.
\]


\textbf{(b)} We have%
\[
\Delta_{A}\circ\left(  g\ast f\right)  =\left(  \left(  g\ast\alpha\right)
\otimes\left(  g\ast f\right)  +\left(  g\ast f\right)  \otimes\left(
g\ast\beta\right)  \right)  \circ\Delta_{C}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma804}.]By the axioms of a bialgebra, $\Delta
_{A}:A\rightarrow A\otimes A$ is a $k$-algebra homomorphism (since $A$ is a
$k$-bialgebra). Thus, $\Delta_{A}\circ\mu_{A}=\mu_{A\otimes A}\circ\left(
\Delta_{A}\otimes\Delta_{A}\right)  $.

In the following, the sign $\ast$ will denote the convolution in
$\mathcal{L}\left(  C,A\right)  $, but also the convolution in $\mathcal{L}%
\left(  C\otimes C,A\otimes A\right)  $ (which is well-defined since $C\otimes
C$ is a $k$-coalgebra and $A\otimes A$ is a $k$-algebra).

\textbf{(a)} By the definition of convolution, $f\ast g=\mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{C}$, so that%
\begin{align}
\Delta_{A}\circ\underbrace{\left(  f\ast g\right)  }_{=\mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{C}}  &  =\underbrace{\Delta_{A}\circ\mu_{A}%
}_{=\mu_{A\otimes A}\circ\left(  \Delta_{A}\otimes\Delta_{A}\right)  }%
\circ\left(  f\otimes g\right)  \circ\Delta_{C}\nonumber\\
&  =\mu_{A\otimes A}\circ\left(  \Delta_{A}\otimes\Delta_{A}\right)
\circ\left(  f\otimes g\right)  \circ\Delta_{C}. \label{804.pf.1}%
\end{align}


But applying (\ref{(X)(X)}) to $C$, $A$, $A\otimes A$, $C$, $A$, $A\otimes A$,
$f$, $\Delta_{A}$, $g$, $\Delta_{A}$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain
\[
\left(  \Delta_{A}\circ f\right)  \otimes\left(  \Delta_{A}\circ g\right)
=\left(  \Delta_{A}\otimes\Delta_{A}\right)  \circ\left(  f\otimes g\right)
,
\]
so that%
\begin{align}
&  \left(  \Delta_{A}\otimes\Delta_{A}\right)  \circ\left(  f\otimes g\right)
\nonumber\\
&  =\underbrace{\left(  \Delta_{A}\circ f\right)  }_{=\left(  \alpha\otimes
f+f\otimes\beta\right)  \circ\Delta_{C}}\otimes\underbrace{\left(  \Delta
_{A}\circ g\right)  }_{\substack{=\left(  g\otimes g\right)  \circ\Delta
_{C}\\\text{(since }g\text{ is a }k\text{-coalgebra}\\\text{homomorphism)}%
}}\nonumber\\
&  =\left(  \left(  \alpha\otimes f+f\otimes\beta\right)  \circ\Delta
_{C}\right)  \otimes\left(  \left(  g\otimes g\right)  \circ\Delta_{C}\right)
\nonumber\\
&  =\underbrace{\left(  \left(  \alpha\otimes f+f\otimes\beta\right)
\otimes\left(  g\otimes g\right)  \right)  }_{\substack{=\alpha\otimes
f\otimes g\otimes g+f\otimes\beta\otimes g\otimes g\\\text{(since the tensor
product of }k\text{-linear maps}\\\text{is distributive)}}}\circ\left(
\Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}) (applied to }C\text{, }C\otimes C\text{, }A\otimes
A\text{, }C\text{, }C\otimes C\text{, }A\otimes A\text{,}\\
\text{ }\Delta_{C}\text{, }\alpha\otimes f+f\otimes\beta\text{, }\Delta
_{C}\text{, }g\otimes g\text{ instead of }U\text{, }V\text{, }W\text{,
}U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }%
\beta\text{, }\alpha^{\prime}\text{, }\beta^{\prime}\text{)}%
\end{array}
\right) \nonumber\\
&  =\left(  \alpha\otimes f\otimes g\otimes g+f\otimes\beta\otimes g\otimes
g\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  . \label{804.pf.4}%
\end{align}


Now, (\ref{804.pf.1}) becomes%
\begin{align*}
&  \Delta_{A}\circ\left(  f\ast g\right) \\
&  =\mu_{A\otimes A}\circ\underbrace{\left(  \Delta_{A}\otimes\Delta
_{A}\right)  \circ\left(  f\otimes g\right)  }_{\substack{=\left(
\alpha\otimes f\otimes g\otimes g+f\otimes\beta\otimes g\otimes g\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  \\\text{(by (\ref{804.pf.4}%
))}}}\circ\Delta_{C}\\
&  =\mu_{A\otimes A}\circ\left(  \alpha\otimes f\otimes g\otimes
g+f\otimes\beta\otimes g\otimes g\right)  \circ\underbrace{\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}}_{\substack{=\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\\\text{(by
(\ref{36.equation}))}}}\\
&  =\mu_{A\otimes A}\circ\left(  \alpha\otimes f\otimes g\otimes
g+f\otimes\beta\otimes g\otimes g\right)  \circ\underbrace{\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  }_{\substack{=\Delta
_{C\otimes C}\\\text{(since the definition of the }k\text{-coalgebra }C\otimes
C\text{ yields}\\\Delta_{C\otimes C}=\left(  \operatorname{id}_{C}\otimes
\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \text{)}}}\circ\Delta_{C}\\
&  =\mu_{A\otimes A}\circ\underbrace{\left(  \alpha\otimes f\otimes g\otimes
g+f\otimes\beta\otimes g\otimes g\right)  \circ\Delta_{C\otimes C}%
}_{\substack{=\left(  \alpha\otimes f\otimes g\otimes g\right)  \circ
\Delta_{C\otimes C}+\left(  f\otimes\beta\otimes g\otimes g\right)
\circ\Delta_{C\otimes C}\\\text{(since composition of }k\text{-linear maps is
distributive)}}}\circ\Delta_{C}\\
&  =\underbrace{\mu_{A\otimes A}\circ\left(  \left(  \alpha\otimes f\otimes
g\otimes g\right)  \circ\Delta_{C\otimes C}+\left(  f\otimes\beta\otimes
g\otimes g\right)  \circ\Delta_{C\otimes C}\right)  }_{\substack{=\mu
_{A\otimes A}\circ\left(  \alpha\otimes f\otimes g\otimes g\right)
\circ\Delta_{C\otimes C}+\mu_{A\otimes A}\circ\left(  f\otimes\beta\otimes
g\otimes g\right)  \circ\Delta_{C\otimes C}\\\text{(since composition of
}k\text{-linear maps is distributive)}}}\circ\Delta_{C}\\
&  =\left(  \underbrace{\mu_{A\otimes A}\circ\left(  \alpha\otimes f\otimes
g\otimes g\right)  \circ\Delta_{C\otimes C}}_{\substack{=\left(  \alpha\otimes
f\right)  \ast\left(  g\otimes g\right)  \\\text{(because the definition of
convolution yields}\\\left(  \alpha\otimes f\right)  \ast\left(  g\otimes
g\right)  =\mu_{A\otimes A}\circ\left(  \alpha\otimes f\otimes g\otimes
g\right)  \circ\Delta_{C\otimes C}\text{)}}}+\underbrace{\mu_{A\otimes A}%
\circ\left(  f\otimes\beta\otimes g\otimes g\right)  \circ\Delta_{C\otimes C}%
}_{\substack{=\left(  f\otimes\beta\right)  \ast\left(  g\otimes g\right)
\\\text{(because the definition of convolution yields}\\\left(  f\otimes
\beta\right)  \ast\left(  g\otimes g\right)  =\mu_{A\otimes A}\circ\left(
f\otimes\beta\otimes g\otimes g\right)  \circ\Delta_{C\otimes C}\text{)}%
}}\right)  \circ\Delta_{C}\\
&  =\left(  \underbrace{\left(  \alpha\otimes f\right)  \ast\left(  g\otimes
g\right)  }_{\substack{=\left(  \alpha\ast g\right)  \otimes\left(  f\ast
g\right)  \\\text{(by Corollary~\ref{Corollary40}, applied to }D=C\text{,
}B=A\text{,}\\p=\alpha\text{, }q=g\text{, }r=f\text{, }s=g\text{)}%
}}+\underbrace{\left(  f\otimes\beta\right)  \ast\left(  g\otimes g\right)
}_{\substack{=\left(  f\ast g\right)  \otimes\left(  \beta\ast g\right)
\\\text{(by Corollary~\ref{Corollary40}, applied to }D=C\text{, }%
B=A\text{,}\\p=f\text{, }q=g\text{, }r=\beta\text{, }s=g\text{)}}}\right)
\circ\Delta_{C}\\
&  =\left(  \left(  \alpha\ast g\right)  \otimes\left(  f\ast g\right)
+\left(  f\ast g\right)  \otimes\left(  \beta\ast g\right)  \right)
\circ\Delta_{C}.
\end{align*}
This proves Lemma~\ref{Lemma804} \textbf{(a)}.

\textbf{(b)} By the definition of convolution, $g\ast f=\mu_{A}\circ\left(
g\otimes f\right)  \circ\Delta_{C}$, so that%
\begin{align}
\Delta_{A}\circ\underbrace{\left(  g\ast f\right)  }_{=\mu_{A}\circ\left(
g\otimes f\right)  \circ\Delta_{C}}  &  =\underbrace{\Delta_{A}\circ\mu_{A}%
}_{=\mu_{A\otimes A}\circ\left(  \Delta_{A}\otimes\Delta_{A}\right)  }%
\circ\left(  g\otimes f\right)  \circ\Delta_{C}\nonumber\\
&  =\mu_{A\otimes A}\circ\left(  \Delta_{A}\otimes\Delta_{A}\right)
\circ\left(  g\otimes f\right)  \circ\Delta_{C}. \label{804.pf.11}%
\end{align}


But applying (\ref{(X)(X)}) to $C$, $A$, $A\otimes A$, $C$, $A$, $A\otimes A$,
$g$, $\Delta_{A}$, $f$, $\Delta_{A}$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain
\[
\left(  \Delta_{A}\circ g\right)  \otimes\left(  \Delta_{A}\circ f\right)
=\left(  \Delta_{A}\otimes\Delta_{A}\right)  \circ\left(  g\otimes f\right)
,
\]
so that%
\begin{align}
&  \left(  \Delta_{A}\otimes\Delta_{A}\right)  \circ\left(  g\otimes f\right)
\nonumber\\
&  =\underbrace{\left(  \Delta_{A}\circ g\right)  }_{\substack{=\left(
g\otimes g\right)  \circ\Delta_{C}\\\text{(since }g\text{ is a }%
k\text{-coalgebra}\\\text{homomorphism)}}}\otimes\underbrace{\left(
\Delta_{A}\circ f\right)  }_{=\left(  \alpha\otimes f+f\otimes\beta\right)
\circ\Delta_{C}}\nonumber\\
&  =\left(  \left(  g\otimes g\right)  \circ\Delta_{C}\right)  \otimes\left(
\left(  \alpha\otimes f+f\otimes\beta\right)  \circ\Delta_{C}\right)
\nonumber\\
&  =\underbrace{\left(  \left(  g\otimes g\right)  \otimes\left(
\alpha\otimes f+f\otimes\beta\right)  \right)  }_{\substack{=g\otimes
g\otimes\alpha\otimes f+g\otimes g\otimes f\otimes\beta\\\text{(since the
tensor product of }k\text{-linear maps}\\\text{is distributive)}}}\circ\left(
\Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}) (applied to }C\text{, }C\otimes C\text{, }A\otimes
A\text{, }C\text{, }C\otimes C\text{, }A\otimes A\text{,}\\
\text{ }\Delta_{C}\text{, }g\otimes g\text{, }\Delta_{C}\text{, }\alpha\otimes
f+f\otimes\beta\text{ instead of }U\text{, }V\text{, }W\text{, }U^{\prime
}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{,
}\alpha^{\prime}\text{, }\beta^{\prime}\text{)}%
\end{array}
\right) \nonumber\\
&  =\left(  g\otimes g\otimes\alpha\otimes f+g\otimes g\otimes f\otimes
\beta\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{804.pf.14}%
\end{align}


Now, (\ref{804.pf.11}) becomes%
\begin{align*}
&  \Delta_{A}\circ\left(  g\ast f\right) \\
&  =\mu_{A\otimes A}\circ\underbrace{\left(  \Delta_{A}\otimes\Delta
_{A}\right)  \circ\left(  g\otimes f\right)  }_{\substack{=\left(  g\otimes
g\otimes\alpha\otimes f+g\otimes g\otimes f\otimes\beta\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right)  \\\text{(by (\ref{804.pf.14}))}}%
}\circ\Delta_{C}\\
&  =\mu_{A\otimes A}\circ\left(  g\otimes g\otimes\alpha\otimes f+g\otimes
g\otimes f\otimes\beta\right)  \circ\underbrace{\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \circ\Delta_{C}}_{\substack{=\left(
\operatorname{id}_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\\\text{(by
(\ref{36.equation}))}}}\\
&  =\mu_{A\otimes A}\circ\left(  g\otimes g\otimes\alpha\otimes f+g\otimes
g\otimes f\otimes\beta\right)  \circ\underbrace{\left(  \operatorname{id}%
_{C}\otimes\tau_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right)  }_{\substack{=\Delta_{C\otimes
C}\\\text{(since the definition of the }k\text{-coalgebra }C\otimes C\text{
yields}\\\Delta_{C\otimes C}=\left(  \operatorname{id}_{C}\otimes\tau
_{C,C}\otimes\operatorname{id}_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  \text{)}}}\circ\Delta_{C}\\
&  =\mu_{A\otimes A}\circ\underbrace{\left(  g\otimes g\otimes\alpha\otimes
f+g\otimes g\otimes f\otimes\beta\right)  \circ\Delta_{C\otimes C}%
}_{\substack{=\left(  g\otimes g\otimes\alpha\otimes f\right)  \circ
\Delta_{C\otimes C}+\left(  g\otimes g\otimes f\otimes\beta\right)
\circ\Delta_{C\otimes C}\\\text{(since composition of }k\text{-linear maps is
distributive)}}}\circ\Delta_{C}\\
&  =\underbrace{\mu_{A\otimes A}\circ\left(  \left(  g\otimes g\otimes
\alpha\otimes f\right)  \circ\Delta_{C\otimes C}+\left(  g\otimes g\otimes
f\otimes\beta\right)  \circ\Delta_{C\otimes C}\right)  }_{\substack{=\mu
_{A\otimes A}\circ\left(  g\otimes g\otimes\alpha\otimes f\right)  \circ
\Delta_{C\otimes C}+\mu_{A\otimes A}\circ\left(  g\otimes g\otimes
f\otimes\beta\right)  \circ\Delta_{C\otimes C}\\\text{(since composition of
}k\text{-linear maps is distributive)}}}\circ\Delta_{C}\\
&  =\left(  \underbrace{\mu_{A\otimes A}\circ\left(  g\otimes g\otimes
\alpha\otimes f\right)  \circ\Delta_{C\otimes C}}_{\substack{=\left(  g\otimes
g\right)  \ast\left(  \alpha\otimes f\right)  \\\text{(because the definition
of convolution yields}\\\left(  g\otimes g\right)  \ast\left(  \alpha\otimes
f\right)  =\mu_{A\otimes A}\circ\left(  g\otimes g\otimes\alpha\otimes
f\right)  \circ\Delta_{C\otimes C}\text{)}}}+\underbrace{\mu_{A\otimes A}%
\circ\left(  g\otimes g\otimes f\otimes\beta\right)  \circ\Delta_{C\otimes C}%
}_{\substack{=\left(  g\otimes g\right)  \ast\left(  f\otimes\beta\right)
\\\text{(because the definition of convolution yields}\\\left(  g\otimes
g\right)  \ast\left(  f\otimes\beta\right)  =\mu_{A\otimes A}\circ\left(
g\otimes g\otimes f\otimes\beta\right)  \circ\Delta_{C\otimes C}\text{)}%
}}\right)  \circ\Delta_{C}\\
&  =\left(  \underbrace{\left(  g\otimes g\right)  \ast\left(  \alpha\otimes
f\right)  }_{\substack{=\left(  g\ast\alpha\right)  \otimes\left(  g\ast
f\right)  \\\text{(by Corollary~\ref{Corollary40}, applied to }D=C\text{,
}B=A\text{,}\\p=g\text{, }q=\alpha\text{, }r=g\text{, }s=f\text{)}%
}}+\underbrace{\left(  g\otimes g\right)  \ast\left(  f\otimes\beta\right)
}_{\substack{=\left(  g\ast f\right)  \otimes\left(  g\ast\beta\right)
\\\text{(by Corollary~\ref{Corollary40}, applied to }D=C\text{, }%
B=A\text{,}\\p=g\text{, }q=f\text{, }r=g\text{, }s=\beta\text{)}}}\right)
\circ\Delta_{C}\\
&  =\left(  \left(  g\ast\alpha\right)  \otimes\left(  g\ast f\right)
+\left(  g\ast f\right)  \otimes\left(  g\ast\beta\right)  \right)
\circ\Delta_{C}.
\end{align*}
This proves Lemma~\ref{Lemma804} \textbf{(b)}.
\end{proof}

Now comes the reason why we proved Lemma~\ref{Lemma804}:

\begin{theorem}
\label{Theorem805} Let $k$ be a field. Let $H$ be a cocommutative
$k$-bialgebra. Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two
$k$-coalgebra homomorphisms satisfying $P\ast\operatorname{id}_{H}\ast
Q=e_{H,H}$. Let $K:H\rightarrow H$ be a coderivation. Then, $P\ast K\ast
Q:H\rightarrow H$ is an $\left(  \varepsilon,\varepsilon\right)
$-coderivation. Here, the map $e_{H,H}$ is defined to be the map $\eta
_{H}\circ\varepsilon_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$
is identical with the definition of the map $e_{H,A}$ in
Definition~\ref{def.eHA}).
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem805}.]Since $K$ is a coderivation, we have
$\Delta_{H}\circ K=\left(  K\otimes\operatorname{id}_{H}+\operatorname{id}%
_{H}\otimes K\right)  \circ\Delta_{H}$ (because Definition~\ref{Definition803}
yields that $K$ is a coderivation if and only if $\Delta_{H}\circ K=\left(
K\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes K\right)
\circ\Delta_{H}$). Thus,%
\[
\Delta_{H}\circ K=\underbrace{\left(  K\otimes\operatorname{id}_{H}%
+\operatorname{id}_{H}\otimes K\right)  }_{=\operatorname{id}_{H}\otimes
K+K\otimes\operatorname{id}_{H}}\circ\Delta_{H}=\left(  \operatorname{id}%
_{H}\otimes K+K\otimes\operatorname{id}_{H}\right)  \circ\Delta_{H}.
\]
Hence, Lemma~\ref{Lemma804} \textbf{(b)} (applied to $C=H$, $A=H$,
$\alpha=\operatorname{id}_{H}$, $\beta=\operatorname{id}_{H}$, $f=K$ and
$g=P$) yields%
\[
\Delta_{H}\circ\left(  P\ast K\right)  =\left(  \left(  P\ast\operatorname{id}%
_{H}\right)  \otimes\left(  P\ast K\right)  +\left(  P\ast K\right)
\otimes\left(  P\ast\operatorname{id}_{H}\right)  \right)  \circ\Delta_{H}.
\]
Thus, Lemma~\ref{Lemma804} \textbf{(a)} (applied to $C=H$, $A=H$,
$\alpha=P\ast\operatorname{id}_{H}$, $\beta=P\ast\operatorname{id}_{H}$,
$f=P\ast K$ and $g=Q$) yields%
\begin{align*}
\Delta_{H}\circ\left(  P\ast K\ast Q\right)   &  =\left(  \underbrace{\left(
P\ast\operatorname{id}_{H}\ast Q\right)  }_{=e_{H,H}}\otimes\left(  P\ast
K\ast Q\right)  +\left(  P\ast K\ast Q\right)  \otimes\underbrace{\left(
P\ast\operatorname{id}_{H}\ast Q\right)  }_{=e_{H,H}}\right)  \circ\Delta
_{H}\\
&  =\underbrace{\left(  e_{H,H}\otimes\left(  P\ast K\ast Q\right)  +\left(
P\ast K\ast Q\right)  \otimes e_{H,H}\right)  }_{=\left(  P\ast K\ast
Q\right)  \otimes e_{H,H}+e_{H,H}\otimes\left(  P\ast K\ast Q\right)  }%
\circ\Delta_{H}\\
&  =\left(  \left(  P\ast K\ast Q\right)  \otimes e_{H,H}+e_{H,H}%
\otimes\left(  P\ast K\ast Q\right)  \right)  \circ\Delta_{H}.
\end{align*}
Thus, $P\ast K\ast Q$ is an $\left(  \varepsilon,\varepsilon\right)
$-coderivation (because by Definition~\ref{Definition22}, the map $P\ast K\ast
Q$ is an $\left(  \varepsilon,\varepsilon\right)  $-coderivation if and only
if $\Delta_{H}\circ\left(  P\ast K\ast Q\right)  =\left(  \left(  P\ast K\ast
Q\right)  \otimes e_{H,H}+e_{H,H}\otimes\left(  P\ast K\ast Q\right)  \right)
\circ\Delta_{H}$). This proves Theorem~\ref{Theorem805}.
\end{proof}

\begin{corollary}
\label{Corollary806} Let $k$ be a field. Let $H$ be a cocommutative $k$-Hopf
algebra. Let $S$ be the antipode of $H$. Let $K:H\rightarrow H$ be a coderivation.

\textbf{(a)} Then, $S\ast K$ is an $\left(  \varepsilon,\varepsilon\right)  $-coderivation.

\textbf{(b)} Then, $K\ast S$ is an $\left(  \varepsilon,\varepsilon\right)  $-coderivation.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary806}.]By Definition~\ref{Definition552}, we
have $H^{\operatorname*{cop}}=\left(  H,\tau_{H,H}\circ\Delta_{H}%
,\varepsilon_{H}\right)  $. Since $H$ is cocommutative, we have $\tau
_{H,H}\circ\Delta_{H}=\Delta_{H}$. Thus, $H^{\operatorname*{cop}}=\left(
H,\underbrace{\tau_{H,H}\circ\Delta_{H}}_{=\Delta_{H}},\varepsilon_{H}\right)
=\left(  H,\Delta_{H},\varepsilon_{H}\right)  =H$.

Proposition~\ref{Proposition553} yields that the antipode of $H$ is a
$k$-coalgebra homomorphism from $H^{\operatorname*{cop}}$ to $H$. Since the
antipode of $H$ is the map $S$, whereas $H^{\operatorname*{cop}}$ is $H$, this
rewrites as follows: The map $S$ is a $k$-coalgebra homomorphism from $H$ to
$H$.

Corollary~\ref{Corollary46} (applied to $n=0$, $C=H$ and $f=\operatorname{id}%
_{H}$) yields that $\operatorname{id}_{H}^{\ast0}$ is a $k$-coalgebra
homomorphism. Since $\operatorname{id}_{H}^{\ast0}=e_{H,H}$, this shows that
$e_{H,H}$ is a $k$-coalgebra homomorphism (where $e_{H,H}$ is defined as
according to Definition~\ref{def.eHA}).

The antipode of a Hopf algebra is defined as the $\ast$-inverse of its
identity map. Thus, the antipode of $H$ is the map $\operatorname{id}%
_{H}^{\ast\left(  -1\right)  }$. Since the antipode of $H$ is $S$, this yields
that $S$ is the map $\operatorname{id}_{H}^{\ast\left(  -1\right)  }$. In
other words, $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$. Hence,
$S\ast\operatorname{id}_{H}\ast e_{H,H}=\underbrace{\operatorname{id}%
_{H}^{\ast\left(  -1\right)  }\ast\operatorname{id}_{H}}_{=e_{H,H}}\ast
e_{H,H}=e_{H,H}$. Thus, Theorem~\ref{Theorem805} (applied to $P=S$ and
$Q=e_{H,H}$) yields that $S\ast K\ast e_{H,H}$ is an $\left(  \varepsilon
,\varepsilon\right)  $-coderivation. In other words, $S\ast K$ is an $\left(
\varepsilon,\varepsilon\right)  $-coderivation (since $S\ast K\ast
e_{H,H}=S\ast K$). This proves Corollary~\ref{Corollary806} \textbf{(a)}.

Since $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$, we have
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}\ast
\underbrace{\operatorname{id}_{H}\ast\operatorname{id}_{H}^{\ast\left(
-1\right)  }}_{=e_{H,H}}=e_{H,H}$. Thus, Theorem~\ref{Theorem805} (applied to
$P=e_{H,H}$ and $Q=S$) yields that $e_{H,H}\ast K\ast S$ is an $\left(
\varepsilon,\varepsilon\right)  $-coderivation. In other words, $K\ast S$ is
an $\left(  \varepsilon,\varepsilon\right)  $-coderivation (since $e_{H,H}\ast
K\ast S=K\ast S$). This proves Corollary~\ref{Corollary806} \textbf{(b)}.
\end{proof}

Next, something easy:

\begin{proposition}
\label{Proposition807} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let
$P:H\rightarrow H$, $Q:H\rightarrow H$ and $K:H\rightarrow H$ be three
$k$-linear maps such that $P\left(  1_{H}\right)  =1_{H}$, $Q\left(
1_{H}\right)  =1_{H}$ and $K\left(  1_{H}\right)  =0$. Then, every
$x\in\operatorname*{Prim}H$ satisfies $\left(  P\ast K\ast Q\right)  \left(
x\right)  =K\left(  x\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition807}.]Let $x\in\operatorname*{Prim}H$.

Since $x\in\operatorname*{Prim}H=\left(  \text{the set of primitive elements
of }H\right)  $, the element $x$ of $H$ is primitive. Thus, $\Delta\left(
x\right)  =x\otimes1_{H}+1_{H}\otimes x$ (because
Definition~\ref{Definition18} yields that $x$ is primitive if and only if
$\Delta\left(  x\right)  =x\otimes1_{H}+1_{H}\otimes x$). Since $\Delta
=\Delta_{H}$, this rewrites as $\Delta_{H}\left(  x\right)  =x\otimes
1_{H}+1_{H}\otimes x$.

But $P\ast K=\mu_{H}\circ\left(  P\otimes K\right)  \circ\Delta_{H}$ (by the
definition of convolution). Thus,
\begin{align*}
\left(  P\ast K\right)  \left(  x\right)   &  =\left(  \mu_{H}\circ\left(
P\otimes K\right)  \circ\Delta_{H}\right)  \left(  x\right)  =\left(  \mu
_{H}\circ\left(  P\otimes K\right)  \right)  \underbrace{\left(  \Delta
_{H}\left(  x\right)  \right)  }_{=x\otimes1_{H}+1_{H}\otimes x}\\
&  =\left(  \mu_{H}\circ\left(  P\otimes K\right)  \right)  \left(
x\otimes1_{H}+1_{H}\otimes x\right) \\
&  =\underbrace{\left(  \mu_{H}\circ\left(  P\otimes K\right)  \right)
\left(  x\otimes1_{H}\right)  }_{=\mu_{H}\left(  \left(  P\otimes K\right)
\left(  x\otimes1_{H}\right)  \right)  }+\underbrace{\left(  \mu_{H}%
\circ\left(  P\otimes K\right)  \right)  \left(  1_{H}\otimes x\right)
}_{=\mu_{H}\left(  \left(  P\otimes K\right)  \left(  1_{H}\otimes x\right)
\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{H}\circ\left(  P\otimes
K\right)  \text{ is }k\text{-linear}\right) \\
&  =\mu_{H}\underbrace{\left(  \left(  P\otimes K\right)  \left(
x\otimes1_{H}\right)  \right)  }_{=P\left(  x\right)  \otimes K\left(
1_{H}\right)  }+\mu_{H}\underbrace{\left(  \left(  P\otimes K\right)  \left(
1_{H}\otimes x\right)  \right)  }_{=P\left(  1_{H}\right)  \otimes K\left(
x\right)  }\\
&  =\underbrace{\mu_{H}\left(  P\left(  x\right)  \otimes K\left(
1_{H}\right)  \right)  }_{\substack{=P\left(  x\right)  K\left(  1_{H}\right)
\\\text{(since }\mu_{H}\text{ is the}\\\text{multiplication map)}%
}}+\underbrace{\mu_{H}\left(  P\left(  1_{H}\right)  \otimes K\left(
x\right)  \right)  }_{\substack{=P\left(  1_{H}\right)  K\left(  x\right)
\\\text{(since }\mu_{H}\text{ is the}\\\text{multiplication map)}}}\\
&  =P\left(  x\right)  \underbrace{K\left(  1_{H}\right)  }_{=0}%
+\underbrace{P\left(  1_{H}\right)  }_{=1_{H}}K\left(  x\right)
=\underbrace{P\left(  x\right)  0}_{=0}+\underbrace{1_{H}K\left(  x\right)
}_{=K\left(  x\right)  }=K\left(  x\right)  .
\end{align*}
On the other hand, by the axioms of a bialgebra, we have $\Delta_{H}\left(
1_{H}\right)  =1_{H}\otimes1_{H}$ (since $H$ is a $k$-bialgebra), so that%
\begin{align*}
\underbrace{\left(  P\ast K\right)  }_{=\mu_{H}\circ\left(  P\otimes K\right)
\circ\Delta_{H}}\left(  1_{H}\right)   &  =\left(  \mu_{H}\circ\left(
P\otimes K\right)  \circ\Delta_{H}\right)  \left(  1_{H}\right)  =\left(
\mu_{H}\circ\left(  P\otimes K\right)  \right)  \underbrace{\left(  \Delta
_{H}\left(  1_{H}\right)  \right)  }_{=1_{H}\otimes1_{H}}\\
&  =\left(  \mu_{H}\circ\left(  P\otimes K\right)  \right)  \left(
1_{H}\otimes1_{H}\right)  =\mu_{H}\underbrace{\left(  \left(  P\otimes
K\right)  \left(  1_{H}\otimes1_{H}\right)  \right)  }_{=P\left(
1_{H}\right)  \otimes K\left(  1_{H}\right)  }\\
&  =\mu_{H}\left(  \underbrace{P\left(  1_{H}\right)  }_{=1_{H}}%
\otimes\underbrace{K\left(  1_{H}\right)  }_{=0}\right)  =\mu_{H}%
\underbrace{\left(  1_{H}\otimes0\right)  }_{=0}=\mu_{H}\left(  0\right)  =0
\end{align*}
(since $\mu_{H}$ is $K$-linear). Now, by the definition of convolution,
$\left(  P\ast K\right)  \ast Q=\mu_{H}\circ\left(  \left(  P\ast K\right)
\otimes Q\right)  \circ\Delta$, so that%
\begin{align*}
&  \left(  \left(  P\ast K\right)  \ast Q\right)  \left(  x\right) \\
&  =\left(  \mu_{H}\circ\left(  \left(  P\ast K\right)  \otimes Q\right)
\circ\Delta_{H}\right)  \left(  x\right)  =\left(  \mu_{H}\circ\left(  \left(
P\ast K\right)  \otimes Q\right)  \right)  \underbrace{\left(  \Delta
_{H}\left(  x\right)  \right)  }_{=x\otimes1_{H}+1_{H}\otimes x}\\
&  =\left(  \mu_{H}\circ\left(  \left(  P\ast K\right)  \otimes Q\right)
\right)  \left(  x\otimes1_{H}+1_{H}\otimes x\right) \\
&  =\underbrace{\left(  \mu_{H}\circ\left(  \left(  P\ast K\right)  \otimes
Q\right)  \right)  \left(  x\otimes1_{H}\right)  }_{=\mu_{H}\left(  \left(
\left(  P\ast K\right)  \otimes Q\right)  \left(  x\otimes1_{H}\right)
\right)  }+\underbrace{\left(  \mu_{H}\circ\left(  \left(  P\ast K\right)
\otimes Q\right)  \right)  \left(  1_{H}\otimes x\right)  }_{=\mu_{H}\left(
\left(  \left(  P\ast K\right)  \otimes Q\right)  \left(  1_{H}\otimes
x\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{H}\circ\left(  \left(  P\ast
K\right)  \otimes Q\right)  \text{ is }k\text{-linear}\right) \\
&  =\mu_{H}\underbrace{\left(  \left(  \left(  P\ast K\right)  \otimes
Q\right)  \left(  x\otimes1_{H}\right)  \right)  }_{=\left(  P\ast K\right)
\left(  x\right)  \otimes Q\left(  1_{H}\right)  }+\mu_{H}\underbrace{\left(
\left(  \left(  P\ast K\right)  \otimes Q\right)  \left(  1_{H}\otimes
x\right)  \right)  }_{=\left(  P\ast K\right)  \left(  1_{H}\right)  \otimes
Q\left(  x\right)  }\\
&  =\underbrace{\mu_{H}\left(  \left(  P\ast K\right)  \left(  x\right)
\otimes Q\left(  1_{H}\right)  \right)  }_{\substack{=\left(  P\ast K\right)
\left(  x\right)  Q\left(  1_{H}\right)  \\\text{(since }\mu_{H}\text{ is
the}\\\text{multiplication map)}}}+\underbrace{\mu_{H}\left(  \left(  P\ast
K\right)  \left(  1_{H}\right)  \otimes Q\left(  x\right)  \right)
}_{\substack{=\left(  P\ast K\right)  \left(  1_{H}\right)  Q\left(  x\right)
\\\text{(since }\mu_{H}\text{ is the}\\\text{multiplication map)}}}\\
&  =\underbrace{\left(  P\ast K\right)  \left(  x\right)  }_{=K\left(
x\right)  }\underbrace{Q\left(  1_{H}\right)  }_{=1_{H}}+\underbrace{\left(
P\ast K\right)  \left(  1_{H}\right)  }_{=0}Q\left(  x\right)
=\underbrace{K\left(  x\right)  1_{H}}_{=K\left(  x\right)  }%
+\underbrace{0Q\left(  x\right)  }_{=0}=K\left(  x\right)  .
\end{align*}
Since $\left(  P\ast K\right)  \ast Q=P\ast K\ast Q$, this rewrites as
$\left(  P\ast K\ast Q\right)  \left(  x\right)  =K\left(  x\right)  $. This
proves Proposition~\ref{Proposition807}.
\end{proof}

Now, let us study a particular coderivation that any graded coalgebra has: the
Euler operator:

\begin{proposition}
\label{Proposition811} Let $k$ be a field. Let $H$ be a graded $k$-coalgebra.
Let $E_{H}$ be defined according to Definition~\ref{Definition701}. Then,
$E_{H}:H\rightarrow H$ is a coderivation.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition811}.]Since $H$ is a graded
$k$-coalgebra, the map $\Delta_{H}:H\rightarrow H\otimes H$ is graded. Thus,
Proposition~\ref{Proposition704} (applied to $V=H$, $W=H\otimes H$ and
$f=\Delta_{H}$) yields%
\[
\Delta_{H}\circ E_{H}=\underbrace{E_{H\otimes H}}_{\substack{=E_{H}%
\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes E_{H}\\\text{(by
Proposition~\ref{Proposition703}, applied to }V=H\text{ and }W=H\text{)}%
}}\circ\Delta_{H}=\left(  E_{H}\otimes\operatorname{id}_{H}+\operatorname{id}%
_{H}\otimes E_{H}\right)  \circ\Delta_{H}.
\]
Thus, $E_{H}$ is a coderivation (because Definition~\ref{Definition803} yields
that $E_{H}$ is a coderivation if and only if $\Delta_{H}\circ E_{H}=\left(
E_{H}\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes E_{H}\right)
\circ\Delta_{H}$). Proposition~\ref{Proposition811} is proven.
\end{proof}

Now, let us come as close as possible to Theorem~\ref{Theorem801} without
requiring $k$ to be of characteristic $0$:

\begin{theorem}
\label{Theorem812} Let $k$ be a field. Let $H$ be a cocommutative graded
$k$-bialgebra. Let $E_{H}$ be defined according to
Definition~\ref{Definition701}. Let $\operatorname*{Prim}H$ denote the
subspace of $H$ consisting of all primitive elements of $H$. Let $\left(
\operatorname*{Prim}H\right)  ^{+}$ denote the intersection $\left(
\operatorname*{Prim}H\right)  \cap\left(  \bigoplus_{n\geq1}H_{n}\right)  $.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-coalgebra
homomorphisms satisfying $P\left(  1_{H}\right)  =1_{H}$, $Q\left(
1_{H}\right)  =1_{H}$ and $P\ast\operatorname{id}_{H}\ast Q=e_{H,H}$. Here,
the map $e_{H,H}$ is defined to be the map $\eta_{H}\circ\varepsilon
_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is identical with
the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

\textbf{(a)} Then, $\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)
\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}$.

\textbf{(b)} Besides, $E_{H}\left(  \left(  \operatorname*{Prim}H\right)
^{+}\right)  \subseteq\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)  $.

\textbf{(c)} We have $\left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast
E_{H}\ast Q\right)  =E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  =\left(
P\ast E_{H}\ast Q\right)  \circ E_{H}$.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem812}.]Notice that $H$ is a $k$-bialgebra, thus a
unital coalgebra.

By Proposition~\ref{Proposition811}, the map $E_{H}$ is a coderivation. Thus,
Theorem~\ref{Theorem805} (applied to $K=E_{H}$) yields that $P\ast E_{H}\ast
Q:H\rightarrow H$ is an $\left(  \varepsilon,\varepsilon\right)
$-coderivation. Thus,
\begin{equation}
\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)  \subseteq
\operatorname*{Prim}H \label{812.pf.1}%
\end{equation}
(because Theorem~\ref{Theorem23} (applied to $C=H$ and $f=P\ast E_{H}\ast Q$)
yields that $P\ast E_{H}\ast Q$ is an $\left(  \varepsilon,\varepsilon\right)
$-coderivation if and only if $\left(  P\ast E_{H}\ast Q\right)  \left(
H\right)  \subseteq\operatorname*{Prim}H$).

On the other hand, since $H$ is a graded $k$-coalgebra, we have%
\[
\Delta_{H}\left(  H_{\ell}\right)  \subseteq\bigoplus_{\substack{\left(
i,j\right)  \in\mathbb{N}^{\times2};\\i+j=\ell}}H_{i}\otimes H_{j}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{N}.
\]
Applied to $\ell=0$, this yields $\Delta_{H}\left(  H_{0}\right)
\subseteq\bigoplus_{\substack{\left(  i,j\right)  \in\mathbb{N}^{\times
2};\\i+j=0}}H_{i}\otimes H_{j}$.

There exists only one $\left(  i,j\right)  \in\mathbb{N}^{\times2}$ satisfying
$i+j=0$: namely, $\left(  i,j\right)  =\left(  0,0\right)  $. Thus, the direct
sum $\bigoplus_{\substack{\left(  i,j\right)  \in\mathbb{N}^{\times2}%
;\\i+j=0}}H_{i}\otimes H_{j}$ has only one addend, namely the one for $\left(
i,j\right)  =\left(  0,0\right)  $. As a consequence, $\bigoplus
_{\substack{\left(  i,j\right)  \in\mathbb{N}^{\times2};\\i+j=0}}H_{i}\otimes
H_{j}=H_{0}\otimes H_{0}$. Thus, $\Delta_{H}\left(  H_{0}\right)
\subseteq\bigoplus_{\substack{\left(  i,j\right)  \in\mathbb{N}^{\times
2};\\i+j=0}}H_{i}\otimes H_{j}=H_{0}\otimes H_{0}$.

Notice that $E_{H}\left(  H_{0}\right)  =0$\ \ \ \ \footnote{This is because
every $v\in H_{0}$ satisfies%
\begin{align*}
E_{H}\left(  v\right)   &  =0v \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{Proposition702} \textbf{(a)}, applied to }n=0\right) \\
&  =0.
\end{align*}
}. By the definition of convolution, $P\ast E_{H}=\mu_{H}\circ\left(  P\otimes
E_{H}\right)  \circ\Delta_{H}$. Thus,%
\begin{align*}
\left(  P\ast E_{H}\right)  \left(  H_{0}\right)   &  =\left(  \mu_{H}%
\circ\left(  P\otimes E_{H}\right)  \circ\Delta_{H}\right)  \left(
H_{0}\right)  =\mu_{H}\left(  \left(  P\otimes E_{H}\right)
\underbrace{\left(  \Delta_{H}\left(  H_{0}\right)  \right)  }_{\subseteq
H_{0}\otimes H_{0}}\right) \\
&  \subseteq\mu_{H}\underbrace{\left(  \left(  P\otimes E_{H}\right)  \left(
H_{0}\otimes H_{0}\right)  \right)  }_{=P\left(  H_{0}\right)  \otimes
E_{H}\left(  H_{0}\right)  }=\mu_{H}\left(  P\left(  H_{0}\right)
\otimes\underbrace{E_{H}\left(  H_{0}\right)  }_{=0}\right) \\
&  =\mu_{H}\left(  \underbrace{P\left(  H_{0}\right)  \otimes0}_{=0}\right)
=\mu_{H}\left(  0\right)  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu
_{H}\text{ is }k\text{-linear}\right)  .
\end{align*}
By the definition of convolution, $\left(  P\ast E_{H}\right)  \ast Q=\mu
_{H}\circ\left(  \left(  P\ast E_{H}\right)  \otimes Q\right)  \circ\Delta
_{H}$. Thus,%
\begin{align*}
\left(  \left(  P\ast E_{H}\right)  \ast Q\right)  \left(  H_{0}\right)   &
=\left(  \mu_{H}\circ\left(  \left(  P\ast E_{H}\right)  \otimes Q\right)
\circ\Delta_{H}\right)  \left(  H_{0}\right)  =\mu_{H}\left(  \left(  \left(
P\ast E_{H}\right)  \otimes Q\right)  \underbrace{\left(  \Delta_{H}\left(
H_{0}\right)  \right)  }_{\subseteq H_{0}\otimes H_{0}}\right) \\
&  \subseteq\mu_{H}\underbrace{\left(  \left(  \left(  P\ast E_{H}\right)
\otimes Q\right)  \left(  H_{0}\otimes H_{0}\right)  \right)  }_{=\left(
P\ast E_{H}\right)  \left(  H_{0}\right)  \otimes Q\left(  H_{0}\right)  }%
=\mu_{H}\left(  \underbrace{\left(  P\ast E_{H}\right)  \left(  H_{0}\right)
}_{\subseteq0}\otimes Q\left(  H_{0}\right)  \right) \\
&  \subseteq\mu_{H}\left(  \underbrace{0\otimes Q\left(  H_{0}\right)  }%
_{=0}\right)  =\mu_{H}\left(  0\right)  =0\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mu_{H}\text{ is }k\text{-linear}\right)  ,
\end{align*}
so that $\left(  \left(  P\ast E_{H}\right)  \ast Q\right)  \left(
H_{0}\right)  =0$. Since $\left(  P\ast E_{H}\right)  \ast Q=P\ast E_{H}\ast
Q$, this rewrites as
\begin{equation}
\left(  P\ast E_{H}\ast Q\right)  \left(  H_{0}\right)  =0. \label{812.pf.2}%
\end{equation}


Since $E_{H}$ is graded (by Proposition~\ref{Proposition702} \textbf{(b)},
applied to $V=H$) and $Q$ is graded, we conclude (by
Proposition~\ref{Proposition224} \textbf{(a)}, applied to $C=H$, $A=H$,
$f=E_{H}$ and $g=Q$) that $E_{H}\ast Q$ is graded.

Since $P$ is graded and $E_{H}\ast Q$ is graded, we conclude (by
Proposition~\ref{Proposition224} \textbf{(a)}, applied to $C=H$, $A=H$, $f=P$
and $g=E_{H}\ast Q$) that $P\ast E_{H}\ast Q$ is graded. Thus, every
$n\in\mathbb{N}$ satisfies $\left(  P\ast E_{H}\ast Q\right)  \left(
H_{n}\right)  \subseteq H_{n}$. Now, since $H$ is graded, we have
$H=\bigoplus_{n\in\mathbb{N}}H_{n}=\sum_{n\in\mathbb{N}}H_{n}$ (since direct
sums are sums), so that%
\begin{align*}
\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)   &  =\left(  P\ast
E_{H}\ast Q\right)  \left(  \sum_{n\in\mathbb{N}}H_{n}\right)  =\sum
_{n\in\mathbb{N}}\left(  P\ast E_{H}\ast Q\right)  \left(  H_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }P\ast E_{H}\ast Q\text{ is
}k\text{-linear}\right) \\
&  =\underbrace{\left(  P\ast E_{H}\ast Q\right)  \left(  H_{0}\right)
}_{\substack{=0\\\text{(by (\ref{812.pf.2}))}}}+\sum_{n\geq1}%
\underbrace{\left(  P\ast E_{H}\ast Q\right)  \left(  H_{n}\right)
}_{\substack{\subseteq H_{n}\\\text{(since }P\ast E_{H}\ast Q\text{ is
graded)}}}\\
&  \subseteq\sum_{n\geq1}H_{n}=\bigoplus_{n\geq1}H_{n}%
\end{align*}
(because the sum $\sum_{n\geq1}H_{n}$ is a direct sum (since it is a subsum of
the direct sum $\bigoplus_{n\in\mathbb{N}}H_{n}$)). Combining (\ref{812.pf.1})
with $\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)  \subseteq
\bigoplus_{n\geq1}H_{n}$, we obtain%
\[
\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)  \subseteq\left(
\operatorname*{Prim}H\right)  \cap\left(  \bigoplus_{n\geq1}H_{n}\right)
=\left(  \operatorname*{Prim}H\right)  ^{+}.
\]
This proves Theorem~\ref{Theorem812} \textbf{(a)}.

\textbf{(b)} We have $1_{H}\in H_{0}$ (since $H$ is a graded $k$-algebra).
Thus, Proposition~\ref{Proposition702} \textbf{(a)} (applied to $V=H$, $n=0$
and $v=1_{H}$) yields $E_{H}\left(  1_{H}\right)  =0\cdot1_{H}=0$. Thus,
Proposition~\ref{Proposition807} (applied to $K=E_{H}$) yields that%
\begin{equation}
\text{every }x\in\operatorname*{Prim}H\text{ satisfies }\left(  P\ast
E_{H}\ast Q\right)  \left(  x\right)  =E_{H}\left(  x\right)  .
\label{812.pf.5}%
\end{equation}
Thus, every $x\in\operatorname*{Prim}H$ satisfies $E_{H}\left(  x\right)
=\left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  \in\left(  P\ast
E_{H}\ast Q\right)  \left(  H\right)  $ (since $x\in H$). In other words,
$E_{H}\left(  \operatorname*{Prim}H\right)  \subseteq\left(  P\ast E_{H}\ast
Q\right)  \left(  H\right)  $. Since $\left(  \operatorname*{Prim}H\right)
^{+}=\left(  \operatorname*{Prim}H\right)  \cap\left(  \bigoplus_{n\geq1}%
H_{n}\right)  \subseteq\operatorname*{Prim}H$, we have $E_{H}\left(  \left(
\operatorname*{Prim}H\right)  ^{+}\right)  \subseteq E_{H}\left(
\operatorname*{Prim}H\right)  \subseteq\left(  P\ast E_{H}\ast Q\right)
\left(  H\right)  $. This proves Theorem~\ref{Theorem812} \textbf{(b)}.

\textbf{(c)} Let $x\in H$. Then, $\left(  P\ast E_{H}\ast Q\right)  \left(
x\right)  \in\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)
\subseteq\operatorname*{Prim}H$ (by (\ref{812.pf.1})). Thus, (\ref{812.pf.5})
(applied to $\left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  $ instead of
$x$) yields%
\[
\left(  P\ast E_{H}\ast Q\right)  \left(  \left(  P\ast E_{H}\ast Q\right)
\left(  x\right)  \right)  =E_{H}\left(  \left(  P\ast E_{H}\ast Q\right)
\left(  x\right)  \right)  .
\]
Thus,%
\begin{align*}
\left(  \left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast E_{H}\ast
Q\right)  \right)  \left(  x\right)   &  =\left(  P\ast E_{H}\ast Q\right)
\left(  \left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  \right) \\
&  =E_{H}\left(  \left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  \right)
=\left(  E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  \right)  \left(
x\right)  .
\end{align*}


Now forget that we fixed $x$. We thus have proven that every $x\in H$
satisfies $\left(  \left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast
E_{H}\ast Q\right)  \right)  \left(  x\right)  =\left(  E_{H}\circ\left(
P\ast E_{H}\ast Q\right)  \right)  \left(  x\right)  $. In other words,
\begin{equation}
\left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast E_{H}\ast Q\right)
=E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  . \label{812.pf.6}%
\end{equation}


Since $P\ast E_{H}\ast Q$ is graded, we have $\left(  P\ast E_{H}\ast
Q\right)  \circ E_{H}=E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  $ (by
Proposition~\ref{Proposition704}, applied to $V=H$, $W=H$ and $f=P\ast
E_{H}\ast Q$). Combined with (\ref{812.pf.6}), this yields%
\[
\left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast E_{H}\ast Q\right)
=E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  =\left(  P\ast E_{H}\ast
Q\right)  \circ E_{H}.
\]
This proves Theorem~\ref{Theorem812} \textbf{(c)}.
\end{proof}

\begin{corollary}
\label{Corollary813} Let $k$ be a field. Let $H$ be a cocommutative graded
$k$-Hopf algebra. Let $S$ be the antipode of $H$. Let $E_{H}$ be defined
according to Definition~\ref{Definition701}. Let $\operatorname*{Prim}H$
denote the subspace of $H$ consisting of all primitive elements of $H$. Let
$\left(  \operatorname*{Prim}H\right)  ^{+}$ denote the intersection $\left(
\operatorname*{Prim}H\right)  \cap\left(  \bigoplus_{n\geq1}H_{n}\right)  $.

\textbf{(a)} Then, $\left(  E_{H}\ast S\right)  \left(  H\right)
\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}$.

\textbf{(b)} Besides, $E_{H}\left(  \left(  \operatorname*{Prim}H\right)
^{+}\right)  \subseteq\left(  E_{H}\ast S\right)  \left(  H\right)  $.

\textbf{(c)} We have $\left(  E_{H}\ast S\right)  \circ\left(  E_{H}\ast
S\right)  =E_{H}\circ\left(  E_{H}\ast S\right)  =\left(  E_{H}\ast S\right)
\circ E_{H}$.

\textbf{(d)} Also, $\left(  S\ast E_{H}\right)  \left(  H\right)
\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}$.

\textbf{(e)} Besides, $E_{H}\left(  \left(  \operatorname*{Prim}H\right)
^{+}\right)  \subseteq\left(  S\ast E_{H}\right)  \left(  H\right)  $.

\textbf{(f)} We have $\left(  S\ast E_{H}\right)  \circ\left(  S\ast
E_{H}\right)  =E_{H}\circ\left(  S\ast E_{H}\right)  =\left(  S\ast
E_{H}\right)  \circ E_{H}$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary813}.]Just as in the proof of
Corollary~\ref{Corollary806}, we can prove the following facts:

\begin{itemize}
\item The map $S$ is a $k$-coalgebra homomorphism from $H$ to $H$.

\item The map $e_{H,H}$ is a $k$-coalgebra homomorphism (where $e_{H,H}$ is
defined as according to Definition~\ref{def.eHA}).

\item We have $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$.

\item We have $S\ast\operatorname{id}_{H}\ast e_{H,H}=e_{H,H}$ and
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}$.
\end{itemize}

Notice that $e_{H,H}=\eta_{H}\circ\varepsilon_{H}$ (by the definition of
$e_{H,H}$), so that%
\begin{align*}
e_{H,H}\left(  1_{H}\right)   &  =\left(  \eta_{H}\circ\varepsilon_{H}\right)
\left(  1_{H}\right)  =\eta_{H}\underbrace{\left(  \varepsilon_{H}\left(
1_{H}\right)  \right)  }_{\substack{=1\\\text{(by the axioms of a
bialgebra,}\\\text{since }H\text{ is a }k\text{-bialgebra)}}}\\
&  =\eta_{H}\left(  1\right)  =1\cdot1_{H}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\eta_{H}\right) \\
&  =1_{H}.
\end{align*}


Since $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$, we have
$S\ast\operatorname{id}_{H}=e_{H,H}$. But the definition of convolution yields
$S\ast\operatorname{id}_{H}=\mu_{H}\circ\left(  S\otimes\operatorname{id}%
_{H}\right)  \circ\Delta_{H}$. Since $H$ is a $k$-bialgebra, we have
$\Delta_{H}\left(  1_{H}\right)  =1_{H}\otimes1_{H}$. Thus,%
\begin{align*}
\underbrace{\left(  S\ast\operatorname{id}_{H}\right)  }_{=\mu_{H}\circ\left(
S\otimes\operatorname{id}_{H}\right)  \circ\Delta_{H}}\left(  1_{H}\right)
&  =\left(  \mu_{H}\circ\left(  S\otimes\operatorname{id}_{H}\right)
\circ\Delta_{H}\right)  \left(  1_{H}\right)  =\mu_{H}\left(  \left(
S\otimes\operatorname{id}_{H}\right)  \underbrace{\left(  \Delta_{H}\left(
1_{H}\right)  \right)  }_{=1_{H}\otimes1_{H}}\right) \\
&  =\mu_{H}\left(  \underbrace{\left(  S\otimes\operatorname{id}_{H}\right)
\left(  1_{H}\otimes1_{H}\right)  }_{=S\left(  1_{H}\right)  \otimes
\operatorname{id}_{H}\left(  1_{H}\right)  }\right)  =\mu_{H}\left(  S\left(
1_{H}\right)  \otimes\operatorname{id}_{H}\left(  1_{H}\right)  \right) \\
&  =S\left(  1_{H}\right)  \cdot\underbrace{\operatorname{id}_{H}\left(
1_{H}\right)  }_{=1_{H}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{H}\text{
is the multiplication map}\right) \\
&  =S\left(  1_{H}\right)  .
\end{align*}
Compared with $\underbrace{\left(  S\ast\operatorname{id}_{H}\right)
}_{=e_{H,H}}\left(  1_{H}\right)  =e_{H,H}\left(  1_{H}\right)  =1_{H}$, this
yields $S\left(  1_{H}\right)  =1_{H}$. Also, $e_{H,H}\left(  1_{H}\right)
=1_{H}$.

Also, the map $e_{H,H}$ is graded (by Proposition~\ref{Proposition224}
\textbf{(b)}, applied to $C=H$ and $A=H$), and the map $S$ is graded (since
$S$ is the antipode of $H$, while $H$ is a graded $k$-Hopf algebra).

Thus, we can apply Theorem~\ref{Theorem812} to $P=e_{H,H}$ and $Q=S$.

Applying Theorem~\ref{Theorem812} \textbf{(a)} to $P=e_{H,H}$ and $Q=S$, we
obtain $\left(  e_{H,H}\ast E_{H}\ast S\right)  \left(  H\right)
\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}$. Since $e_{H,H}\ast
E_{H}\ast S=E_{H}\ast S$, this rewrites as $\left(  E_{H}\ast S\right)
\left(  H\right)  \subseteq\left(  \operatorname*{Prim}H\right)  ^{+}$. This
proves Corollary~\ref{Corollary813} \textbf{(a)}.

Applying Theorem~\ref{Theorem812} \textbf{(b)} to $P=e_{H,H}$ and $Q=S$, we
obtain $E_{H}\left(  \left(  \operatorname*{Prim}H\right)  ^{+}\right)
\subseteq\left(  e_{H,H}\ast E_{H}\ast S\right)  \left(  H\right)  $. Since
$e_{H,H}\ast E_{H}\ast S=E_{H}\ast S$, this rewrites as $E_{H}\left(  \left(
\operatorname*{Prim}H\right)  ^{+}\right)  \subseteq\left(  E_{H}\ast
S\right)  \left(  H\right)  $. This proves Corollary~\ref{Corollary813}
\textbf{(b)}.

Applying Theorem~\ref{Theorem812} \textbf{(c)} to $P=e_{H,H}$ and $Q=S$, we
obtain $\left(  e_{H,H}\ast E_{H}\ast S\right)  \circ\left(  e_{H,H}\ast
E_{H}\ast S\right)  =E_{H}\circ\left(  e_{H,H}\ast E_{H}\ast S\right)
=\left(  e_{H,H}\ast E_{H}\ast S\right)  \circ E_{H}$. Since $e_{H,H}\ast
E_{H}\ast S=E_{H}\ast S$, this rewrites as $\left(  E_{H}\ast S\right)
\circ\left(  E_{H}\ast S\right)  =E_{H}\circ\left(  E_{H}\ast S\right)
=\left(  E_{H}\ast S\right)  \circ E_{H}$. This proves
Corollary~\ref{Corollary813} \textbf{(c)}.

But we can also apply Theorem~\ref{Theorem812} to $P=S$ and $Q=e_{H,H}$.

Applying Theorem~\ref{Theorem812} \textbf{(a)} to $P=S$ and $Q=e_{H,H}$, we
obtain $\left(  S\ast E_{H}\ast e_{H,H}\right)  \left(  H\right)
\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}$. Since $S\ast E_{H}\ast
e_{H,H}=S\ast E_{H}$, this rewrites as $\left(  S\ast E_{H}\right)  \left(
H\right)  \subseteq\left(  \operatorname*{Prim}H\right)  ^{+}$. This proves
Corollary~\ref{Corollary813} \textbf{(d)}.

Applying Theorem~\ref{Theorem812} \textbf{(b)} to $P=S$ and $Q=e_{H,H}$, we
obtain $E_{H}\left(  \left(  \operatorname*{Prim}H\right)  ^{+}\right)
\subseteq\left(  S\ast E_{H}\ast e_{H,H}\right)  \left(  H\right)  $. Since
$S\ast E_{H}\ast e_{H,H}=S\ast E_{H}$, this rewrites as $E_{H}\left(  \left(
\operatorname*{Prim}H\right)  ^{+}\right)  \subseteq\left(  S\ast
E_{H}\right)  \left(  H\right)  $. This proves Corollary~\ref{Corollary813}
\textbf{(e)}.

Applying Theorem~\ref{Theorem812} \textbf{(c)} to $P=S$ and $Q=e_{H,H}$, we
obtain $\left(  S\ast E_{H}\ast e_{H,H}\right)  \circ\left(  S\ast E_{H}\ast
e_{H,H}\right)  =E_{H}\circ\left(  S\ast E_{H}\ast e_{H,H}\right)  =\left(
S\ast E_{H}\ast e_{H,H}\right)  \circ E_{H}$. Since $S\ast E_{H}\ast
e_{H,H}=S\ast E_{H}$, this rewrites as $\left(  S\ast E_{H}\right)
\circ\left(  S\ast E_{H}\right)  =E_{H}\circ\left(  S\ast E_{H}\right)
=\left(  S\ast E_{H}\right)  \circ E_{H}$. This proves
Corollary~\ref{Corollary813} \textbf{(f)}.
\end{proof}

In order to specialize the above results to connected graded Hopf algebras, we
first observe a simple fact:

\begin{proposition}
\label{Proposition821} Let $k$ be a field. Let $H$ be a connected graded
$k$-bialgebra. Then, $\operatorname*{Prim}H\subseteq\bigoplus_{n\geq1}H_{n}$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition821}.]Let $x\in\operatorname*{Prim}H$.
Since $H$ is graded, we have $H=\bigoplus_{n\in\mathbb{N}}H_{n}$. Now,
$x\in\operatorname*{Prim}H\subseteq H=\bigoplus_{n\in\mathbb{N}}H_{n}%
=H_{0}\oplus\left(  \bigoplus_{n\geq1}H_{n}\right)  $. Thus, there exists an
$\alpha\in H_{0}$ and a $\beta\in\bigoplus_{n\geq1}H_{n}$ such that
$x=\alpha+\beta$. Consider these $\alpha$ and $\beta$.

Since $H$ is a graded $k$-coalgebra, we have $\varepsilon_{H}\left(
H_{n}\right)  =0$ for every integer $n\geq1$. Since $\beta\in\bigoplus
_{n\geq1}H_{n}=\sum_{n\geq1}H_{n}$ (since direct sums are sums), we have%
\begin{align*}
\varepsilon_{H}\left(  \beta\right)   &  \in\varepsilon_{H}\left(  \sum
_{n\geq1}H_{n}\right)  =\sum_{n\geq1}\underbrace{\varepsilon_{H}\left(
H_{n}\right)  }_{\substack{=0\\\text{(since }n\geq1\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon_{H}\text{ is
}k\text{-linear}\right) \\
&  =\sum_{n\geq1}0=0,
\end{align*}
so that $\varepsilon_{H}\left(  \beta\right)  =0$. But Remark~\ref{Remark21}
shows that $\varepsilon\left(  x\right)  =0$ (since $H$ is a $k$-bialgebra,
thus a unital coalgebra). Since $x=\alpha+\beta$, we have
\[
\varepsilon_{H}\left(  x\right)  =\varepsilon_{H}\left(  \alpha+\beta\right)
=\varepsilon_{H}\left(  \alpha\right)  +\underbrace{\varepsilon_{H}\left(
\beta\right)  }_{=0}=\varepsilon_{H}\left(  \alpha\right)  .
\]
Compared with $\underbrace{\varepsilon_{H}}_{=\varepsilon}\left(  x\right)
=\varepsilon\left(  x\right)  =0$, this yields $\varepsilon_{H}\left(
\alpha\right)  =0$. Since $\alpha\in H_{0}$, we have $\varepsilon_{H}\left(
\alpha\right)  =\left(  \varepsilon_{H}\mid_{H_{0}}\right)  \left(
\alpha\right)  $. Thus, the equality $\varepsilon_{H}\left(  \alpha\right)
=0$ (which we just proved) rewrites as $\left(  \varepsilon_{H}\mid_{H_{0}%
}\right)  \left(  \alpha\right)  =0$.

But since $H$ is connected, the map $\varepsilon_{H}\mid_{H_{0}}%
:H_{0}\rightarrow k$ is a $k$-vector space isomorphism (because
Definition~\ref{Definition212} yields that $H$ is connected if and only if the
map $\varepsilon_{H}\mid_{H_{0}}:H_{0}\rightarrow k$ is a $k$-vector space
isomorphism). Thus, from the equality $\left(  \varepsilon_{H}\mid_{H_{0}%
}\right)  \left(  \alpha\right)  =0$ (which we proved above), we can conclude
that $\alpha=0$. Now, $x=\underbrace{\alpha}_{=0}+\beta=\beta\in
\bigoplus_{n\geq1}H_{n}$.

Now, forget that we fixed $x$. We have thus proven that every $x\in
\operatorname*{Prim}H$ satisfies $x\in\bigoplus_{n\geq1}H_{n}$. In other
words, $\operatorname*{Prim}H\subseteq\bigoplus_{n\geq1}H_{n}$.
Proposition~\ref{Proposition821} is proven.
\end{proof}

Here come equivalent versions of parts \textbf{(a)} and \textbf{(b)} of
Theorem~\ref{Theorem812}:

\begin{corollary}
\label{Corollary822} Let $k$ be a field. Let $H$ be a cocommutative graded
$k$-bialgebra. Let $E_{H}$ be defined according to
Definition~\ref{Definition701}. Let $\operatorname*{Prim}H$ denote the
subspace of $H$ consisting of all primitive elements of $H$.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-coalgebra
homomorphisms satisfying $P\left(  1_{H}\right)  =1_{H}$, $Q\left(
1_{H}\right)  =1_{H}$ and $P\ast\operatorname{id}_{H}\ast Q=e_{H,H}$. Here,
the map $e_{H,H}$ is defined to be the map $\eta_{H}\circ\varepsilon
_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is identical with
the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

\textbf{(a)} Then, $\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)
\subseteq\operatorname*{Prim}H$.

\textbf{(b)} Besides, $E_{H}\left(  \operatorname*{Prim}H\right)
\subseteq\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary822}.]Define $\left(  \operatorname*{Prim}%
H\right)  ^{+}$ as in Theorem~\ref{Theorem812}. Then, $\left(
\operatorname*{Prim}H\right)  ^{+}=\left(  \operatorname*{Prim}H\right)
\cap\left(  \bigoplus_{n\geq1}H_{n}\right)  \subseteq\operatorname*{Prim}H$.

Theorem~\ref{Theorem812} \textbf{(a)} states that $\left(  P\ast E_{H}\ast
Q\right)  \left(  H\right)  \subseteq\left(  \operatorname*{Prim}H\right)
^{+}$. Combined with $\left(  \operatorname*{Prim}H\right)  ^{+}%
\subseteq\operatorname*{Prim}H$, this yields $\left(  P\ast E_{H}\ast
Q\right)  \left(  H\right)  \subseteq\operatorname*{Prim}H$. This proves
Corollary~\ref{Corollary822} \textbf{(a)}.

In the proof of Theorem~\ref{Theorem812} \textbf{(b)}, we showed that
$E_{H}\left(  \operatorname*{Prim}H\right)  \subseteq\left(  P\ast E_{H}\ast
Q\right)  \left(  H\right)  $. This proves Corollary~\ref{Corollary822}
\textbf{(b)}.

(Note that we referred to the proof of Theorem~\ref{Theorem812} \textbf{(b)}
here; but this was not strictly necessary since we could have just as well
derived Corollary~\ref{Corollary822} \textbf{(b)} from
Theorem~\ref{Theorem812} \textbf{(b)} without recourse to its proof. This
would have taken a bit more work, though.)
\end{proof}

Now, let us do the same to parts \textbf{(a)}, \textbf{(b)}, \textbf{(d)} and
\textbf{(e)} of Corollary~\ref{Corollary813}:

\begin{corollary}
\label{Corollary823} Let $k$ be a field. Let $H$ be a cocommutative graded
$k$-Hopf algebra. Let $S$ be the antipode of $H$. Let $E_{H}$ be defined
according to Definition~\ref{Definition701}. Let $\operatorname*{Prim}H$
denote the subspace of $H$ consisting of all primitive elements of $H$.

\textbf{(a)} Then, $\left(  E_{H}\ast S\right)  \left(  H\right)
\subseteq\operatorname*{Prim}H$.

\textbf{(b)} Besides, $E_{H}\left(  \operatorname*{Prim}H\right)
\subseteq\left(  E_{H}\ast S\right)  \left(  H\right)  $.

\textbf{(c)} Also, $\left(  S\ast E_{H}\right)  \left(  H\right)
\subseteq\operatorname*{Prim}H$.

\textbf{(d)} Besides, $E_{H}\left(  \operatorname*{Prim}H\right)
\subseteq\left(  S\ast E_{H}\right)  \left(  H\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary823}.]Just as in the proof of
Corollary~\ref{Corollary806}, we can prove the following facts:

\begin{itemize}
\item The map $S$ is a $k$-coalgebra homomorphism from $H$ to $H$.

\item The map $e_{H,H}$ is a $k$-coalgebra homomorphism (where $e_{H,H}$ is
defined as according to Definition~\ref{def.eHA}).

\item We have $S\ast\operatorname{id}_{H}\ast e_{H,H}=e_{H,H}$ and
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}$.
\end{itemize}

Just as in the proof of Corollary~\ref{Corollary813}, we can prove the
following facts:

\begin{itemize}
\item We have $e_{H,H}\left(  1_{H}\right)  =1_{H}$ and $S\left(
1_{H}\right)  =1_{H}$.

\item The maps $e_{H,H}$ and $S$ are graded.
\end{itemize}

As a consequence, we can apply Corollary~\ref{Corollary822} to $P=e_{H,H}$ and
$Q=S$.

Applying Corollary~\ref{Corollary822} \textbf{(a)} to $P=e_{H,H}$ and $Q=S$,
we obtain that $\left(  e_{H,H}\ast E_{H}\ast S\right)  \left(  H\right)
\subseteq\operatorname*{Prim}H$. Since $e_{H,H}\ast E_{H}\ast S=E_{H}\ast S$,
this rewrites as $\left(  E_{H}\ast S\right)  \left(  H\right)  \subseteq
\operatorname*{Prim}H$. This proves Corollary~\ref{Corollary823} \textbf{(a)}.

Applying Corollary~\ref{Corollary822} \textbf{(b)} to $P=e_{H,H}$ and $Q=S$,
we obtain that $E_{H}\left(  \operatorname*{Prim}H\right)  \subseteq\left(
e_{H,H}\ast E_{H}\ast S\right)  \left(  H\right)  $. Since $e_{H,H}\ast
E_{H}\ast S=E_{H}\ast S$, this rewrites as $E_{H}\left(  \operatorname*{Prim}%
H\right)  \subseteq\left(  E_{H}\ast S\right)  \left(  H\right)  $. This
proves Corollary~\ref{Corollary823} \textbf{(b)}.

But we can also apply Corollary~\ref{Corollary822} to $P=S$ and $Q=e_{H,H}$.

Applying Corollary~\ref{Corollary822} \textbf{(a)} to $P=S$ and $Q=e_{H,H}$,
we obtain that $\left(  S\ast E_{H}\ast e_{H,H}\right)  \left(  H\right)
\subseteq\operatorname*{Prim}H$. Since $S\ast E_{H}\ast e_{H,H}=S\ast E_{H}$,
this rewrites as $\left(  S\ast E_{H}\right)  \left(  H\right)  \subseteq
\operatorname*{Prim}H$. This proves Corollary~\ref{Corollary823} \textbf{(c)}.

Applying Corollary~\ref{Corollary822} \textbf{(b)} to $P=S$ and $Q=e_{H,H}$,
we obtain that $E_{H}\left(  \operatorname*{Prim}H\right)  \subseteq\left(
S\ast E_{H}\ast e_{H,H}\right)  \left(  H\right)  $. Since $S\ast E_{H}\ast
e_{H,H}=S\ast E_{H}$, this rewrites as $E_{H}\left(  \operatorname*{Prim}%
H\right)  \subseteq\left(  S\ast E_{H}\right)  \left(  H\right)  $. This
proves Corollary~\ref{Corollary823} \textbf{(d)}.
\end{proof}

We now come to the case of fields of characteristic $0$. The following
generalizes Theorem~\ref{Theorem801}:

\begin{theorem}
\label{Theorem831} Let $k$ be a field of characteristic $0$. Let $H$ be a
cocommutative graded $k$-bialgebra. Let $E_{H}$ be defined according to
Definition~\ref{Definition701}. Let $\operatorname*{Prim}H$ denote the
subspace of $H$ consisting of all primitive elements of $H$. Let $\left(
\operatorname*{Prim}H\right)  ^{+}$ denote the intersection $\left(
\operatorname*{Prim}H\right)  \cap\left(  \bigoplus_{n\geq1}H_{n}\right)  $.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-coalgebra
homomorphisms satisfying $P\left(  1_{H}\right)  =1_{H}$, $Q\left(
1_{H}\right)  =1_{H}$ and $P\ast\operatorname{id}_{H}\ast Q=e_{H,H}$. Here,
the map $e_{H,H}$ is defined to be the map $\eta_{H}\circ\varepsilon
_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is identical with
the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

Then, the map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  $ is a projection from $H$ to the subspace $\left(
\operatorname*{Prim}H\right)  ^{+}$.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem831}.]Recall that $\left(  \operatorname*{Prim}%
H\right)  ^{+}=\left(  \operatorname*{Prim}H\right)  \cap\left(
\bigoplus_{n\geq1}H_{n}\right)  =\left(  \operatorname*{Prim}H\right)
\cap\left(  \bigoplus_{\ell\geq1}H_{\ell}\right)  $ (here, we renamed the
index $n$ as $\ell$ in the direct sum). Notice that $\left(
\operatorname*{Prim}H\right)  ^{+}=\left(  \operatorname*{Prim}H\right)
\cap\left(  \bigoplus_{n\geq1}H_{n}\right)  \subseteq\operatorname*{Prim}H$.

Proposition~\ref{Proposition302} yields that $\operatorname*{Prim}H$ is a
homogeneous subspace of $H$. In other words, $\operatorname*{Prim}%
H=\bigoplus_{n\in\mathbb{N}}\left(  \left(  \operatorname*{Prim}H\right)  \cap
H_{n}\right)  $ (because Definition~\ref{Definition300} yields that
$\operatorname*{Prim}H$ is a homogeneous subspace of $H$ if and only if
$\operatorname*{Prim}H=\bigoplus_{n\in\mathbb{N}}\left(  \left(
\operatorname*{Prim}H\right)  \cap H_{n}\right)  $).

Now, it is easy to see that every $n\in\mathbb{N}$ satisfies%
\begin{equation}
E_{H}^{\operatorname*{inv}}\left(  \left(  \operatorname*{Prim}H\right)  \cap
H_{n}\right)  \subseteq\left(  \operatorname*{Prim}H\right)  ^{+}.
\label{831.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{831.pf.2}).} Let $n\in\mathbb{N}$. Let
$v\in\left(  \operatorname*{Prim}H\right)  \cap H_{n}$. We will prove that
$E_{H}^{\operatorname*{inv}}\left(  v\right)  \in\left(  \operatorname*{Prim}%
H\right)  ^{+}$ now.
\par
In fact, define $\left(  b_{\ell}\right)  _{\ell\in\mathbb{N}}$ as in
Definition~\ref{Definition721}. Note that this definition yields
$b_{0}=\left\{
\begin{array}
[c]{c}%
\dfrac{1}{0},\ \ \ \ \ \ \ \ \ \ \text{if }0>0;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }0=0
\end{array}
\right.  =0$ (since $0=0$).
\par
By Remark~\ref{Remark722} (applied to $V=H$), we have $E_{H}%
^{\operatorname*{inv}}=E_{H}^{\left(  b_{\ell}\right)  _{\ell\in\mathbb{N}}}$.
\par
We have $v\in\left(  \operatorname*{Prim}H\right)  \cap H_{n}\subseteq H_{n}$
and thus $E_{H}^{\left(  b_{\ell}\right)  _{\ell\in\mathbb{N}}}\left(
v\right)  =b_{n}v$ (by Proposition~\ref{Proposition707} \textbf{(a)}, applied
to $\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  b_{\ell}\right)
_{\ell\in\mathbb{N}}$). Since $E_{H}^{\left(  b_{\ell}\right)  _{\ell
\in\mathbb{N}}}=E_{H}^{\operatorname*{inv}}$, this rewrites as $E_{H}%
^{\operatorname*{inv}}\left(  v\right)  =b_{n}v$.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} We have $n=0$.
\par
\textit{Case 2:} We have $n>0$.
\par
Let us consider Case 1 first. In this case, $n=0$, so that $b_{n}%
v=\underbrace{b_{0}}_{=0}v=0v=0$. Thus, $E_{H}^{\operatorname*{inv}}\left(
v\right)  =b_{n}v=0\in\left(  \operatorname*{Prim}H\right)  ^{+}$. Thus,
$E_{H}^{\operatorname*{inv}}\left(  v\right)  \in\left(  \operatorname*{Prim}%
H\right)  ^{+}$ is proven in Case 1.
\par
Next, let us consider Case 2. In this case, $n>0$, so that $n\geq1$ and thus
$H_{n}\subseteq\bigoplus_{\ell\geq1}H_{\ell}$. Thus, $v\in H_{n}%
\subseteq\bigoplus_{\ell\geq1}H_{\ell}$. Combined with $v\in\left(
\operatorname*{Prim}H\right)  \cap H_{n}\subseteq\operatorname*{Prim}H$, this
yields $v\in\left(  \operatorname*{Prim}H\right)  \cap\left(  \bigoplus
_{\ell\geq1}H_{\ell}\right)  =\left(  \operatorname*{Prim}H\right)  ^{+}$.
Now, $E_{H}^{\operatorname*{inv}}\left(  v\right)  =b_{n}\underbrace{v}%
_{\in\left(  \operatorname*{Prim}H\right)  ^{+}}\in b_{n}\left(
\operatorname*{Prim}H\right)  ^{+}\subseteq\left(  \operatorname*{Prim}%
H\right)  ^{+}$ (since $\left(  \operatorname*{Prim}H\right)  ^{+}$ is a
$k$-vector space). Thus, $E_{H}^{\operatorname*{inv}}\left(  v\right)
\in\left(  \operatorname*{Prim}H\right)  ^{+}$ is proven in Case 2.
\par
We have therefore proven $E_{H}^{\operatorname*{inv}}\left(  v\right)
\in\left(  \operatorname*{Prim}H\right)  ^{+}$ in both cases 1 and 2. Since
these cases cover all possibilities, we conclude that $E_{H}%
^{\operatorname*{inv}}\left(  v\right)  \in\left(  \operatorname*{Prim}%
H\right)  ^{+}$ always holds.
\par
Now forget that we fixed $v$. We have thus shown that every $v\in\left(
\operatorname*{Prim}H\right)  \cap H_{n}$ satisfies $E_{H}%
^{\operatorname*{inv}}\left(  v\right)  \in\left(  \operatorname*{Prim}%
H\right)  ^{+}$. In other words, $E_{H}^{\operatorname*{inv}}\left(  \left(
\operatorname*{Prim}H\right)  \cap H_{n}\right)  \subseteq\left(
\operatorname*{Prim}H\right)  ^{+}$, thus proving (\ref{831.pf.2}).} Now,
since $\operatorname*{Prim}H=\bigoplus_{n\in\mathbb{N}}\left(  \left(
\operatorname*{Prim}H\right)  \cap H_{n}\right)  =\sum_{n\in\mathbb{N}}\left(
\operatorname*{Prim}H\right)  \cap H_{n}$ (since direct sums are sums), we
have%
\begin{align}
E_{H}^{\operatorname*{inv}}\left(  \operatorname*{Prim}H\right)   &
=E_{H}^{\operatorname*{inv}}\left(  \sum_{n\in\mathbb{N}}\left(
\operatorname*{Prim}H\right)  \cap H_{n}\right)  =\sum_{n\in\mathbb{N}%
}\underbrace{E_{H}^{\operatorname*{inv}}\left(  \left(  \operatorname*{Prim}%
H\right)  \cap H_{n}\right)  }_{\substack{\subseteq\left(
\operatorname*{Prim}H\right)  ^{+}\\\text{(by (\ref{831.pf.2}))}}}\nonumber\\
&  \subseteq\sum_{n\in\mathbb{N}}\left(  \operatorname*{Prim}H\right)
^{+}\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \operatorname*{Prim}H\right)
^{+}\text{ is a }k\text{-vector space}\right)  . \label{831.pf.3}%
\end{align}


Now,%
\begin{align}
\left(  E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)
\right)  \left(  H\right)   &  =E_{H}^{\operatorname*{inv}}\underbrace{\left(
\left(  P\ast E_{H}\ast Q\right)  \left(  H\right)  \right)  }%
_{\substack{\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}\\\text{(by
Theorem~\ref{Theorem812} \textbf{(a)})}}}\subseteq E_{H}^{\operatorname*{inv}%
}\left(  \underbrace{\left(  \operatorname*{Prim}H\right)  ^{+}}%
_{\subseteq\operatorname*{Prim}H}\right) \nonumber\\
&  \subseteq E_{H}^{\operatorname*{inv}}\left(  \operatorname*{Prim}H\right)
\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{831.pf.3})}\right)  . \label{831.pf.4}%
\end{align}


On the other hand,
\begin{equation}
\left(  E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)
\right)  \mid_{\left(  \operatorname*{Prim}H\right)  ^{+}}=\operatorname{id}%
_{\left(  \operatorname*{Prim}H\right)  ^{+}}. \label{831.pf.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{831.pf.6}).} We have $1_{H}\in H_{0}$ (since
$H$ is a graded $k$-algebra). Thus, Proposition~\ref{Proposition702}
\textbf{(a)} (applied to $n=0$ and $v=1_{H}$) yields $E_{H}\left(
1_{H}\right)  =0\cdot1_{H}=0$. Thus, Proposition~\ref{Proposition807} (applied
to $K=E_{H}$) yields that%
\begin{equation}
\text{every }x\in\operatorname*{Prim}H\text{ satisfies }\left(  P\ast
E_{H}\ast Q\right)  \left(  x\right)  =E_{H}\left(  x\right)  .
\label{831.pf.7}%
\end{equation}
\par
Now, let $x\in\left(  \operatorname*{Prim}H\right)  ^{+}$. Then, $x\in\left(
\operatorname*{Prim}H\right)  ^{+}=\left(  \operatorname*{Prim}H\right)
\cap\left(  \bigoplus_{n\geq1}H_{n}\right)  \subseteq\bigoplus_{n\geq1}H_{n}$.
Hence, Corollary~\ref{Corollary724} \textbf{(b)} (applied to $V=H$ and $v=x$)
yields $\left(  E_{H}^{\operatorname*{inv}}\circ E_{H}\right)  \left(
x\right)  =x$.
\par
On the other hand, $x\in\left(  \operatorname*{Prim}H\right)  ^{+}%
\subseteq\operatorname*{Prim}H$, so that $\left(  P\ast E_{H}\ast Q\right)
\left(  x\right)  =E_{H}\left(  x\right)  $ (by (\ref{831.pf.7})). Now,%
\begin{align*}
&  \left(  \left(  E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  \right)  \mid_{\left(  \operatorname*{Prim}H\right)  ^{+}}\right)
\left(  x\right) \\
&  =\left(  E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)
\right)  \left(  x\right)  =E_{H}^{\operatorname*{inv}}\underbrace{\left(
\left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  \right)  }_{=E_{H}\left(
x\right)  }=E_{H}^{\operatorname*{inv}}\left(  E_{H}\left(  x\right)  \right)
\\
&  =\left(  E_{H}^{\operatorname*{inv}}\circ E_{H}\right)  \left(  x\right)
=x=\operatorname{id}_{\left(  \operatorname*{Prim}H\right)  ^{+}}\left(
x\right)  .
\end{align*}
\par
Now forget that we fixed $x$. We have thus proven that every $x\in\left(
\operatorname*{Prim}H\right)  ^{+}$ satisfies $\left(  \left(  E_{H}%
^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  \right)
\mid_{\left(  \operatorname*{Prim}H\right)  ^{+}}\right)  \left(  x\right)
=\operatorname{id}_{\left(  \operatorname*{Prim}H\right)  ^{+}}\left(
x\right)  $. In other words, $\left(  E_{H}^{\operatorname*{inv}}\circ\left(
P\ast E_{H}\ast Q\right)  \right)  \mid_{\left(  \operatorname*{Prim}H\right)
^{+}}=\operatorname{id}_{\left(  \operatorname*{Prim}H\right)  ^{+}}$. This
proves (\ref{831.pf.6}).}

So we know that $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  $ is a $k$-linear map satisfying $\left(  E_{H}^{\operatorname*{inv}%
}\circ\left(  P\ast E_{H}\ast Q\right)  \right)  \left(  H\right)
\subseteq\left(  \operatorname*{Prim}H\right)  ^{+}$ (by (\ref{831.pf.4})) and
$\left(  E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)
\right)  \mid_{\left(  \operatorname*{Prim}H\right)  ^{+}}=\operatorname{id}%
_{\left(  \operatorname*{Prim}H\right)  ^{+}}$ (by (\ref{831.pf.6})). In other
words, $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  $ is
a projection from $H$ to the subspace $\left(  \operatorname*{Prim}H\right)
^{+}$. This proves Theorem~\ref{Theorem831}.
\end{proof}

We can now get Theorem~\ref{Theorem801} as a corollary:

\begin{proof}
[Proof of Theorem~\ref{Theorem801}.]Just as in the proof of
Corollary~\ref{Corollary806}, we can prove the following facts:

\begin{itemize}
\item The map $S$ is a $k$-coalgebra homomorphism from $H$ to $H$.

\item The map $e_{H,H}$ is a $k$-coalgebra homomorphism (where $e_{H,H}$ is
defined as according to Definition~\ref{def.eHA}).

\item We have $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$.

\item We have $S\ast\operatorname{id}_{H}\ast e_{H,H}=e_{H,H}$ and
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}$.
\end{itemize}

Just as in the proof of Corollary~\ref{Corollary813}, we can prove the
following facts:

\begin{itemize}
\item We have $e_{H,H}\left(  1_{H}\right)  =1_{H}$ and $S\left(
1_{H}\right)  =1_{H}$.

\item The maps $e_{H,H}$ and $S$ are graded.
\end{itemize}

As a consequence, we can apply Theorem~\ref{Theorem831} to $P=e_{H,H}$ and
$Q=S$. As a result, we obtain that the map $E_{H}^{\operatorname*{inv}}%
\circ\left(  e_{H,H}\ast E_{H}\ast S\right)  $ is a projection from $H$ to the
subspace $\left(  \operatorname*{Prim}H\right)  ^{+}$. Since $e_{H,H}\ast
E_{H}\ast S=E_{H}\ast S$, this rewrites as follows: The map $E_{H}%
^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  $ is a projection from
$H$ to the subspace $\left(  \operatorname*{Prim}H\right)  ^{+}$. This proves
Theorem~\ref{Theorem801} \textbf{(a)}.

But we can also apply Theorem~\ref{Theorem831} to $P=S$ and $Q=e_{H,H}$. As a
result, we obtain that the map $E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\ast e_{H,H}\right)  $ is a projection from $H$ to the subspace $\left(
\operatorname*{Prim}H\right)  ^{+}$. Since $S\ast E_{H}\ast e_{H,H}=S\ast
E_{H}$, this rewrites as follows: The map $E_{H}^{\operatorname*{inv}}%
\circ\left(  S\ast E_{H}\right)  $ is a projection from $H$ to the subspace
$\left(  \operatorname*{Prim}H\right)  ^{+}$. This proves
Theorem~\ref{Theorem801} \textbf{(b)}.
\end{proof}

Next, we specialize our results to connected graded bialgebras. First, the
specialization of Theorem~\ref{Theorem831}:

\begin{corollary}
\label{Corollary832} Let $k$ be a field of characteristic $0$. Let $H$ be a
cocommutative connected graded $k$-bialgebra. Let $E_{H}$ be defined according
to Definition~\ref{Definition701}. Let $\operatorname*{Prim}H$ denote the
subspace of $H$ consisting of all primitive elements of $H$.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-coalgebra
homomorphisms satisfying $P\left(  1_{H}\right)  =1_{H}$, $Q\left(
1_{H}\right)  =1_{H}$ and $P\ast\operatorname{id}_{H}\ast Q=e_{H,H}$. Here,
the map $e_{H,H}$ is defined to be the map $\eta_{H}\circ\varepsilon
_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is identical with
the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

Then, the map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  $ is a projection from $H$ to the subspace $\operatorname*{Prim}H$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary832}.]Define $\left(  \operatorname*{Prim}%
H\right)  ^{+}$ as in Theorem~\ref{Theorem812}. Then, $\left(
\operatorname*{Prim}H\right)  ^{+}=\left(  \operatorname*{Prim}H\right)
\cap\left(  \bigoplus_{n\geq1}H_{n}\right)  =\operatorname*{Prim}H$ (since
Proposition~\ref{Proposition821} yields $\operatorname*{Prim}H\subseteq
\bigoplus_{n\geq1}H_{n}$).

Theorem~\ref{Theorem831} says that the map $E_{H}^{\operatorname*{inv}}%
\circ\left(  P\ast E_{H}\ast Q\right)  $ is a projection from $H$ to the
subspace $\left(  \operatorname*{Prim}H\right)  ^{+}$. Since $\left(
\operatorname*{Prim}H\right)  ^{+}=\operatorname*{Prim}H$, this rewrites as
follows: The map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  $ is a projection from $H$ to the subspace $\operatorname*{Prim}H$.
This proves Corollary~\ref{Corollary832}.
\end{proof}

The specialization of Theorem~\ref{Theorem801} is Theorem~\ref{Theorem802},
and here is its (obvious) proof:

\begin{proof}
[Proof of Theorem~\ref{Theorem802}.]Define $\left(  \operatorname*{Prim}%
H\right)  ^{+}$ as in Theorem~\ref{Theorem801}. Then, $\left(
\operatorname*{Prim}H\right)  ^{+}=\left(  \operatorname*{Prim}H\right)
\cap\left(  \bigoplus_{n\geq1}H_{n}\right)  =\operatorname*{Prim}H$ (since
Proposition~\ref{Proposition821} yields $\operatorname*{Prim}H\subseteq
\bigoplus_{n\geq1}H_{n}$).

Theorem~\ref{Theorem801} \textbf{(a)} says that the map $E_{H}%
^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  $ is a projection from
$H$ to the subspace $\left(  \operatorname*{Prim}H\right)  ^{+}$. Since
$\left(  \operatorname*{Prim}H\right)  ^{+}=\operatorname*{Prim}H$, this
rewrites as follows: The map $E_{H}^{\operatorname*{inv}}\circ\left(
E_{H}\ast S\right)  $ is a projection from $H$ to the subspace
$\operatorname*{Prim}H$. This proves Theorem~\ref{Theorem802} \textbf{(a)}.

Theorem~\ref{Theorem801} \textbf{(b)} says that the map $E_{H}%
^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  $ is a projection from
$H$ to the subspace $\left(  \operatorname*{Prim}H\right)  ^{+}$. Since
$\left(  \operatorname*{Prim}H\right)  ^{+}=\operatorname*{Prim}H$, this
rewrites as follows: The map $E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\right)  $ is a projection from $H$ to the subspace $\operatorname*{Prim}%
H$. This proves Theorem~\ref{Theorem802} \textbf{(b)}.
\end{proof}

Finally, let us prove a relation between the maps $S\ast E_{H}$ and $E_{H}\ast
S$:

\begin{theorem}
\label{Theorem841} Let $k$ be a field. Let $H$ be a cocommutative graded
$k$-Hopf algebra. Let $S$ be the antipode of $H$. Let $E_{H}$ be defined
according to Definition~\ref{Definition701}.

\textbf{(a)} We have%
\[
\left(  E_{H}\ast S\right)  \circ S=S\circ\left(  S\ast E_{H}\right)  =-S\ast
E_{H}.
\]


\textbf{(b)} We have%
\[
\left(  S\ast E_{H}\right)  \circ S=S\circ\left(  E_{H}\ast S\right)
=-E_{H}\ast S.
\]

\end{theorem}

To prove Theorem~\ref{Theorem841}, we need a (famous) fact:

\begin{theorem}
\label{Theorem842} Let $k$ be a field. Let $H$ be a $k$-Hopf algebra. Let $S$
be the antipode of $H$.

\textbf{(a)} We have $\varepsilon_{H}\circ S=\varepsilon_{H}$.

\textbf{(b)} If $H$ is cocommutative, we have $\Delta_{H}\circ S=\left(
S\otimes S\right)  \circ\Delta_{H}$.

\textbf{(c)} If $H$ is cocommutative, we have $S\circ S=\operatorname{id}_{H}$.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem842}.]Let us define $H^{\operatorname*{cop}}$ as
according to Definition~\ref{Definition552}. Then, $H^{\operatorname*{cop}%
}=\left(  H,\tau_{H,H}\circ\Delta_{H},\varepsilon_{H}\right)  $, so that
$\varepsilon_{H^{\operatorname*{cop}}}=\varepsilon_{H}$.

But Proposition~\ref{Proposition553} yields that the antipode of $H$ is a
$k$-coalgebra homomorphism from $H^{\operatorname*{cop}}$ to $H$. Since the
antipode of $H$ is the map $S$, this rewrites as follows: The map $S$ is a
$k$-coalgebra homomorphism from $H^{\operatorname*{cop}}$ to $H$. Thus,
$\varepsilon_{H^{\operatorname*{cop}}}\circ S=\varepsilon_{H}$. Since
$\varepsilon_{H^{\operatorname*{cop}}}=\varepsilon_{H}$, this rewrites as
$\varepsilon_{H}\circ S=\varepsilon_{H}$. This proves Theorem~\ref{Theorem842}
\textbf{(a)}.

\textbf{(b)} Assume that $H$ is cocommutative. Then, $\tau_{H,H}\circ
\Delta_{H}=\Delta_{H}$, so that $H^{\operatorname*{cop}}=\left(
H,\underbrace{\tau_{H,H}\circ\Delta_{H}}_{=\Delta_{H}},\varepsilon_{H}\right)
=\left(  H,\Delta_{H},\varepsilon_{H}\right)  =H$.

We have showed above that the map $S$ is a $k$-coalgebra homomorphism from
$H^{\operatorname*{cop}}$ to $H$. Since $H^{\operatorname*{cop}}=H$, this
rewrites as follows: The map $S$ is a $k$-coalgebra homomorphism from $H$ to
$H$. Thus, $\Delta_{H}\circ S=\left(  S\otimes S\right)  \circ\Delta_{H}$.
This proves Theorem~\ref{Theorem842} \textbf{(b)}.

\textbf{(c)} The antipode of $H$ is the $\ast$-inverse of the map
$\operatorname{id}_{H}$ (because the antipode of a Hopf algebra is defined as
the $\ast$-inverse of its identity map). Since the antipode of $H$ is $S$, we
thus have shown that $S$ is the $\ast$-inverse of the map $\operatorname{id}%
_{H}$. Thus, $S\ast\operatorname{id}_{H}=\operatorname{id}_{H}\ast S=e_{H,H}$
(where $e_{H,H}$ is defined as according to Definition~\ref{def.eHA}).

Applying (\ref{(X)(X)}) to $H$, $H$, $H$, $H$, $H$, $H$, $S$, $S$, $S$,
$\operatorname{id}_{H}$ instead of $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$,
$W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$, $\beta^{\prime}$, we
obtain%
\begin{equation}
\left(  S\circ S\right)  \otimes\left(  \operatorname{id}_{H}\circ S\right)
=\left(  S\otimes\operatorname{id}_{H}\right)  \circ\left(  S\otimes S\right)
. \label{842.pf.1}%
\end{equation}
Now, by the definition of convolution,%
\begin{align*}
\left(  S\circ S\right)  \ast S  &  =\mu_{H}\circ\left(  \left(  S\circ
S\right)  \otimes\underbrace{S}_{=\operatorname{id}_{H}\circ S}\right)
\circ\Delta_{H}=\mu_{H}\circ\underbrace{\left(  \left(  S\circ S\right)
\otimes\left(  \operatorname{id}_{H}\circ S\right)  \right)  }%
_{\substack{=\left(  S\otimes\operatorname{id}_{H}\right)  \circ\left(
S\otimes S\right)  \\\text{(by (\ref{842.pf.1}))}}}\circ\Delta_{H}\\
&  =\mu_{H}\circ\left(  S\otimes\operatorname{id}_{H}\right)  \circ
\underbrace{\left(  S\otimes S\right)  \circ\Delta_{H}}_{\substack{=\Delta
_{H}\circ S\\\text{(by Theorem~\ref{Theorem842} \textbf{(b)})}}%
}=\underbrace{\mu_{H}\circ\left(  S\otimes\operatorname{id}_{H}\right)
\circ\Delta_{H}}_{\substack{=S\ast\operatorname{id}_{H}\\\text{(because}%
\\S\ast\operatorname{id}_{H}=\mu_{H}\circ\left(  S\otimes\operatorname{id}%
_{H}\right)  \circ\Delta_{H}\\\text{by the definition of convolution)}}}\circ
S\\
&  =\underbrace{\left(  S\ast\operatorname{id}_{H}\right)  }%
_{\substack{=e_{H,H}=\eta_{H}\circ\varepsilon_{H}\\\text{(by the definition of
}e_{H,H}\text{)}}}\circ S=\eta_{H}\circ\underbrace{\varepsilon_{H}\circ
S}_{\substack{=\varepsilon_{H}\\\text{(by Theorem~\ref{Theorem842}
\textbf{(a)})}}}=\eta_{H}\circ\varepsilon_{H}=e_{H,H}%
\end{align*}
(since $e_{H,H}=\eta_{H}\circ\varepsilon_{H}$ by the definition of $e_{H,H}$).
Thus,%
\[
\underbrace{\left(  S\circ S\right)  \ast S}_{=e_{H,H}}\ast\operatorname{id}%
_{H}=e_{H,H}\ast\operatorname{id}_{H}=\operatorname{id}_{H}.
\]
Comparing this with%
\[
\left(  S\circ S\right)  \ast\underbrace{S\ast\operatorname{id}_{H}}%
_{=e_{H,H}}=\left(  S\circ S\right)  \ast e_{H,H}=S\circ S,
\]
we obtain $S\circ S=\operatorname{id}_{H}$. This proves
Theorem~\ref{Theorem842} \textbf{(c)}.
\end{proof}

Furthermore, we need something very easy and well-known:

\begin{proposition}
\label{Proposition843} Let $k$ be a field. Let $H$ be a $k$-Hopf algebra. Let
$S$ be the antipode of $H$. Let $x\in\operatorname*{Prim}H$. Then, $S\left(
x\right)  =-x$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition843}.]Since $x\in\operatorname*{Prim}%
H=\left(  \text{the set of all primitive elements of }H\right)  $, the element
$x$ of $H$ is primitive. Thus, $\Delta\left(  x\right)  =x\otimes1_{H}%
+1_{H}\otimes x$ (since Definition~\ref{Definition18} yields that $x$ is
primitive if and only if $\Delta\left(  x\right)  =x\otimes1_{H}+1_{H}\otimes
x$).

The antipode of $H$ is the $\ast$-inverse of the map $\operatorname{id}_{H}$
(because the antipode of a Hopf algebra is defined as the $\ast$-inverse of
its identity map). Since the antipode of $H$ is $S$, we thus have shown that
$S$ is the $\ast$-inverse of the map $\operatorname{id}_{H}$. Thus,
$S\ast\operatorname{id}_{H}=\operatorname{id}_{H}\ast S=e_{H,H}$ (where
$e_{H,H}$ is defined as according to Definition~\ref{def.eHA}) and
$S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$.

Since $e_{H,H}=\eta_{H}\circ\varepsilon_{H}$ (by the definition of $e_{H,H}$),
we have%
\[
e_{H,H}\left(  x\right)  =\left(  \eta_{H}\circ\varepsilon_{H}\right)  \left(
x\right)  =\eta_{H}\left(  \underbrace{\varepsilon_{H}}_{=\varepsilon}\left(
x\right)  \right)  =\eta_{H}\underbrace{\left(  \varepsilon\left(  x\right)
\right)  }_{\substack{=0\\\text{(by Remark~\ref{Remark21})}}}=\eta_{H}\left(
0\right)  =0
\]
(since $\eta_{H}$ is $k$-linear).

Just as in the proof of Corollary~\ref{Corollary813}, we can now show that
$S\left(  1_{H}\right)  =1_{H}$. Now, since $e_{H,H}=S\ast\operatorname{id}%
_{H}=\mu_{H}\circ\left(  S\otimes\operatorname{id}_{H}\right)  \circ\Delta
_{H}$ (by the definition of convolution), we have%
\begin{align*}
e_{H,H}\left(  x\right)   &  =\left(  \mu_{H}\circ\left(  S\otimes
\operatorname{id}_{H}\right)  \circ\Delta_{H}\right)  \left(  x\right)
=\mu_{H}\left(  \left(  S\otimes\operatorname{id}_{H}\right)  \left(
\underbrace{\Delta_{H}}_{=\Delta}\left(  x\right)  \right)  \right) \\
&  =\mu_{H}\left(  \left(  S\otimes\operatorname{id}_{H}\right)
\underbrace{\left(  \Delta\left(  x\right)  \right)  }_{=x\otimes1_{H}%
+1_{H}\otimes x}\right)  =\mu_{H}\left(  \underbrace{\left(  S\otimes
\operatorname{id}_{H}\right)  \left(  x\otimes1_{H}+1_{H}\otimes x\right)
}_{\substack{=\left(  S\otimes\operatorname{id}_{H}\right)  \left(
x\otimes1_{H}\right)  +\left(  S\otimes\operatorname{id}_{H}\right)  \left(
1_{H}\otimes x\right)  \\\text{(since }S\otimes\operatorname{id}_{H}\text{ is
}k\text{-linear)}}}\right) \\
&  =\mu_{H}\left(  \underbrace{\left(  S\otimes\operatorname{id}_{H}\right)
\left(  x\otimes1_{H}\right)  }_{=S\left(  x\right)  \otimes\operatorname{id}%
_{H}\left(  1_{H}\right)  }+\underbrace{\left(  S\otimes\operatorname{id}%
_{H}\right)  \left(  1_{H}\otimes x\right)  }_{=S\left(  1_{H}\right)
\otimes\operatorname{id}_{H}\left(  x\right)  }\right) \\
&  =\mu_{H}\left(  S\left(  x\right)  \otimes\underbrace{\operatorname{id}%
_{H}\left(  1_{H}\right)  }_{=1_{H}}+\underbrace{S\left(  1_{H}\right)
}_{=1_{H}}\otimes\underbrace{\operatorname{id}_{H}\left(  x\right)  }%
_{=x}\right)  =\mu_{H}\left(  S\left(  x\right)  \otimes1_{H}+1_{H}\otimes
x\right) \\
&  =\underbrace{\mu_{H}\left(  S\left(  x\right)  \otimes1_{H}\right)
}_{\substack{=S\left(  x\right)  1_{H}\\\text{(since }\mu_{H}\text{ is
the}\\\text{multiplication map)}}}+\underbrace{\mu_{H}\left(  1_{H}\otimes
x\right)  }_{\substack{=1_{H}x\\\text{(since }\mu_{H}\text{ is the}%
\\\text{multiplication map)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu
_{H}\text{ is }k\text{-linear}\right) \\
&  =S\left(  x\right)  1_{H}+1_{H}x=S\left(  x\right)  +x.
\end{align*}
Compared with $e_{H,H}\left(  x\right)  =0$, this yields $S\left(  x\right)
+x=0$, and thus $S\left(  x\right)  =-x$. This proves
Proposition~\ref{Proposition843}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem841}.]Let us first notice that
\begin{equation}
E_{H}\circ e_{H,H}=0. \label{841.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{841.pf.1}):} Let $x\in H$. Then,
$e_{H,H}=\eta_{H}\circ\varepsilon_{H}$ (by the definition of $e_{H,H}$), so
that
\begin{align*}
e_{H,H}\left(  x\right)   &  =\left(  \eta_{H}\circ\varepsilon_{H}\right)
\left(  x\right)  =\eta_{H}\left(  \varepsilon_{H}\left(  x\right)  \right)
=\underbrace{\varepsilon_{H}\left(  x\right)  }_{\in k}\cdot\underbrace{1_{H}%
}_{\substack{\in H_{0}\\\text{(since }H\text{ is a graded}\\k\text{-algebra)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\eta_{H}\right) \\
&  \in kH_{0}\subseteq H_{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
H_{0}\text{ is a }k\text{-vector space}\right)  ,
\end{align*}
and thus $E_{H}\left(  e_{H,H}\left(  x\right)  \right)  =0e_{H,H}\left(
x\right)  $ (by Proposition~\ref{Proposition702} \textbf{(a)}, applied to
$V=H$, $n=0$ and $v=e_{H,H}\left(  x\right)  $). Thus, $\left(  E_{H}\circ
e_{H,H}\right)  \left(  x\right)  =E_{H}\left(  e_{H,H}\left(  x\right)
\right)  =0e_{H,H}\left(  x\right)  =0$.
\par
Now, forget that we fixed $x$. We thus have proven that every $x\in H$
satisfies $\left(  E_{H}\circ e_{H,H}\right)  \left(  x\right)  =0$. Thus,
$E_{H}\circ e_{H,H}=0$, so that (\ref{841.pf.1}) is proven.}

Since $H$ is a graded $k$-algebra, the multiplication map $\mu_{H}:H\otimes
H\rightarrow H$ of $H$ is graded. Thus, Proposition~\ref{Proposition704}
(applied to $V=H\otimes H$ and $W=H$) yields $\mu_{H}\circ E_{H\otimes
H}=E_{H}\circ\mu_{H}$. Thus,%
\begin{align*}
E_{H}\circ\mu_{H}  &  =\mu_{H}\circ\underbrace{E_{H\otimes H}}%
_{\substack{=E_{H}\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes
E_{H}\\\text{(by Proposition~\ref{Proposition703},}\\\text{applied to
}V=H\text{ and }W=H\text{)}}}=\mu_{H}\circ\left(  E_{H}\otimes
\operatorname{id}_{H}+\operatorname{id}_{H}\otimes E_{H}\right) \\
&  =\mu_{H}\circ\left(  E_{H}\otimes\operatorname{id}_{H}\right)  +\mu
_{H}\circ\left(  \operatorname{id}_{H}\otimes E_{H}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right)  .
\end{align*}
This equality can be rewritten in two ways: first,%
\begin{equation}
\mu_{H}\circ\left(  E_{H}\otimes\operatorname{id}_{H}\right)  =E_{H}\circ
\mu_{H}-\mu_{H}\circ\left(  \operatorname{id}_{H}\otimes E_{H}\right)  ;
\label{841.pf.2a}%
\end{equation}
second,%
\begin{equation}
\mu_{H}\circ\left(  \operatorname{id}_{H}\otimes E_{H}\right)  =E_{H}\circ
\mu_{H}-\mu_{H}\circ\left(  E_{H}\otimes\operatorname{id}_{H}\right)  .
\label{841.pf.2b}%
\end{equation}


The antipode of $H$ is the $\ast$-inverse of the map $\operatorname{id}_{H}$
(because the antipode of a Hopf algebra is defined as the $\ast$-inverse of
its identity map). Since the antipode of $H$ is $S$, we thus have shown that
$S$ is the $\ast$-inverse of the map $\operatorname{id}_{H}$. Thus,
$S\ast\operatorname{id}_{H}=\operatorname{id}_{H}\ast S=e_{H,H}$ (where
$e_{H,H}$ is defined as according to Definition~\ref{def.eHA}).

\textbf{(a)} By the definition of convolution, $E_{H}\ast S=\mu_{H}%
\circ\left(  E_{H}\otimes S\right)  \circ\Delta_{H}$, so that%
\begin{align}
&  \left(  E_{H}\ast S\right)  \circ S\nonumber\\
&  =\mu_{H}\circ\left(  E_{H}\otimes S\right)  \circ\underbrace{\Delta
_{H}\circ S}_{\substack{=\left(  S\otimes S\right)  \circ\Delta_{H}\\\text{(by
}\\\text{Theorem~\ref{Theorem842} \textbf{(b)})}}}=\mu_{H}\circ
\underbrace{\left(  E_{H}\otimes S\right)  \circ\left(  S\otimes S\right)
}_{\substack{=\left(  E_{H}\circ S\right)  \otimes\left(  S\circ S\right)
\\\text{(because (\ref{(X)(X)}) (applied to }H\text{, }H\text{, }H\text{,
}H\text{, }H\text{, }H\text{,}\\S\text{, }E_{H}\text{, }S\text{, }S\text{
instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{, }V^{\prime}\text{,
}W^{\prime}\text{, }\alpha\text{, }\beta\text{, }\alpha^{\prime}\text{, }%
\beta^{\prime}\text{)}\\\text{yields }\left(  E_{H}\circ S\right)
\otimes\left(  S\circ S\right)  =\left(  E_{H}\otimes S\right)  \circ\left(
S\otimes S\right)  \text{)}}}\circ\Delta_{H}\nonumber\\
&  =\mu_{H}\circ\left(  \left(  E_{H}\circ S\right)  \otimes
\underbrace{\left(  S\circ S\right)  }_{\substack{=\operatorname{id}%
_{H}\\\text{(by Theorem~\ref{Theorem842} \textbf{(c)})}}}\right)  \circ
\Delta_{H}=\mu_{H}\circ\left(  \left(  E_{H}\circ S\right)  \otimes
\underbrace{\operatorname{id}_{H}}_{=\operatorname{id}_{H}\circ
\operatorname{id}_{H}}\right)  \circ\Delta_{H}\nonumber\\
&  =\mu_{H}\circ\underbrace{\left(  \left(  E_{H}\circ S\right)
\otimes\left(  \operatorname{id}_{H}\circ\operatorname{id}_{H}\right)
\right)  }_{\substack{=\left(  E_{H}\otimes\operatorname{id}_{H}\right)
\circ\left(  S\otimes\operatorname{id}_{H}\right)  \\\text{(by (\ref{(X)(X)})
(applied to }H\text{, }H\text{, }H\text{, }H\text{, }H\text{, }H\text{,}%
\\S\text{, }E_{H}\text{, }\operatorname{id}_{H}\text{, }\operatorname{id}%
_{H}\text{ instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{,
}V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }%
\alpha^{\prime}\text{, }\beta^{\prime}\text{))}}}\circ\Delta_{H}\nonumber\\
&  =\underbrace{\mu_{H}\circ\left(  E_{H}\otimes\operatorname{id}_{H}\right)
}_{\substack{=E_{H}\circ\mu_{H}-\mu_{H}\circ\left(  \operatorname{id}%
_{H}\otimes E_{H}\right)  \\\text{(by (\ref{841.pf.2a}))}}}\circ\left(
S\otimes\operatorname{id}_{H}\right)  \circ\Delta_{H}=\left(  E_{H}\circ
\mu_{H}-\mu_{H}\circ\left(  \operatorname{id}_{H}\otimes E_{H}\right)
\right)  \circ\left(  S\otimes\operatorname{id}_{H}\right)  \circ\Delta
_{H}\nonumber\\
&  =E_{H}\circ\underbrace{\mu_{H}\circ\left(  S\otimes\operatorname{id}%
_{H}\right)  \circ\Delta_{H}}_{\substack{=S\ast\operatorname{id}%
_{H}\\\text{(since }S\ast\operatorname{id}_{H}=\mu_{H}\circ\left(
S\otimes\operatorname{id}_{H}\right)  \circ\Delta_{H}\\\text{by the definition
of convolution)}}}-\mu_{H}\circ\underbrace{\left(  \operatorname{id}%
_{H}\otimes E_{H}\right)  \circ\left(  S\otimes\operatorname{id}_{H}\right)
}_{\substack{=\left(  \operatorname{id}_{H}\circ S\right)  \otimes\left(
E_{H}\circ\operatorname{id}_{H}\right)  \\\text{(because (\ref{(X)(X)})
(applied to }H\text{, }H\text{, }H\text{, }H\text{, }H\text{, }H\text{,}%
\\S\text{, }\operatorname{id}_{H}\text{, }\operatorname{id}_{H}\text{, }%
E_{H}\text{ instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{,
}V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }%
\alpha^{\prime}\text{, }\beta^{\prime}\text{)}\\\text{yields }\left(
\operatorname{id}_{H}\circ S\right)  \otimes\left(  E_{H}\circ
\operatorname{id}_{H}\right)  =\left(  \operatorname{id}_{H}\otimes
E_{H}\right)  \circ\left(  S\otimes\operatorname{id}_{H}\right)  \text{)}%
}}\circ\Delta_{H}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right) \nonumber\\
&  =E_{H}\circ\underbrace{\left(  S\ast\operatorname{id}_{H}\right)
}_{=e_{H,H}}-\mu_{H}\circ\left(  \underbrace{\left(  \operatorname{id}%
_{H}\circ S\right)  }_{=S}\otimes\underbrace{\left(  E_{H}\circ
\operatorname{id}_{H}\right)  }_{=E_{H}}\right)  \circ\Delta_{H}\nonumber\\
&  =\underbrace{E_{H}\circ e_{H,H}}_{\substack{=0\\\text{(by (\ref{841.pf.1}%
))}}}-\underbrace{\mu_{H}\circ\left(  S\otimes E_{H}\right)  \circ\Delta_{H}%
}_{\substack{=S\ast E_{H}\\\text{(since }S\ast E_{H}=\mu_{H}\circ\left(
S\otimes E_{H}\right)  \circ\Delta_{H}\\\text{by the definition of
convolution)}}}=-S\ast E_{H}. \label{841.pf.5}%
\end{align}


Let $\left(  \operatorname*{Prim}H\right)  ^{+}$ denote the intersection
$\left(  \operatorname*{Prim}H\right)  \cap\left(  \bigoplus_{n\geq1}%
H_{n}\right)  $. Then, $\left(  \operatorname*{Prim}H\right)  ^{+}%
\subseteq\operatorname*{Prim}H$.

Now, let $y\in H$. Then,
\begin{align*}
\left(  S\ast E_{H}\right)  \left(  y\right)   &  \in\left(  S\ast
E_{H}\right)  \left(  H\right)  \subseteq\left(  \operatorname*{Prim}H\right)
^{+}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary~\ref{Corollary813}
\textbf{(d)}}\right) \\
&  \subseteq\operatorname*{Prim}H,
\end{align*}
so that Proposition~\ref{Proposition843} (applied to $x=\left(  S\ast
E_{H}\right)  \left(  y\right)  $) yields $S\left(  \left(  S\ast
E_{H}\right)  \left(  y\right)  \right)  =-\left(  S\ast E_{H}\right)  \left(
y\right)  $. Thus,%
\[
\left(  S\circ\left(  S\ast E_{H}\right)  \right)  \left(  y\right)  =S\left(
\left(  S\ast E_{H}\right)  \left(  y\right)  \right)  =-\left(  S\ast
E_{H}\right)  \left(  y\right)  =\left(  -S\ast E_{H}\right)  \left(
y\right)  .
\]


Now, forget that we fixed $y$. We have thus proven that every $y\in H$
satisfies $\left(  S\circ\left(  S\ast E_{H}\right)  \right)  \left(
y\right)  =\left(  -S\ast E_{H}\right)  \left(  y\right)  $. In other words,
$S\circ\left(  S\ast E_{H}\right)  =-S\ast E_{H}$. Combined with
(\ref{841.pf.5}), this yields $\left(  E_{H}\ast S\right)  \circ
S=S\circ\left(  S\ast E_{H}\right)  =-S\ast E_{H}$. This proves
Theorem~\ref{Theorem841} \textbf{(a)}.

\textbf{(b)} By the definition of convolution, $S\ast E_{H}=\mu_{H}%
\circ\left(  S\otimes E_{H}\right)  \circ\Delta_{H}$, so that%
\begin{align}
&  \left(  S\ast E_{H}\right)  \circ S\nonumber\\
&  =\mu_{H}\circ\left(  S\otimes E_{H}\right)  \circ\underbrace{\Delta
_{H}\circ S}_{\substack{=\left(  S\otimes S\right)  \circ\Delta_{H}\\\text{(by
}\\\text{Theorem~\ref{Theorem842} \textbf{(b)})}}}=\mu_{H}\circ
\underbrace{\left(  S\otimes E_{H}\right)  \circ\left(  S\otimes S\right)
}_{\substack{=\left(  S\circ S\right)  \otimes\left(  E_{H}\circ S\right)
\\\text{(because (\ref{(X)(X)}) (applied to }H\text{, }H\text{, }H\text{,
}H\text{, }H\text{, }H\text{,}\\S\text{, }S\text{, }S\text{, }E_{H}\text{
instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{, }V^{\prime}\text{,
}W^{\prime}\text{, }\alpha\text{, }\beta\text{, }\alpha^{\prime}\text{, }%
\beta^{\prime}\text{)}\\\text{yields }\left(  S\circ S\right)  \otimes\left(
E_{H}\circ S\right)  =\left(  S\otimes E_{H}\right)  \circ\left(  S\otimes
S\right)  \text{)}}}\circ\Delta_{H}\nonumber\\
&  =\mu_{H}\circ\left(  \underbrace{\left(  S\circ S\right)  }%
_{\substack{=\operatorname{id}_{H}\\\text{(by Theorem~\ref{Theorem842}
\textbf{(c)})}}}\otimes\left(  E_{H}\circ S\right)  \right)  \circ\Delta
_{H}=\mu_{H}\circ\left(  \underbrace{\operatorname{id}_{H}}%
_{=\operatorname{id}_{H}\circ\operatorname{id}_{H}}\otimes\left(  E_{H}\circ
S\right)  \right)  \circ\Delta_{H}\nonumber\\
&  =\mu_{H}\circ\underbrace{\left(  \left(  \operatorname{id}_{H}%
\circ\operatorname{id}_{H}\right)  \otimes\left(  E_{H}\circ S\right)
\right)  }_{\substack{=\left(  \operatorname{id}_{H}\otimes E_{H}\right)
\circ\left(  \operatorname{id}_{H}\otimes S\right)  \\\text{(by (\ref{(X)(X)})
(applied to }H\text{, }H\text{, }H\text{, }H\text{, }H\text{, }H\text{,}%
\\\operatorname{id}_{H}\text{, }\operatorname{id}_{H}\text{, }S\text{, }%
E_{H}\text{ instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{,
}V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }%
\alpha^{\prime}\text{, }\beta^{\prime}\text{))}}}\circ\Delta_{H}\nonumber\\
&  =\underbrace{\mu_{H}\circ\left(  \operatorname{id}_{H}\otimes E_{H}\right)
}_{\substack{=E_{H}\circ\mu_{H}-\mu_{H}\circ\left(  E_{H}\otimes
\operatorname{id}_{H}\right)  \\\text{(by (\ref{841.pf.2b}))}}}\circ\left(
\operatorname{id}_{H}\otimes S\right)  \circ\Delta_{H}=\left(  E_{H}\circ
\mu_{H}-\mu_{H}\circ\left(  E_{H}\otimes\operatorname{id}_{H}\right)  \right)
\circ\left(  \operatorname{id}_{H}\otimes S\right)  \circ\Delta_{H}\nonumber\\
&  =E_{H}\circ\underbrace{\mu_{H}\circ\left(  \operatorname{id}_{H}\otimes
S\right)  \circ\Delta_{H}}_{\substack{=\operatorname{id}_{H}\ast
S\\\text{(since }\operatorname{id}_{H}\ast S=\mu_{H}\circ\left(
\operatorname{id}_{H}\otimes S\right)  \circ\Delta_{H}\\\text{by the
definition of convolution)}}}-\mu_{H}\circ\underbrace{\left(  E_{H}%
\otimes\operatorname{id}_{H}\right)  \circ\left(  \operatorname{id}_{H}\otimes
S\right)  }_{\substack{=\left(  E_{H}\circ\operatorname{id}_{H}\right)
\otimes\left(  \operatorname{id}_{H}\circ S\right)  \\\text{(because
(\ref{(X)(X)}) (applied to }H\text{, }H\text{, }H\text{, }H\text{, }H\text{,
}H\text{,}\\\operatorname{id}_{H}\text{, }E_{H}\text{, }S\text{,
}\operatorname{id}_{H}\text{ instead of }U\text{, }V\text{, }W\text{,
}U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }%
\beta\text{, }\alpha^{\prime}\text{, }\beta^{\prime}\text{)}\\\text{yields
}\left(  E_{H}\circ\operatorname{id}_{H}\right)  \otimes\left(
\operatorname{id}_{H}\circ S\right)  =\left(  E_{H}\otimes\operatorname{id}%
_{H}\right)  \circ\left(  \operatorname{id}_{H}\otimes S\right)  \text{)}%
}}\circ\Delta_{H}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right) \nonumber\\
&  =E_{H}\circ\underbrace{\left(  \operatorname{id}_{H}\ast S\right)
}_{=e_{H,H}}-\mu_{H}\circ\left(  \underbrace{\left(  E_{H}\circ
\operatorname{id}_{H}\right)  }_{=E_{H}}\otimes\underbrace{\left(
\operatorname{id}_{H}\circ S\right)  }_{=S}\right)  \circ\Delta_{H}\nonumber\\
&  =\underbrace{E_{H}\circ e_{H,H}}_{\substack{=0\\\text{(by (\ref{841.pf.1}%
))}}}-\underbrace{\mu_{H}\circ\left(  E_{H}\otimes S\right)  \circ\Delta_{H}%
}_{\substack{=E_{H}\ast S\\\text{(since }E_{H}\ast S=\mu_{H}\circ\left(
E_{H}\otimes S\right)  \circ\Delta_{H}\\\text{by the definition of
convolution)}}}=-E_{H}\ast S. \label{841.pf.15}%
\end{align}


Let $\left(  \operatorname*{Prim}H\right)  ^{+}$ denote the intersection
$\left(  \operatorname*{Prim}H\right)  \cap\left(  \bigoplus_{n\geq1}%
H_{n}\right)  $. Then, $\left(  \operatorname*{Prim}H\right)  ^{+}%
\subseteq\operatorname*{Prim}H$.

Now, let $y\in H$. Then,
\begin{align*}
\left(  E_{H}\ast S\right)  \left(  y\right)   &  \in\left(  E_{H}\ast
S\right)  \left(  H\right)  \subseteq\left(  \operatorname*{Prim}H\right)
^{+}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary~\ref{Corollary813}
\textbf{(a)}}\right) \\
&  \subseteq\operatorname*{Prim}H,
\end{align*}
so that Proposition~\ref{Proposition843} (applied to $x=\left(  E_{H}\ast
S\right)  \left(  y\right)  $) yields $S\left(  \left(  E_{H}\ast S\right)
\left(  y\right)  \right)  =-\left(  E_{H}\ast S\right)  \left(  y\right)  $.
Thus,%
\[
\left(  S\circ\left(  E_{H}\ast S\right)  \right)  \left(  y\right)  =S\left(
\left(  E_{H}\ast S\right)  \left(  y\right)  \right)  =-\left(  E_{H}\ast
S\right)  \left(  y\right)  =\left(  -E_{H}\ast S\right)  \left(  y\right)  .
\]


Now, forget that we fixed $y$. We have thus proven that every $y\in H$
satisfies $\left(  S\circ\left(  E_{H}\ast S\right)  \right)  \left(
y\right)  =\left(  -E_{H}\ast S\right)  \left(  y\right)  $. In other words,
$S\circ\left(  E_{H}\ast S\right)  =-E_{H}\ast S$. Combined with
(\ref{841.pf.15}), this yields $\left(  S\ast E_{H}\right)  \circ
S=S\circ\left(  E_{H}\ast S\right)  =-E_{H}\ast S$. This proves
Theorem~\ref{Theorem841} \textbf{(b)}.
\end{proof}

\section{\label{S26}The Dynkin idempotents in commutative Hopf algebras}

This section, \S \ref{S26}, is devoted to the dual statements of the ones from
\S \ref{S25}. Here comes the dual of Theorem~\ref{Theorem801}:

\begin{theorem}
\label{Theorem901} Let $k$ be a field of characteristic $0$. Let $H$ be a
commutative graded $k$-Hopf algebra. Let $S$ be the antipode of $H$. Define a
map $E_{H}:H\rightarrow H$ according to Definition~\ref{Definition701}. Define
a map $E_{H}^{\operatorname*{inv}}:H\rightarrow H$ according to
Definition~\ref{Definition721}. Then:

\textbf{(a)} The map $E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast
S\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  \right)
=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$.\ \ \ \ \footnote{Recall that the notation $\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}$ is to be understood according
to Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of $H$.}

\textbf{(b)} The map $E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  \right)
=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$.
\end{theorem}

Note that in the case when $H$ is connected, it is easy to see that
$H_{0}=k\cdot1_{H}$, and thus this yields:

\begin{theorem}
\label{Theorem902} Let $k$ be a field of characteristic $0$. Let $H$ be a
commutative connected graded $k$-Hopf algebra. Let $S$ be the antipode of $H$.
Define a map $E_{H}:H\rightarrow H$ according to
Definition~\ref{Definition701}. Define a map $E_{H}^{\operatorname*{inv}%
}:H\rightarrow H$ according to Definition~\ref{Definition721}. Then:

\textbf{(a)} The map $E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast
S\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  \right)
=k\cdot1_{H}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$.\ \ \ \ \footnote{Recall that the notation $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ is to be
understood according to Convention~\ref{Convention100}. Hence, $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ means the
subspace $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of
$H$.}

\textbf{(b)} The map $E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  \right)
=k\cdot1_{H}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$.
\end{theorem}

The maps $E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  $ and
$E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  $ are called the
\textit{Dynkin idempotents} of $H$.

We will prove these theorems through a generalization, which requires us to
define the notion of a \textit{derivation}:

\begin{definition}
\label{Definition903} Let $k$ be a field. Let $H$ be a $k$-algebra. Let
$f:H\rightarrow H$ be a $k$-linear map. Then, $f$ is said to be a
\textit{derivation} if and only if $f\circ\mu_{H}=\mu_{H}\circ\left(
f\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes f\right)  $.
\end{definition}

Keep in mind that \textbf{a derivation is not the same as an }$\left(
\varepsilon_{H},\varepsilon_{H}\right)  $\textbf{-derivation}. (The latter has
been defined in Definitions~\ref{Definition108} and~\ref{Definition109}.) We
will, however, connect these two notions in the following results.

Definition~\ref{Definition903} could be generalized to $k$-linear maps
$f:H\rightarrow M$ with $M$ being a $\left(  H,H\right)  $-bimodule; but we
will not need this generalization and we will not even define the notion of a
$\left(  H,H\right)  $-bimodule. Let us, however, give an equivalent rewriting
of Definition~\ref{Definition903}:

\begin{definition}
\label{Definition904} Let $k$ be a field. Let $H$ be a $k$-algebra. Let
$f:H\rightarrow H$ be a $k$-linear map. Then, $f$ is said to be a
\textit{derivation} if and only if every $\left(  a,b\right)  \in H\times H$
satisfies $f\left(  ab\right)  =f\left(  a\right)  b+af\left(  b\right)  $.
\end{definition}

We will be able to use both Definitions~\ref{Definition903}
and~\ref{Definition904} in parallel as soon as we have shown the following proposition:

\begin{proposition}
\label{Proposition905} Definition~\ref{Definition903} and
Definition~\ref{Definition904} are equivalent.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition905}.]Let $k$ be a field. Let $H$ be a
$k$-algebra. Let $f:H\rightarrow H$ be a $k$-linear map.

It is well-known that two $k$-linear maps from a tensor product are equal if
and only if they are equal on each pure tensor. Applying this fact to the two
$k$-linear maps $f\circ\mu_{H}:H\otimes H\rightarrow H$ and $\mu_{H}%
\circ\left(  f\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes
f\right)  :H\otimes H\rightarrow H$, we conclude that we have the following
equivalence:%
\begin{align}
&  \ \left(  \text{the two maps }f\circ\mu_{H}\text{ and }\mu_{H}\circ\left(
f\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes f\right)  \text{
are equal}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the two maps }f\circ\mu_{H}\text{ and
}\mu_{H}\circ\left(  f\otimes\operatorname{id}_{H}+\operatorname{id}%
_{H}\otimes f\right)  \text{ are equal on each pure tensor}\right)  .
\label{905.pf.1}%
\end{align}
Since pure tensors in $H\otimes H$ are tensors of the form $a\otimes b$ with
$\left(  a,b\right)  \in H\times H$, we have the equivalence%
\begin{align}
&  \ \left(  \text{the two maps }f\circ\mu_{H}\text{ and }\mu_{H}\circ\left(
f\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes f\right)  \text{
are equal on each pure tensor}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the two maps }f\circ\mu_{H}\text{ and
}\mu_{H}\circ\left(  f\otimes\operatorname{id}_{H}+\operatorname{id}%
_{H}\otimes f\right)  \text{ are}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  \text{equal on }a\otimes
b\text{ for each }\left(  a,b\right)  \in H\times H\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  f\circ\mu_{H}\right)  \left(  a\otimes
b\right)  =\left(  \mu_{H}\circ\left(  f\otimes\operatorname{id}%
_{H}+\operatorname{id}_{H}\otimes f\right)  \right)  \left(  a\otimes
b\right)  \text{ for each }\left(  a,b\right)  \in H\times H\right)  .
\label{905.pf.3}%
\end{align}
But every $\left(  a,b\right)  \in H\times H$ satisfies%
\[
\left(  f\circ\mu_{H}\right)  \left(  a\otimes b\right)  =f\left(
\underbrace{\mu_{H}\left(  a\otimes b\right)  }_{\substack{=ab\\\text{(since
}\mu_{H}\text{ is the multiplication map)}}}\right)  =f\left(  ab\right)
\]
and%
\begin{align*}
&  \left(  \mu_{H}\circ\left(  f\otimes\operatorname{id}_{H}+\operatorname{id}%
_{H}\otimes f\right)  \right)  \left(  a\otimes b\right) \\
&  =\mu_{H}\left(  \underbrace{\left(  f\otimes\operatorname{id}%
_{H}+\operatorname{id}_{H}\otimes f\right)  \left(  a\otimes b\right)
}_{=\left(  f\otimes\operatorname{id}_{H}\right)  \left(  a\otimes b\right)
+\left(  \operatorname{id}_{H}\otimes f\right)  \left(  a\otimes b\right)
}\right) \\
&  =\mu_{H}\left(  \underbrace{\left(  f\otimes\operatorname{id}_{H}\right)
\left(  a\otimes b\right)  }_{=f\left(  a\right)  \otimes\operatorname{id}%
_{H}\left(  b\right)  }+\underbrace{\left(  \operatorname{id}_{H}\otimes
f\right)  \left(  a\otimes b\right)  }_{=\operatorname{id}_{H}\left(
a\right)  \otimes f\left(  b\right)  }\right) \\
&  =\mu_{H}\left(  f\left(  a\right)  \otimes\underbrace{\operatorname{id}%
_{H}\left(  b\right)  }_{=b}+\underbrace{\operatorname{id}_{H}\left(
a\right)  }_{=a}\otimes f\left(  b\right)  \right)  =\mu_{H}\left(  f\left(
a\right)  \otimes b+a\otimes f\left(  b\right)  \right) \\
&  =\underbrace{\mu_{H}\left(  f\left(  a\right)  \otimes b\right)
}_{\substack{=f\left(  a\right)  b\\\text{(since }\mu_{H}\text{ is
the}\\\text{multiplication map)}}}+\underbrace{\mu_{H}\left(  a\otimes
f\left(  b\right)  \right)  }_{\substack{=af\left(  b\right)  \\\text{(since
}\mu_{H}\text{ is the}\\\text{multiplication map)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{H}\text{ is }k\text{-linear}%
\right) \\
&  =f\left(  a\right)  b+af\left(  b\right)  .
\end{align*}


Now, we have the following chain of equivalences:%
\begin{align*}
&  \ \left(  \text{the map }f\text{ is a derivation in the sense of
Definition~\ref{Definition903}}\right) \\
&  \Longleftrightarrow\ \left(  f\circ\mu_{H}=\mu_{H}\circ\left(
f\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes f\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Definition~\ref{Definition903}}\right) \\
&  \Longleftrightarrow\ \left(  \text{the two maps }f\circ\mu_{H}\text{ and
}\mu_{H}\circ\left(  f\otimes\operatorname{id}_{H}+\operatorname{id}%
_{H}\otimes f\right)  \text{ are equal}\right) \\
&  \Longleftrightarrow\ \left(  \text{the two maps }f\circ\mu_{H}\text{ and
}\mu_{H}\circ\left(  f\otimes\operatorname{id}_{H}+\operatorname{id}%
_{H}\otimes f\right)  \text{ are equal on each pure tensor}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{905.pf.1})}\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  f\circ\mu_{H}\right)
\left(  a\otimes b\right)  }_{=f\left(  ab\right)  }=\underbrace{\left(
\mu_{H}\circ\left(  f\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes
f\right)  \right)  \left(  a\otimes b\right)  }_{=f\left(  a\right)
b+af\left(  b\right)  }\text{ for each }\left(  a,b\right)  \in H\times
H\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{905.pf.3})}\right) \\
&  \Longleftrightarrow\ \left(  f\left(  ab\right)  =f\left(  a\right)
b+af\left(  b\right)  \text{ for each }\left(  a,b\right)  \in H\times
H\right) \\
&  \Longleftrightarrow\ \left(  \text{the map }f\text{ is a derivation in the
sense of Definition~\ref{Definition904}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Definition~\ref{Definition904}%
}\right)  .
\end{align*}


Now, forget that we fixed $f$. We have thus proven that, for every $k$-linear
map $f:H\rightarrow H$, we have the equivalence%
\begin{align*}
&  \left(  \text{the map }f\text{ is a derivation in the sense of
Definition~\ref{Definition903}}\right) \\
&  \Longleftrightarrow\ \left(  \text{the map }f\text{ is a derivation in the
sense of Definition~\ref{Definition904}}\right)  .
\end{align*}
In other words, Definition~\ref{Definition903} and
Definition~\ref{Definition904} are equivalent. This proves
Proposition~\ref{Proposition905}.
\end{proof}

Before we prove anything more serious about derivations, let us show a
technical lemma:

\begin{lemma}
\label{Lemma906} Let $k$ be a field. Let $C$ be a $k$-bialgebra. Let $A$ be a
commutative $k$-algebra. Let $\alpha:C\rightarrow A$ and $\beta:C\rightarrow
A$ be any $k$-linear maps. Let $g:C\rightarrow A$ be a $k$-algebra
homomorphism. Let $f:C\rightarrow A$ be a $k$-linear map satisfying%
\[
f\circ\mu_{C}=\mu_{A}\circ\left(  \alpha\otimes f+f\otimes\beta\right)  .
\]
Then:

\textbf{(a)} We have%
\[
\left(  f\ast g\right)  \circ\mu_{C}=\mu_{A}\circ\left(  \left(  \alpha\ast
g\right)  \otimes\left(  f\ast g\right)  +\left(  f\ast g\right)
\otimes\left(  \beta\ast g\right)  \right)  .
\]


\textbf{(b)} We have%
\[
\left(  g\ast f\right)  \circ\mu_{C}=\mu_{A}\circ\left(  \left(  g\ast
\alpha\right)  \otimes\left(  g\ast f\right)  +\left(  g\ast f\right)
\otimes\left(  g\ast\beta\right)  \right)  .
\]

\end{lemma}

This is a dual of Lemma~\ref{Lemma804}, and unsurprisingly the proof will be analogous.

\begin{proof}
[Proof of Lemma~\ref{Lemma906}.]By the axioms of a bialgebra, $\mu
_{C}:C\otimes C\rightarrow C$ is a $k$-coalgebra homomorphism (since $C$ is a
$k$-bialgebra). Thus, $\Delta_{C}\circ\mu_{C}=\left(  \mu_{C}\otimes\mu
_{C}\right)  \circ\Delta_{C\otimes C}$.

In the following, the sign $\ast$ will denote the convolution in
$\mathcal{L}\left(  C,A\right)  $, but also the convolution in $\mathcal{L}%
\left(  C\otimes C,A\otimes A\right)  $ (which is well-defined since $C\otimes
C$ is a $k$-coalgebra and $A\otimes A$ is a $k$-algebra).

\textbf{(a)} By the definition of convolution, $f\ast g=\mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{C}$, so that%
\begin{align}
\underbrace{\left(  f\ast g\right)  }_{=\mu_{A}\circ\left(  f\otimes g\right)
\circ\Delta_{C}}\circ\mu_{C}  &  =\mu_{A}\circ\left(  f\otimes g\right)
\circ\underbrace{\Delta_{C}\circ\mu_{C}}_{=\left(  \mu_{C}\otimes\mu
_{C}\right)  \circ\Delta_{C\otimes C}}\nonumber\\
&  =\mu_{A}\circ\left(  f\otimes g\right)  \circ\left(  \mu_{C}\otimes\mu
_{C}\right)  \circ\Delta_{C\otimes C}. \label{906.pf.1}%
\end{align}


But applying (\ref{(X)(X)}) to $C\otimes C$, $C$, $A$, $C\otimes C$, $C$, $A$,
$\mu_{C}$, $f$, $\mu_{C}$, $g$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain
\[
\left(  f\circ\mu_{C}\right)  \otimes\left(  g\circ\mu_{C}\right)  =\left(
f\otimes g\right)  \circ\left(  \mu_{C}\otimes\mu_{C}\right)  ,
\]
so that%
\begin{align}
&  \left(  f\otimes g\right)  \circ\left(  \mu_{C}\otimes\mu_{C}\right)
\nonumber\\
&  =\underbrace{\left(  f\circ\mu_{C}\right)  }_{=\mu_{A}\circ\left(
\alpha\otimes f+f\otimes\beta\right)  }\otimes\underbrace{\left(  g\circ
\mu_{C}\right)  }_{\substack{=\mu_{A}\circ\left(  g\otimes g\right)
\\\text{(since }g\text{ is a }k\text{-algebra}\\\text{homomorphism)}%
}}\nonumber\\
&  =\left(  \mu_{A}\circ\left(  \alpha\otimes f+f\otimes\beta\right)  \right)
\otimes\left(  \mu_{A}\circ\left(  g\otimes g\right)  \right) \nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\underbrace{\left(  \left(
\alpha\otimes f+f\otimes\beta\right)  \otimes\left(  g\otimes g\right)
\right)  }_{\substack{=\alpha\otimes f\otimes g\otimes g+f\otimes\beta\otimes
g\otimes g\\\text{(since the tensor product of }k\text{-linear maps}\\\text{is
distributive)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}) (applied to }C\otimes C\text{, }A\otimes A\text{,
}A\text{, }C\otimes C\text{, }A\otimes A\text{, }A\text{, }\\
\alpha\otimes f+f\otimes\beta\text{, }\mu_{A}\text{, }g\otimes g\text{, }%
\mu_{A}\text{ instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{,
}V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }%
\alpha^{\prime}\text{, }\beta^{\prime}\text{)}%
\end{array}
\right) \nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\left(  \alpha\otimes f\otimes
g\otimes g+f\otimes\beta\otimes g\otimes g\right)  . \label{906.pf.4}%
\end{align}


Now, (\ref{906.pf.1}) becomes%
\begin{align*}
&  \left(  f\ast g\right)  \circ\mu_{C}\\
&  =\mu_{A}\circ\underbrace{\left(  f\otimes g\right)  \circ\left(  \mu
_{C}\otimes\mu_{C}\right)  }_{\substack{=\left(  \mu_{A}\otimes\mu_{A}\right)
\circ\left(  \alpha\otimes f\otimes g\otimes g+f\otimes\beta\otimes g\otimes
g\right)  \\\text{(by (\ref{906.pf.4}))}}}\circ\Delta_{C\otimes C}\\
&  =\underbrace{\mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)
}_{\substack{=\mu_{A}\circ\mu_{A\otimes A}\\\text{(since }\mu_{A}:A\otimes
A\rightarrow A\text{ is a }k\text{-algebra}\\\text{homomorphism (by
Lemma~\ref{Lemma512}))}}}\circ\left(  \alpha\otimes f\otimes g\otimes
g+f\otimes\beta\otimes g\otimes g\right)  \circ\Delta_{C\otimes C}\\
&  =\mu_{A}\circ\underbrace{\mu_{A\otimes A}\circ\left(  \alpha\otimes
f\otimes g\otimes g+f\otimes\beta\otimes g\otimes g\right)  }_{\substack{=\mu
_{A\otimes A}\circ\left(  \alpha\otimes f\otimes g\otimes g\right)
+\mu_{A\otimes A}\circ\left(  f\otimes\beta\otimes g\otimes g\right)
\\\text{(since composition of }k\text{-linear maps is distributive)}}%
}\circ\Delta_{C\otimes C}\\
&  =\mu_{A}\circ\underbrace{\left(  \mu_{A\otimes A}\circ\left(  \alpha\otimes
f\otimes g\otimes g\right)  +\mu_{A\otimes A}\circ\left(  f\otimes\beta\otimes
g\otimes g\right)  \right)  \circ\Delta_{C\otimes C}}_{\substack{=\mu
_{A\otimes A}\circ\left(  \alpha\otimes f\otimes g\otimes g\right)
\circ\Delta_{C\otimes C}+\mu_{A\otimes A}\circ\left(  f\otimes\beta\otimes
g\otimes g\right)  \circ\Delta_{C\otimes C}\\\text{(since composition of
}k\text{-linear maps is distributive)}}}\\
&  =\mu_{A}\circ\left(  \underbrace{\mu_{A\otimes A}\circ\left(  \alpha\otimes
f\otimes g\otimes g\right)  \circ\Delta_{C\otimes C}}_{\substack{=\left(
\alpha\otimes f\right)  \ast\left(  g\otimes g\right)  \\\text{(because the
definition of convolution yields}\\\left(  \alpha\otimes f\right)  \ast\left(
g\otimes g\right)  =\mu_{A\otimes A}\circ\left(  \alpha\otimes f\otimes
g\otimes g\right)  \circ\Delta_{C\otimes C}\text{)}}}+\underbrace{\mu
_{A\otimes A}\circ\left(  f\otimes\beta\otimes g\otimes g\right)  \circ
\Delta_{C\otimes C}}_{\substack{=\left(  f\otimes\beta\right)  \ast\left(
g\otimes g\right)  \\\text{(because the definition of convolution
yields}\\\left(  f\otimes\beta\right)  \ast\left(  g\otimes g\right)
=\mu_{A\otimes A}\circ\left(  f\otimes\beta\otimes g\otimes g\right)
\circ\Delta_{C\otimes C}\text{)}}}\right) \\
&  =\mu_{A}\circ\left(  \underbrace{\left(  \alpha\otimes f\right)
\ast\left(  g\otimes g\right)  }_{\substack{=\left(  \alpha\ast g\right)
\otimes\left(  f\ast g\right)  \\\text{(by Corollary~\ref{Corollary40},
applied to }D=C\text{, }B=A\text{,}\\p=\alpha\text{, }q=g\text{, }r=f\text{,
}s=g\text{)}}}+\underbrace{\left(  f\otimes\beta\right)  \ast\left(  g\otimes
g\right)  }_{\substack{=\left(  f\ast g\right)  \otimes\left(  \beta\ast
g\right)  \\\text{(by Corollary~\ref{Corollary40}, applied to }D=C\text{,
}B=A\text{,}\\p=f\text{, }q=g\text{, }r=\beta\text{, }s=g\text{)}}}\right) \\
&  =\mu_{A}\circ\left(  \left(  \alpha\ast g\right)  \otimes\left(  f\ast
g\right)  +\left(  f\ast g\right)  \otimes\left(  \beta\ast g\right)  \right)
.
\end{align*}
This proves Lemma~\ref{Lemma906} \textbf{(a)}.

\textbf{(b)} By the definition of convolution, $g\ast f=\mu_{A}\circ\left(
g\otimes f\right)  \circ\Delta_{C}$, so that%
\begin{align}
\underbrace{\left(  g\ast f\right)  }_{=\mu_{A}\circ\left(  g\otimes f\right)
\circ\Delta_{C}}\circ\mu_{C}  &  =\mu_{A}\circ\left(  g\otimes f\right)
\circ\underbrace{\Delta_{C}\circ\mu_{C}}_{=\left(  \mu_{C}\otimes\mu
_{C}\right)  \circ\Delta_{C\otimes C}}\nonumber\\
&  =\mu_{A}\circ\left(  g\otimes f\right)  \circ\left(  \mu_{C}\otimes\mu
_{C}\right)  \circ\Delta_{C\otimes C}. \label{906.pf.11}%
\end{align}


But applying (\ref{(X)(X)}) to $C\otimes C$, $C$, $A$, $C\otimes C$, $C$, $A$,
$\mu_{C}$, $g$, $\mu_{C}$, $f$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain
\[
\left(  g\circ\mu_{C}\right)  \otimes\left(  f\circ\mu_{C}\right)  =\left(
g\otimes f\right)  \circ\left(  \mu_{C}\otimes\mu_{C}\right)  ,
\]
so that%
\begin{align}
&  \left(  g\otimes f\right)  \circ\left(  \mu_{C}\otimes\mu_{C}\right)
\nonumber\\
&  =\underbrace{\left(  g\circ\mu_{C}\right)  }_{\substack{=\mu_{A}%
\circ\left(  g\otimes g\right)  \\\text{(since }g\text{ is a }k\text{-algebra}%
\\\text{homomorphism)}}}\otimes\underbrace{\left(  f\circ\mu_{C}\right)
}_{=\mu_{A}\circ\left(  \alpha\otimes f+f\otimes\beta\right)  }\nonumber\\
&  =\left(  \mu_{A}\circ\left(  g\otimes g\right)  \right)  \otimes\left(
\mu_{A}\circ\left(  \alpha\otimes f+f\otimes\beta\right)  \right) \nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\underbrace{\left(  \left(
g\otimes g\right)  \otimes\left(  \alpha\otimes f+f\otimes\beta\right)
\right)  }_{\substack{=g\otimes g\otimes\alpha\otimes f+g\otimes g\otimes
f\otimes\beta\\\text{(since the tensor product of }k\text{-linear
maps}\\\text{is distributive)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{(X)(X)}) (applied to }C\otimes C\text{, }A\otimes A\text{,
}A\text{, }C\otimes C\text{, }A\otimes A\text{, }A\text{,}\\
\text{ }g\otimes g\text{, }\mu_{A}\text{, }\alpha\otimes f+f\otimes
\beta\text{, }\mu_{A}\text{ instead of }U\text{, }V\text{, }W\text{,
}U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }%
\beta\text{, }\alpha^{\prime}\text{, }\beta^{\prime}\text{)}%
\end{array}
\right) \nonumber\\
&  =\left(  \mu_{A}\otimes\mu_{A}\right)  \circ\left(  g\otimes g\otimes
\alpha\otimes f+g\otimes g\otimes f\otimes\beta\right)  . \label{906.pf.14}%
\end{align}


Now, (\ref{906.pf.11}) becomes%
\begin{align*}
&  \left(  g\ast f\right)  \circ\mu_{C}\\
&  =\mu_{A}\circ\underbrace{\left(  g\otimes f\right)  \circ\left(  \mu
_{C}\otimes\mu_{C}\right)  }_{\substack{=\left(  \mu_{A}\otimes\mu_{A}\right)
\circ\left(  g\otimes g\otimes\alpha\otimes f+g\otimes g\otimes f\otimes
\beta\right)  \\\text{(by (\ref{906.pf.14}))}}}\circ\Delta_{C\otimes C}\\
&  =\underbrace{\mu_{A}\circ\left(  \mu_{A}\otimes\mu_{A}\right)
}_{\substack{=\mu_{A}\circ\mu_{A\otimes A}\\\text{(since }\mu_{A}:A\otimes
A\rightarrow A\text{ is a }k\text{-algebra}\\\text{homomorphism (by
Lemma~\ref{Lemma512}))}}} \circ\left(  g\otimes g\otimes\alpha\otimes
f+g\otimes g\otimes f\otimes\beta\right)  \circ\Delta_{C\otimes C}\\
&  =\mu_{A}\circ\underbrace{\mu_{A\otimes A}\circ\left(  g\otimes
g\otimes\alpha\otimes f+g\otimes g\otimes f\otimes\beta\right)  }%
_{\substack{=\mu_{A\otimes A}\circ\left(  g\otimes g\otimes\alpha\otimes
f\right)  +\mu_{A\otimes A}\circ\left(  g\otimes g\otimes f\otimes
\beta\right)  \\\text{(since composition of }k\text{-linear maps is
distributive)}}}\circ\Delta_{C\otimes C}\\
&  =\mu_{A}\circ\underbrace{\left(  \mu_{A\otimes A}\circ\left(  g\otimes
g\otimes\alpha\otimes f\right)  +\mu_{A\otimes A}\circ\left(  g\otimes
g\otimes f\otimes\beta\right)  \right)  \circ\Delta_{C\otimes C}%
}_{\substack{=\mu_{A\otimes A}\circ\left(  g\otimes g\otimes\alpha\otimes
f\right)  \circ\Delta_{C\otimes C}+\mu_{A\otimes A}\circ\left(  g\otimes
g\otimes f\otimes\beta\right)  \circ\Delta_{C\otimes C}\\\text{(since
composition of }k\text{-linear maps is distributive)}}}\\
&  =\mu_{A}\circ\left(  \underbrace{\mu_{A\otimes A}\circ\left(  g\otimes
g\otimes\alpha\otimes f\right)  \circ\Delta_{C\otimes C}}_{\substack{=\left(
g\otimes g\right)  \ast\left(  \alpha\otimes f\right)  \\\text{(because the
definition of convolution yields}\\\left(  g\otimes g\right)  \ast\left(
\alpha\otimes f\right)  =\mu_{A\otimes A}\circ\left(  g\otimes g\otimes
\alpha\otimes f\right)  \circ\Delta_{C\otimes C}\text{)}}}+\underbrace{\mu
_{A\otimes A}\circ\left(  g\otimes g\otimes f\otimes\beta\right)  \circ
\Delta_{C\otimes C}}_{\substack{=\left(  g\otimes g\right)  \ast\left(
f\otimes\beta\right)  \\\text{(because the definition of convolution
yields}\\\left(  g\otimes g\right)  \ast\left(  f\otimes\beta\right)
=\mu_{A\otimes A}\circ\left(  g\otimes g\otimes f\otimes\beta\right)
\circ\Delta_{C\otimes C}\text{)}}}\right) \\
&  =\mu_{A}\circ\left(  \underbrace{\left(  g\otimes g\right)  \ast\left(
\alpha\otimes f\right)  }_{\substack{=\left(  g\ast\alpha\right)
\otimes\left(  g\ast f\right)  \\\text{(by Corollary~\ref{Corollary40},
applied to }D=C\text{, }B=A\text{,}\\p=g\text{, }q=\alpha\text{, }r=g\text{,
}s=f\text{)}}}+\underbrace{\left(  g\otimes g\right)  \ast\left(
f\otimes\beta\right)  }_{\substack{=\left(  g\ast f\right)  \otimes\left(
g\ast\beta\right)  \\\text{(by Corollary~\ref{Corollary40}, applied to
}D=C\text{, }B=A\text{,}\\p=g\text{, }q=f\text{, }r=g\text{, }s=\beta\text{)}%
}}\right) \\
&  =\mu_{A}\circ\left(  \left(  g\ast\alpha\right)  \otimes\left(  g\ast
f\right)  +\left(  g\ast f\right)  \otimes\left(  g\ast\beta\right)  \right)
.
\end{align*}
This proves Lemma~\ref{Lemma906} \textbf{(b)}.
\end{proof}

Now, let us show the reason why we proved Lemma~\ref{Lemma906}:

\begin{theorem}
\label{Theorem907} Let $k$ be a field. Let $H$ be a commutative $k$-bialgebra.
Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two $k$-algebra homomorphisms
satisfying $P\ast\operatorname{id}_{H}\ast Q=e_{H,H}$. Let $K:H\rightarrow H$
be a derivation. Then, $P\ast K\ast Q:H\rightarrow H$ is an $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivation. Here, the map $e_{H,H}$
is defined to be the map $\eta_{H}\circ\varepsilon_{H}:H\rightarrow H$ (this
definition of the map $e_{H,H}$ is identical with the definition of the map
$e_{H,A}$ in Definition~\ref{def.eHA}).
\end{theorem}

Theorem~\ref{Theorem907} is the dual of Theorem~\ref{Theorem805}.

\begin{proof}
[Proof of Theorem~\ref{Theorem907}.]Since $K$ is a derivation, we have
$K\circ\mu_{H}=\mu_{H}\circ\left(  K\otimes\operatorname{id}_{H}%
+\operatorname{id}_{H}\otimes K\right)  $ (because
Definition~\ref{Definition903} yields that $K$ is a derivation if and only if
$K\circ\mu_{H}=\mu_{H}\circ\left(  K\otimes\operatorname{id}_{H}%
+\operatorname{id}_{H}\otimes K\right)  $). Thus,%
\[
K\circ\mu_{H}=\mu_{H}\circ\underbrace{\left(  K\otimes\operatorname{id}%
_{H}+\operatorname{id}_{H}\otimes K\right)  }_{=\operatorname{id}_{H}\otimes
K+K\otimes\operatorname{id}_{H}}=\mu_{H}\circ\left(  \operatorname{id}%
_{H}\otimes K+K\otimes\operatorname{id}_{H}\right)  .
\]
Hence, Lemma~\ref{Lemma906} \textbf{(b)} (applied to $C=H$, $A=H$,
$\alpha=\operatorname{id}_{H}$, $\beta=\operatorname{id}_{H}$, $f=K$ and
$g=P$) yields%
\[
\left(  P\ast K\right)  \circ\mu_{H}=\mu_{H}\circ\left(  \left(
P\ast\operatorname{id}_{H}\right)  \otimes\left(  P\ast K\right)  +\left(
P\ast K\right)  \otimes\left(  P\ast\operatorname{id}_{H}\right)  \right)  .
\]
Thus, Lemma~\ref{Lemma906} \textbf{(a)} (applied to $C=H$, $A=H$,
$\alpha=P\ast\operatorname{id}_{H}$, $\beta=P\ast\operatorname{id}_{H}$,
$f=P\ast K$ and $g=Q$) yields%
\begin{align*}
\left(  P\ast K\ast Q\right)  \circ\mu_{H}  &  =\mu_{H}\circ\left(
\underbrace{\left(  P\ast\operatorname{id}_{H}\ast Q\right)  }_{=e_{H,H}%
}\otimes\left(  P\ast K\ast Q\right)  +\left(  P\ast K\ast Q\right)
\otimes\underbrace{\left(  P\ast\operatorname{id}_{H}\ast Q\right)
}_{=e_{H,H}}\right) \\
&  =\mu_{H}\circ\underbrace{\left(  e_{H,H}\otimes\left(  P\ast K\ast
Q\right)  +\left(  P\ast K\ast Q\right)  \otimes e_{H,H}\right)  }_{=\left(
P\ast K\ast Q\right)  \otimes e_{H,H}+e_{H,H}\otimes\left(  P\ast K\ast
Q\right)  }\\
&  =\mu_{H}\circ\left(  \left(  P\ast K\ast Q\right)  \otimes e_{H,H}%
+e_{H,H}\otimes\left(  P\ast K\ast Q\right)  \right)  .
\end{align*}
Thus, $P\ast K\ast Q$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation (because by Definition~\ref{Definition108}, the map $P\ast K\ast
Q$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and
only if \newline$\left(  P\ast K\ast Q\right)  \circ\mu_{H}=\mu_{H}%
\circ\left(  \left(  P\ast K\ast Q\right)  \otimes e_{H,H}+e_{H,H}%
\otimes\left(  P\ast K\ast Q\right)  \right)  $). This proves
Theorem~\ref{Theorem907}.
\end{proof}

The analogue of Corollary~\ref{Corollary806} is the following fact:

\begin{corollary}
\label{Corollary908} Let $k$ be a field. Let $H$ be a commutative $k$-Hopf
algebra. Let $S$ be the antipode of $H$. Let $K:H\rightarrow H$ be a derivation.

\textbf{(a)} Then, $S\ast K$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation.

\textbf{(b)} Then, $K\ast S$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary908}.]By Definition~\ref{Definition602}, we
have $H^{\operatorname*{op}}=\left(  H,\mu_{H}\circ\tau_{H,H},\eta_{H}\right)
$. Since $H$ is commutative, we have $\mu_{H}\circ\tau_{H,H}=\mu_{H}$. Thus,
$H^{\operatorname*{op}}=\left(  H,\underbrace{\mu_{H}\circ\tau_{H,H}}%
_{=\mu_{H}},\eta_{H}\right)  =\left(  H,\mu_{H},\eta_{H}\right)  =H$.

Proposition~\ref{Proposition603} yields that the antipode of $H$ is a
$k$-algebra homomorphism from $H$ to $H^{\operatorname*{op}}$. Since the
antipode of $H$ is the map $S$, whereas $H^{\operatorname*{op}}$ is $H$, this
rewrites as follows: The map $S$ is a $k$-algebra homomorphism from $H$ to $H$.

Corollary~\ref{Corollary136} (applied to $n=0$, $A=H$ and $f=\operatorname{id}%
_{H}$) yields that $\operatorname{id}_{H}^{\ast0}$ is a $k$-algebra
homomorphism. Since $\operatorname{id}_{H}^{\ast0}=e_{H,H}$, this shows that
$e_{H,H}$ is a $k$-algebra homomorphism (where $e_{H,H}$ is defined as
according to Definition~\ref{def.eHA}).

The antipode of a Hopf algebra is defined as the $\ast$-inverse of its
identity map. Thus, the antipode of $H$ is the map $\operatorname{id}%
_{H}^{\ast\left(  -1\right)  }$. Since the antipode of $H$ is $S$, this yields
that $S$ is the map $\operatorname{id}_{H}^{\ast\left(  -1\right)  }$. In
other words, $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$. Hence,
$S\ast\operatorname{id}_{H}\ast e_{H,H}=\underbrace{\operatorname{id}%
_{H}^{\ast\left(  -1\right)  }\ast\operatorname{id}_{H}}_{=e_{H,H}}\ast
e_{H,H}=e_{H,H}$. Thus, Theorem~\ref{Theorem907} (applied to $P=S$ and
$Q=e_{H,H}$) yields that $S\ast K\ast e_{H,H}$ is an $\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivation. In other words, $S\ast K$ is an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation (since $S\ast
K\ast e_{H,H}=S\ast K$). This proves Corollary~\ref{Corollary908} \textbf{(a)}.

Since $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$, we have
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}\ast
\underbrace{\operatorname{id}_{H}\ast\operatorname{id}_{H}^{\ast\left(
-1\right)  }}_{=e_{H,H}}=e_{H,H}$. Thus, Theorem~\ref{Theorem907} (applied to
$P=e_{H,H}$ and $Q=S$) yields that $e_{H,H}\ast K\ast S$ is an $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivation. In other words, $K\ast
S$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation (since
$e_{H,H}\ast K\ast S=K\ast S$). This proves Corollary~\ref{Corollary908}
\textbf{(b)}.
\end{proof}

Next, the dual of Proposition~\ref{Proposition807}:

\begin{proposition}
\label{Proposition909} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let
$P:H\rightarrow H$, $Q:H\rightarrow H$ and $K:H\rightarrow H$ be three
$k$-linear maps such that $\varepsilon_{H}\circ P=\varepsilon_{H}$,
$\varepsilon_{H}\circ Q=\varepsilon_{H}$ and $\varepsilon_{H}\circ K=0$. Then,
every $x\in H$ satisfies $\left(  P\ast K\ast Q\right)  \left(  x\right)
-K\left(  x\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}$. \ \ \ \ \footnote{Recall that the notation
$\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ is
to be understood according to Convention~\ref{Convention100}. Hence, $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ means the
subspace $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of
$H$.}
\end{proposition}

In order to prove this, we invoke a simple fact:

\begin{lemma}
\label{Lemma910} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let $A$ be a
$k$-algebra. Let $f:C\rightarrow A$ and $g:C\rightarrow A$ be two $k$-linear
maps. Then, $\left(  f\ast g\right)  \left(  C\right)  \subseteq f\left(
C\right)  \cdot g\left(  C\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma910}.]By the definition of convolution, $f\ast
g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}$. Thus,%
\begin{align*}
\left(  f\ast g\right)  \left(  C\right)   &  =\left(  \mu_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{C}\right)  \left(  C\right)  =\left(  \mu
_{A}\circ\left(  f\otimes g\right)  \right)  \underbrace{\left(  \Delta
_{C}\left(  C\right)  \right)  }_{\subseteq C\otimes C}\subseteq\left(
\mu_{A}\circ\left(  f\otimes g\right)  \right)  \left(  C\otimes C\right) \\
&  =\mu_{A}\underbrace{\left(  \left(  f\otimes g\right)  \left(  C\otimes
C\right)  \right)  }_{\substack{=f\left(  C\right)  \otimes g\left(  C\right)
\\\text{(where we consider }f\left(  C\right)  \otimes g\left(  C\right)
\text{ as a}\\\text{vector subspace of }A\otimes A\text{ by tensoring
the}\\\text{inclusion maps }f\left(  C\right)  \rightarrow A\text{ and
}g\left(  C\right)  \rightarrow A\text{)}}}=\mu_{A}\left(  f\left(  C\right)
\otimes g\left(  C\right)  \right)  =f\left(  C\right)  \cdot g\left(
C\right)
\end{align*}
(by (\ref{104.UV}), applied to $U=f\left(  C\right)  $ and $V=g\left(
C\right)  $). This proves Lemma~\ref{Lemma910}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition909}.]Let $x\in H$. Define $e_{H,H}$
according to Definition~\ref{def.eHA}.

Notice that%
\begin{equation}
\left(  P-e_{H,H}\right)  \left(  H\right)  \subseteq\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \label{909.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{909.pf.1}):} Let $x\in H$. Then,
$e_{H,H}=\eta_{H}\circ\varepsilon_{H}$ (by the definition of $e_{H,H}$), so
that $e_{H,H}\left(  x\right)  =\left(  \eta_{H}\circ\varepsilon_{H}\right)
\left(  x\right)  =\eta_{H}\left(  \varepsilon_{H}\left(  x\right)  \right)
=\varepsilon_{H}\left(  x\right)  \cdot1_{H}$ (by the definition of $\eta_{H}%
$). Thus,
\begin{align*}
\varepsilon_{H}\left(  e_{H,H}\left(  x\right)  \right)   &  =\varepsilon
_{H}\left(  \varepsilon_{H}\left(  x\right)  \cdot1_{H}\right)  =\varepsilon
_{H}\left(  x\right)  \cdot\underbrace{\varepsilon_{H}\left(  1_{H}\right)
}_{\substack{=1\\\text{(by the axioms of}\\\text{a bialgebra, since
}H\\\text{is a }k\text{-bialgebra)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\varepsilon_{H}\text{ is }k\text{-linear}\right) \\
&  =\varepsilon_{H}\left(  x\right)  .
\end{align*}
On the other hand, $\varepsilon_{H}\left(  P\left(  x\right)  \right)
=\underbrace{\left(  \varepsilon_{H}\circ P\right)  }_{=\varepsilon_{H}%
}\left(  x\right)  =\varepsilon_{H}\left(  x\right)  $. Now,%
\[
\varepsilon_{H}\underbrace{\left(  \left(  P-e_{H,H}\right)  \left(  x\right)
\right)  }_{=P\left(  x\right)  -e_{H,H}\left(  x\right)  }=\varepsilon
_{H}\left(  P\left(  x\right)  -e_{H,H}\left(  x\right)  \right)
=\underbrace{\varepsilon_{H}\left(  P\left(  x\right)  \right)  }%
_{=\varepsilon_{H}\left(  x\right)  }-\underbrace{\varepsilon_{H}\left(
e_{H,H}\left(  x\right)  \right)  }_{=\varepsilon_{H}\left(  x\right)
}=\varepsilon_{H}\left(  x\right)  -\varepsilon_{H}\left(  x\right)  =0,
\]
so that $\left(  P-e_{H,H}\right)  \left(  x\right)  \in\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $.
\par
Now, forget that we fixed $x$. We thus have proven that every $x\in H$
satisfies $\left(  P-e_{H,H}\right)  \left(  x\right)  \in\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $. In other words, $\left(  P-e_{H,H}\right)
\left(  H\right)  \subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)
$. This proves (\ref{909.pf.1}).} and similarly%
\begin{equation}
\left(  Q-e_{H,H}\right)  \left(  H\right)  \subseteq\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \label{909.pf.2}%
\end{equation}


Also,%
\begin{equation}
K\left(  H\right)  \subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)
\label{909.pf.3}%
\end{equation}
(because every $x\in H$ satisfies $K\left(  x\right)  \in\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $ (since $\varepsilon_{H}\left(  K\left(
x\right)  \right)  =\underbrace{\left(  \varepsilon_{H}\circ K\right)  }%
_{=0}\left(  x\right)  =0\left(  x\right)  =0$)).

Since $H$ is a $k$-bialgebra, the map $\varepsilon_{H}$ is a $k$-algebra
homomorphism (by the axioms of a bialgebra), so that $\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $ is an ideal of $H$. Thus,
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \cdot H\subseteq
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ and $H\cdot
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \subseteq
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $.

Applying Lemma~\ref{Lemma910} to $C=H$, $A=H$, $f=P-e_{H,H}$ and $g=K\ast Q$,
we obtain%
\begin{align}
\left(  \left(  P-e_{H,H}\right)  \ast K\ast Q\right)  \left(  H\right)   &
\subseteq\underbrace{\left(  P-e_{H,H}\right)  \left(  H\right)
}_{\substack{\subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)
\\\text{(by (\ref{909.pf.1}))}}}\cdot\underbrace{\left(  K\ast Q\right)
\left(  H\right)  }_{\substack{\subseteq K\left(  H\right)  \cdot Q\left(
H\right)  \\\text{(by Lemma~\ref{Lemma910}, applied to}\\C=H\text{,
}A=H\text{, }f=K\text{ and }g=Q\text{)}}}\nonumber\\
&  \subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \cdot
\underbrace{K\left(  H\right)  }_{\substack{\subseteq\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \\\text{(by (\ref{909.pf.3}))}}}\cdot
\underbrace{Q\left(  H\right)  }_{\subseteq H}\nonumber\\
&  \subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \cdot
\underbrace{\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \cdot
H}_{\subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)  }%
\subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \cdot
\operatorname*{Ker}\left(  \varepsilon_{H}\right) \nonumber\\
&  =\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}.
\label{909.pf.4}%
\end{align}


Applying Lemma~\ref{Lemma910} to $C=H$, $A=H$, $f=K$ and $g=Q-e_{H,H}$, we
obtain%
\begin{align}
\left(  K\ast\left(  Q-e_{H,H}\right)  \right)  \left(  H\right)   &
\subseteq\underbrace{K\left(  H\right)  }_{\substack{\subseteq
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \\\text{(by (\ref{909.pf.3}%
))}}}\cdot\underbrace{\left(  Q-e_{H,H}\right)  \left(  H\right)
}_{\substack{\subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)
\\\text{(by (\ref{909.pf.2}))}}}\nonumber\\
&  \subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \cdot
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  =\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}.
\label{909.pf.5}%
\end{align}


Since
\begin{align}
&  \underbrace{\left(  P-e_{H,H}\right)  \ast K\ast Q}_{=P\ast K\ast
Q-e_{H,H}\ast K\ast Q}+\underbrace{K\ast\left(  Q-e_{H,H}\right)  }_{=K\ast
Q-K\ast e_{H,H}}\nonumber\\
&  =P\ast K\ast Q-\underbrace{e_{H,H}\ast K\ast Q}_{=K\ast Q}+K\ast
Q-\underbrace{K\ast e_{H,H}}_{=K}\nonumber\\
&  =P\ast K\ast Q-K\ast Q+K\ast Q-K=P\ast K\ast Q-K, \label{909.pf.6}%
\end{align}
we have%
\begin{align*}
&  \left(  P\ast K\ast Q\right)  \left(  x\right)  -K\left(  x\right) \\
&  =\underbrace{\left(  P\ast K\ast Q-K\right)  }_{=\left(  P-e_{H,H}\right)
\ast K\ast Q+K\ast\left(  Q-e_{H,H}\right)  }\left(  x\right)  =\left(
\left(  P-e_{H,H}\right)  \ast K\ast Q+K\ast\left(  Q-e_{H,H}\right)  \right)
\left(  x\right) \\
&  =\left(  \left(  P-e_{H,H}\right)  \ast K\ast Q\right)  \left(
\underbrace{x}_{\in H}\right)  +\left(  K\ast\left(  Q-e_{H,H}\right)
\right)  \left(  \underbrace{x}_{\in H}\right) \\
&  \in\underbrace{\left(  \left(  P-e_{H,H}\right)  \ast K\ast Q\right)
\left(  H\right)  }_{\substack{\subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\\\text{(by (\ref{909.pf.4}))}%
}}+\underbrace{\left(  K\ast\left(  Q-e_{H,H}\right)  \right)  \left(
H\right)  }_{\substack{\subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\\\text{(by (\ref{909.pf.5}))}}}\\
&  \subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\text{ is a
}k\text{-vector space}\right)  .
\end{align*}
This proves Proposition~\ref{Proposition909}.
\end{proof}

Now, let us study a particular derivation that any graded algebra has: the
Euler operator. We state the dual of Proposition~\ref{Proposition811}:

\begin{proposition}
\label{Proposition911} Let $k$ be a field. Let $H$ be a graded $k$-algebra.
Let $E_{H}$ be defined according to Definition~\ref{Definition701}. Then,
$E_{H}:H\rightarrow H$ is a derivation.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition911}.]Since $H$ is a graded $k$-algebra,
the map $\mu_{H}:H\otimes H\rightarrow H$ is graded. Thus,
Proposition~\ref{Proposition704} (applied to $V=H\otimes H$, $W=H$ and
$f=\mu_{H}$) yields $\mu_{H}\circ E_{H\otimes H}=E_{H}\circ\mu_{H}$, so that%
\[
E_{H}\circ\mu_{H}=\mu_{H}\circ\underbrace{E_{H\otimes H}}_{\substack{=E_{H}%
\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes E_{H}\\\text{(by
Proposition~\ref{Proposition703}, applied to }V=H\text{ and }W=H\text{)}}%
}=\mu_{H}\circ\left(  E_{H}\otimes\operatorname{id}_{H}+\operatorname{id}%
_{H}\otimes E_{H}\right)  .
\]
Thus, $E_{H}$ is a derivation (because Definition~\ref{Definition903} yields
that $E_{H}$ is a derivation if and only if $E_{H}\circ\mu_{H}=\mu_{H}%
\circ\left(  E_{H}\otimes\operatorname{id}_{H}+\operatorname{id}_{H}\otimes
E_{H}\right)  $). Proposition~\ref{Proposition911} is proven.
\end{proof}

Now, let us come as close as possible to Theorem~\ref{Theorem901} without
requiring $k$ to be of characteristic $0$:

\begin{theorem}
\label{Theorem912} Let $k$ be a field. Let $H$ be a commutative graded
$k$-bialgebra. Let $E_{H}$ be defined according to
Definition~\ref{Definition701}.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-algebra
homomorphisms satisfying $\varepsilon_{H}\circ P=\varepsilon_{H}$,
$\varepsilon_{H}\circ Q=\varepsilon_{H}$ and $P\ast\operatorname{id}_{H}\ast
Q=e_{H,H}$. Here, the map $e_{H,H}$ is defined to be the map $\eta_{H}%
\circ\varepsilon_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is
identical with the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

\textbf{(a)} Then, $\left(  P\ast E_{H}\ast Q\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
\ \ \ \ \footnote{Recall that the notation $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ is to be understood according to
Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of $H$.}

\textbf{(b)} Besides, every $x\in H$ satisfies $\left(  P\ast E_{H}\ast
Q\right)  \left(  x\right)  -E_{H}\left(  x\right)  \in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$.

\textbf{(c)} We have $\left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast
E_{H}\ast Q\right)  =E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  =\left(
P\ast E_{H}\ast Q\right)  \circ E_{H}$.
\end{theorem}

This Theorem~\ref{Theorem912}, of course, is the dual of
Theorem~\ref{Theorem812}, although not precisely (part \textbf{(b)} of
Theorem~\ref{Theorem912} is slightly stronger than the dual of part
\textbf{(b)} of Theorem~\ref{Theorem812}, but the difference is negligible).

\begin{proof}
[Proof of Theorem~\ref{Theorem912}.]Since $H$ is a $k$-bialgebra, the map
$\varepsilon_{H}:H\rightarrow k$ is a $k$-algebra homomorphism (by the axioms
of a bialgebra).

By Proposition~\ref{Proposition911}, the map $E_{H}$ is a derivation. Thus,
Theorem~\ref{Theorem907} (applied to $K=E_{H}$) yields that $P\ast E_{H}\ast
Q:H\rightarrow H$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation. Thus,
\begin{equation}
\left(  P\ast E_{H}\ast Q\right)  \left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0 \label{912.pf.1}%
\end{equation}
(because Theorem~\ref{Theorem115} (applied to $A=H$ and $f=P\ast E_{H}\ast Q$)
yields that $P\ast E_{H}\ast Q$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation if and only if $\left(  P\ast E_{H}\ast Q\right)
\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  =0$). Thus,%
\[
\left(  P\ast E_{H}\ast Q\right)  \underbrace{\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)
}_{\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}}\subseteq\left(  P\ast E_{H}\ast Q\right)  \left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  =0
\]
(by (\ref{912.pf.1})), so that%
\begin{equation}
\left(  P\ast E_{H}\ast Q\right)  \left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\right)  =0. \label{912.pf.2}%
\end{equation}
On the other hand, just as in the proof of Theorem~\ref{Theorem812}, we can
prove that (\ref{812.pf.2}) holds. In other words,%

\begin{equation}
\left(  P\ast E_{H}\ast Q\right)  \left(  H_{0}\right)  =0. \label{912.pf.3}%
\end{equation}


Now, since $P\ast E_{H}\ast Q$ is $k$-linear, we have%
\begin{align*}
\left(  P\ast E_{H}\ast Q\right)  \left(  H_{0}+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)   &  =\underbrace{\left(
P\ast E_{H}\ast Q\right)  \left(  H_{0}\right)  }_{\substack{=0 \\\text{(by
(\ref{912.pf.3}))}}}+\underbrace{\left(  P\ast E_{H}\ast Q\right)  \left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\right)  }_{\substack{=0 \\\text{(by (\ref{912.pf.2}))}}}\\
&  =0+0=0.
\end{align*}
This proves Theorem~\ref{Theorem912} \textbf{(a)}.

Recall that the grading on the $k$-vector space $k$ satisfies $k_{0}=k$. Thus,
$E_{k}=0$\ \ \ \ \footnote{\textit{Proof.} Every $v\in k$ satisfies
$E_{k}\left(  v\right)  =0$ (since $v\in k=k_{0}$, and thus
Proposition~\ref{Proposition702} \textbf{(a)} (applied to $V=k$ and $n=0$)
yields $E_{k}\left(  v\right)  =0v=0$). Thus, $E_{k}=0$, qed.}. Since
$\varepsilon_{H}$ is a graded map (because $H$ is a graded $k$-coalgebra), we
have $\varepsilon_{H}\circ E_{H}=E_{k}\circ\varepsilon_{H}$ (by
Proposition~\ref{Proposition704}, applied to $V=H$, $W=k$ and $f=\varepsilon
_{H}$). Thus, $\varepsilon_{H}\circ E_{H}=\underbrace{E_{k}}_{=0}%
\circ\varepsilon_{H}=0\circ\varepsilon_{H}=0$. Combined with the fact that the
map $E_{H}$ is graded (by Proposition~\ref{Proposition702} \textbf{(b)},
applied to $V=H$), this shows that we can apply
Proposition~\ref{Proposition909} to $K=E_{H}$. As a result, we conclude that
every $x\in H$ satisfies $\left(  P\ast E_{H}\ast Q\right)  \left(  x\right)
-E_{H}\left(  x\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}$. This proves Theorem~\ref{Theorem912} \textbf{(b)}.

\textbf{(c)} Let $x\in H$. Then, Theorem~\ref{Theorem912} \textbf{(b)} yields%
\[
\mathbf{\ }\left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  -E_{H}\left(
x\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}\subseteq H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2},
\]
so that%
\[
\left(  P\ast E_{H}\ast Q\right)  \left(  \left(  P\ast E_{H}\ast Q\right)
\left(  x\right)  -E_{H}\left(  x\right)  \right)  \in\left(  P\ast E_{H}\ast
Q\right)  \left(  H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)  =0
\]
(by Theorem~\ref{Theorem912} \textbf{(a)}), thus $\left(  P\ast E_{H}\ast
Q\right)  \left(  \left(  P\ast E_{H}\ast Q\right)  \left(  x\right)
-E_{H}\left(  x\right)  \right)  =0$. Thus,%
\begin{align*}
0  &  =\left(  P\ast E_{H}\ast Q\right)  \left(  \left(  P\ast E_{H}\ast
Q\right)  \left(  x\right)  -E_{H}\left(  x\right)  \right) \\
&  =\underbrace{\left(  P\ast E_{H}\ast Q\right)  \left(  \left(  P\ast
E_{H}\ast Q\right)  \left(  x\right)  \right)  }_{=\left(  \left(  P\ast
E_{H}\ast Q\right)  \circ\left(  P\ast E_{H}\ast Q\right)  \right)  \left(
x\right)  }-\underbrace{\left(  P\ast E_{H}\ast Q\right)  \left(  E_{H}\left(
x\right)  \right)  }_{=\left(  \left(  P\ast E_{H}\ast Q\right)  \circ
E_{H}\right)  \left(  x\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}P\ast E_{H}\ast Q\text{ is }k\text{-linear}\right) \\
&  =\left(  \left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast E_{H}\ast
Q\right)  \right)  \left(  x\right)  -\left(  \left(  P\ast E_{H}\ast
Q\right)  \circ E_{H}\right)  \left(  x\right)  ,
\end{align*}
so that $\left(  \left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast
E_{H}\ast Q\right)  \right)  \left(  x\right)  =\left(  \left(  P\ast
E_{H}\ast Q\right)  \circ E_{H}\right)  \left(  x\right)  $.

Now forget that we fixed $x$. We thus have proven that every $x\in H$
satisfies $\left(  \left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast
E_{H}\ast Q\right)  \right)  \left(  x\right)  =\left(  \left(  P\ast
E_{H}\ast Q\right)  \circ E_{H}\right)  \left(  x\right)  $. In other words,
\begin{equation}
\left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast E_{H}\ast Q\right)
=\left(  P\ast E_{H}\ast Q\right)  \circ E_{H}. \label{912.pf.6}%
\end{equation}


Since $E_{H}$ is graded (by Proposition~\ref{Proposition702} \textbf{(b)},
applied to $V=H$) and $Q$ is graded, we conclude (by
Proposition~\ref{Proposition224} \textbf{(a)}, applied to $C=H$, $A=H$,
$f=E_{H}$ and $g=Q$) that $E_{H}\ast Q$ is graded.

Since $P$ is graded and $E_{H}\ast Q$ is graded, we conclude (by
Proposition~\ref{Proposition224} \textbf{(a)}, applied to $C=H$, $A=H$, $f=P$
and $g=E_{H}\ast Q$) that $P\ast E_{H}\ast Q$ is graded. Thus, $\left(  P\ast
E_{H}\ast Q\right)  \circ E_{H}=E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  $
(by Proposition~\ref{Proposition704}, applied to $V=H$, $W=H$ and $f=P\ast
E_{H}\ast Q$). Combined with (\ref{912.pf.6}), this yields%
\[
\left(  P\ast E_{H}\ast Q\right)  \circ\left(  P\ast E_{H}\ast Q\right)
=E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  =\left(  P\ast E_{H}\ast
Q\right)  \circ E_{H}.
\]
This proves Theorem~\ref{Theorem912} \textbf{(c)}.
\end{proof}

The dual of Corollary~\ref{Corollary813} (not the exact dual, though, but an
almost equivalent statement) takes the following form:

\begin{corollary}
\label{Corollary913} Let $k$ be a field. Let $H$ be a commutative graded
$k$-Hopf algebra. Let $S$ be the antipode of $H$. Let $E_{H}$ be defined
according to Definition~\ref{Definition701}.

\textbf{(a)} Then, $\left(  E_{H}\ast S\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
\ \ \ \ \footnote{Recall that the notation $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ is to be understood according to
Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of $H$.}

\textbf{(b)} Every $x\in H$ satisfies $\left(  E_{H}\ast S\right)  \left(
x\right)  -E_{H}\left(  x\right)  \in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$.

\textbf{(c)} We have $\left(  E_{H}\ast S\right)  \circ\left(  E_{H}\ast
S\right)  =E_{H}\circ\left(  E_{H}\ast S\right)  =\left(  E_{H}\ast S\right)
\circ E_{H}$.

\textbf{(d)} Also, $\left(  S\ast E_{H}\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.

\textbf{(e)} Every $x\in H$ satisfies $\left(  S\ast E_{H}\right)  \left(
x\right)  -E_{H}\left(  x\right)  \in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$.

\textbf{(f)} We have $\left(  S\ast E_{H}\right)  \circ\left(  S\ast
E_{H}\right)  =E_{H}\circ\left(  S\ast E_{H}\right)  =\left(  S\ast
E_{H}\right)  \circ E_{H}$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary913}.]Just as in the proof of
Corollary~\ref{Corollary908}, we can prove the following facts:

\begin{itemize}
\item The map $S$ is a $k$-algebra homomorphism from $H$ to $H$.

\item The map $e_{H,H}$ is a $k$-algebra homomorphism (where $e_{H,H}$ is
defined as according to Definition~\ref{def.eHA}).

\item We have $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$.

\item We have $S\ast\operatorname{id}_{H}\ast e_{H,H}=e_{H,H}$ and
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}$.
\end{itemize}

Notice that $e_{H,H}=\eta_{H}\circ\varepsilon_{H}$ (by the definition of
$e_{H,H}$) and $\varepsilon_{H}\circ\eta_{H}=\operatorname{id}_{k}$ (by the
axioms of a bialgebra, since $H$ is a $k$-bialgebra), so that $\varepsilon
_{H}\circ\underbrace{e_{H,H}}_{=\varepsilon_{H}\circ\eta_{H}}%
=\underbrace{\varepsilon_{H}\circ\eta_{H}}_{=\operatorname{id}_{k}}%
\circ\varepsilon_{H}=\varepsilon_{H}$. Also, Theorem~\ref{Theorem842}
\textbf{(a)} yields $\varepsilon_{H}\circ S=\varepsilon_{H}$.

Also, the map $e_{H,H}$ is graded (by Proposition~\ref{Proposition224}
\textbf{(b)}, applied to $C=H$ and $A=H$), and the map $S$ is graded (since
$S$ is the antipode of $H$, while $H$ is a graded $k$-Hopf algebra).

Thus, we can apply Theorem~\ref{Theorem912} to $P=e_{H,H}$ and $Q=S$.

Applying Theorem~\ref{Theorem912} \textbf{(a)} to $P=e_{H,H}$ and $Q=S$, we
obtain \newline$\left(  e_{H,H}\ast E_{H}\ast S\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
Since $e_{H,H}\ast E_{H}\ast S=E_{H}\ast S$, this rewrites as $\left(
E_{H}\ast S\right)  \left(  H_{0}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\right)  =0$. This proves
Corollary~\ref{Corollary913} \textbf{(a)}.

Applying Theorem~\ref{Theorem912} \textbf{(b)} to $P=e_{H,H}$ and $Q=S$, we
conclude that every $x\in H$ satisfies $\left(  e_{H,H}\ast E_{H}\ast
S\right)  \left(  x\right)  -E_{H}\left(  x\right)  \in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$. Since
$e_{H,H}\ast E_{H}\ast S=E_{H}\ast S$, this rewrites as follows: Every $x\in
H$ satisfies $\left(  E_{H}\ast S\right)  \left(  x\right)  -E_{H}\left(
x\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$. This proves Corollary~\ref{Corollary913} \textbf{(b)}.

Applying Theorem~\ref{Theorem912} \textbf{(c)} to $P=e_{H,H}$ and $Q=S$, we
obtain $\left(  e_{H,H}\ast E_{H}\ast S\right)  \circ\left(  e_{H,H}\ast
E_{H}\ast S\right)  =E_{H}\circ\left(  e_{H,H}\ast E_{H}\ast S\right)
=\left(  e_{H,H}\ast E_{H}\ast S\right)  \circ E_{H}$. Since $e_{H,H}\ast
E_{H}\ast S=E_{H}\ast S$, this rewrites as $\left(  E_{H}\ast S\right)
\circ\left(  E_{H}\ast S\right)  =E_{H}\circ\left(  E_{H}\ast S\right)
=\left(  E_{H}\ast S\right)  \circ E_{H}$. This proves
Corollary~\ref{Corollary913} \textbf{(c)}.

But we can also apply Theorem~\ref{Theorem912} to $P=S$ and $Q=e_{H,H}$.

Applying Theorem~\ref{Theorem912} \textbf{(a)} to $P=S$ and $Q=e_{H,H}$, we
obtain \newline$\left(  S\ast E_{H}\ast e_{H,H}\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
Since $S\ast E_{H}\ast e_{H,H}=S\ast E_{H}$, this rewrites as $\left(  S\ast
E_{H}\right)  \left(  H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)  =0$. This proves
Corollary~\ref{Corollary913} \textbf{(d)}.

Applying Theorem~\ref{Theorem912} \textbf{(b)} to $P=S$ and $Q=e_{H,H}$, we
conclude that every $x\in H$ satisfies $\left(  S\ast E_{H}\ast e_{H,H}%
\right)  \left(  x\right)  -E_{H}\left(  x\right)  \in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$. Since
$S\ast E_{H}\ast e_{H,H}=S\ast E_{H}$, this rewrites as follows: Every $x\in
H$ satisfies $\left(  S\ast E_{H}\right)  \left(  x\right)  -E_{H}\left(
x\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$. This proves Corollary~\ref{Corollary913} \textbf{(e)}.

Applying Theorem~\ref{Theorem912} \textbf{(c)} to $P=S$ and $Q=e_{H,H}$, we
obtain $\left(  S\ast E_{H}\ast e_{H,H}\right)  \circ\left(  S\ast E_{H}\ast
e_{H,H}\right)  =E_{H}\circ\left(  S\ast E_{H}\ast e_{H,H}\right)  =\left(
S\ast E_{H}\ast e_{H,H}\right)  \circ E_{H}$. Since $S\ast E_{H}\ast
e_{H,H}=S\ast E_{H}$, this rewrites as $\left(  S\ast E_{H}\right)
\circ\left(  S\ast E_{H}\right)  =E_{H}\circ\left(  S\ast E_{H}\right)
=\left(  S\ast E_{H}\right)  \circ E_{H}$. This proves
Corollary~\ref{Corollary913} \textbf{(f)}.
\end{proof}

In order to specialize the above results to connected graded Hopf algebras, we
first observe a simple fact:

\begin{proposition}
\label{Proposition921} Let $k$ be a field. Let $H$ be a connected graded
$k$-bialgebra. Then, $H_{0}=k\cdot1_{H}$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition921}.]By the definition of $H_{\leq0}$,
we have $H_{\leq0}=\bigoplus_{\ell=0}^{0}H_{\ell}=H_{0}$. But since $H$ is
connected, we have $H_{\leq0}=k\cdot1_{H}$ (because Remark~\ref{Remark7m}
yields that $H$ is connected if and only if $H_{\leq0}=k\cdot1_{H}$), so that
$H_{0}=H_{\leq0}=k\cdot1_{H}$. This proves Proposition~\ref{Proposition921}.
\end{proof}

Here comes an equivalent version of part \textbf{(a)} of
Theorem~\ref{Theorem912}:

\begin{corollary}
\label{Corollary922} Let $k$ be a field. Let $H$ be a commutative graded
$k$-bialgebra. Let $E_{H}$ be defined according to
Definition~\ref{Definition701}.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-algebra
homomorphisms satisfying $\varepsilon_{H}\circ P=\varepsilon_{H}$,
$\varepsilon_{H}\circ Q=\varepsilon_{H}$ and $P\ast\operatorname{id}_{H}\ast
Q=e_{H,H}$. Here, the map $e_{H,H}$ is defined to be the map $\eta_{H}%
\circ\varepsilon_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is
identical with the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

Then, $\left(  P\ast E_{H}\ast Q\right)  \left(  k\cdot1_{H}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
\ \ \ \ \footnote{Recall that the notation $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ is to be understood according to
Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of $H$.}
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary922}.]Since $1_{H}\in H_{0}$ (because $H$ is
a graded $k$-algebra), we have $k\cdot1_{H}\subseteq kH_{0}\subseteq H_{0}$
(since $H_{0}$ is a $k$-vector space). Thus,%
\[
\left(  P\ast E_{H}\ast Q\right)  \left(  \underbrace{k\cdot1_{H}}_{\subseteq
H_{0}}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\right)  \subseteq\left(  P\ast E_{H}\ast Q\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0
\]
(by Theorem~\ref{Theorem912} \textbf{(a)}). Thus, $\left(  P\ast E_{H}\ast
Q\right)  \left(  k\cdot1_{H}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\right)  =0$. This proves
Corollary~\ref{Corollary922}.
\end{proof}

Now, let us do the same to parts \textbf{(a)} and \textbf{(d)} of
Corollary~\ref{Corollary913}:

\begin{corollary}
\label{Corollary923} Let $k$ be a field. Let $H$ be a commutative graded
$k$-Hopf algebra. Let $S$ be the antipode of $H$. Let $E_{H}$ be defined
according to Definition~\ref{Definition701}.

\textbf{(a)} Then, $\left(  E_{H}\ast S\right)  \left(  k\cdot1_{H}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
\ \ \ \ \footnote{Recall that the notation $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ is to be understood according to
Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of $H$.}

\textbf{(b)} Also, $\left(  S\ast E_{H}\right)  \left(  k\cdot1_{H}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary923}.]Since $1_{H}\in H_{0}$ (because $H$ is
a graded $k$-algebra), we have $k\cdot1_{H}\subseteq kH_{0}\subseteq H_{0}$
(since $H_{0}$ is a $k$-vector space).

\textbf{(a)} We have%
\[
\left(  E_{H}\ast S\right)  \left(  \underbrace{k\cdot1_{H}}_{\subseteq H_{0}%
}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\right)  \subseteq\left(  E_{H}\ast S\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0
\]
(by Corollary~\ref{Corollary913} \textbf{(a)}), so that $\left(  E_{H}\ast
S\right)  \left(  k\cdot1_{H}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
Corollary~\ref{Corollary923} \textbf{(a)} is proven.

\textbf{(b)} We have%
\[
\left(  S\ast E_{H}\right)  \left(  \underbrace{k\cdot1_{H}}_{\subseteq H_{0}%
}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\right)  \subseteq\left(  S\ast E_{H}\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0
\]
(by Corollary~\ref{Corollary913} \textbf{(d)}), so that $\left(  S\ast
E_{H}\right)  \left(  k\cdot1_{H}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\right)  =0$.
Corollary~\ref{Corollary923} \textbf{(b)} is proven.
\end{proof}

We now come to the case of fields of characteristic $0$. The following
generalizes Theorem~\ref{Theorem901} and is a dual of Theorem~\ref{Theorem831}:

\begin{theorem}
\label{Theorem931} Let $k$ be a field of characteristic $0$. Let $H$ be a
commutative graded $k$-bialgebra. Let $E_{H}$ be defined according to
Definition~\ref{Definition701}.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-algebra
homomorphisms satisfying $\varepsilon_{H}\circ P=\varepsilon_{H}$,
$\varepsilon_{H}\circ Q=\varepsilon_{H}$ and $P\ast\operatorname{id}_{H}\ast
Q=e_{H,H}$. Here, the map $e_{H,H}$ is defined to be the map $\eta_{H}%
\circ\varepsilon_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is
identical with the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

Then, the map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  \right)
=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$. \ \ \ \ \footnote{Recall that the notation $\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}$ is to be understood according
to Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of $H$.}
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem931}.]Let $M$ denote the map $E_{H}%
^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  $. Then,
\begin{align}
\underbrace{M}_{=E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  }\left(  H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)   &  =\left(  E_{H}^{\operatorname*{inv}%
}\circ\left(  P\ast E_{H}\ast Q\right)  \right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)
\nonumber\\
&  =E_{H}^{\operatorname*{inv}}\left(  \underbrace{\left(  P\ast E_{H}\ast
Q\right)  \left(  H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)  }_{\substack{=0\\\text{(by
Theorem~\ref{Theorem912} \textbf{(a)})}}}\right) \nonumber\\
&  =E_{H}^{\operatorname*{inv}}\left(  0\right)  =0 \label{931.pf.1}%
\end{align}
(since the map $E_{H}^{\operatorname*{inv}}$ is $k$-linear), so that%
\begin{equation}
H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\subseteq\operatorname*{Ker}M. \label{931.pf.2}%
\end{equation}


Since $E_{H}$ is graded (by Proposition~\ref{Proposition702} \textbf{(b)},
applied to $V=H$) and $Q$ is graded, we conclude (by
Proposition~\ref{Proposition224} \textbf{(a)}, applied to $C=H$, $A=H$,
$f=E_{H}$ and $g=Q$) that $E_{H}\ast Q$ is graded.

Since $P$ is graded and $E_{H}\ast Q$ is graded, we conclude (by
Proposition~\ref{Proposition224} \textbf{(a)}, applied to $C=H$, $A=H$, $f=P$
and $g=E_{H}\ast Q$) that $P\ast E_{H}\ast Q$ is graded. Thus,
\begin{equation}
\left(  P\ast E_{H}\ast Q\right)  \circ E_{H}=E_{H}\circ\left(  P\ast
E_{H}\ast Q\right)  \label{931.pf.commute}%
\end{equation}
(by Proposition~\ref{Proposition704}, applied to $V=H$, $W=H$ and $f=P\ast
E_{H}\ast Q$).

Since the map $E_{H}^{\operatorname*{inv}}$ is graded\footnote{\textit{Proof.}
Define $\left(  b_{\ell}\right)  _{\ell\in\mathbb{N}}$ as in
Definition~\ref{Definition721}. By Remark~\ref{Remark722} (applied to $V=H$),
we have $E_{H}^{\operatorname*{inv}}=E_{H}^{\left(  b_{\ell}\right)  _{\ell
\in\mathbb{N}}}$. Since the map $E_{H}^{\left(  b_{\ell}\right)  _{\ell
\in\mathbb{N}}}$ is graded (by Proposition~\ref{Proposition707} \textbf{(b)},
applied to $\left(  a_{\ell}\right)  _{\ell\in\mathbb{N}}=\left(  b_{\ell
}\right)  _{\ell\in\mathbb{N}}$), this yields that the map $E_{H}%
^{\operatorname*{inv}}$ is graded, qed.} and the map $P\ast E_{H}\ast Q$ is
graded, the map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  $ is graded as well (because the composition of two graded maps must
always be graded). Since $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast
E_{H}\ast Q\right)  =M$, we have thus shown that the map $M$ is graded. Since
the maps $M$ and $\operatorname{id}_{H}$ are graded, the map
$M-\operatorname{id}_{H}$ must also be graded (because the difference of two
graded maps must always be graded).

We are now going to prove that%
\begin{equation}
\left(  M-\operatorname{id}_{H}\right)  \left(  H_{m}\right)  \subseteq
H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}\text{.}
\label{931.pf.3}%
\end{equation}


\textit{Proof of (\ref{931.pf.3}):} Let $m\in\mathbb{N}$. Then, we must be in
one of the following cases:

\textit{Case 1:} We have $m=0$.

\textit{Case 2:} We have $m>0$.

First, let us consider Case 1. In this case, $m=0$. Thus,%
\begin{align*}
\left(  M-\operatorname{id}_{H}\right)  \left(  H_{m}\right)   &  =\left(
M-\operatorname{id}_{H}\right)  \left(  H_{0}\right)  \subseteq H_{0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }M-\operatorname{id}_{H}\text{ is
graded}\right) \\
&  \subseteq H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}.
\end{align*}
Hence, (\ref{931.pf.3}) is proven in Case 1.

Now, let us consider Case 2. In this case, $m>0$, so that $\dfrac{1}{m}$ is
well-defined in $k$ (since $k$ has characteristic $0$). Also, since $m>0$, we
have $m\geq1$, and thus $H_{m}$ is an addend in the direct sum $\bigoplus
_{n\geq1}H_{n}$. Thus, $H_{m}\subseteq\bigoplus_{n\geq1}H_{n}$.

Let $x\in H_{m}$. Then, $E_{H}\left(  x\right)  =mx$ (by
Proposition~\ref{Proposition702} \textbf{(a)}, applied to $V=H$, $n=m$ and
$v=x$). On the other hand, $\left(  P\ast E_{H}\ast Q\right)  \left(
H_{m}\right)  \subseteq H_{m}$ (since $P\ast E_{H}\ast Q$ is graded). Since
$x\in H_{m}$, we have
\[
\left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  \in\left(  P\ast E_{H}\ast
Q\right)  \left(  H_{m}\right)  \subseteq H_{m}\subseteq\bigoplus_{n\geq
1}H_{n}.
\]
Thus, Corollary~\ref{Corollary724} \textbf{(b)} (applied to $V=H$ and
$v=\left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  $) yields
\begin{equation}
\left(  E_{H}^{\operatorname*{inv}}\circ E_{H}\right)  \left(  \left(  P\ast
E_{H}\ast Q\right)  \left(  x\right)  \right)  =\left(  P\ast E_{H}\ast
Q\right)  \left(  x\right)  . \label{931.pf.4}%
\end{equation}
But since
\begin{align*}
\left(  M-\operatorname{id}_{H}\right)  \circ E_{H}  &  =\underbrace{M}%
_{=E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  }\circ
E_{H}-\underbrace{\operatorname{id}_{H}\circ E_{H}}_{=E_{H}}=E_{H}%
^{\operatorname*{inv}}\circ\underbrace{\left(  P\ast E_{H}\ast Q\right)  \circ
E_{H}}_{\substack{=E_{H}\circ\left(  P\ast E_{H}\ast Q\right)  \\\text{(by
(\ref{931.pf.commute}))}}}-E_{H}\\
&  =E_{H}^{\operatorname*{inv}}\circ E_{H}\circ\left(  P\ast E_{H}\ast
Q\right)  -E_{H},
\end{align*}
we have%
\begin{align*}
\left(  \left(  M-\operatorname{id}_{H}\right)  \circ E_{H}\right)  \left(
x\right)   &  =\left(  E_{H}^{\operatorname*{inv}}\circ E_{H}\circ\left(
P\ast E_{H}\ast Q\right)  -E_{H}\right)  \left(  x\right) \\
&  =\underbrace{\left(  E_{H}^{\operatorname*{inv}}\circ E_{H}\circ\left(
P\ast E_{H}\ast Q\right)  \right)  \left(  x\right)  }_{\substack{=\left(
E_{H}^{\operatorname*{inv}}\circ E_{H}\right)  \left(  \left(  P\ast E_{H}\ast
Q\right)  \left(  x\right)  \right)  =\left(  P\ast E_{H}\ast Q\right)
\left(  x\right)  \\\text{(by (\ref{931.pf.4}))}}}-E_{H}\left(  x\right) \\
&  =\left(  P\ast E_{H}\ast Q\right)  \left(  x\right)  -E_{H}\left(
x\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}%
\end{align*}
(by Theorem~\ref{Theorem912} \textbf{(b)}). Since%
\[
\left(  \left(  M-\operatorname{id}_{H}\right)  \circ E_{H}\right)  \left(
x\right)  =\left(  M-\operatorname{id}_{H}\right)  \underbrace{\left(
E_{H}\left(  x\right)  \right)  }_{=mx}=\left(  M-\operatorname{id}%
_{H}\right)  \left(  mx\right)  =m\left(  M-\operatorname{id}_{H}\right)
\left(  x\right)
\]
(since $M-\operatorname{id}_{H}$ is $k$-linear), this rewrites as $m\left(
M-\operatorname{id}_{H}\right)  \left(  x\right)  \in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$. Hence,
$\left(  M-\operatorname{id}_{H}\right)  \left(  x\right)  \in\dfrac{1}%
{m}\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$ (since $\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}$ is a $k$-vector space).

Now, forget that we fixed $x$. We thus have proven that every $x\in H_{m}$
satisfies $\left(  M-\operatorname{id}_{H}\right)  \left(  x\right)
\in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$.
In other words, $\left(  M-\operatorname{id}_{H}\right)  \left(  H_{m}\right)
\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$. Thus,%
\[
\left(  M-\operatorname{id}_{H}\right)  \left(  H_{m}\right)  \subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\subseteq
H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}.
\]
Hence, (\ref{931.pf.3}) is proven in Case 2.

We have thus proven (\ref{931.pf.3}) in each of the two cases 1 and 2. Since
these two cases cover all possibilities, this yields that (\ref{931.pf.3})
always holds.

Now, since $H$ is graded, we have $H=\bigoplus_{m\in\mathbb{N}}H_{m}%
=\sum_{m\in\mathbb{N}}H_{m}$ (since direct sums are sums). Thus,%
\begin{align}
\left(  M-\operatorname{id}_{H}\right)  \left(  H\right)   &  =\left(
M-\operatorname{id}_{H}\right)  \left(  \sum_{m\in\mathbb{N}}H_{m}\right)
=\sum_{m\in\mathbb{N}}\underbrace{\left(  M-\operatorname{id}_{H}\right)
\left(  H_{m}\right)  }_{\substack{\subseteq H_{0}+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}\\\text{(by (\ref{931.pf.3}))}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }M-\operatorname{id}_{H}\text{ is
}k\text{-linear}\right) \nonumber\\
&  \subseteq\sum_{m\in\mathbb{N}}\left(  H_{0}+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  \subseteq H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}
\label{931.pf.10}%
\end{align}
(since $H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$ is a $k$-vector space). Thus,%
\begin{align*}
\left(  M\circ M-\underbrace{M}_{=M\circ\operatorname{id}_{H}}\right)  \left(
H\right)   &  =\underbrace{\left(  M\circ M-M\circ\operatorname{id}%
_{H}\right)  }_{\substack{=M\circ\left(  M-\operatorname{id}_{H}\right)
\\\text{(since composition of }k\text{-linear maps}\\\text{is distributive)}%
}}\left(  H\right) \\
&  =\left(  M\circ\left(  M-\operatorname{id}_{H}\right)  \right)  \left(
H\right)  =M\left(  \underbrace{\left(  M-\operatorname{id}_{H}\right)
\left(  H\right)  }_{\substack{\subseteq H_{0}+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}\\\text{(by (\ref{931.pf.10}))}%
}}\right) \\
&  \subseteq M\left(  H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{931.pf.1})}\right)  ,
\end{align*}
so that $\left(  M\circ M-M\right)  \left(  H\right)  =0$. Thus, $M\circ
M-M=0$, so that $M\circ M=M$. In other words, $M$ is a projection.

We will now prove that $\operatorname*{Ker}M=H_{0}+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}$.

In fact, let $x\in\operatorname*{Ker}M$ be arbitrary. Then, $M\left(
x\right)  =0$, so that $\left(  M-\operatorname{id}_{H}\right)  \left(
x\right)  =\underbrace{M\left(  x\right)  }_{=0}-\underbrace{\operatorname{id}%
_{H}\left(  x\right)  }_{=x}=-x$. Thus,%
\[
-x=\left(  M-\operatorname{id}_{H}\right)  \left(  \underbrace{x}_{\in
H}\right)  \in\left(  M-\operatorname{id}_{H}\right)  \left(  H\right)
\subseteq H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}%
\]
(by (\ref{931.pf.10})). Consequently, $x\in-\left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)
\subseteq H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$ (since $H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}$ is a $k$-vector space).

Now forget that we fixed $x$. We have thus proven that every $x\in
\operatorname*{Ker}M$ satisfies $x\in H_{0}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. In other words, $\operatorname*{Ker}%
M\subseteq H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$. Combined with (\ref{931.pf.2}), this yields
$\operatorname*{Ker}M=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}$.

So we know that the map $M$ is a projection such that $\operatorname*{Ker}%
M=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$. Since $M=E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  $, this rewrites as follows: The map $E_{H}^{\operatorname*{inv}%
}\circ\left(  P\ast E_{H}\ast Q\right)  $ is a projection such that
$\operatorname*{Ker}\left(  E_{H}^{\operatorname*{inv}}\circ\left(  P\ast
E_{H}\ast Q\right)  \right)  =H_{0}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. This proves Theorem~\ref{Theorem931}.
\end{proof}

We can now get Theorem~\ref{Theorem901} as a corollary:

\begin{proof}
[Proof of Theorem~\ref{Theorem901}.]Just as in the proof of
Corollary~\ref{Corollary908}, we can prove the following facts:

\begin{itemize}
\item The map $S$ is a $k$-algebra homomorphism from $H$ to $H$.

\item The map $e_{H,H}$ is a $k$-algebra homomorphism (where $e_{H,H}$ is
defined as according to Definition~\ref{def.eHA}).

\item We have $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$.

\item We have $S\ast\operatorname{id}_{H}\ast e_{H,H}=e_{H,H}$ and
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}$.
\end{itemize}

Just as in the proof of Corollary~\ref{Corollary913}, we can prove the
following facts:

\begin{itemize}
\item We have $\varepsilon_{H}\circ e_{H,H}=\varepsilon_{H}$ and
$\varepsilon_{H}\circ S=\varepsilon_{H}$.

\item The maps $e_{H,H}$ and $S$ are graded.
\end{itemize}

As a consequence, we can apply Theorem~\ref{Theorem931} to $P=e_{H,H}$ and
$Q=S$. As a result, we obtain that the map $E_{H}^{\operatorname*{inv}}%
\circ\left(  e_{H,H}\ast E_{H}\ast S\right)  $ is a projection such that
$\operatorname*{Ker}\left(  E_{H}^{\operatorname*{inv}}\circ\left(
e_{H,H}\ast E_{H}\ast S\right)  \right)  =H_{0}+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}$. Since $e_{H,H}\ast E_{H}\ast
S=E_{H}\ast S$, this rewrites as follows: The map $E_{H}^{\operatorname*{inv}%
}\circ\left(  E_{H}\ast S\right)  $ is a projection such that
$\operatorname*{Ker}\left(  E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast
S\right)  \right)  =H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}$. This proves Theorem~\ref{Theorem901} \textbf{(a)}.

But we can also apply Theorem~\ref{Theorem931} to $P=S$ and $Q=e_{H,H}$. As a
result, we obtain that the map $E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\ast e_{H,H}\right)  $ is a projection such that $\operatorname*{Ker}%
\left(  E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\ast e_{H,H}%
\right)  \right)  =H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}$. Since $S\ast E_{H}\ast e_{H,H}=S\ast E_{H}$, this
rewrites as follows: The map $E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  \right)
=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$. This proves Theorem~\ref{Theorem901} \textbf{(b)}.
\end{proof}

Next, we specialize our results to connected graded bialgebras. The following
fact is the specialization of Theorem~\ref{Theorem931} and the dual of
Corollary~\ref{Corollary832}:

\begin{corollary}
\label{Corollary932} Let $k$ be a field of characteristic $0$. Let $H$ be a
commutative connected graded $k$-bialgebra. Let $E_{H}$ be defined according
to Definition~\ref{Definition701}.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-algebra
homomorphisms satisfying $\varepsilon_{H}\circ P=\varepsilon_{H}$,
$\varepsilon_{H}\circ Q=\varepsilon_{H}$ and $P\ast\operatorname{id}_{H}\ast
Q=e_{H,H}$. Here, the map $e_{H,H}$ is defined to be the map $\eta_{H}%
\circ\varepsilon_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is
identical with the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

Then, the map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  \right)
=k\cdot1_{H}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$. \ \ \ \ \footnote{Recall that the notation $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ is to be
understood according to Convention~\ref{Convention100}. Hence, $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ means the
subspace $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of
$H$.}
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary932}.]According to Theorem~\ref{Theorem931},
the map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  $
is a projection such that $\operatorname*{Ker}\left(  E_{H}%
^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  \right)
=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$. Since $H_{0}=k\cdot1_{H}$ (by Proposition~\ref{Proposition921}), this
rewrites as follows: The map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast
E_{H}\ast Q\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  \right)
=k\cdot1_{H}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}$. This proves Corollary~\ref{Corollary932}.
\end{proof}

The specialization of Theorem~\ref{Theorem901} is Theorem~\ref{Theorem902},
and here is its (obvious) proof:

\begin{proof}
[Proof of Theorem~\ref{Theorem902}.]Theorem~\ref{Theorem901} \textbf{(a)} says
that the map $E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  $ is
a projection such that $\operatorname*{Ker}\left(  E_{H}^{\operatorname*{inv}%
}\circ\left(  E_{H}\ast S\right)  \right)  =H_{0}+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}$. Since $H_{0}=k\cdot1_{H}$ (by
Proposition~\ref{Proposition921}), this rewrites as follows: The map
$E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  $ is a projection
such that $\operatorname*{Ker}\left(  E_{H}^{\operatorname*{inv}}\circ\left(
E_{H}\ast S\right)  \right)  =k\cdot1_{H}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. This proves Theorem~\ref{Theorem902}
\textbf{(a)}.

Theorem~\ref{Theorem901} \textbf{(b)} says that the map $E_{H}%
^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  $ is a projection such
that $\operatorname*{Ker}\left(  E_{H}^{\operatorname*{inv}}\circ\left(  S\ast
E_{H}\right)  \right)  =H_{0}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. Since $H_{0}=k\cdot1_{H}$ (by
Proposition~\ref{Proposition921}), this rewrites as follows: The map
$E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  $ is a projection
such that $\operatorname*{Ker}\left(  E_{H}^{\operatorname*{inv}}\circ\left(
S\ast E_{H}\right)  \right)  =k\cdot1_{H}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. This proves Theorem~\ref{Theorem902}
\textbf{(b)}.
\end{proof}

Finally, let us prove a relation between the maps $S\ast E_{H}$ and $E_{H}\ast
S$ which is dual to Theorem~\ref{Theorem841}:

\begin{theorem}
\label{Theorem941} Let $k$ be a field. Let $H$ be a commutative graded
$k$-Hopf algebra. Let $S$ be the antipode of $H$. Let $E_{H}$ be defined
according to Definition~\ref{Definition701}.

\textbf{(a)} We have%
\[
S\circ\left(  E_{H}\ast S\right)  =\left(  S\ast E_{H}\right)  \circ S=-S\ast
E_{H}.
\]


\textbf{(b)} We have%
\[
S\circ\left(  S\ast E_{H}\right)  =\left(  E_{H}\ast S\right)  \circ
S=-E_{H}\ast S.
\]

\end{theorem}

We choose to prove this not in the same way as we proved
Theorem~\ref{Theorem841}, but differently. But still, let us formulate and
verify the dual of Theorem~\ref{Theorem842}:

\begin{theorem}
\label{Theorem942} Let $k$ be a field. Let $H$ be a $k$-Hopf algebra. Let $S$
be the antipode of $H$.

\textbf{(a)} We have $S\circ\eta_{H}=\eta_{H}$.

\textbf{(b)} If $H$ is commutative, we have $S\circ\mu_{H}=\mu_{H}\circ\left(
S\otimes S\right)  $.

\textbf{(c)} If $H$ is commutative, we have $S\circ S=\operatorname{id}_{H}$.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem942}.]Let us define $H^{\operatorname*{op}}$ as
according to Definition~\ref{Definition602}. Then, $H^{\operatorname*{op}%
}=\left(  H,\mu_{H}\circ\tau_{H,H},\eta_{H}\right)  $, so that $\eta
_{H^{\operatorname*{op}}}=\eta_{H}$.

But Proposition~\ref{Proposition603} yields that the antipode of $H$ is a
$k$-algebra homomorphism from $H$ to $H^{\operatorname*{op}}$. Since the
antipode of $H$ is the map $S$, this rewrites as follows: The map $S$ is a
$k$-algebra homomorphism from $H$ to $H^{\operatorname*{op}}$. Thus,
$S\circ\eta_{H}=\eta_{H^{\operatorname*{op}}}$. Since $\eta
_{H^{\operatorname*{op}}}=\eta_{H}$, this rewrites as $S\circ\eta_{H}=\eta
_{H}$. This proves Theorem~\ref{Theorem942} \textbf{(a)}.

\textbf{(b)} Assume that $H$ is commutative. Then, $\mu_{H}\circ\tau_{H,H}%
=\mu_{H}$, so that $H^{\operatorname*{op}}=\left(  H,\underbrace{\mu_{H}%
\circ\tau_{H,H}}_{=\mu_{H}},\eta_{H}\right)  =\left(  H,\mu_{H},\eta
_{H}\right)  =H$.

We have showed above that the map $S$ is a $k$-algebra homomorphism from $H$
to $H^{\operatorname*{op}}$. Since $H^{\operatorname*{op}}=H$, this rewrites
as follows: The map $S$ is a $k$-algebra homomorphism from $H$ to $H$. Thus,
$S\circ\mu_{H}=\mu_{H}\circ\left(  S\otimes S\right)  $. This proves
Theorem~\ref{Theorem942} \textbf{(b)}.

\textbf{(c)} The antipode of $H$ is the $\ast$-inverse of the map
$\operatorname{id}_{H}$ (because the antipode of a Hopf algebra is defined as
the $\ast$-inverse of its identity map). Since the antipode of $H$ is $S$, we
thus have shown that $S$ is the $\ast$-inverse of the map $\operatorname{id}%
_{H}$. Thus, $S\ast\operatorname{id}_{H}=\operatorname{id}_{H}\ast S=e_{H,H}$
(where $e_{H,H}$ is defined as according to Definition~\ref{def.eHA}).

Applying (\ref{(X)(X)}) to $H$, $H$, $H$, $H$, $H$, $H$, $S$, $S$,
$\operatorname{id}_{H}$, $S$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$,
$\beta^{\prime}$, we obtain%
\begin{equation}
\left(  S\circ S\right)  \otimes\left(  S\circ\operatorname{id}_{H}\right)
=\left(  S\otimes S\right)  \circ\left(  S\otimes\operatorname{id}_{H}\right)
. \label{942.pf.1}%
\end{equation}
Now, by the definition of convolution,%
\begin{align*}
\left(  S\circ S\right)  \ast S  &  =\mu_{H}\circ\left(  \left(  S\circ
S\right)  \otimes\underbrace{S}_{=S\circ\operatorname{id}_{H}}\right)
\circ\Delta_{H}=\mu_{H}\circ\underbrace{\left(  \left(  S\circ S\right)
\otimes\left(  S\circ\operatorname{id}_{H}\right)  \right)  }%
_{\substack{=\left(  S\otimes S\right)  \circ\left(  S\otimes\operatorname{id}%
_{H}\right)  \\\text{(by (\ref{942.pf.1}))}}}\circ\Delta_{H}\\
&  =\underbrace{\mu_{H}\circ\left(  S\otimes S\right)  }_{\substack{=S\circ
\mu_{H} \\\text{(by Theorem~\ref{Theorem942} \textbf{(b)})}}}\circ\left(
S\otimes\operatorname{id}_{H}\right)  \circ\Delta_{H}=s\circ\underbrace{\mu
_{H}\circ\left(  S\otimes\operatorname{id}_{H}\right)  \circ\Delta_{H}}
_{\substack{=S\ast\operatorname{id}_{H} \\\text{(because} \\S\ast
\operatorname{id}_{H}=\mu_{H}\circ\left(  S\otimes\operatorname{id}%
_{H}\right)  \circ\Delta_{H} \\\text{by the definition of convolution)}}}\\
&  =S\circ\underbrace{\left(  S\ast\operatorname{id}_{H}\right)  }
_{\substack{=e_{H,H}=\eta_{H}\circ\varepsilon_{H} \\\text{(by the definition
of }e_{H,H}\text{)}}}=\underbrace{S\circ\eta_{H}}_{\substack{=\eta_{H}
\\\text{(by Theorem~\ref{Theorem942} \textbf{(a)})}}}\circ\varepsilon_{H}%
=\eta_{H}\circ\varepsilon_{H}=e_{H,H}%
\end{align*}
(since $e_{H,H}=\eta_{H}\circ\varepsilon_{H}$ by the definition of $e_{H,H}$).
Thus,%
\[
\underbrace{\left(  S\circ S\right)  \ast S}_{=e_{H,H}}\ast\operatorname{id}%
_{H}=e_{H,H}\ast\operatorname{id}_{H}=\operatorname{id}_{H}.
\]
Comparing this with%
\[
\left(  S\circ S\right)  \ast\underbrace{S\ast\operatorname{id}_{H}}%
_{=e_{H,H}}=\left(  S\circ S\right)  \ast e_{H,H}=S\circ S,
\]
we obtain $S\circ S=\operatorname{id}_{H}$. This proves
Theorem~\ref{Theorem942} \textbf{(c)}.
\end{proof}

The following fact is (more or less) a dual of
Proposition~\ref{Proposition843} and will be used in our proof of
Theorem~\ref{Theorem941}:

\begin{proposition}
\label{Proposition943} Let $k$ be a field. Let $H$ be a $k$-Hopf algebra. Let
$S$ be the antipode of $H$. Let $x\in H$.

\textbf{(a)} Then, $S\left(  x\right)  +x\in k\cdot1_{H}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$.

\textbf{(b)} Assume that $H$ is a graded $k$-algebra. Then, $S\left(
x\right)  +x\in H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition943}.]The antipode of $H$ is the $\ast
$-inverse of the map $\operatorname{id}_{H}$ (because the antipode of a Hopf
algebra is defined as the $\ast$-inverse of its identity map). Since the
antipode of $H$ is $S$, we thus have shown that $S$ is the $\ast$-inverse of
the map $\operatorname{id}_{H}$. Thus, $S\ast\operatorname{id}_{H}%
=\operatorname{id}_{H}\ast S=e_{H,H}$ (where $e_{H,H}$ is defined as according
to Definition~\ref{def.eHA}) and $S=\operatorname{id}_{H}^{\ast\left(
-1\right)  }$.

Since $H$ is a bialgebra, we have $\varepsilon_{H}\circ\eta_{H}%
=\operatorname{id}_{k}$ (by the axioms of a bialgebra). But $e_{H,H}=\eta
_{H}\circ\varepsilon_{H}$ (by the definition of $e_{H,H}$). Thus,
$\varepsilon_{H}\circ e_{H,H}=\underbrace{\varepsilon_{H}\circ\eta_{H}%
}_{=\operatorname{id}_{k}}\circ\varepsilon_{H}=\varepsilon_{H}$.

Theorem~\ref{Theorem842} \textbf{(a)} yields $\varepsilon_{H}\circ
S=\varepsilon_{H}$. Now,%
\begin{align*}
\varepsilon_{H}\circ\left(  S-e_{H,H}\right)   &  =\underbrace{\varepsilon
_{H}\circ S}_{=\varepsilon_{H}}-\underbrace{\varepsilon_{H}\circ e_{H,H}%
}_{=\varepsilon_{H}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of
}k\text{-linear maps is distributive}\right) \\
&  =\varepsilon_{H}-\varepsilon_{H}=0.
\end{align*}
Thus,%
\[
\varepsilon_{H}\left(  \left(  S-e_{H,H}\right)  \left(  H\right)  \right)
=\underbrace{\left(  \varepsilon_{H}\circ\left(  S-e_{H,H}\right)  \right)
}_{=0}\left(  H\right)  =0\left(  H\right)  =0,
\]
so that%
\begin{equation}
\left(  S-e_{H,H}\right)  \left(  H\right)  \subseteq\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  . \label{943.pf.1}%
\end{equation}


Also,%
\begin{align*}
\varepsilon_{H}\circ\left(  \operatorname{id}_{H}-e_{H,H}\right)   &
=\underbrace{\varepsilon_{H}\circ\operatorname{id}_{H}}_{=\varepsilon_{H}%
}-\underbrace{\varepsilon_{H}\circ e_{H,H}}_{=\varepsilon_{H}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is distributive}\right) \\
&  =\varepsilon_{H}-\varepsilon_{H}=0.
\end{align*}
Thus,%
\[
\varepsilon_{H}\left(  \left(  \operatorname{id}_{H}-e_{H,H}\right)  \left(
H\right)  \right)  =\underbrace{\left(  \varepsilon_{H}\circ\left(
\operatorname{id}_{H}-e_{H,H}\right)  \right)  }_{=0}\left(  H\right)
=0\left(  H\right)  =0,
\]
so that%
\begin{equation}
\left(  \operatorname{id}_{H}-e_{H,H}\right)  \left(  H\right)  \subseteq
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  . \label{943.pf.2}%
\end{equation}


On the other hand,
\begin{align}
\underbrace{e_{H,H}}_{=\eta_{H}\circ\varepsilon_{H}}\left(  H\right)   &
=\left(  \eta_{H}\circ\varepsilon_{H}\right)  \left(  H\right)  =\eta
_{H}\underbrace{\left(  \varepsilon_{H}\left(  H\right)  \right)  }_{\subseteq
k}\nonumber\\
&  \subseteq\eta_{H}\left(  k\right)  =\left\{  \underbrace{\eta_{H}\left(
\lambda\right)  }_{\substack{=\lambda\cdot1_{H}\\\text{(by the definition of
}\eta_{H}\text{)}}}\ \mid\ \lambda\in k\right\}  =\left\{  \lambda\cdot
1_{H}\ \mid\ \lambda\in k\right\} \nonumber\\
&  =k\cdot1_{H}. \label{943.pf.7}%
\end{align}


Now, Lemma~\ref{Lemma910} (applied to $C=H$, $A=H$, $f=S-e_{H,H}$ and
$g=\operatorname{id}_{H}-e_{H,H}$) yields%
\begin{align}
\left(  \left(  S-e_{H,H}\right)  \ast\left(  \operatorname{id}_{H}%
-e_{H,H}\right)  \right)  \left(  H\right)   &  \subseteq\underbrace{\left(
S-e_{H,H}\right)  \left(  H\right)  }_{\substack{\subseteq\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \\\text{(by (\ref{943.pf.1}))}}}\cdot
\underbrace{\left(  \operatorname{id}_{H}-e_{H,H}\right)  \left(  H\right)
}_{\substack{\subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)
\\\text{(by (\ref{943.pf.2}))}}}\nonumber\\
&  \subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \cdot
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  =\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}.
\label{943.pf.3}%
\end{align}
But%
\begin{align*}
\left(  S-e_{H,H}\right)  \ast\left(  \operatorname{id}_{H}-e_{H,H}\right)
&  =\underbrace{S\ast\operatorname{id}_{H}}_{=e_{H,H}}-\underbrace{S\ast
e_{H,H}}_{=S}-\underbrace{e_{H,H}\ast\operatorname{id}_{H}}%
_{=\operatorname{id}_{H}}+\underbrace{e_{H,H}\ast e_{H,H}}_{=e_{H,H}}\\
&  =e_{H,H}-S-\operatorname{id}_{H}+e_{H,H}=2e_{H,H}-\left(
S+\operatorname{id}_{H}\right)  ,
\end{align*}
so that
\[
S+\operatorname{id}_{H}=2e_{H,H}-\left(  S-e_{H,H}\right)  \ast\left(
\operatorname{id}_{H}-e_{H,H}\right)  .
\]
Thus,%
\begin{align*}
\left(  S+\operatorname{id}_{H}\right)  \left(  x\right)   &  =\left(
2e_{H,H}-\left(  S-e_{H,H}\right)  \ast\left(  \operatorname{id}_{H}%
-e_{H,H}\right)  \right)  \left(  x\right) \\
&  =2e_{H,H}\left(  \underbrace{x}_{\in H}\right)  -\left(  \left(
S-e_{H,H}\right)  \ast\left(  \operatorname{id}_{H}-e_{H,H}\right)  \right)
\left(  \underbrace{x}_{\in H}\right) \\
&  \in2\underbrace{e_{H,H}\left(  H\right)  }_{\substack{\subseteq k\cdot
1_{H}\\\text{(by (\ref{943.pf.7}))}}}-\underbrace{\left(  \left(
S-e_{H,H}\right)  \ast\left(  \operatorname{id}_{H}-e_{H,H}\right)  \right)
\left(  H\right)  }_{\substack{\subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\\\text{(by (\ref{943.pf.3}))}}}\\
&  \subseteq2k\cdot1_{H}-\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}=\underbrace{2k\cdot1_{H}}_{\substack{\subseteq
k\cdot1_{H}\\\text{(since }k\cdot1_{H}\text{ is a}\\k\text{-vector space)}%
}}+\underbrace{\left(  -\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)  }_{\substack{\subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\\\text{(since
}\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\text{ is a}\\k\text{-vector space)}}}\\
&  \subseteq k\cdot1_{H}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}.
\end{align*}
Since $\left(  S+\operatorname{id}_{H}\right)  \left(  x\right)  =S\left(
x\right)  +\underbrace{\operatorname{id}_{H}\left(  x\right)  }_{=x}=S\left(
x\right)  +x$, this rewrites as $S\left(  x\right)  +x\in k\cdot1_{H}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$. This proves
Proposition~\ref{Proposition943} \textbf{(a)}.

\textbf{(b)} The grading on $k$ is such that $k_{0}=k$. Thus, $\eta_{H}\left(
\underbrace{k}_{=k_{0}}\right)  =\eta_{H}\left(  k_{0}\right)  \subseteq
H_{0}$ (since $\eta_{H}$ is a graded map (because $H$ is a graded
$k$-algebra)). But we have shown above that $\eta_{H}\left(  k\right)
=k\cdot1_{H}$. Hence, $k\cdot1_{H}=\eta_{H}\left(  k\right)  \subseteq H_{0}$.

Proposition~\ref{Proposition943} \textbf{(a)} yields%
\[
S\left(  x\right)  +x\in\underbrace{k\cdot1_{H}}_{\subseteq H_{0}}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\subseteq
H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}.
\]
This proves Proposition~\ref{Proposition943} \textbf{(b)}.
\end{proof}

We will now generalize part of Theorem~\ref{Theorem941} to graded Hopf
algebras which are not necessarily commutative or cocommutative:

\begin{theorem}
\label{Theorem944} Let $k$ be a field. Let $H$ be a graded $k$-Hopf algebra.
Let $S$ be the antipode of $H$. Let $E_{H}$ be defined according to
Definition~\ref{Definition701}.

\textbf{(a)} We have%
\[
S\circ\left(  E_{H}\ast S\right)  =\left(  S\ast E_{H}\right)  \circ S.
\]


\textbf{(b)} We have%
\[
S\circ\left(  S\ast E_{H}\right)  =\left(  E_{H}\ast S\right)  \circ S.
\]

\end{theorem}

Something even stronger holds:

\begin{theorem}
\label{Theorem945} Let $k$ be a field. Let $H$ be a $k$-Hopf algebra. Let $S$
be the antipode of $H$. Let $f:H\rightarrow H$ and $g:H\rightarrow H$ be two
$k$-linear maps such that $f\circ S=S\circ f$ and $g\circ S=S\circ g$. Then,%
\[
S\circ\left(  f\ast g\right)  =\left(  g\ast f\right)  \circ S.
\]

\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem945}.]Proposition~\ref{Proposition35}
\textbf{(a)} (applied to $H$, $H$, $H$, $H$, $S$ and $S$ instead of $V$, $W$,
$V^{\prime}$, $W^{\prime}$, $f$ and $g$) yields
\begin{equation}
\left(  S\otimes S\right)  \circ\tau_{H,H}=\tau_{H,H}\circ\left(  S\otimes
S\right)  . \label{945.pf.SS}%
\end{equation}
Proposition~\ref{Proposition35} \textbf{(a)} (applied to $H$, $H$, $H$ and $H$
instead of $V$, $W$, $V^{\prime}$ and $W^{\prime}$) yields
\begin{equation}
\left(  g\otimes f\right)  \circ\tau_{H,H}=\tau_{H,H}\circ\left(  f\otimes
g\right)  . \label{945.pf.gf}%
\end{equation}


Let us define $H^{\operatorname*{op}}$ as according to
Definition~\ref{Definition602}. Then, $H^{\operatorname*{op}}=\left(
H,\mu_{H}\circ\tau_{H,H},\eta_{H}\right)  $, so that $\mu
_{H^{\operatorname*{op}}}=\mu_{H}\circ\tau_{H,H}$.

But Proposition~\ref{Proposition603} yields that the antipode of $H$ is a
$k$-algebra homomorphism from $H$ to $H^{\operatorname*{op}}$. In other words,
$S$ is a $k$-algebra homomorphism from $H$ to $H^{\operatorname*{op}}$ (since
$S$ is the antipode of $H$). Thus,%
\begin{align}
S\circ\mu_{H}  &  =\underbrace{\mu_{H^{\operatorname*{op}}}}_{=\mu_{H}%
\circ\tau_{H,H}}\circ\left(  S\otimes S\right)  =\mu_{H}\circ\underbrace{\tau
_{H,H}\circ\left(  S\otimes S\right)  }_{\substack{=\left(  S\otimes S\right)
\circ\tau_{H,H} \\\text{(by (\ref{945.pf.SS}))}}}\nonumber\\
&  =\mu_{H}\circ\left(  S\otimes S\right)  \circ\tau_{H,H}. \label{945.pf.1}%
\end{align}


Let us define $H^{\operatorname*{cop}}$ as according to
Definition~\ref{Definition552}. Then, $H^{\operatorname*{cop}}=\left(
H,\tau_{H,H}\circ\Delta_{H},\varepsilon_{H}\right)  $, so that $\Delta
_{H^{\operatorname*{cop}}}=\tau_{H,H}\circ\Delta_{H}$.

But Proposition~\ref{Proposition553} yields that the antipode of $H$ is a
$k$-coalgebra homomorphism from $H^{\operatorname*{cop}}$ to $H$. In other
words, $S$ is a $k$-coalgebra homomorphism from $H^{\operatorname*{cop}}$ to
$H$ (since $S$ is the antipode of $H$). Thus,%
\begin{align}
\Delta_{H}\circ S  &  =\left(  S\otimes S\right)  \circ\underbrace{\Delta
_{H^{\operatorname*{cop}}}}_{=\tau_{H,H}\circ\Delta_{H}}=\underbrace{\left(
S\otimes S\right)  \circ\tau_{H,H}}_{\substack{=\tau_{H,H}\circ\left(
S\otimes S\right)  \\\text{(by (\ref{945.pf.SS}))}}}\circ\Delta_{H}\nonumber\\
&  =\tau_{H,H}\circ\left(  S\otimes S\right)  \circ\Delta_{H}.
\label{945.pf.2}%
\end{align}


By the definition of convolution, we have $f\ast g=\mu_{H}\circ\left(
f\otimes g\right)  \circ\Delta_{H}$ and $g\ast f=\mu_{H}\circ\left(  g\otimes
f\right)  \circ\Delta_{H}$.

Now,%
\begin{align*}
\underbrace{\left(  g\ast f\right)  }_{=\mu_{H}\circ\left(  g\otimes f\right)
\circ\Delta_{H}}\circ S  &  =\mu_{H}\circ\left(  g\otimes f\right)
\circ\underbrace{\Delta_{H}\circ S}_{\substack{=\tau_{H,H}\circ\left(
S\otimes S\right)  \circ\Delta_{H} \\\text{(by (\ref{945.pf.2}))}}}=\mu
_{H}\circ\underbrace{\left(  g\otimes f\right)  \circ\tau_{H,H}}%
_{\substack{=\tau_{H,H}\circ\left(  f\otimes g\right)  \\\text{(by
(\ref{945.pf.gf}))} }}\circ\left(  S\otimes S\right)  \circ\Delta_{H}\\
&  =\mu_{H}\circ\tau_{H,H}\circ\underbrace{\left(  f\otimes g\right)
\circ\left(  S\otimes S\right)  }_{\substack{=\left(  f\circ S\right)
\otimes\left(  g\circ S\right)  \\\text{(since (\ref{(X)(X)}) (applied to
}H\text{, }H\text{, }H\text{, }H\text{, }H\text{, }H\text{,} \\S\text{,
}f\text{, }S\text{, }g\text{ instead of }U\text{, }V\text{, }W\text{,
}U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }%
\beta\text{, }\alpha^{\prime}\text{, }\beta^{\prime}\text{)} \\\text{yields
}\left(  f\circ S\right)  \otimes\left(  g\circ S\right)  =\left(  f\otimes
g\right)  \circ\left(  S\otimes S\right)  \text{)}}}\circ\Delta_{H}\\
&  =\mu_{H}\circ\tau_{H,H}\circ\left(  \underbrace{\left(  f\circ S\right)
}_{=S\circ f}\otimes\underbrace{\left(  g\circ S\right)  }_{=S\circ g}\right)
\circ\Delta_{H}\\
&  =\mu_{H}\circ\tau_{H,H}\circ\underbrace{\left(  \left(  S\circ f\right)
\otimes\left(  S\circ g\right)  \right)  }_{\substack{=\left(  S\otimes
S\right)  \circ\left(  f\otimes g\right)  \\\text{(by (\ref{(X)(X)}), applied
to }H\text{, }H\text{, }H\text{, }H\text{, }H\text{, }H\text{,} \\f\text{,
}S\text{, }g\text{, }S\text{ instead of }U\text{, }V\text{, }W\text{,
}U^{\prime}\text{, }V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }%
\beta\text{, }\alpha^{\prime}\text{, }\beta^{\prime}\text{)}}}\circ\Delta
_{H}\\
&  =\mu_{H}\circ\underbrace{\tau_{H,H}\circ\left(  S\otimes S\right)  }
_{\substack{=\left(  S\otimes S\right)  \circ\tau_{H,H} \\\text{(by
(\ref{945.pf.SS}))}}}\circ\left(  f\otimes g\right)  \circ\Delta_{H}\\
&  =\underbrace{\mu_{H}\circ\left(  S\otimes S\right)  \circ\tau_{H,H}}
_{\substack{=S\circ\mu_{H} \\\text{(by (\ref{945.pf.1}))}}}\circ\left(
f\otimes g\right)  \circ\Delta_{H}=S\circ\mu_{H}\circ\left(  f\otimes
g\right)  \circ\Delta_{H}.
\end{align*}
Compared with%
\[
S\circ\underbrace{\left(  f\ast g\right)  }_{=\mu_{H}\circ\left(  f\otimes
g\right)  \circ\Delta_{H}}=S\circ\mu_{H}\circ\left(  f\otimes g\right)
\circ\Delta_{H},
\]
this yields $\left(  g\ast f\right)  \circ S=S\circ\left(  f\ast g\right)  $.
This proves Theorem~\ref{Theorem945}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem944}.]Since $H$ is a graded $k$-Hopf algebra,
the antipode of $H$ is graded. Since the antipode of $H$ is $S$, this shows
that $S$ is graded. Thus, $S\circ E_{H}=E_{H}\circ S$ (by
Proposition~\ref{Proposition704}, applied to $V=H$, $W=H$ and $f=S$). In other
words, $E_{H}\circ S=S\circ E_{H}$. Also, clearly, $S\circ S=S\circ S$. Thus,
we can apply Theorem~\ref{Theorem945} to $f=E_{H}$ and $g=S$. As a
consequence, we obtain $S\circ\left(  E_{H}\ast S\right)  =\left(  S\ast
E_{H}\right)  \circ S$. This proves Theorem~\ref{Theorem944} \textbf{(a)}.

But we can also apply Theorem~\ref{Theorem945} to $f=S$ and $g=E_{H}$. As a
consequence, we obtain $S\circ\left(  S\ast E_{H}\right)  =\left(  E_{H}\ast
S\right)  \circ S$. This proves Theorem~\ref{Theorem944} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem941}.]\textbf{(a)} Let $x\in H$. Then,%
\[
\left(  S\ast E_{H}\right)  \underbrace{\left(  S\left(  x\right)  +x\right)
}_{\substack{\in H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\\\text{(by Proposition~\ref{Proposition943}
\textbf{(b)})}}}\in\left(  S\ast E_{H}\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0
\]
(by Corollary~\ref{Corollary913} \textbf{(d)}). Thus, $\left(  S\ast
E_{H}\right)  \left(  S\left(  x\right)  +x\right)  =0$. In other words,%
\begin{align*}
0  &  =\left(  S\ast E_{H}\right)  \left(  S\left(  x\right)  +x\right)
=\underbrace{\left(  S\ast E_{H}\right)  \left(  S\left(  x\right)  \right)
}_{=\left(  \left(  S\ast E_{H}\right)  \circ S\right)  \left(  x\right)
}+\left(  S\ast E_{H}\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }S\ast E_{H}\text{ is }k\text{-linear}\right) \\
&  =\left(  \left(  S\ast E_{H}\right)  \circ S\right)  \left(  x\right)
+\left(  S\ast E_{H}\right)  \left(  x\right)  .
\end{align*}
Hence, $\left(  \left(  S\ast E_{H}\right)  \circ S\right)  \left(  x\right)
=-\left(  S\ast E_{H}\right)  \left(  x\right)  =\left(  -S\ast E_{H}\right)
\left(  x\right)  $.

Now, forget that we fixed $x$. We thus have proven that every $x\in H$
satisfies $\left(  \left(  S\ast E_{H}\right)  \circ S\right)  \left(
x\right)  =\left(  -S\ast E_{H}\right)  \left(  x\right)  $. In other words,
$\left(  S\ast E_{H}\right)  \circ S=-S\ast E_{H}$. Combined with
$S\circ\left(  E_{H}\ast S\right)  =\left(  S\ast E_{H}\right)  \circ S$
(which follows from Theorem~\ref{Theorem944} \textbf{(a)}), this yields
\[
S\circ\left(  E_{H}\ast S\right)  =\left(  S\ast E_{H}\right)  \circ S=-S\ast
E_{H}.
\]
This proves Theorem~\ref{Theorem941} \textbf{(a)}.

\textbf{(b)} Let $x\in H$. Then,%
\[
\left(  E_{H}\ast S\right)  \underbrace{\left(  S\left(  x\right)  +x\right)
}_{\substack{\in H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\\\text{(by Proposition~\ref{Proposition943}
\textbf{(b)})}}}\in\left(  E_{H}\ast S\right)  \left(  H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)  =0
\]
(by Corollary~\ref{Corollary913} \textbf{(a)}). Thus, $\left(  E_{H}\ast
S\right)  \left(  S\left(  x\right)  +x\right)  =0$. In other words,%
\begin{align*}
0  &  =\left(  E_{H}\ast S\right)  \left(  S\left(  x\right)  +x\right)
=\underbrace{\left(  E_{H}\ast S\right)  \left(  S\left(  x\right)  \right)
}_{=\left(  \left(  E_{H}\ast S\right)  \circ S\right)  \left(  x\right)
}+\left(  E_{H}\ast S\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }E_{H}\ast S\text{ is }k\text{-linear}\right) \\
&  =\left(  \left(  E_{H}\ast S\right)  \circ S\right)  \left(  x\right)
+\left(  E_{H}\ast S\right)  \left(  x\right)  .
\end{align*}
Hence, $\left(  \left(  E_{H}\ast S\right)  \circ S\right)  \left(  x\right)
=-\left(  E_{H}\ast S\right)  \left(  x\right)  =\left(  -E_{H}\ast S\right)
\left(  x\right)  $.

Now, forget that we fixed $x$. We thus have proven that every $x\in H$
satisfies $\left(  \left(  E_{H}\ast S\right)  \circ S\right)  \left(
x\right)  =\left(  -E_{H}\ast S\right)  \left(  x\right)  $. In other words,
$\left(  E_{H}\ast S\right)  \circ S=-E_{H}\ast S$. Combined with
$S\circ\left(  S\ast E_{H}\right)  =\left(  E_{H}\ast S\right)  \circ S$
(which follows from Theorem~\ref{Theorem944} \textbf{(b)}), this yields
\[
S\circ\left(  S\ast E_{H}\right)  =\left(  E_{H}\ast S\right)  \circ
S=-E_{H}\ast S.
\]
This proves Theorem~\ref{Theorem941} \textbf{(b)}.
\end{proof}

Note that we could have used this method (along with
Proposition~\ref{Proposition843}) to give an alternative proof of
Theorem~\ref{Theorem841} as well.

\section{\label{S27}Non-integer convolution powers and Dynkin idempotents}

In the following, we will introduce and study non-integer powers of linear
maps with respect to convolution. These powers are defined for any $k$-linear
map $f\in G\left(  C,A\right)  $, where $k$ is a field of characteristic $0$
(the condition that the characteristic of $k$ be $0$ cannot be dispensed
with!), $C$ is a connected filtered $k$-coalgebra and $A$ is a $k$-algebra.
The goal will be to obtain ``interpolations'' between the Dynkin idempotents
$E_{H}\ast S$ and $S\ast E_{H}$ found by Reutenauer and Procesi and mentioned
in the Remark in \S 3 of \cite{4}.

First of all, we define a generalization of the familiar notion of binomial coefficients.

\begin{definition}
\label{Definition1000} Let $k$ be a field of characteristic $0$. Let
$\gamma\in k$. Let $i\in\mathbb{N}$. Then, we define $\dbinom{\gamma}{i}$ to
denote the element $\dfrac{\gamma\left(  \gamma-1\right)  \cdots\left(
\gamma-i+1\right)  }{i!}$ of $k$.
\end{definition}

Of course, when $k=\mathbb{Q}$ and $\gamma\in\mathbb{Z}$, this definition
agrees with the standard definition of binomial coefficients.

Next, we define fractional powers of linear maps with respect to convolution:

\begin{definition}
\label{Definition1001} Let $k$ be a field of characteristic $0$. Let
$\gamma\in k$. Let $A$ be a $k$-algebra, and let $H$ be a connected filtered
$k$-coalgebra. For every $f\in\mathfrak{g}\left(  H,A\right)  $, let us define
a map $\operatorname*{Pow}\nolimits_{\gamma}f:H\rightarrow A$ by the formula%
\begin{equation}
\left(  \left(  \operatorname*{Pow}\nolimits_{\gamma}f\right)  \left(
x\right)  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}f^{\ast i}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in H\right)  \label{1001.Pow}%
\end{equation}
(where $\dbinom{\gamma}{i}$ is defined as in Definition~\ref{Definition1000}).
This map $\operatorname*{Pow}\nolimits_{\gamma}f$ is well-defined, because for
every $x\in H$ the infinite sum $\sum_{i\in\mathbb{N}}\dbinom{\gamma}%
{i}f^{\ast i}\left(  x\right)  $ converges with respect to the discrete
topology\footnote{\textit{Proof.} Let $x\in H$. Then, there exists some
$n\in\mathbb{N}$ such that $x\in H_{\leq n}$ (since $H$ is filtered). Consider
this $n$. Then, every integer $i>n$ satisfies $f^{\ast i}\left(  x\right)  =0$
(this is proven just as in Definition~\ref{Definition10}). Therefore, every
integer $i>n$ satisfies $\dbinom{\gamma}{i}\underbrace{f^{\ast i}\left(
x\right)  }_{=0}=0$. In other words, for every integer $i>n$, the $i$-th
addend of the infinite sum $\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}f^{\ast
i}\left(  x\right)  $ is zero. Hence, this infinite sum $\sum_{i\in\mathbb{N}%
}\dbinom{\gamma}{i}f^{\ast i}\left(  x\right)  $ has only finitely many
nonzero addends. Thus, this sum converges with respect to the discrete
topology, qed.}. Besides, $\operatorname*{Pow}\nolimits_{\gamma}f$ is a
$k$-linear map\footnote{\textit{Proof.} Let $\alpha\in k$, $\beta\in k$, $x\in
H$ and $y\in H$ be arbitrary. Then, (\ref{1001.Pow}) (applied to $y$ instead
of $x$) yields $\left(  \operatorname*{Pow}\nolimits_{\gamma}f\right)  \left(
y\right)  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}f^{\ast i}\left(  y\right)
$. But (\ref{1001.Pow}) (applied to $\alpha x+\beta y$ instead of $x$) yields%
\begin{align*}
\left(  \operatorname*{Pow}\nolimits_{\gamma}f\right)  \left(  \alpha x+\beta
y\right)   &  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\underbrace{f^{\ast
i}\left(  \alpha x+\beta y\right)  }_{\substack{=\alpha f^{\ast i}\left(
x\right)  +\beta f^{\ast i}\left(  y\right)  \\\text{(since }f^{\ast i}\text{
is a }k\text{-linear map)}}}=\sum_{i\in\mathbb{N}}\underbrace{\dbinom{\gamma
}{i}\left(  \alpha f^{\ast i}\left(  x\right)  +\beta f^{\ast i}\left(
y\right)  \right)  }_{=\alpha\dbinom{\gamma}{i}f^{\ast i}\left(  x\right)
+\beta\dbinom{\gamma}{i}f^{\ast i}\left(  y\right)  }\\
&  =\alpha\underbrace{\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}f^{\ast i}\left(
x\right)  }_{=\left(  \operatorname*{Pow}\nolimits_{\gamma}f\right)  \left(
x\right)  }+\beta\underbrace{\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}f^{\ast
i}\left(  y\right)  }_{=\left(  \operatorname*{Pow}\nolimits_{\gamma}f\right)
\left(  y\right)  }=\alpha\left(  \operatorname*{Pow}\nolimits_{\gamma
}f\right)  \left(  x\right)  +\beta\left(  \operatorname*{Pow}%
\nolimits_{\gamma}f\right)  \left(  y\right)  .
\end{align*}
Since this holds for all $\alpha\in k$, $\beta\in k$, $x\in H$ and $y\in H$,
we thus see that $\operatorname*{Pow}\nolimits_{\gamma}f$ is $k$-linear,
qed.}, so that $\operatorname*{Pow}\nolimits_{\gamma}f\in\mathcal{L}\left(
H,A\right)  $. More precisely, $\operatorname*{Pow}\nolimits_{\gamma}f\in
G\left(  H,A\right)  $.\ \ \ \ \footnote{\textit{Proof.} Every integer $i>0$
satisfies $f^{\ast i}\left(  H_{\leq0}\right)  =0$ (by Remark~\ref{Remark9},
applied to $n=0$) and thus $f^{\ast i}\left(  \underbrace{1_{H}}_{\in
H_{\leq0}}\right)  \in f^{\ast i}\left(  H_{\leq0}\right)  =0$, so that
$f^{\ast i}\left(  1_{H}\right)  =0$. Also, $f^{\ast0}=e_{H,A}$ (where
$e_{H,A}$ is defined according to Definition~\ref{def.eHA}), so that
\begin{align*}
f^{\ast0}\left(  1_{H}\right)   &  =\underbrace{e_{H,A}}_{\substack{=\eta
_{A}\circ\varepsilon_{H}\\\text{(by the definition}\\\text{of }e_{H,A}%
\text{)}}}\left(  1_{H}\right)  =\left(  \eta_{A}\circ\varepsilon_{H}\right)
\left(  1_{H}\right)  =\eta_{A}\left(  \varepsilon_{H}\left(  1_{H}\right)
\right) \\
&  =\underbrace{\varepsilon_{H}\left(  1_{H}\right)  }%
_{\substack{=1\\\text{(by the axioms of a bialgebra,}\\\text{since }H\text{ is
a }k\text{-bialgebra)}}}\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\eta_{A}\right) \\
&  =1_{A}.
\end{align*}
\par
But applying (\ref{1001.Pow}) to $x=1_{H}$, we get%
\begin{align*}
\left(  \operatorname*{Pow}\nolimits_{\gamma}f\right)  \left(  1_{H}\right)
&  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}f^{\ast i}\left(  1_{H}\right)
=\underbrace{\dbinom{\gamma}{0}}_{=1}\underbrace{f^{\ast0}\left(
1_{H}\right)  }_{=1_{A}}+\sum_{i>0}\dbinom{\gamma}{i}\underbrace{f^{\ast
i}\left(  1_{H}\right)  }_{\substack{=0\\\text{(since }i>0\text{)}}}\\
&  =1_{A}+\underbrace{\sum_{i>0}\dbinom{\gamma}{i}0}_{=0}=1_{A}.
\end{align*}
Thus, $\operatorname*{Pow}\nolimits_{\gamma}f\in G\left(  H,A\right)  $ (by
the definition of $G\left(  H,A\right)  $).}
\end{definition}

\begin{definition}
\label{Definition1002} Let $k$ be a field of characteristic $0$. Let
$\gamma\in k$. Let $A$ be a $k$-algebra, and let $H$ be a connected filtered
$k$-coalgebra. For every $F\in G\left(  H,A\right)  $, let us define an
element $\operatorname*{pow}\nolimits_{\gamma}F\in G\left(  H,A\right)  $ by
$\operatorname*{pow}\nolimits_{\gamma}F=\operatorname*{Pow}\nolimits_{\gamma
}\left(  F-e_{H,A}\right)  $.\ \ \ \ \footnote{This is well-defined for the
following reason:
\par
From Definition~\ref{Definition1001}, we know that $\operatorname*{Pow}%
\nolimits_{\gamma}f\in G\left(  H,A\right)  $ for every $f\in\mathfrak{g}%
\left(  H,A\right)  $. Since%
\[
\underbrace{F}_{\in G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(
H,A\right)  }-e_{H,A}\in e_{H,A}+\mathfrak{g}\left(  H,A\right)
-e_{H,A}=\mathfrak{g}\left(  H,A\right)  ,
\]
we can apply this to $f=F-e_{H,A}$, and conclude that $\operatorname*{Pow}%
\nolimits_{\gamma}\left(  F-e_{H,A}\right)  \in G\left(  H,A\right)  $. Hence,
$\operatorname*{pow}\nolimits_{\gamma}F$ is well-defined, qed.}
\end{definition}

The notation $\operatorname*{pow}\nolimits_{\gamma}F$ is suggestive: it
generalizes the powers of $F$ with respect to convolution. In fact, we have:

\begin{theorem}
\label{Theorem1003} Let $k$ be a field of characteristic $0$. Let $A$ be a
$k$-algebra, and let $H$ be a connected filtered $k$-coalgebra. Let $F\in
G\left(  H,A\right)  $.

\textbf{(a)} The map $F$ is $\ast$-invertible.

\textbf{(b)} Every $n\in\mathbb{Z}$ satisfies $\operatorname*{pow}%
\nolimits_{n}F=F^{\ast n}$ (where the $n$ in $\operatorname*{pow}%
\nolimits_{n}F$ means the element $n\cdot1$ of $k$).
\end{theorem}

We will not prove Theorem~\ref{Theorem1003} directly, but obtain this as a
consequence of a stronger result:

\begin{theorem}
\label{Theorem1004} Let $k$ be a field of characteristic $0$. Let $\gamma\in
k$. Let $A$ be a $k$-algebra, and let $H$ be a connected filtered
$k$-coalgebra. For every $F\in G\left(  H,A\right)  $, we have
$\operatorname*{pow}\nolimits_{\gamma}F=e^{\ast\left(  \gamma
\operatorname*{Log}F\right)  }$.
\end{theorem}

The proof of this theorem will (like that of Proposition~\ref{Proposition13})
proceed by applying identities for formal power series. Just as we used
Theorem~\ref{Theorem14} to prove Proposition~\ref{Proposition13}, here we need
the following fact:

\begin{theorem}
\label{Theorem1005} Let $k$ be a field of characteristic $0$. Let $\gamma\in
k$.

Then, every power series $P\in k\left[  \left[  X\right]  \right]  $ whose
coefficient before $X^{0}$ is $0$ satisfies $\sum_{i\in\mathbb{N}}%
\dbinom{\gamma}{i}P^{i}=\exp\left(  \gamma\log\left(  1+P\right)  \right)  $.
(Here, $\dbinom{\gamma}{i}$ is defined as in Definition~\ref{Definition1000},
whereas $\log\left(  1+P\right)  $ and $\exp\left(  \gamma\left(  \log\left(
1+P\right)  \right)  \right)  $ are defined as in
Definition~\ref{def.exp-log.powser}.)
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1005}.]Here is something very easy to prove:
Every $i\in\mathbb{N}$ satisfies%
\begin{equation}
\dbinom{\gamma}{i+1}\cdot\left(  i+1\right)  =\dbinom{\gamma}{i}\cdot\left(
\gamma-i\right)  . \label{1005.pf.ident}%
\end{equation}
\footnote{\textit{Proof of (\ref{1005.pf.ident}):} Every $i\in\mathbb{N}$
satisfies%
\begin{align*}
&  \underbrace{\dbinom{\gamma}{i+1}}_{\substack{=\dfrac{\gamma\left(
\gamma-1\right)  \cdots\left(  \gamma-\left(  i+1\right)  +1\right)  }{\left(
i+1\right)  !}\\\text{(by the definition of }\dbinom{\gamma}{i}\text{)}}%
}\cdot\left(  i+1\right) \\
&  =\dfrac{\gamma\left(  \gamma-1\right)  \cdots\left(  \gamma-\left(
i+1\right)  +1\right)  }{\left(  i+1\right)  !}\cdot\left(  i+1\right)
=\left(  \gamma\left(  \gamma-1\right)  \cdots\underbrace{\left(
\gamma-\left(  i+1\right)  +1\right)  }_{=\gamma-i}\right)  \cdot
\underbrace{\dfrac{i+1}{\left(  i+1\right)  !}}_{\substack{=\dfrac
{i+1}{\left(  i+1\right)  i!}\\\text{(since }\left(  i+1\right)  !=\left(
i+1\right)  i!\text{)}}}\\
&  =\underbrace{\left(  \gamma\left(  \gamma-1\right)  \cdots\left(
\gamma-i\right)  \right)  }_{=\left(  \gamma\left(  \gamma-1\right)
\cdots\left(  \gamma-i+1\right)  \right)  \cdot\left(  \gamma-i\right)  }%
\cdot\underbrace{\dfrac{i+1}{\left(  i+1\right)  i!}}_{=\dfrac{1}{i!}}=\left(
\gamma\left(  \gamma-1\right)  \cdots\left(  \gamma-i+1\right)  \right)
\cdot\left(  \gamma-i\right)  \cdot\dfrac{1}{i!}\\
&  =\underbrace{\dfrac{\gamma\left(  \gamma-1\right)  \cdots\left(
\gamma-i+1\right)  }{i!}}_{\substack{=\dbinom{\gamma}{i}\\\text{(since
}\dbinom{\gamma}{i}=\dfrac{\gamma\left(  \gamma-1\right)  \cdots\left(
\gamma-i+1\right)  }{i!}\\\text{by the definition of }\dbinom{\gamma}%
{i}\text{)}}}\cdot\left(  \gamma-i\right)  =\dbinom{\gamma}{i}\cdot\left(
\gamma-i\right)  .
\end{align*}
This proves (\ref{1005.pf.ident}).}

Let $P\in k\left[  \left[  X\right]  \right]  $ be a power series whose
coefficient before $X^{0}$ is $0$. Write $P$ in the form $P=\sum
_{i\in\mathbb{N}}p_{i}X^{i}$, where $p_{0}$, $p_{1}$, $p_{2}$, $\ldots$ are
elements of $k$. Then, $p_{0}$ is the coefficient of $P$ before $X^{0}$, and
thus equals $0$ (since the coefficient of $P$ before $X^{0}$ is $0$). We have
thus shown that $p_{0}=0$. Now,%
\begin{align*}
P  &  =\sum_{i\in\mathbb{N}}p_{i}X^{i}=\underbrace{p_{0}}_{=0}X^{0}%
+\sum_{\substack{i\in\mathbb{N};\\i\geq1}}p_{i}\underbrace{X^{i}}_{=X^{i-1}%
X}=\underbrace{0X^{0}}_{=0}+\sum_{\substack{i\in\mathbb{N};\\i\geq1}%
}p_{i}X^{i-1}X\\
&  =\sum_{\substack{i\in\mathbb{N};\\i\geq1}}p_{i}X^{i-1}X=\left(
\sum_{\substack{i\in\mathbb{N};\\i\geq1}}p_{i}X^{i-1}\right)  X.
\end{align*}
Hence, $X\mid P$ in $k\left[  \left[  X\right]  \right]  $. Thus, every
$i\in\mathbb{N}$ satisfies $X^{i}\mid P^{i}$. Hence, every positive
$i\in\mathbb{N}$ satisfies $X\mid P^{i}$ (since $X\mid X^{i}$ (because $i$ is
positive) and $X^{i}\mid P^{i}$). In other words,%
\begin{equation}
\text{every positive }i\in\mathbb{N}\text{ satisfies }P^{i}\equiv
0\operatorname{mod}Xk\left[  \left[  X\right]  \right]  . \label{1005.pf.1}%
\end{equation}


Now,%
\begin{align*}
\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}  &  =\underbrace{\dbinom{\gamma
}{0}}_{=1}\underbrace{P^{0}}_{=1}+\sum_{i>0}\dbinom{\gamma}{i}%
\underbrace{P^{i}}_{\substack{\equiv0\operatorname{mod}Xk\left[  \left[
X\right]  \right]  \\\text{(by (\ref{1005.pf.1}), since}\\i\text{ is
positive)}}}\\
&  \equiv1+\underbrace{\sum_{i>0}\dbinom{\gamma}{i}0}_{=0}=1\operatorname{mod}%
Xk\left[  \left[  X\right]  \right]  .
\end{align*}
In other words, $X\mid\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}-1$ in
$k\left[  \left[  X\right]  \right]  $. This means that the coefficient of the
power series $\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}-1$ before $X^{0}$
is $0$. In other words,%
\begin{align*}
0  &  =\left(  \text{the coefficient of the power series }\sum_{i\in
\mathbb{N}}\dbinom{\gamma}{i}P^{i}-1\text{ before }X^{0}\right) \\
&  =\left(  \text{the coefficient of the power series }\sum_{i\in\mathbb{N}%
}\dbinom{\gamma}{i}P^{i}\text{ before }X^{0}\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\left(  \text{the coefficient of the power
series }1\text{ before }X^{0}\right)  }_{=1}\\
&  =\left(  \text{the coefficient of the power series }\sum_{i\in\mathbb{N}%
}\dbinom{\gamma}{i}P^{i}\text{ before }X^{0}\right)  -1.
\end{align*}
In other words, the coefficient of the power series $\sum_{i\in\mathbb{N}%
}\dbinom{\gamma}{i}P^{i}$ before $X^{0}$ equals $1$. Hence, $\sum
_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}$ is a power series whose coefficient
before $X^{0}$ is $1$. This yields that the power series $\log\left(
\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  \in k\left[  \left[
X\right]  \right]  $ is well-defined.

Since $\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}$ is a power series whose
coefficient before $X^{0}$ is $1$, we can apply
Proposition~\ref{prop.deriv-log} to $Q=\sum_{i\in\mathbb{N}}\dbinom{\gamma}%
{i}P^{i}$. Thus, Proposition~\ref{prop.deriv-log} \textbf{(a)} (applied to
$Q=\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}$) shows that the power series
$\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}$ has a multiplicative inverse
$\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  ^{-1}$.
Furthermore, Proposition~\ref{prop.deriv-log} \textbf{(a)} (applied to
$Q=\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}$) shows that it satisfies%
\begin{equation}
\dfrac{d}{dX}\log\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)
=\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  ^{-1}\cdot
\dfrac{d}{dX}\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  .
\label{1005.pf.2}%
\end{equation}


But%
\begin{align}
\dfrac{d}{dX}\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)   &
=\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\dfrac{d}{dX}\left(  P^{i}\right)
=\underbrace{\dbinom{\gamma}{0}}_{=0}\underbrace{\dfrac{d}{dX}\left(
P^{0}\right)  }_{\substack{=0\\\text{(since }P^{0}=1\text{)}}}+\sum
_{i>0}\dbinom{\gamma}{i}\underbrace{\dfrac{d}{dX}\left(  P^{i}\right)
}_{\substack{=iP^{i-1}\cdot\dfrac{d}{dX}P\\\text{(by
Proposition~\ref{prop.deriv-leibnizPn}}\\\text{(applied to }i\text{ instead of
}n\text{))}}}\nonumber\\
&  =\sum_{i>0}\dbinom{\gamma}{i}\cdot iP^{i-1}\cdot\left(  \dfrac{d}%
{dX}P\right) \label{1005.pf.5}\\
&  =\sum_{i\in\mathbb{N}}\underbrace{\dbinom{\gamma}{i+1}\cdot\left(
i+1\right)  }_{\substack{=\dbinom{\gamma}{i}\cdot\left(  \gamma-i\right)
\\\text{(by (\ref{1005.pf.ident}))}}}\underbrace{P^{\left(  i+1\right)  -1}%
}_{=P^{i}}\cdot\left(  \dfrac{d}{dX}P\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i+1\text{ for
}i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot\left(  \gamma-i\right)
P^{i}\cdot\left(  \dfrac{d}{dX}P\right)  , \label{1005.pf.6}%
\end{align}
and%
\begin{align}
&  \left(  \dfrac{d}{dX}\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}%
P^{i}\right)  \right)  \cdot\left(  1+P\right) \nonumber\\
&  =\underbrace{\dfrac{d}{dX}\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}%
{i}P^{i}\right)  }_{\substack{=\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}%
\cdot\left(  \gamma-i\right)  P^{i}\cdot\left(  \dfrac{d}{dX}P\right)
\\\text{(by (\ref{1005.pf.6}))}}}+\underbrace{\left(  \dfrac{d}{dX}\left(
\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  \right)  }%
_{\substack{=\sum_{i>0}\dbinom{\gamma}{i}\cdot iP^{i-1}\cdot\left(  \dfrac
{d}{dX}P\right)  \\\text{(by (\ref{1005.pf.5}))}}}\cdot P\nonumber\\
&  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot\left(  \gamma-i\right)
P^{i}\cdot\left(  \dfrac{d}{dX}P\right)  +\underbrace{\left(  \sum
_{i>0}\dbinom{\gamma}{i}\cdot iP^{i-1}\cdot\left(  \dfrac{d}{dX}P\right)
\right)  \cdot P}_{=\sum_{i>0}\dbinom{\gamma}{i}\cdot iP^{i-1}P\cdot\left(
\dfrac{d}{dX}P\right)  }\nonumber\\
&  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot\left(  \gamma-i\right)
P^{i}\cdot\left(  \dfrac{d}{dX}P\right)  +\sum_{i>0}\dbinom{\gamma}{i}\cdot
i\underbrace{P^{i-1}P}_{=P^{i}}\cdot\left(  \dfrac{d}{dX}P\right) \nonumber\\
&  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot\left(  \gamma-i\right)
P^{i}\cdot\left(  \dfrac{d}{dX}P\right)  +\sum_{i>0}\dbinom{\gamma}{i}\cdot
iP^{i}\cdot\left(  \dfrac{d}{dX}P\right)  . \label{1005.pf.7}%
\end{align}


Since
\begin{align}
\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot iP^{i}\cdot\left(  \dfrac{d}%
{dX}P\right)   &  =\underbrace{\dbinom{\gamma}{0}\cdot0P^{0}\cdot\left(
\dfrac{d}{dX}P\right)  }_{=0}+\sum_{i>0}\dbinom{\gamma}{i}\cdot iP^{i}%
\cdot\left(  \dfrac{d}{dX}P\right) \nonumber\\
&  =\sum_{i>0}\dbinom{\gamma}{i}\cdot iP^{i}\cdot\left(  \dfrac{d}%
{dX}P\right)  , \label{1005.pf.8}%
\end{align}
the equality (\ref{1005.pf.7}) becomes%
\begin{align}
&  \left(  \dfrac{d}{dX}\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}%
P^{i}\right)  \right)  \cdot\left(  1+P\right) \nonumber\\
&  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot\left(  \gamma-i\right)
P^{i}\cdot\left(  \dfrac{d}{dX}P\right)  +\underbrace{\sum_{i>0}\dbinom
{\gamma}{i}\cdot iP^{i}\cdot\left(  \dfrac{d}{dX}P\right)  }_{\substack{=\sum
_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot iP^{i}\cdot\left(  \dfrac{d}%
{dX}P\right)  \\\text{(by (\ref{1005.pf.8}))}}}\nonumber\\
&  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot\left(  \gamma-i\right)
P^{i}\cdot\left(  \dfrac{d}{dX}P\right)  +\sum_{i\in\mathbb{N}}\dbinom{\gamma
}{i}\cdot iP^{i}\cdot\left(  \dfrac{d}{dX}P\right) \nonumber\\
&  =\sum_{i\in\mathbb{N}}\underbrace{\left(  \dbinom{\gamma}{i}\cdot\left(
\gamma-i\right)  P^{i}\cdot\left(  \dfrac{d}{dX}P\right)  +\dbinom{\gamma}%
{i}\cdot iP^{i}\cdot\left(  \dfrac{d}{dX}P\right)  \right)  }_{=\dbinom
{\gamma}{i}\cdot\left(  \left(  \gamma-i\right)  +i\right)  P^{i}\cdot\left(
\dfrac{d}{dX}P\right)  }\nonumber\\
&  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot\underbrace{\left(  \left(
\gamma-i\right)  +i\right)  }_{=\gamma}P^{i}\cdot\left(  \dfrac{d}%
{dX}P\right)  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}\cdot\gamma P^{i}%
\cdot\left(  \dfrac{d}{dX}P\right) \nonumber\\
&  =\gamma\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)
\cdot\left(  \dfrac{d}{dX}P\right)  . \label{1005.pf.10}%
\end{align}


On the other hand,%
\begin{align*}
&  \left(  \text{the coefficient of the power series }1+P\text{ before }%
X^{0}\right) \\
&  =\underbrace{\left(  \text{the coefficient of the power series }1\text{
before }X^{0}\right)  }_{=1}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the coefficient of the power
series }P\text{ before }X^{0}\right)  }_{=0}\\
&  =1.
\end{align*}
In other words, the coefficient of the power series $1+P$ before $X^{0}$
equals $1$. Hence, $1+P$ is a power series whose coefficient before $X^{0}$ is
$1$. This yields that the power series $\log\left(  1+P\right)  \in k\left[
\left[  X\right]  \right]  $ is well-defined.

Since $1+P$ is a power series whose coefficient before $X^{0}$ is $1$, we can
apply by Proposition~\ref{prop.deriv-log} to $Q=1+P$. Thus,
Proposition~\ref{prop.deriv-log} \textbf{(a)} (applied to $Q=1+P$) shows that
the power series $1+P$ has a multiplicative inverse $\left(  1+P\right)
^{-1}$. Furthermore, Proposition~\ref{prop.deriv-log} \textbf{(b)} (applied to
$Q=1+P$) shows that it satisfies%
\begin{align}
\dfrac{d}{dX}\left(  \log\left(  1+P\right)  \right)   &  =\left(  1+P\right)
^{-1}\cdot\underbrace{\dfrac{d}{dX}\left(  1+P\right)  }_{=\dfrac{d}%
{dX}1+\dfrac{d}{dX}P}=\left(  1+P\right)  ^{-1}\cdot\left(  \underbrace{\dfrac
{d}{dX}1}_{=0}+\dfrac{d}{dX}P\right) \nonumber\\
&  =\left(  1+P\right)  ^{-1}\cdot\dfrac{d}{dX}P. \label{1005.pf.16}%
\end{align}


Since $1+P$ has a multiplicative inverse, we can divide the equality
(\ref{1005.pf.10}) by $1+P$, and obtain%
\begin{equation}
\dfrac{d}{dX}\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)
=\left(  1+P\right)  ^{-1}\cdot\gamma\left(  \sum_{i\in\mathbb{N}}%
\dbinom{\gamma}{i}P^{i}\right)  \cdot\left(  \dfrac{d}{dX}P\right)  .
\label{1005.pf.20}%
\end{equation}
Now, (\ref{1005.pf.2}) becomes%
\begin{align}
\dfrac{d}{dX}\log\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)
&  =\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  ^{-1}%
\cdot\underbrace{\dfrac{d}{dX}\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}%
{i}P^{i}\right)  }_{\substack{=\left(  1+P\right)  ^{-1}\cdot\gamma\left(
\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  \cdot\left(  \dfrac
{d}{dX}P\right)  \\\text{(by (\ref{1005.pf.20}))}}}\nonumber\\
&  =\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  ^{-1}%
\cdot\left(  1+P\right)  ^{-1}\cdot\gamma\left(  \sum_{i\in\mathbb{N}}%
\dbinom{\gamma}{i}P^{i}\right)  \cdot\left(  \dfrac{d}{dX}P\right) \nonumber\\
&  =\gamma\cdot\underbrace{\left(  1+P\right)  ^{-1}\cdot\dfrac{d}{dX}%
P}_{\substack{=\dfrac{d}{dX}\left(  \log\left(  1+P\right)  \right)
\\\text{(by (\ref{1005.pf.16}))}}}=\gamma\cdot\dfrac{d}{dX}\left(  \log\left(
1+P\right)  \right) \nonumber\\
&  =\dfrac{d}{dX}\left(  \gamma\log\left(  1+P\right)  \right)  .
\label{1005.pf.22}%
\end{align}


Since $1+P$ is a power series whose coefficient before $X^{0}$ is $1$, we can
apply Proposition~\ref{Proposition1009} \textbf{(b)} to $Q=1+P$, and conclude
that $\log\left(  1+P\right)  $ is a power series whose coefficient before
$X^{0}$ is $0$. In other words,%
\[
\left(  \text{the coefficient of }\log\left(  1+P\right)  \text{ before }%
X^{0}\right)  =0.
\]
Thus,%
\begin{align*}
&  \left(  \text{the coefficient of }\gamma\log\left(  1+P\right)  \text{
before }X^{0}\right) \\
&  =\gamma\cdot\underbrace{\left(  \text{the coefficient of }\log\left(
1+P\right)  \text{ before }X^{0}\right)  }_{=0}=0.
\end{align*}


Since $\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}$ is a power series whose
coefficient before $X^{0}$ is $1$, we can apply
Proposition~\ref{Proposition1009} \textbf{(b)} to $Q=\sum_{i\in\mathbb{N}%
}\dbinom{\gamma}{i}P^{i}$, and conclude that $\log\left(  \sum_{i\in
\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  $ is a power series whose
coefficient before $X^{0}$ is $0$. In other words,%
\[
\left(  \text{the coefficient of }\log\left(  \sum_{i\in\mathbb{N}}%
\dbinom{\gamma}{i}P^{i}\right)  \text{ before }X^{0}\right)  =0.
\]
Thus,%
\begin{align*}
&  \left(  \text{the coefficient of }\log\left(  \sum_{i\in\mathbb{N}}%
\dbinom{\gamma}{i}P^{i}\right)  \text{ before }X^{0}\right) \\
&  =0=\left(  \text{the coefficient of }\gamma\log\left(  1+P\right)  \text{
before }X^{0}\right)  .
\end{align*}
In other words, the coefficient of $\log\left(  \sum_{i\in\mathbb{N}}%
\dbinom{\gamma}{i}P^{i}\right)  $ before $X^{0}$ equals the coefficient of
$\gamma\log\left(  1+P\right)  $ before $X^{0}$. Combined with
(\ref{1005.pf.22}), this yields that we can apply
Proposition~\ref{prop.deriv-uniq-det} to $U=\log\left(  \sum_{i\in\mathbb{N}%
}\dbinom{\gamma}{i}P^{i}\right)  $ and $V=\gamma\log\left(  1+P\right)  $. As
a result, we obtain%
\begin{equation}
\log\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  =\gamma
\log\left(  1+P\right)  . \label{1005.pf.40}%
\end{equation}


But since $\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}$ is a power series
whose coefficient before $X^{0}$ is $1$, we can apply Theorem~\ref{Theorem14}
\textbf{(b)} to $Q=\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}$. As a result,
we obtain%
\[
\exp\left(  \log\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)
\right)  =\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}.
\]
Thus,%
\[
\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}=\exp\left(  \underbrace{\log
\left(  \sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}P^{i}\right)  }%
_{\substack{=\gamma\log\left(  1+P\right)  \\\text{(by (\ref{1005.pf.40}))}%
}}\right)  =\exp\left(  \gamma\log\left(  1+P\right)  \right)  .
\]
This proves Theorem~\ref{Theorem1005}.
\end{proof}

Just as Corollary~\ref{Corollary15} was derived from Theorem~\ref{Theorem14},
we can derive from Theorem~\ref{Theorem1005} the following corollary:

\begin{corollary}
\label{Corollary1021} Let $k$ be a field of characteristic $0$. Let
$n\in\mathbb{N}$. Let $\gamma\in k$.

Let $b$ be an element of a $k$-algebra such that $b^{n+1}=0$. Then,%
\[
\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}b^{i}\right)  ^{j}}{j!}=\sum_{i=0}^{n}\dbinom{\gamma}{i}b^{i}%
\]
(where $\dbinom{\gamma}{i}$ is defined as in Definition~\ref{Definition1000}).
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1021}.]Consider the ring of power series
$k\left[  \left[  X\right]  \right]  $. Clearly, $X$ is a power series whose
coefficient before $X^{0}$ is $0$. Thus, applying Theorem~\ref{Theorem1005} to
$P=X$, we obtain $\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}X^{i}=\exp\left(
\gamma\log\left(  1+X\right)  \right)  $. Now, in the ring $k\left[  \left[
X\right]  \right]  $, we have%
\begin{align*}
\log\left(  1+X\right)   &  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  \underbrace{\left(  1+X\right)  -1}_{=X}\right)  ^{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\log\right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}=\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}+\sum_{i=n+1}^{\infty
}\underbrace{\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}}_{\substack{\equiv
0\operatorname{mod}X^{n+1}k\left[  \left[  X\right]  \right]  \\\text{(since
}i\geq n+1\text{ leads to}\\X^{i}\equiv0\operatorname{mod}X^{n+1}k\left[
\left[  X\right]  \right]  \text{)}}}\\
&  \equiv\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}%
+\underbrace{\sum_{i=n+1}^{\infty}0}_{=0}=\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}X^{i}\operatorname{mod}X^{n+1}k\left[  \left[  X\right]
\right]  .
\end{align*}


On the other hand, since $X\mid\log\left(  1+X\right)  $ in $k\left[  \left[
X\right]  \right]  $ (because $\log\left(  1+X\right)  =\sum_{i\geq1}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{X^{i}}_{=XX^{i-1}}%
=X\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i-1}$ is divisible by
$X$), we have $X\mid\gamma\log\left(  1+X\right)  $ in $k\left[  \left[
X\right]  \right]  $. Thus, $X^{j}\mid\left(  \gamma\log\left(  1+X\right)
\right)  ^{j}$ for every $j\in\mathbb{N}$. Thus every $j\in\mathbb{N}$ such
that $j\geq n+1$ satisfies
\begin{equation}
\left(  \gamma\log\left(  1+X\right)  \right)  ^{j}\equiv0\operatorname{mod}%
X^{n+1}k\left[  \left[  X\right]  \right]  \label{1021.pf.3}%
\end{equation}
(since $j\geq n+1$ leads to $X^{n+1}\mid X^{j}\mid\left(  \gamma\log\left(
1+X\right)  \right)  ^{j}$). Now,%

\begin{align}
\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}X^{i}  &  =\exp\left(  \gamma
\log\left(  1+X\right)  \right)  =\sum_{j=0}^{\infty}\dfrac{\left(  \gamma
\log\left(  1+X\right)  \right)  ^{j}}{j!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\exp\right) \nonumber\\
&  =\sum_{j=0}^{n}\dfrac{\left(  \gamma\log\left(  1+X\right)  \right)  ^{j}%
}{j!}+\sum_{j=n+1}^{\infty}\underbrace{\dfrac{\left(  \gamma\log\left(
1+X\right)  \right)  ^{j}}{j!}}_{\substack{\equiv0\operatorname{mod}%
X^{n+1}k\left[  \left[  X\right]  \right]  \\\text{(since }j\geq n+1\text{ and
thus}\\\left(  \gamma\log\left(  1+X\right)  \right)  ^{j}\equiv
0\operatorname{mod}X^{n+1}k\left[  \left[  X\right]  \right]  \\\text{(by
(\ref{1021.pf.3})))}}}\nonumber\\
&  \equiv\sum_{j=0}^{n}\dfrac{\left(  \gamma\log\left(  1+X\right)  \right)
^{j}}{j!}+\underbrace{\sum_{j=n+1}^{\infty}0}_{=0}=\sum_{j=0}^{n}%
\dfrac{\left(  \gamma\log\left(  1+X\right)  \right)  ^{j}}{j!}\nonumber\\
&  \equiv\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}\operatorname{mod}X^{n+1}k\left[
\left[  X\right]  \right]  \label{1021.pf.4}%
\end{align}
(since $\log\left(  1+X\right)  \equiv\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}X^{i}\operatorname{mod}X^{n+1}k\left[  \left[  X\right]  \right]
$). On the other hand,%
\begin{align*}
\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}X^{i}  &  =\sum_{i=0}^{n}\dbinom
{\gamma}{i}X^{i}+\sum_{i=n+1}^{\infty}\dbinom{\gamma}{i}\underbrace{X^{i}%
}_{\substack{\equiv0\operatorname{mod}X^{n+1}k\left[  \left[  X\right]
\right]  \\\text{(since }i\geq n+1\text{, and thus }X^{n+1}\mid X^{i}\text{)}%
}}\\
&  \equiv\sum_{i=0}^{n}\dbinom{\gamma}{i}X^{i}+\underbrace{\sum_{i=n+1}%
^{\infty}\dbinom{\gamma}{i}0}_{=0}=\sum_{i=0}^{n}\dbinom{\gamma}{i}%
X^{i}\operatorname{mod}X^{n+1}k\left[  \left[  X\right]  \right]  ,
\end{align*}
so that%
\[
\sum_{i=0}^{n}\dbinom{\gamma}{i}X^{i}\equiv\sum_{i\in\mathbb{N}}\dbinom
{\gamma}{i}X^{i}\equiv\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}\operatorname{mod}%
X^{n+1}k\left[  \left[  X\right]  \right]
\]
(by (\ref{1021.pf.4})). Thus, $X^{n+1}\mid\sum_{i=0}^{n}\dbinom{\gamma}%
{i}X^{i}-\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$ in $k\left[  \left[  X\right]
\right]  $. This means that the coefficient of the power series $\sum
_{i=0}^{n}\dbinom{\gamma}{i}X^{i}-\sum_{j=0}^{n}\dfrac{\left(  \gamma
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$
before $X^{\lambda}$ is $0$ for every $\lambda\in\left\{  0,1,\ldots
,n\right\}  $. But the power series $\sum_{i=0}^{n}\dbinom{\gamma}{i}%
X^{i}-\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$ is actually a polynomial, so
this rewrites as follows: The coefficient of the polynomial $\sum_{i=0}%
^{n}\dbinom{\gamma}{i}X^{i}-\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}$ before
$X^{\lambda}$ is $0$ for every $\lambda\in\left\{  0,1,\ldots,n\right\}  $. In
other words, $X^{n+1}\mid\sum_{i=0}^{n}\dbinom{\gamma}{i}X^{i}-\sum_{j=0}%
^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}X^{i}\right)  ^{j}}{j!}$ in $k\left[  X\right]  $. Hence, there exists a
polynomial $\mathbf{Q}\in k\left[  X\right]  $ such that $\sum_{i=0}%
^{n}\dbinom{\gamma}{i}X^{i}-\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}X^{i}\right)  ^{j}}{j!}=X^{n+1}%
\mathbf{Q}$. Consider this polynomial $\mathbf{Q}$.

Applying the polynomial identity $\sum_{i=0}^{n}\dbinom{\gamma}{i}X^{i}%
-\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}X^{i}\right)  ^{j}}{j!}=X^{n+1}\mathbf{Q}$ to $b$ instead of $X$, we
get%
\[
\sum_{i=0}^{n}\dbinom{\gamma}{i}b^{i}-\sum_{j=0}^{n}\dfrac{\left(  \gamma
\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}b^{i}\right)  ^{j}}%
{j!}=\underbrace{b^{n+1}}_{=0}\mathbf{Q}\left(  b\right)  =0,
\]
so that $\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}b^{i}\right)  ^{j}}{j!}=\sum_{i=0}^{n}\dbinom{\gamma}%
{i}b^{i}$. This proves Corollary~\ref{Corollary1021}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem1004}.]Due to how we defined $\mathcal{L}%
^{n+1}\left(  H,A\right)  $ (in Definition~\ref{Definition8} \textbf{(b)}), we
have%
\begin{align}
\mathcal{L}^{n+1}\left(  H,A\right)   &  =\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ \underbrace{f\mid_{H_{\leq n+1-1}}}_{=f\mid_{H_{\leq n}}%
}=0\right\}  =\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid
\ f\mid_{H_{\leq n}}=0\right\} \nonumber\\
&  =\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid_{H_{\leq n}%
}=0\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }f\text{ as
}h\right)  \label{1004.pf.0}%
\end{align}
for every $n\in\mathbb{N}$.

Let $F\in G\left(  H,A\right)  $. Let $n\in\mathbb{N}$.

According to Proposition~\ref{Proposition81} (applied to $n+1$ instead of
$n$), the subset $\mathcal{L}^{n+1}\left(  H,A\right)  $ of $\mathcal{L}%
\left(  H,A\right)  $ is an ideal of $\mathcal{L}\left(  H,A\right)  $. Thus,
there is a factor algebra $\mathcal{L}\left(  H,A\right)  \diagup
\mathcal{L}^{n+1}\left(  H,A\right)  $. For every $p\in\mathcal{L}\left(
H,A\right)  $, we are going to denote by $\overline{p}$ the projection of $p$
to this factor algebra (i. e., the residue class of $p$ modulo the ideal
$\mathcal{L}^{n+1}\left(  H,A\right)  $). We are going to denote the
multiplication in the factor algebra $\mathcal{L}\left(  H,A\right)
\diagup\mathcal{L}^{n+1}\left(  H,A\right)  $ by the $\ast$ sign, and we are
going to write $q^{\ast i}$ for the $i$-th power of $q$ (in the factor algebra
$\mathcal{L}\left(  H,A\right)  \diagup\mathcal{L}^{n+1}\left(  H,A\right)  $)
whenever $q\in\mathcal{L}\left(  H,A\right)  \diagup\mathcal{L}^{n+1}\left(
H,A\right)  $ and $i\in\mathbb{N}$.

Let $g=F-e_{H,A}$. Then, $g\in\mathfrak{g}\left(  H,A\right)  $ (since $F\in
G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(  H,A\right)  $, so that
$F-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  $). Hence, Remark~\ref{Remark9}
(applied to $g$ instead of $f$) yields $g^{\ast i}\left(  H_{\leq n}\right)
=0$ for every $i>n$.

Let $\varphi=\operatorname*{Log}F$. Then, $\varphi=\operatorname*{Log}%
F\in\mathfrak{g}\left(  H,A\right)  $, so that $\gamma\varphi\in
\gamma\mathfrak{g}\left(  H,A\right)  \subseteq\mathfrak{g}\left(  H,A\right)
$ (because $\mathfrak{g}\left(  H,A\right)  $ is a $k$-vector space). Thus,
Remark~\ref{Remark9} (applied to $\gamma\varphi$ instead of $f$) yields
$\left(  \gamma\varphi\right)  ^{\ast i}\left(  H_{\leq n}\right)  =0$ for
every $i>n$.

By the definition of $\operatorname*{Log}$, we have $\operatorname*{Log}%
F=\operatorname{Log}_{1}\underbrace{\left(  F-e_{H,A}\right)  }_{=g}%
=\operatorname{Log}_{1}g$. Hence, $\varphi=\operatorname*{Log}%
f=\operatorname{Log}_{1}g$.

Thus, every $x\in H_{\leq n}$ satisfies%
\begin{align*}
\varphi\left(  x\right)   &  =\left(  \operatorname{Log}_{1}g\right)  \left(
x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname{Log}_{1}\right) \\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  +\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\underbrace{g^{\ast i}\left(  x\right)  }_{\substack{=0\text{ (since
}\\x\in H_{\leq n}\text{ and thus}\\g^{\ast i}\left(  x\right)  \in g^{\ast
i}\left(  H_{\leq n}\right)  =0\\\text{(since }i>n\text{))}}}=\sum_{i=1}%
^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(  x\right)
+\underbrace{\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}}{i}0}%
_{=0}\\
&  =\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\left(
x\right)  .
\end{align*}
In other words,%
\[
\varphi\mid_{H_{\leq n}}=\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}.
\]
Now,%
\begin{align*}
\left(  \varphi-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\right)  \mid_{H_{\leq n}}  &  =\underbrace{\varphi\mid_{H_{\leq n}}%
}_{=\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\right)  \mid_{H_{\leq n}}}-\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}\\
&  =\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\right)  \mid_{H_{\leq n}}-\left(  \sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}=0.
\end{align*}
In other words,%
\[
\varphi-\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}%
\in\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid_{H_{\leq n}%
}=0\right\}  =\mathcal{L}^{n+1}\left(  H,A\right)
\]
(by (\ref{1004.pf.0})). This rewrites as%
\begin{equation}
\varphi\equiv\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast
i}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  . \label{1004.pf.6}%
\end{equation}


But every $x\in H_{\leq n}$ satisfies%
\begin{align*}
e^{\ast\left(  \gamma\varphi\right)  }\left(  x\right)   &  =\sum_{i\geq
0}\dfrac{\left(  \gamma\varphi\right)  ^{\ast i}\left(  x\right)  }%
{i!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }e^{\ast\left(
\gamma\varphi\right)  }\right) \\
&  =\sum_{i=0}^{n}\dfrac{\left(  \gamma\varphi\right)  ^{\ast i}\left(
x\right)  }{i!}+\sum_{i=n+1}^{\infty}\underbrace{\dfrac{\left(  \gamma
\varphi\right)  ^{\ast i}\left(  x\right)  }{i!}}_{\substack{=0\text{ (since
}\\x\in H_{\leq n}\text{ and thus}\\\left(  \gamma\varphi\right)  ^{\ast
i}\left(  x\right)  \in\left(  \gamma\varphi\right)  ^{\ast i}\left(  H_{\leq
n}\right)  =0\\\text{(since }i>n\text{), so that}\\\left(  \gamma
\varphi\right)  ^{\ast i}\left(  x\right)  =0\text{)}}}=\sum_{i=0}^{n}%
\dfrac{\left(  \gamma\varphi\right)  ^{\ast i}\left(  x\right)  }%
{i!}+\underbrace{\sum_{i=n+1}^{\infty}0}_{=0}=\sum_{i=0}^{n}\dfrac{\left(
\gamma\varphi\right)  ^{\ast i}\left(  x\right)  }{i!}\\
&  =\sum_{i=0}^{n}\dfrac{\left(  \gamma\varphi\right)  ^{\ast i}}{i!}\left(
x\right)  =\sum_{j=0}^{n}\dfrac{\left(  \gamma\varphi\right)  ^{\ast j}}%
{j!}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted
}j\text{ for }i\text{ in the sum}\right)  .
\end{align*}
In other words,
\[
e^{\ast\left(  \gamma\varphi\right)  }\mid_{H_{\leq n}}=\left(  \sum_{j=0}%
^{n}\dfrac{\left(  \gamma\varphi\right)  ^{\ast j}}{j!}\right)  \mid_{H_{\leq
n}}.
\]
But now,%
\begin{align*}
&  \left(  e^{\ast\left(  \gamma\varphi\right)  }-\sum_{j=0}^{n}\dfrac{\left(
\gamma\varphi\right)  ^{\ast j}}{j!}\right)  \mid_{H_{\leq n}}\\
&  =\underbrace{e^{\ast\left(  \gamma\varphi\right)  }\mid_{H_{\leq n}}%
}_{=\left(  \sum_{j=0}^{n}\dfrac{\left(  \gamma\varphi\right)  ^{\ast j}}%
{j!}\right)  \mid_{H_{\leq n}}}-\left(  \sum_{j=0}^{n}\dfrac{\left(
\gamma\varphi\right)  ^{\ast j}}{j!}\right)  \mid_{H_{\leq n}}\\
&  =\left(  \sum_{j=0}^{n}\dfrac{\left(  \gamma\varphi\right)  ^{\ast j}}%
{j!}\right)  \mid_{H_{\leq n}}-\left(  \sum_{j=0}^{n}\dfrac{\left(
\gamma\varphi\right)  ^{\ast j}}{j!}\right)  \mid_{H_{\leq n}}=0.
\end{align*}
In other words,%
\[
e^{\ast\left(  \gamma\varphi\right)  }-\sum_{j=0}^{n}\dfrac{\left(
\gamma\varphi\right)  ^{\ast j}}{j!}\in\left\{  h\in\mathcal{L}\left(
H,A\right)  \ \mid\ h\mid_{H_{\leq n}}=0\right\}  =\mathcal{L}^{n+1}\left(
H,A\right)
\]
(by (\ref{1004.pf.0})). This rewrites as%
\[
e^{\ast\left(  \gamma\varphi\right)  }\equiv\sum_{j=0}^{n}\dfrac{\left(
\gamma\varphi\right)  ^{\ast j}}{j!}\operatorname{mod}\mathcal{L}^{n+1}\left(
H,A\right)  .
\]
Substituting (\ref{1004.pf.6}) into the right hand side of this congruence, we
get%
\begin{equation}
e^{\ast\left(  \gamma\varphi\right)  }\equiv\sum_{j=0}^{n}\dfrac{\left(
\gamma\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\right)
^{\ast j}}{j!}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  .
\label{1004.pf.8}%
\end{equation}


But since $g^{\ast\left(  n+1\right)  }\mid_{H_{\leq n}}=0$ (since
Remark~\ref{Remark9} (applied to $n+1$ and $g$ instead of $i$ and $f$) yields
$g^{\ast\left(  n+1\right)  }\left(  H_{\leq n}\right)  =0$), we have
$g^{\ast\left(  n+1\right)  }\in\left\{  h\in\mathcal{L}\left(  H,A\right)
\ \mid\ h\mid_{H_{\leq n}}=0\right\}  =\mathcal{L}^{n+1}\left(  H,A\right)  $
(by (\ref{1004.pf.0})). In other words, $g^{\ast\left(  n+1\right)  }%
\equiv0\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  $, so that
$\overline{g^{\ast\left(  n+1\right)  }}=0$. Thus, $\overline{g}^{\ast\left(
n+1\right)  }=\overline{g^{\ast\left(  n+1\right)  }}=0$. Hence, we can apply
Corollary~\ref{Corollary1021} to $b=\overline{g}$ and obtain $\sum_{j=0}%
^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\overline{g}^{\ast i}\right)  ^{\ast j}}{j!}=\sum_{i=0}^{n}\dbinom{\gamma
}{i}\overline{g}^{\ast i}$ (where $\dbinom{\gamma}{i}$ is defined as in
Definition~\ref{Definition1000}). Thus,%
\[
\overline{\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(
-1\right)  ^{i-1}}{i}g^{\ast i}\right)  ^{\ast j}}{j!}}=\sum_{j=0}^{n}%
\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\overline{g}^{\ast i}\right)  ^{\ast j}}{j!}=\sum_{i=0}^{n}\dbinom{\gamma
}{i}\overline{g}^{\ast i}=\overline{\sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast
i}}.
\]
In other words,%
\begin{equation}
\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}g^{\ast i}\right)  ^{\ast j}}{j!}\equiv\sum_{i=0}^{n}\dbinom{\gamma
}{i}g^{\ast i}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  .
\label{1004.pf.10}%
\end{equation}


Next, we notice that Definition~\ref{Definition1002} yields%
\[
\operatorname*{pow}\nolimits_{\gamma}F=\operatorname*{Pow}\nolimits_{\gamma
}\underbrace{\left(  F-e_{H,A}\right)  }_{=g}=\operatorname*{Pow}%
\nolimits_{\gamma}g.
\]
Thus, every $x\in H_{\leq n}$ satisfies%
\begin{align*}
\left(  \operatorname*{pow}\nolimits_{\gamma}F\right)  \left(  x\right)   &
=\left(  \operatorname*{Pow}\nolimits_{\gamma}g\right)  \left(  x\right)
=\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}g^{\ast i}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1001.Pow}), applied to }g\text{
instead of }f\right) \\
&  =\sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\left(  x\right)  +\sum
_{i=n+1}^{\infty}\dbinom{\gamma}{i}\underbrace{g^{\ast i}\left(  x\right)
}_{\substack{=0\text{ (since }\\x\in H_{\leq n}\text{ and thus}\\g^{\ast
i}\left(  x\right)  \in g^{\ast i}\left(  H_{\leq n}\right)  =0\\\text{(since
}i>n\text{))}}}\\
&  =\underbrace{\sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\left(  x\right)
}_{=\left(  \sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\right)  \left(
x\right)  }+\underbrace{\sum_{i=n+1}^{\infty}\dbinom{\gamma}{i}0}_{=0}=\left(
\sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\right)  \left(  x\right)  .
\end{align*}
Thus,%
\[
\left(  \operatorname*{pow}\nolimits_{\gamma}F\right)  \mid_{H_{\leq n}%
}=\left(  \sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\right)  \mid_{H_{\leq n}%
}.
\]
Now,%
\begin{align*}
\left(  \operatorname*{pow}\nolimits_{\gamma}F-\sum_{i=0}^{n}\dbinom{\gamma
}{i}g^{\ast i}\right)   &  \mid_{H_{\leq n}}=\underbrace{\left(
\operatorname*{pow}\nolimits_{\gamma}F\right)  \mid_{H_{\leq n}}}_{=\left(
\sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}}-\left(
\sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\right)  \mid_{H_{\leq n}}\\
&  =\left(  \sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\right)  \mid_{H_{\leq
n}}-\left(  \sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast i}\right)  \mid_{H_{\leq
n}}=0.
\end{align*}
In other words,%
\[
\operatorname*{pow}\nolimits_{\gamma}F-\sum_{i=0}^{n}\dbinom{\gamma}{i}g^{\ast
i}\in\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\mid_{H_{\leq n}%
}=0\right\}  =\mathcal{L}^{n+1}\left(  H,A\right)
\]
(by (\ref{1004.pf.0})). This rewrites as%
\begin{equation}
\operatorname*{pow}\nolimits_{\gamma}F\equiv\sum_{i=0}^{n}\dbinom{\gamma}%
{i}g^{\ast i}\operatorname{mod}\mathcal{L}^{n+1}\left(  H,A\right)  .
\label{1004.pf.25}%
\end{equation}
Hence,%
\begin{align*}
\operatorname*{pow}\nolimits_{\gamma}F  &  \equiv\sum_{i=0}^{n}\dbinom{\gamma
}{i}g^{\ast i}\equiv\sum_{j=0}^{n}\dfrac{\left(  \gamma\sum_{i=1}^{n}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}g^{\ast i}\right)  ^{\ast j}}%
{j!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1004.pf.10})}\right) \\
&  \equiv e^{\ast\left(  \gamma\varphi\right)  }\operatorname{mod}%
\mathcal{L}^{n+1}\left(  H,A\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1004.pf.8})}\right)  .
\end{align*}
In other words,
\[
\operatorname*{pow}\nolimits_{\gamma}F-e^{\ast\left(  \gamma\varphi\right)
}\in\mathcal{L}^{n+1}\left(  H,A\right)  =\left\{  h\in\mathcal{L}\left(
H,A\right)  \ \mid\ h\mid_{H_{\leq n}}=0\right\}
\]
(by (\ref{1004.pf.0})). In other words, $\left(  \operatorname*{pow}%
\nolimits_{\gamma}F-e^{\ast\left(  \gamma\varphi\right)  }\right)
\mid_{H_{\leq n}}=0$. Thus,%
\[
0=\left(  \operatorname*{pow}\nolimits_{\gamma}F-e^{\ast\left(  \gamma
\varphi\right)  }\right)  \mid_{H_{\leq n}}=\left(  \operatorname*{pow}%
\nolimits_{\gamma}F\right)  \mid_{H_{\leq n}}-e^{\ast\left(  \gamma
\varphi\right)  }\mid_{H_{\leq n}},
\]
so that $\left(  \operatorname*{pow}\nolimits_{\gamma}F\right)  \mid_{H_{\leq
n}}=e^{\ast\left(  \gamma\varphi\right)  }\mid_{H_{\leq n}}$. Since
$\varphi=\operatorname*{Log}F$, this becomes $\left(  \operatorname*{pow}%
\nolimits_{\gamma}F\right)  \mid_{H_{\leq n}}=e^{\ast\left(  \gamma
\operatorname*{Log}F\right)  }\mid_{H_{\leq n}}$.

Now, forget that we fixed $n$. We have thus proven that
\begin{equation}
\left(  \operatorname*{pow}\nolimits_{\gamma}F\right)  \mid_{H_{\leq n}%
}=e^{\ast\left(  \gamma\operatorname*{Log}F\right)  }\mid_{H_{\leq n}%
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}. \label{1004.pf.31}%
\end{equation}


Now, let $x\in H$ be arbitrary. Since $H$ is filtered, there must exist some
$n\in\mathbb{N}$ such that $x\in H_{\leq n}$. Then,%
\[
\left(  \operatorname*{pow}\nolimits_{\gamma}F\right)  \left(  x\right)
=\underbrace{\left(  \left(  \operatorname*{pow}\nolimits_{\gamma}F\right)
\mid_{H_{\leq n}}\right)  }_{\substack{=e^{\ast\left(  \gamma
\operatorname*{Log}F\right)  }\mid_{H_{\leq n}}\\\text{(by (\ref{1004.pf.31}%
))}}}\left(  x\right)  =\left(  e^{\ast\left(  \gamma\operatorname*{Log}%
F\right)  }\mid_{H_{\leq n}}\right)  \left(  x\right)  =e^{\ast\left(
\gamma\operatorname*{Log}F\right)  }\left(  x\right)  .
\]


Now, forget that we fixed $x$. We have thus proven that every $x\in H$
satisfies $\left(  \operatorname*{pow}\nolimits_{\gamma}F\right)  \left(
x\right)  =e^{\ast\left(  \gamma\operatorname*{Log}F\right)  }\left(
x\right)  $. In other words, $\operatorname*{pow}\nolimits_{\gamma}%
F=e^{\ast\left(  \gamma\operatorname*{Log}F\right)  }$. We are thus done
proving Theorem~\ref{Theorem1004}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem1003}.]Let $f=\operatorname*{Log}F$. Let $g=-f$.
Then, $f+g=f+\left(  -f\right)  =0$. Besides, from $f=\operatorname*{Log}F$,
we obtain $e^{\ast f}=e^{\ast\left(  \operatorname*{Log}F\right)  }=F$ (by
Proposition~\ref{Proposition13} \textbf{(b)}).

On the other hand, $f=\operatorname*{Log}F\in\mathfrak{g}\left(  H,A\right)  $
and thus $g=-\underbrace{f}_{\in\mathfrak{g}\left(  H,A\right)  }%
\in-\mathfrak{g}\left(  H,A\right)  \subseteq\mathfrak{g}\left(  H,A\right)  $
(since $\mathfrak{g}\left(  H,A\right)  $ is a $k$-vector space). Also,
$f\ast\underbrace{g}_{=-f}=f\ast\left(  -f\right)  =-f\ast f$ and
$\underbrace{g}_{=-f}\ast f=\left(  -f\right)  \ast f=-f\ast f$, so that
$f\ast g=-f\ast f=g\ast f$. Hence, Proposition~\ref{Proposition47} (applied to
$H$ and $A$ instead of $C$ and $H$) yields $e^{\ast\left(  f+g\right)
}=e^{\ast f}\ast e^{\ast g}$. Since $f+g=0$ and $e^{\ast f}=F$, this rewrites
as $e^{\ast0}=F\ast e^{\ast g}$.

Corollary~\ref{Corollary48} (applied to $H$, $A$ and $0$ instead of $C$, $H$
and $n$) yields $e^{\ast\left(  0f\right)  }=\left(  e^{\ast f}\right)
^{\ast0}$. Since $0f=0$ and $\left(  e^{\ast f}\right)  ^{\ast0}=e_{H,A}$
(where $e_{H,A}$ is defined as in Definition~\ref{def.eHA}), this rewrites as
$e^{\ast0}=e_{H,A}$. Compared with $e^{\ast0}=F\ast e^{\ast g}$, this yields
$F\ast e^{\ast g}=e_{H,A}$.

On the other hand, Proposition~\ref{Proposition47} (applied to $g$, $f$, $H$
and $A$ instead of $f$, $g$, $C$ and $H$) yields $e^{\ast\left(  g+f\right)
}=e^{\ast g}\ast e^{\ast f}$. Since $g+f=f+g=0$ and $e^{\ast f}=F$, this
rewrites as $e^{\ast0}=e^{\ast g}\ast F$. Compared with $e^{\ast0}=e_{H,A}$,
this yields $e^{\ast g}\ast F=e_{H,A}$.

Now, the map $F$ has the $\ast$-inverse $e^{\ast g}$ (since $F\ast e^{\ast
g}=e_{H,A}$ and $e^{\ast g}\ast F=e_{H,A}$). Thus, $F$ is $\ast$-invertible.
This proves Theorem~\ref{Theorem1003} \textbf{(a)}.

\textbf{(b)} Let $n\in\mathbb{Z}$. Since $f=\operatorname*{Log}F$, we have%
\begin{equation}
e^{\ast\left(  nf\right)  }=e^{\ast\left(  n\operatorname*{Log}F\right)
}=\operatorname*{pow}\nolimits_{n}F \label{1003.pf.10}%
\end{equation}
(because Theorem~\ref{Theorem1004} (applied to $\gamma=n$) yields
$\operatorname*{pow}\nolimits_{n}F=e^{\ast\left(  n\operatorname*{Log}%
F\right)  }$).

We must be in one of the following two cases:

\textit{Case 1:} We have $n\geq0$.

\textit{Case 2:} We have $n<0$.

First, let us consider Case 1. In this case, $n\geq0$, so that $n\in
\mathbb{N}$, and thus Corollary~\ref{Corollary48} (applied to $H$ and $A$
instead of $C$ and $H$) yields $e^{\ast\left(  nf\right)  }=\left(  e^{\ast
f}\right)  ^{\ast n}$. Since $e^{\ast\left(  nf\right)  }=\operatorname*{pow}%
\nolimits_{n}F$ (by (\ref{1003.pf.10})) and $e^{\ast f}=F$, this rewrites as
$\operatorname*{pow}\nolimits_{n}F=F^{\ast n}$. Thus,
Theorem~\ref{Theorem1003} \textbf{(b)} is proven in Case 1.

Next, let us consider Case 2. In this case, $n<0$, so that $-n>0$ and thus
$-n\in\mathbb{N}$. Hence, Corollary~\ref{Corollary48} (applied to $-n$, $H$
and $A$ instead of $n$, $C$ and $H$) yields $e^{\ast\left(  \left(  -n\right)
f\right)  }=\left(  e^{\ast f}\right)  ^{\ast\left(  -n\right)  }$. Since
$e^{\ast f}=F$, this rewrites as $e^{\ast\left(  \left(  -n\right)  f\right)
}=F^{\ast\left(  -n\right)  }$. Thus, $\underbrace{e^{\ast\left(  \left(
-n\right)  f\right)  }}_{=F^{\ast\left(  -n\right)  }}\ast e^{\ast\left(
nf\right)  }=F^{\ast\left(  -n\right)  }\ast e^{\ast\left(  nf\right)  }$.

Clearly, $\left(  -n\right)  \underbrace{f}_{\in\mathfrak{g}\left(
H,A\right)  }\in\left(  -n\right)  \mathfrak{g}\left(  H,A\right)
\subseteq\mathfrak{g}\left(  H,A\right)  $ (since $\mathfrak{g}\left(
H,A\right)  $ is a $k$-vector space) and $n\underbrace{f}_{\in\mathfrak{g}%
\left(  H,A\right)  }\in n\mathfrak{g}\left(  H,A\right)  \subseteq
\mathfrak{g}\left(  H,A\right)  $ (since $\mathfrak{g}\left(  H,A\right)  $ is
a $k$-vector space). Also, $\left(  \left(  -n\right)  f\right)  \ast\left(
nf\right)  =-n^{2}\left(  f\ast f\right)  =\left(  nf\right)  \ast\left(
\left(  -n\right)  f\right)  $. Thus, Proposition~\ref{Proposition47} (applied
to $\left(  -n\right)  f$, $nf$, $H$ and $A$ instead of $f$, $g$, $C$ and $H$)
yields $e^{\ast\left(  \left(  -n\right)  f+nf\right)  }=e^{\ast\left(
-nf\right)  }\ast e^{\ast\left(  nf\right)  }$. Since $\left(  -n\right)
f+nf=0$, this rewrites as $e^{\ast0}=e^{\ast\left(  -nf\right)  }\ast
e^{\ast\left(  nf\right)  }$. Since $e^{\ast0}=e_{H,A}$, this rewrites as
$e_{H,A}=e^{\ast\left(  -nf\right)  }\ast e^{\ast\left(  nf\right)  }$. Thus,%
\[
e_{H,A}=e^{\ast\left(  -nf\right)  }\ast e^{\ast\left(  nf\right)  }%
=F^{\ast\left(  -n\right)  }\ast e^{\ast\left(  nf\right)  },
\]
so that
\[
F^{\ast n}\ast\underbrace{e_{H,A}}_{=F^{\ast\left(  -n\right)  }\ast
e^{\ast\left(  nf\right)  }}=\underbrace{F^{\ast n}\ast F^{\ast\left(
-n\right)  }}_{=e_{H,A}}\ast e^{\ast\left(  nf\right)  }=e_{H,A}\ast
e^{\ast\left(  nf\right)  }=e^{\ast\left(  nf\right)  }=\operatorname*{pow}%
\nolimits_{n}F
\]
(by (\ref{1003.pf.10})). Since $F^{\ast n}\ast e_{H,A}=F^{\ast n}$, this
rewrites as $F^{\ast n}=\operatorname*{pow}\nolimits_{n}F$. In other words,
$\operatorname*{pow}\nolimits_{n}F=F^{\ast n}$. Thus,
Theorem~\ref{Theorem1003} \textbf{(b)} is proven in Case 2.

We have thus proven Theorem~\ref{Theorem1003} \textbf{(b)} in each of the
cases 1 and 2. Since these two cases cover all possibilities, this yields that
Theorem~\ref{Theorem1003} \textbf{(b)} always holds. This completes the proof
of Theorem~\ref{Theorem1003}.
\end{proof}

Another consequence of Theorem~\ref{Theorem1004} is the following:

\begin{theorem}
\label{Theorem1031} Let $k$ be a field of characteristic $0$. Let $\gamma\in
k$. Let $A$ be a $k$-bialgebra, and let $H$ be a connected filtered
cocommutative $k$-coalgebra. Let $F\in G\left(  H,A\right)  $ be a
$k$-coalgebra homomorphism. Then, $\operatorname*{pow}\nolimits_{\gamma}F$ is
a $k$-coalgebra homomorphism.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1031}.]Let $f=\operatorname*{Log}F$. Then,
$f=\operatorname*{Log}F\in\mathfrak{g}\left(  H,A\right)  $ and $e^{\ast
f}=e^{\ast\left(  \operatorname*{Log}F\right)  }=F$ (by
Proposition~\ref{Proposition13} \textbf{(b)}). Since $F$ is a $k$-coalgebra
homomorphism, we know that $e^{\ast f}$ is a $k$-coalgebra homomorphism
(because $e^{\ast f}=F$). Thus, $f$ is an $\left(  \varepsilon,\varepsilon
\right)  $-coderivation (since Theorem~\ref{Theorem30} (applied to $H$ and $A$
instead of $C$ and $H$) shows that $f$ is an $\left(  \varepsilon
,\varepsilon\right)  $-coderivation if and only if $e^{\ast f}$ is a
$k$-coalgebra homomorphism). Thus,%
\[
\Delta_{A}\circ f=\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)
\circ\Delta_{H}%
\]
(because Definition~\ref{Definition22} (applied to $H$ and $A$ instead of $C$
and $H$) shows that $f$ is an $\left(  \varepsilon,\varepsilon\right)
$-coderivation if and only if $\Delta_{A}\circ f=\left(  f\otimes
e_{H,A}+e_{H,A}\otimes f\right)  \circ\Delta_{H}$). Hence,%
\begin{align*}
\Delta_{A}\circ\left(  \gamma f\right)   &  =\gamma\cdot\underbrace{\left(
\Delta_{A}\circ f\right)  }_{=\left(  f\otimes e_{H,A}+e_{H,A}\otimes
f\right)  \circ\Delta_{H}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition
of }k\text{-linear maps is }k\text{-bilinear}\right) \\
&  =\gamma\cdot\left(  \left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)
\circ\Delta_{H}\right) \\
&  =\underbrace{\left(  \gamma\cdot\left(  f\otimes e_{H,A}+e_{H,A}\otimes
f\right)  \right)  }_{=\gamma\cdot\left(  f\otimes e_{H,A}\right)
+\gamma\cdot\left(  e_{H,A}\otimes f\right)  }\circ\Delta_{H}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps is
}k\text{-bilinear}\right) \\
&  =\left(  \underbrace{\gamma\cdot\left(  f\otimes e_{H,A}\right)
}_{\substack{=\left(  \gamma f\right)  \otimes e_{H,A}\\\text{(since tensoring
of }k\text{-linear}\\\text{maps is }k\text{-bilinear)}}}+\underbrace{\gamma
\cdot\left(  e_{H,A}\otimes f\right)  }_{\substack{=e_{H,A}\otimes\left(
\gamma f\right)  \\\text{(since tensoring of }k\text{-linear}\\\text{maps is
}k\text{-bilinear)}}}\right)  \circ\Delta_{H}\\
&  =\left(  \left(  \gamma f\right)  \otimes e_{H,A}+e_{H,A}\otimes\left(
\gamma f\right)  \right)  \circ\Delta_{H}.
\end{align*}
Hence, $\gamma f$ is an $\left(  \varepsilon,\varepsilon\right)
$-coderivation (because Definition~\ref{Definition22} (applied to $\gamma f$,
$H$ and $A$ instead of $f$, $C$ and $H$) shows that $\gamma f$ is an $\left(
\varepsilon,\varepsilon\right)  $-coderivation if and only if $\Delta_{A}%
\circ\left(  \gamma f\right)  =\left(  \left(  \gamma f\right)  \otimes
e_{H,A}+e_{H,A}\otimes\left(  \gamma f\right)  \right)  \circ\Delta_{H}$).
Since $\gamma\underbrace{f}_{\in\mathfrak{g}\left(  H,A\right)  }\in
\gamma\mathfrak{g}\left(  H,A\right)  \subseteq\mathfrak{g}\left(  H,A\right)
$ (since $\mathfrak{g}\left(  H,A\right)  $ is a $k$-vector space), this
yields that we can apply Theorem~\ref{Theorem30} to $\gamma f$, $H$ and $A$
instead of $f$, $C$ and $H$. As a result, we obtain that $\gamma f$ is an
$\left(  \varepsilon,\varepsilon\right)  $-coderivation if and only if
$e^{\ast\left(  \gamma f\right)  }$ is a $k$-coalgebra homomorphism. Since we
know that $\gamma f$ is an $\left(  \varepsilon,\varepsilon\right)
$-coderivation, we thus conclude that $e^{\ast\left(  \gamma f\right)  }$ is a
$k$-coalgebra homomorphism. Since
\begin{align*}
e^{\ast\left(  \gamma f\right)  }  &  =e^{\ast\left(  \gamma
\operatorname*{Log}F\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{because
}f=\operatorname*{Log}F\right) \\
&  =\operatorname*{pow}\nolimits_{\gamma}F\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Theorem~\ref{Theorem1004}}\right)  ,
\end{align*}
this rewrites as follows: The map $\operatorname*{pow}\nolimits_{\gamma}F$ is
a $k$-coalgebra homomorphism. This proves Theorem~\ref{Theorem1031}.
\end{proof}

Here is the dual statement to Theorem~\ref{Theorem1031} (although not
precisely dual due to the presence of filtrations):

\begin{theorem}
\label{Theorem1032} Let $k$ be a field of characteristic $0$. Let $\gamma\in
k$. Let $H$ be a connected filtered $k$-coalgebra and, at the same time, a
$k$-bialgebra with the same underlying $k$-coalgebra structure. Let $A$ be a
commutative $k$-algebra. Let $F\in G\left(  H,A\right)  $ be a $k$-algebra
homomorphism. Then, $\operatorname*{pow}\nolimits_{\gamma}F$ is a $k$-algebra homomorphism.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1032}.]Let $f=\operatorname*{Log}F$. Then,
$f=\operatorname*{Log}F\in\mathfrak{g}\left(  H,A\right)  $ and $e^{\ast
f}=e^{\ast\left(  \operatorname*{Log}F\right)  }=F$ (by
Proposition~\ref{Proposition13} \textbf{(b)}). Since $F$ is a $k$-algebra
homomorphism, we know that $e^{\ast f}$ is a $k$-algebra homomorphism (because
$e^{\ast f}=F$). Thus, $f$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation (since Theorem~\ref{Theorem30} shows that $f$ is an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and only if
$e^{\ast f}$ is a $k$-algebra homomorphism). Thus,%
\[
f\circ\mu_{H}=\mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)
\]
(because Definition~\ref{Definition108} shows that $f$ is an $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and only if $f\circ
\mu_{H}=\mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes f\right)  $).
Hence,%
\begin{align*}
\left(  \gamma f\right)  \circ\mu_{H}  &  =\gamma\cdot\underbrace{\left(
f\circ\mu_{H}\right)  }_{=\mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes
f\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of
}k\text{-linear maps is }k\text{-bilinear}\right) \\
&  =\gamma\cdot\left(  \mu_{A}\circ\left(  f\otimes e_{H,A}+e_{H,A}\otimes
f\right)  \right) \\
&  =\mu_{A}\circ\underbrace{\left(  \gamma\cdot\left(  f\otimes e_{H,A}%
+e_{H,A}\otimes f\right)  \right)  }_{=\gamma\cdot\left(  f\otimes
e_{H,A}\right)  +\gamma\cdot\left(  e_{H,A}\otimes f\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps is
}k\text{-bilinear}\right) \\
&  =\mu_{A}\circ\left(  \underbrace{\gamma\cdot\left(  f\otimes e_{H,A}%
\right)  }_{\substack{=\left(  \gamma f\right)  \otimes e_{H,A}\\\text{(since
tensoring of }k\text{-linear}\\\text{maps is }k\text{-bilinear)}%
}}+\underbrace{\gamma\cdot\left(  e_{H,A}\otimes f\right)  }%
_{\substack{=e_{H,A}\otimes\left(  \gamma f\right)  \\\text{(since tensoring
of }k\text{-linear}\\\text{maps is }k\text{-bilinear)}}}\right) \\
&  =\mu_{A}\circ\left(  \left(  \gamma f\right)  \otimes e_{H,A}%
+e_{H,A}\otimes\left(  \gamma f\right)  \right)  .
\end{align*}
Hence, $\gamma f$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation (because Definition~\ref{Definition108} (applied to $\gamma f$
instead of $f$) shows that $\gamma f$ is an $\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivation if and only if $\left(  \gamma
f\right)  \circ\mu_{H}=\mu_{A}\circ\left(  \left(  \gamma f\right)  \otimes
e_{H,A}+e_{H,A}\otimes\left(  \gamma f\right)  \right)  $). Since
$\gamma\underbrace{f}_{\in\mathfrak{g}\left(  H,A\right)  }\in\gamma
\mathfrak{g}\left(  H,A\right)  \subseteq\mathfrak{g}\left(  H,A\right)  $
(since $\mathfrak{g}\left(  H,A\right)  $ is a $k$-vector space), this yields
that we can apply Theorem~\ref{Theorem120} to $\gamma f$ instead of $f$. As a
result, we obtain that $\gamma f$ is an $\left(  \varepsilon_{H}%
,\varepsilon_{H}\right)  $-derivation if and only if $e^{\ast\left(  \gamma
f\right)  }$ is a $k$-algebra homomorphism. Since we know that $\gamma f$ is
an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation, we thus
conclude that $e^{\ast\left(  \gamma f\right)  }$ is a $k$-algebra
homomorphism. Since
\begin{align*}
e^{\ast\left(  \gamma f\right)  }  &  =e^{\ast\left(  \gamma
\operatorname*{Log}F\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{because
}f=\operatorname*{Log}F\right) \\
&  =\operatorname*{pow}\nolimits_{\gamma}F\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Theorem~\ref{Theorem1004}}\right)  ,
\end{align*}
this rewrites as follows: The map $\operatorname*{pow}\nolimits_{\gamma}F$ is
a $k$-algebra homomorphism. This proves Theorem~\ref{Theorem1032}.
\end{proof}

Next, we prove something fundamental:

\begin{theorem}
\label{Theorem1033} Let $k$ be a field of characteristic $0$. Let $A$ be a
graded $k$-algebra, and let $H$ be a connected graded $k$-coalgebra.

\textbf{(a)} For every graded map $f\in\mathfrak{g}\left(  H,A\right)  $, the
map $e^{\ast f}$ is also graded.

\textbf{(b)} For every graded map $F\in G\left(  H,A\right)  $, the map
$\operatorname*{Log}F$ is also graded.

\textbf{(c)} For every $\gamma\in k$ and every graded map $F\in G\left(
H,A\right)  $, the map $\operatorname*{pow}\nolimits_{\gamma}F$ is also graded.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1033}.]First, we notice that the graded
$k$-coalgebra $H$ is connected. Thus, the filtered $k$-coalgebra $H$ (defined
as according to Convention~\ref{Convention207}) is connected (because
Remark~\ref{Remark213} (applied to $C=H$) yields that the graded $k$-coalgebra
$H$ is connected if and only if the filtered $k$-coalgebra $H$ is connected).

Define $e_{H,A}$ as according to Definition~\ref{def.eHA}.

\textbf{(a)} Let $f\in\mathfrak{g}\left(  H,A\right)  $ be a graded map. For
every $i\in\mathbb{N}$, the map $f^{\ast i}$ is graded (by
Proposition~\ref{Proposition224} \textbf{(c)}, applied to $n=i$), and thus
satisfies%
\begin{equation}
f^{\ast i}\left(  H_{n}\right)  \subseteq A_{n}\ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N}. \label{1033.pf.a.1}%
\end{equation}


Let $n\in\mathbb{N}$. Let $x\in H_{n}$.

By the definition of $H_{\leq n}$, we have $H_{\leq n}=\bigoplus_{\ell=0}%
^{n}H_{\ell}$. But since $H_{n}$ is an addend of the direct sum $\bigoplus
_{\ell=0}^{n}H_{\ell}$, we have $H_{n}\subseteq\bigoplus_{\ell=0}^{n}H_{\ell
}=H_{\leq n}$. Thus, $x\in H_{n}\subseteq H_{\leq n}$.

Notice that%
\begin{equation}
\text{every }i\in\mathbb{N}\text{ such that }i>n\text{ satisfies }f^{\ast
i}\left(  x\right)  =0 \label{1033.pf.a.2}%
\end{equation}
(because every $i\in\mathbb{N}$ such that $i>n$ satisfies $x\in H_{\leq n}$
and thus $f^{\ast i}\left(  x\right)  \in f^{\ast i}\left(  H_{\leq n}\right)
=0$ (by Remark~\ref{Remark9}), so that $f^{\ast i}\left(  x\right)  =0$).

From (\ref{exp}), we have%
\begin{align*}
e^{\ast f}\left(  x\right)   &  =\sum_{i\geq0}\underbrace{\dfrac{f^{\ast
i}\left(  x\right)  }{i!}}_{=\dfrac{1}{i!}f^{\ast i}\left(  x\right)  }%
=\sum_{i\geq0}\dfrac{1}{i!}f^{\ast i}\left(  x\right)  =\sum_{i=0}^{n}%
\dfrac{1}{i!}f^{\ast i}\left(  \underbrace{x}_{\in H_{n}}\right)
+\sum_{i=n+1}^{\infty}\dfrac{1}{i!}\underbrace{f^{\ast i}\left(  x\right)
}_{\substack{=0\text{ (because }i>n\text{, and thus}\\\text{(\ref{1033.pf.a.2}%
) yields }f^{\ast i}\left(  x\right)  =0\text{)}}}\\
&  \in\sum_{i=0}^{n}\dfrac{1}{i!}\underbrace{f^{\ast i}\left(  H_{n}\right)
}_{\substack{\subseteq A_{n}\\\text{(by (\ref{1033.pf.a.1}))}}%
}+\underbrace{\sum_{i=n+1}^{\infty}\dfrac{1}{i!}0}_{=0}\subseteq\sum_{i=0}%
^{n}\dfrac{1}{i!}A_{n}\subseteq A_{n}%
\end{align*}
(since $A_{n}$ is a $k$-vector space).

Now forget that we fixed $x$. We have thus proven that every $x\in H_{n}$
satisfies $e^{\ast f}\left(  x\right)  \in A_{n}$. In other words, $e^{\ast
f}\left(  H_{n}\right)  \subseteq A_{n}$.

Now forget that we fixed $n$. We thus have proven that every $n\in\mathbb{N}$
satisfies $e^{\ast f}\left(  H_{n}\right)  \subseteq A_{n}$. In other words,
$e^{\ast f}$ is graded. This proves Theorem~\ref{Theorem1033} \textbf{(a)}.

\textbf{(b)} Let $F\in G\left(  H,A\right)  $ be a graded map. Let
$f=F-e_{H,A}$. Since $F\in G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(
H,A\right)  $, we have $F-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  $, so that
$f=F-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  $.

For every $i\in\mathbb{N}$, the map $f^{\ast i}$ is graded (by
Proposition~\ref{Proposition224} \textbf{(c)}, applied to $n=i$), and thus
satisfies%
\begin{equation}
f^{\ast i}\left(  H_{n}\right)  \subseteq A_{n}\ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N}. \label{1033.pf.b.1}%
\end{equation}


Let $n\in\mathbb{N}$. Let $x\in H_{n}$.

By the definition of $H_{\leq n}$, we have $H_{\leq n}=\bigoplus_{\ell=0}%
^{n}H_{\ell}$. But since $H_{n}$ is an addend of the direct sum $\bigoplus
_{\ell=0}^{n}H_{\ell}$, we have $H_{n}\subseteq\bigoplus_{\ell=0}^{n}H_{\ell
}=H_{\leq n}$. Thus, $x\in H_{n}\subseteq H_{\leq n}$.

Notice that%
\begin{equation}
\text{every }i\in\mathbb{N}\text{ such that }i>n\text{ satisfies }f^{\ast
i}\left(  x\right)  =0 \label{1033.pf.b.2}%
\end{equation}
(because every $i\in\mathbb{N}$ such that $i>n$ satisfies $x\in H_{\leq n}$
and thus $f^{\ast i}\left(  x\right)  \in f^{\ast i}\left(  H_{\leq n}\right)
=0$ (by Remark~\ref{Remark9}), so that $f^{\ast i}\left(  x\right)  =0$).

From (\ref{log}), we have%
\begin{align*}
\left(  \operatorname{Log}_{1}f\right)  \left(  x\right)   &  =\sum_{i\geq
1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)  =\sum
_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  \underbrace{x}%
_{\in H_{n}}\right)  +\sum_{i=n+1}^{\infty}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}\underbrace{f^{\ast i}\left(  x\right)  }_{\substack{=0\text{ (because
}i>n\text{, and thus}\\\text{(\ref{1033.pf.b.2}) yields }f^{\ast i}\left(
x\right)  =0\text{)}}}\\
&  \in\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{f^{\ast
i}\left(  H_{n}\right)  }_{\substack{\subseteq A_{n}\\\text{(by
(\ref{1033.pf.b.1}))}}}+\underbrace{\sum_{i=n+1}^{\infty}\dfrac{\left(
-1\right)  ^{i-1}}{i}0}_{=0}\subseteq\sum_{i=1}^{n}\dfrac{\left(  -1\right)
^{i-1}}{i}A_{n}\subseteq A_{n}%
\end{align*}
(since $A_{n}$ is a $k$-vector space).

Now forget that we fixed $x$. We have thus proven that every $x\in H_{n}$
satisfies $\left(  \operatorname{Log}_{1}f\right)  \left(  x\right)  \in
A_{n}$. In other words, $\left(  \operatorname{Log}_{1}f\right)  \left(
H_{n}\right)  \subseteq A_{n}$.

Definition~\ref{Definition12} yields $\operatorname*{Log}F=\operatorname{Log}%
_{1}\underbrace{\left(  F-e_{H,A}\right)  }_{=f}=\operatorname{Log}_{1}f$.
Thus, $\left(  \operatorname*{Log}F\right)  \left(  H_{n}\right)  =\left(
\operatorname{Log}_{1}f\right)  \left(  H_{n}\right)  \subseteq A_{n}$.

Now forget that we fixed $n$. We thus have proven that every $n\in\mathbb{N}$
satisfies $\left(  \operatorname*{Log}F\right)  \left(  H_{n}\right)
\subseteq A_{n}$. In other words, $\operatorname*{Log}F$ is graded. This
proves Theorem~\ref{Theorem1033} \textbf{(b)}.

\textbf{(c)} Let $F\in G\left(  H,A\right)  $ be a graded map. Let
$f=F-e_{H,A}$. Since $F\in G\left(  H,A\right)  =e_{H,A}+\mathfrak{g}\left(
H,A\right)  $, we have $F-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  $, so that
$f=F-e_{H,A}\in\mathfrak{g}\left(  H,A\right)  $.

For every $i\in\mathbb{N}$, the map $f^{\ast i}$ is graded (by
Proposition~\ref{Proposition224} \textbf{(c)}, applied to $n=i$), and thus
satisfies%
\begin{equation}
f^{\ast i}\left(  H_{n}\right)  \subseteq A_{n}\ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N}. \label{1033.pf.c.1}%
\end{equation}


Let $n\in\mathbb{N}$. Let $x\in H_{n}$.

By the definition of $H_{\leq n}$, we have $H_{\leq n}=\bigoplus_{\ell=0}%
^{n}H_{\ell}$. But since $H_{n}$ is an addend of the direct sum $\bigoplus
_{\ell=0}^{n}H_{\ell}$, we have $H_{n}\subseteq\bigoplus_{\ell=0}^{n}H_{\ell
}=H_{\leq n}$. Thus, $x\in H_{n}\subseteq H_{\leq n}$.

Notice that%
\begin{equation}
\text{every }i\in\mathbb{N}\text{ such that }i>n\text{ satisfies }f^{\ast
i}\left(  x\right)  =0 \label{1033.pf.c.2}%
\end{equation}
(because every $i\in\mathbb{N}$ such that $i>n$ satisfies $x\in H_{\leq n}$
and thus $f^{\ast i}\left(  x\right)  \in f^{\ast i}\left(  H_{\leq n}\right)
=0$ (by Remark~\ref{Remark9}), so that $f^{\ast i}\left(  x\right)  =0$).

From (\ref{1001.Pow}), we have%
\begin{align*}
\left(  \operatorname*{Pow}\nolimits_{\gamma}f\right)  \left(  x\right)   &
=\sum_{i\in\mathbb{N}}\dbinom{\gamma}{i}f^{\ast i}\left(  x\right)
=\sum_{i=0}^{\infty}\dbinom{\gamma}{i}f^{\ast i}\left(  \underbrace{x}_{\in
H_{n}}\right)  +\sum_{i=n+1}^{\infty}\dbinom{\gamma}{i}\underbrace{f^{\ast
i}\left(  x\right)  }_{\substack{=0\text{ (because }i>n\text{, and
thus}\\\text{(\ref{1033.pf.c.2}) yields }f^{\ast i}\left(  x\right)
=0\text{)}}}\\
&  \in\sum_{i=0}^{n}\dbinom{\gamma}{i}\underbrace{f^{\ast i}\left(
H_{n}\right)  }_{\substack{\subseteq A_{n}\\\text{(by (\ref{1033.pf.c.1}))}%
}}+\underbrace{\sum_{i=n+1}^{\infty}\dbinom{\gamma}{i}0}_{=0}\subseteq
\sum_{i=0}^{n}\dbinom{\gamma}{i}A_{n}\subseteq A_{n}%
\end{align*}
(since $A_{n}$ is a $k$-vector space).

Now forget that we fixed $x$. We have thus proven that every $x\in H_{n}$
satisfies $\left(  \operatorname*{Pow}\nolimits_{\gamma}f\right)  \left(
x\right)  \in A_{n}$. In other words, $\left(  \operatorname*{Pow}%
\nolimits_{\gamma}f\right)  \left(  H_{n}\right)  \subseteq A_{n}$.

Definition~\ref{Definition1002} yields $\operatorname*{pow}\nolimits_{\gamma
}F=\operatorname*{Pow}\nolimits_{\gamma}\underbrace{\left(  F-e_{H,A}\right)
}_{=f}=\operatorname*{Pow}\nolimits_{\gamma}f$. Thus, $\left(
\operatorname*{pow}\nolimits_{\gamma}F\right)  \left(  H_{n}\right)  =\left(
\operatorname*{Pow}\nolimits_{\gamma}f\right)  \left(  H_{n}\right)  \subseteq
A_{n}$.

Now forget that we fixed $n$. We thus have proven that every $n\in\mathbb{N}$
satisfies $\left(  \operatorname*{pow}\nolimits_{\gamma}F\right)  \left(
H_{n}\right)  \subseteq A_{n}$. In other words, $\operatorname*{pow}%
\nolimits_{\gamma}F$ is graded. This proves Theorem~\ref{Theorem1033}
\textbf{(c)}.
\end{proof}

\begin{definition}
\label{Definition1041} Let $k$ be a field of characteristic $0$. Let $A$ be a
$k$-algebra, and let $H$ be a connected filtered $k$-algebra. Let $F\in
G\left(  H,A\right)  $.

Whenever $\gamma$ is an element of $k$, we will write $F^{\ast\gamma}$ for
$\operatorname*{pow}\nolimits_{\gamma}F$. This notation does not conflict with
the notation $F^{\ast\gamma}$ for the $\gamma$-th power of $F$ with respect to
convolution when $\gamma$ is an integer. (In fact, the only case in which the
former notation could conflict with the latter is when $\gamma$ is an integer,
but in this case Theorem~\ref{Theorem1003} \textbf{(b)} shows that the two
notations are equivalent.)
\end{definition}

We can now generalize Theorem~\ref{Theorem802}:

\begin{theorem}
\label{Theorem1051} Let $k$ be a field of characteristic $0$. Let $H$ be a
cocommutative connected graded $k$-Hopf algebra. Let $S$ be the antipode of
$H$. Define a map $E_{H}:H\rightarrow H$ according to
Definition~\ref{Definition701}. Define a map $E_{H}^{\operatorname*{inv}%
}:H\rightarrow H$ according to Definition~\ref{Definition721}. Let
$\operatorname*{Prim}H$ denote the subspace of $H$ consisting of all primitive
elements of $H$. Let $\alpha\in k$ and $\beta\in k$ satisfy $\alpha+\beta=1$.
Then, the map $E_{H}^{\operatorname*{inv}}\circ\left(  S^{\ast\alpha}\ast
E_{H}\ast S^{\ast\beta}\right)  $ is a projection from $H$ to the subspace
$\operatorname*{Prim}H$.
\end{theorem}

Theorem~\ref{Theorem802} \textbf{(a)} easily follows from
Theorem~\ref{Theorem1051} when $\alpha=0$ and $\beta=1$.
Theorem~\ref{Theorem801} \textbf{(b)} easily follows from
Theorem~\ref{Theorem1051} when $\alpha=1$ and $\beta=0$.

\begin{proof}
[Proof of Theorem~\ref{Theorem1051}.]Define the map $e_{H,H}$ as according to
Definition~\ref{def.eHA}.

It is easy to see that $S\in G\left(  H,H\right)  $%
\ \ \ \ \footnote{\textit{Proof.} As in the proof of
Corollary~\ref{Corollary813}, we can show that $S\left(  1_{H}\right)  =1_{H}%
$. Thus,%
\[
S\in\left\{  f\in\mathcal{L}\left(  H,H\right)  \ \mid\ f\left(  1_{H}\right)
=1_{H}\right\}  =G\left(  H,H\right)
\]
(since $G\left(  H,H\right)  =\left\{  f\in\mathcal{L}\left(  H,H\right)
\ \mid\ f\left(  1_{H}\right)  =1_{H}\right\}  $ by the definition of
$G\left(  H,H\right)  $), qed.}.

By the definition of $S^{\ast\alpha}$, we have $S^{\ast\alpha}%
=\operatorname*{pow}\nolimits_{\alpha}S$.

Since $H$ is a graded $k$-Hopf algebra, the antipode of $H$ is graded. Since
the antipode of $H$ is $S$, this yields that $S$ is graded. Combined with
$S\in G\left(  H,H\right)  $, this yields that we can apply
Theorem~\ref{Theorem1033} \textbf{(c)} to $A=H$, $\gamma=\alpha$ and $F=S$. As
a result, we obtain that the map $\operatorname*{pow}\nolimits_{\alpha}S$ is graded.

By Definition~\ref{Definition552}, we have $H^{\operatorname*{cop}}=\left(
H,\tau_{H,H}\circ\Delta_{H},\varepsilon_{H}\right)  $. Since $H$ is
cocommutative, we have $\tau_{H,H}\circ\Delta_{H}=\Delta_{H}$. Thus,
$H^{\operatorname*{cop}}=\left(  H,\underbrace{\tau_{H,H}\circ\Delta_{H}%
}_{=\Delta_{H}},\varepsilon_{H}\right)  =\left(  H,\Delta_{H},\varepsilon
_{H}\right)  =H$.

Proposition~\ref{Proposition553} yields that the antipode of $H$ is a
$k$-coalgebra homomorphism from $H^{\operatorname*{cop}}$ to $H$. Since the
antipode of $H$ is the map $S$, whereas $H^{\operatorname*{cop}}$ is $H$, this
rewrites as follows: The map $S$ is a $k$-coalgebra homomorphism from $H$ to
$H$. Thus, Theorem~\ref{Theorem1031} (applied to $A=H$, $\gamma=\alpha$ and
$F=S$) yields that $\operatorname*{pow}\nolimits_{\alpha}S$ is a $k$-coalgebra homomorphism.

Altogether, we have now shown that the map $\operatorname*{pow}%
\nolimits_{\alpha}S$ is a graded $k$-coalgebra homomorphism. Since
$\operatorname*{pow}\nolimits_{\alpha}S=S^{\ast\alpha}$, this rewrites as
follows: The map $S^{\ast\alpha}$ is a graded $k$-coalgebra homomorphism. The
same argument, but with $\alpha$ replaced by $\beta$, shows that the map
$S^{\ast\beta}$ is a graded $k$-coalgebra homomorphism.

We have $S^{\ast\alpha}=\operatorname*{pow}\nolimits_{\alpha}S\in G\left(
H,H\right)  $ (because $\operatorname*{pow}\nolimits_{\alpha}F\in G\left(
H,H\right)  $ for every $F\in G\left(  H,H\right)  $ (due to the definition of
$\operatorname*{pow}\nolimits_{\alpha}F$)). Thus,%
\[
S^{\ast\alpha}\in G\left(  H,H\right)  =\left\{  f\in\mathcal{L}\left(
H,H\right)  \ \mid\ f\left(  1_{H}\right)  =1_{H}\right\}
\]
(by the definition of $G\left(  H,H\right)  $), so that $S^{\ast\alpha}\left(
1_{H}\right)  =1_{H}$. The same argument, but with $\alpha$ replaced by
$\beta$, shows that $S^{\ast\beta}\left(  1_{H}\right)  =1_{H}$.

Recall that $S$ is the antipode of $H$. Thus, $S$ is the $\ast$-inverse of
$\operatorname{id}_{H}$ (because the antipode of $H$ is defined as the $\ast
$-inverse of $\operatorname{id}_{H}$). In other words, $S=\operatorname{id}%
_{H}^{\ast\left(  -1\right)  }$. Hence, $S$ itself is $\ast$-invertible and
satisfies $S^{\ast\left(  -1\right)  }=\operatorname{id}_{H}$. But
Theorem~\ref{Theorem1003} \textbf{(b)} (applied to $A=H$, $n=-1$ and $F=S$)
yields $\operatorname*{pow}\nolimits_{-1}S=S^{\ast\left(  -1\right)  }$.

Now, Theorem~\ref{Theorem1004} (applied to $A=H$, $\gamma=\alpha$ and $F=S$)
yields $\operatorname*{pow}\nolimits_{\alpha}S=e^{\ast\left(  \alpha
\operatorname*{Log}S\right)  }$. By the definition of $S^{\ast\alpha}$, we
have $S^{\ast\alpha}=\operatorname*{pow}\nolimits_{\alpha}S=e^{\ast\left(
\alpha\operatorname*{Log}S\right)  }$.

Also, Theorem~\ref{Theorem1004} (applied to $A=H$, $\gamma=-1$ and $F=S$)
yields $\operatorname*{pow}\nolimits_{-1}S=e^{\ast\left(  \left(  -1\right)
\operatorname*{Log}S\right)  }$. Since $\operatorname*{pow}\nolimits_{-1}%
S=S^{\ast\left(  -1\right)  }=\operatorname{id}_{H}$ and $\left(  -1\right)
\operatorname*{Log}S=-\operatorname*{Log}S$, this rewrites as follows:
$\operatorname{id}_{H}=e^{\ast\left(  -\operatorname*{Log}S\right)  }$.

Also, Theorem~\ref{Theorem1004} (applied to $A=H$, $\gamma=\beta$ and $F=S$)
yields $\operatorname*{pow}\nolimits_{\beta}S=e^{\ast\left(  \beta
\operatorname*{Log}S\right)  }$. By the definition of $S^{\ast\beta}$, we have
$S^{\ast\beta}=\operatorname*{pow}\nolimits_{\beta}S=e^{\ast\left(
\beta\operatorname*{Log}S\right)  }$.

Since $\operatorname*{Log}S\in\mathfrak{g}\left(  H,H\right)  $ (because
$\operatorname*{Log}F\in\mathfrak{g}\left(  H,H\right)  $ for every $F\in
G\left(  H,H\right)  $ (due to the definition of $\operatorname*{Log}F$)), the
elements $\alpha\operatorname*{Log}S$, $-\operatorname*{Log}S$, $-\beta
\operatorname*{Log}S$ and $\beta\operatorname*{Log}S$ must also lie in
$\mathfrak{g}\left(  H,H\right)  $ (since $\mathfrak{g}\left(  H,H\right)  $
is a $k$-vector space).

Since $\left(  \alpha\operatorname*{Log}S\right)  \ast\left(
-\operatorname*{Log}S\right)  =-\alpha\left(  \left(  \operatorname*{Log}%
S\right)  \ast\left(  \operatorname*{Log}S\right)  \right)  =\left(
-\operatorname*{Log}S\right)  \ast\left(  \alpha\operatorname*{Log}S\right)
$, we can apply Proposition~\ref{Proposition47} to $C=H$, $f=\alpha
\operatorname*{Log}S$ and $g=-\operatorname*{Log}S$. As a result, we obtain%
\[
e^{\ast\left(  \alpha\operatorname*{Log}S+\left(  -\operatorname*{Log}%
S\right)  \right)  }=\underbrace{e^{\ast\left(  \alpha\operatorname*{Log}%
S\right)  }}_{=S^{\ast\alpha}}\ast\underbrace{e^{\ast\left(
-\operatorname*{Log}S\right)  }}_{=\operatorname{id}_{H}}=S^{\ast\alpha}%
\ast\operatorname{id}_{H}.
\]
Since $\alpha\operatorname*{Log}S+\left(  -\operatorname*{Log}S\right)
=\underbrace{\left(  \alpha-1\right)  }_{\substack{=-\beta\\\text{(since
}\alpha+\beta=1\text{)}}}\operatorname*{Log}S=-\beta\operatorname*{Log}S$,
this rewrites as follows:%
\begin{equation}
e^{\ast\left(  -\beta\operatorname*{Log}S\right)  }=S^{\ast\alpha}%
\ast\operatorname{id}_{H}. \label{1051.pf.1}%
\end{equation}


On the other hand, since $\left(  -\beta\operatorname*{Log}S\right)
\ast\left(  \beta\operatorname*{Log}S\right)  =-\beta^{2}\left(  \left(
\operatorname*{Log}S\right)  \ast\left(  \operatorname*{Log}S\right)  \right)
=\left(  \beta\operatorname*{Log}S\right)  \ast\left(  -\beta
\operatorname*{Log}S\right)  $, we can apply Proposition~\ref{Proposition47}
to $C=H$, $f=-\beta\operatorname*{Log}S$ and $g=\beta\operatorname*{Log}S$. As
a result, we obtain%
\[
e^{\ast\left(  -\beta\operatorname*{Log}S+\beta\operatorname*{Log}S\right)
}=\underbrace{e^{\ast\left(  -\beta\operatorname*{Log}S\right)  }%
}_{\substack{=S^{\ast\alpha}\ast\operatorname{id}_{H}\\\text{(by
(\ref{1051.pf.1}))}}}\ast\underbrace{e^{\ast\left(  \beta\operatorname*{Log}%
S\right)  }}_{=S^{\ast\beta}}=S^{\ast\alpha}\ast\operatorname{id}_{H}\ast
S^{\ast\beta}.
\]
Since $-\beta\operatorname*{Log}S+\beta\operatorname*{Log}S=0$, this rewrites
as%
\begin{equation}
e^{\ast0}=S^{\ast\alpha}\ast\operatorname{id}_{H}\ast S^{\ast\beta}.
\label{1051.pf.2}%
\end{equation}


Corollary~\ref{Corollary48} (applied to $H$, $A$, $0$ and $0$ instead of $C$,
$H$, $f$ and $n$) yields $e^{\ast\left(  0\cdot0\right)  }=\left(  e^{\ast
0}\right)  ^{\ast0}$. Since $0\cdot0=0$ and $\left(  e^{\ast0}\right)
^{\ast0}=e_{H,H}$, this rewrites as $e^{\ast0}=e_{H,H}$. Compared with
(\ref{1051.pf.2}), this yields
\[
S^{\ast\alpha}\ast\operatorname{id}_{H}\ast S^{\ast\beta}=e_{H,H}.
\]


Thus, we can apply Corollary~\ref{Corollary832} to $P=S^{\ast\alpha}$ and
$Q=S^{\ast\beta}$. As a result, we can conclude that the map $E_{H}%
^{\operatorname*{inv}}\circ\left(  S^{\ast\alpha}\ast E_{H}\ast S^{\ast\beta
}\right)  $ is a projection from $H$ to the subspace $\operatorname*{Prim}H$.
This proves Theorem~\ref{Theorem1051}.
\end{proof}

Here is the expectable dual of Theorem~\ref{Theorem1051}:

\begin{theorem}
\label{Theorem1061} Let $k$ be a field of characteristic $0$. Let $H$ be a
commutative connected graded $k$-Hopf algebra. Let $S$ be the antipode of $H$.
Define a map $E_{H}:H\rightarrow H$ according to
Definition~\ref{Definition701}. Define a map $E_{H}^{\operatorname*{inv}%
}:H\rightarrow H$ according to Definition~\ref{Definition721}. Let $\alpha\in
k$ and $\beta\in k$ satisfy $\alpha+\beta=1$. Then, the map $E_{H}%
^{\operatorname*{inv}}\circ\left(  S^{\ast\alpha}\ast E_{H}\ast S^{\ast\beta
}\right)  $ is a projection from $H$ such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  S^{\ast\alpha}\ast E_{H}\ast
S^{\ast\beta}\right)  \right)  =k\cdot1_{H}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. \ \ \ \ \footnote{Recall that the
notation $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$ is to be understood according to Convention~\ref{Convention100}. Hence,
$\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$
means the subspace $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  $ of $H$.}
\end{theorem}

Theorem~\ref{Theorem902} \textbf{(a)} easily follows from
Theorem~\ref{Theorem1051} when $\alpha=0$ and $\beta=1$.
Theorem~\ref{Theorem901} \textbf{(b)} easily follows from
Theorem~\ref{Theorem1051} when $\alpha=1$ and $\beta=0$.

\begin{proof}
[Proof of Theorem~\ref{Theorem1061}.]Define the map $e_{H,H}$ as according to
Definition~\ref{def.eHA}.

Just as in the proof of Theorem~\ref{Theorem1051}, we can see that $S\in
G\left(  H,H\right)  $.

By the definition of $S^{\ast\alpha}$, we have $S^{\ast\alpha}%
=\operatorname*{pow}\nolimits_{\alpha}S$.

By Definition~\ref{Definition602}, we have $H^{\operatorname*{op}}=\left(
H,\mu_{H}\circ\tau_{H,H},\eta_{H}\right)  $. Since $H$ is commutative, we have
$\mu_{H}\circ\tau_{H,H}=\mu_{H}$. Thus, $H^{\operatorname*{op}}=\left(
H,\underbrace{\mu_{H}\circ\tau_{H,H}}_{=\mu_{H}},\eta_{H}\right)  =\left(
H,\mu_{H},\eta_{H}\right)  =H$.

Proposition~\ref{Proposition603} yields that the antipode of $H$ is a
$k$-algebra homomorphism from $H$ to $H^{\operatorname*{op}}$. Since the
antipode of $H$ is the map $S$, whereas $H^{\operatorname*{op}}$ is $H$, this
rewrites as follows: The map $S$ is a $k$-algebra homomorphism from $H$ to $H$.

Just as in the proof of Theorem~\ref{Theorem1051}, we can see that
$\operatorname*{pow}\nolimits_{\alpha}S$ is graded.

Altogether, we have now shown that the map $\operatorname*{pow}%
\nolimits_{\alpha}S$ is a graded $k$-algebra homomorphism. Since
$\operatorname*{pow}\nolimits_{\alpha}S=S^{\ast\alpha}$, this rewrites as
follows: The map $S^{\ast\alpha}$ is a graded $k$-algebra homomorphism. The
same argument, but with $\alpha$ replaced by $\beta$, shows that the map
$S^{\ast\beta}$ is a graded $k$-algebra homomorphism.

Just as in the proof of Theorem~\ref{Theorem1051}, we can see that
$S^{\ast\alpha}\in G\left(  H,H\right)  $ and $S^{\ast\alpha}\left(
1_{H}\right)  =1_{H}$. The same argument, but with $\alpha$ replaced by
$\beta$, shows that $S^{\ast\beta}\in G\left(  H,H\right)  $ and $S^{\ast
\beta}\left(  1_{H}\right)  =1_{H}$.

It is now easy to see that
\begin{equation}
\left(  \varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)  \left(
H_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}.
\label{1061.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{1061.pf.1}):} Let $n\in\mathbb{N}$. Let $x\in
H_{n}$. We are going to show that $\left(  \varepsilon_{H}\circ S^{\ast\alpha
}-\varepsilon_{H}\right)  \left(  x\right)  =0$.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} We have $n=0$.
\par
\textit{Case 2:} We have $n>0$.
\par
Let us first consider Case 1. In this case, $n=0$, so that $H_{n}=H_{0}%
=k\cdot1_{H}$ (by Proposition~\ref{Proposition921}). Thus, $x\in H_{n}%
=k\cdot1_{H}$. In other words, there exists some $\lambda\in k$ such that
$x=\lambda\cdot1_{H}$. Consider this $\lambda$. Then,%
\begin{align*}
\left(  \varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)
\underbrace{\left(  x\right)  }_{=\lambda\cdot1_{H}}  &  =\left(
\varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)  \left(
\lambda\cdot1_{H}\right)  =\lambda\cdot\underbrace{\left(  \varepsilon
_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)  \left(  1_{H}\right)
}_{=\left(  \varepsilon_{H}\circ S^{\ast\alpha}\right)  \left(  1_{H}\right)
-\varepsilon_{H}\left(  1_{H}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon_{H}\circ S^{\ast
\alpha}-\varepsilon_{H}\text{ is }k\text{-linear}\right) \\
&  =\lambda\cdot\left(  \underbrace{\left(  \varepsilon_{H}\circ S^{\ast
\alpha}\right)  \left(  1_{H}\right)  }_{=\varepsilon_{H}\left(  S^{\ast
\alpha}\left(  1_{H}\right)  \right)  }-\varepsilon_{H}\left(  1_{H}\right)
\right) \\
&  =\lambda\cdot\left(  \varepsilon_{H}\left(  \underbrace{S^{\ast\alpha
}\left(  1_{H}\right)  }_{=1_{H}}\right)  -\varepsilon_{H}\left(
1_{H}\right)  \right)  =\lambda\cdot\underbrace{\left(  \varepsilon_{H}\left(
1_{H}\right)  -\varepsilon_{H}\left(  1_{H}\right)  \right)  }_{=0}=0.
\end{align*}
We have thus proven that $\left(  \varepsilon_{H}\circ S^{\ast\alpha
}-\varepsilon_{H}\right)  \left(  x\right)  =0$ in Case 1.
\par
Next, let us consider Case 2. In this case, $n>0$, so that $\varepsilon
_{H}\left(  H_{n}\right)  =0$ (because $H$ is a graded $k$-coalgebra). On the
other hand, $S^{\ast\alpha}\left(  H_{n}\right)  \subseteq H_{n}$ (since
$S^{\ast\alpha}$ is graded). Thus,%
\begin{align*}
\left(  \varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)  \left(
x\right)   &  =\left(  \varepsilon_{H}\circ S^{\ast\alpha}\right)  \left(
x\right)  -\varepsilon_{H}\left(  x\right)  =\varepsilon_{H}\left(
S^{\ast\alpha}\left(  \underbrace{x}_{\in H_{n}}\right)  \right)
-\varepsilon_{H}\left(  \underbrace{x}_{\in H_{n}}\right) \\
&  \in\varepsilon_{H}\left(  \underbrace{S^{\ast\alpha}\left(  H_{n}\right)
}_{\subseteq H_{n}}\right)  -\underbrace{\varepsilon_{H}\left(  H_{n}\right)
}_{=0}\subseteq\varepsilon_{H}\left(  H_{n}\right)  =0,
\end{align*}
so that $\left(  \varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)
\left(  x\right)  =0$. We have thus proven that $\left(  \varepsilon_{H}\circ
S^{\ast\alpha}-\varepsilon_{H}\right)  \left(  x\right)  =0$ in Case 2.
\par
Hence, $\left(  \varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)
\left(  x\right)  =0$ is proven in both cases 1 and 2. Since these two cases
cover all possibilities, this yields that $\left(  \varepsilon_{H}\circ
S^{\ast\alpha}-\varepsilon_{H}\right)  \left(  x\right)  =0$ always holds.
\par
Now forget that we fixed $x$. We thus have shown that every $x\in H_{n}$
satisfies $\left(  \varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)
\left(  x\right)  =0$. In other words, $\left(  \varepsilon_{H}\circ
S^{\ast\alpha}-\varepsilon_{H}\right)  \left(  H_{n}\right)  =0$. This proves
(\ref{1061.pf.1}).} Now, it is easy to conclude that $\varepsilon_{H}\circ
S^{\ast\alpha}=\varepsilon_{H}$.\ \ \ \ \footnote{\textit{Proof.} Since $H$ is
graded, we have $H=\bigoplus_{n\in\mathbb{N}}H_{n}=\sum_{n\in\mathbb{N}}H_{n}$
(since direct sums are sums). Thus,%
\begin{align*}
\left(  \varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\right)  \left(
H\right)   &  =\left(  \varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon
_{H}\right)  \left(  \sum_{n\in\mathbb{N}}H_{n}\right) \\
&  =\sum_{n\in\mathbb{N}}\underbrace{\left(  \varepsilon_{H}\circ
S^{\ast\alpha}-\varepsilon_{H}\right)  \left(  H_{n}\right)  }%
_{\substack{=0\\\text{(by (\ref{1061.pf.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}\text{ is
}k\text{-linear}\right) \\
&  =\sum_{n\in\mathbb{N}}0=0,
\end{align*}
so that $\varepsilon_{H}\circ S^{\ast\alpha}-\varepsilon_{H}=0$, and thus
$\varepsilon_{H}\circ S^{\ast\alpha}=\varepsilon_{H}$, qed.} The same
argument, but with $\alpha$ replaced by $\beta$, shows that $\varepsilon
_{H}\circ S^{\ast\beta}=\varepsilon_{H}$.

Finally, just as in the proof of Theorem~\ref{Theorem1051}, we can prove that
$S^{\ast\alpha}\ast\operatorname{id}_{H}\ast S^{\ast\beta}=e_{H,H}$.

Thus, we can apply Corollary~\ref{Corollary932} to $P=S^{\ast\alpha}$ and
$Q=S^{\ast\beta}$. As a result, we can conclude that the map $E_{H}%
^{\operatorname*{inv}}\circ\left(  S^{\ast\alpha}\ast E_{H}\ast S^{\ast\beta
}\right)  $ is a projection such that $\operatorname*{Ker}\left(
E_{H}^{\operatorname*{inv}}\circ\left(  S^{\ast\alpha}\ast E_{H}\ast
S^{\ast\beta}\right)  \right)  =k\cdot1_{H}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. This proves Theorem~\ref{Theorem1061}.
\end{proof}

\section{\label{S28}On convolution and composition}

The following innocuous fact about convolution and composition will help us
later in proving the Cartier-Milnor-Moore theorem:

\begin{proposition}
\label{Proposition1101} Let $k$ be a field. Let $C$ and $D$ be $k$-coalgebras.
Let $A$ be a $k$-algebra. Let $\varphi:D\rightarrow C$ be a $k$-coalgebra homomorphism.

\textbf{(a)} Every $f\in\mathcal{L}\left(  C,A\right)  $ and $g\in
\mathcal{L}\left(  C,A\right)  $ satisfy $\left(  f\circ\varphi\right)
\ast\left(  g\circ\varphi\right)  =\left(  f\ast g\right)  \circ\varphi$.

\textbf{(b)} We have $e_{C,A}\circ\varphi=e_{D,A}$.

\textbf{(c)} Every $f\in\mathcal{L}\left(  C,A\right)  $ and $i\in\mathbb{N}$
satisfy $\left(  f\circ\varphi\right)  ^{\ast i}=f^{\ast i}\circ\varphi$.

\textbf{(d)} Assume that the $k$-coalgebras $C$ and $D$ are connected filtered
$k$-coalgebras. Assume further that the map $\varphi:D\rightarrow C$ satisfies
$\varphi\left(  1_{D}\right)  =1_{C}$. Assume finally that $k$ is a field of
characteristic $0$. Then, every $f\in\mathfrak{g}\left(  C,A\right)  $
satisfies $f\circ\varphi\in\mathfrak{g}\left(  D,A\right)  $ and
$e^{\ast\left(  f\circ\varphi\right)  }=e^{\ast f}\circ\varphi$.

\textbf{(e)} Assume that the $k$-coalgebras $C$ and $D$ are connected filtered
$k$-coalgebras. Assume further that the map $\varphi:D\rightarrow C$ satisfies
$\varphi\left(  1_{D}\right)  =1_{C}$. Assume finally that $k$ is a field of
characteristic $0$. Then, every $F\in G\left(  C,A\right)  $ satisfies
$F\circ\varphi\in G\left(  D,A\right)  $ and $\operatorname*{Log}\left(
F\circ\varphi\right)  =\left(  \operatorname*{Log}F\right)  \circ\varphi$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1101}.]\textbf{(a)} Let $f\in
\mathcal{L}\left(  C,A\right)  $ and $g\in\mathcal{L}\left(  C,A\right)  $. By
the definition of the convolution $f\ast g$, we have%
\[
f\ast g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}.
\]
By the definition of the convolution $\left(  f\circ\varphi\right)
\ast\left(  g\circ\varphi\right)  $, we have%
\begin{align*}
\left(  f\circ\varphi\right)  \ast\left(  g\circ\varphi\right)   &  =\mu
_{A}\circ\underbrace{\left(  \left(  f\circ\varphi\right)  \otimes\left(
g\circ\varphi\right)  \right)  }_{\substack{=\left(  f\otimes g\right)
\circ\left(  \varphi\otimes\varphi\right)  \\\text{(by (\ref{(X)(X)}), applied
to }U=D\text{, }V=C\text{, }W=A\text{,}\\U^{\prime}=D\text{, }V^{\prime
}=C\text{, }W^{\prime}=A\text{, }\alpha=\varphi\text{, }\beta=f\text{, }%
\alpha^{\prime}=\varphi\text{ and }\beta^{\prime}=g\text{)}}}\circ\Delta_{D}\\
&  =\mu_{A}\circ\left(  f\otimes g\right)  \circ\underbrace{\left(
\varphi\otimes\varphi\right)  \circ\Delta_{D}}_{\substack{=\Delta_{C}%
\circ\varphi\\\text{(since }\varphi\text{ is a }k\text{-coalgebra}%
\\\text{homomorphism)}}}=\underbrace{\mu_{A}\circ\left(  f\otimes g\right)
\circ\Delta_{C}}_{=f\ast g}\circ\varphi\\
&  =\left(  f\ast g\right)  \circ\varphi.
\end{align*}
This proves Proposition~\ref{Proposition1101} \textbf{(a)}.

\textbf{(b)} By the definition of $e_{C,A}$, we have $e_{C,A}=\eta_{A}%
\circ\varepsilon_{C}$. By the definition of $e_{D,A}$, we have $e_{D,A}%
=\eta_{A}\circ\varepsilon_{D}$. Thus,%
\[
\underbrace{e_{C,A}}_{=\eta_{A}\circ\varepsilon_{C}}\circ\varphi=\eta_{A}%
\circ\underbrace{\varepsilon_{C}\circ\varphi}_{\substack{=\varepsilon
_{D}\\\text{(since }\varphi\text{ is a }k\text{-coalgebra}%
\\\text{homomorphism)}}}=\eta_{A}\circ\varepsilon_{D}=e_{D,A}.
\]
This proves Proposition~\ref{Proposition1101} \textbf{(b)}.

\textbf{(c)} We are going to prove Proposition~\ref{Proposition1101}
\textbf{(c)} by induction over $i$:

\textit{Induction base:} Every $f\in\mathcal{L}\left(  C,A\right)  $ satisfies%
\begin{align*}
\left(  f\circ\varphi\right)  ^{\ast0}  &  =e_{D,A}=\underbrace{e_{C,A}%
}_{=f^{\ast0}}\circ\varphi\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{Proposition1101} \textbf{(b)}}\right) \\
&  =f^{\ast0}\circ\varphi.
\end{align*}
In other words, Proposition~\ref{Proposition1101} \textbf{(c)} holds for
$i=0$. This completes the induction base.

\textit{Induction step:} Let $I\in\mathbb{N}$. Assume that
Proposition~\ref{Proposition1101} \textbf{(c)} holds for $i=I$. We now must
prove that Proposition~\ref{Proposition1101} \textbf{(c)} holds for $i=I+1$.

Every $f\in\mathcal{L}\left(  C,A\right)  $ satisfies $\left(  f\circ
\varphi\right)  ^{\ast I}=f^{\ast I}\circ\varphi$ (since
Proposition~\ref{Proposition1101} \textbf{(c)} holds for $i=I$). Now, every
$f\in\mathcal{L}\left(  C,A\right)  $ satisfies%
\begin{align*}
\left(  f\circ\varphi\right)  ^{\ast\left(  I+1\right)  }  &  =\left(
f\circ\varphi\right)  ^{\ast\left(  1+I\right)  }=\left(  f\circ
\varphi\right)  \ast\underbrace{\left(  f\circ\varphi\right)  ^{\ast I}%
}_{=f^{\ast I}\circ\varphi}=\left(  f\circ\varphi\right)  \ast\left(  f^{\ast
I}\circ\varphi\right) \\
&  =\underbrace{\left(  f\ast f^{\ast I}\right)  }_{=f^{\ast\left(
1+I\right)  }=f^{\ast\left(  I+1\right)  }}\circ\varphi
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition~\ref{Proposition1101}
\textbf{(a)}, applied to }g=f^{\ast I}\right) \\
&  =f^{\ast\left(  I+1\right)  }\circ\varphi.
\end{align*}
In other words, Proposition~\ref{Proposition1101} \textbf{(c)} holds for
$i=I+1$. This completes the induction step. The induction proof of
Proposition~\ref{Proposition1101} \textbf{(c)} is thus finished.

\textbf{(d)} For every connected filtered $k$-coalgebra $H$, we have%
\begin{align}
\mathfrak{g}\left(  H,A\right)   &  =\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathfrak{g}\left(
H,A\right)  \right) \nonumber\\
&  =\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\left(  1_{H}\right)
=0\right\}  \label{1101.pf.6}%
\end{align}
(here, we renamed the index $f$ as $h$).

Now, let $f\in\mathfrak{g}\left(  C,A\right)  $. Then,%
\[
f\in\mathfrak{g}\left(  C,A\right)  =\left\{  h\in\mathcal{L}\left(
C,A\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1101.pf.6}), applied to
}H=C\right)  .
\]
Hence, $f\in\mathcal{L}\left(  C,A\right)  $ and $f\left(  1_{C}\right)  =0$.
We have $f\circ\varphi\in\mathcal{L}\left(  D,A\right)  $ and%
\[
\left(  f\circ\varphi\right)  \left(  1_{D}\right)  =f\left(
\underbrace{\varphi\left(  1_{D}\right)  }_{=1_{C}}\right)  =f\left(
1_{C}\right)  =0.
\]
Hence, $f\circ\varphi\in\left\{  h\in\mathcal{L}\left(  D,A\right)
\ \mid\ h\left(  1_{D}\right)  =0\right\}  $. Applying (\ref{1101.pf.6}) to
$H=D$, we obtain $\mathfrak{g}\left(  D,A\right)  =\left\{  h\in
\mathcal{L}\left(  D,A\right)  \ \mid\ h\left(  1_{D}\right)  =0\right\}  $.
Thus, $f\circ\varphi\in\left\{  h\in\mathcal{L}\left(  D,A\right)
\ \mid\ h\left(  1_{D}\right)  =0\right\}  =\mathfrak{g}\left(  D,A\right)  $.
Hence, $e^{\ast\left(  f\circ\varphi\right)  }$ is well-defined.

Let $x\in D$. By the definition of $e^{\ast\left(  f\circ\varphi\right)  }$,
we have $e^{\ast\left(  f\circ\varphi\right)  }\left(  x\right)  =\sum
_{i\geq0}\dfrac{\left(  f\circ\varphi\right)  ^{\ast i}\left(  x\right)  }%
{i!}$.

By the definition of $e^{\ast f}$, we have $e^{\ast f}\left(  y\right)
=\sum_{i\geq0}\dfrac{f^{\ast i}\left(  y\right)  }{i!}$ for every $y\in C$.
Applying this to $y=\varphi\left(  x\right)  $, we obtain
\begin{align*}
&  e^{\ast f}\left(  \varphi\left(  x\right)  \right) \\
&  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(  \varphi\left(  x\right)  \right)
}{i!}=\sum_{i\geq0}\dfrac{\left(  f\circ\varphi\right)  ^{\ast i}\left(
x\right)  }{i!}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }i\in\mathbb{N}\text{
satisfies }f^{\ast i}\left(  \varphi\left(  x\right)  \right)
=\underbrace{\left(  f^{\ast i}\circ\varphi\right)  }_{\substack{=\left(
f\circ\varphi\right)  ^{\ast i}\\\text{(by Proposition~\ref{Proposition1101}
\textbf{(c)})}}}\left(  x\right)  =\left(  f\circ\varphi\right)  ^{\ast
i}\left(  x\right)  \right) \\
&  =e^{\ast\left(  f\circ\varphi\right)  }\left(  x\right)  .
\end{align*}
Thus, $e^{\ast\left(  f\circ\varphi\right)  }\left(  x\right)  =e^{\ast
f}\left(  \varphi\left(  x\right)  \right)  =\left(  e^{\ast f}\circ
\varphi\right)  \left(  x\right)  $.

Now, forget that we fixed $x$. We thus have proven that $e^{\ast\left(
f\circ\varphi\right)  }\left(  x\right)  =\left(  e^{\ast f}\circ
\varphi\right)  \left(  x\right)  $ for every $x\in D$. In other words,
$e^{\ast\left(  f\circ\varphi\right)  }=e^{\ast f}\circ\varphi$. This proves
Proposition~\ref{Proposition1101} \textbf{(d)}.

\textbf{(e)} For every connected filtered $k$-coalgebra $H$, we have%
\begin{align}
G\left(  H,A\right)   &  =\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\left(  1_{H}\right)  =1_{A}\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }G\left(  H,A\right)  \right) \nonumber\\
&  =\left\{  h\in\mathcal{L}\left(  H,A\right)  \ \mid\ h\left(  1_{H}\right)
=1_{A}\right\}  \label{1101.pf.12}%
\end{align}
(here, we renamed the index $f$ as $h$).

Now, let $F\in G\left(  C,A\right)  $. Then,%
\[
F\in G\left(  C,A\right)  =\left\{  h\in\mathcal{L}\left(  C,A\right)
\ \mid\ h\left(  1_{C}\right)  =1_{A}\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{1101.pf.12}), applied to }H=C\right)  .
\]
Hence, $F\in\mathcal{L}\left(  C,A\right)  $ and $F\left(  1_{C}\right)
=1_{A}$. We have $F\circ\varphi\in\mathcal{L}\left(  D,A\right)  $ and%
\[
\left(  F\circ\varphi\right)  \left(  1_{D}\right)  =F\left(
\underbrace{\varphi\left(  1_{D}\right)  }_{=1_{C}}\right)  =F\left(
1_{C}\right)  =1_{A}.
\]
Hence, $F\circ\varphi\in\left\{  h\in\mathcal{L}\left(  D,A\right)
\ \mid\ h\left(  1_{D}\right)  =1_{A}\right\}  $. Applying (\ref{1101.pf.12})
to $H=D$, we obtain $G\left(  D,A\right)  =\left\{  h\in\mathcal{L}\left(
D,A\right)  \ \mid\ h\left(  1_{D}\right)  =1_{A}\right\}  $. Thus,
$F\circ\varphi\in\left\{  h\in\mathcal{L}\left(  D,A\right)  \ \mid\ h\left(
1_{D}\right)  =1_{A}\right\}  =G\left(  D,A\right)  $. Hence,
$\operatorname*{Log}\left(  F\circ\varphi\right)  $ is well-defined.

Let $f=F-e_{C,A}$. Since $f=F-e_{C,A}$, we have%
\begin{align*}
f\circ\varphi &  =\left(  F-e_{C,A}\right)  \circ\varphi=F\circ\varphi
-\underbrace{e_{C,A}}_{\substack{=\eta_{A}\circ\varepsilon_{C}\\\text{(by the
definition of }e_{C,A}\text{)}}}\circ\varphi\\
&  =F\circ\varphi-\eta_{A}\circ\underbrace{\varepsilon_{C}\circ\varphi
}_{\substack{=\varepsilon_{D}\\\text{(since }\varphi\text{ is a }%
k\text{-coalgebra homomorphism)}}}=F\circ\varphi-\underbrace{\eta_{A}%
\circ\varepsilon_{D}}_{\substack{=e_{D,A}\\\text{(since }e_{D,A}=\eta_{A}%
\circ\varepsilon_{D}\\\text{(by the definition of }e_{D,A}\text{))}}}\\
&  =F\circ\varphi-e_{D,A}.
\end{align*}


Let $x\in D$. By the definition of $\operatorname*{Log}\left(  F\circ
\varphi\right)  $, we have $\operatorname*{Log}\left(  F\circ\varphi\right)
=\operatorname{Log}_{1}\left(  \underbrace{F\circ\varphi-e_{D,A}}%
_{=f\circ\varphi}\right)  =\operatorname{Log}_{1}\left(  f\circ\varphi\right)
$. Thus,%
\[
\left(  \operatorname*{Log}\left(  F\circ\varphi\right)  \right)  \left(
x\right)  =\left(  \operatorname{Log}_{1}\left(  f\circ\varphi\right)
\right)  \left(  x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}\left(  f\circ\varphi\right)  ^{\ast i}\left(  x\right)
\]
(by the definition of $\operatorname{Log}_{1}\left(  f\circ\varphi\right)  $).

By the definition of $\operatorname*{Log}F$, we have $\operatorname*{Log}%
F=\operatorname{Log}_{1}\left(  \underbrace{F-e_{C,A}}_{=f}\right)
=\operatorname{Log}_{1}f$. Hence, for every $y\in C$, we have%
\[
\left(  \operatorname*{Log}F\right)  \left(  y\right)  =\left(
\operatorname{Log}_{1}f\right)  \left(  y\right)  =\sum_{i\geq1}\dfrac{\left(
-1\right)  ^{i-1}}{i}f^{\ast i}\left(  y\right)
\]
(by the definition of $\operatorname{Log}_{1}f$). Applying this to
$y=\varphi\left(  x\right)  $, we obtain
\begin{align*}
&  \left(  \operatorname*{Log}F\right)  \left(  \varphi\left(  x\right)
\right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
\varphi\left(  x\right)  \right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}\left(  f\circ\varphi\right)  ^{\ast i}\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }i\in\mathbb{N}\text{
satisfies }f^{\ast i}\left(  \varphi\left(  x\right)  \right)
=\underbrace{\left(  f^{\ast i}\circ\varphi\right)  }_{\substack{=\left(
f\circ\varphi\right)  ^{\ast i}\\\text{(by Proposition~\ref{Proposition1101}
\textbf{(c)})}}}\left(  x\right)  =\left(  f\circ\varphi\right)  ^{\ast
i}\left(  x\right)  \right) \\
&  =\left(  \operatorname*{Log}\left(  F\circ\varphi\right)  \right)  \left(
x\right)  .
\end{align*}
Thus, $\left(  \operatorname*{Log}\left(  F\circ\varphi\right)  \right)
\left(  x\right)  =\left(  \operatorname*{Log}F\right)  \left(  \varphi\left(
x\right)  \right)  =\left(  \left(  \operatorname*{Log}F\right)  \circ
\varphi\right)  \left(  x\right)  $.

Now, forget that we fixed $x$. We thus have proven that $\left(
\operatorname*{Log}\left(  F\circ\varphi\right)  \right)  \left(  x\right)
=\left(  \left(  \operatorname*{Log}F\right)  \circ\varphi\right)  \left(
x\right)  $ for every $x\in D$. In other words, $\operatorname*{Log}\left(
F\circ\varphi\right)  =\left(  \operatorname*{Log}F\right)  \circ\varphi$.
This proves Proposition~\ref{Proposition1101} \textbf{(e)}.
\end{proof}

The following proposition is ``more or less'' a dual of
Proposition~\ref{Proposition1101}:

\begin{proposition}
\label{Proposition1102} Let $k$ be a field. Let $A$ and $B$ be $k$-algebras.
Let $C$ be a $k$-coalgebra. Let $\psi:A\rightarrow B$ be a $k$-algebra homomorphism.

\textbf{(a)} Every $f\in\mathcal{L}\left(  C,A\right)  $ and $g\in
\mathcal{L}\left(  C,A\right)  $ satisfy $\left(  \psi\circ f\right)
\ast\left(  \psi\circ g\right)  =\psi\circ\left(  f\ast g\right)  $.

\textbf{(b)} We have $\psi\circ e_{C,A}=e_{C,B}$.

\textbf{(c)} Every $f\in\mathcal{L}\left(  C,A\right)  $ and $i\in\mathbb{N}$
satisfy $\left(  \psi\circ f\right)  ^{\ast i}=\psi\circ f^{\ast i}$.

\textbf{(d)} Assume that the $k$-coalgebra $C$ is a connected filtered
$k$-coalgebra. Assume finally that $k$ is a field of characteristic $0$. Then,
every $f\in\mathfrak{g}\left(  C,A\right)  $ satisfies $\psi\circ
f\in\mathfrak{g}\left(  C,B\right)  $ and $e^{\ast\left(  \psi\circ f\right)
}=\psi\circ e^{\ast f}$.

\textbf{(e)} Assume that the $k$-coalgebra $C$ is a connected filtered
$k$-coalgebra. Assume finally that $k$ is a field of characteristic $0$. Then,
every $F\in G\left(  C,A\right)  $ satisfies $\psi\circ F\in G\left(
C,B\right)  $ and $\operatorname*{Log}\left(  \psi\circ F\right)  =\psi
\circ\left(  \operatorname*{Log}F\right)  $.
\end{proposition}

Notice that the conditions in Proposition~\ref{Proposition1102} \textbf{(d)}
and \textbf{(e)} are a bit more liberal than those in
Proposition~\ref{Proposition1101} \textbf{(d)} and \textbf{(e)}, whence it
would not be proper to call Proposition~\ref{Proposition1102} a precise dual
of Proposition~\ref{Proposition1101}; but these conditions were technical in
the first place, allowing one to define $e^{\ast\left(  f\circ\varphi\right)
}$ respectively $\operatorname*{Log}\left(  F\circ\varphi\right)  $.

\begin{proof}
[Proof of Proposition~\ref{Proposition1102}.]\textbf{(a)} Let $f\in
\mathcal{L}\left(  C,A\right)  $ and $g\in\mathcal{L}\left(  C,A\right)  $. By
the definition of the convolution $f\ast g$, we have%
\[
f\ast g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}.
\]
By the definition of the convolution $\left(  \psi\circ f\right)  \ast\left(
\psi\circ g\right)  $, we have%
\begin{align*}
\left(  \psi\circ f\right)  \ast\left(  \psi\circ g\right)   &  =\mu_{B}%
\circ\underbrace{\left(  \left(  \psi\circ f\right)  \otimes\left(  \psi\circ
g\right)  \right)  }_{\substack{=\left(  \psi\otimes\psi\right)  \circ\left(
f\otimes g\right)  \\\text{(by (\ref{(X)(X)}), applied to }U=C\text{,
}V=A\text{, }W=B\text{,}\\U^{\prime}=C\text{, }V^{\prime}=A\text{, }W^{\prime
}=B\text{, }\alpha=f\text{, }\beta=\psi\text{, }\alpha^{\prime}=g\text{ and
}\beta^{\prime}=\psi\text{)}}}\circ\Delta_{C}\\
&  =\underbrace{\mu_{B}\circ\left(  \psi\otimes\psi\right)  }_{\substack{=\psi
\circ\mu_{A}\\\text{(since }\psi\text{ is a }k\text{-algebra}%
\\\text{homomorphism)}}}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
=\psi\circ\underbrace{\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
}_{=f\ast g}=\psi\circ\left(  f\ast g\right)  .
\end{align*}
This proves Proposition~\ref{Proposition1102} \textbf{(a)}.

\textbf{(b)} By the definition of $e_{C,A}$, we have $e_{C,A}=\eta_{A}%
\circ\varepsilon_{C}$. By the definition of $e_{C,B}$, we have $e_{C,B}%
=\eta_{B}\circ\varepsilon_{C}$. Thus,%
\[
\psi\circ\underbrace{e_{C,A}}_{=\eta_{A}\circ\varepsilon_{C}}=\underbrace{\psi
\circ\eta_{A}}_{\substack{=\eta_{B}\\\text{(since }\psi\text{ is a
}k\text{-algebra}\\\text{homomorphism)}}}\circ\varepsilon_{C}=\eta_{B}%
\circ\varepsilon_{C}=e_{C,B}.
\]
This proves Proposition~\ref{Proposition1102} \textbf{(b)}.

\textbf{(c)} We are going to prove Proposition~\ref{Proposition1102}
\textbf{(c)} by induction over $i$:

\textit{Induction base:} Every $f\in\mathcal{L}\left(  C,A\right)  $ satisfies%
\begin{align*}
\left(  f\circ\psi\right)  ^{\ast0}  &  =e_{C,B}=\psi\circ\underbrace{e_{C,A}%
}_{=f^{\ast0}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{Proposition1102} \textbf{(b)}}\right) \\
&  =\psi\circ f^{\ast0}.
\end{align*}
In other words, Proposition~\ref{Proposition1102} \textbf{(c)} holds for
$i=0$. This completes the induction base.

\textit{Induction step:} Let $I\in\mathbb{N}$. Assume that
Proposition~\ref{Proposition1102} \textbf{(c)} holds for $i=I$. We now must
prove that Proposition~\ref{Proposition1102} \textbf{(c)} holds for $i=I+1$.

Every $f\in\mathcal{L}\left(  C,A\right)  $ satisfies $\left(  \psi\circ
f\right)  ^{\ast I}=\psi\circ f^{\ast I}$ (since
Proposition~\ref{Proposition1102} \textbf{(c)} holds for $i=I$). Now, every
$f\in\mathcal{L}\left(  C,A\right)  $ satisfies%
\begin{align*}
\left(  \psi\circ f\right)  ^{\ast\left(  I+1\right)  }  &  =\left(  \psi\circ
f\right)  ^{\ast\left(  1+I\right)  }=\left(  \psi\circ f\right)
\ast\underbrace{\left(  \psi\circ f\right)  ^{\ast I}}_{=\psi\circ f^{\ast I}%
}=\left(  \psi\circ f\right)  \ast\left(  \psi\circ f^{\ast I}\right) \\
&  =\psi\circ\underbrace{\left(  f\ast f^{\ast I}\right)  }_{=f^{\ast\left(
1+I\right)  }=f^{\ast\left(  I+1\right)  }}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition~\ref{Proposition1102} \textbf{(a)}, applied to }g=f^{\ast
I}\right) \\
&  =\psi\circ f^{\ast\left(  I+1\right)  }.
\end{align*}
In other words, Proposition~\ref{Proposition1102} \textbf{(c)} holds for
$i=I+1$. This completes the induction step. The induction proof of
Proposition~\ref{Proposition1102} \textbf{(c)} is thus finished.

\textbf{(d)} For every connected filtered $k$-coalgebra $H$ and any
$k$-algebra $\mathfrak{A}$, we have%
\begin{align}
\mathfrak{g}\left(  H,\mathfrak{A}\right)   &  =\left\{  f\in\mathcal{L}%
\left(  H,\mathfrak{A}\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathfrak{g}\left(
H,\mathfrak{A}\right)  \right) \nonumber\\
&  =\left\{  h\in\mathcal{L}\left(  H,\mathfrak{A}\right)  \ \mid\ h\left(
1_{H}\right)  =0\right\}  \label{1201.pf.6}%
\end{align}
(here, we renamed the index $f$ as $h$).

Now, let $f\in\mathfrak{g}\left(  C,A\right)  $. Then,%
\[
f\in\mathfrak{g}\left(  C,A\right)  =\left\{  h\in\mathcal{L}\left(
C,A\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1201.pf.6}), applied to }H=C\text{
and }\mathfrak{A}=A\right)  .
\]
Hence, $f\in\mathcal{L}\left(  C,A\right)  $ and $f\left(  1_{C}\right)  =0$.
We have $\psi\circ f\in\mathcal{L}\left(  C,B\right)  $ and%
\[
\left(  \psi\circ f\right)  \left(  1_{C}\right)  =\psi\left(
\underbrace{f\left(  1_{C}\right)  }_{=0}\right)  =\psi\left(  0\right)  =0
\]
(since $\psi$ is $k$-linear). Hence, $\psi\circ f\in\left\{  h\in
\mathcal{L}\left(  C,B\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}  $.
Applying (\ref{1201.pf.6}) to $H=C$ and $\mathfrak{A}=B$, we obtain
$\mathfrak{g}\left(  C,B\right)  =\left\{  h\in\mathcal{L}\left(  C,B\right)
\ \mid\ h\left(  1_{C}\right)  =0\right\}  $. Thus, $\psi\circ f\in\left\{
h\in\mathcal{L}\left(  C,B\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}
=\mathfrak{g}\left(  C,B\right)  $. Hence, $e^{\ast\left(  \psi\circ f\right)
}$ is well-defined.

Let $x\in C$. By the definition of $e^{\ast\left(  \psi\circ f\right)  }$, we
have $e^{\ast\left(  \psi\circ f\right)  }\left(  x\right)  =\sum_{i\geq
0}\dfrac{\left(  \psi\circ f\right)  ^{\ast i}\left(  x\right)  }{i!}$.

By the definition of $e^{\ast f}$, we have $e^{\ast f}\left(  x\right)
=\sum_{i\geq0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}$. Notice that the sum
$\sum_{i\geq0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}$ converges with respect
to the discrete topology. Now,%
\begin{align*}
\psi\left(  e^{\ast f}\left(  x\right)  \right)   &  =\psi\left(  \sum
_{i\geq0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }e^{\ast f}\left(  x\right)
=\sum_{i\geq0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}\right) \\
&  =\sum_{i\geq0}\dfrac{\psi\left(  f^{\ast i}\left(  x\right)  \right)  }%
{i!}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\psi\text{ is }k\text{-linear, and since the sum }\sum_{i\geq
0}\dfrac{f^{\ast i}\left(  x\right)  }{i!}\\
\text{converges with respect to the discrete topology}%
\end{array}
\right) \\
&  =\sum_{i\geq0}\dfrac{\left(  \psi\circ f\right)  ^{\ast i}\left(  x\right)
}{i!}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }i\in\mathbb{N}\text{
satisfies }\psi\left(  f^{\ast i}\left(  x\right)  \right)
=\underbrace{\left(  \psi\circ f^{\ast i}\right)  }_{\substack{=\left(
\psi\circ f\right)  ^{\ast i}\\\text{(by Proposition~\ref{Proposition1102}
\textbf{(c)})}}}\left(  x\right)  =\left(  \psi\circ f\right)  ^{\ast
i}\left(  x\right)  \right) \\
&  =e^{\ast\left(  \psi\circ f\right)  }\left(  x\right)  .
\end{align*}


Thus, $e^{\ast\left(  \psi\circ f\right)  }\left(  x\right)  =\psi\left(
e^{\ast f}\left(  x\right)  \right)  =\left(  \psi\circ e^{\ast f}\right)
\left(  x\right)  $.

Now, forget that we fixed $x$. We thus have proven that $e^{\ast\left(
\psi\circ f\right)  }\left(  x\right)  =\left(  \psi\circ e^{\ast f}\right)
\left(  x\right)  $ for every $x\in C$. In other words, $e^{\ast\left(
\psi\circ f\right)  }=\psi\circ e^{\ast f}$. This proves
Proposition~\ref{Proposition1102} \textbf{(d)}.

\textbf{(e)} For every connected filtered $k$-coalgebra $H$ and any
$k$-algebra $\mathfrak{A}$, we have%
\begin{align}
G\left(  H,\mathfrak{A}\right)   &  =\left\{  f\in\mathcal{L}\left(
H,\mathfrak{A}\right)  \ \mid\ f\left(  1_{H}\right)  =1_{\mathfrak{A}%
}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }G\left(
H,\mathfrak{A}\right)  \right) \nonumber\\
&  =\left\{  h\in\mathcal{L}\left(  H,\mathfrak{A}\right)  \ \mid\ h\left(
1_{H}\right)  =1_{\mathfrak{A}}\right\}  \label{1201.pf.12}%
\end{align}
(here, we renamed the index $f$ as $h$).

Now, let $F\in G\left(  C,A\right)  $. Then,%
\[
F\in G\left(  C,A\right)  =\left\{  h\in\mathcal{L}\left(  C,A\right)
\ \mid\ h\left(  1_{C}\right)  =1_{A}\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{1201.pf.12}), applied to }H=C\text{ and }\mathfrak{A}=A\right)
.
\]
Hence, $F\in\mathcal{L}\left(  C,A\right)  $ and $F\left(  1_{C}\right)
=1_{A}$. We have $\psi\circ F\in\mathcal{L}\left(  C,B\right)  $ and%
\[
\left(  \psi\circ F\right)  \left(  1_{C}\right)  =\psi\left(
\underbrace{F\left(  1_{C}\right)  }_{=1_{A}}\right)  =\psi\left(
1_{A}\right)  =1_{B}%
\]
(since $\psi$ is a $k$-algebra homomorphism). Hence, $\psi\circ F\in\left\{
h\in\mathcal{L}\left(  C,B\right)  \ \mid\ h\left(  1_{C}\right)
=1_{B}\right\}  $. Applying (\ref{1201.pf.12}) to $H=C$ and $\mathfrak{A}=B$,
we obtain $G\left(  C,B\right)  =\left\{  h\in\mathcal{L}\left(  C,B\right)
\ \mid\ h\left(  1_{C}\right)  =1_{B}\right\}  $. Thus, $\psi\circ
F\in\left\{  h\in\mathcal{L}\left(  C,B\right)  \ \mid\ h\left(  1_{C}\right)
=1_{B}\right\}  =G\left(  C,B\right)  $. Hence, $\operatorname*{Log}\left(
\psi\circ F\right)  $ is well-defined.

Let $f=F-e_{C,A}$. Since $f=F-e_{C,A}$, we have%
\begin{align*}
\psi\circ f  &  =\psi\circ\left(  F-e_{C,A}\right)  =\psi\circ F-\psi
\circ\underbrace{e_{C,A}}_{\substack{=\eta_{A}\circ\varepsilon_{C}\\\text{(by
the definition of }e_{C,A}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is }k\text{-bilinear}\right) \\
&  =\psi\circ F-\underbrace{\psi\circ\eta_{A}}_{\substack{=\eta_{B}%
\\\text{(since }\psi\text{ is a }k\text{-algebra homomorphism)}}%
}\circ\varepsilon_{C}=\psi\circ F-\underbrace{\eta_{B}\circ\varepsilon_{C}%
}_{\substack{=e_{C,B}\\\text{(since }e_{C,B}=\eta_{B}\circ\varepsilon
_{C}\\\text{(by the definition of }e_{C,B}\text{))}}}\\
&  =\psi\circ F-e_{C,B}.
\end{align*}


Let $x\in C$. By the definition of $\operatorname*{Log}\left(  \psi\circ
F\right)  $, we have $\operatorname*{Log}\left(  \psi\circ F\right)
=\operatorname{Log}_{1}\left(  \underbrace{\psi\circ F-e_{C,B}}_{=\psi\circ
f}\right)  =\operatorname{Log}_{1}\left(  \psi\circ f\right)  $. Thus,%
\[
\left(  \operatorname*{Log}\left(  \psi\circ F\right)  \right)  \left(
x\right)  =\left(  \operatorname{Log}_{1}\left(  \psi\circ f\right)  \right)
\left(  x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(
\psi\circ f\right)  ^{\ast i}\left(  x\right)
\]
(by the definition of $\operatorname{Log}_{1}\left(  \psi\circ f\right)  $).

By the definition of $\operatorname*{Log}F$, we have $\operatorname*{Log}%
F=\operatorname{Log}_{1}\left(  \underbrace{F-e_{C,A}}_{=f}\right)
=\operatorname{Log}_{1}f$. Hence,%
\[
\left(  \operatorname*{Log}F\right)  \left(  x\right)  =\left(
\operatorname{Log}_{1}f\right)  \left(  x\right)  =\sum_{i\geq1}\dfrac{\left(
-1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)
\]
(by the definition of $\operatorname{Log}_{1}f$). Notice that the sum
$\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)
$ converges with respect to the discrete topology. Now,%
\begin{align*}
&  \psi\left(  \left(  \operatorname*{Log}F\right)  \left(  x\right)  \right)
\\
&  =\psi\left(  \sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast
i}\left(  x\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\operatorname*{Log}F\right)  \left(  x\right)  =\sum_{i\geq1}\dfrac{\left(
-1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)  \right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\psi\left(  f^{\ast
i}\left(  x\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\psi\text{ is }k\text{-linear, and since the sum }\sum_{i\geq
1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right) \\
\text{converges with respect to the discrete topology}%
\end{array}
\right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\left(  \psi\circ
f\right)  ^{\ast i}\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }i\in\mathbb{N}\text{
satisfies }\psi\left(  f^{\ast i}\left(  x\right)  \right)
=\underbrace{\left(  \psi\circ f^{\ast i}\right)  }_{\substack{=\left(
\psi\circ f\right)  ^{\ast i}\\\text{(by Proposition~\ref{Proposition1102}
\textbf{(c)})}}}\left(  x\right)  =\left(  \psi\circ f\right)  ^{\ast
i}\left(  x\right)  \right) \\
&  =\left(  \operatorname*{Log}\left(  \psi\circ F\right)  \right)  \left(
x\right)  .
\end{align*}
Thus, $\left(  \operatorname*{Log}\left(  \psi\circ F\right)  \right)  \left(
x\right)  =\psi\left(  \left(  \operatorname*{Log}F\right)  \left(  x\right)
\right)  =\left(  \psi\circ\left(  \operatorname*{Log}F\right)  \right)
\left(  x\right)  $.

Now, forget that we fixed $x$. We thus have proven that $\left(
\operatorname*{Log}\left(  \psi\circ F\right)  \right)  \left(  x\right)
=\left(  \psi\circ\left(  \operatorname*{Log}F\right)  \right)  \left(
x\right)  $ for every $x\in C$. In other words, $\operatorname*{Log}\left(
\psi\circ F\right)  =\psi\circ\left(  \operatorname*{Log}F\right)  $. This
proves Proposition~\ref{Proposition1102} \textbf{(e)}.
\end{proof}

We record, for future use, a consequence of Lemma~\ref{Lemma122}:

\begin{corollary}
\label{Corollary1103} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $A$ be a commutative
$k$-algebra. Let $F:H\rightarrow A$ be a $k$-algebra homomorphism. Let $x\in
H$ and $y\in H$ satisfy $\varepsilon_{H}\left(  x\right)  =0$ and
$\varepsilon_{H}\left(  y\right)  =0$. Then, $\operatorname*{Log}F$ is a
well-defined element of $\mathfrak{g}\left(  H,A\right)  $ and satisfies
$\left(  \operatorname*{Log}F\right)  \left(  xy\right)  =0$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1103}.]Recall that $G\left(  H,A\right)
=\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(  1_{H}\right)
=1_{A}\right\}  $ (by the definition of $G\left(  H,A\right)  $). But $F$ is a
$k$-algebra homomorphism, and thus satisfies $F\left(  1_{H}\right)  =1_{A}$.
Now,%
\begin{align*}
F  &  \in\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(
1_{H}\right)  =1_{A}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
F\in\mathcal{L}\left(  H,A\right)  \text{ and }F\left(  1_{H}\right)
=1_{A}\right) \\
&  =G\left(  H,A\right)  .
\end{align*}
Hence, $\operatorname*{Log}F$ is a well-defined element of $\mathfrak{g}%
\left(  H,A\right)  $.

It remains to prove that $\left(  \operatorname*{Log}F\right)  \left(
xy\right)  =0$.

Let $f=\operatorname*{Log}F$. Thus, $e^{\ast f}=e^{\ast\left(
\operatorname*{Log}F\right)  }=F$ (by Proposition~\ref{Proposition13}
\textbf{(b)}). Hence, $e^{\ast f}$ is a $k$-algebra homomorphism (since we
know that $F$ is a $k$-algebra homomorphism).

We have $f=\operatorname*{Log}F\in\mathfrak{g}\left(  H,A\right)  $. Hence,
Lemma~\ref{Lemma122} yields that $e^{\ast f}$ is a $k$-algebra homomorphism if
and only if $f$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation. Thus, $f$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation (since we know that $e^{\ast f}$ is a $k$-algebra homomorphism).

From Definition~\ref{Definition109}, we know that $f$ is an $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and only if every
$\left(  a,b\right)  \in H\times H$ satisfies $f\left(  ab\right)  =f\left(
a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)
f\left(  b\right)  $. Hence, every $\left(  a,b\right)  \in H\times H$
satisfies $f\left(  ab\right)  =f\left(  a\right)  \varepsilon_{H}\left(
b\right)  +\varepsilon_{H}\left(  a\right)  f\left(  b\right)  $ (because $f$
is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation). Applying
this to $\left(  a,b\right)  =\left(  x,y\right)  $, we obtain%
\[
f\left(  xy\right)  =f\left(  x\right)  \underbrace{\varepsilon_{H}\left(
y\right)  }_{=0}+\underbrace{\varepsilon_{H}\left(  x\right)  }_{=0}f\left(
y\right)  =0+0=0.
\]
Since $f=\operatorname*{Log}F$, this rewrites as $\left(  \operatorname*{Log}%
F\right)  \left(  xy\right)  =0$. This completes the proof of
Corollary~\ref{Corollary1103}.
\end{proof}

As the following proposition shows, the commutativity requirement on $A$ in
Corollary~\ref{Corollary1103} can be replaced by a commutativity requirement
on $H$:

\begin{corollary}
\label{Corollary1104} Let $k$ be a field of characteristic $0$. Let $H$ be a
commutative connected filtered $k$-coalgebra and, at the same time, a
$k$-bialgebra with the same underlying $k$-coalgebra structure. Let $A$ be a
$k$-algebra. Let $F:H\rightarrow A$ be a $k$-algebra homomorphism. Let $x\in
H$ and $y\in H$ satisfy $\varepsilon_{H}\left(  x\right)  =0$ and
$\varepsilon_{H}\left(  y\right)  =0$. Then, $\operatorname*{Log}F$ is a
well-defined element of $\mathfrak{g}\left(  H,A\right)  $ and satisfies
$\left(  \operatorname*{Log}F\right)  \left(  xy\right)  =0$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1104}.]First of all, $F\in G\left(
H,A\right)  $ (this can be proven just as in the proof of
Corollary~\ref{Corollary1103}). Thus, $\operatorname*{Log}F$ is a well-defined
element of $\mathfrak{g}\left(  H,A\right)  $.

We now need to prove that $\left(  \operatorname*{Log}F\right)  \left(
xy\right)  =0$.

Since $F:H\rightarrow A$ is a $k$-algebra homomorphism, its image $F\left(
H\right)  $ is a $k$-subalgebra of $A$. This $k$-subalgebra $F\left(
H\right)  $ is commutative\footnote{\textit{Proof.} Let $\alpha\in F\left(
H\right)  $ and $\beta\in F\left(  H\right)  $.
\par
Since $\alpha\in F\left(  H\right)  $, there exists an $\widetilde{\alpha}\in
H$ such that $\alpha=F\left(  \widetilde{\alpha}\right)  $. Consider this
$\widetilde{\alpha}$.
\par
Since $\beta\in F\left(  H\right)  $, there exists a $\widetilde{\beta}\in H$
such that $\beta=F\left(  \widetilde{\beta}\right)  $. Consider this
$\widetilde{\beta}$.
\par
Since $H$ is commutative, we have $\widetilde{\alpha}\widetilde{\beta
}=\widetilde{\beta}\widetilde{\alpha}$. Now, multiplying the equalities
$\alpha=F\left(  \widetilde{\alpha}\right)  $ and $\beta=F\left(
\widetilde{\beta}\right)  $, we obtain $\alpha\beta=F\left(  \widetilde{\alpha
}\right)  F\left(  \widetilde{\beta}\right)  =F\left(  \widetilde{\alpha
}\widetilde{\beta}\right)  $ (since $F$ is a $k$-algebra homomorphism).
Multiplying the equalities $\beta=F\left(  \widetilde{\beta}\right)  $ and
$\alpha=F\left(  \widetilde{\alpha}\right)  $, we obtain $\beta\alpha=F\left(
\widetilde{\beta}\right)  F\left(  \widetilde{\alpha}\right)  =F\left(
\widetilde{\beta}\widetilde{\alpha}\right)  $ (since $F$ is a $k$-algebra
homomorphism). Thus, $\alpha\beta=F\left(  \underbrace{\widetilde{\alpha
}\widetilde{\beta}}_{=\widetilde{\beta}\widetilde{\alpha}}\right)  =F\left(
\widetilde{\beta}\widetilde{\alpha}\right)  =\beta\alpha$.
\par
Now, forget that we fixed $\alpha$ and $\beta$. We thus have shown that every
$\alpha\in F\left(  H\right)  $ and $\beta\in F\left(  H\right)  $ satisfy
$\alpha\beta=\beta\alpha$. In other words, the $k$-algebra $F\left(  H\right)
$ is commutative, qed.}.

Every $x\in H$ satisfies $F\left(  \underbrace{x}_{\in H}\right)  \in F\left(
H\right)  $. Thus, we can define a map $\widetilde{F}:H\rightarrow F\left(
H\right)  $ by%
\[
\left(  \widetilde{F}\left(  x\right)  =F\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in H\right)  .
\]
This map $\widetilde{F}$ is a $k$-algebra
homomorphism\footnote{\textit{Proof.} Let $\lambda\in k$, $\mu\in k$, $a\in H$
and $b\in H$. By the definition of $\widetilde{F}$, we have the equalities
$\widetilde{F}\left(  \lambda a+\mu b\right)  =F\left(  \lambda a+\mu
b\right)  $, $\widetilde{F}\left(  a\right)  =F\left(  a\right)  $ and
$\widetilde{F}\left(  b\right)  =F\left(  b\right)  $. Now,%
\begin{align*}
\widetilde{F}\left(  \lambda a+\mu b\right)   &  =F\left(  \lambda a+\mu
b\right)  =\lambda\underbrace{F\left(  a\right)  }_{=\widetilde{F}\left(
a\right)  }+\mu\underbrace{F\left(  b\right)  }_{=\widetilde{F}\left(
b\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  =\lambda\widetilde{F}\left(  a\right)  +\mu\widetilde{F}\left(  b\right)  .
\end{align*}
\par
Now, forget that we fixed $\lambda$, $\mu$, $a$ and $b$. We thus have shown
that $\widetilde{F}\left(  \lambda a+\mu b\right)  =\lambda\widetilde{F}%
\left(  a\right)  +\mu\widetilde{F}\left(  b\right)  $ for all $\lambda\in k$,
$\mu\in k$, $a\in H$ and $b\in H$. In other words, the map $\widetilde{F}$ is
$k$-linear.
\par
Since $F\left(  H\right)  $ is a $k$-subalgebra of $A$, we have $1_{F\left(
H\right)  }=1_{A}$. Now, by the definition of $\widetilde{F}$, we have%
\begin{align*}
\widetilde{F}\left(  1_{H}\right)   &  =F\left(  1_{H}\right)  =1_{A}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  =1_{F\left(  H\right)  }.
\end{align*}
\par
Moreover, any $a\in H$ and $b\in H$ satisfy%
\begin{align*}
\widetilde{F}\left(  ab\right)   &  =F\left(  ab\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\widetilde{F}\right)
\\
&  =\underbrace{F\left(  a\right)  }_{\substack{=\widetilde{F}\left(
a\right)  \\\text{(since }\widetilde{F}\left(  a\right)  =F\left(  a\right)
\\\text{(by the definition of }\widetilde{F}\text{))}}}\underbrace{F\left(
b\right)  }_{\substack{=\widetilde{F}\left(  b\right)  \\\text{(since
}\widetilde{F}\left(  b\right)  =F\left(  b\right)  \\\text{(by the definition
of }\widetilde{F}\text{))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{
is a }k\text{-algebra homomorphism}\right) \\
&  =\widetilde{F}\left(  a\right)  \widetilde{F}\left(  b\right)  .
\end{align*}
Combining this with the fact that $\widetilde{F}\left(  1_{H}\right)
=1_{F\left(  H\right)  }$, we conclude that $\widetilde{F}$ is a $k$-algebra
homomorphism (since $\widetilde{F}$ is $k$-linear), qed.}. Hence,
$\widetilde{F}\left(  1_{H}\right)  =1_{F\left(  H\right)  }$. As a
consequence, $\widetilde{F}\in G\left(  H,F\left(  H\right)  \right)
$\ \ \ \ \footnote{\textit{Proof.} Recall that $G\left(  H,F\left(  H\right)
\right)  =\left\{  f\in\mathcal{L}\left(  H,F\left(  H\right)  \right)
\ \mid\ f\left(  1_{H}\right)  =1_{F\left(  H\right)  }\right\}  $ (by the
definition of $G\left(  H,F\left(  H\right)  \right)  $). Now,%
\begin{align*}
\widetilde{F}  &  \in\left\{  f\in\mathcal{L}\left(  H,F\left(  H\right)
\right)  \ \mid\ f\left(  1_{H}\right)  =1_{F\left(  H\right)  }\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\in\mathcal{L}\left(  H,F\left(
H\right)  \right)  \text{ and }F\left(  1_{H}\right)  =1_{F\left(  H\right)
}\right) \\
&  =G\left(  H,F\left(  H\right)  \right)  ,
\end{align*}
qed.}.

Let $\iota$ be the canonical inclusion $F\left(  H\right)  \rightarrow A$.
Clearly, $\iota$ is a $k$-algebra homomorphism. Moreover, $\iota
\circ\widetilde{F}=F$\ \ \ \ \footnote{\textit{Proof.} Every $x\in H$
satisfies%
\begin{align*}
\left(  \iota\circ\widetilde{F}\right)  \left(  x\right)   &  =\iota\left(
\widetilde{F}\left(  x\right)  \right)  =\widetilde{F}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\iota\text{ is just an inclusion
map}\right) \\
&  =F\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\widetilde{F}\right)  .
\end{align*}
In other words, $\iota\circ\widetilde{F}=F$, qed.}.

Proposition~\ref{Proposition1102} \textbf{(e)} (applied to $H$, $F\left(
H\right)  $, $A$, $\widetilde{F}$ and $\iota$ instead of $C$, $A$, $B$, $F$
and $\psi$) yields that $\iota\circ\widetilde{F}\in G\left(  H,A\right)  $ and
$\operatorname*{Log}\left(  \iota\circ\widetilde{F}\right)  =\iota\circ\left(
\operatorname*{Log}\widetilde{F}\right)  $.

Now, Corollary~\ref{Corollary1103} (applied to $F\left(  H\right)  $ and
$\widetilde{F}$ instead of $A$ and $F$) yields that $\operatorname*{Log}%
\widetilde{F}$ is a well-defined element of $\mathfrak{g}\left(  H,F\left(
H\right)  \right)  $ and satisfies $\left(  \operatorname*{Log}\widetilde{F}%
\right)  \left(  xy\right)  =0$.

Since $F=\iota\circ\widetilde{F}$, we have $\operatorname*{Log}%
F=\operatorname*{Log}\left(  \iota\circ\widetilde{F}\right)  =\iota
\circ\left(  \operatorname*{Log}\widetilde{F}\right)  $, so that%
\[
\left(  \operatorname*{Log}F\right)  \left(  xy\right)  =\left(  \iota
\circ\left(  \operatorname*{Log}\widetilde{F}\right)  \right)  \left(
xy\right)  =\iota\left(  \underbrace{\left(  \operatorname*{Log}%
\widetilde{F}\right)  \left(  xy\right)  }_{=0}\right)  =\iota\left(
0\right)  =0
\]
(since $\iota$ is just an inclusion map). This completes the proof of
Corollary~\ref{Corollary1104}.
\end{proof}

\section{\label{S29}The spaces $\operatorname{symp}_{n}V$ are spanned by
$n$-th powers}

We are now going to take a closer look at the vector spaces
$\operatorname{symp}_{n}V$ defined in Definition~\ref{Definition279}
\textbf{(c)}. The backbone of this will be the following algebraic identity:

\begin{theorem}
\label{Theorem1141} Let $A$ be a ring. Let $n\in\mathbb{N}$. Let $v_{1}$,
$v_{2}$, $\ldots$, $v_{n}$ be $n$ elements of $A$. For every set $X$, let
$\mathcal{P}\left(  X\right)  $ denote the power set of $X$. Then,%
\[
\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }=\sum_{S\in\mathcal{P}\left(  \left\{
1,2,\ldots,n\right\}  \right)  }\left(  -1\right)  ^{n-\left\vert S\right\vert
}\left(  \sum_{s\in S}v_{s}\right)  ^{n}.
\]

\end{theorem}

It is possible to derive Theorem~\ref{Theorem1141} from known facts (such as
Lemma 1 in \cite{7}), but it is more in the spirit of this note to give a
self-contained proof.\footnote{Notice that Theorem~\ref{Theorem1141} also
appears as Exercise 6.50 \textbf{(c)} in \cite{detnotes}.} This is what we are
going to do now, after we have shown the following simple lemma:

\begin{lemma}
\label{Lemma1142} Let $A$ be a finite set. Let $B$ be a subset of $A$. Let
$n\in\mathbb{Z}$.

\textbf{(a)} If $B\neq A$, then%
\[
\sum_{\substack{S\in\mathcal{P}\left(  A\right)  ;\\B\subseteq S}}\left(
-1\right)  ^{n-\left\vert S\right\vert }=0.
\]


\textbf{(b)} If $B=A$ and $n=\left\vert A\right\vert $, then%
\[
\sum_{\substack{S\in\mathcal{P}\left(  A\right)  ;\\B\subseteq S}}\left(
-1\right)  ^{n-\left\vert S\right\vert }=1.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1142}.]\textbf{(a)} Assume that $B\neq A$. If we had
$A\subseteq B$, then we would have $B=A$ (because $A\subseteq B$ and
$B\subseteq A$), contradicting $B\neq A$. Hence, we cannot have $A\subseteq
B$. Thus, $A\not \subseteq B$. Hence, there exists an $x\in A$ such that
$x\notin B$. Consider such an $x$.

Let $\mathbf{P}_{1}=\left\{  W\in\mathcal{P}\left(  A\right)  \ \mid
\ B\subseteq W;\ x\notin W\right\}  $ and $\mathbf{P}_{2}=\left\{
W\in\mathcal{P}\left(  A\right)  \ \mid\ B\subseteq W;\ x\in W\right\}  $.

For every $S\in\mathbf{P}_{1}$, we have $S\cup\left\{  x\right\}
\in\mathbf{P}_{2}$\ \ \ \ \footnote{\textit{Proof.} Let $S\in\mathbf{P}_{1}$.
\par
Since $S\in\mathbf{P}_{1}=\left\{  W\in\mathcal{P}\left(  A\right)
\ \mid\ B\subseteq W;\ x\notin W\right\}  $, we have $S\in\mathcal{P}\left(
A\right)  $ and $B\subseteq S$ and $x\notin S$. Since $S\in\mathcal{P}\left(
A\right)  $, we have $S\subseteq A$. Combining $S\subseteq A$ with $\left\{
x\right\}  \subseteq A$ (because $x\in A$), we obtain $S\cup\left\{
x\right\}  \subseteq A$. In other words, $S\cup\left\{  x\right\}
\in\mathcal{P}\left(  A\right)  $. Also, $B\subseteq S\subseteq S\cup\left\{
x\right\}  $ and $x\in\left\{  x\right\}  \subseteq S\cup\left\{  x\right\}
$.
\par
Altogether, we have thus shown that $S\cup\left\{  x\right\}  \in
\mathcal{P}\left(  A\right)  $, that $B\subseteq S\cup\left\{  x\right\}  $
and $x\in S\cup\left\{  x\right\}  $. In other words, $S\cup\left\{
x\right\}  \in\left\{  W\in\mathcal{P}\left(  A\right)  \ \mid\ B\subseteq
W;\ x\in W\right\}  =\mathbf{P}_{2}$, qed.}. Hence, we can define a map
$\rho_{1}:\mathbf{P}_{1}\rightarrow\mathbf{P}_{2}$ by%
\[
\left(  \rho_{1}\left(  S\right)  =S\cup\left\{  x\right\}
\ \ \ \ \ \ \ \ \ \ \text{for all }S\in\mathbf{P}_{1}\right)  .
\]
Consider this map $\rho_{1}$.

For every $S\in\mathbf{P}_{2}$, we have $S\setminus\left\{  x\right\}
\in\mathbf{P}_{1}$\ \ \ \ \footnote{\textit{Proof.} Let $S\in\mathbf{P}_{2}$.
\par
Since $S\in\mathbf{P}_{2}=\left\{  W\in\mathcal{P}\left(  A\right)
\ \mid\ B\subseteq W;\ x\in W\right\}  $, we have $S\in\mathcal{P}\left(
A\right)  $ and $B\subseteq S$ and $x\in S$. Since $S\in\mathcal{P}\left(
A\right)  $, we have $S\subseteq A$. Now, $S\setminus\left\{  x\right\}
\subseteq S\subseteq A$. In other words, $S\setminus\left\{  x\right\}
\in\mathcal{P}\left(  A\right)  $. Also, $B\setminus\left\{  x\right\}  =B$
(since $x\notin B$), so that $B=\underbrace{B}_{\subseteq S}\setminus\left\{
x\right\}  \subseteq S\setminus\left\{  x\right\}  $. Finally, $x\notin
S\setminus\left\{  x\right\}  $.
\par
Altogether, we have thus shown that $S\setminus\left\{  x\right\}
\in\mathcal{P}\left(  A\right)  $, that $B\subseteq S\setminus\left\{
x\right\}  $ and $x\notin S\setminus\left\{  x\right\}  $. In other words,
$S\setminus\left\{  x\right\}  \in\left\{  W\in\mathcal{P}\left(  A\right)
\ \mid\ B\subseteq W;\ x\notin W\right\}  =\mathbf{P}_{1}$, qed.}. Hence, we
can define a map $\rho_{2}:\mathbf{P}_{2}\rightarrow\mathbf{P}_{1}$ by%
\[
\left(  \rho_{2}\left(  S\right)  =S\setminus\left\{  x\right\}
\ \ \ \ \ \ \ \ \ \ \text{for all }S\in\mathbf{P}_{2}\right)  .
\]
Consider this map $\rho_{2}$.

We have $\rho_{1}\circ\rho_{2}=\operatorname*{id}$%
\ \ \ \ \footnote{\textit{Proof.} Let $S\in\mathbf{P}_{2}$ be arbitrary. Then,
$S\in\mathbf{P}_{2}=\left\{  W\in\mathcal{P}\left(  A\right)  \ \mid
\ B\subseteq W;\ x\in W\right\}  $. In other words, $S\in\mathcal{P}\left(
A\right)  $ and $B\subseteq S$ and $x\in S$. Since $x\in S$, we have $\left\{
x\right\}  \subseteq S$.
\par
Any two sets $X$ and $Y$ with $Y\subseteq X$ satisfy $\left(  X\setminus
Y\right)  \cup Y=X$. Applying this to $X=S$ and $Y=\left\{  x\right\}  $, we
obtain $\left(  S\setminus\left\{  x\right\}  \right)  \cup\left\{  x\right\}
=S$ (since $\left\{  x\right\}  \subseteq S$).
\par
By the definition of $\rho_{1}$, we have $\rho_{1}\left(  \rho_{2}\left(
S\right)  \right)  =\underbrace{\rho_{2}\left(  S\right)  }_{=S\setminus
\left\{  x\right\}  }\cup\left\{  x\right\}  =\left(  S\setminus\left\{
x\right\}  \right)  \cup\left\{  x\right\}  =S$. Thus, $\left(  \rho_{1}%
\circ\rho_{2}\right)  \left(  S\right)  =\rho_{1}\left(  \rho_{2}\left(
S\right)  \right)  =S=\operatorname*{id}\left(  S\right)  $.
\par
Now, forget that we fixed $S$. We thus have shown that $\left(  \rho_{1}%
\circ\rho_{2}\right)  \left(  S\right)  =\operatorname*{id}\left(  S\right)  $
for every $S\in\mathbf{P}_{2}$. In other words, $\rho_{1}\circ\rho
_{2}=\operatorname*{id}$, qed.} and $\rho_{2}\circ\rho_{1}=\operatorname*{id}%
$\ \ \ \ \footnote{\textit{Proof.} Let $S\in\mathbf{P}_{1}$ be arbitrary.
Then, $S\in\mathbf{P}_{1}=\left\{  W\in\mathcal{P}\left(  A\right)
\ \mid\ B\subseteq W;\ x\notin W\right\}  $. In other words, $S\in
\mathcal{P}\left(  A\right)  $ and $B\subseteq S$ and $x\notin S$. Since
$x\notin S$, we have $S\setminus\left\{  x\right\}  =S$.
\par
By the definition of $\rho_{2}$, we have $\rho_{2}\left(  \rho_{1}\left(
S\right)  \right)  =\underbrace{\rho_{1}\left(  S\right)  }_{=S\cup\left\{
x\right\}  }\setminus\left\{  x\right\}  =\left(  S\cup\left\{  x\right\}
\right)  \setminus\left\{  x\right\}  =\left(  S\setminus\left\{  x\right\}
\right)  \cup\underbrace{\left(  \left\{  x\right\}  \setminus\left\{
x\right\}  \right)  }_{=\varnothing}=S\setminus\left\{  x\right\}  =S$. Thus,
$\left(  \rho_{2}\circ\rho_{1}\right)  \left(  S\right)  =\rho_{2}\left(
\rho_{1}\left(  S\right)  \right)  =S=\operatorname*{id}\left(  S\right)  $.
\par
Now, forget that we fixed $S$. We thus have shown that $\left(  \rho_{2}%
\circ\rho_{1}\right)  \left(  S\right)  =\operatorname*{id}\left(  S\right)  $
for every $S\in\mathbf{P}_{1}$. In other words, $\rho_{2}\circ\rho
_{1}=\operatorname*{id}$, qed.}. Hence, the maps $\rho_{1}$ and $\rho_{2}$ are
mutually inverse. Thus, $\rho_{1}$ is a bijection from $\mathbf{P}_{1}$ to
$\mathbf{P}_{2}$.

Notice that%
\begin{equation}
\left(  -1\right)  ^{n-\left\vert \rho_{1}\left(  S\right)  \right\vert
}=-\left(  -1\right)  ^{n-\left\vert S\right\vert }%
\ \ \ \ \ \ \ \ \ \ \text{for every }S\in\mathbf{P}_{1} \label{1142.pf.sign}%
\end{equation}
\footnote{\textit{Proof of (\ref{1142.pf.sign}):} Let $S\in\mathbf{P}_{1}$.
Then, $S\in\mathbf{P}_{1}=\left\{  W\in\mathcal{P}\left(  A\right)
\ \mid\ B\subseteq W;\ x\notin W\right\}  $. In other words, $S\in
\mathcal{P}\left(  A\right)  $ and $B\subseteq S$ and $x\notin S$. The set $S$
is disjoint from $\left\{  x\right\}  $ (since $x\notin S$).
\par
Any two disjoint finite sets $X$ and $Y$ satisfy $\left\vert X\cup
Y\right\vert =\left\vert X\right\vert +\left\vert Y\right\vert $. Applying
this to $X=S$ and $Y=\left\{  x\right\}  $, we obtain $\left\vert
S\cup\left\{  x\right\}  \right\vert =\left\vert S\right\vert +\left\vert
\left\{  x\right\}  \right\vert $ (since $S$ is disjoint from $\left\{
x\right\}  $).
\par
Now, $\rho_{1}\left(  S\right)  =S\cup\left\{  x\right\}  $, so that
$\left\vert \rho_{1}\left(  S\right)  \right\vert =\left\vert S\cup\left\{
x\right\}  \right\vert =\left\vert S\right\vert +\underbrace{\left\vert
\left\{  x\right\}  \right\vert }_{=1}=\left\vert S\right\vert +1$, hence%
\[
\left(  -1\right)  ^{n-\left\vert \rho_{1}\left(  S\right)  \right\vert
}=\left(  -1\right)  ^{n-\left(  \left\vert S\right\vert +1\right)  }=\left(
-1\right)  ^{n-\left\vert S\right\vert -1}=\dfrac{\left(  -1\right)
^{n-\left\vert S\right\vert }}{-1}=-\left(  -1\right)  ^{n-\left\vert
S\right\vert }.
\]
This proves (\ref{1142.pf.sign}).}.

Now, every $S\in\mathcal{P}\left(  A\right)  $ satisfies either $x\in S$ or
$x\notin S$. Hence,%
\begin{align*}
\sum_{\substack{S\in\mathcal{P}\left(  A\right)  ;\\B\subseteq S}}\left(
-1\right)  ^{n-\left\vert S\right\vert }  &  =\underbrace{\sum_{\substack{S\in
\mathcal{P}\left(  A\right)  ;\\B\subseteq S;\\x\in S}}}_{\substack{=\sum
_{S\in\left\{  W\in\mathcal{P}\left(  A\right)  \ \mid\ B\subseteq W;\ x\in
W\right\}  }\\=\sum_{S\in\mathbf{P}_{2}}\\\text{(since }\left\{
W\in\mathcal{P}\left(  A\right)  \ \mid\ B\subseteq W;\ x\in W\right\}
=\mathbf{P}_{2}\text{)}}}\left(  -1\right)  ^{n-\left\vert S\right\vert
}+\underbrace{\sum_{\substack{S\in\mathcal{P}\left(  A\right)  ;\\B\subseteq
S;\\x\notin S}}}_{\substack{=\sum_{S\in\left\{  W\in\mathcal{P}\left(
A\right)  \ \mid\ B\subseteq W;\ x\notin W\right\}  }\\=\sum_{S\in
\mathbf{P}_{1}}\\\text{(since }\left\{  W\in\mathcal{P}\left(  A\right)
\ \mid\ B\subseteq W;\ x\notin W\right\}  =\mathbf{P}_{1}\text{)}}}\left(
-1\right)  ^{n-\left\vert S\right\vert }\\
&  =\sum_{S\in\mathbf{P}_{2}}\left(  -1\right)  ^{n-\left\vert S\right\vert
}+\sum_{S\in\mathbf{P}_{1}}\left(  -1\right)  ^{n-\left\vert S\right\vert }\\
&  =\sum_{S\in\mathbf{P}_{1}}\underbrace{\left(  -1\right)  ^{n-\left\vert
\rho_{1}\left(  S\right)  \right\vert }}_{\substack{=-\left(  -1\right)
^{n-\left\vert S\right\vert }\\\text{(by (\ref{1142.pf.sign}))}}}+\sum
_{S\in\mathbf{P}_{1}}\left(  -1\right)  ^{n-\left\vert S\right\vert }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\rho_{1}\left(  S\right)  \text{ for }S\text{ in
the first sum, since }\rho_{1}\\
\text{is a bijection from }\mathbf{P}_{1}\text{ to }\mathbf{P}_{2}%
\end{array}
\right) \\
&  =\sum_{S\in\mathbf{P}_{1}}\left(  -\left(  -1\right)  ^{n-\left\vert
S\right\vert }\right)  +\sum_{S\in\mathbf{P}_{1}}\left(  -1\right)
^{n-\left\vert S\right\vert }\\
&  =\sum_{S\in\mathbf{P}_{1}}\underbrace{\left(  -\left(  -1\right)
^{n-\left\vert S\right\vert }+\left(  -1\right)  ^{n-\left\vert S\right\vert
}\right)  }_{=0}=\sum_{S\in\mathbf{P}_{1}}0=0.
\end{align*}
This proves Lemma~\ref{Lemma1142} \textbf{(a)}.

\textbf{(b)} Assume that $B=A$ and $n=\left\vert A\right\vert $. For every
$S\in\mathcal{P}\left(  A\right)  $, the assertions $B\subseteq S$ and $S=A$
are equivalent\footnote{\textit{Proof.} Let $S\in\mathcal{P}\left(  A\right)
$. We need to prove that the assertions $B\subseteq S$ and $S=A$ are
equivalent.
\par
First, assume that $B\subseteq S$ holds. Combining $S\subseteq A$ (since
$S\in\mathcal{P}\left(  A\right)  $) with $A=B\subseteq S$, we obtain $S=A$.
Now, forget that we assumed that $B\subseteq S$ holds. We thus have shown that
if $B\subseteq S$, then $S=A$.
\par
On the other hand, if $S=A$, then $B\subseteq S$ (because if $S=A$, then
$B=A=S$).
\par
Altogether, we know that if $S=A$, then $B\subseteq S$, and we know that if
$B\subseteq S$, then $S=A$. In other words, each of the two assertions
$B\subseteq S$ and $S=A$ implies the other. In other words, the assertions
$B\subseteq S$ and $S=A$ are equivalent, qed.}. Hence, we can replace the sign
$\sum_{\substack{S\in\mathcal{P}\left(  A\right)  ;\\B\subseteq S}}$ by
$\sum_{\substack{S\in\mathcal{P}\left(  A\right)  ;\\S=A}}$ in $\sum
_{\substack{S\in\mathcal{P}\left(  A\right)  ;\\B\subseteq S}}\left(
-1\right)  ^{n-\left\vert S\right\vert }$. Consequently,%
\begin{align*}
\sum_{\substack{S\in\mathcal{P}\left(  A\right)  ;\\B\subseteq S}}\left(
-1\right)  ^{n-\left\vert S\right\vert }  &  =\sum_{\substack{S\in
\mathcal{P}\left(  A\right)  ;\\S=A}}\left(  -1\right)  ^{n-\left\vert
S\right\vert }=\left(  -1\right)  ^{n-\left\vert A\right\vert }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A\in\mathcal{P}\left(  A\right)
\right) \\
&  =\left(  -1\right)  ^{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\underbrace{n}_{=\left\vert A\right\vert }-\left\vert A\right\vert
=\left\vert A\right\vert -\left\vert A\right\vert =0\right) \\
&  =1.
\end{align*}
This proves Lemma~\ref{Lemma1142} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem1141}.]Let us first introduce some notations and
make some preliminary observations.

For any sets $X$ and $Y$, let $\operatorname*{Map}\left(  X,Y\right)  $ denote
the set of all maps from $X$ to $Y$.

It is known that for any nonnegative integer $m$ and any set $T$, the
$m$-tuples of elements of $T$ are in bijection with the maps from $\left\{
1,2,\ldots,m\right\}  $ to $T$. More precisely: If $m$ is a nonnegative
integer and $T$ is a set, then there is a bijection%
\[
\operatorname*{Map}\left(  \left\{  1,2,\ldots,m\right\}  ,T\right)
\rightarrow T^{\times m}%
\]
which sends every map $f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots
,m\right\}  ,T\right)  $ to the $m$-tuple $\left(  f\left(  1\right)
,f\left(  2\right)  ,\ldots,f\left(  m\right)  \right)  \in T^{\times m}$.
This bijection will be denoted by $\operatorname*{tuple}\nolimits_{T,m}%
$.\ \ \ \ \footnote{Some authors actually \textbf{define} $m$-tuples of
elements of $T$ as maps from $\left\{  1,2,\ldots,m\right\}  $ to $T$. In that
case, this bijection $\operatorname*{tuple}\nolimits_{T,m}$ is simply
$\operatorname{id}_{T^{\times m}}$.} Thus,%
\begin{equation}
\operatorname*{tuple}\nolimits_{T,m}\left(  f\right)  =\left(  f\left(
1\right)  ,f\left(  2\right)  ,\ldots,f\left(  m\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }f\in\operatorname*{Map}\left(  \left\{
1,2,\ldots,m\right\}  ,T\right)  . \label{1141.pf.tuple}%
\end{equation}


For any nonnegative integer $m$ and any two sets $X$ and $Y$ satisfying
$Y\subseteq X$, we regard $Y^{\times m}$ as a subset of $X^{\times m}$ in the
obvious way.

For any nonnegative integer $m$ and any $m$-tuple $\mathbf{t}$ and any
$i\in\left\{  1,2,\ldots,m\right\}  $, we denote by $\mathbf{t}\left[
i\right]  $ the $i$-th entry of the $m$-tuple $\mathbf{t}$. Hence, any
nonnegative integer $m$ and any $m$-tuple $\mathbf{t}$ satisfy%
\[
\mathbf{t}=\left(  \mathbf{t}\left[  1\right]  ,\mathbf{t}\left[  2\right]
,\ldots,\mathbf{t}\left[  m\right]  \right)  .
\]
Every set $T$, every nonnegative integer $m$, every $i\in\left\{
1,2,\ldots,m\right\}  $ and every $f\in\operatorname*{Map}\left(  \left\{
1,2,\ldots,m\right\}  ,T\right)  $ satisfy%
\begin{equation}
\left(  \operatorname*{tuple}\nolimits_{T,m}\left(  f\right)  \right)  \left[
i\right]  =f\left(  i\right)  \label{1141.pf.tuple3}%
\end{equation}
\footnote{\textit{Proof of (\ref{1141.pf.tuple3}):} Let $T$ be a set. Let $m$
be a nonnegative integer. Let $i\in\left\{  1,2,\ldots,m\right\}  $. Let
$f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots,m\right\}  ,T\right)  $.
Then, (\ref{1141.pf.tuple}) shows that $\operatorname*{tuple}\nolimits_{T,m}%
\left(  f\right)  =\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  m\right)  \right)  $. Hence,%
\begin{align*}
\left(  \operatorname*{tuple}\nolimits_{T,m}\left(  f\right)  \right)  \left[
i\right]   &  =\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(
m\right)  \right)  \left[  i\right] \\
&  =\left(  \text{the }i\text{-th entry of the }m\text{-tuple }\left(
f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  m\right)  \right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(  f\left(
1\right)  ,f\left(  2\right)  ,\ldots,f\left(  m\right)  \right)  \left[
i\right]  \right) \\
&  =f\left(  i\right)  ,
\end{align*}
qed.}. Every set $T$, every nonnegative integer $m$, every $i\in\left\{
1,2,\ldots,m\right\}  $ and any $m$ elements $t_{1}$, $t_{2}$, $\ldots$,
$t_{m}$ of $T$ satisfy%
\begin{equation}
\left(  t_{1},t_{2},\ldots,t_{m}\right)  \left[  i\right]  =t_{i}.
\label{1141.pf.tuple4}%
\end{equation}
\footnote{\textit{Proof of (\ref{1141.pf.tuple4}):} Let $T$ be a set. Let $m$
be a nonnegative integer. Let $t_{1}$, $t_{2}$, $\ldots$, $t_{m}$ be $m$
elements of $T$. Let $i\in\left\{  1,2,\ldots,m\right\}  $. Then, by the
definition of $\left(  t_{1},t_{2},\ldots,t_{m}\right)  \left[  i\right]  $,
we have%
\[
\left(  t_{1},t_{2},\ldots,t_{m}\right)  \left[  i\right]  =\left(  \text{the
}i\text{-th entry of the }m\text{-tuple }\left(  t_{1},t_{2},\ldots
,t_{m}\right)  \right)  =t_{i},
\]
qed.}

Here is one last piece of notation I want to introduce: If $x$ and $y$ are two
integers such that $y\geq x-1$, and if $B$ is a ring, and if $b_{x}$,
$b_{x+1}$, $\ldots$, $b_{y}$ are elements of $B$, then $\overrightarrow{\prod
_{i=x}^{y}}b_{i}$ will mean the product $b_{x}b_{x+1}\cdots b_{y}\in B$. When
the ring $B$ is commutative, this product $\overrightarrow{\prod_{i=x}^{y}%
}b_{i}$ is identical with the standard product $\prod_{i=x}^{y}b_{i}$, but
when $B$ is not supposed to be commutative, the notation $\prod_{i=x}^{y}%
b_{i}$ makes no sense (unless $b_{x}$, $b_{x+1}$, $\ldots$, $b_{y}$ commute).

For every $S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)  $,
we have%
\begin{align*}
&  \left(  \sum_{s\in S}v_{s}\right)  ^{n}\\
&  =\sum_{\left(  s_{1},s_{2},\ldots,s_{n}\right)  \in S^{\times n}%
}\underbrace{v_{s_{1}}v_{s_{2}}\cdots v_{s_{n}}}_{=\overrightarrow{\prod
_{i=1}^{n}}v_{s_{i}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product
rule}\right) \\
&  =\sum_{\left(  s_{1},s_{2},\ldots,s_{n}\right)  \in S^{\times n}%
}\overrightarrow{\prod_{i=1}^{n}}\underbrace{v_{s_{i}}}_{\substack{=v_{\left(
s_{1},s_{2},\ldots,s_{n}\right)  \left[  i\right]  }\\\text{(because
(\ref{1141.pf.tuple4}) (applied to }m=n\text{, }T=S\text{ and }t_{i}%
=s_{i}\text{)}\\\text{yields }\left(  s_{1},s_{2},\ldots,s_{n}\right)  \left[
i\right]  =s_{i}\text{, so that }v_{\left(  s_{1},s_{2},\ldots,s_{n}\right)
\left[  i\right]  }=v_{s_{i}}\text{)}}}\\
&  =\sum_{\left(  s_{1},s_{2},\ldots,s_{n}\right)  \in S^{\times n}%
}\overrightarrow{\prod_{i=1}^{n}}v_{\left(  s_{1},s_{2},\ldots,s_{n}\right)
\left[  i\right]  }=\underbrace{\sum_{\mathbf{s}\in S^{\times n}}%
}_{\substack{=\sum_{\substack{\mathbf{s}\in\left\{  1,2,\ldots,n\right\}
^{\times n};\\\mathbf{s}\in S^{\times n}}}\\\text{(since }S^{\times
n}\subseteq\left\{  1,2,\ldots,n\right\}  ^{\times n}\text{)}}%
}\overrightarrow{\prod_{i=1}^{n}}v_{\mathbf{s}\left[  i\right]  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  s_{1},s_{2},\ldots,s_{n}\right)  \text{ as }\mathbf{s}\right) \\
&  =\sum_{\substack{\mathbf{s}\in\left\{  1,2,\ldots,n\right\}  ^{\times
n};\\\mathbf{s}\in S^{\times n}}}\overrightarrow{\prod_{i=1}^{n}}%
v_{\mathbf{s}\left[  i\right]  }=\sum_{\substack{f\in\operatorname*{Map}%
\left(  \left\{  1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}
\right)  ;\\\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}
,n}\left(  f\right)  \in S^{\times n}}}\overrightarrow{\prod_{i=1}^{n}%
}\underbrace{v_{\left(  \operatorname*{tuple}\nolimits_{\left\{
1,2,\ldots,n\right\}  ,n}\left(  f\right)  \right)  \left[  i\right]  }%
}_{\substack{=v_{f\left(  i\right)  }\\\text{(since (\ref{1141.pf.tuple3})
(applied to }T=\left\{  1,2,\ldots,n\right\}  \\\text{and }m=n\text{) yields
}\left(  \operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}
,n}\left(  f\right)  \right)  \left[  i\right]  =f\left(  i\right)  \text{)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\operatorname*{tuple}\nolimits_{\left\{
1,2,\ldots,n\right\}  ,n}\left(  f\right)  \text{ for }\mathbf{s}\text{ in the
sum,}\\
\text{since }\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}
,n}\text{ is a bijection}\\
\text{from }\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}
,\left\{  1,2,\ldots,n\right\}  \right)  \text{ to }\left\{  1,2,\ldots
,n\right\}  ^{\times n}%
\end{array}
\right) \\
&  =\sum_{\substack{f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots
,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  ;\\\operatorname*{tuple}%
\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(  f\right)  \in S^{\times
n}}}\underbrace{\overrightarrow{\prod_{i=1}^{n}}v_{f\left(  i\right)  }%
}_{=v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)
}}=\sum_{\substack{f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots
,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  ;\\\operatorname*{tuple}%
\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(  f\right)  \in S^{\times
n}}}v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)
}.
\end{align*}
Hence,%
\begin{align}
&  \sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
}\left(  -1\right)  ^{n-\left\vert S\right\vert }\underbrace{\left(
\sum_{s\in S}v_{s}\right)  ^{n}}_{=\sum_{\substack{f\in\operatorname*{Map}%
\left(  \left\{  1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}
\right)  ;\\\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}
,n}\left(  f\right)  \in S^{\times n}}}v_{f\left(  1\right)  }v_{f\left(
2\right)  }\cdots v_{f\left(  n\right)  }}\nonumber\\
&  =\sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
}\left(  -1\right)  ^{n-\left\vert S\right\vert }\sum_{\substack{f\in
\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}  ,\left\{
1,2,\ldots,n\right\}  \right)  ;\\\operatorname*{tuple}\nolimits_{\left\{
1,2,\ldots,n\right\}  ,n}\left(  f\right)  \in S^{\times n}}}v_{f\left(
1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)  }\nonumber\\
&  =\underbrace{\sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}
\right)  }\sum_{\substack{f\in\operatorname*{Map}\left(  \left\{
1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)
;\\\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(
f\right)  \in S^{\times n}}}}_{=\sum_{f\in\operatorname*{Map}\left(  \left\{
1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  }%
\sum_{\substack{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
;\\\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(
f\right)  \in S^{\times n}}}}\left(  -1\right)  ^{n-\left\vert S\right\vert
}v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)
}\nonumber\\
&  =\sum_{f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}
,\left\{  1,2,\ldots,n\right\}  \right)  }\sum_{\substack{S\in\mathcal{P}%
\left(  \left\{  1,2,\ldots,n\right\}  \right)  ;\\\operatorname*{tuple}%
\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(  f\right)  \in S^{\times
n}}}\left(  -1\right)  ^{n-\left\vert S\right\vert }v_{f\left(  1\right)
}v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)  }.
\label{1141.pf.half-there}%
\end{align}


Now, for every $f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}
,\left\{  1,2,\ldots,n\right\}  \right)  $ and every $S\in\mathcal{P}\left(
\left\{  1,2,\ldots,n\right\}  \right)  $, we have the following equivalence
of assertions:%
\begin{equation}
\left(  \operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}
,n}\left(  f\right)  \in S^{\times n}\right)  \ \Longleftrightarrow\ \left(
f\left(  \left\{  1,2,\ldots,n\right\}  \right)  \subseteq S\right)  .
\label{1141.pf.eq}%
\end{equation}
\footnote{\textit{Proof of (\ref{1141.pf.eq}):} Let $f\in\operatorname*{Map}%
\left(  \left\{  1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}
\right)  $ and $S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
$. Since $\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}
,n}\left(  f\right)  $ is an $n$-tuple, we have $\operatorname*{tuple}%
\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(  f\right)  \in S^{\times
n}$ if and only if every of the $n$ entries of $\operatorname*{tuple}%
\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(  f\right)  $ belongs to
$S$. In other words, we have $\operatorname*{tuple}\nolimits_{\left\{
1,2,\ldots,n\right\}  ,n}\left(  f\right)  \in S^{\times n}$ if and only if
every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies $\left(
\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(
f\right)  \right)  \left[  i\right]  \in S$. Thus, we have the following
equivalence of assertions:%
\begin{align*}
&  \ \left(  \operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}
,n}\left(  f\right)  \in S^{\times n}\right) \\
&  \Longleftrightarrow\ \left(  \text{every }i\in\left\{  1,2,\ldots
,n\right\}  \text{ satisfies }\underbrace{\left(  \operatorname*{tuple}%
\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(  f\right)  \right)
\left[  i\right]  }_{\substack{=f\left(  i\right)  \\\text{(by
(\ref{1141.pf.tuple3}) (applied to }T=\left\{  1,2,\ldots,n\right\}  \text{
and }m=n\text{))}}}\in S\right) \\
&  \Longleftrightarrow\ \left(  \text{every }i\in\left\{  1,2,\ldots
,n\right\}  \text{ satisfies }f\left(  i\right)  \in S\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left\{  f\left(  i\right)
\ \mid\ i\in\left\{  1,2,\ldots,n\right\}  \right\}  }_{=f\left(  \left\{
1,2,\ldots,n\right\}  \right)  }\subseteq S\right)  \ \Longleftrightarrow
\ \left(  f\left(  \left\{  1,2,\ldots,n\right\}  \right)  \subseteq S\right)
.
\end{align*}
This proves (\ref{1141.pf.eq}).} Thus, we can replace the $\sum
_{\substack{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
;\\\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(
f\right)  \in S^{\times n}}}$ sign by a $\sum_{\substack{S\in\mathcal{P}%
\left(  \left\{  1,2,\ldots,n\right\}  \right)  ;\\f\left(  \left\{
1,2,\ldots,n\right\}  \right)  \subseteq S}}$ sign in
(\ref{1141.pf.half-there}). Hence, (\ref{1141.pf.half-there}) becomes%
\begin{align}
&  \sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
}\left(  -1\right)  ^{n-\left\vert S\right\vert }\left(  \sum_{s\in S}%
v_{s}\right)  ^{n}\nonumber\\
&  =\sum_{f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}
,\left\{  1,2,\ldots,n\right\}  \right)  }\underbrace{\sum_{\substack{S\in
\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
;\\\operatorname*{tuple}\nolimits_{\left\{  1,2,\ldots,n\right\}  ,n}\left(
f\right)  \in S^{\times n}}}}_{\substack{=\sum_{\substack{S\in\mathcal{P}%
\left(  \left\{  1,2,\ldots,n\right\}  \right)  ;\\f\left(  \left\{
1,2,\ldots,n\right\}  \right)  \subseteq S}}\\\text{(due to the equivalence
(\ref{1141.pf.eq}))}}}\left(  -1\right)  ^{n-\left\vert S\right\vert
}v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)
}\nonumber\\
&  =\sum_{f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}
,\left\{  1,2,\ldots,n\right\}  \right)  }\sum_{\substack{S\in\mathcal{P}%
\left(  \left\{  1,2,\ldots,n\right\}  \right)  ;\\f\left(  \left\{
1,2,\ldots,n\right\}  \right)  \subseteq S}}\left(  -1\right)  ^{n-\left\vert
S\right\vert }v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(
n\right)  }\nonumber\\
&  =\sum_{\substack{f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots
,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  ;\\f\text{ is bijective}%
}}\sum_{\substack{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}
\right)  ;\\f\left(  \left\{  1,2,\ldots,n\right\}  \right)  \subseteq
S}}\left(  -1\right)  ^{n-\left\vert S\right\vert }v_{f\left(  1\right)
}v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{f\in\operatorname*{Map}\left(
\left\{  1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)
;\\f\text{ is not bijective}}}\sum_{\substack{S\in\mathcal{P}\left(  \left\{
1,2,\ldots,n\right\}  \right)  ;\\f\left(  \left\{  1,2,\ldots,n\right\}
\right)  \subseteq S}}\left(  -1\right)  ^{n-\left\vert S\right\vert
}v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)  }.
\label{1141.pf.bij-nonbij}%
\end{align}


Now, every element $f$ of $\operatorname*{Map}\left(  \left\{  1,2,\ldots
,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  $ which is not bijective
satisfies
\begin{equation}
\sum_{\substack{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
;\\f\left(  \left\{  1,2,\ldots,n\right\}  \right)  \subseteq S}}\left(
-1\right)  ^{n-\left\vert S\right\vert }=0 \label{1141.pf.nonbij}%
\end{equation}
\footnote{\textit{Proof of (\ref{1141.pf.nonbij}):} Let $f$ be an element of
$\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}  ,\left\{
1,2,\ldots,n\right\}  \right)  $ such that $f$ is not bijective.
\par
We know that $f$ is a map from $\left\{  1,2,\ldots,n\right\}  $ to $\left\{
1,2,\ldots,n\right\}  $ (since $f\in\operatorname*{Map}\left(  \left\{
1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  $). Thus, $f$
is a map from a finite set to itself (since $\left\{  1,2,\ldots,n\right\}  $
is a finite set).
\par
But it is known that every surjective map from a finite set to itself must be
bijective. Hence, if $f$ was surjective, then $f$ would be bijective (because
$f$ is a map from a finite set to itself), which would contradict the fact
that $f$ is not bijective. Hence, $f$ cannot be surjective.
\par
We have $f\left(  \left\{  1,2,\ldots,n\right\}  \right)  \subseteq\left\{
1,2,\ldots,n\right\}  $ (since $f\in\operatorname*{Map}\left(  \left\{
1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  $).
Furthermore, $f\left(  \left\{  1,2,\ldots,n\right\}  \right)  \neq\left\{
1,2,\ldots,n\right\}  $ (because otherwise, we would have $f\left(  \left\{
1,2,\ldots,n\right\}  \right)  =\left\{  1,2,\ldots,n\right\}  $, so that $f$
would be surjective, which contradicts the fact that $f$ is not surjective).
Hence, Lemma~\ref{Lemma1142} \textbf{(a)} (applied to $A=\left\{
1,2,\ldots,n\right\}  $ and $B=f\left(  \left\{  1,2,\ldots,n\right\}
\right)  $) yields $\sum_{\substack{S\in\mathcal{P}\left(  \left\{
1,2,\ldots,n\right\}  \right)  ;\\f\left(  \left\{  1,2,\ldots,n\right\}
\right)  \subseteq S}}\left(  -1\right)  ^{n-\left\vert S\right\vert }=0$.
This proves (\ref{1141.pf.nonbij}).}. Furthermore, every element $f$ of
$\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}  ,\left\{
1,2,\ldots,n\right\}  \right)  $ which is bijective satisfies%
\begin{equation}
\sum_{\substack{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
;\\f\left(  \left\{  1,2,\ldots,n\right\}  \right)  \subseteq S}}\left(
-1\right)  ^{n-\left\vert S\right\vert }=1 \label{1141.pf.bij}%
\end{equation}
\footnote{\textit{Proof of (\ref{1141.pf.bij}):} Let $f$ be an element of
$\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}  ,\left\{
1,2,\ldots,n\right\}  \right)  $ such that $f$ is bijective.
\par
We know that $f$ is a map from $\left\{  1,2,\ldots,n\right\}  $ to $\left\{
1,2,\ldots,n\right\}  $ (since $f\in\operatorname*{Map}\left(  \left\{
1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  $). Since $f$
is surjective (because $f$ is bijective), we have $f\left(  \left\{
1,2,\ldots,n\right\}  \right)  =\left\{  1,2,\ldots,n\right\}  $. Moreover,
$n=\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert $. Thus, we can apply
Lemma~\ref{Lemma1142} \textbf{(b)} to $A=\left\{  1,2,\ldots,n\right\}  $ and
$B=f\left(  \left\{  1,2,\ldots,n\right\}  \right)  $. As a consequence, we
obtain $\sum_{\substack{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}
\right)  ;\\f\left(  \left\{  1,2,\ldots,n\right\}  \right)  \subseteq
S}}\left(  -1\right)  ^{n-\left\vert S\right\vert }=1$. This proves
(\ref{1141.pf.bij}).}.

Now, (\ref{1141.pf.bij-nonbij}) becomes%
\begin{align*}
&  \sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
}\left(  -1\right)  ^{n-\left\vert S\right\vert }\left(  \sum_{s\in S}%
v_{s}\right)  ^{n}\\
&  =\sum_{\substack{f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots
,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  ;\\f\text{ is bijective}%
}}\underbrace{\sum_{\substack{S\in\mathcal{P}\left(  \left\{  1,2,\ldots
,n\right\}  \right)  ;\\f\left(  \left\{  1,2,\ldots,n\right\}  \right)
\subseteq S}}\left(  -1\right)  ^{n-\left\vert S\right\vert }}%
_{\substack{=1\\\text{(by (\ref{1141.pf.bij}))}}}v_{f\left(  1\right)
}v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{f\in\operatorname*{Map}\left(
\left\{  1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)
;\\f\text{ is not bijective}}}\underbrace{\sum_{\substack{S\in\mathcal{P}%
\left(  \left\{  1,2,\ldots,n\right\}  \right)  ;\\f\left(  \left\{
1,2,\ldots,n\right\}  \right)  \subseteq S}}\left(  -1\right)  ^{n-\left\vert
S\right\vert }}_{\substack{=0\\\text{(by (\ref{1141.pf.nonbij}))}}}v_{f\left(
1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)  }\\
&  =\sum_{\substack{f\in\operatorname*{Map}\left(  \left\{  1,2,\ldots
,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  ;\\f\text{ is bijective}%
}}\underbrace{1v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots
v_{f\left(  n\right)  }}_{=v_{f\left(  1\right)  }v_{f\left(  2\right)
}\cdots v_{f\left(  n\right)  }}+\underbrace{\sum_{\substack{f\in
\operatorname*{Map}\left(  \left\{  1,2,\ldots,n\right\}  ,\left\{
1,2,\ldots,n\right\}  \right)  ;\\f\text{ is not bijective}}}0v_{f\left(
1\right)  }v_{f\left(  2\right)  }\cdots v_{f\left(  n\right)  }}_{=0}\\
&  =\underbrace{\sum_{\substack{f\in\operatorname*{Map}\left(  \left\{
1,2,\ldots,n\right\}  ,\left\{  1,2,\ldots,n\right\}  \right)  ;\\f\text{ is
bijective}}}}_{\substack{=\sum_{\substack{f\text{ is a map from }\left\{
1,2,\ldots,n\right\}  \text{ to }\left\{  1,2,\ldots,n\right\}  \text{;}%
\\f\text{ is bijective}}}=\sum_{f\text{ is a bijective map from }\left\{
1,2,\ldots,n\right\}  \text{ to }\left\{  1,2,\ldots,n\right\}  }%
\\=\sum_{f\text{ is a permutation of }\left\{  1,2,\ldots,n\right\}
}\\\text{(since the bijective maps from }\left\{  1,2,\ldots,n\right\}  \text{
to }\left\{  1,2,\ldots,n\right\}  \\\text{are precisely the permutations of
}\left\{  1,2,\ldots,n\right\}  \text{)}}}v_{f\left(  1\right)  }v_{f\left(
2\right)  }\cdots v_{f\left(  n\right)  }\\
&  =\underbrace{\sum_{f\text{ is a permutation of }\left\{  1,2,\ldots
,n\right\}  }}_{\substack{=\sum_{f\in\left(  \text{the set of permutations of
}\left\{  1,2,\ldots,n\right\}  \right)  }=\sum_{f\in S_{n}}\\\text{(since the
set of permutations of }\left\{  1,2,\ldots,n\right\}  \text{ is }%
S_{n}\text{)}}}v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots
v_{f\left(  n\right)  }\\
&  =\sum_{f\in S_{n}}v_{f\left(  1\right)  }v_{f\left(  2\right)  }\cdots
v_{f\left(  n\right)  }=\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\end{align*}
(here, we renamed the summation index $f$ as $\sigma$). This proves
Theorem~\ref{Theorem1141}.
\end{proof}

Theorem~\ref{Theorem1141} has various useful corollaries; here is probably the
simplest one:\footnote{Notice that Corollary~\ref{Corollary1143} also appears
as Exercise 6.50 \textbf{(d)} in \cite{detnotes}.}

\begin{corollary}
\label{Corollary1143} Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$v_{1}$, $v_{2}$, $\ldots$, $v_{n}$ be $n$ elements of $A$. For every set $X$,
let $\mathcal{P}\left(  X\right)  $ denote the power set of $X$. Then,%
\[
n!\cdot v_{1}v_{2}\cdots v_{n}=\sum_{S\in\mathcal{P}\left(  \left\{
1,2,\ldots,n\right\}  \right)  }\left(  -1\right)  ^{n-\left\vert S\right\vert
}\left(  \sum_{s\in S}v_{s}\right)  ^{n}.
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1143}.]Let $\sigma\in S_{n}$. Then, $\sigma$
is a permutation of $\left\{  1,2,\ldots,n\right\}  $. In other words,
$\sigma$ is a bijection from $\left\{  1,2,\ldots,n\right\}  $ to $\left\{
1,2,\ldots,n\right\}  $.

Since the ring $A$ is commutative, terms like $\prod_{i\in\left\{
1,2,\ldots,n\right\}  }v_{i}$ and $\prod_{i\in\left\{  1,2,\ldots,n\right\}
}v_{\sigma\left(  i\right)  }$ make sense. We have%
\begin{align*}
v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }  &  =\prod_{i\in\left\{  1,2,\ldots,n\right\}  }%
v_{\sigma\left(  i\right)  }=\prod_{i\in\left\{  1,2,\ldots,n\right\}  }%
v_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }i\text{ for }\sigma\left(  i\right)  \text{
in the product,}\\
\text{since }\sigma\text{ is a bijection from }\left\{  1,2,\ldots,n\right\}
\text{ to }\left\{  1,2,\ldots,n\right\}
\end{array}
\right) \\
&  =v_{1}v_{2}\cdots v_{n}.
\end{align*}


Now, forget that we fixed $\sigma$. We thus have proven that $v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}=v_{1}v_{2}\cdots v_{n}$ for every $\sigma\in S_{n}$. Thus,%
\[
\sum_{\sigma\in S_{n}}\underbrace{v_{\sigma\left(  1\right)  }v_{\sigma\left(
2\right)  }\cdots v_{\sigma\left(  n\right)  }}_{=v_{1}v_{2}\cdots v_{n}}%
=\sum_{\sigma\in S_{n}}v_{1}v_{2}\cdots v_{n}=\underbrace{\left\vert
S_{n}\right\vert }_{=n!}\cdot v_{1}v_{2}\cdots v_{n}=n!\cdot v_{1}v_{2}\cdots
v_{n}.
\]
Comparing this with%
\[
\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }=\sum_{S\in\mathcal{P}\left(  \left\{
1,2,\ldots,n\right\}  \right)  }\left(  -1\right)  ^{n-\left\vert S\right\vert
}\left(  \sum_{s\in S}v_{s}\right)  ^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Theorem~\ref{Theorem1141}}\right)  ,
\]
we obtain%
\[
n!\cdot v_{1}v_{2}\cdots v_{n}=\sum_{S\in\mathcal{P}\left(  \left\{
1,2,\ldots,n\right\}  \right)  }\left(  -1\right)  ^{n-\left\vert S\right\vert
}\left(  \sum_{s\in S}v_{s}\right)  ^{n}.
\]
This proves Corollary~\ref{Corollary1143}.
\end{proof}

A more interesting application of Theorem~\ref{Theorem1141} concerns the
vector spaces $\operatorname{symp}_{n}V$ defined in
Definition~\ref{Definition279} \textbf{(c)}:

\begin{theorem}
\label{Theorem1144} Let $k$ be a field of characteristic $0$. Let $A$ be a
$k$-algebra. Let $V$ be a $k$-vector subspace of $A$. Let $n\in\mathbb{N}$.
Consider the $k$-vector subspace $\operatorname{symp}_{n}V$ of $A$ defined in
Definition~\ref{Definition279} \textbf{(c)}.

We have%
\[
\operatorname{symp}_{n}V=\left\langle v^{n}\ \mid\ v\in V\right\rangle .
\]

\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1144}.]Recall that $\operatorname{symp}_{n}V$
was defined by
\[
\operatorname{symp}_{n}V=\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\rangle .
\]


\textbf{a)} We have $\operatorname{symp}_{n}V\subseteq\left\langle v^{n}%
\ \mid\ v\in V\right\rangle $.

\textit{Proof.} Let $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}$. Then, $v_{s}\in V$ for every $s\in\left\{  1,2,\ldots,n\right\}  $. For
every $S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)  $, we
have
\[
\sum_{s\in S}\underbrace{v_{s}}_{\in V}\in\sum_{s\in S}V\subseteq V
\]
(since $V$ is a $k$-vector space). Hence, for every $S\in\mathcal{P}\left(
\left\{  1,2,\ldots,n\right\}  \right)  $, we have
\[
\left(  \sum_{s\in S}v_{s}\right)  ^{n}\in\left\{  v^{n}\ \mid\ v\in
V\right\}
\]
(because $\left(  \sum_{s\in S}v_{s}\right)  ^{n}=v^{n}$ for $v=\sum_{s\in
S}v_{s}$, and because $\sum_{s\in S}v_{s}\in V$). Thus, for every
$S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)  $, we have
\begin{align}
\left(  \sum_{s\in S}v_{s}\right)  ^{n}  &  \in\left\{  v^{n}\ \mid\ v\in
V\right\}  \subseteq\left\langle \left\{  v^{n}\ \mid\ v\in V\right\}
\right\rangle \nonumber\\
&  =\left\langle v^{n}\ \mid\ v\in V\right\rangle . \label{1144.pf.a.3}%
\end{align}


Now,%
\begin{align*}
&  \dfrac{1}{n!}\underbrace{\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
}_{\substack{=\sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}
\right)  }\left(  -1\right)  ^{n-\left\vert S\right\vert }\left(  \sum_{s\in
S}v_{s}\right)  ^{n}\\\text{(by Theorem~\ref{Theorem1141})}}}\\
&  =\dfrac{1}{n!}\sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}
\right)  }\left(  -1\right)  ^{n-\left\vert S\right\vert }\underbrace{\left(
\sum_{s\in S}v_{s}\right)  ^{n}}_{\substack{\in\left\langle v^{n}\ \mid\ v\in
V\right\rangle \\\text{(by (\ref{1144.pf.a.3}))}}}\\
&  \in\dfrac{1}{n!}\sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}
\right)  }\left(  -1\right)  ^{n-\left\vert S\right\vert }\left\langle
v^{n}\ \mid\ v\in V\right\rangle \subseteq\left\langle v^{n}\ \mid\ v\in
V\right\rangle
\end{align*}
(since $\left\langle v^{n}\ \mid\ v\in V\right\rangle $ is a $k$-vector space).

Now, forget that we fixed $\left(  v_{1},v_{2},\ldots,v_{n}\right)  $. We thus
have shown that $\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\in\left\langle v^{n}\ \mid\ v\in V\right\rangle $ for every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$. In other words,%
\[
\left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\subseteq\left\langle v^{n}\ \mid\ v\in V\right\rangle .
\]
Thus, (\ref{284.pf.b.2}) (applied to $M=A$, $S=\left\{  \dfrac{1}{n!}%
\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}\right\}  $ and $Q=\left\langle v^{n}%
\ \mid\ v\in V\right\rangle $) yields that%
\begin{align*}
&  \left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle \\
&  \subseteq\left\langle v^{n}\ \mid\ v\in V\right\rangle .
\end{align*}
Now,%
\begin{align*}
\operatorname{symp}_{n}V  &  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\rangle \\
&  =\left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle \\
&  \subseteq\left\langle v^{n}\ \mid\ v\in V\right\rangle .
\end{align*}
We have thus proven that $\operatorname{symp}_{n}V\subseteq\left\langle
v^{n}\ \mid\ v\in V\right\rangle $.

\textbf{b)} We have $\left\langle v^{n}\ \mid\ v\in V\right\rangle
\subseteq\operatorname{symp}_{n}V$.

\textit{Proof.} Let $v\in V$. Then, $\left(  \underbrace{v,v,\ldots
,v}_{n\text{ times }v}\right)  \in V^{\times n}$ and $\dfrac{1}{n!}%
\sum_{\sigma\in S_{n}}\underbrace{vv\cdots v}_{n\text{ times }v}=\dfrac{1}%
{n!}\underbrace{\left\vert S_{n}\right\vert }_{=n!}%
\underbrace{\underbrace{vv\cdots v}_{n\text{ times }v}}_{=v^{n}}=\dfrac{1}%
{n!}n!v^{n}=v^{n}$. Thus, $v^{n}=\dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}\underbrace{vv\cdots v}_{n\text{ times }v}$. Hence, the element $v^{n}$ has
the form $\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }$ for some
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ (namely, for
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  =\left(  \underbrace{v,v,\ldots
,v}_{n\text{ times }v}\right)  $). In other words,%
\begin{align*}
v^{n}  &  \in\left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\} \\
&  \subseteq\left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}%
v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\}  \right\rangle \\
&  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle
\\
&  =\operatorname{symp}_{n}V.
\end{align*}


Now, forget that we fixed $v$. We thus have shown that $v^{n}\in
\operatorname{symp}_{n}V$ for every $v\in V$. In other words, $\left\{
v^{n}\ \mid\ v\in V\right\}  \subseteq\operatorname{symp}_{n}V$. Thus,
(\ref{284.pf.b.2}) (applied to $M=A$, $S=\left\{  v^{n}\ \mid\ v\in V\right\}
$ and $Q=\operatorname{symp}_{n}V$) yields that $\left\langle \left\{
v^{n}\ \mid\ v\in V\right\}  \right\rangle \subseteq\operatorname{symp}_{n}V$.
Thus, $\left\langle v^{n}\ \mid\ v\in V\right\rangle =\left\langle \left\{
v^{n}\ \mid\ v\in V\right\}  \right\rangle \subseteq\operatorname{symp}_{n}V$.
We have thus proven that $\left\langle v^{n}\ \mid\ v\in V\right\rangle
\subseteq\operatorname{symp}_{n}V$.

\textbf{c)} We now know that $\operatorname{symp}_{n}V\subseteq\left\langle
v^{n}\ \mid\ v\in V\right\rangle $ and $\left\langle v^{n}\ \mid\ v\in
V\right\rangle \subseteq\operatorname{symp}_{n}V$. Combining these two
inclusions, we obtain $\operatorname{symp}_{n}V=\left\langle v^{n}\ \mid\ v\in
V\right\rangle $. Theorem~\ref{Theorem1144} is thus proven.
\end{proof}

Theorem~\ref{Theorem1144} takes a simpler form in the particular case when the
algebra is commutative:

\begin{corollary}
\label{Corollary1145} Let $k$ be a field of characteristic $0$. Let $A$ be a
commutative $k$-algebra. Let $V$ be a $k$-vector subspace of $A$. Let
$n\in\mathbb{N}$. Then,%
\[
V^{n}=\left\langle v^{n}\ \mid\ v\in V\right\rangle .
\]

\end{corollary}

Note that this corollary is (generally) wrong when $k$ doesn't have
characteristic $0$.

\begin{proof}
[Proof of Corollary~\ref{Corollary1145}.]Consider the $k$-vector subspace
$\operatorname{symp}_{n}V$ of $A$ defined in Definition~\ref{Definition279}
\textbf{(c)}. Recall that $\operatorname{symp}_{n}V$ was defined by
\[
\operatorname{symp}_{n}V=\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\rangle .
\]


Every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfies
$\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }=n!\cdot v_{1}v_{2}\cdots v_{n}%
$\ \ \ \ \footnote{This can be proven in the same way as we did in the proof
of Corollary~\ref{Corollary1143}.}. Hence,%
\begin{align*}
&  \left\langle \dfrac{1}{n!}\underbrace{\sum_{\sigma\in S_{n}}v_{\sigma
\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(
n\right)  }}_{=n!\cdot v_{1}v_{2}\cdots v_{n}}\ \mid\ \left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle \\
&  =\left\langle \underbrace{\dfrac{1}{n!}n!\cdot v_{1}v_{2}\cdots v_{n}%
}_{=v_{1}v_{2}\cdots v_{n}}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{\times n}\right\rangle \\
&  =\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\rangle =V^{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{100.Vn})}\right)
\end{align*}
But Theorem~\ref{Theorem1144} yields $\operatorname{symp}_{n}V=\left\langle
v^{n}\ \mid\ v\in V\right\rangle $, so that%
\begin{align*}
\left\langle v^{n}\ \mid\ v\in V\right\rangle  &  =\operatorname{symp}_{n}V\\
&  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle
=V^{n}.
\end{align*}
This proves Corollary~\ref{Corollary1145}.
\end{proof}

A corollary of this corollary is the following property of symmetric powers:

\begin{corollary}
\label{Corollary1146} Let $k$ be a field of characteristic $0$. Let $V$ be a
$k$-vector space. Let $n\in\mathbb{N}$. Consider the $n$-th symmetric power
$\operatorname{Sym}^{n}\left(  V\right)  $ of the vector space $V$. Then,%
\[
\operatorname{Sym}^{n}\left(  V\right)  =\left\langle v^{n}\ \mid\ v\in
V\right\rangle ,
\]
where $v^{n}$ denotes the projection of $\underbrace{v\otimes v\otimes
\cdots\otimes v}_{n\text{ times }v}\in V^{\otimes n}$ onto the $n$-th
symmetric power $\operatorname{Sym}^{n}\left(  V\right)  $.
\end{corollary}

Corollary~\ref{Corollary1146} appears as Lemma 4.56 (i) in \cite{6} (where it
is proven using representation theory).

\begin{proof}
[Proof of Corollary~\ref{Corollary1146}.]Let $\operatorname{Sym}\left(
V\right)  $ be the symmetric algebra of the $k$-vector space $V$. It is known
that both $V$ and $\operatorname{Sym}^{n}\left(  V\right)  $ canonically
inject into $\operatorname{Sym}\left(  V\right)  $. We shall identify $V$ and
$\operatorname*{Sym}\nolimits^{n}\left(  V\right)  $ with $k$-vector subspaces
of $\operatorname*{Sym}\left(  V\right)  $ using these injections.

For every $v\in V$, the element $v^{n}$ of $\operatorname{Sym}^{n}\left(
V\right)  $ is the actual $n$-th power of the element $v\in V\subseteq
\operatorname{Sym}\left(  V\right)  $ taken in $\operatorname{Sym}\left(
V\right)  $.\ \ \ \ \footnote{\textit{Proof.} Let $v\in V$. Consider the
tensor algebra $\otimes V$ of the $k$-vector space $V$. Let $\pi$ be the
canonical projection from $\otimes V$ to $\operatorname{Sym}\left(  V\right)
$. Then, $\pi$ is a $k$-algebra homomorphism. Moreover, $\pi\left(  w\right)
=w$ for every $w\in V$. Applied to $w=v$, this yields $\pi\left(  v\right)
=v$.
\par
We have%
\begin{align*}
&  \left(  \text{the element }v^{n}\text{ of }\operatorname{Sym}^{n}\left(
V\right)  \right) \\
&  =\left(  \text{the projection of }\underbrace{v\otimes v\otimes
\cdots\otimes v}_{n\text{ times }v}\in V^{\otimes n}\text{ onto the
}n\text{-th symmetric power }\operatorname{Sym}^{n}\left(  V\right)  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the element }%
v^{n}\text{ of }\operatorname{Sym}^{n}\left(  V\right)  \right) \\
&  =\left(  \text{the projection of }\underbrace{v\otimes v\otimes
\cdots\otimes v}_{n\text{ times }v}\in\otimes V\text{ onto the symmetric
algebra }\operatorname{Sym}\left(  V\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the canonical projection from }V^{\otimes n}\text{ onto the
}n\text{-th symmetric power }\operatorname{Sym}^{n}\left(  V\right) \\
\text{is a restriction of the canonical projection from }\left.  \otimes
V\right.  \text{ onto the symmetric algebra }\operatorname{Sym}\left(
V\right)
\end{array}
\right) \\
&  =\pi\left(  \underbrace{\underbrace{v\otimes v\otimes\cdots\otimes
v}_{n\text{ times }v}}_{=\left(  \text{the }n\text{-th power of }v\text{ in
the algebra }\otimes V\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the canonical projection from
}\left.  \otimes V\right.  \text{ onto the symmetric algebra }%
\operatorname{Sym}\left(  V\right)  \text{ is }\pi\right) \\
&  =\pi\left(  \text{the }n\text{-th power of }v\text{ in the algebra }\otimes
V\right) \\
&  =\left(  \text{the }n\text{-th power of }\underbrace{\pi\left(  v\right)
}_{=v}\text{ in the algebra }\operatorname{Sym}\left(  V\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  =\left(  \text{the }n\text{-th power of }v\text{ in the algebra
}\operatorname{Sym}\left(  V\right)  \right)  .
\end{align*}
In other words, the element $v^{n}$ of $\operatorname{Sym}^{n}\left(
V\right)  $ is the $n$-th power of $v$ in the algebra $\operatorname{Sym}%
\left(  V\right)  $, qed.} Hence, the notation $v^{n}$ in
Corollary~\ref{Corollary1146} does not conflict with the notation $v^{n}$ in
Corollary~\ref{Corollary1145} (applied to $A=\operatorname{Sym}\left(
V\right)  $). Since the symmetric algebra $\operatorname{Sym}\left(  V\right)
$ is commutative, we can apply Corollary~\ref{Corollary1145} to
$A=\operatorname{Sym}\left(  V\right)  $. We obtain $V^{n}=\left\langle
v^{n}\ \mid\ v\in V\right\rangle $. But it is easy to see that $V^{n}%
=\operatorname{Sym}^{n}\left(  V\right)  $ as subspaces of $\operatorname{Sym}%
\left(  V\right)  $\ \ \ \ \footnote{\textit{Proof.} Consider the tensor
algebra $\otimes V$ of the $k$-vector space $V$. Let $\pi$ be the canonical
projection from $\otimes V$ to $\operatorname{Sym}\left(  V\right)  $. Then,
$\pi$ is a $k$-algebra homomorphism. Moreover,
\begin{equation}
\pi\left(  w\right)  =w\ \ \ \ \ \ \ \ \ \ \text{for every }w\in V.
\label{1146.pf.10}%
\end{equation}
\par
We know that $\operatorname{Sym}^{n}\left(  V\right)  $ is the image of
$V^{\otimes n}$ under the canonical projection $\pi$. In other words,
$\operatorname{Sym}^{n}\left(  V\right)  =\pi\left(  V^{\otimes n}\right)  $.
\par
Let $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$. Then, in the
$k$-algebra $\otimes V$, we have%
\begin{equation}
v_{1}v_{2}\cdots v_{n}=v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}
\label{1146.pf.11}%
\end{equation}
(since the multiplication on the $k$-algebra $\otimes V$ is the tensor
product). But we have $v_{i}\in V$ for every $i\in\left\{  1,2,\ldots
,n\right\}  $. Thus, $\pi\left(  v_{i}\right)  =v_{i}$ for every $i\in\left\{
1,2,\ldots,n\right\}  $ (by (\ref{1146.pf.10}), applied to $w=v_{i}$). In
other words, $\left(  \pi\left(  v_{1}\right)  ,\pi\left(  v_{2}\right)
,\ldots,\pi\left(  v_{n}\right)  \right)  =\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $. Hence, $\pi\left(  v_{1}\right)  \pi\left(  v_{2}\right)
\cdots\pi\left(  v_{n}\right)  =v_{1}v_{2}\cdots v_{n}$, where both products
in this equality are taken in the $k$-algebra $\operatorname*{Sym}\left(
V\right)  $. Now,%
\begin{align}
&  \pi\left(  \underbrace{v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
}_{\substack{_{\substack{=v_{1}v_{2}\cdots v_{n}}}\\\text{(by
(\ref{1146.pf.11}))}}}\right) \nonumber\\
&  =\pi\left(  v_{1}v_{2}\cdots v_{n}\right) \nonumber\\
&  =\pi\left(  v_{1}\right)  \pi\left(  v_{2}\right)  \cdots\pi\left(
v_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a
}k\text{-algebra homomorphism}\right) \nonumber\\
&  =v_{1}v_{2}\cdots v_{n}, \label{1146.pf.12}%
\end{align}
where the product $v_{1}v_{2}\ldots v_{n}$ is taken in the $k$-algebra
$\operatorname*{Sym}\left(  V\right)  $.
\par
Now, forget that we fixed $\left(  v_{1},v_{2},\ldots,v_{n}\right)  $. We thus
have proven the equality (\ref{1146.pf.12}) for each $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{\times n}$.
\par
But the $n$-th tensor power $V^{\otimes n}$ is spanned by all pure tensors. In
other words,%
\[
V^{\otimes n}=\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle
.
\]
Hence,%
\begin{align*}
V^{\otimes n}  &  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\right\rangle \\
&  =\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle .
\end{align*}
Applying the map $\pi$ to both sides of this equality, we obtain%
\begin{align*}
\pi\left(  V^{\otimes n}\right)   &  =\pi\left(  \left\langle \left\{
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\}  \right\rangle \right) \\
&  =\left\langle \underbrace{\pi\left(  \left\{  v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\}  \right)  }_{=\left\{  \pi\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\}  }\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{316.pf.3}), applied to }M=\otimes V\text{, }%
R=\operatorname{Sym}\left(  V\right)  \text{, }\phi=\pi\\
\text{and }S=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\end{array}
\right) \\
&  =\left\langle \left\{  \pi\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  \ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\right\}  \right\rangle \\
&  =\left\langle \underbrace{\pi\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  }_{\substack{=v_{1}v_{2}\cdots v_{n}\\\text{(by
(\ref{1146.pf.12}))}}}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\rangle \\
&  =\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\rangle =V^{n}.
\end{align*}
Thus, $V^{n}=\pi\left(  V^{\otimes n}\right)  =\operatorname*{Sym}%
\nolimits^{n}\left(  V\right)  $.}. Hence, $\operatorname{Sym}^{n}\left(
V\right)  =V^{n}=\left\langle v^{n}\ \mid\ v\in V\right\rangle $. This proves
Corollary~\ref{Corollary1146}.
\end{proof}

The following consequence of Theorem~\ref{Theorem1144} will be of most
importance to us:

\begin{corollary}
\label{Corollary1147} Let $k$ be a field of characteristic $0$. Let $A$ be a
$k$-algebra. Let $V$ be a $k$-vector subspace of $A$. Assume that%
\[
xy-yx\in V\ \ \ \ \ \ \ \ \ \ \text{for any }x\in V\text{ and }y\in V.
\]
Assume further that $A=\sum_{n\in\mathbb{N}}V^{n}$\ \ \ \ \footnote{This is an
equivalent way to say that the $k$-algebra $A$ is generated by the subset $V$.
But it is easier for us to word it as $\sum_{n\in\mathbb{N}}V^{n}$.}.

Let $W$ be a $k$-vector space. Let $f:A\rightarrow W$ and $g:A\rightarrow W$
be two $k$-linear maps. Assume that $f\left(  v^{n}\right)  =g\left(
v^{n}\right)  $ for every $v\in V$ and $n\in\mathbb{N}$. Then, $f=g$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1147}.]Since $f$ and $g$ are $k$-linear
maps, their difference $f-g$ is also a $k$-linear map. Hence,
$\operatorname*{Ker}\left(  f-g\right)  $ is a $k$-vector subspace of $A$.

Now, let $n\in\mathbb{N}$. We know that every $v\in V$ satisfies%
\[
\left(  f-g\right)  \left(  v^{n}\right)  =\underbrace{f\left(  v^{n}\right)
}_{=g\left(  v^{n}\right)  }-g\left(  v^{n}\right)  =g\left(  v^{n}\right)
-g\left(  v^{n}\right)  =0.
\]
In other words, every $v\in V$ satisfies $v^{n}\in\operatorname*{Ker}\left(
f-g\right)  $. In other words, $\left\{  v^{n}\ \mid\ v\in V\right\}
\subseteq\operatorname*{Ker}\left(  f-g\right)  $. Hence, (\ref{284.pf.b.2})
(applied to $M=A$, $S=\left\{  v^{n}\ \mid\ v\in V\right\}  $ and
$Q=\operatorname*{Ker}\left(  f-g\right)  $) yields that $\left\langle
\left\{  v^{n}\ \mid\ v\in V\right\}  \right\rangle \subseteq
\operatorname*{Ker}\left(  f-g\right)  $. Now, consider the $k$-vector
subspace $\operatorname{symp}_{n}V$ of $A$ defined in
Definition~\ref{Definition279} \textbf{(c)}. Theorem~\ref{Theorem1144} yields
$\operatorname{symp}_{n}V=\left\langle v^{n}\ \mid\ v\in V\right\rangle
\subseteq\operatorname*{Ker}\left(  f-g\right)  $.

Now, forget that we have fixed $n$. We thus have proven that%
\begin{equation}
\operatorname{symp}_{n}V\subseteq\operatorname*{Ker}\left(  f-g\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}. \label{1147.pf.1}%
\end{equation}


Now, every $\ell\in\mathbb{N}$ satisfies%
\begin{align}
V^{\ell}  &  \subseteq\sum_{i=0}^{\ell}V^{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }V^{\ell}\text{ is an addend in the sum }\sum_{i=0}^{\ell}%
V^{i}\right) \nonumber\\
&  =\sum_{i=0}^{\ell}\underbrace{\operatorname{symp}_{i}V}%
_{\substack{\subseteq\operatorname*{Ker}\left(  f-g\right)  \\\text{(by
(\ref{1147.pf.1}), applied to }n=i\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition~\ref{Proposition284} \textbf{(b)}}\right) \nonumber\\
&  \subseteq\sum_{i=0}^{\ell}\operatorname*{Ker}\left(  f-g\right)
\subseteq\operatorname*{Ker}\left(  f-g\right)  \label{1147.pf.4}%
\end{align}
(since $\operatorname*{Ker}\left(  f-g\right)  $ is a $k$-vector space). Now,%
\[
A=\sum_{n\in\mathbb{N}}\underbrace{V^{n}}_{\substack{\subseteq
\operatorname*{Ker}\left(  f-g\right)  \\\text{(by (\ref{1147.pf.1}), applied
to }\ell=n\text{)}}}\subseteq\sum_{n\in\mathbb{N}}\operatorname*{Ker}\left(
f-g\right)  \subseteq\operatorname*{Ker}\left(  f-g\right)
\]
(since $\operatorname*{Ker}\left(  f-g\right)  $ is a $k$-vector space). In
other words, $f-g=0$. In other words, $f=g$. This proves
Corollary~\ref{Corollary1147}.
\end{proof}

\section{\label{S30}$\operatorname*{Log}\operatorname*{id}$ on powers of
primitives}

Our next goal is to prove the following fact:

\begin{theorem}
\label{Theorem1201} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $v\in\operatorname*{Prim}H$
and $n\in\mathbb{N}$ be such that $n>1$. Then,%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(  v^{n}\right)
=0.
\]

\end{theorem}

This is a rather light variation on Lemma 2 ii) in \cite{5}, and our proof is
an adaptation of the proof given in \cite{5}. Actually we will show a slightly
more general fact:

\begin{theorem}
\label{Theorem1202} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $A$ be a $k$-algebra. Let
$F:H\rightarrow A$ be a $k$-algebra homomorphism. Let $v\in
\operatorname*{Prim}H$ and $n\in\mathbb{N}$ be such that $n>1$. Then, $F\in
G\left(  H,A\right)  $ and%
\[
\left(  \operatorname*{Log}F\right)  \left(  v^{n}\right)  =0.
\]

\end{theorem}

In order to prove these theorems, we will use the universal property of the
polynomial ring $k\left[  X\right]  $ (a trick I learnt from \cite{5}). Here
is this universal property:

\begin{theorem}
\label{Theorem1203} Let $k$ be a field. Let $X$ be a new symbol. Consider the
polynomial ring $k\left[  X\right]  $ over $k$ in one indeterminate $X$.

For every commutative $k$-algebra $A$ and every $a\in A$, there exists a
unique $k$-algebra homomorphism $\varphi:k\left[  X\right]  \rightarrow A$
satisfying $\varphi\left(  X\right)  =a$.
\end{theorem}

This Theorem~\ref{Theorem1203} is just the universal property of the
polynomial ring $k\left[  X\right]  $. This theorem also holds if $A$ is not
required to be commutative:

\begin{theorem}
\label{Theorem1204} Let $k$ be a field. Let $X$ be a new symbol. Consider the
polynomial ring $k\left[  X\right]  $ over $k$ in one indeterminate $X$.

For every $k$-algebra $A$ and every $a\in A$, there exists a unique
$k$-algebra homomorphism $\varphi:k\left[  X\right]  \rightarrow A$ satisfying
$\varphi\left(  X\right)  =a$.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1204}.]Let $A$ be a $k$-algebra. Let $a\in A$.
Let $B$ be the $k$-subalgebra of $A$ generated by the element $a$. Hence, the
$k$-algebra $B$ is generated by the element $a$. Consequently, the $k$-algebra
$B$ is generated by pairwise commuting elements (because the element $a$ is
clearly pairwise commuting (since it is only one element)). Thus, the
$k$-algebra $B$ is commutative (since any $k$-algebra generated by pairwise
commuting elements must be commutative). Moreover, we have $a\in B$ (since $B$
is the $k$-subalgebra of $A$ generated by the element $a$). Hence, we can
apply Theorem~\ref{Theorem1203} to $B$ instead of $A$. As a consequence, we
conclude that there exists a unique $k$-algebra homomorphism $\varphi:k\left[
X\right]  \rightarrow B$ satisfying $\varphi\left(  X\right)  =a$. Denote this
homomorphism $\varphi$ by $\Phi$. Thus, $\Phi$ is a $k$-algebra homomorphism
$\varphi:k\left[  X\right]  \rightarrow B$ satisfying $\varphi\left(
X\right)  =a$. In other words, $\Phi$ is a $k$-algebra homomorphism $k\left[
X\right]  \rightarrow B$ and satisfies $\Phi\left(  X\right)  =a$.

Let $\iota$ be the canonical inclusion $B\rightarrow A$. Clearly, $\iota$ is a
$k$-algebra homomorphism. Since $\Phi$ and $\iota$ are $k$-algebra
homomorphisms, their composition $\iota\circ\Phi$ is a $k$-algebra
homomorphism (because the composition of two $k$-algebra homomorphisms must
always be a $k$-algebra homomorphism). Also, $\left(  \iota\circ\Phi\right)
\left(  X\right)  =\iota\left(  \underbrace{\Phi\left(  X\right)  }%
_{=a}\right)  =\iota\left(  a\right)  =a$ (since $\iota$ is an inclusion map).
Thus, there exists a $k$-algebra homomorphism $\varphi:k\left[  X\right]
\rightarrow A$ satisfying $\varphi\left(  X\right)  =a$ (namely,
$\varphi=\iota\circ\Phi$).

Now, let $\psi_{1}$ and $\psi_{2}$ be any two $k$-algebra homomorphisms
$\varphi:k\left[  X\right]  \rightarrow A$ satisfying $\varphi\left(
X\right)  =a$. We are going to show that $\psi_{1}=\psi_{2}$.

Let $p\in k\left[  X\right]  $. Then, $p$ is a polynomial in the indeterminate
$X$ over $k$. Hence, there exists some family $\left(  \lambda_{i}\right)
_{i\in\mathbb{N}}\in k^{\mathbb{N}}$ such that $\left(  \text{all but finitely
many }i\in\mathbb{N}\text{ satisfy }\lambda_{i}=0\right)  $ and $p=\sum
_{i\in\mathbb{N}}\lambda_{i}X^{i}$. Consider this family $\left(  \lambda
_{i}\right)  _{i\in\mathbb{N}}$.

We know that $\psi_{1}$ is a $k$-algebra homomorphism $\varphi:k\left[
X\right]  \rightarrow A$ satisfying $\varphi\left(  X\right)  =a$. In other
words, $\psi_{1}$ is a $k$-algebra homomorphism $k\left[  X\right]
\rightarrow A$ and satisfies $\psi_{1}\left(  X\right)  =a$. Since
$p=\sum_{i\in\mathbb{N}}\lambda_{i}X^{i}$, we have%
\begin{align*}
\psi_{1}\left(  p\right)   &  =\psi_{1}\left(  \sum_{i\in\mathbb{N}}%
\lambda_{i}X^{i}\right)  =\sum_{i\in\mathbb{N}}\lambda_{i}\left(
\underbrace{\psi_{1}\left(  X\right)  }_{=a}\right)  ^{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\psi_{1}\text{ is a }%
k\text{-algebra homomorphism}\right) \\
&  =\sum_{i\in\mathbb{N}}\lambda_{i}a^{i}.
\end{align*}
The same argument, with $\psi_{1}$ replaced by $\psi_{2}$, yields that
$\psi_{2}\left(  p\right)  =\sum_{i\in\mathbb{N}}\lambda_{i}a^{i}$. Hence,
$\psi_{1}\left(  p\right)  =\sum_{i\in\mathbb{N}}\lambda_{i}a^{i}=\psi
_{2}\left(  p\right)  $.

Now, forget that we fixed $p$. We thus have proven that every $p\in k\left[
X\right]  $ satisfies $\psi_{1}\left(  p\right)  =\psi_{2}\left(  p\right)  $.
In other words, $\psi_{1}=\psi_{2}$.

Now, forget that we fixed $\psi_{1}$ and $\psi_{2}$. We thus have shown that
if $\psi_{1}$ and $\psi_{2}$ are any two $k$-algebra homomorphisms
$\varphi:k\left[  X\right]  \rightarrow A$ satisfying $\varphi\left(
X\right)  =a$, then $\psi_{1}=\psi_{2}$. In other words, there exists at most
one $k$-algebra homomorphism $\varphi:k\left[  X\right]  \rightarrow A$
satisfying $\varphi\left(  X\right)  =a$. Combining this with the fact that
there exists a $k$-algebra homomorphism $\varphi:k\left[  X\right]
\rightarrow A$ satisfying $\varphi\left(  X\right)  =a$, we conclude the
following: There exists a unique $k$-algebra homomorphism $\varphi:k\left[
X\right]  \rightarrow A$ satisfying $\varphi\left(  X\right)  =a$. This proves
Theorem~\ref{Theorem1204}.
\end{proof}

\begin{definition}
\label{Definition1205} Let $k$ be a field. Let $X$ be a new symbol. Consider
the polynomial ring $k\left[  X\right]  $ over $k$ in one indeterminate $X$.

For every $k$-algebra $A$ and every $a\in A$, there exists a unique
$k$-algebra homomorphism $\varphi:k\left[  X\right]  \rightarrow A$ satisfying
$\varphi\left(  X\right)  =a$ (according to Theorem~\ref{Theorem1204}). This
homomorphism $\varphi$ will be denoted by $\operatorname*{ev}\nolimits_{A,\ a}%
$ or by $\operatorname*{ev}\nolimits_{a}$ (when $A$ is clear from the
context). Thus, $\operatorname*{ev}\nolimits_{A,\ a}$ is the unique
$k$-algebra homomorphism $\varphi:k\left[  X\right]  \rightarrow A$ satisfying
$\varphi\left(  X\right)  =a$. Hence, $\operatorname*{ev}\nolimits_{A,\ a}$ is
a $k$-algebra homomorphism $k\left[  X\right]  \rightarrow A$ and satisfies
\begin{equation}
\operatorname*{ev}\nolimits_{A,\ a}\left(  X\right)  =a. \label{1205}%
\end{equation}

\end{definition}

\begin{corollary}
\label{Corollary1206} Let $X$ be a new symbol. Consider the polynomial ring
$k\left[  X\right]  $ over $k$ in one indeterminate $X$.

Let $A$ be a $k$-algebra. Let $f:k\left[  X\right]  \rightarrow A$ and
$g:k\left[  X\right]  \rightarrow A$ be two $k$-algebra homomorphisms such
that $f\left(  X\right)  =g\left(  X\right)  $. Then, $f=g$.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1206}.]Let $a=f\left(  X\right)  $. Then,
$a=f\left(  X\right)  =g\left(  X\right)  $.

We know that $\operatorname*{ev}\nolimits_{A,\ a}$ is the unique $k$-algebra
homomorphism $\varphi:k\left[  X\right]  \rightarrow A$ satisfying
$\varphi\left(  X\right)  =a$ (by the definition of $\operatorname*{ev}%
\nolimits_{A,\ a}$). Hence, every $k$-algebra homomorphism $\varphi:k\left[
X\right]  \rightarrow A$ satisfying $\varphi\left(  X\right)  =a$ must be
equal to $\operatorname*{ev}\nolimits_{A,\ a}$. In other words, if
$\varphi:k\left[  X\right]  \rightarrow A$ is a $k$-algebra homomorphism
satisfying $\varphi\left(  X\right)  =a$, then $\varphi=\operatorname*{ev}%
\nolimits_{A,\ a}$. Applying this to $\varphi=f$, we obtain
$f=\operatorname*{ev}\nolimits_{A,\ a}$ (since $f\left(  X\right)  =a$).

On the other hand, if $\varphi:k\left[  X\right]  \rightarrow A$ is a
$k$-algebra homomorphism satisfying $\varphi\left(  X\right)  =a$, then
$\varphi=\operatorname*{ev}\nolimits_{A,\ a}$. Applying this to $\varphi=g$,
we obtain $g=\operatorname*{ev}\nolimits_{A,\ a}$ (since $g\left(  X\right)
=a$). Hence, $f=\operatorname*{ev}\nolimits_{A,\ a}=g$. This proves
Corollary~\ref{Corollary1206}.
\end{proof}

Here are a number of properties of the $k$-bialgebra $k\left[  X\right]  $
that we are going to need:

\begin{theorem}
\label{Theorem1211} Let $k$ be a field. Let $X$ be a new symbol. Consider the
polynomial ring $k\left[  X\right]  $ over $k$ in one indeterminate $X$. Equip
$k\left[  X\right]  $ with the usual grading (by degree).

Let $\Delta_{X}$ be the $k$-algebra homomorphism $\operatorname*{ev}%
\nolimits_{k\left[  X\right]  \otimes k\left[  X\right]  ,\ X\otimes1+1\otimes
X}:k\left[  X\right]  \rightarrow k\left[  X\right]  \otimes k\left[
X\right]  $ (where $1$ means $1_{k\left[  X\right]  }$).

Let $\varepsilon_{X}$ be the $k$-algebra homomorphism $\operatorname*{ev}%
\nolimits_{k,\ 0}:k\left[  X\right]  \rightarrow k$.

Then, $\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $ is a
commutative connected graded $k$-bialgebra.
\end{theorem}

\begin{remark}
\label{Remark1212} The $k$-bialgebra $\left(  k\left[  X\right]  ,\Delta
_{X},\varepsilon_{X}\right)  $ of Theorem~\ref{Theorem1211} is also
cocommutative, but we won't need this in the following.
\end{remark}

\begin{proof}
[Proof of Theorem~\ref{Theorem1211}.]Recall that $k\left[  X\right]  $ is the
free $k$-module with basis $\left(  X^{n}\right)  _{n\in\mathbb{N}}$. Recall
also that the grading on $k\left[  X\right]  $ is such that $\left(  k\left[
X\right]  \right)  _{n}$ is a free $k$-module with basis $\left(
X^{n}\right)  $ for every $n\in\mathbb{N}$. (Note the difference between
$\left(  X^{n}\right)  _{n\in\mathbb{N}}$ and $\left(  X^{n}\right)  $: The
former is a family indexed by nonnegative integer, while the latter is a
one-element family.)

It is known that $k\left[  X\right]  $ is a commutative graded algebra.

For every $k$-vector space $V$, let $\operatorname*{kanl}\nolimits_{V}$ be the
canonical isomorphism $V\rightarrow V\otimes k$ which sends every $v\in V$ to
$v\otimes1\in V\otimes k$.

For every $k$-vector space $V$, let $\operatorname*{kanr}\nolimits_{V}$ be the
canonical isomorphism $V\rightarrow k\otimes V$ which sends every $v\in V$ to
$1\otimes v\in k\otimes V$.

Notice that if $A$ is a $k$-algebra, then $\operatorname*{kanl}\nolimits_{A}$
and $\operatorname*{kanr}\nolimits_{A}$ are $k$-algebra homomorphisms.
Applying this to $A=k\left[  X\right]  $, we conclude that
$\operatorname*{kanl}\nolimits_{k\left[  X\right]  }$ and
$\operatorname*{kanr}\nolimits_{k\left[  X\right]  }$ are $k$-algebra homomorphisms.

Since $\Delta_{X}=\operatorname*{ev}\nolimits_{k\left[  X\right]  \otimes
k\left[  X\right]  ,\ X\otimes1+1\otimes X}$, we have%
\[
\Delta_{X}\left(  X\right)  =\operatorname*{ev}\nolimits_{k\left[  X\right]
\otimes k\left[  X\right]  ,\ X\otimes1+1\otimes X}\left(  X\right)
=X\otimes1+1\otimes X
\]
(by (\ref{1205}), applied to $A=k\left[  X\right]  \otimes k\left[  X\right]
$ and $a=X\otimes1+1\otimes X$). Since $\varepsilon_{X}=\operatorname*{ev}%
\nolimits_{k,\ 0}$, we have%
\[
\varepsilon_{X}\left(  X\right)  =\operatorname*{ev}\nolimits_{k,\ 0}\left(
X\right)  =0
\]
(by (\ref{1205}), applied to $A=k$ and $a=0$).

We know that $\operatorname{id}_{k\left[  X\right]  }$ and $\Delta_{X}$ are
$k$-algebra homomorphisms. Thus, their tensor product $\operatorname{id}%
_{k\left[  X\right]  }\otimes\Delta_{X}$ is a $k$-algebra homomorphism (since
the tensor product of two $k$-algebra homomorphisms is always a $k$-algebra homomorphism).

We know that $\Delta_{X}$ and $\operatorname{id}_{k\left[  X\right]  }$ are
$k$-algebra homomorphisms. Thus, their tensor product $\Delta_{X}%
\otimes\operatorname{id}_{k\left[  X\right]  }$ is a $k$-algebra homomorphism
(since the tensor product of two $k$-algebra homomorphisms is always a
$k$-algebra homomorphism).

We know that $\operatorname{id}_{k\left[  X\right]  }$ and $\varepsilon_{X}$
are $k$-algebra homomorphisms. Thus, their tensor product $\operatorname{id}%
_{k\left[  X\right]  }\otimes\varepsilon_{X}$ is a $k$-algebra homomorphism
(since the tensor product of two $k$-algebra homomorphisms is always a
$k$-algebra homomorphism).

We know that $\varepsilon_{X}$ and $\operatorname{id}_{k\left[  X\right]  }$
are $k$-algebra homomorphisms. Thus, their tensor product $\varepsilon
_{X}\otimes\operatorname{id}_{k\left[  X\right]  }$ is a $k$-algebra
homomorphism (since the tensor product of two $k$-algebra homomorphisms is
always a $k$-algebra homomorphism).

We know that $\operatorname{id}_{k\left[  X\right]  }\otimes\Delta_{X}$ and
$\Delta_{X}$ are $k$-algebra homomorphisms. Thus, their composition $\left(
\operatorname{id}_{k\left[  X\right]  }\otimes\Delta_{X}\right)  \circ
\Delta_{X}$ is a $k$-algebra homomorphism (since the composition of two
$k$-algebra homomorphisms is always a $k$-algebra homomorphism).

We know that $\Delta_{X}\otimes\operatorname{id}_{k\left[  X\right]  }$ and
$\Delta_{X}$ are $k$-algebra homomorphisms. Thus, their composition $\left(
\Delta_{X}\otimes\operatorname{id}_{k\left[  X\right]  }\right)  \circ
\Delta_{X}$ is a $k$-algebra homomorphism (since the composition of two
$k$-algebra homomorphisms is always a $k$-algebra homomorphism).

We know that $\operatorname{id}_{k\left[  X\right]  }\otimes\varepsilon_{X}$
and $\Delta_{X}$ are $k$-algebra homomorphisms. Thus, their composition
$\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\varepsilon_{X}\right)
\circ\Delta_{X}$ is a $k$-algebra homomorphism (since the composition of two
$k$-algebra homomorphisms is always a $k$-algebra homomorphism).

We know that $\varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]  }$
and $\Delta_{X}$ are $k$-algebra homomorphisms. Thus, their composition
$\left(  \varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]  }\right)
\circ\Delta_{X}$ is a $k$-algebra homomorphism (since the composition of two
$k$-algebra homomorphisms is always a $k$-algebra homomorphism).

Comparing%
\begin{align*}
&  \left(  \left(  \operatorname{id}_{k\left[  X\right]  }\otimes\Delta
_{X}\right)  \circ\Delta_{X}\right)  \left(  X\right) \\
&  =\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\Delta_{X}\right)
\left(  \underbrace{\Delta_{X}\left(  X\right)  }_{=X\otimes1+1\otimes
X}\right)  =\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\Delta
_{X}\right)  \left(  X\otimes1+1\otimes X\right) \\
&  =\underbrace{\operatorname{id}_{k\left[  X\right]  }\left(  X\right)
}_{=X}\otimes\underbrace{\Delta_{X}\left(  1\right)  }_{\substack{=1_{k\left[
X\right]  \otimes k\left[  X\right]  }\\\text{(since }\Delta_{X}\text{ is a
}k\text{-algebra}\\\text{homomorphism)}}}+\underbrace{\operatorname{id}%
_{k\left[  X\right]  }\left(  1\right)  }_{=1}\otimes\underbrace{\Delta
_{X}\left(  X\right)  }_{=X\otimes1+1\otimes X}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname{id}%
_{k\left[  X\right]  }\otimes\Delta_{X}\right) \\
&  =X\otimes\underbrace{1_{k\left[  X\right]  \otimes k\left[  X\right]  }%
}_{=1\otimes1}+\underbrace{1\otimes\left(  X\otimes1+1\otimes X\right)
}_{\substack{=1\otimes X\otimes1+1\otimes1\otimes X\\\text{(since the tensor
product is distributive)}}}\\
&  =X\otimes1\otimes1+1\otimes X\otimes1+1\otimes1\otimes X
\end{align*}
with%
\begin{align*}
&  \left(  \left(  \Delta_{X}\otimes\operatorname{id}_{k\left[  X\right]
}\right)  \circ\Delta_{X}\right)  \left(  X\right) \\
&  =\left(  \Delta_{X}\otimes\operatorname{id}_{k\left[  X\right]  }\right)
\left(  \underbrace{\Delta_{X}\left(  X\right)  }_{=X\otimes1+1\otimes
X}\right)  =\left(  \Delta_{X}\otimes\operatorname{id}_{k\left[  X\right]
}\right)  \left(  X\otimes1+1\otimes X\right) \\
&  =\underbrace{\Delta_{X}\left(  X\right)  }_{=X\otimes1+1\otimes X}%
\otimes\underbrace{\operatorname{id}_{k\left[  X\right]  }\left(  1\right)
}_{=1}+\underbrace{\Delta_{X}\left(  1\right)  }_{\substack{=1_{k\left[
X\right]  \otimes k\left[  X\right]  }\\\text{(since }\Delta_{X}\text{ is a
}k\text{-algebra}\\\text{homomorphism)}}}\otimes\underbrace{\operatorname{id}%
_{k\left[  X\right]  }\left(  X\right)  }_{=X}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Delta_{X}%
\otimes\operatorname{id}_{k\left[  X\right]  }\right) \\
&  =\underbrace{\left(  X\otimes1+1\otimes X\right)  \otimes1}%
_{\substack{=X\otimes1\otimes1+1\otimes X\otimes1\\\text{(since the tensor
product is distributive)}}}+\underbrace{1_{k\left[  X\right]  \otimes k\left[
X\right]  }}_{=1\otimes1}\otimes X\\
&  =X\otimes1\otimes1+1\otimes X\otimes1+1\otimes1\otimes X,
\end{align*}
we obtain%
\[
\left(  \left(  \operatorname{id}_{k\left[  X\right]  }\otimes\Delta
_{X}\right)  \circ\Delta_{X}\right)  \left(  X\right)  =\left(  \left(
\Delta_{X}\otimes\operatorname{id}_{k\left[  X\right]  }\right)  \circ
\Delta_{X}\right)  \left(  X\right)  .
\]
Thus, Corollary~\ref{Corollary1206} (applied to $A=k\left[  X\right]  \otimes
k\left[  X\right]  \otimes k\left[  X\right]  $, $f=\left(  \operatorname{id}%
_{k\left[  X\right]  }\otimes\Delta_{X}\right)  \circ\Delta_{X}$ and
$g=\left(  \Delta_{X}\otimes\operatorname{id}_{k\left[  X\right]  }\right)
\circ\Delta_{X}$) yields
\[
\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\Delta_{X}\right)
\circ\Delta_{X}=\left(  \Delta_{X}\otimes\operatorname{id}_{k\left[  X\right]
}\right)  \circ\Delta_{X}%
\]
(since $\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\Delta
_{X}\right)  \circ\Delta_{X}$ and $\left(  \Delta_{X}\otimes\operatorname{id}%
_{k\left[  X\right]  }\right)  \circ\Delta_{X}$ are $k$-algebra homomorphisms).

Comparing%
\begin{align*}
&  \left(  \left(  \operatorname{id}_{k\left[  X\right]  }\otimes
\varepsilon_{X}\right)  \circ\Delta_{X}\right)  \left(  X\right) \\
&  =\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\varepsilon
_{X}\right)  \left(  \underbrace{\Delta_{X}\left(  X\right)  }_{=X\otimes
1+1\otimes X}\right)  =\left(  \operatorname{id}_{k\left[  X\right]  }%
\otimes\varepsilon_{X}\right)  \left(  X\otimes1+1\otimes X\right) \\
&  =\underbrace{\operatorname{id}_{k\left[  X\right]  }\left(  X\right)
}_{=X}\otimes\underbrace{\varepsilon_{X}\left(  1\right)  }%
_{\substack{=1\\\text{(since }\varepsilon_{X}\text{ is a }k\text{-algebra}%
\\\text{homomorphism)}}}+\operatorname{id}_{k\left[  X\right]  }\left(
1\right)  \otimes\underbrace{\varepsilon_{X}\left(  X\right)  }_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname{id}%
_{k\left[  X\right]  }\otimes\varepsilon_{X}\right) \\
&  =X\otimes1+\underbrace{\operatorname{id}_{k\left[  X\right]  }\left(
1\right)  \otimes0}_{=0}=X\otimes1
\end{align*}
with%
\[
\operatorname*{kanl}\nolimits_{k\left[  X\right]  }\left(  X\right)
=X\otimes1\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{kanl}\nolimits_{k\left[  X\right]  }\right)  ,
\]
we obtain
\[
\left(  \left(  \operatorname{id}_{k\left[  X\right]  }\otimes\varepsilon
_{X}\right)  \circ\Delta_{X}\right)  \left(  X\right)  =\operatorname*{kanl}%
\nolimits_{k\left[  X\right]  }\left(  X\right)  .
\]
Thus, Corollary~\ref{Corollary1206} (applied to $A=k\left[  X\right]  \otimes
k$, $f=\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\varepsilon
_{X}\right)  \circ\Delta_{X}$ and $g=\operatorname*{kanl}\nolimits_{k\left[
X\right]  }$) yields
\[
\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\varepsilon_{X}\right)
\circ\Delta_{X}=\operatorname*{kanl}\nolimits_{k\left[  X\right]  }%
\]
(since $\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\varepsilon
_{X}\right)  \circ\Delta_{X}$ and $\operatorname*{kanl}\nolimits_{k\left[
X\right]  }$ are $k$-algebra homomorphisms).

Comparing%
\begin{align*}
&  \left(  \left(  \varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]
}\right)  \circ\Delta_{X}\right)  \left(  X\right) \\
&  =\left(  \varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]
}\right)  \left(  \underbrace{\Delta_{X}\left(  X\right)  }_{=X\otimes
1+1\otimes X}\right)  =\left(  \varepsilon_{X}\otimes\operatorname{id}%
_{k\left[  X\right]  }\right)  \left(  X\otimes1+1\otimes X\right) \\
&  =\underbrace{\varepsilon_{X}\left(  X\right)  }_{=0}\otimes
\operatorname{id}_{k\left[  X\right]  }\left(  1\right)
+\underbrace{\varepsilon_{X}\left(  1\right)  }_{\substack{=1\\\text{(since
}\varepsilon_{X}\text{ is a }k\text{-algebra}\\\text{homomorphism)}}%
}\otimes\underbrace{\operatorname{id}_{k\left[  X\right]  }\left(  X\right)
}_{=X}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\varepsilon
_{X}\otimes\operatorname{id}_{k\left[  X\right]  }\right) \\
&  =\underbrace{0\otimes\operatorname{id}_{k\left[  X\right]  }\left(
1\right)  }_{=0}+1\otimes X=1\otimes X
\end{align*}
with%
\[
\operatorname*{kanr}\nolimits_{k\left[  X\right]  }\left(  X\right)  =1\otimes
X\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{kanr}%
\nolimits_{k\left[  X\right]  }\right)  ,
\]
we obtain
\[
\left(  \left(  \varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]
}\right)  \circ\Delta_{X}\right)  \left(  X\right)  =\operatorname*{kanr}%
\nolimits_{k\left[  X\right]  }\left(  X\right)  .
\]
Thus, Corollary~\ref{Corollary1206} (applied to $A=k\otimes k\left[  X\right]
$, $f=\left(  \varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]
}\right)  \circ\Delta_{X}$ and $g=\operatorname*{kanr}\nolimits_{k\left[
X\right]  }$) yields
\[
\left(  \varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]  }\right)
\circ\Delta_{X}=\operatorname*{kanr}\nolimits_{k\left[  X\right]  }%
\]
(since $\left(  \varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]
}\right)  \circ\Delta_{X}$ and $\operatorname*{kanr}\nolimits_{k\left[
X\right]  }$ are $k$-algebra homomorphisms).

Recall the definition of a $k$-coalgebra: If $C$ is a $k$-vector space, and
$\Delta:C\rightarrow C\otimes C$ and $\varepsilon:C\rightarrow k$ are two
$k$-linear maps, then $\left(  C,\Delta,\varepsilon\right)  $ is a
$k$-coalgebra if and only if the equalities%
\begin{align*}
\left(  \operatorname{id}_{C}\otimes\Delta\right)  \circ\Delta &  =\left(
\Delta\otimes\operatorname{id}_{C}\right)  \circ\Delta,\\
\left(  \operatorname{id}_{C}\otimes\varepsilon\right)  \circ\Delta &
=\operatorname*{kanl}\nolimits_{C},\\
\left(  \varepsilon\otimes\operatorname{id}_{C}\right)  \circ\Delta &
=\operatorname*{kanr}\nolimits_{C}%
\end{align*}
hold. Applying this to $\left(  C,\Delta,\varepsilon\right)  =\left(  k\left[
X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $, we conclude that $\left(
k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $ is a $k$-coalgebra if
and only if the equalities%
\begin{align}
\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\Delta_{X}\right)
\circ\Delta_{X}  &  =\left(  \Delta_{X}\otimes\operatorname{id}_{k\left[
X\right]  }\right)  \circ\Delta_{X},\label{1211.pf.coalg1}\\
\left(  \operatorname{id}_{k\left[  X\right]  }\otimes\varepsilon_{X}\right)
\circ\Delta_{X}  &  =\operatorname*{kanl}\nolimits_{k\left[  X\right]
},\label{1211.pf.coalg2}\\
\left(  \varepsilon_{X}\otimes\operatorname{id}_{k\left[  X\right]  }\right)
\circ\Delta_{X}  &  =\operatorname*{kanr}\nolimits_{k\left[  X\right]  }
\label{1211.pf.coalg3}%
\end{align}
hold. Since we know that these equalities (\ref{1211.pf.coalg1}),
(\ref{1211.pf.coalg2}) and (\ref{1211.pf.coalg3}) hold, this yields that
$\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $ is a $k$-coalgebra.

Now, let us recall the definition of a $k$-bialgebra, or at least one possible
definition of it: If $C$ is a $k$-algebra, and $\Delta:C\rightarrow C\otimes
C$ and $\varepsilon:C\rightarrow k$ are two $k$-linear maps such that $\left(
C,\Delta,\varepsilon\right)  $ is a $k$-coalgebra, then $\left(
C,\Delta,\varepsilon\right)  $ (endowed with the given $k$-algebra structure
on $C$) is a $k$-bialgebra if and only if the maps $\Delta$ and $\varepsilon$
are $k$-algebra homomorphisms. Applying this to $\left(  C,\Delta
,\varepsilon\right)  =\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon
_{X}\right)  $, we conclude that $\left(  k\left[  X\right]  ,\Delta
_{X},\varepsilon_{X}\right)  $ is a $k$-bialgebra if and only if the maps
$\Delta_{X}$ and $\varepsilon_{X}$ are $k$-algebra homomorphisms (because we
already know that $\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon
_{X}\right)  $ is a $k$-coalgebra). Thus, $\left(  k\left[  X\right]
,\Delta_{X},\varepsilon_{X}\right)  $ is a $k$-bialgebra (since we know that
the maps $\Delta_{X}$ and $\varepsilon_{X}$ are $k$-algebra homomorphisms).

The map $\Delta_{X}$ is graded\footnote{\textit{Proof.} Recall that $k\left[
X\right]  $ is a graded $k$-algebra. Hence, $k\left[  X\right]  \otimes
k\left[  X\right]  $ is a graded $k$-algebra as well (since the tensor product
of two graded $k$-algebras always is a graded $k$-algebra). Denote this graded
$k$-algebra $k\left[  X\right]  \otimes k\left[  X\right]  $ by $A$.
\par
We know that the map $\Delta_{X}$ is a $k$-algebra homomorphism $k\left[
X\right]  \rightarrow k\left[  X\right]  \otimes k\left[  X\right]  $. Since
$k\left[  X\right]  \otimes k\left[  X\right]  =A$, this rewrites as follows:
The map $\Delta_{X}$ is a $k$-algebra homomorphism $k\left[  X\right]
\rightarrow A$.
\par
By the definition of the grading on $k\left[  X\right]  \otimes k\left[
X\right]  $, we have $\left(  k\left[  X\right]  \otimes k\left[  X\right]
\right)  _{\ell}=\sum_{i=0}^{\ell}\left(  k\left[  X\right]  \right)
_{i}\otimes\left(  k\left[  X\right]  \right)  _{\ell-i}$ for every $\ell
\in\mathbb{N}$. Applying this to $\ell=1$, we obtain%
\begin{align*}
\left(  k\left[  X\right]  \otimes k\left[  X\right]  \right)  _{1}  &
=\sum_{i=0}^{1}\left(  k\left[  X\right]  \right)  _{i}\otimes\left(  k\left[
X\right]  \right)  _{1-i}=\left(  k\left[  X\right]  \right)  _{0}%
\otimes\underbrace{\left(  k\left[  X\right]  \right)  _{1-0}}_{=\left(
k\left[  X\right]  \right)  _{1}}+\left(  k\left[  X\right]  \right)
_{1}\otimes\underbrace{\left(  k\left[  X\right]  \right)  _{1-1}}_{=\left(
k\left[  X\right]  \right)  _{0}}\\
&  =\left(  k\left[  X\right]  \right)  _{0}\otimes\left(  k\left[  X\right]
\right)  _{1}+\left(  k\left[  X\right]  \right)  _{1}\otimes\left(  k\left[
X\right]  \right)  _{0}.
\end{align*}
Now,
\begin{align*}
\Delta_{X}\left(  X\right)   &  =X\otimes1+1\otimes X=\underbrace{1}%
_{\in\left(  k\left[  X\right]  \right)  _{0}}\otimes\underbrace{X}%
_{\in\left(  k\left[  X\right]  \right)  _{1}}+\underbrace{X}_{\in\left(
k\left[  X\right]  \right)  _{1}}\otimes\underbrace{1}_{\in\left(  k\left[
X\right]  \right)  _{0}}\\
&  \in\left(  k\left[  X\right]  \right)  _{0}\otimes\left(  k\left[
X\right]  \right)  _{1}+\left(  k\left[  X\right]  \right)  _{1}\otimes\left(
k\left[  X\right]  \right)  _{0}=\left(  \underbrace{k\left[  X\right]
\otimes k\left[  X\right]  }_{=A}\right)  _{1}=A_{1}.
\end{align*}
\par
Now, let us show that%
\begin{equation}
\Delta_{X}\left(  X^{n}\right)  \in A_{n}\ \ \ \ \ \ \ \ \ \ \text{for every
}n\in\mathbb{N}. \label{1211.pf.Del.grad.1}%
\end{equation}
\par
\textit{Proof of (\ref{1211.pf.Del.grad.1}):} We will prove
(\ref{1211.pf.Del.grad.1}) by induction over $n$:
\par
\textit{Induction base:} Since $X^{0}=1$, we have
\begin{align*}
\Delta_{X}\left(  X^{0}\right)   &  =\Delta_{X}\left(  1\right)
=1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{X}\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  \in A_{0}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }A\text{ is a graded }k\text{-algebra, and since any}\\
\text{graded }k\text{-algebra }B\text{ satisfies }1_{B}\in B_{0}%
\end{array}
\right)  .
\end{align*}
In other words, (\ref{1211.pf.Del.grad.1}) is proven for $n=0$. This completes
the induction base.
\par
\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
(\ref{1211.pf.Del.grad.1}) holds for $n=N$. We need to show that
(\ref{1211.pf.Del.grad.1}) holds for $n=N+1$.
\par
We have $\Delta_{X}\left(  X^{N}\right)  \in A_{N}$ (since
(\ref{1211.pf.Del.grad.1}) holds for $n=N$). But every graded $k$-algebra $B$
and any two nonnegative integers $u$ and $v$ satisfy $B_{u}B_{v}\subseteq
B_{u+v}$. Applying this to $B=A$, $u=N$ and $v=1$, we obtain $A_{N}%
A_{1}\subseteq A_{N+1}$. Now, $X^{N+1}=X^{N}X$, so that%
\begin{align*}
\Delta_{X}\left(  X^{N+1}\right)   &  =\Delta_{X}\left(  X^{N}X\right)
=\underbrace{\Delta_{X}\left(  X^{N}\right)  }_{\in A_{N}}\cdot
\underbrace{\Delta_{X}\left(  X\right)  }_{\in A_{1}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{X}\text{ is a }%
k\text{-algebra homomorphism}\right) \\
&  \in A_{N}A_{1}\subseteq A_{N+1}.
\end{align*}
In other words, (\ref{1211.pf.Del.grad.1}) holds for $n=N+1$. This completes
the induction step. The induction proof of (\ref{1211.pf.Del.grad.1}) is thus
complete.
\par
Now, let $n\in\mathbb{N}$. We know that $\left(  k\left[  X\right]  \right)
_{n}$ is a free $k$-module with basis $\left(  X^{n}\right)  $. Thus, the
$k$-module $\left(  k\left[  X\right]  \right)  _{n}$ is spanned by $\left(
X^{n}\right)  $. In other words, $\left(  k\left[  X\right]  \right)
_{n}=k\cdot X^{n}$. Thus,%
\begin{align*}
\Delta_{X}\left(  \left(  k\left[  X\right]  \right)  _{n}\right)   &
=\Delta_{X}\left(  k\cdot X^{n}\right)  =k\cdot\underbrace{\Delta_{X}\left(
X^{n}\right)  }_{\substack{\in A_{n}\\\text{(by (\ref{1211.pf.Del.grad.1}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{X}\text{ is }%
k\text{-linear}\right) \\
&  \subseteq k\cdot A_{n}\subseteq A_{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }A_{n}\text{ is a }k\text{-vector space}\right)  .
\end{align*}
Now, forget that we fixed $n$. We thus have proven that $\Delta_{X}\left(
\left(  k\left[  X\right]  \right)  _{n}\right)  \subseteq A_{n}$ for every
$n\in\mathbb{N}$. In other words, the map $\Delta_{X}$ is graded, qed.}. The
map $\varepsilon_{X}$ is graded\footnote{\textit{Proof.} Let us show that%
\begin{equation}
\varepsilon_{X}\left(  X^{n}\right)  \in k_{n}\ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N}. \label{1211.pf.eps.grad.1}%
\end{equation}
\par
\textit{Proof of (\ref{1211.pf.eps.grad.1}):} Let $n\in\mathbb{N}$. If $n>0$,
then%
\begin{align*}
\varepsilon_{X}\left(  X^{n}\right)   &  =\left(  \underbrace{\varepsilon
_{X}\left(  X\right)  }_{=0}\right)  ^{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\varepsilon_{X}\text{ is a }k\text{-algebra homomorphism}\right)
\\
&  =0^{n}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n>0\right) \\
&  \in k_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k_{n}\text{ is a
}k\text{-vector space}\right)  .
\end{align*}
Hence, if $n>0$, then (\ref{1211.pf.eps.grad.1}) holds. Thus, for the rest of
the proof of (\ref{1211.pf.eps.grad.1}), we can WLOG assume that we don't have
$n>0$. Assume this.
\par
We have $n\in\mathbb{N}$, but we don't have $n>0$. Hence, $n=0$. Thus,
$k_{n}=k_{0}=k$, so that $\varepsilon_{X}\left(  X^{n}\right)  \in k=k_{n}$.
Hence, (\ref{1211.pf.eps.grad.1}) is proven.
\par
Now, let $n\in\mathbb{N}$. We know that $\left(  k\left[  X\right]  \right)
_{n}$ is a free $k$-module with basis $\left(  X^{n}\right)  $. Thus, the
$k$-module $\left(  k\left[  X\right]  \right)  _{n}$ is spanned by $\left(
X^{n}\right)  $. In other words, $\left(  k\left[  X\right]  \right)
_{n}=k\cdot X^{n}$. Thus,%
\begin{align*}
\varepsilon_{X}\left(  \left(  k\left[  X\right]  \right)  _{n}\right)   &
=\varepsilon_{X}\left(  k\cdot X^{n}\right)  =k\cdot\underbrace{\varepsilon
_{X}\left(  X^{n}\right)  }_{\substack{\in k_{n}\\\text{(by
(\ref{1211.pf.eps.grad.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\varepsilon_{X}\text{ is }k\text{-linear}\right) \\
&  \subseteq k\cdot k_{n}\subseteq k_{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }k_{n}\text{ is a }k\text{-vector space}\right)  .
\end{align*}
Now, forget that we fixed $n$. We thus have proven that $\varepsilon
_{X}\left(  \left(  k\left[  X\right]  \right)  _{n}\right)  \subseteq k_{n}$
for every $n\in\mathbb{N}$. In other words, the map $\varepsilon_{X}$ is
graded, qed.}.

Let us now recall one possible definition of a graded $k$-coalgebra: If $C$ is
a graded $k$-vector space, and $\Delta:C\rightarrow C\otimes C$ and
$\varepsilon:C\rightarrow k$ are two $k$-linear maps such that $\left(
C,\Delta,\varepsilon\right)  $ is a $k$-coalgebra, then the $k$-coalgebra
$\left(  C,\Delta,\varepsilon\right)  $ is graded if and only if the maps
$\Delta$ and $\varepsilon$ are graded. Applying this to $\left(
C,\Delta,\varepsilon\right)  =\left(  k\left[  X\right]  ,\Delta
_{X},\varepsilon_{X}\right)  $, we conclude that the $k$-coalgebra $\left(
k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $ is graded if and only
if the maps $\Delta_{X}$ and $\varepsilon_{X}$ are graded (because we already
know that $\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $ is
a $k$-coalgebra). Thus, the $k$-coalgebra $\left(  k\left[  X\right]
,\Delta_{X},\varepsilon_{X}\right)  $ is graded (since we know that the maps
$\Delta_{X}$ and $\varepsilon_{X}$ are graded). Combining this with the fact
that the $k$-algebra $k\left[  X\right]  $ is graded, we conclude that the
$k$-bialgebra $\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)
$ is graded.

Denote the $k$-coalgebra $\left(  k\left[  X\right]  ,\Delta_{X}%
,\varepsilon_{X}\right)  $ by $k\left[  X\right]  $. Then, $\Delta_{k\left[
X\right]  }=\Delta_{X}$ and $\varepsilon_{k\left[  X\right]  }=\varepsilon
_{X}$.

The graded $k$-coalgebra $k\left[  X\right]  $ is
connected\footnote{\textit{Proof.} Recall that $\left(  k\left[  X\right]
\right)  _{n}$ is a free $k$-module with basis $\left(  X^{n}\right)  $ for
every $n\in\mathbb{N}$. Applied to $n=0$, this yields that $\left(  k\left[
X\right]  \right)  _{0}$ is a free $k$-module with basis $\left(
X^{0}\right)  $. Hence, the $k$-module $\left(  k\left[  X\right]  \right)
_{0}$ is spanned by $\left(  X^{0}\right)  $. In other words, $\left(
k\left[  X\right]  \right)  _{0}=k\cdot X^{0}$. Thus, $\left(  k\left[
X\right]  \right)  _{0}=k\cdot\underbrace{X^{0}}_{=1}=k\cdot1$. Moreover,
$\underbrace{\varepsilon_{k\left[  X\right]  }}_{=\varepsilon_{X}}\left(
1\right)  =\varepsilon_{X}\left(  1\right)  =1$. Hence, Remark~\ref{Remark214}
(applied to $C=k\left[  X\right]  $ and $\lambda=1$) yields that the graded
$k$-coalgebra $k\left[  X\right]  $ is connected, qed.}. Since the graded
$k$-coalgebra $k\left[  X\right]  $ is the graded $k$-coalgebra $\left(
k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $, this rewrites as
follows: The graded $k$-coalgebra $\left(  k\left[  X\right]  ,\Delta
_{X},\varepsilon_{X}\right)  $ is connected. Combining this with the fact that
$\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $ is a graded
$k$-bialgebra, we conclude that $\left(  k\left[  X\right]  ,\Delta
_{X},\varepsilon_{X}\right)  $ is a connected graded $k$-bialgebra. Combining
this with the fact that $k\left[  X\right]  $ is a commutative $k$-algebra, we
obtain that $\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $
is a commutative connected graded $k$-bialgebra. This proves
Theorem~\ref{Theorem1211}.
\end{proof}

\begin{proposition}
\label{Proposition1213} Let $k$ be a field. Let $X$ be a new symbol. Denote
the commutative connected graded $k$-bialgebra $\left(  k\left[  X\right]
,\Delta_{X},\varepsilon_{X}\right)  $ defined in Theorem~\ref{Theorem1211}
simply by $k\left[  X\right]  $.

Let $A$ be a $k$-bialgebra. Let $a\in A$ be a primitive element of $A$. Then,
$\operatorname*{ev}\nolimits_{A,\ a}:k\left[  X\right]  \rightarrow A$ is a
$k$-bialgebra homomorphism.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1213}.]By the definition of
$\operatorname*{ev}\nolimits_{A,\ a}$, we know that $\operatorname*{ev}%
\nolimits_{A,\ a}$ is a $k$-algebra homomorphism. Thus, $\operatorname*{ev}%
\nolimits_{A,\ a}\left(  1\right)  =1_{A}$.

We will use the notations defined in Theorem~\ref{Theorem1211}. The
$k$-bialgebra $k\left[  X\right]  $ is defined as $\left(  k\left[  X\right]
,\Delta_{X},\varepsilon_{X}\right)  $. Thus, $\Delta_{k\left[  X\right]
}=\Delta_{X}$ and $\varepsilon_{k\left[  X\right]  }=\varepsilon_{X}$.

As in the proof of Theorem~\ref{Theorem1211}, we can show that%
\[
\Delta_{X}\left(  X\right)  =X\otimes1+1\otimes
X\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \varepsilon_{X}\left(
X\right)  =0.
\]


Since $A$ is a $k$-bialgebra, we know that $\Delta_{A}$ and $\varepsilon_{A}$
are $k$-algebra homomorphisms (due to the axioms of a $k$-bialgebra).

We know that $\operatorname*{ev}\nolimits_{A,\ a}$ and $\operatorname*{ev}%
\nolimits_{A,\ a}$ are $k$-algebra homomorphisms. Thus, their tensor product
$\operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}\nolimits_{A,\ a}%
$ is a $k$-algebra homomorphism (since the tensor product of two $k$-algebra
homomorphisms is always a $k$-algebra homomorphism).

We know that $\operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}%
\nolimits_{A,\ a}$ and $\Delta_{k\left[  X\right]  }$ are $k$-algebra
homomorphisms. Thus, their composition $\left(  \operatorname*{ev}%
\nolimits_{A,\ a}\otimes\operatorname*{ev}\nolimits_{A,\ a}\right)
\circ\Delta_{k\left[  X\right]  }$ is a $k$-algebra homomorphism (since the
composition of two $k$-algebra homomorphisms is always a $k$-algebra homomorphism).

We know that $\Delta_{A}$ and $\operatorname*{ev}\nolimits_{A,\ a}$ are
$k$-algebra homomorphisms. Thus, their composition $\Delta_{A}\circ
\operatorname*{ev}\nolimits_{A,\ a}$ is a $k$-algebra homomorphism (since the
composition of two $k$-algebra homomorphisms is always a $k$-algebra homomorphism).

We know that $\varepsilon_{A}$ and $\operatorname*{ev}\nolimits_{A,\ a}$ are
$k$-algebra homomorphisms. Thus, their composition $\varepsilon_{A}%
\circ\operatorname*{ev}\nolimits_{A,\ a}$ is a $k$-algebra homomorphism (since
the composition of two $k$-algebra homomorphisms is always a $k$-algebra homomorphism).

Comparing%
\begin{align*}
&  \left(  \left(  \operatorname*{ev}\nolimits_{A,\ a}\otimes
\operatorname*{ev}\nolimits_{A,\ a}\right)  \circ\Delta_{k\left[  X\right]
}\right)  \left(  X\right) \\
&  =\left(  \operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}%
\nolimits_{A,\ a}\right)  \left(  \underbrace{\Delta_{k\left[  X\right]  }%
}_{=\Delta_{X}}\left(  X\right)  \right)  =\left(  \operatorname*{ev}%
\nolimits_{A,\ a}\otimes\operatorname*{ev}\nolimits_{A,\ a}\right)  \left(
\underbrace{\Delta_{X}\left(  X\right)  }_{=X\otimes1+1\otimes X}\right) \\
&  =\left(  \operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}%
\nolimits_{A,\ a}\right)  \left(  X\otimes1+1\otimes X\right)
=\underbrace{\operatorname*{ev}\nolimits_{A,\ a}\left(  X\right)
}_{\substack{=a\\\text{(by (\ref{1205}))}}}\otimes
\underbrace{\operatorname*{ev}\nolimits_{A,\ a}\left(  1\right)  }_{=1_{A}%
}+\underbrace{\operatorname*{ev}\nolimits_{A,\ a}\left(  1\right)  }_{=1_{A}%
}\otimes\underbrace{\operatorname*{ev}\nolimits_{A,\ a}\left(  X\right)
}_{\substack{=a\\\text{(by (\ref{1205}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{ev}%
\nolimits_{A,\ a}\otimes\operatorname*{ev}\nolimits_{A,\ a}\right) \\
&  =a\otimes1_{A}+1_{A}\otimes a
\end{align*}
with%
\begin{align*}
\left(  \Delta_{A}\circ\operatorname*{ev}\nolimits_{A,\ a}\right)  \left(
X\right)   &  =\Delta_{A}\left(  \underbrace{\operatorname*{ev}%
\nolimits_{A,\ a}\left(  X\right)  }_{\substack{=a\\\text{(by (\ref{1205}))}%
}}\right)  =\Delta_{A}\left(  a\right) \\
&  =a\otimes1_{A}+1_{A}\otimes a\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}a\text{ is primitive}\right)  ,
\end{align*}
we obtain%
\[
\left(  \left(  \operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}%
\nolimits_{A,\ a}\right)  \circ\Delta_{k\left[  X\right]  }\right)  \left(
X\right)  =\left(  \Delta_{A}\circ\operatorname*{ev}\nolimits_{A,\ a}\right)
\left(  X\right)  .
\]
Thus, Corollary~\ref{Corollary1206} (applied to $A\otimes A$, $\left(
\operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}\nolimits_{A,\ a}%
\right)  \circ\Delta_{k\left[  X\right]  }$ and $\Delta_{A}\circ
\operatorname*{ev}\nolimits_{A,\ a}$ instead of $A$, $f$ and $g$) yields
\[
\left(  \operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}%
\nolimits_{A,\ a}\right)  \circ\Delta_{k\left[  X\right]  }=\Delta_{A}%
\circ\operatorname*{ev}\nolimits_{A,\ a}%
\]
(since $\left(  \operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}%
\nolimits_{A,\ a}\right)  \circ\Delta_{k\left[  X\right]  }$ and $\Delta
_{A}\circ\operatorname*{ev}\nolimits_{A,\ a}$ are $k$-algebra homomorphisms).

But $A$ is a $k$-bialgebra. Hence, Proposition~\ref{Proposition7c} (applied to
$C=A$) yields that $\left(  A,1_{A}\right)  $ is a unital coalgebra. Since $a$
is a primitive element of $A$, we have $a\in\operatorname*{Prim}A$ (since
$\operatorname*{Prim}A$ is the set of all primitive elements of $A$). Thus,
Remark~\ref{Remark21} (applied to $H=A$ and $x=a$) yields that $\varepsilon
\left(  a\right)  =0$.

Comparing%
\[
\underbrace{\varepsilon_{k\left[  X\right]  }}_{=\varepsilon_{X}}\left(
X\right)  =\varepsilon_{X}\left(  X\right)  =0
\]
with%
\[
\left(  \varepsilon_{A}\circ\operatorname*{ev}\nolimits_{A,\ a}\right)
\left(  X\right)  =\varepsilon_{A}\left(  \underbrace{\operatorname*{ev}%
\nolimits_{A,\ a}\left(  X\right)  }_{\substack{=a\\\text{(by (\ref{1205}))}%
}}\right)  =\varepsilon_{A}\left(  a\right)  =\varepsilon\left(  a\right)
=0,
\]
we obtain%
\[
\varepsilon_{k\left[  X\right]  }\left(  X\right)  =\left(  \varepsilon
_{A}\circ\operatorname*{ev}\nolimits_{A,\ a}\right)  \left(  X\right)  .
\]
Thus, Corollary~\ref{Corollary1206} (applied to $k$, $\varepsilon_{k\left[
X\right]  }$ and $\varepsilon_{A}\circ\operatorname*{ev}\nolimits_{A,\ a}$
instead of $A$, $f$ and $g$) yields
\[
\varepsilon_{k\left[  X\right]  }=\varepsilon_{A}\circ\operatorname*{ev}%
\nolimits_{A,\ a}%
\]
(since $\varepsilon_{k\left[  X\right]  }$ and $\varepsilon_{A}\circ
\operatorname*{ev}\nolimits_{A,\ a}$ are $k$-algebra homomorphisms).

Now, recall the definition of a $k$-coalgebra homomorphism: If $C$ and $D$ are
two $k$-coalgebras and $f:C\rightarrow D$ is a $k$-linear map, then the map
$f$ is a $k$-coalgebra homomorphism if and only if it satisfies the equalities
$\left(  f\otimes f\right)  \circ\Delta_{C}=\Delta_{D}\circ f$ and
$\varepsilon_{C}=\varepsilon_{D}\circ f$. Applying this to $C=k\left[
X\right]  $, $D=A$ and $f=\operatorname*{ev}\nolimits_{A,\ a}$, we obtain the
following: The map $\operatorname*{ev}\nolimits_{A,\ a}$ is a $k$-coalgebra
homomorphism if and only if it satisfies the equalities $\left(
\operatorname*{ev}\nolimits_{A,\ a}\otimes\operatorname*{ev}\nolimits_{A,\ a}%
\right)  \circ\Delta_{k\left[  X\right]  }=\Delta_{A}\circ\operatorname*{ev}%
\nolimits_{A,\ a}$ and $\varepsilon_{k\left[  X\right]  }=\varepsilon_{A}%
\circ\operatorname*{ev}\nolimits_{A,\ a}$. Thus, the map $\operatorname*{ev}%
\nolimits_{A,\ a}$ is a $k$-coalgebra homomorphism (because it satisfies the
equalities $\left(  \operatorname*{ev}\nolimits_{A,\ a}\otimes
\operatorname*{ev}\nolimits_{A,\ a}\right)  \circ\Delta_{k\left[  X\right]
}=\Delta_{A}\circ\operatorname*{ev}\nolimits_{A,\ a}$ and $\varepsilon
_{k\left[  X\right]  }=\varepsilon_{A}\circ\operatorname*{ev}\nolimits_{A,\ a}%
$). Combining this with the fact that $\operatorname*{ev}\nolimits_{A,\ a}$ is
a $k$-algebra homomorphism, we obtain that $\operatorname*{ev}%
\nolimits_{A,\ a}$ is a $k$-bialgebra homomorphism. This proves
Proposition~\ref{Proposition1213}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem1202}.]Since $F$ is a $k$-algebra homomorphism,
we have $F\left(  1_{H}\right)  =1_{A}$. Thus, $F\in G\left(  H,A\right)
$\ \ \ \ \footnote{\textit{Proof.} By the definition of $G\left(  H,A\right)
$, we have $G\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\left(  1_{H}\right)  =1_{A}\right\}  $.
\par
Now, since $F\in\mathcal{L}\left(  H,A\right)  $ and $F\left(  1_{H}\right)
=1_{A}$, we have $F\in\left\{  f\in\mathcal{L}\left(  H,A\right)
\ \mid\ f\left(  1_{H}\right)  =1_{A}\right\}  =G\left(  H,A\right)  $, qed.}.
Hence, $\operatorname*{Log}F$ is a well-defined element of $\mathfrak{g}%
\left(  H,A\right)  $.

We have $v\in\operatorname*{Prim}H$. This shows that $v$ is a primitive
element of $H$ (since $\operatorname*{Prim}H$ is the set of all primitive
elements of $H$).

Let $X$ be a new symbol. Denote the commutative connected graded $k$-bialgebra
$\left(  k\left[  X\right]  ,\Delta_{X},\varepsilon_{X}\right)  $ defined in
Theorem~\ref{Theorem1211} simply by $k\left[  X\right]  $. Then,
$\Delta_{k\left[  X\right]  }=\Delta_{X}$ and $\varepsilon_{k\left[  X\right]
}=\varepsilon_{X}$.

As in the proof of Theorem~\ref{Theorem1211}, we can show that $\varepsilon
_{X}$ is a $k$-algebra homomorphism, and that $\varepsilon_{X}\left(
X\right)  =0$.

Recall that $k\left[  X\right]  $ is a connected graded $k$-bialgebra, thus a
connected graded $k$-coalgebra. Hence, $k\left[  X\right]  $ canonically
becomes a filtered $k$-coalgebra (according to Convention~\ref{Convention207},
applied to $C=k\left[  X\right]  $). Remark~\ref{Remark213} (applied to
$C=k\left[  X\right]  $) shows that the graded $k$-coalgebra $k\left[
X\right]  $ is connected if and only if the filtered $k$-coalgebra $k\left[
X\right]  $ is connected. Thus, the filtered $k$-coalgebra $k\left[  X\right]
$ is connected (since we know that the graded $k$-coalgebra $k\left[
X\right]  $ is connected).

Proposition~\ref{Proposition1213} (applied to $H$ and $v$ instead of $A$ and
$a$) yields that $\operatorname*{ev}\nolimits_{H,\ v}:k\left[  X\right]
\rightarrow H$ is a $k$-bialgebra homomorphism. In particular, this shows that
$\operatorname*{ev}\nolimits_{H,\ v}:k\left[  X\right]  \rightarrow H$ is a
$k$-coalgebra homomorphism.

Since $n>1$, we have $n-1>0$. Thus,%
\begin{align*}
\underbrace{\varepsilon_{k\left[  X\right]  }}_{=\varepsilon_{X}}\left(
X^{n-1}\right)   &  =\varepsilon_{X}\left(  X^{n-1}\right) \\
&  =\left(  \underbrace{\varepsilon_{X}\left(  X\right)  }_{=0}\right)
^{n-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon_{X}\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  =0^{n-1}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-1>0\right)  .
\end{align*}
Also, $\underbrace{\varepsilon_{k\left[  X\right]  }}_{=\varepsilon_{X}%
}\left(  X\right)  =\varepsilon_{X}\left(  X\right)  =0$.

We know that $\operatorname*{ev}\nolimits_{H,\ v}$ and $F$ are $k$-algebra
homomorphisms. Hence, their composition $F\circ\operatorname*{ev}%
\nolimits_{H,\ v}$ also is a $k$-algebra homomorphism (since the composition
of any two $k$-algebra homomorphisms must be a $k$-algebra homomorphism).
Thus, we can apply Corollary~\ref{Corollary1104} to $k\left[  X\right]  $,
$F\circ\operatorname*{ev}\nolimits_{H,\ v}$, $X^{n-1}$ and $X$ instead of $H$,
$F$, $x$ and $y$ (because $\varepsilon_{k\left[  X\right]  }\left(
X^{n-1}\right)  =0$ and $\varepsilon_{k\left[  X\right]  }\left(  X\right)
=0$). As a consequence, we obtain that $\operatorname*{Log}\left(
F\circ\operatorname*{ev}\nolimits_{H,\ v}\right)  $ is a well-defined element
of $\mathfrak{g}\left(  k\left[  X\right]  ,A\right)  $ and satisfies
\begin{equation}
\left(  \operatorname*{Log}\left(  F\circ\operatorname*{ev}\nolimits_{H,\ v}%
\right)  \right)  \left(  X^{n-1}\cdot X\right)  =0. \label{1202.pf.Xn-1X}%
\end{equation}


Recall that $\operatorname*{ev}\nolimits_{H,\ v}:k\left[  X\right]
\rightarrow H$ is a $k$-algebra homomorphism, so that $\operatorname*{ev}%
\nolimits_{H,\ v}\left(  1_{k\left[  X\right]  }\right)  =1_{H}$. Hence,
Proposition~\ref{Proposition1101} \textbf{(e)} (applied to $k\left[  X\right]
$, $H$ and $\operatorname*{ev}\nolimits_{H,\ v}$ instead of $D$, $C$ and
$\varphi$) yields that $F\circ\operatorname*{ev}\nolimits_{H,\ v}\in G\left(
k\left[  X\right]  ,A\right)  $ and that $\operatorname*{Log}\left(
F\circ\operatorname*{ev}\nolimits_{H,\ v}\right)  =\left(  \operatorname*{Log}%
F\right)  \circ\operatorname*{ev}\nolimits_{H,\ v}$. Thus,%
\begin{align*}
&  \left(  \underbrace{\operatorname*{Log}\left(  F\circ\operatorname*{ev}%
\nolimits_{H,\ v}\right)  }_{=\left(  \operatorname*{Log}F\right)
\circ\operatorname*{ev}\nolimits_{H,\ v}}\right)  \left(  \underbrace{X^{n-1}%
\cdot X}_{=X^{\left(  n-1\right)  +1}=X^{n}}\right) \\
&  =\left(  \left(  \operatorname*{Log}F\right)  \circ\operatorname*{ev}%
\nolimits_{H,\ v}\right)  \left(  X^{n}\right) \\
&  =\left(  \operatorname*{Log}F\right)  \left(
\underbrace{\operatorname*{ev}\nolimits_{H,\ v}\left(  X^{n}\right)
}_{\substack{=\left(  \operatorname*{ev}\nolimits_{H,\ v}\left(  X\right)
\right)  ^{n}\\\text{(since }\operatorname*{ev}\nolimits_{H,\ v}\text{ is a
}k\text{-algebra homomorphism)}}}\right) \\
&  =\left(  \operatorname*{Log}F\right)  \left(  \left(
\underbrace{\operatorname*{ev}\nolimits_{H,\ v}\left(  X\right)
}_{\substack{=v\\\text{(by (\ref{1205}), applied to }H\text{ and
}v\\\text{instead of }A\text{ and }a\text{)}}}\right)  ^{n}\right)  =\left(
\operatorname*{Log}F\right)  \left(  v^{n}\right)  .
\end{align*}
Comparing this with (\ref{1202.pf.Xn-1X}), we obtain $\left(
\operatorname*{Log}F\right)  \left(  v^{n}\right)  =0$. This completes the
proof of Theorem~\ref{Theorem1202}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem1201}.]We know that $\operatorname*{id}%
:H\rightarrow H$ is a $k$-algebra homomorphism. Hence, we can apply
Theorem~\ref{Theorem1202} to $A=H$ and $F=\operatorname*{id}$. Thus, we
conclude that $\operatorname*{id}\in G\left(  H,H\right)  $ and that $\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  v^{n}\right)  =0$. This
proves Theorem~\ref{Theorem1201}.
\end{proof}

Using Theorem~\ref{Theorem1202} and Theorem~\ref{Theorem1141}, we can prove a
``symmetrized-product'' version of Theorem~\ref{Theorem1202}:

\begin{theorem}
\label{Theorem1215} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $A$ be a $k$-algebra. Let
$F:H\rightarrow A$ be a $k$-algebra homomorphism. Let $n\in\mathbb{N}$ be such
that $n>1$. Let $v_{1}$, $v_{2}$, $\ldots$, $v_{n}$ be $n$ elements of
$\operatorname*{Prim}H$. Then, $F\in G\left(  H,A\right)  $ and%
\[
\left(  \operatorname*{Log}F\right)  \left(  \sum_{\sigma\in S_{n}}%
v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\right)  =0.
\]

\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1215}.]We have $F\in G\left(  H,A\right)  $
(this can be proven just as in the proof of Theorem~\ref{Theorem1202}).

We have $v_{s}\in\operatorname*{Prim}H$ for every $s\in\left\{  1,2,\ldots
,n\right\}  $. For every $S\in\mathcal{P}\left(  \left\{  1,2,\ldots
,n\right\}  \right)  $, we have
\[
\sum_{s\in S}\underbrace{v_{s}}_{\in\operatorname*{Prim}H}\in\sum_{s\in
S}\operatorname*{Prim}H\subseteq\operatorname*{Prim}H
\]
(since $\operatorname*{Prim}H$ is a $k$-vector space). Hence, for every
$S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)  $, we have
\begin{equation}
\left(  \operatorname*{Log}F\right)  \left(  \left(  \sum_{s\in S}%
v_{s}\right)  ^{n}\right)  =0 \label{1215.pf.1}%
\end{equation}
(by Theorem~\ref{Theorem1202}, applied to $v=\sum_{s\in S}v_{s}$). Now,
Theorem~\ref{Theorem1141} (applied to $H$ instead of $A$) yields%
\[
\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }=\sum_{S\in\mathcal{P}\left(  \left\{
1,2,\ldots,n\right\}  \right)  }\left(  -1\right)  ^{n-\left\vert S\right\vert
}\left(  \sum_{s\in S}v_{s}\right)  ^{n}.
\]
Applying $\operatorname*{Log}F$ to this equality, we obtain%
\begin{align*}
&  \left(  \operatorname*{Log}F\right)  \left(  \sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\right) \\
&  =\left(  \operatorname*{Log}F\right)  \left(  \sum_{S\in\mathcal{P}\left(
\left\{  1,2,\ldots,n\right\}  \right)  }\left(  -1\right)  ^{n-\left\vert
S\right\vert }\left(  \sum_{s\in S}v_{s}\right)  ^{n}\right) \\
&  =\sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
}\left(  -1\right)  ^{n-\left\vert S\right\vert }\underbrace{\left(
\operatorname*{Log}F\right)  \left(  \left(  \sum_{s\in S}v_{s}\right)
^{n}\right)  }_{\substack{=0\\\text{(by (\ref{1215.pf.1}))}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{Log}F\text{ is
}k\text{-linear}\right) \\
&  =\sum_{S\in\mathcal{P}\left(  \left\{  1,2,\ldots,n\right\}  \right)
}\left(  -1\right)  ^{n-\left\vert S\right\vert }0=0.
\end{align*}
This proves Theorem~\ref{Theorem1215}.
\end{proof}

Of course, we can specialize Theorem~\ref{Theorem1215} to a
``symmetrized-product'' version of Theorem~\ref{Theorem1201}:

\begin{theorem}
\label{Theorem1216} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $n\in\mathbb{N}$ be such that
$n>1$. Let $v_{1}$, $v_{2}$, $\ldots$, $v_{n}$ be $n$ elements of
$\operatorname*{Prim}H$. Then,%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(  \sum_{\sigma\in
S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots
v_{\sigma\left(  n\right)  }\right)  =0.
\]

\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1216}.]We know that $\operatorname*{id}%
:H\rightarrow H$ is a $k$-algebra homomorphism. Hence, we can apply
Theorem~\ref{Theorem1215} to $A=H$ and $F=\operatorname*{id}$. Thus, we
conclude that $\operatorname*{id}\in G\left(  H,H\right)  $ and that $\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  \sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\right)  =0$. This proves Theorem~\ref{Theorem1216}.
\end{proof}

\section{\label{S31}Finishing the proof of Cartier-Milnor-Moore}

We are now going to prepare for finishing the proof of
Theorem~\ref{Theorem271}. The final result we need is the following universal
property of the universal enveloping algebra as a Hopf algebra. Before we
state it, let us recall the universal property of the universal enveloping
algebra as an algebra:

\begin{proposition}
\label{Proposition1302} Let $k$ be a field. Let $\mathfrak{g}$ be a Lie
algebra. Consider the universal enveloping algebra $U\left(  \mathfrak{g}%
\right)  $ of $\mathfrak{g}$.

Let $A$ be any $k$-algebra. Consider $A$ as a Lie algebra under the commutator
of the multiplication.

Let $\iota_{\mathfrak{g}}:\mathfrak{g}\rightarrow U\left(  \mathfrak{g}%
\right)  $ be the canonical map from the Lie algebra $\mathfrak{g}$ into its
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $.

Let $f:\mathfrak{g}\rightarrow A$ be a homomorphism of Lie algebras. Then,
there exists a unique $k$-algebra homomorphism $F:U\left(  \mathfrak{g}%
\right)  \rightarrow A$ satisfying $F\circ\iota_{\mathfrak{g}}=f$.
\end{proposition}

\begin{definition}
\label{Definition1303} In the situation of Proposition~\ref{Proposition1302},
the unique $k$-algebra homomorphism $F:U\left(  \mathfrak{g}\right)
\rightarrow A$ satisfying $F\circ\iota_{\mathfrak{g}}=f$ will be denoted by
$\operatorname*{Ulift}f$.
\end{definition}

\begin{proposition}
\label{Proposition1304} Let $k$ be a field. Let $\mathfrak{g}$ be a Lie
algebra. Consider the universal enveloping algebra $U\left(  \mathfrak{g}%
\right)  $ of $\mathfrak{g}$ equipped with its canonical Hopf algebra structure.

Let $H$ be any $k$-bialgebra. Consider $H$ as a Lie algebra under the
commutator of the multiplication.

Let $f:\mathfrak{g}\rightarrow H$ be a homomorphism of Lie algebras such that
$f\left(  \mathfrak{g}\right)  \subseteq\operatorname*{Prim}H$. Then,
$\operatorname*{Ulift}f:U\left(  \mathfrak{g}\right)  \rightarrow H$ is a
$k$-bialgebra homomorphism.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1304}.]Let $\iota_{\mathfrak{g}%
}:\mathfrak{g}\rightarrow U\left(  \mathfrak{g}\right)  $ be the canonical map
from the Lie algebra $\mathfrak{g}$ into its universal enveloping algebra
$U\left(  \mathfrak{g}\right)  $.

By the definition of $\operatorname*{Ulift}f$, we know that
$\operatorname*{Ulift}f$ is the unique $k$-algebra homomorphism $F:U\left(
\mathfrak{g}\right)  \rightarrow H$ satisfying $F\circ\iota_{\mathfrak{g}}=f$.
Hence, $\operatorname*{Ulift}f$ is a $k$-algebra homomorphism $U\left(
\mathfrak{g}\right)  \rightarrow H$ and satisfies $\left(
\operatorname*{Ulift}f\right)  \circ\iota_{\mathfrak{g}}=f$. Since
$\operatorname*{Ulift}f$ is a $k$-algebra homomorphism, it satisfies $\left(
\operatorname*{Ulift}f\right)  \left(  1_{U\left(  \mathfrak{g}\right)
}\right)  =1_{H}$.

Clearly, $\left(  \operatorname*{Ulift}f\right)  \otimes\left(
\operatorname*{Ulift}f\right)  $ is a $k$-algebra homomorphism (since
$\operatorname*{Ulift}f$ is a $k$-algebra homomorphism, and since the tensor
product of two $k$-algebra homomorphisms is a $k$-algebra homomorphism).

By the axioms of a $k$-bialgebra, we know that $\Delta_{B}$ and $\varepsilon
_{B}$ are $k$-algebra homomorphisms for every $k$-bialgebra $B$. Applying this
to $B=U\left(  \mathfrak{g}\right)  $, we see that $\Delta_{U\left(
\mathfrak{g}\right)  }$ and $\varepsilon_{U\left(  \mathfrak{g}\right)  }$ are
$k$-algebra homomorphisms.

Since $\Delta_{U\left(  \mathfrak{g}\right)  }$ and $\left(
\operatorname*{Ulift}f\right)  \otimes\left(  \operatorname*{Ulift}f\right)  $
are $k$-algebra homomorphisms, we conclude that $\left(  \left(
\operatorname*{Ulift}f\right)  \otimes\left(  \operatorname*{Ulift}f\right)
\right)  \circ\Delta_{U\left(  \mathfrak{g}\right)  }$ is a $k$-algebra
homomorphism (because the composition of two $k$-algebra homomorphisms is a
$k$-algebra homomorphism).

On the other hand, by the axioms of a $k$-bialgebra, we know that $\Delta_{B}$
and $\varepsilon_{B}$ are $k$-algebra homomorphisms for every $k$-bialgebra
$B$. Applying this to $B=H$, we see that $\Delta_{H}$ and $\varepsilon_{H}$
are $k$-algebra homomorphisms.

Since $\operatorname*{Ulift}f$ and $\Delta_{H}$ are $k$-algebra homomorphisms,
we conclude that $\Delta_{H}\circ\left(  \operatorname*{Ulift}f\right)  $ is a
$k$-algebra homomorphism (because the composition of two $k$-algebra
homomorphisms is a $k$-algebra homomorphism).

Consider $H\otimes H$ as a Lie algebra under the commutator of the
multiplication. Then, $\Delta_{H}$ is a Lie algebra homomorphism (since
$\Delta_{H}$ is a $k$-algebra homomorphism). Since $\Delta_{H}$ and $f$ are
Lie algebra homomorphisms, the composition $\Delta_{H}\circ f$ is a Lie
algebra homomorphism (since the composition of two Lie algebra homomorphisms
is a Lie algebra homomorphism). Hence, Proposition~\ref{Proposition1302}
(applied to $H\otimes H$ and $\Delta_{H}\circ f$ instead of $A$ and $f$)
yields that there exists a unique $k$-algebra homomorphism $F:U\left(
\mathfrak{g}\right)  \rightarrow H\otimes H$ satisfying $F\circ\iota
_{\mathfrak{g}}=\Delta_{H}\circ f$. In particular, there exists \textbf{at
most one} such homomorphism. In other words,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{any two }k\text{-algebra homomorphisms }F:U\left(  \mathfrak{g}\right)
\rightarrow H\otimes H\\
\text{satisfying }F\circ\iota_{\mathfrak{g}}=\Delta_{H}\circ f\text{ must be
equal}%
\end{array}
\right)  . \label{1104.pf.1}%
\end{equation}


Let $x\in\mathfrak{g}$. Then, $\Delta_{U\left(  \mathfrak{g}\right)  }\left(
\iota_{\mathfrak{g}}\left(  x\right)  \right)  =\iota_{\mathfrak{g}}\left(
x\right)  \otimes1_{U\left(  \mathfrak{g}\right)  }+1_{U\left(  \mathfrak{g}%
\right)  }\otimes\iota_{\mathfrak{g}}\left(  x\right)  $ due to the definition
of the comultiplication on the universal enveloping algebra $U\left(
\mathfrak{g}\right)  $. Now,%
\begin{align*}
&  \left(  \left(  \left(  \operatorname*{Ulift}f\right)  \otimes\left(
\operatorname*{Ulift}f\right)  \right)  \circ\Delta_{U\left(  \mathfrak{g}%
\right)  }\circ\iota_{\mathfrak{g}}\right)  \left(  x\right) \\
&  =\left(  \left(  \operatorname*{Ulift}f\right)  \otimes\left(
\operatorname*{Ulift}f\right)  \right)  \underbrace{\left(  \Delta_{U\left(
\mathfrak{g}\right)  }\left(  \iota_{\mathfrak{g}}\left(  x\right)  \right)
\right)  }_{=\iota_{\mathfrak{g}}\left(  x\right)  \otimes1_{U\left(
\mathfrak{g}\right)  }+1_{U\left(  \mathfrak{g}\right)  }\otimes
\iota_{\mathfrak{g}}\left(  x\right)  }\\
&  =\left(  \left(  \operatorname*{Ulift}f\right)  \otimes\left(
\operatorname*{Ulift}f\right)  \right)  \left(  \iota_{\mathfrak{g}}\left(
x\right)  \otimes1_{U\left(  \mathfrak{g}\right)  }+1_{U\left(  \mathfrak{g}%
\right)  }\otimes\iota_{\mathfrak{g}}\left(  x\right)  \right) \\
&  =\underbrace{\left(  \operatorname*{Ulift}f\right)  \left(  \iota
_{\mathfrak{g}}\left(  x\right)  \right)  }_{\substack{=\left(  \left(
\operatorname*{Ulift}f\right)  \circ\iota_{g}\right)  \left(  x\right)
=f\left(  x\right)  \\\text{(since }\left(  \operatorname*{Ulift}f\right)
\circ\iota_{g}=f\text{)}}}\otimes\underbrace{\left(  \operatorname*{Ulift}%
f\right)  \left(  1_{U\left(  \mathfrak{g}\right)  }\right)  }_{=1_{H}%
}+\underbrace{\left(  \operatorname*{Ulift}f\right)  \left(  1_{U\left(
\mathfrak{g}\right)  }\right)  }_{=1_{H}}\otimes\underbrace{\left(
\operatorname*{Ulift}f\right)  \left(  \iota_{\mathfrak{g}}\left(  x\right)
\right)  }_{\substack{=\left(  \left(  \operatorname*{Ulift}f\right)
\circ\iota_{g}\right)  \left(  x\right)  =f\left(  x\right)  \\\text{(since
}\left(  \operatorname*{Ulift}f\right)  \circ\iota_{g}=f\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
\operatorname*{Ulift}f\right)  \otimes\left(  \operatorname*{Ulift}f\right)
\right) \\
&  =f\left(  x\right)  \otimes1_{H}+1_{H}\otimes f\left(  x\right)
=\Delta_{H}\left(  f\left(  x\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }f\left(  \underbrace{x}_{\in\mathfrak{g}}\right)  \in f\left(
\mathfrak{g}\right)  \subseteq\operatorname*{Prim}H\text{, so that }f\left(
x\right)  \text{ is primitive, and thus}\\
\Delta_{H}\left(  f\left(  x\right)  \right)  =f\left(  x\right)  \otimes
1_{H}+1_{H}\otimes f\left(  x\right)
\end{array}
\right) \\
&  =\left(  \Delta_{H}\circ f\right)  \left(  x\right)  .
\end{align*}


Now, forget that we fixed $x$. We thus have shown that
\[
\left(  \left(  \left(  \operatorname*{Ulift}f\right)  \otimes\left(
\operatorname*{Ulift}f\right)  \right)  \circ\Delta_{U\left(  \mathfrak{g}%
\right)  }\circ\iota_{\mathfrak{g}}\right)  \left(  x\right)  =\left(
\Delta_{H}\circ f\right)  \left(  x\right)
\]
for every $x\in\mathfrak{g}$. In other words, $\left(  \left(
\operatorname*{Ulift}f\right)  \otimes\left(  \operatorname*{Ulift}f\right)
\right)  \circ\Delta_{U\left(  \mathfrak{g}\right)  }\circ\iota_{\mathfrak{g}%
}=\Delta_{H}\circ f$. Hence, $\left(  \left(  \operatorname*{Ulift}f\right)
\otimes\left(  \operatorname*{Ulift}f\right)  \right)  \circ\Delta_{U\left(
\mathfrak{g}\right)  }$ is a $k$-algebra homomorphism $F:U\left(
\mathfrak{g}\right)  \rightarrow H\otimes H$ satisfying $F\circ\iota
_{\mathfrak{g}}=\Delta_{H}\circ f$ (since we know that $\left(  \left(
\operatorname*{Ulift}f\right)  \otimes\left(  \operatorname*{Ulift}f\right)
\right)  \circ\Delta_{U\left(  \mathfrak{g}\right)  }$ is a $k$-algebra homomorphism).

On the other hand, $\Delta_{H}\circ\left(  \operatorname*{Ulift}f\right)  $ is
a $k$-algebra homomorphism and satisfies $\Delta_{H}\circ\underbrace{\left(
\operatorname*{Ulift}f\right)  \circ\iota_{\mathfrak{g}}}_{=f}=\Delta_{H}\circ
f$. Hence, $\Delta_{H}\circ\left(  \operatorname*{Ulift}f\right)  $ is a
$k$-algebra homomorphism $F:U\left(  \mathfrak{g}\right)  \rightarrow H\otimes
H$ satisfying $F\circ\iota_{\mathfrak{g}}=\Delta_{H}\circ f$.

Now, we know that $\left(  \left(  \operatorname*{Ulift}f\right)
\otimes\left(  \operatorname*{Ulift}f\right)  \right)  \circ\Delta_{U\left(
\mathfrak{g}\right)  }$ and $\Delta_{H}\circ\left(  \operatorname*{Ulift}%
f\right)  $ are two $k$-algebra homomorphisms $F:U\left(  \mathfrak{g}\right)
\rightarrow H\otimes H$ satisfying $F\circ\iota_{\mathfrak{g}}=\Delta_{H}\circ
f$. Hence, $\left(  \left(  \operatorname*{Ulift}f\right)  \otimes\left(
\operatorname*{Ulift}f\right)  \right)  \circ\Delta_{U\left(  \mathfrak{g}%
\right)  }$ and $\Delta_{H}\circ\left(  \operatorname*{Ulift}f\right)  $ must
be equal (because (\ref{1104.pf.1}) says that any two $k$-algebra
homomorphisms $F:U\left(  \mathfrak{g}\right)  \rightarrow H\otimes H$
satisfying $F\circ\iota_{\mathfrak{g}}=\Delta_{H}\circ f$ must be equal). In
other words,
\begin{equation}
\left(  \left(  \operatorname*{Ulift}f\right)  \otimes\left(
\operatorname*{Ulift}f\right)  \right)  \circ\Delta_{U\left(  \mathfrak{g}%
\right)  }=\Delta_{H}\circ\left(  \operatorname*{Ulift}f\right)  .
\label{1104.pf.3}%
\end{equation}


On the other hand, consider $k$ endowed with the trivial Lie algebra structure
(which is also given by the commutator of the multiplication on $k$). Then,
$0:U\left(  \mathfrak{g}\right)  \rightarrow k$ is a Lie algebra homomorphism
(since the zero map between two Lie algebras always is a Lie algebra
homomorphism). Hence, Proposition~\ref{Proposition1302} (applied to $k$ and
$0$ instead of $A$ and $f$) yields that there exists a unique $k$-algebra
homomorphism $F:U\left(  \mathfrak{g}\right)  \rightarrow k$ satisfying
$F\circ\iota_{\mathfrak{g}}=0$. In particular, there exists \textbf{at most
one} such homomorphism. In other words,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{any two }k\text{-algebra homomorphisms }F:U\left(  \mathfrak{g}\right)
\rightarrow k\\
\text{satisfying }F\circ\iota_{\mathfrak{g}}=0\text{ must be equal}%
\end{array}
\right)  . \label{1104.pf.5}%
\end{equation}


Now, every $x\in\mathfrak{g}$ satisfies $\left(  \varepsilon_{U\left(
\mathfrak{g}\right)  }\circ\iota_{\mathfrak{g}}\right)  \left(  x\right)
=\varepsilon_{U\left(  \mathfrak{g}\right)  }\left(  \iota_{\mathfrak{g}%
}\left(  x\right)  \right)  =0$ (by the definition of the counit on the
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $). Thus, every
$x\in\mathfrak{g}$ satisfies $\left(  \varepsilon_{U\left(  \mathfrak{g}%
\right)  }\circ\iota_{\mathfrak{g}}\right)  \left(  x\right)  =0=0\left(
x\right)  $. In other words, $\varepsilon_{U\left(  \mathfrak{g}\right)
}\circ\iota_{\mathfrak{g}}=0$. Since $\varepsilon_{U\left(  \mathfrak{g}%
\right)  }$ is a $k$-algebra homomorphism, this yields that $\varepsilon
_{U\left(  \mathfrak{g}\right)  }$ is a $k$-algebra homomorphism $F:U\left(
\mathfrak{g}\right)  \rightarrow k$ satisfying $F\circ\iota_{\mathfrak{g}}=0$.

On the other hand, let $x\in\mathfrak{g}$. Then, $f\left(  \underbrace{x}%
_{\in\mathfrak{g}}\right)  \in f\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Prim}H$, so that $f\left(  x\right)  $ is a primitive element
of $H$. Since $H$ is a unital coalgebra (by Proposition~\ref{Proposition7c}),
we can apply Remark~\ref{Remark21} to $f\left(  x\right)  $ instead of $x$. As
a result, we obtain $\varepsilon\left(  f\left(  x\right)  \right)  =0$. In
other words, $\varepsilon_{H}\left(  f\left(  x\right)  \right)  =0$. Now,%
\[
\left(  \varepsilon_{H}\circ\underbrace{\left(  \operatorname*{Ulift}f\right)
\circ\iota_{\mathfrak{g}}}_{=f}\right)  \left(  x\right)  =\left(
\varepsilon_{H}\circ f\right)  \left(  x\right)  =\varepsilon_{H}\left(
f\left(  x\right)  \right)  =0=0\left(  x\right)  .
\]


Now, forget that we fixed $x$. We thus have proven that $\left(
\varepsilon_{H}\circ\left(  \operatorname*{Ulift}f\right)  \circ
\iota_{\mathfrak{g}}\right)  \left(  x\right)  =0\left(  x\right)  $ for every
$x\in\mathfrak{g}$. In other words, $\varepsilon_{H}\circ\left(
\operatorname*{Ulift}f\right)  \circ\iota_{\mathfrak{g}}=0$. Since
$\varepsilon_{H}$ and $\operatorname*{Ulift}f$ are $k$-algebra homomorphisms,
their composition $\varepsilon_{H}\circ\left(  \operatorname*{Ulift}f\right)
$ is a $k$-algebra homomorphism (since the composition of any two $k$-algebra
homomorphisms is a $k$-algebra homomorphism). Since $\varepsilon_{H}%
\circ\left(  \operatorname*{Ulift}f\right)  \circ\iota_{\mathfrak{g}}=0$, this
yields that $\varepsilon_{H}\circ\left(  \operatorname*{Ulift}f\right)  $ is a
$k$-algebra homomorphism $F:U\left(  \mathfrak{g}\right)  \rightarrow k$
satisfying $F\circ\iota_{\mathfrak{g}}=0$.

Now, we know that $\varepsilon_{U\left(  \mathfrak{g}\right)  }$ and
$\varepsilon_{H}\circ\left(  \operatorname*{Ulift}f\right)  $ are two
$k$-algebra homomorphisms $F:U\left(  \mathfrak{g}\right)  \rightarrow k$
satisfying $F\circ\iota_{\mathfrak{g}}=0$. Hence, $\varepsilon_{U\left(
\mathfrak{g}\right)  }$ and $\varepsilon_{H}\circ\left(  \operatorname*{Ulift}%
f\right)  $ must be equal (since (\ref{1104.pf.5}) says that any two
$k$-algebra homomorphisms $F:U\left(  \mathfrak{g}\right)  \rightarrow k$
satisfying $F\circ\iota_{\mathfrak{g}}=0$ must be equal). In other words,%
\[
\varepsilon_{U\left(  \mathfrak{g}\right)  }=\varepsilon_{H}\circ\left(
\operatorname*{Ulift}f\right)  .
\]
Combined with (\ref{1104.pf.3}), this yields that $\operatorname*{Ulift}f$ is
a $k$-coalgebra homomorphism. Combining this with the fact that
$\operatorname*{Ulift}f$ is a $k$-algebra homomorphism, we conclude that
$\operatorname*{Ulift}f$ is a $k$-bialgebra homomorphism. This proves
Proposition~\ref{Proposition1304}.
\end{proof}

We are getting closer to the proof of the Cartier-Milnor-Moore theorem. Let us
recall the definition of the tensor Hopf algebra:

\begin{theorem}
\label{Theorem1305} Let $k$ be a field. Let $V$ be a $k$-vector space. Then,
there exists a unique $k$-bialgebra structure on the tensor algebra $\otimes
V$ whose comultiplication $\Delta_{\otimes V}$ and counity $\varepsilon
_{\otimes V}$ satisfy%
\[
\left(
\begin{array}
[c]{r}%
\Delta_{\otimes V}\left(  v\right)  =v\otimes1_{\otimes V}+1_{\otimes
V}\otimes v\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \varepsilon
_{\otimes V}\left(  v\right)  =0\\
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V
\end{array}
\right)
\]
(where we consider $V$ as a subspace of $\otimes V$).
\end{theorem}

\begin{definition}
\label{Definition1306} Let $k$ be a field. Let $V$ be a $k$-vector space.
Then, when we speak of the \textquotedblleft$k$-bialgebra $\otimes
V$\textquotedblright, we mean the tensor algebra $\otimes V$ endowed with the
unique $k$-bialgebra structure whose comultiplication $\Delta_{\otimes V}$ and
counity $\varepsilon_{\otimes V}$ satisfy%
\[
\left(
\begin{array}
[c]{r}%
\Delta_{\otimes V}\left(  v\right)  =v\otimes1_{\otimes V}+1_{\otimes
V}\otimes v\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \varepsilon
_{\otimes V}\left(  v\right)  =0\\
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V
\end{array}
\right)
\]
(where we consider $V$ as a subspace of $\otimes V$). (This is well-defined
due to Theorem~\ref{Theorem1305}.)
\end{definition}

\begin{theorem}
\label{Theorem1307} Let $k$ be a field. Let $V$ be a $k$-vector space.
Consider $V$ as a subspace of $\otimes V$.

\textbf{(a)} The $k$-bialgebra $\otimes V$ is a Hopf algebra, whose antipode
$S_{\otimes V}$ satisfies $S_{\otimes V}\left(  v\right)  =-v$ for every $v\in
V$.

\textbf{(b)} The $k$-bialgebra $\otimes V$ is cocommutative.

\textbf{(c)} The standard grading on the $k$-vector space $\otimes V$ (the one
where $\left(  \otimes V\right)  _{n}=V^{\otimes n}$ for every $n\in
\mathbb{N}$) makes the $k$-bialgebra $\otimes V$ into a graded $k$-bialgebra.

\textbf{(d)} This graded $k$-bialgebra $\otimes V$ is connected.
\end{theorem}

We are now going to state similar properties of the universal enveloping Hopf algebra:

\begin{theorem}
\label{Theorem1311} Let $k$ be a field. Let $\mathfrak{g}$ be a Lie algebra.
Let $U\left(  \mathfrak{g}\right)  $ denote the universal enveloping algebra
of $\mathfrak{g}$.

Let $\iota_{\mathfrak{g}}:\mathfrak{g}\rightarrow U\left(  \mathfrak{g}%
\right)  $ be the canonical map from the Lie algebra $\mathfrak{g}$ into its
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $.

Then, there exists a unique $k$-bialgebra structure on the $k$-algebra
$U\left(  \mathfrak{g}\right)  $ whose comultiplication $\Delta_{U\left(
\mathfrak{g}\right)  }$ and counity $\varepsilon_{U\left(  \mathfrak{g}%
\right)  }$ satisfy%
\[
\left(
\begin{array}
[c]{r}%
\Delta_{U\left(  \mathfrak{g}\right)  }\left(  v\right)  =v\otimes1_{U\left(
\mathfrak{g}\right)  }+1_{U\left(  \mathfrak{g}\right)  }\otimes
v\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \varepsilon_{U\left(
\mathfrak{g}\right)  }\left(  v\right)  =0\\
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)
\end{array}
\right)  .
\]

\end{theorem}

\begin{definition}
\label{Definition1312} Let $k$ be a field. Let $\mathfrak{g}$ be a Lie
algebra. Let $U\left(  \mathfrak{g}\right)  $ denote the universal enveloping
algebra of $\mathfrak{g}$.

Let $\iota_{\mathfrak{g}}:\mathfrak{g}\rightarrow U\left(  \mathfrak{g}%
\right)  $ be the canonical map from the Lie algebra $\mathfrak{g}$ into its
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $.

Then, when we speak of the \textquotedblleft$k$-bialgebra $U\left(
\mathfrak{g}\right)  $\textquotedblright, we mean the universal enveloping
algebra $U\left(  \mathfrak{g}\right)  $ endowed with the unique $k$-bialgebra
structure whose comultiplication $\Delta_{U\left(  \mathfrak{g}\right)  }$ and
counity $\varepsilon_{U\left(  \mathfrak{g}\right)  }$ satisfy%
\begin{equation}
\left(
\begin{array}
[c]{r}%
\Delta_{U\left(  \mathfrak{g}\right)  }\left(  v\right)  =v\otimes1_{U\left(
\mathfrak{g}\right)  }+1_{U\left(  \mathfrak{g}\right)  }\otimes
v\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \varepsilon_{U\left(
\mathfrak{g}\right)  }\left(  v\right)  =0\\
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)
\end{array}
\right)  . \label{1312}%
\end{equation}
(This is well-defined due to Theorem~\ref{Theorem1311}.)
\end{definition}

\begin{theorem}
\label{Theorem1313} Let $k$ be a field. Let $\mathfrak{g}$ be a Lie algebra.
Let $U\left(  \mathfrak{g}\right)  $ denote the universal enveloping algebra
of $\mathfrak{g}$.

Let $\iota_{\mathfrak{g}}:\mathfrak{g}\rightarrow U\left(  \mathfrak{g}%
\right)  $ be the canonical map from the Lie algebra $\mathfrak{g}$ into its
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $.

\textbf{(a)} We have $U\left(  \mathfrak{g}\right)  =\sum_{n\in\mathbb{N}%
}\left(  \iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  \right)  ^{n}$.

\textbf{(b)} The family $\left(  \sum_{n=0}^{i}\left(  \iota_{\mathfrak{g}%
}\left(  \mathfrak{g}\right)  \right)  ^{n}\right)  _{i\in\mathbb{N}}$ is a
filtration of the $k$-vector space $U\left(  \mathfrak{g}\right)  $.

\textbf{(c)} The $k$-bialgebra $U\left(  \mathfrak{g}\right)  $ endowed with
the filtration $\left(  \sum_{n=0}^{i}\left(  \iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  \right)  ^{n}\right)  _{i\in\mathbb{N}}$ is a connected
filtered $k$-bialgebra.

\textbf{(d)} We have $\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)
\subseteq\operatorname*{Prim}\left(  U\left(  \mathfrak{g}\right)  \right)  $.

\textbf{(e)} The $k$-bialgebra $U\left(  \mathfrak{g}\right)  $ is cocommutative.

\textbf{(f)} The $k$-bialgebra $U\left(  \mathfrak{g}\right)  $ is a Hopf algebra.
\end{theorem}

A simple way to prove Theorem~\ref{Theorem1313} is to deduce it from analogous
properties of the tensor algebra $\otimes\mathfrak{g}$ (which are either part
of Theorem~\ref{Theorem1307} or trivial), because the $k$-bialgebra $U\left(
\mathfrak{g}\right)  $ is a quotient of $\otimes\mathfrak{g}$ (as a
$k$-bialgebra, not just as a $k$-algebra). We will use a different strategy.
We will only prove the parts of Theorem~\ref{Theorem1313} that we need, though.

Before we prove Theorem~\ref{Theorem1313} proper, here is an auxiliary result:

\begin{proposition}
\label{Proposition1314} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let
$V$ be a $k$-vector subspace of $H$ such that $H=\sum_{n\in\mathbb{N}}V^{n}$
and $\Delta_{H}\left(  V\right)  \subseteq V\otimes\left(  k\cdot1_{H}\right)
+\left(  k\cdot1_{H}\right)  \otimes V$. Then:

\textbf{(a)} The family $\left(  \sum_{n=0}^{i}V^{n}\right)  _{i\in\mathbb{N}%
}$ is a filtration of the $k$-vector space $H$.

\textbf{(b)} The $k$-bialgebra $H$ endowed with the filtration $\left(
\sum_{n=0}^{i}V^{n}\right)  _{i\in\mathbb{N}}$ is a connected filtered $k$-bialgebra.

\textbf{(c)} If $V\subseteq\operatorname*{Prim}H$, then the $k$-bialgebra $H$
is cocommutative.
\end{proposition}

Before we prove this, let us collect some auxiliary facts. The following lemma
is a basic fact about algebras:

\begin{lemma}
\label{Lemma1315} Let $k$ be a field. Let $U$ and $V$ be two $k$-algebras. Let
$A$ and $C$ be two $k$-vector subspaces of $U$. Let $B$ and $D$ be two
$k$-vector subspaces of $V$. Then,%
\[
\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right)  =\left(  AC\right)
\otimes\left(  BD\right)
\]
(as $k$-vector subspaces of the $k$-algebra $U\otimes V$).
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1315}.]\textbf{a)} We have
\begin{equation}
\alpha\gamma\in\left(  AC\right)  \otimes\left(  BD\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\alpha\in A\otimes B\text{ and }\gamma\in
C\otimes D. \label{1315.pf.a}%
\end{equation}


\textit{Proof of (\ref{1315.pf.a}):} Let $\alpha\in A\otimes B$ and $\gamma\in
C\otimes D$.

We know that every tensor in a tensor product is a $k$-linear combination of
pure tensors. Applying this to the tensor $\alpha$ in the tensor product
$A\otimes B$, we conclude that the tensor $\alpha$ is a $k$-linear combination
of pure tensors. In other words, there exists a $p\in\mathbb{N}$ and elements
$\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{p}$ of $k$ and elements $a_{1}$, $a_{2}%
$, $\ldots$, $a_{p}$ of $A$ and elements $b_{1}$, $b_{2}$, $\ldots$, $b_{p}$
of $B$ such that $\alpha=\sum_{i=1}^{p}\mu_{i}a_{i}\otimes b_{i}$. Consider
this $p$, these $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{p}$, these $a_{1}$,
$a_{2}$, $\ldots$, $a_{p}$ and these $b_{1}$, $b_{2}$, $\ldots$, $b_{p}$.

We know that every tensor in a tensor product is a $k$-linear combination of
pure tensors. Applying this to the tensor $\beta$ in the tensor product
$C\otimes D$, we conclude that the tensor $\gamma$ is a $k$-linear combination
of pure tensors. In other words, there exists a $q\in\mathbb{N}$ and elements
$\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu_{q}$ of $k$ and elements $c_{1}$, $c_{2}%
$, $\ldots$, $c_{q}$ of $C$ and elements $d_{1}$, $d_{2}$, $\ldots$, $d_{q}$
of $D$ such that $\gamma=\sum_{j=1}^{q}\nu_{j}c_{j}\otimes d_{j}$. Consider
this $q$, these $\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu_{q}$, these $c_{1}$,
$c_{2}$, $\ldots$, $c_{q}$ and these $d_{1}$, $d_{2}$, $\ldots$, $d_{q}$.

Multiplying the equality $\alpha=\sum_{i=1}^{p}\mu_{i}a_{i}\otimes b_{i}$ with
the equality $\gamma=\sum_{j=1}^{q}\nu_{j}c_{j}\otimes d_{j}$, we obtain%
\begin{align*}
\alpha\gamma &  =\left(  \sum_{i=1}^{p}\mu_{i}a_{i}\otimes b_{i}\right)
\left(  \sum_{j=1}^{q}\nu_{j}c_{j}\otimes d_{j}\right) \\
&  =\sum_{i=1}^{p}\mu_{i}\sum_{j=1}^{q}\nu_{j}\underbrace{\left(  a_{i}\otimes
b_{i}\right)  \left(  c_{j}\otimes d_{j}\right)  }_{=a_{i}c_{j}\otimes
b_{i}d_{j}}\\
&  =\sum_{i=1}^{p}\mu_{i}\sum_{j=1}^{q}\nu_{j}\underbrace{a_{i}}_{\in
A}\underbrace{c_{j}}_{\in C}\otimes\underbrace{b_{i}}_{\in B}\underbrace{d_{j}%
}_{\in D}\\
&  \in\sum_{i=1}^{p}\mu_{i}\sum_{j=1}^{q}\nu_{j}\left(  AC\right)
\otimes\left(  BD\right)  \subseteq\left(  AC\right)  \otimes\left(
BD\right)
\end{align*}
(since $\left(  AC\right)  \otimes\left(  BD\right)  $ is a $k$-vector space).
This proves (\ref{1315.pf.a}).

\textbf{b)} We have%
\begin{equation}
\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right)  \subseteq\left(
AC\right)  \otimes\left(  BD\right)  . \label{1315.pf.b}%
\end{equation}


\textit{Proof of (\ref{1315.pf.b}):} Let $\xi\in\left(  A\otimes B\right)
\cdot\left(  C\otimes D\right)  $. We know that $\left(  A\otimes B\right)
\cdot\left(  C\otimes D\right)  $ is the $k$-linear span of the set of all
products of the form $\alpha\gamma$ with $\alpha\in A\otimes B$ and $\gamma\in
C\otimes D$ (by the definition of $\left(  A\otimes B\right)  \cdot\left(
C\otimes D\right)  $). In other words,
\begin{align*}
&  \left(  A\otimes B\right)  \cdot\left(  C\otimes D\right) \\
&  =\left(  \text{the }k\text{-linear span of the set of all products
of}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  \text{the form }%
\alpha\gamma\text{ with }\alpha\in A\otimes B\text{ and }\gamma\in C\otimes
D\right) \\
&  =\left(  \text{the set of all }k\text{-linear combinations of products
of}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ }\left.  \text{the form
}\alpha\gamma\text{ with }\alpha\in A\otimes B\text{ and }\gamma\in C\otimes
D\right)  .
\end{align*}
Hence,
\begin{align*}
\xi &  \in\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right) \\
&  =\left(  \text{the set of all }k\text{-linear combinations of products
of}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ }\left.  \text{the form
}\alpha\gamma\text{ with }\alpha\in A\otimes B\text{ and }\gamma\in C\otimes
D\right)  .
\end{align*}
Thus, $\xi$ is a $k$-linear combination of products of the form $\alpha\gamma$
with $\alpha\in A\otimes B$ and $\gamma\in C\otimes D$. In other words, there
exists an $n\in\mathbb{N}$ and elements $\lambda_{1}$, $\lambda_{2}$, $\ldots
$, $\lambda_{n}$ of $k$ and elements $\alpha_{1}$, $\alpha_{2}$, $\ldots$,
$\alpha_{n}$ of $A\otimes B$ and elements $\gamma_{1}$, $\gamma_{2}$, $\ldots
$, $\gamma_{n}$ of $C\otimes D$ such that $\xi=\sum_{\ell=1}^{n}\lambda_{\ell
}\alpha_{\ell}\gamma_{\ell}$. Consider this $n$, these $\lambda_{1}$,
$\lambda_{2}$, $\ldots$, $\lambda_{n}$, these $\alpha_{1}$, $\alpha_{2}$,
$\ldots$, $\alpha_{n}$ and these $\gamma_{1}$, $\gamma_{2}$, $\ldots$,
$\gamma_{n}$. Then,%
\[
\xi=\sum_{\ell=1}^{n}\lambda_{\ell}\underbrace{\alpha_{\ell}\gamma_{\ell}%
}_{\substack{\in\left(  AC\right)  \otimes\left(  BD\right)  \\\text{(by
(\ref{1315.pf.a}), applied to }\alpha=\alpha_{\ell}\text{ and }\gamma
=\gamma_{\ell}\text{)}}}\in\sum_{\ell=1}^{n}\lambda_{\ell}\left(  AC\right)
\otimes\left(  BD\right)  \subseteq\left(  AC\right)  \otimes\left(
BD\right)
\]
(since $\left(  AC\right)  \otimes\left(  BD\right)  $ is a $k$-vector space).

Now, forget that we fixed $\xi$. We thus have proven that every $\xi\in\left(
A\otimes B\right)  \cdot\left(  C\otimes D\right)  $ satisfies $\xi\in\left(
AC\right)  \otimes\left(  BD\right)  $. In other words, $\left(  A\otimes
B\right)  \cdot\left(  C\otimes D\right)  \subseteq\left(  AC\right)
\otimes\left(  BD\right)  $. This proves (\ref{1315.pf.b}).

\textbf{c)} We have
\begin{equation}
x\otimes y\in\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in AC\text{ and }y\in BD.
\label{1315.pf.c}%
\end{equation}


\textit{Proof of (\ref{1315.pf.c}):} Let $x\in AB$ and $y\in CD$.

We know that $AC$ is the $k$-linear span of the set of all products of the
form $ac$ with $a\in A$ and $c\in C$ (by the definition of $AC$). In other
words,
\begin{align*}
&  AC\\
&  =\left(  \text{the }k\text{-linear span of the set of all products of the
form }ac\text{ with }a\in A\text{ and }c\in C\right) \\
&  =\left(  \text{the set of all }k\text{-linear combinations of products of
the form }ac\text{ with }a\in A\text{ and }c\in C\right)  .
\end{align*}
Hence,
\begin{align*}
x  &  \in AC\\
&  =\left(  \text{the set of all }k\text{-linear combinations of products of
the form }ac\text{ with }a\in A\text{ and }c\in C\right)  .
\end{align*}
Thus, $x$ is a $k$-linear combination of products of the form $ac$ with $a\in
A$ and $c\in C$. In other words, there exists an $n\in\mathbb{N}$ and elements
$\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$ of $k$ and elements $a_{1}$, $a_{2}%
$, $\ldots$, $a_{n}$ of $A$ and elements $c_{1}$, $c_{2}$, $\ldots$, $c_{n}$
of $C$ such that $x=\sum_{i=1}^{n}\mu_{i}a_{i}c_{i}$. Consider this $n$, these
$\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$, these $a_{1}$, $a_{2}$, $\ldots$,
$a_{n}$ and these $c_{1}$, $c_{2}$, $\ldots$, $c_{n}$.

We know that $BD$ is the $k$-linear span of the set of all products of the
form $bd$ with $b\in B$ and $d\in D$ (by the definition of $BD$). In other
words,
\begin{align*}
&  BD\\
&  =\left(  \text{the }k\text{-linear span of the set of all products of the
form }bd\text{ with }b\in B\text{ and }d\in D\right) \\
&  =\left(  \text{the set of all }k\text{-linear combinations of products of
the form }bd\text{ with }b\in B\text{ and }d\in D\right)  .
\end{align*}
Hence,
\begin{align*}
y  &  \in BD\\
&  =\left(  \text{the set of all }k\text{-linear combinations of products of
the form }bd\text{ with }b\in B\text{ and }d\in D\right)  .
\end{align*}
Thus, $y$ is a $k$-linear combination of products of the form $bd$ with $b\in
B$ and $d\in D$. In other words, there exists an $m\in\mathbb{N}$ and elements
$\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu_{m}$ of $k$ and elements $b_{1}$, $b_{2}%
$, $\ldots$, $b_{m}$ of $B$ and elements $d_{1}$, $d_{2}$, $\ldots$, $d_{m}$
of $D$ such that $y=\sum_{j=1}^{m}\nu_{j}b_{j}d_{j}$. Consider this $m$, these
$\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu_{m}$, these $b_{1}$, $b_{2}$, $\ldots$,
$b_{m}$ and these $d_{1}$, $d_{2}$, $\ldots$, $d_{m}$.

Taking the tensor product of the equalities $x=\sum_{i=1}^{n}\mu_{i}a_{i}%
c_{i}$ and $y=\sum_{j=1}^{m}\nu_{j}b_{j}d_{j}$, we obtain%
\begin{align*}
x\otimes y  &  =\left(  \sum_{i=1}^{n}\mu_{i}a_{i}c_{i}\right)  \otimes\left(
\sum_{j=1}^{m}\nu_{j}b_{j}d_{j}\right) \\
&  =\sum_{i=1}^{n}\mu_{i}\sum_{j=1}^{m}\nu_{j}\underbrace{\left(  a_{i}%
c_{i}\right)  \otimes\left(  b_{j}d_{j}\right)  }_{=\left(  a_{i}\otimes
b_{j}\right)  \cdot\left(  c_{i}\otimes d_{j}\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the tensor product is }%
k\text{-bilinear}\right) \\
&  =\sum_{i=1}^{n}\mu_{i}\sum_{j=1}^{m}\nu_{j}\left(  \underbrace{a_{i}}_{\in
A}\otimes\underbrace{b_{j}}_{\in B}\right)  \cdot\left(  \underbrace{c_{i}%
}_{\in C}\otimes\underbrace{d_{j}}_{\in D}\right) \\
&  \in\sum_{i=1}^{n}\mu_{i}\sum_{j=1}^{m}\nu_{j}\left(  A\otimes B\right)
\cdot\left(  C\otimes D\right)  \subseteq\left(  A\otimes B\right)
\cdot\left(  C\otimes D\right)
\end{align*}
(since $\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right)  $ is a
$k$-vector space). This proves (\ref{1315.pf.c}).

\textbf{d)} We have%
\begin{equation}
\left(  AC\right)  \otimes\left(  BD\right)  \subseteq\left(  A\otimes
B\right)  \cdot\left(  C\otimes D\right)  . \label{1315.pf.d}%
\end{equation}


\textit{Proof.} Let $\xi\in\left(  AC\right)  \otimes\left(  BD\right)  $.

We know that every tensor in a tensor product is a $k$-linear combination of
pure tensors. Applying this to the tensor $\xi$ in the tensor product $\left(
AC\right)  \otimes\left(  BD\right)  $, we conclude that the tensor $\xi$ is a
$k$-linear combination of pure tensors. In other words, there exists a
$p\in\mathbb{N}$ and elements $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{p}$ of $k$ and elements $x_{1}$, $x_{2}$, $\ldots$, $x_{p}$ of $AC$
and elements $y_{1}$, $y_{2}$, $\ldots$, $y_{p}$ of $BD$ such that $\xi
=\sum_{i=1}^{p}\lambda_{i}x_{i}\otimes y_{i}$. Consider this $p$, these
$\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{p}$, these $x_{1}$, $x_{2}$,
$\ldots$, $x_{p}$ and these $y_{1}$, $y_{2}$, $\ldots$, $y_{p}$. Now,%
\[
\xi=\sum_{i=1}^{p}\lambda_{i}\underbrace{x_{i}\otimes y_{i}}_{\substack{\in
\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right)  \\\text{(by
(\ref{1315.pf.c}), applied to }x=x_{i}\text{ and }y=y_{i}\text{)}}}\in
\sum_{i=1}^{p}\lambda_{i}\left(  A\otimes B\right)  \cdot\left(  C\otimes
D\right)  \subseteq\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right)
\]
(since $\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right)  $ is a
$k$-vector space).

Now, forget that we fixed $\xi$. We thus have shown that every $\xi\in\left(
AC\right)  \otimes\left(  BD\right)  $ satisfies $\xi\in\left(  A\otimes
B\right)  \cdot\left(  C\otimes D\right)  $. In other words, $\left(
AC\right)  \otimes\left(  BD\right)  \subseteq\left(  A\otimes B\right)
\cdot\left(  C\otimes D\right)  $. This proves (\ref{1315.pf.d}).

\textbf{e)} Combining (\ref{1315.pf.b}) with (\ref{1315.pf.d}), we obtain
$\left(  A\otimes B\right)  \cdot\left(  C\otimes D\right)  =\left(
AC\right)  \otimes\left(  BD\right)  $. This proves Lemma~\ref{Lemma1315}.
\end{proof}

Here is another very simple property of algebras:

\begin{lemma}
\label{Lemma1316} Let $k$ be a field. Let $A$ and $B$ be two $k$-algebras. Let
$f:A\rightarrow B$ and $g:A\rightarrow B$ be two $k$-algebra homomorphisms.
Then, $\operatorname*{Ker}\left(  f-g\right)  $ is a $k$-subalgebra of $A$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1316}.]The maps $f$ and $g$ are $k$-algebra
homomorphisms, therefore $k$-linear. Hence, their difference $f-g$ is also
$k$-linear. Thus, $\operatorname*{Ker}\left(  f-g\right)  $ is a $k$-vector
subspace of $A$.

Since $f$ is a $k$-algebra homomorphism, we have $f\left(  1_{A}\right)
=1_{B}$. Since $g$ is a $k$-algebra homomorphism, we have $g\left(
1_{A}\right)  =1_{B}$. Thus, $\left(  f-g\right)  \left(  1_{A}\right)
=\underbrace{f\left(  1_{A}\right)  }_{=1_{B}}-\underbrace{g\left(
1_{A}\right)  }_{=1_{B}}=1_{B}-1_{B}=0$. Hence, $1_{A}\in\operatorname*{Ker}%
\left(  f-g\right)  $.

Now, let $a\in\operatorname*{Ker}\left(  f-g\right)  $ and $b\in
\operatorname*{Ker}\left(  f-g\right)  $. Then, $\left(  f-g\right)  \left(
a\right)  =0$ (since $a\in\operatorname*{Ker}\left(  f-g\right)  $) and
$\left(  f-g\right)  \left(  b\right)  =0$ (since $b\in\operatorname*{Ker}%
\left(  f-g\right)  $).

Since $f$ is a $k$-algebra homomorphism, we have $f\left(  ab\right)
=f\left(  a\right)  f\left(  b\right)  $. Since $g$ is a $k$-algebra
homomorphism, we have $g\left(  ab\right)  =g\left(  a\right)  g\left(
b\right)  $. Since $0=\left(  f-g\right)  \left(  a\right)  =f\left(
a\right)  -g\left(  a\right)  $, we have $g\left(  a\right)  =f\left(
a\right)  $. Since $0=\left(  f-g\right)  \left(  b\right)  =f\left(
b\right)  -g\left(  b\right)  $, we have $g\left(  b\right)  =f\left(
b\right)  $. Now, $g\left(  ab\right)  =\underbrace{g\left(  a\right)
}_{=f\left(  a\right)  }\underbrace{g\left(  b\right)  }_{=f\left(  b\right)
}=f\left(  a\right)  f\left(  b\right)  =f\left(  ab\right)  $, so that
$0=f\left(  ab\right)  -g\left(  ab\right)  =\left(  f-g\right)  \left(
ab\right)  $. In other words, $\left(  f-g\right)  \left(  ab\right)  =0$, so
that $ab\in\operatorname*{Ker}\left(  f-g\right)  $.

Now, forget that we fixed $a$ and $b$. We thus have proven that $ab\in
\operatorname*{Ker}\left(  f-g\right)  $ for any $a\in\operatorname*{Ker}%
\left(  f-g\right)  $ and $b\in\operatorname*{Ker}\left(  f-g\right)  $.
Combined with the fact that $1_{A}\in\operatorname*{Ker}\left(  f-g\right)  $,
this yields that $\operatorname*{Ker}\left(  f-g\right)  $ is a $k$-subalgebra
of $A$ (since $\operatorname*{Ker}\left(  f-g\right)  $ is a $k$-vector
subspace of $A$). This proves Lemma~\ref{Lemma1316}.
\end{proof}

The next two propositions we state are completely elementary:

\begin{proposition}
\label{Proposition1317} Let $k$ be a field. Let $C$ and $D$ be two
$k$-coalgebras. Let $f:C\rightarrow D$ be an invertible $k$-coalgebra
homomorphism. Then, $f$ is a $k$-coalgebra isomorphism.
\end{proposition}

\begin{proposition}
\label{Proposition1318} Let $k$ be a field. Let $C$ and $D$ be two
$k$-bialgebras. Let $f:C\rightarrow D$ be an invertible $k$-bialgebra
homomorphism. Then, $f$ is a $k$-bialgebra isomorphism.
\end{proposition}

Propositions~\ref{Proposition1317} and~\ref{Proposition1318} are obvious
\textquotedblleft from the right viewpoint\textquotedblright, but let us give
down-to-earth proofs for them.

\begin{proof}
[Proof of Proposition~\ref{Proposition1317}.]The map $f$ is an invertible
$k$-coalgebra homomorphism, therefore an invertible $k$-linear map. Thus, its
inverse $f^{-1}$ is also a $k$-linear map.

Since $f$ is a $k$-coalgebra homomorphism, we have $\Delta_{D}\circ f=\left(
f\otimes f\right)  \circ\Delta_{C}$ and $\varepsilon_{D}\circ f=\varepsilon
_{C}$. The second of these two equalities yields $\varepsilon_{C}%
=\varepsilon_{D}\circ f$. Now, (\ref{(X)(X)}) (applied to $U=C$, $V=D$, $W=C$,
$U^{\prime}=C$, $V^{\prime}=D$, $W^{\prime}=C$, $\alpha=f$, $\beta=f^{-1}$,
$\gamma=f$ and $\delta=f^{-1}$) yields $\left(  f^{-1}\circ f\right)
\otimes\left(  f^{-1}\circ f\right)  =\left(  f^{-1}\otimes f^{-1}\right)
\circ\left(  f\otimes f\right)  $. Thus,%
\[
\left(  f^{-1}\otimes f^{-1}\right)  \circ\left(  f\otimes f\right)
=\underbrace{\left(  f^{-1}\circ f\right)  }_{=\operatorname{id}_{C}}%
\otimes\underbrace{\left(  f^{-1}\circ f\right)  }_{=\operatorname{id}_{C}%
}=\operatorname{id}_{C}\otimes\operatorname{id}_{C}=\operatorname{id}%
_{C\otimes C}.
\]
Now,%
\[
\left(  f^{-1}\otimes f^{-1}\right)  \circ\underbrace{\Delta_{D}\circ
f}_{=\left(  f\otimes f\right)  \circ\Delta_{C}}=\underbrace{\left(
f^{-1}\otimes f^{-1}\right)  \circ\left(  f\otimes f\right)  }%
_{=\operatorname{id}_{C\otimes C}}\circ\Delta_{C}=\Delta_{C}.
\]
Hence, $\underbrace{\left(  f^{-1}\otimes f^{-1}\right)  \circ\Delta_{D}\circ
f}_{=\Delta_{C}}\circ f^{-1}=\Delta_{C}\circ f^{-1}$, so that%
\[
\Delta_{C}\circ f^{-1}=\left(  f^{-1}\otimes f^{-1}\right)  \circ\Delta
_{D}\circ\underbrace{f\circ f^{-1}}_{=\operatorname{id}_{D}}=\left(
f^{-1}\otimes f^{-1}\right)  \circ\Delta_{D}.
\]
Combined with $\varepsilon_{C}\circ f^{-1}=\varepsilon_{D}$ (this follows from
$\underbrace{\varepsilon_{C}}_{=\varepsilon_{D}\circ f}\circ f^{-1}%
=\varepsilon_{D}\circ\underbrace{f\circ f^{-1}}_{=\operatorname{id}_{D}%
}=\varepsilon_{D}$), this shows that $f^{-1}$ is a $k$-coalgebra homomorphism.

Now, we know that $f$ is a $k$-coalgebra homomorphism which has an inverse,
and this inverse $f^{-1}$ is also a $k$-coalgebra homomorphism. Hence, the
homomorphism $f$ is invertible in the category of $k$-coalgebras. In other
words, $f$ is a $k$-coalgebra isomorphism. Proposition~\ref{Proposition1317}
is thus proven.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition1318}.]The map $f$ is a $k$-bialgebra
homomorphism, thus a $k$-coalgebra homomorphism. Hence, $f$ is an invertible
$k$-coalgebra homomorphism. Therefore, Proposition~\ref{Proposition1317} shows
that $f$ is a $k$-coalgebra isomorphism. Thus, the inverse $f^{-1}$ of $f$ is
also a $k$-coalgebra homomorphism.

Also, $f$ is a $k$-algebra homomorphism (since $f$ is a $k$-bialgebra
homomorphism). Thus, $f$ is an invertible $k$-algebra homomorphism. It is
well-known that this yields that $f$ is a $k$-algebra isomorphism. Thus, we
know that $f$ is a $k$-algebra isomorphism. Hence, the inverse $f^{-1}$ of $f$
is also a $k$-algebra homomorphism.

Now, $f^{-1}$ is both a $k$-algebra homomorphism and a $k$-coalgebra
homomorphism. In other words, $f^{-1}$ is a $k$-bialgebra homomorphism. So we
know that $f$ is a $k$-bialgebra homomorphism which has an inverse, and this
inverse $f^{-1}$ is also a $k$-bialgebra homomorphism. Hence, the homomorphism
$f$ is invertible in the category of $k$-bialgebras. In other words, $f$ is a
$k$-bialgebra isomorphism. Proposition~\ref{Proposition1318} is thus proven.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{Proposition1314}.]\textbf{(a)} Clearly, every
$i\in\mathbb{N}$ satisfies $\sum_{n=0}^{i}V^{n}\subseteq\sum_{n=0}^{i+1}V^{n}$
(since $\sum_{n=0}^{i+1}V^{n}=\sum_{n=0}^{i}V^{n}+V^{i+1}\supseteq\sum
_{n=0}^{i}V^{n}$). In other words,%
\begin{equation}
\sum_{n=0}^{0}V^{n}\subseteq\sum_{n=0}^{1}V^{n}\subseteq\sum_{n=0}^{2}%
V^{n}\subseteq\cdots. \label{1314.pf.a.1}%
\end{equation}


Now, let $x\in H$. Then, $x\in H=\sum_{n\in\mathbb{N}}V^{n}$. Hence, there
exists a family $\left(  x_{n}\right)  _{n\in\mathbb{N}}\in\prod
_{n\in\mathbb{N}}V^{n}$ such that $\left(  \text{all but finitely many }%
n\in\mathbb{N}\text{ satisfy }x_{n}=0\right)  $ and $x=\sum_{n\in\mathbb{N}%
}x_{n}$. Consider this family $\left(  x_{n}\right)  _{n\in\mathbb{N}}$. We
know that all but finitely many $n\in\mathbb{N}$ satisfy $x_{n}=0$. Hence,
there exists an $N\in\mathbb{N}$ such that every nonnegative integer $n>N$
satisfies $x_{n}=0$. Consider this $N$. We have%
\begin{align*}
x  &  =\sum_{n\in\mathbb{N}}x_{n}=\underbrace{\sum_{\substack{n\in
\mathbb{N};\\n\leq N}}}_{=\sum_{n=0}^{N}}x_{n}+\sum_{\substack{n\in
\mathbb{N};\\n>N}}\underbrace{x_{n}}_{\substack{=0\\\text{(since }n\geq
N\text{)}}}=\sum_{n=0}^{N}x_{n}+\underbrace{\sum_{\substack{n\in
\mathbb{N};\\n>N}}0}_{=0}\\
&  =\sum_{n=0}^{N}\underbrace{x_{n}}_{\in V^{n}}\in\sum_{n=0}^{N}%
V^{n}\subseteq\bigcup_{i\in\mathbb{N}}\left(  \sum_{n=0}^{i}V^{n}\right)
\end{align*}
(since $N\in\mathbb{N}$, and since $\sum_{n=0}^{N}V^{n}=\sum_{n=0}^{i}V^{n}$
for $i=N$).

Now, forget that we fixed $x$. We thus have shown that every $x\in H$
satisfies $x\in\bigcup_{i\in\mathbb{N}}\left(  \sum_{n=0}^{i}V^{n}\right)  $.
In other words, $H\subseteq\bigcup_{i\in\mathbb{N}}\left(  \sum_{n=0}^{i}%
V^{n}\right)  $. Combining this with $\bigcup_{i\in\mathbb{N}}\left(
\sum_{n=0}^{i}V^{n}\right)  \subseteq H$ (which is obvious), we obtain
$H=\bigcup_{i\in\mathbb{N}}\left(  \sum_{n=0}^{i}V^{n}\right)  $. Combined
with (\ref{1314.pf.a.1}), this shows that $\left(  \sum_{n=0}^{i}V^{n}\right)
_{i\in\mathbb{N}}$ is a filtration of the $k$-vector space $H$. This proves
Proposition~\ref{Proposition1314} \textbf{(a)}.

\textbf{(b)} Let us denote the $k$-bialgebra $H$ endowed with the filtration
$\left(  \sum_{n=0}^{i}V^{n}\right)  _{i\in\mathbb{N}}$ simply by $H$. Then,
\begin{equation}
H_{\leq i}=\sum_{n=0}^{i}V^{n}\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\mathbb{N}. \label{1314.pf.b.Hi}%
\end{equation}
for every $i\in\mathbb{N}$.

We have
\begin{align*}
H_{\leq0}  &  =\sum_{n=0}^{0}V^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1314.pf.b.Hi}), applied to }i=0\right) \\
&  =V^{0}=k\cdot1_{H},
\end{align*}
and $1_{H}\in k\cdot1_{H}=V^{0}=H_{\leq0}$.

Moreover, every $j\in\mathbb{N}$ satisfies%
\begin{align}
H_{\leq j}  &  =\sum_{n=0}^{j}V^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1314.pf.b.Hi}), applied to }j\text{ instead of }i\right) \nonumber\\
&  =\sum_{m=0}^{j}V^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }n\text{ as }m\right)  . \label{1314.pf.b.Hj}%
\end{align}


Now, every $i\in\mathbb{N}$ and $j\in\mathbb{N}$ satisfy $H_{\leq i}H_{\leq
j}\subseteq H_{\leq i+j}$\ \ \ \ \footnote{\textit{Proof.} Let $i\in
\mathbb{N}$ and $j\in\mathbb{N}$. Let $n\in\left\{  0,1,\ldots,i\right\}  $
and $m\in\left\{  0,1,\ldots,j\right\}  $. Then, $0\leq n$ and $0\leq m$, so
that $0=\underbrace{0}_{\leq n}+\underbrace{0}_{\leq m}\leq n+m$. Also, $n\leq
i$ and $m\leq j$, so that $\underbrace{n}_{\leq i}+\underbrace{m}_{\leq j}\leq
i+j$. Hence, $0\leq n+m\leq i+j$, so that $n+m\in\left\{  0,1,\ldots
,i+j\right\}  $. Hence, $V^{n+m}$ is an addend in the sum $\sum_{N=0}%
^{i+j}V^{N}$. Thus,
\begin{equation}
V^{n+m}\subseteq\sum_{N=0}^{i+j}V^{N}. \label{1314.pf.b.HiHj.1}%
\end{equation}
\par
Now, forget that we fixed $n$ and $m$. We thus have shown that
(\ref{1314.pf.b.HiHj.1}) holds for all $n\in\left\{  0,1,\ldots,i\right\}  $
and $m\in\left\{  0,1,\ldots,j\right\}  $. Now,
\begin{align*}
&  \underbrace{H_{\leq i}}_{\substack{=\sum_{n=0}^{i}V^{n}\\\text{(by
(\ref{1314.pf.b.Hi}))}}}\cdot\underbrace{H_{\leq j}}_{\substack{=\sum
_{m=0}^{j}V^{m}\\\text{(by (\ref{1314.pf.b.Hj}))}}}\\
&  =\left(  \sum_{n=0}^{i}V^{n}\right)  \cdot\left(  \sum_{m=0}^{j}%
V^{m}\right)  =\sum_{n=0}^{i}\sum_{m=0}^{j}\underbrace{V^{n}\cdot V^{m}%
}_{\substack{_{\substack{=V^{n+m}\subseteq\sum_{N=0}^{i+j}V^{N}}}\\\text{(by
(\ref{1314.pf.b.HiHj.1}))}}}\subseteq\sum_{n=0}^{i}\sum_{m=0}^{j}\sum
_{N=0}^{i+j}V^{N}\\
&  \subseteq\sum_{N=0}^{i+j}V^{N}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\sum_{N=0}^{i+j}V^{N}\text{ is a }k\text{-vector space}\right) \\
&  =\sum_{n=0}^{i+j}V^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed
the summation index }N\text{ as }n\right) \\
&  =H_{\leq i+j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }H_{\leq i+j}%
=\sum_{n=0}^{i+j}V^{n}\text{ (by (\ref{1314.pf.b.Hi}), applied to }i+j\text{
instead of }i\text{)}\right)  .
\end{align*}
This proves $H_{\leq i}H_{\leq j}\subseteq H_{\leq i+j}$.}. Combining this
observation with the fact that $1_{H}\in H_{\leq0}$, we conclude that $H$ is a
filtered $k$-algebra.

On the other hand, every $m\in\mathbb{N}$ satisfies%
\begin{equation}
\Delta_{H}\left(  V^{m}\right)  \subseteq\sum_{u=0}^{m}V^{u}\otimes V^{m-u}.
\label{1314.pf.b.coal.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{1314.pf.b.coal.1}):} We will prove
(\ref{1314.pf.b.coal.1}) by induction over $m$:
\par
\textit{Induction base:} By the axioms of a $k$-bialgebra, the map $\Delta
_{H}$ is a $k$-algebra homomorphism (since $H$ is a $k$-bialgebra). Hence,
$\Delta_{H}\left(  1_{H}\right)  =1_{H\otimes H}=1_{H}\otimes1_{H}$.
\par
We have $V^{0}=k\cdot1_{H}$, so that
\begin{align*}
\Delta_{H}\left(  V^{0}\right)   &  =\Delta_{H}\left(  k\cdot1_{H}\right)
=k\cdot\underbrace{\Delta_{H}\left(  1_{H}\right)  }_{=1_{H}\otimes1_{H}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{H}\text{ is }k\text{-linear}%
\right) \\
&  =k\cdot\underbrace{1_{H}}_{\in V^{0}}\otimes\underbrace{1_{H}}_{\in V^{0}%
}\subseteq k\cdot V^{0}\otimes V^{0}\subseteq V^{0}\otimes V^{0}%
\end{align*}
(since $V^{0}\otimes V^{0}$ is a $k$-vector space). On the other hand,%
\[
\sum_{u=0}^{0}V^{u}\otimes V^{0-u}=V^{0}\otimes V^{0-0}=V^{0}\otimes V^{0}.
\]
Thus, $\Delta_{H}\left(  V^{0}\right)  \subseteq V^{0}\otimes V^{0}=\sum
_{u=0}^{0}V^{u}\otimes V^{0-u}$. In other words, (\ref{1314.pf.b.coal.1})
holds for $m=0$. This completes the induction base.
\par
\textit{Induction step:} Let $M\in\mathbb{N}$. Assume that
(\ref{1314.pf.b.coal.1}) holds for $m=M$. We must now show that
(\ref{1314.pf.b.coal.1}) also holds for $m=M+1$.
\par
We have
\[
\Delta_{H}\left(  V^{M}\right)  \subseteq\sum_{u=0}^{M}V^{u}\otimes V^{M-u}%
\]
(since (\ref{1314.pf.b.coal.1}) holds for $m=M$). Now, $V^{M+1}=V^{M}\cdot V$,
so that%
\begin{align*}
\Delta_{H}\left(  V^{M+1}\right)   &  =\Delta_{H}\left(  V^{M}\cdot V\right)
=\underbrace{\Delta_{H}\left(  V^{M}\right)  }_{\subseteq\sum_{u=0}^{M}%
V^{u}\otimes V^{M-u}}\cdot\underbrace{\Delta_{H}\left(  V\right)  }_{\subseteq
V\otimes\left(  k\cdot1_{H}\right)  +\left(  k\cdot1_{H}\right)  \otimes V}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{H}\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  \subseteq\left(  \sum_{u=0}^{M}V^{u}\otimes V^{M-u}\right)  \cdot\left(
V\otimes\left(  k\cdot1_{H}\right)  +\left(  k\cdot1_{H}\right)  \otimes
V\right) \\
&  =\left(  \sum_{u=0}^{M}V^{u}\otimes V^{M-u}\right)  \cdot\left(
V\otimes\left(  k\cdot1_{H}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \sum_{u=0}^{M}V^{u}\otimes V^{M-u}\right)
\cdot\left(  \left(  k\cdot1_{H}\right)  \otimes V\right) \\
&  =\sum_{u=0}^{M}\underbrace{\left(  V^{u}\otimes V^{M-u}\right)
\cdot\left(  V\otimes\left(  k\cdot1_{H}\right)  \right)  }%
_{\substack{=\left(  V^{u}\cdot V\right)  \otimes\left(  V^{M-u}\cdot\left(
k\cdot1_{H}\right)  \right)  \\\text{(by Lemma~\ref{Lemma1315}, applied to
}H\text{, }H\text{, }V^{u}\text{, }V^{M-u}\text{, }V\text{ and }k\cdot
1_{H}\\\text{instead of }U\text{, }V\text{, }A\text{, }B\text{, }C\text{ and
}D\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{u=0}^{M}\underbrace{\left(  V^{u}\otimes
V^{M-u}\right)  \cdot\left(  \left(  k\cdot1_{H}\right)  \otimes V\right)
}_{\substack{=\left(  V^{u}\cdot\left(  k\cdot1_{H}\right)  \right)
\otimes\left(  V^{M-u}\cdot V\right)  \\\text{(by Lemma~\ref{Lemma1315},
applied to }H\text{, }H\text{, }V^{u}\text{, }V^{M-u}\text{, }k\cdot
1_{H}\text{ and }V\\\text{instead of }U\text{, }V\text{, }A\text{, }B\text{,
}C\text{ and }D\text{)}}}\\
&  =\sum_{u=0}^{M}\underbrace{\left(  V^{u}\cdot V\right)  }_{=V^{u+1}}%
\otimes\underbrace{\left(  V^{M-u}\cdot\left(  k\cdot1_{H}\right)  \right)
}_{\substack{=V^{M-u}\cdot k\cdot1_{H}=V^{M-u}\cdot k\subseteq V^{M-u}%
\\\text{(since }V^{M-u}\text{ is a }k\text{-vector space)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{u=0}^{M}\underbrace{\left(  V^{u}\cdot\left(
k\cdot1_{H}\right)  \right)  }_{\substack{=V^{u}\cdot k\cdot1_{H}=V^{u}\cdot
k\subseteq V^{u}\\\text{(since }V^{u}\text{ is a }k\text{-vector space)}%
}}\otimes\underbrace{\left(  V^{M-u}\cdot V\right)  }_{=V^{M-u+1}=V^{M+1-u}}\\
&  \subseteq\sum_{u=0}^{M}V^{u+1}\otimes V^{M-u}+\sum_{u=0}^{M}V^{u}\otimes
V^{M+1-u}%
\end{align*}%
\begin{align*}
&  =\sum_{u=1}^{M+1}\underbrace{V^{\left(  u-1\right)  +1}}_{=V^{u}}%
\otimes\underbrace{V^{M-\left(  u-1\right)  }}_{=V^{M-u+1}=V^{M+1-u}}%
+\sum_{u=0}^{M}V^{u}\otimes V^{M+1-u}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }u-1\text{ for
}u\text{ in the first sum}\right) \\
&  \subseteq\underbrace{\sum_{u=1}^{M+1}V^{u}\otimes V^{M+1-u}}%
_{\substack{\subseteq\sum_{u=0}^{M+1}V^{u}\otimes V^{M+1-u}\\\text{(since
}1\geq0\text{)}}}+\underbrace{\sum_{u=0}^{M}V^{u}\otimes V^{M+1-u}%
}_{\substack{\subseteq\sum_{u=0}^{M+1}V^{u}\otimes V^{M+1-u}\\\text{(since
}M\leq M+1\text{)}}}\\
&  \subseteq\sum_{u=0}^{M+1}V^{u}\otimes V^{M+1-u}+\sum_{u=0}^{M+1}%
V^{u}\otimes V^{M+1-u}\subseteq\sum_{u=0}^{M+1}V^{u}\otimes V^{M+1-u}%
\end{align*}
(since $\sum_{u=0}^{M+1}V^{u}\otimes V^{M+1-u}$ is a $k$-vector space). In
other words, (\ref{1314.pf.b.coal.1}) also holds for $m=M+1$. This completes
the induction step. The induction proof of (\ref{1314.pf.b.coal.1}) is thus
finished.} Furthermore, every $m\in\mathbb{N}$ satisfies
\begin{equation}
V^{i}\subseteq H_{\leq m}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
0,1,\ldots,m\right\}  . \label{1314.pf.b.coal.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{1314.pf.b.coal.3}):} Let $m\in\mathbb{N}$ and
$i\in\left\{  0,1,\ldots,m\right\}  $. We have $H_{\leq m}=\sum_{n=0}^{m}%
V^{n}$ (by (\ref{1314.pf.b.Hi}), applied to $i=m$). But $V^{i}$ is an addend
in the sum $\sum_{n=0}^{m}V^{n}$ (since $i\in\left\{  0,1,\ldots,m\right\}
$). Hence, $V^{i}\subseteq\sum_{n=0}^{m}V^{n}=H_{\leq m}$. This proves
(\ref{1314.pf.b.coal.3}).} Finally, every $m\in\mathbb{N}$ satisfies%

\begin{equation}
\Delta_{H}\left(  H_{\leq m}\right)  \subseteq\sum_{u=0}^{m}H_{\leq u}\otimes
H_{\leq m-u}. \label{1314.pf.b.coal.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{1314.pf.b.coal.5}):} Let $m\in\mathbb{N}$.
Applying (\ref{1314.pf.b.Hi}) to $i=m$, we obtain $H_{\leq m}=\sum_{n=0}%
^{m}V^{n}$. Thus,%
\begin{align*}
\Delta_{H}\left(  H_{\leq m}\right)   &  =\Delta_{H}\left(  \sum_{n=0}%
^{m}V^{n}\right)  =\sum_{n=0}^{m}\underbrace{\Delta_{H}\left(  V^{n}\right)
}_{\substack{\subseteq\sum_{u=0}^{n}V^{u}\otimes V^{n-u}\\\text{(by
(\ref{1314.pf.b.coal.1}),}\\\text{applied to }n\text{ instead of }m\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{H}\text{ is }%
k\text{-linear}\right) \\
&  \subseteq\sum_{n=0}^{m}\sum_{u=0}^{n}\underbrace{V^{u}}%
_{\substack{\subseteq H_{\leq u}\\\text{(by (\ref{1314.pf.b.coal.3}), applied
to}\\u\text{ and }u\text{ instead of }m\text{ and }i\\\text{(since }%
u\in\left\{  0,1,\ldots,u\right\}  \text{))}}}\otimes\underbrace{V^{n-u}%
}_{\substack{\subseteq H_{\leq m-u}\\\text{(by (\ref{1314.pf.b.coal.3}),
applied to}\\m-u\text{ and }n-u\text{ instead of }m\text{ and }i\\\text{(since
}n-u\in\left\{  0,1,\ldots,m-u\right\}  \\\text{(because }n-u\geq0\text{
(since }u\leq n\text{) and}\\n-u\leq m-u\text{ (since }n\leq m\text{))))}}}\\
&  \subseteq\sum_{n=0}^{m}\underbrace{\sum_{u=0}^{n}H_{\leq u}\otimes H_{\leq
m-u}}_{\substack{\subseteq\sum_{u=0}^{m}H_{\leq u}\otimes H_{\leq
m-u}\\\text{(since }n\leq m\text{)}}}\subseteq\sum_{n=0}^{m}\sum_{u=0}%
^{m}H_{\leq u}\otimes H_{\leq m-u}\subseteq\sum_{u=0}^{m}H_{\leq u}\otimes
H_{\leq m-u}%
\end{align*}
(since $\sum_{u=0}^{m}H_{\leq u}\otimes H_{\leq m-u}$ is a $k$-vector space).
This proves (\ref{1314.pf.b.coal.5}).} Hence, $H$ is a filtered $k$-coalgebra.

By the axioms of a $k$-bialgebra, we have $\varepsilon_{H}\left(
1_{H}\right)  =1$ (since $H$ is a $k$-bialgebra).

We have now shown that $H$ is a filtered $k$-algebra, and that $H$ is a
filtered $k$-coalgebra. Combining these two observations, we conclude that $H$
is a filtered $k$-bialgebra (since we know that $H$ is a $k$-bialgebra). Thus,
we can apply Remark~\ref{Remark215} to $C=H$ and $\lambda=1_{H}$ (since
$H_{\leq0}=k\cdot1_{H}$ and $\varepsilon_{H}\left(  1_{H}\right)  =1$). As a
result, we conclude that the filtered $k$-coalgebra $H$ is connected. Thus,
the filtered $k$-bialgebra $H$ is connected.

We thus have shown that $H$ is a connected filtered $k$-bialgebra. Since the
filtration on $H$ is $\left(  \sum_{n=0}^{i}V^{n}\right)  _{i\in\mathbb{N}}$,
this rewrites as follows: The $k$-bialgebra $H$ endowed with the filtration
$\left(  \sum_{n=0}^{i}V^{n}\right)  _{i\in\mathbb{N}}$ is a connected
filtered $k$-bialgebra. Thus, Proposition~\ref{Proposition1314} \textbf{(b)}
is proven.

\textbf{(c)} Assume that $V\subseteq\operatorname*{Prim}H$.

Consider the $\left(  H,H\right)  $-flip $\tau_{H,H}:H\otimes H\rightarrow
H\otimes H$ (defined according to Definition~\ref{Definition34}).

It is known that for any two $k$-algebras $A$ and $B$, the $\left(
A,B\right)  $-flip $\tau_{A,B}:A\otimes B\rightarrow B\otimes A$ is a
$k$-algebra homomorphism. Applying this to $A=H$ and $B=H$, we conclude that
the $\left(  H,H\right)  $-flip $\tau_{H,H}:H\otimes H\rightarrow H\otimes H$
is a $k$-algebra homomorphism.

Also, $H$ is a $k$-bialgebra. Thus, $\Delta_{H}$ is a $k$-algebra homomorphism
(by the axioms of a $k$-bialgebra).

We know that $\tau_{H,H}$ and $\Delta_{H}$ are $k$-algebra homomorphisms.
Hence, their composition $\tau_{H,H}\circ\Delta_{H}$ is a $k$-algebra
homomorphism (because the composition of two $k$-algebra homomorphisms must
always be a $k$-algebra homomorphism).

Since $\Delta_{H}$ and $\tau_{H,H}\circ\Delta_{H}$ are $k$-algebra
homomorphisms, we know that $\operatorname*{Ker}\left(  \Delta_{H}-\tau
_{H,H}\circ\Delta_{H}\right)  $ is a $k$-subalgebra of $H$ (by
Lemma~\ref{Lemma1316}, applied to $H$, $H\otimes H$, $\Delta_{H}$ and
$\tau_{H,H}\circ\Delta_{H}$ instead of $A$, $B$, $f$ and $g$).

Clearly, every $k$-subalgebra $\mathfrak{A}$ of $H$ satisfies $\mathfrak{A}%
^{n}\subseteq\mathfrak{A}$ for every $n\in\mathbb{N}$. Applying this to
$\mathfrak{A}=\operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}\circ\Delta
_{H}\right)  $, we conclude that $\left(  \operatorname*{Ker}\left(
\Delta_{H}-\tau_{H,H}\circ\Delta_{H}\right)  \right)  ^{n}\subseteq
\operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}\circ\Delta_{H}\right)  $ for
every $n\in\mathbb{N}$ (because $\operatorname*{Ker}\left(  \Delta_{H}%
-\tau_{H,H}\circ\Delta_{H}\right)  $ is a $k$-subalgebra of $H$).

Now, it is easy to see that $V\subseteq\operatorname*{Ker}\left(  \Delta
_{H}-\tau_{H,H}\circ\Delta_{H}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$v\in V$. Then, $v\in V\subseteq\operatorname*{Prim}H$. In other words, $v$ is
a primitive element of $H$ (since $\operatorname*{Prim}H$ is the set of all
primitive elements of $H$). In other words, $\Delta_{H}\left(  x\right)
=x\otimes1_{H}+1_{H}\otimes x$. Now,
\begin{align*}
\left(  \tau_{H,H}\circ\Delta_{H}\right)  \left(  x\right)   &  =\tau
_{H,H}\left(  \underbrace{\Delta_{H}\left(  x\right)  }_{=x\otimes1_{H}%
+1_{H}\otimes x}\right)  =\tau_{H,H}\left(  x\otimes1_{H}+1_{H}\otimes
x\right) \\
&  =\underbrace{\tau_{H,H}\left(  x\otimes1_{H}\right)  }_{\substack{=1_{H}%
\otimes x\\\text{(by the definition of }\tau_{H,H}\text{)}}}+\underbrace{\tau
_{H,H}\left(  1_{H}\otimes x\right)  }_{\substack{=x\otimes1_{H}\\\text{(by
the definition of }\tau_{H,H}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\tau_{H,H}\text{ is }k\text{-linear}\right) \\
&  =1_{H}\otimes x+x\otimes1_{H}=x\otimes1_{H}+1_{H}\otimes x=\Delta
_{H}\left(  x\right)  .
\end{align*}
Thus,%
\[
\left(  \Delta_{H}-\tau_{H,H}\circ\Delta_{H}\right)  \left(  x\right)
=\Delta_{H}\left(  x\right)  -\underbrace{\left(  \tau_{H,H}\circ\Delta
_{H}\right)  \left(  x\right)  }_{=\Delta_{H}\left(  x\right)  }=\Delta
_{H}\left(  x\right)  -\Delta_{H}\left(  x\right)  =0.
\]
In other words, $x\in\operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}%
\circ\Delta_{H}\right)  $.
\par
Now, forget that we fixed $x$. We thus have shown that every $x\in V$
satisfies $x\in\operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}\circ\Delta
_{H}\right)  $. In other words, $V\subseteq\operatorname*{Ker}\left(
\Delta_{H}-\tau_{H,H}\circ\Delta_{H}\right)  $, qed.}. Thus, every
$n\in\mathbb{N}$ satisfies%
\[
V^{n}\subseteq\left(  \operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}%
\circ\Delta_{H}\right)  \right)  ^{n}\subseteq\operatorname*{Ker}\left(
\Delta_{H}-\tau_{H,H}\circ\Delta_{H}\right)  .
\]
Now,%
\[
H=\sum_{n\in\mathbb{N}}\underbrace{V^{n}}_{\subseteq\operatorname*{Ker}\left(
\Delta_{H}-\tau_{H,H}\circ\Delta_{H}\right)  }\subseteq\sum_{n\in\mathbb{N}%
}\operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}\circ\Delta_{H}\right)
\subseteq\operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}\circ\Delta
_{H}\right)
\]
(since $\operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}\circ\Delta
_{H}\right)  $ is a $k$-vector space). In other words, every $x\in H$
satisfies $x\in\operatorname*{Ker}\left(  \Delta_{H}-\tau_{H,H}\circ\Delta
_{H}\right)  $. In other words, every $x\in H$ satisfies $\left(  \Delta
_{H}-\tau_{H,H}\circ\Delta_{H}\right)  \left(  x\right)  =0$. In other words,
$\Delta_{H}-\tau_{H,H}\circ\Delta_{H}=0$. Thus, $\Delta_{H}=\tau_{H,H}%
\circ\Delta_{H}$. In other words, $\tau_{H,H}\circ\Delta_{H}=\Delta_{H}$.

Now, let us recall the definition of a cocommutative $k$-coalgebra: A
$k$-coalgebra $C$ is cocommutative if and only if $\tau_{C,C}\circ\Delta
_{C}=\Delta_{C}$. Applying this to $C=H$, we conclude that the $k$-coalgebra
$H$ is cocommutative if and only if $\tau_{H,H}\circ\Delta_{H}=\Delta_{H}$.
Thus, the $k$-coalgebra $H$ is cocommutative (since $\tau_{H,H}\circ\Delta
_{H}=\Delta_{H}$). In other words, the $k$-bialgebra $H$ is cocommutative.
This proves Proposition~\ref{Proposition1314} \textbf{(c)}.
\end{proof}

\begin{proof}
[Partial proof of Theorem~\ref{Theorem1313}.]First of all, we have
$\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Prim}\left(  U\left(  \mathfrak{g}\right)  \right)
$\ \ \ \ \footnote{\textit{Proof.} Let $v\in\iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  $. Then, $\Delta_{U\left(  \mathfrak{g}\right)  }\left(
v\right)  =v\otimes1_{U\left(  \mathfrak{g}\right)  }+1_{U\left(
\mathfrak{g}\right)  }\otimes v$ (according to (\ref{1312})). In other words,
the element $v$ of $U\left(  \mathfrak{g}\right)  $ is primitive. In other
words, $v\in\operatorname*{Prim}\left(  U\left(  \mathfrak{g}\right)  \right)
$ (since $\operatorname*{Prim}\left(  U\left(  \mathfrak{g}\right)  \right)  $
is the set of all primitive elements of $U\left(  \mathfrak{g}\right)  $).
\par
Now, forget that we fixed $v$. We thus have shown that every $v\in
\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  $ satisfies $v\in
\operatorname*{Prim}\left(  U\left(  \mathfrak{g}\right)  \right)  $. In other
words, $\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Prim}\left(  U\left(  \mathfrak{g}\right)  \right)  $, qed.}.
This proves Theorem~\ref{Theorem1313} \textbf{(d)}.

Let $\pi$ be the canonical projection $\otimes\mathfrak{g}\rightarrow U\left(
\mathfrak{g}\right)  $. Then, $\pi$ is a surjective $k$-algebra homomorphism.
[Actually, $\pi$ is a $k$-bialgebra homomorphism, but we don't need to know this.]

The canonical map $\iota_{\mathfrak{g}}$ from $\mathfrak{g}$ to $U\left(
\mathfrak{g}\right)  $ factors through the projection $\pi:\otimes
\mathfrak{g}\rightarrow U\left(  \mathfrak{g}\right)  $. More precisely,
$\iota_{\mathfrak{g}}\left(  x\right)  =\pi\left(  x\right)  $ for every
$x\in\mathfrak{g}$ (this follows from the definition of $\iota_{\mathfrak{g}}%
$). Thus,%
\begin{equation}
\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  =\left\{  \underbrace{\iota
_{\mathfrak{g}}\left(  x\right)  }_{=\pi\left(  x\right)  }\ \mid
\ x\in\mathfrak{g}\right\}  =\left\{  \pi\left(  x\right)  \ \mid
\ x\in\mathfrak{g}\right\}  =\pi\left(  \mathfrak{g}\right)  .
\label{1313.pf.4}%
\end{equation}


By the definition of the tensor algebra, we have $\otimes\mathfrak{g}%
=\bigoplus_{n\in\mathbb{N}}\mathfrak{g}^{\otimes n}$. Thus, $\otimes
\mathfrak{g}=\bigoplus_{n\in\mathbb{N}}\mathfrak{g}^{\otimes n}=\sum
_{n\in\mathbb{N}}\mathfrak{g}^{\otimes n}$ (since direct sums are sums).

For every $n\in\mathbb{N}$, we have $\mathfrak{g}^{\otimes n}=\mathfrak{g}%
^{n}$ as $k$-vector subspaces of $\otimes\mathfrak{g}$ (where $\mathfrak{g}%
^{n}$ means $\underbrace{\mathfrak{g}\cdot\mathfrak{g}\cdot\cdots
\cdot\mathfrak{g}}_{n\text{ times}}$, as
usual)\ \ \ \ \footnote{\textit{Proof.} Let $n\in\mathbb{N}$. The $n$-th
tensor power $\mathfrak{g}^{\otimes n}$ is spanned by all pure tensors. In
other words,%
\begin{align*}
\mathfrak{g}^{\otimes n}  &  =\left\langle \underbrace{v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}}_{\substack{=v_{1}v_{2}\cdots v_{n}%
\\\text{(because the multiplication on }\otimes\mathfrak{g}\text{ is
the}\\\text{tensor product, so we have}\\v_{1}v_{2}\cdots v_{n}=v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\text{)}}}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in\mathfrak{g}^{\times n}\right\rangle \\
&  =\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in\mathfrak{g}^{\times n}\right\rangle .
\end{align*}
Compared with%
\[
\mathfrak{g}^{n}=\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in\mathfrak{g}^{\times n}\right\rangle ,
\]
this yields $\mathfrak{g}^{\otimes n}=\mathfrak{g}^{n}$, qed.}. Hence,
$\sum_{n\in\mathbb{N}}\mathfrak{g}^{\otimes n}=\sum_{n\in\mathbb{N}%
}\mathfrak{g}^{n}$.

Now, $\otimes\mathfrak{g}=\sum_{n\in\mathbb{N}}\mathfrak{g}^{\otimes n}%
=\sum_{n\in\mathbb{N}}\mathfrak{g}^{n}$. Applying the map $\pi$ to this
equality, we obtain%
\begin{align*}
\pi\left(  \otimes\mathfrak{g}\right)   &  =\pi\left(  \sum_{n\in\mathbb{N}%
}\mathfrak{g}^{n}\right)  =\sum_{n\in\mathbb{N}}\left(  \underbrace{\pi\left(
\mathfrak{g}\right)  }_{\substack{=\iota_{\mathfrak{g}}\left(  \mathfrak{g}%
\right)  \\\text{(by (\ref{1313.pf.4}))}}}\right)  ^{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  =\sum_{n\in\mathbb{N}}\left(  \iota_{\mathfrak{g}}\left(  \mathfrak{g}%
\right)  \right)  ^{n}.
\end{align*}
Since $\pi\left(  \otimes\mathfrak{g}\right)  =U\left(  \mathfrak{g}\right)  $
(because $\pi$ is surjective), this rewrites as $U\left(  \mathfrak{g}\right)
=\sum_{n\in\mathbb{N}}\left(  \iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)
\right)  ^{n}$. Thus, Theorem~\ref{Theorem1313} \textbf{(a)} is proven.

We have
\[
\Delta_{U\left(  \mathfrak{g}\right)  }\left(  \iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  \right)  \subseteq\iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  \otimes\left(  k\cdot1_{U\left(  \mathfrak{g}\right)
}\right)  +\left(  k\cdot1_{U\left(  \mathfrak{g}\right)  }\right)
\otimes\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  .
\]
\footnote{\textit{Proof.} Every $v\in\iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  $ satisfies%
\begin{align*}
\Delta_{U\left(  \mathfrak{g}\right)  }\left(  v\right)   &  =\underbrace{v}%
_{\in\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  }\otimes
\underbrace{1_{U\left(  \mathfrak{g}\right)  }}_{\in k\cdot1_{U\left(
\mathfrak{g}\right)  }}+\underbrace{1_{U\left(  \mathfrak{g}\right)  }}_{\in
k\cdot1_{U\left(  \mathfrak{g}\right)  }}\otimes\underbrace{v}_{\in
\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{according to (\ref{1312})}\right) \\
&  \in\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  \otimes\left(
k\cdot1_{U\left(  \mathfrak{g}\right)  }\right)  +\left(  k\cdot1_{U\left(
\mathfrak{g}\right)  }\right)  \otimes\iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  .
\end{align*}
In other words, $\Delta_{U\left(  \mathfrak{g}\right)  }\left(  \iota
_{\mathfrak{g}}\left(  \mathfrak{g}\right)  \right)  \subseteq\iota
_{\mathfrak{g}}\left(  \mathfrak{g}\right)  \otimes\left(  k\cdot1_{U\left(
\mathfrak{g}\right)  }\right)  +\left(  k\cdot1_{U\left(  \mathfrak{g}\right)
}\right)  \otimes\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)  $, qed.}
Combining this with the equality $U\left(  \mathfrak{g}\right)  =\sum
_{n\in\mathbb{N}}\left(  \iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)
\right)  ^{n}$, we see that we can apply Proposition~\ref{Proposition1314} to
$H=U\left(  \mathfrak{g}\right)  $ and $V=\iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  $.

Proposition~\ref{Proposition1314} \textbf{(a)} (applied to $H=U\left(
\mathfrak{g}\right)  $ and $V=\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)
$) yields that the family $\left(  \sum_{n=0}^{i}\left(  \iota_{\mathfrak{g}%
}\left(  \mathfrak{g}\right)  \right)  ^{n}\right)  _{i\in\mathbb{N}}$ is a
filtration of the $k$-vector space $U\left(  \mathfrak{g}\right)  $. This
proves Theorem~\ref{Theorem1313} \textbf{(b)}.

Proposition~\ref{Proposition1314} \textbf{(b)} (applied to $H=U\left(
\mathfrak{g}\right)  $ and $V=\iota_{\mathfrak{g}}\left(  \mathfrak{g}\right)
$) yields that the $k$-bialgebra $U\left(  \mathfrak{g}\right)  $ endowed with
the filtration $\left(  \sum_{n=0}^{i}\left(  \iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  \right)  ^{n}\right)  _{i\in\mathbb{N}}$ is a connected
filtered $k$-bialgebra. This proves Theorem~\ref{Theorem1313} \textbf{(c)}.

Finally, Proposition~\ref{Proposition1314} \textbf{(c)} (applied to
$H=U\left(  \mathfrak{g}\right)  $ and $V=\iota_{\mathfrak{g}}\left(
\mathfrak{g}\right)  $) yields that the $k$-bialgebra $U\left(  \mathfrak{g}%
\right)  $ is cocommutative (since $\iota_{\mathfrak{g}}\left(  \mathfrak{g}%
\right)  \subseteq\operatorname*{Prim}\left(  U\left(  \mathfrak{g}\right)
\right)  $). This proves Theorem~\ref{Theorem1313} \textbf{(e)}.

We are not going to prove Theorem~\ref{Theorem1313} \textbf{(f)} (as we are
not going to need it).
\end{proof}

Now, we can finally prove the Cartier-Milnor-Moore theorem:

\begin{proof}
[Proof of Theorem~\ref{Theorem271}.]Consider $H$ as a Lie algebra under the
commutator of the multiplication of $H$. Then, $\operatorname*{Prim}H$ is a
Lie subalgebra of $H$\ \ \ \ \footnote{since every $x\in\operatorname*{Prim}H$
and $y\in\operatorname*{Prim}H$ satisfy $xy-yx\in\operatorname*{Prim}H$
(according to Proposition~\ref{Proposition283})}.

In the following, the symbol ``$\operatorname*{id}$'' without a subscript will
always mean $\operatorname{id}_{H}$.

Let $\mathfrak{i}$ be the inclusion map $\operatorname*{Prim}H\rightarrow H$.
Clearly, $\mathfrak{i}$ is a Lie algebra homomorphism. Hence, a $k$-algebra
homomorphism $\operatorname*{Ulift}\mathfrak{i}:U\left(  \operatorname*{Prim}%
H\right)  \rightarrow H$ is well-defined (according to
Definition~\ref{Definition1303}).

Since $\mathfrak{i}$ is the inclusion map $\operatorname*{Prim}H\rightarrow
H$, we have $\mathfrak{i}\left(  \operatorname*{Prim}H\right)
=\operatorname*{Prim}H$. Thus, Proposition~\ref{Proposition1304} (applied to
$\mathfrak{g}=\operatorname*{Prim}H$ and $f=\mathfrak{i}$) yields that
$\operatorname*{Ulift}\mathfrak{i}:U\left(  \operatorname*{Prim}H\right)
\rightarrow H$ is a $k$-bialgebra homomorphism.

Let $\iota_{\operatorname*{Prim}H}:\operatorname*{Prim}H\rightarrow U\left(
\operatorname*{Prim}H\right)  $ be the canonical map from the Lie algebra
$\operatorname*{Prim}H$ into its universal enveloping algebra $U\left(
\operatorname*{Prim}H\right)  $. By the definition of $\operatorname*{Ulift}%
\mathfrak{i}$, we know that $\operatorname*{Ulift}\mathfrak{i}$ is the unique
$k$-algebra homomorphism $F:U\left(  \operatorname*{Prim}H\right)  \rightarrow
H$ satisfying $F\circ\iota_{\operatorname*{Prim}H}=\mathfrak{i}$. Hence,
$\operatorname*{Ulift}\mathfrak{i}$ is a $k$-algebra homomorphism $U\left(
\operatorname*{Prim}H\right)  \rightarrow H$ and satisfies $\left(
\operatorname*{Ulift}\mathfrak{i}\right)  \circ\iota_{\operatorname*{Prim}%
H}=\mathfrak{i}$.

\textbf{a)} The map $\operatorname*{Ulift}\mathfrak{i}$ is surjective.

\textit{Proof.} Since $\operatorname*{Ulift}\mathfrak{i}$ is a $k$-algebra
homomorphism, its image $\left(  \operatorname*{Ulift}\mathfrak{i}\right)
\left(  U\left(  \operatorname*{Prim}H\right)  \right)  $ is a $k$-subalgebra
of $H$. We have%
\[
\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \left(  \iota
_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)  \right)
=\underbrace{\left(  \left(  \operatorname*{Ulift}\mathfrak{i}\right)
\circ\iota_{\operatorname*{Prim}H}\right)  }_{=\mathfrak{i}}\left(
\operatorname*{Prim}H\right)  =\mathfrak{i}\left(  \operatorname*{Prim}%
H\right)  =\operatorname*{Prim}H.
\]
Hence,%
\[
\operatorname*{Prim}H=\left(  \operatorname*{Ulift}\mathfrak{i}\right)
\left(  \underbrace{\iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}%
H\right)  }_{\subseteq U\left(  \operatorname*{Prim}H\right)  }\right)
\subseteq\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \left(  U\left(
\operatorname*{Prim}H\right)  \right)  .
\]
Hence, $\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \left(  U\left(
\operatorname*{Prim}H\right)  \right)  $ contains $\operatorname*{Prim}H$ as a
subset. Thus, $\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \left(
U\left(  \operatorname*{Prim}H\right)  \right)  $ is a $k$-subalgebra of $H$
containing $\operatorname*{Prim}H$ as a subset (since $\left(
\operatorname*{Ulift}\mathfrak{i}\right)  \left(  U\left(
\operatorname*{Prim}H\right)  \right)  $ is a $k$-subalgebra of $H$).

Now, Theorem~\ref{Theorem270} yields%
\[
H=\operatorname{AlgGen}_{k}\left(  \operatorname*{Prim}H\right)  =\left(
\text{the }k\text{-subalgebra of }H\text{ generated by }\operatorname*{Prim}%
H\right)
\]
(by the definition of $\operatorname{AlgGen}_{k}\left(  \operatorname*{Prim}%
H\right)  $). In other words, the $k$-subalgebra of $H$ generated by
$\operatorname*{Prim}H$ is the algebra $H$.

But the $k$-subalgebra of $H$ generated by $\operatorname*{Prim}H$ is the
smallest $k$-subalgebra of $H$ containing $\operatorname*{Prim}H$ as a subset
(by the definition of the $k$-subalgebra of $H$ generated by
$\operatorname*{Prim}H$). Since the $k$-subalgebra of $H$ generated by
$\operatorname*{Prim}H$ is the algebra $H$, this rewrites as follows: The
algebra $H$ is the smallest $k$-subalgebra of $H$ containing
$\operatorname*{Prim}H$ as a subset. This means that whenever $V$ is a
$k$-subalgebra of $H$ containing $\operatorname*{Prim}H$ as a subset, we must
necessarily have $H\subseteq V$. Applying this to $V=\left(
\operatorname*{Ulift}\mathfrak{i}\right)  \left(  U\left(
\operatorname*{Prim}H\right)  \right)  $, we obtain $H\subseteq\left(
\operatorname*{Ulift}\mathfrak{i}\right)  \left(  U\left(
\operatorname*{Prim}H\right)  \right)  $ (since $\left(  \operatorname*{Ulift}%
\mathfrak{i}\right)  \left(  U\left(  \operatorname*{Prim}H\right)  \right)  $
is a $k$-subalgebra of $H$ containing $\operatorname*{Prim}H$ as a subset). In
other words, the map $\operatorname*{Ulift}\mathfrak{i}$ is surjective.

\textbf{b)} Theorem~\ref{Theorem1} yields that the map $\operatorname*{Log}%
\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $ is a projection from $H$
to the subspace $\operatorname*{Prim}H$ of all primitive elements of $H$.
Thus, $\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(  H\right)
\subseteq\operatorname*{Prim}H$. Hence, every $x\in H$ satisfies $\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  \underbrace{x}_{\in
H}\right)  \in\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(
H\right)  \subseteq\operatorname*{Prim}H$. Thus, we can define a map
$\operatorname*{eul}:H\rightarrow\operatorname*{Prim}H$ by
\[
\left(  \operatorname*{eul}x=\left(  \operatorname*{Log}\operatorname*{id}%
\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }x\in H\right)
.
\]
This map $\operatorname*{eul}$ is $k$-linear\footnote{\textit{Proof.} Let
$\lambda\in k$, $\mu\in k$, $a\in H$ and $b\in H$. By the definition of
$\operatorname*{eul}$, we have the equalities $\operatorname*{eul}\left(
\lambda a+\mu b\right)  =\left(  \operatorname*{Log}\operatorname*{id}\right)
\left(  \lambda a+\mu b\right)  $, $\operatorname*{eul}a=\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  a\right)  $ and
$\operatorname*{eul}b=\left(  \operatorname*{Log}\operatorname*{id}\right)
\left(  b\right)  $. Now,%
\begin{align*}
\operatorname*{eul}\left(  \lambda a+\mu b\right)   &  =\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  \lambda a+\mu b\right)
=\lambda\underbrace{\left(  \operatorname*{Log}\operatorname*{id}\right)
\left(  a\right)  }_{=\operatorname*{eul}a}+\mu\underbrace{\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  b\right)  }%
_{=\operatorname*{eul}b}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\operatorname*{Log}\operatorname*{id}\text{ is }k\text{-linear}\right) \\
&  =\lambda\operatorname*{eul}\left(  a\right)  +\mu\operatorname*{eul}\left(
b\right)  .
\end{align*}
\par
Now, forget that we fixed $\lambda$, $\mu$, $a$ and $b$. We thus have shown
that $\operatorname*{eul}\left(  \lambda a+\mu b\right)  =\lambda
\operatorname*{eul}\left(  a\right)  +\mu\operatorname*{eul}\left(  b\right)
$ for all $\lambda\in k$, $\mu\in k$, $a\in H$ and $b\in H$. In other words,
the map $\operatorname*{eul}$ is $k$-linear, qed.}. Moreover,
$\operatorname*{eul}\left(  1_{H}\right)  =0$\ \ \ \ \footnote{\textit{Proof.}
We have $\operatorname*{Log}F\in\mathfrak{g}\left(  H,H\right)  $ for every
$F\in G\left(  H,H\right)  $. Applying this to $F=\operatorname*{id}$, we
obtain
\[
\operatorname*{Log}\operatorname*{id}\in\mathfrak{g}\left(  H,H\right)
=\left\{  f\in\mathcal{L}\left(  H,H\right)  \ \mid\ f\left(  1_{H}\right)
=0\right\}
\]
(by the definition of $\mathfrak{g}\left(  H,H\right)  $). In other words,
$\operatorname*{Log}\operatorname*{id}$ is an element of $\mathcal{L}\left(
H,H\right)  $ and satisfies $\left(  \operatorname*{Log}\operatorname*{id}%
\right)  \left(  1_{H}\right)  =0$.
\par
Now, by the definition of $\operatorname*{eul}$, we have $\operatorname*{eul}%
\left(  1_{H}\right)  =\left(  \operatorname*{Log}\operatorname*{id}\right)
\left(  1_{H}\right)  =0$, qed.}.

Now, let $\widetilde{\iota}$ be the map $\iota_{\operatorname*{Prim}H}%
\circ\operatorname*{eul}:H\rightarrow U\left(  \operatorname*{Prim}H\right)
$. This map $\widetilde{\iota}$ is $k$-linear (since it is the composition of
the $k$-linear maps $\iota_{\operatorname*{Prim}H}$ and $\operatorname*{eul}%
$). Moreover, $\widetilde{\iota}\left(  1_{H}\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} Since $\widetilde{\iota}=\iota
_{\operatorname*{Prim}H}\circ\operatorname*{eul}$, we have $\widetilde{\iota
}\left(  1_{H}\right)  =\left(  \iota_{\operatorname*{Prim}H}\circ
\operatorname*{eul}\right)  \left(  1_{H}\right)  =\iota_{\operatorname*{Prim}%
H}\left(  \underbrace{\operatorname*{eul}\left(  1_{H}\right)  }_{=0}\right)
=\iota_{\operatorname*{Prim}H}\left(  0\right)  =0$ (since $\iota
_{\operatorname*{Prim}H}$ is $k$-linear), qed.}. Thus, $\widetilde{\iota}%
\in\mathfrak{g}\left(  H,U\left(  \operatorname*{Prim}H\right)  \right)
$\ \ \ \ \footnote{\textit{Proof.} By the definition of $\mathfrak{g}\left(
H,U\left(  \operatorname*{Prim}H\right)  \right)  $, we have $\mathfrak{g}%
\left(  H,U\left(  \operatorname*{Prim}H\right)  \right)  =\left\{
f\in\mathcal{L}\left(  H,U\left(  \operatorname*{Prim}H\right)  \right)
\ \mid\ f\left(  1_{H}\right)  =0\right\}  $.
\par
But $\widetilde{\iota}$ satisfies $\widetilde{\iota}\in\mathcal{L}\left(
H,U\left(  \operatorname*{Prim}H\right)  \right)  $ and $\widetilde{\iota
}\left(  1_{H}\right)  =0$. In other words, $\widetilde{\iota}\in\left\{
f\in\mathcal{L}\left(  H,U\left(  \operatorname*{Prim}H\right)  \right)
\ \mid\ f\left(  1_{H}\right)  =0\right\}  =\mathfrak{g}\left(  H,U\left(
\operatorname*{Prim}H\right)  \right)  $, qed.}.

\textbf{c)} Applying Theorem~\ref{Theorem1313} \textbf{(b)} to $\mathfrak{g}%
=\operatorname*{Prim}H$, we see that the family \newline$\left(  \sum
_{n=0}^{i}\left(  \iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}%
H\right)  \right)  ^{n}\right)  _{i\in\mathbb{N}}$ is a filtration of the
$k$-vector space $U\left(  \operatorname*{Prim}H\right)  $. Applying
Theorem~\ref{Theorem1313} \textbf{(c)} to $\mathfrak{g}=\operatorname*{Prim}%
H$, we see that the $k$-bialgebra $U\left(  \operatorname*{Prim}H\right)  $
endowed with the filtration $\left(  \sum_{n=0}^{i}\left(  \iota
_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)  \right)
^{n}\right)  _{i\in\mathbb{N}}$ is a connected filtered $k$-bialgebra. This
connected filtered $k$-bialgebra will be simply denoted by $U\left(
\operatorname*{Prim}H\right)  $ in the following. Thus, $\left(  U\left(
\operatorname*{Prim}H\right)  \right)  _{\leq i}=\sum_{n=0}^{i}\left(
\iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)  \right)
^{n}$ for every $i\in\mathbb{N}$.

We know that $U\left(  \operatorname*{Prim}H\right)  $ is a connected filtered
$k$-bialgebra. Hence, $U\left(  \operatorname*{Prim}H\right)  $ is a connected
filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with the same
underlying $k$-coalgebra structure.

Every $k$-bialgebra $A$ satisfies $\operatorname{id}_{A}\in G\left(
A,A\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $A$ be a $k$-bialgebra. By
the definition of $G\left(  A,A\right)  $, we have $G\left(  A,A\right)
=\left\{  f\in\mathcal{L}\left(  A,A\right)  \ \mid\ f\left(  1_{A}\right)
=1_{A}\right\}  $. Now, $\operatorname{id}_{A}$ satisfies $\operatorname{id}%
_{A}\in\mathcal{L}\left(  A,A\right)  $ and $\operatorname{id}_{A}\left(
1_{A}\right)  =1_{A}$. In other words, $\operatorname{id}_{A}\in\left\{
f\in\mathcal{L}\left(  A,A\right)  \ \mid\ f\left(  1_{A}\right)
=1_{A}\right\}  =G\left(  A,A\right)  $, qed.}. Applying this to $A=U\left(
\operatorname*{Prim}H\right)  $, we obtain $\operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\in G\left(  U\left(  \operatorname*{Prim}%
H\right)  ,U\left(  \operatorname*{Prim}H\right)  \right)  $. Hence,
$\operatorname*{Log}\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}%
H\right)  }\right)  $ is a well-defined element of $\mathfrak{g}\left(
U\left(  \operatorname*{Prim}H\right)  ,U\left(  \operatorname*{Prim}H\right)
\right)  $. Therefore,
\begin{equation}
\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  1_{U\left(
\operatorname*{Prim}H\right)  }\right)  =0 \label{271.pf.c.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{271.pf.c.5}):} We have
\begin{align*}
\operatorname*{Log}\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}%
H\right)  }\right)   &  \in\mathfrak{g}\left(  U\left(  \operatorname*{Prim}%
H\right)  ,U\left(  \operatorname*{Prim}H\right)  \right) \\
&  =\left\{  f\in\mathcal{L}\left(  U\left(  \operatorname*{Prim}H\right)
,U\left(  \operatorname*{Prim}H\right)  \right)  \ \mid\ f\left(  1_{U\left(
\operatorname*{Prim}H\right)  }\right)  =0\right\}
\end{align*}
(by the definition of $\mathfrak{g}\left(  U\left(  \operatorname*{Prim}%
H\right)  ,U\left(  \operatorname*{Prim}H\right)  \right)  $). In other words,
$\operatorname*{Log}\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}%
H\right)  }\right)  \in\mathcal{L}\left(  U\left(  \operatorname*{Prim}%
H\right)  ,U\left(  \operatorname*{Prim}H\right)  \right)  $ and $\left(
\operatorname*{Log}\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}%
H\right)  }\right)  \right)  \left(  1_{U\left(  \operatorname*{Prim}H\right)
}\right)  =0$. This proves (\ref{271.pf.c.5}).}.

\textbf{d)} We have
\begin{equation}
\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  v^{n}\right)
=\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}%
\right)  \right)  \left(  v^{n}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}v\in\iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)  \text{
and }n\in\mathbb{N}. \label{271.pf.d}%
\end{equation}


\textit{Proof of (\ref{271.pf.d}):} Let $v\in\iota_{\operatorname*{Prim}%
H}\left(  \operatorname*{Prim}H\right)  $ and $n\in\mathbb{N}$. Since
$v\in\iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)  $,
there exists a $w\in\operatorname*{Prim}H$ such that $v=\iota
_{\operatorname*{Prim}H}\left(  w\right)  $. Consider this $w$.

Theorem~\ref{Theorem1313} \textbf{(d)} (applied to $\mathfrak{g}%
=\operatorname*{Prim}H$) yields $\iota_{\operatorname*{Prim}H}\left(
\operatorname*{Prim}H\right)  \subseteq\operatorname*{Prim}\left(  U\left(
\operatorname*{Prim}H\right)  \right)  $. Thus, $v\in\iota
_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)  \subseteq
\operatorname*{Prim}\left(  U\left(  \operatorname*{Prim}H\right)  \right)  $.

Recall that $\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \circ
\iota_{\operatorname*{Prim}H}=\mathfrak{i}$. Thus, $\left(  \left(
\operatorname*{Ulift}\mathfrak{i}\right)  \circ\iota_{\operatorname*{Prim}%
H}\right)  \left(  w\right)  =\mathfrak{i}\left(  w\right)  =w$ (since
$\mathfrak{i}$ is just an inclusion map). Thus,%
\[
w=\left(  \left(  \operatorname*{Ulift}\mathfrak{i}\right)  \circ
\iota_{\operatorname*{Prim}H}\right)  \left(  w\right)  =\left(
\operatorname*{Ulift}\mathfrak{i}\right)  \left(  \underbrace{\iota
_{\operatorname*{Prim}H}\left(  w\right)  }_{=v}\right)  =\left(
\operatorname*{Ulift}\mathfrak{i}\right)  \left(  v\right)  .
\]
Now, since $\operatorname*{Ulift}\mathfrak{i}$ is a $k$-algebra homomorphism,
we have
\[
\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \left(  v^{n}\right)
=\left(  \underbrace{\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \left(
v\right)  }_{=w}\right)  ^{n}=w^{n},
\]
so that%
\begin{equation}
\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}%
\right)  \right)  \left(  v^{n}\right)  =\widetilde{\iota}\left(
\underbrace{\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \left(
v^{n}\right)  }_{=w^{n}}\right)  =\widetilde{\iota}\left(  w^{n}\right)  .
\label{271.pf.d.pf.1}%
\end{equation}


Now, we distinguish between three cases:

\textit{Case 1:} We have $n=0$.

\textit{Case 2:} We have $n=1$.

\textit{Case 3:} We have neither $n=0$ nor $n=1$.

Let us first consider Case 1. In this case, we have $n=0$. Thus, $v^{n}%
=v^{0}=1_{U\left(  \operatorname*{Prim}H\right)  }$, so that
\[
\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  v^{n}\right)
=\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  1_{U\left(
\operatorname*{Prim}H\right)  }\right)  =0
\]
(by (\ref{271.pf.c.5})). Comparing this with%
\begin{align*}
\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}%
\right)  \right)  \left(  v^{n}\right)   &  =\widetilde{\iota}\left(
w^{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{271.pf.d.pf.1}%
)}\right) \\
&  =\widetilde{\iota}\left(  1_{H}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }n=0\text{, so that }w^{n}=w^{0}=1_{H}\right) \\
&  =0,
\end{align*}
we obtain $\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  v^{n}\right)
=\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}%
\right)  \right)  \left(  v^{n}\right)  $. Hence, (\ref{271.pf.d}) is proven
in Case 1.

Let us now consider Case 2. In this case, we have $n=1$. Thus, $v^{n}=v^{1}=v$.

Now, recall that $\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)
}\in G\left(  U\left(  \operatorname*{Prim}H\right)  ,U\left(
\operatorname*{Prim}H\right)  \right)  $. Hence,
Proposition~\ref{Proposition19} \textbf{(c)} (applied to $U\left(
\operatorname*{Prim}H\right)  $, $U\left(  \operatorname*{Prim}H\right)  $ and
$\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }$ instead of $H$,
$A$ and $F$) yields that
\begin{equation}
\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \mid_{\operatorname*{Prim}%
\left(  U\left(  \operatorname*{Prim}H\right)  \right)  }=\operatorname{id}%
_{U\left(  \operatorname*{Prim}H\right)  }\mid_{\operatorname*{Prim}\left(
U\left(  \operatorname*{Prim}H\right)  \right)  }. \label{271.pf.d.pf.2}%
\end{equation}


On the other hand, recall that every $k$-bialgebra $A$ satisfies
$\operatorname{id}_{A}\in G\left(  A,A\right)  $. Applying this to $A=H$, we
obtain $\operatorname{id}_{H}\in G\left(  H,H\right)  $. In other words,
$\operatorname*{id}\in G\left(  H,H\right)  $. Thus,
Proposition~\ref{Proposition19} \textbf{(c)} (applied to $H$ and
$\operatorname*{id}$ instead of $A$ and $F$) yields that $\left(
\operatorname*{Log}\operatorname*{id}\right)  \mid_{\operatorname*{Prim}%
H}=\operatorname*{id}\mid_{\operatorname*{Prim}H}$.

Now, we have%
\begin{align*}
\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  \underbrace{v^{n}%
}_{=v}\right)   &  =\left(  \operatorname*{Log}\left(  \operatorname{id}%
_{U\left(  \operatorname*{Prim}H\right)  }\right)  \right)  \left(  v\right)
\\
&  =\left(  \underbrace{\left(  \operatorname*{Log}\left(  \operatorname{id}%
_{U\left(  \operatorname*{Prim}H\right)  }\right)  \right)  \mid
_{\operatorname*{Prim}\left(  U\left(  \operatorname*{Prim}H\right)  \right)
}}_{\substack{=\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }%
\mid_{\operatorname*{Prim}\left(  U\left(  \operatorname*{Prim}H\right)
\right)  }\\\text{(by (\ref{271.pf.d.pf.2}))}}}\right)  \left(  v\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }v\in\operatorname*{Prim}\left(
U\left(  \operatorname*{Prim}H\right)  \right)  \right) \\
&  =\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }%
\mid_{\operatorname*{Prim}\left(  U\left(  \operatorname*{Prim}H\right)
\right)  }\right)  \left(  v\right)  =\operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\left(  v\right)  =v.
\end{align*}
Comparing this with%
\begin{align*}
\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}%
\right)  \right)  \left(  v^{n}\right)   &  =\widetilde{\iota}\left(
w^{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{271.pf.d.pf.1}%
)}\right) \\
&  =\widetilde{\iota}\left(  w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}n=1\text{, so that }w^{n}=w^{1}=w\right) \\
&  =\left(  \iota_{\operatorname*{Prim}H}\circ\operatorname*{eul}\right)
\left(  w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widetilde{\iota
}=\iota_{\operatorname*{Prim}H}\circ\operatorname*{eul}\right) \\
&  =\iota_{\operatorname*{Prim}H}\left(  \underbrace{\operatorname*{eul}%
w}_{\substack{=\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(
w\right)  \\\text{(by the definition of }\operatorname*{eul}\text{)}}}\right)
=\iota_{\operatorname*{Prim}H}\left(  \underbrace{\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  w\right)  }_{\substack{=\left(  \left(
\operatorname*{Log}\operatorname*{id}\right)  \mid_{\operatorname*{Prim}%
H}\right)  \left(  w\right)  \\\text{(since }w\in\operatorname*{Prim}%
H\text{)}}}\right) \\
&  =\iota_{\operatorname*{Prim}H}\left(  \underbrace{\left(  \left(
\operatorname*{Log}\operatorname*{id}\right)  \mid_{\operatorname*{Prim}%
H}\right)  }_{=\operatorname*{id}\mid_{\operatorname*{Prim}H}}\left(
w\right)  \right)  =\iota_{\operatorname*{Prim}H}\left(  \underbrace{\left(
\operatorname*{id}\mid_{\operatorname*{Prim}H}\right)  \left(  w\right)
}_{=\operatorname*{id}\left(  w\right)  =w}\right) \\
&  =\iota_{\operatorname*{Prim}H}\left(  w\right)  =v,
\end{align*}
we obtain $\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  v^{n}\right)
=\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}%
\right)  \right)  \left(  v^{n}\right)  $. Hence, (\ref{271.pf.d}) is proven
in Case 2.

Let us finally consider Case 3. In this case, we have neither $n=0$ nor $n=1$.
Since $n\in\mathbb{N}$, we must thus have $n>1$. We have $v\in
\operatorname*{Prim}\left(  U\left(  \operatorname*{Prim}H\right)  \right)  $.
Thus, Theorem~\ref{Theorem1201} (applied to $U\left(  \operatorname*{Prim}%
H\right)  $ instead of $H$) yields%
\[
\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  v^{n}\right)  =0.
\]
On the other hand, $w\in\operatorname*{Prim}H$. Hence,
Theorem~\ref{Theorem1201} (applied to $w$ instead of $v$) yields $\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  w^{n}\right)  =0$. Now,
comparing%
\[
\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  v^{n}\right)  =0
\]
with%
\begin{align*}
\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}%
\right)  \right)  \left(  v^{n}\right)   &  =\widetilde{\iota}\left(
w^{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{271.pf.d.pf.1}%
)}\right) \\
&  =\left(  \iota_{\operatorname*{Prim}H}\circ\operatorname*{eul}\right)
\left(  w^{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\widetilde{\iota}=\iota_{\operatorname*{Prim}H}\circ\operatorname*{eul}\right)
\\
&  =\iota_{\operatorname*{Prim}H}\left(  \underbrace{\operatorname*{eul}%
\left(  w^{n}\right)  }_{\substack{=\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  w^{n}\right)  \\\text{(by the definition of
}\operatorname*{eul}\text{)}}}\right)  =\iota_{\operatorname*{Prim}H}\left(
\underbrace{\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(
w^{n}\right)  }_{\substack{=0}}\right) \\
&  =\iota_{\operatorname*{Prim}H}\left(  0\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\iota_{\operatorname*{Prim}H}\text{
is }k\text{-linear}\right)  ,
\end{align*}
we obtain $\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  v^{n}\right)
=\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}%
\right)  \right)  \left(  v^{n}\right)  $. Hence, (\ref{271.pf.d}) is proven
in Case 3.

We have thus proven (\ref{271.pf.d}) in each of the three cases 1, 2 and 3.
Since these three cases cover all possibilities, this yields that
(\ref{271.pf.d}) always holds. The proof of (\ref{271.pf.d}) is thus complete.

\textbf{e)} We have%
\begin{equation}
\operatorname*{Log}\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}%
H\right)  }\right)  =\widetilde{\iota}\circ\left(  \operatorname*{Ulift}%
\mathfrak{i}\right)  . \label{271.pf.e}%
\end{equation}


\textit{Proof of (\ref{271.pf.e}):} Let $V$ be the $k$-vector subspace
$\iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)  $ of
$U\left(  \operatorname*{Prim}H\right)  $. Then, $xy-yx\in V$ for any $x\in V$
and $y\in V$\ \ \ \ \footnote{\textit{Proof.} We know that $\iota
_{\operatorname*{Prim}H}$ is the canonical map from the Lie algebra
$\operatorname*{Prim}H$ to the universal enveloping algebra $U\left(
\operatorname*{Prim}H\right)  $. Hence, $\iota_{\operatorname*{Prim}H}$ is a
Lie algebra homomorphism (where the Lie algebra structure on $U\left(
\operatorname*{Prim}H\right)  $ is given by the commutator of the
multiplication). Thus, its image $\iota_{\operatorname*{Prim}H}\left(
\operatorname*{Prim}H\right)  $ is a Lie subalgebra of $U\left(
\operatorname*{Prim}H\right)  $. Since $\iota_{\operatorname*{Prim}H}\left(
\operatorname*{Prim}H\right)  =V$, this rewrites as follows: The set $V$ is a
Lie subalgebra of $U\left(  \operatorname*{Prim}H\right)  $. In other words,
we have $\left[  x,y\right]  \in V$ for any $x\in V$ and $y\in V$. Since
$\left[  x,y\right]  =xy-yx$ for any $x\in V$ and $y\in V$, this rewrites as
follows: We have $xy-yx\in V$ for any $x\in V$ and $y\in V$, qed.}. Moreover,
\begin{align*}
U\left(  \operatorname*{Prim}H\right)   &  =\sum_{n\in\mathbb{N}}\left(
\underbrace{\iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)
}_{=V}\right)  ^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Theorem~\ref{Theorem1313} \textbf{(a)}, applied to }\mathfrak{g}%
=\operatorname*{Prim}H\right) \\
&  =\sum_{n\in\mathbb{N}}V^{n}.
\end{align*}
Finally, we have $\left(  \operatorname*{Log}\left(  \operatorname{id}%
_{U\left(  \operatorname*{Prim}H\right)  }\right)  \right)  \left(
v^{n}\right)  =\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}%
\mathfrak{i}\right)  \right)  \left(  v^{n}\right)  $ for every $v\in V$ and
$n\in\mathbb{N}$\ \ \ \ \footnote{\textit{Proof.} Let $v\in V$ and
$n\in\mathbb{N}$. Then, $v\in V=\iota_{\operatorname*{Prim}H}\left(
\operatorname*{Prim}H\right)  $. Hence, (\ref{271.pf.d}) yields $\left(
\operatorname*{Log}\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}%
H\right)  }\right)  \right)  \left(  v^{n}\right)  =\left(  \widetilde{\iota
}\circ\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \right)  \left(
v^{n}\right)  $, qed.}. Consequently, we can apply
Corollary~\ref{Corollary1147} to $U\left(  \operatorname*{Prim}H\right)  $,
$U\left(  \operatorname*{Prim}H\right)  $, $\operatorname*{Log}\left(
\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }\right)  $ and
$\widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}\right)  $
instead of $A$, $W$, $f$ and $g$. As a result, we obtain $\operatorname*{Log}%
\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }\right)
=\widetilde{\iota}\circ\left(  \operatorname*{Ulift}\mathfrak{i}\right)  $.
This proves (\ref{271.pf.e}).

\textbf{f)} We have%
\begin{equation}
\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }=e^{\ast
\widetilde{\iota}}\circ\left(  \operatorname*{Ulift}\mathfrak{i}\right)  .
\label{271.pf.f}%
\end{equation}


\textit{Proof of (\ref{271.pf.f}):} We know that $\operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\in G\left(  U\left(  \operatorname*{Prim}%
H\right)  ,U\left(  \operatorname*{Prim}H\right)  \right)  $. Hence,
Proposition~\ref{Proposition13} \textbf{(b)} (applied to $U\left(
\operatorname*{Prim}H\right)  $, $U\left(  \operatorname*{Prim}H\right)  $ and
$\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }$ instead of $H$,
$A$ and $F$) yields $e^{\ast\left(  \operatorname*{Log}\left(
\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }\right)  \right)
}=\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }$. Thus,%
\[
\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }=e^{\ast\left(
\operatorname*{Log}\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}%
H\right)  }\right)  \right)  }=e^{\ast\left(  \widetilde{\iota}\circ\left(
\operatorname*{Ulift}\mathfrak{i}\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{271.pf.e})}\right)  .
\]
But recall that $\operatorname*{Ulift}\mathfrak{i}$ is a $k$-bialgebra
homomorphism. Hence, $\operatorname*{Ulift}\mathfrak{i}$ is a $k$-coalgebra
homomorphism. Also, $\left(  \operatorname*{Ulift}\mathfrak{i}\right)  \left(
1_{U\left(  \operatorname*{Prim}H\right)  }\right)  =1_{H}$ (since
$\operatorname*{Ulift}\mathfrak{i}$ is a $k$-algebra homomorphism). Moreover,
$\widetilde{\iota}\in\mathfrak{g}\left(  H,U\left(  \operatorname*{Prim}%
H\right)  \right)  $. Hence, Proposition~\ref{Proposition1101} \textbf{(d)}
(applied to $U\left(  \operatorname*{Prim}H\right)  $, $H$, $U\left(
\operatorname*{Prim}H\right)  $, $\operatorname*{Ulift}\mathfrak{i}$ and
$\widetilde{\iota}$ instead of $D$, $C$, $A$, $\varphi$ and $f$) yields
$e^{\ast\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}%
\mathfrak{i}\right)  \right)  }=e^{\ast\widetilde{\iota}}\circ\left(
\operatorname*{Ulift}\mathfrak{i}\right)  $.

Thus, $\operatorname{id}_{U\left(  \operatorname*{Prim}H\right)  }%
=e^{\ast\left(  \widetilde{\iota}\circ\left(  \operatorname*{Ulift}%
\mathfrak{i}\right)  \right)  }=e^{\ast\widetilde{\iota}}\circ\left(
\operatorname*{Ulift}\mathfrak{i}\right)  $. This proves (\ref{271.pf.f}).

\textbf{g)} Due to (\ref{271.pf.f}), we have $e^{\ast\widetilde{\iota}}%
\circ\left(  \operatorname*{Ulift}\mathfrak{i}\right)  =\operatorname{id}%
_{U\left(  \operatorname*{Prim}H\right)  }$. Hence, the map
$\operatorname*{Ulift}\mathfrak{i}$ has a left inverse. Thus,
$\operatorname*{Ulift}\mathfrak{i}$ is injective. Combining this with the
(already proven) fact that $\operatorname*{Ulift}\mathfrak{i}$ is surjective,
we conclude that $\operatorname*{Ulift}\mathfrak{i}$ is bijective. Since
$\operatorname*{Ulift}\mathfrak{i}$ is a $k$-linear map, this yields that
$\operatorname*{Ulift}\mathfrak{i}$ is a $k$-vector space isomorphism (because
every bijective $k$-linear map is a $k$-vector space isomorphism). Thus, the
map $\operatorname*{Ulift}\mathfrak{i}$ is invertible. Since
$\operatorname*{Ulift}\mathfrak{i}$ is a $k$-bialgebra homomorphism, this
yields that
\begin{equation}
\operatorname*{Ulift}\mathfrak{i}\text{ is a }k\text{-bialgebra isomorphism}
\label{271.pf.g}%
\end{equation}
(because every invertible $k$-bialgebra homomorphism must be a $k$-bialgebra
isomorphism (by Proposition~\ref{Proposition1318})).

Now, recall that the map $\operatorname*{Ulift}\mathfrak{i}$ is the canonical
$k$-algebra homomorphism $U\left(  \operatorname*{Prim}H\right)  \rightarrow
H$ obtained from the Lie algebra homomorphism $\mathfrak{i}$ via the universal
property of the universal enveloping algebra. Since the Lie algebra
homomorphism $\mathfrak{i}$ is the inclusion map $\operatorname*{Prim}%
H\rightarrow H$, this rewrites as follows: The map $\operatorname*{Ulift}%
\mathfrak{i}$ is the canonical $k$-algebra homomorphism $U\left(
\operatorname*{Prim}H\right)  \rightarrow H$ obtained from the inclusion map
$\operatorname*{Prim}H\rightarrow H$ via the universal property of the
universal enveloping algebra. Hence, (\ref{271.pf.g}) rewrites as follows: The
canonical $k$-algebra homomorphism $U\left(  \operatorname*{Prim}H\right)
\rightarrow H$ obtained from the inclusion map $\operatorname*{Prim}%
H\rightarrow H$ via the universal property of the universal enveloping algebra
is a $k$-bialgebra isomorphism. Theorem~\ref{Theorem271} is thus proven.
\end{proof}

Our above proof of Theorem~\ref{Theorem271} allows for a corollary:

\begin{corollary}
\label{Corollary1321} Let $k$ be a field of characteristic $0$, and let $H$ be
a connected filtered cocommutative bialgebra over $k$. Let
$\operatorname*{Prim}H$ denote the set of all primitive elements of $H$. For
every Lie algebra $\mathfrak{g}$, let $U\left(  \mathfrak{g}\right)  $ denote
the universal enveloping algebra of $\mathfrak{g}$. Consider $U\left(
\operatorname*{Prim}H\right)  $ as a $k$-bialgebra. Let $\iota
_{\operatorname*{Prim}H}:\operatorname*{Prim}H\rightarrow U\left(
\operatorname*{Prim}H\right)  $ be the canonical map from the Lie algebra
$\operatorname*{Prim}H$ into its universal enveloping algebra $U\left(
\operatorname*{Prim}H\right)  $. Then, $\operatorname*{Prim}\left(  U\left(
\operatorname*{Prim}H\right)  \right)  =\iota_{\operatorname*{Prim}H}\left(
\operatorname*{Prim}H\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1321}.]Let us work with the notations
introduced in the proof of Theorem~\ref{Theorem271}. The connected filtered
$k$-bialgebra $U\left(  \operatorname*{Prim}H\right)  $ is cocommutative (by
Theorem~\ref{Theorem1313} \textbf{(e)}, applied to $\mathfrak{g}%
=\operatorname*{Prim}H$). Thus, Theorem~\ref{Theorem1} (applied to $U\left(
\operatorname*{Prim}H\right)  $ instead of $H$) yields that the map
$\operatorname*{Log}\left(  \operatorname{id}_{U\left(  \operatorname*{Prim}%
H\right)  }\right)  \in\mathcal{L}\left(  U\left(  \operatorname*{Prim}%
H\right)  ,U\left(  \operatorname*{Prim}H\right)  \right)  $ is a projection
from $U\left(  \operatorname*{Prim}H\right)  $ to the subspace
$\operatorname*{Prim}\left(  U\left(  \operatorname*{Prim}H\right)  \right)  $
of all primitive elements of $U\left(  \operatorname*{Prim}H\right)  $. Hence,
$\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  \left(  U\left(
\operatorname*{Prim}H\right)  \right)  =\operatorname*{Prim}\left(  U\left(
\operatorname*{Prim}H\right)  \right)  $. Thus,%
\begin{align}
\operatorname*{Prim}\left(  U\left(  \operatorname*{Prim}H\right)  \right)
&  =\underbrace{\left(  \operatorname*{Log}\left(  \operatorname{id}_{U\left(
\operatorname*{Prim}H\right)  }\right)  \right)  }%
_{\substack{=\widetilde{\iota}\circ\left(  \operatorname*{Ulift}%
\mathfrak{i}\right)  \\\text{(by (\ref{271.pf.e}))}}}\left(  U\left(
\operatorname*{Prim}H\right)  \right)  =\left(  \widetilde{\iota}\circ\left(
\operatorname*{Ulift}\mathfrak{i}\right)  \right)  \left(  U\left(
\operatorname*{Prim}H\right)  \right) \nonumber\\
&  =\widetilde{\iota}\left(  \underbrace{\left(  \operatorname*{Ulift}%
\mathfrak{i}\right)  \left(  U\left(  \operatorname*{Prim}H\right)  \right)
}_{\subseteq H}\right)  \subseteq\underbrace{\widetilde{\iota}}_{=\iota
_{\operatorname*{Prim}H}\circ\operatorname*{eul}}\left(  H\right)  =\left(
\iota_{\operatorname*{Prim}H}\circ\operatorname*{eul}\right)  \left(  H\right)
\nonumber\\
&  =\iota_{\operatorname*{Prim}H}\left(  \underbrace{\operatorname*{eul}%
\left(  H\right)  }_{\subseteq\operatorname*{Prim}H}\right)  \subseteq
\iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}H\right)  .
\label{1321.pf.1}%
\end{align}
But Theorem~\ref{Theorem1313} \textbf{(d)} (applied to $\mathfrak{g}%
=\operatorname*{Prim}H$) yields $\iota_{\operatorname*{Prim}H}\left(
\operatorname*{Prim}H\right)  \subseteq\operatorname*{Prim}\left(  U\left(
\operatorname*{Prim}H\right)  \right)  $. Combining this with (\ref{1321.pf.1}%
), we obtain $\operatorname*{Prim}\left(  U\left(  \operatorname*{Prim}%
H\right)  \right)  =\iota_{\operatorname*{Prim}H}\left(  \operatorname*{Prim}%
H\right)  $. This proves Corollary~\ref{Corollary1321}.
\end{proof}

Notice that Corollary~\ref{Corollary1321} is merely a particular case of the
following fact:

\begin{corollary}
\label{Corollary1322} Let $k$ be a field of characteristic $0$, and let
$\mathfrak{g}$ be a $k$-Lie algebra. Consider the universal enveloping algebra
$U\left(  \mathfrak{g}\right)  $ as a $k$-bialgebra. Let $\iota_{\mathfrak{g}%
}:\mathfrak{g}\rightarrow U\left(  \mathfrak{g}\right)  $ be the canonical map
from the Lie algebra $\mathfrak{g}$ into its universal enveloping algebra
$U\left(  \mathfrak{g}\right)  $. Then, $\operatorname*{Prim}\left(  U\left(
\mathfrak{g}\right)  \right)  =\iota_{\mathfrak{g}}\left(  \mathfrak{g}%
\right)  $.
\end{corollary}

However, Corollary~\ref{Corollary1322} is not susceptible to the attack we
have led on Corollary~\ref{Corollary1321}; it is commonly proven using the
Poincar\'{e}-Birkhoff-Witt theorem instead.

Corollary~\ref{Corollary1321} and Corollary~\ref{Corollary1322} become false
if the hypothesis that $k$ have characteristic $0$ is lifted. (If $k$ is a
field of characteristic $p$, then the $k$-vector subspace
$\operatorname*{Prim}\left(  k\left[  X\right]  \right)  $ of $k\left[
X\right]  $ is spanned by $X,X^{p},X^{p^{2}},X^{p^{3}},\ldots$.)

\section{\label{S32}Maps in $\mathfrak{g}\left(  H,A\right)  $ and products of
primitives}

We next show some properties of products of primitives with respect to maps in
$\mathfrak{g}\left(  H,A\right)  $ (that is, linear maps from a $k$-bialgebra
$H$ to a $k$-algebra $A$ that annihilate $1_{H}$).

\begin{theorem}
\label{Theorem1401} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be
a $k$-algebra. Let $r\in\mathbb{N}$. Let $f_{1}$, $f_{2}$, $\ldots$, $f_{r}$
be $r$ maps in $\mathfrak{g}\left(  H,A\right)  $.

\textbf{(a)} Every $s\in\left\{  0,1,\ldots,r-1\right\}  $ and every $s$
elements $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ of $\operatorname*{Prim}H$
satisfy%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a_{1}a_{2}\cdots
a_{s}\right)  =0.
\]


\textbf{(b)} Every $r$ elements $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ of
$\operatorname*{Prim}H$ satisfy%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a_{1}a_{2}\cdots
a_{r}\right)  =\sum_{\sigma\in S_{r}}f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{r}\left(  a_{\sigma\left(  r\right)  }\right)
\]
(where $S_{r}$ denotes the $r$-th symmetric group).
\end{theorem}

Before we prove this theorem, let us show some simple facts:

\begin{lemma}
\label{Lemma1402} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $V$ be a
$k$-vector subspace of $\operatorname*{Prim}H$. Let $s\in\mathbb{N}$. Let
$a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$ elements of $V$. Then,%
\begin{align*}
\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)   &  \in\sum_{m=1}^{s}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{s}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{s}\right)
+\sum_{\ell=0}^{s-2}H\otimes V^{\ell}.
\end{align*}
\footnote{Recall that $V^{\ell}$ is defined according to
Convention~\ref{Convention100}. Hence, $V^{\ell}$ means the $\ell$-th power of
the subspace $V$ of the $k$-algebra $H$.}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1402}.]We will prove that every $r\in\left\{
0,1,\ldots,s\right\}  $ satisfies%
\begin{align}
\Delta\left(  a_{1}a_{2}\cdots a_{r}\right)   &  \in\sum_{m=1}^{r}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{r}\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{r}\right)
+\sum_{\ell=0}^{r-2}H\otimes V^{\ell}. \label{1402.pf.goal}%
\end{align}


\textit{Proof of (\ref{1402.pf.goal}):} We will prove (\ref{1402.pf.goal}) by
induction over $r$:

\textit{Induction base:} Let $r=0$. Then, $a_{1}a_{2}\cdots a_{r}=a_{1}%
a_{2}\cdots a_{0}=\left(  \text{empty product}\right)  =1_{H}$. Applying the
map $\Delta$ to this equality, we obtain%
\begin{align*}
\Delta\left(  a_{1}a_{2}\cdots a_{r}\right)   &  =\Delta\left(  1_{H}\right)
=1_{H}\otimes\underbrace{1_{H}}_{=a_{1}a_{2}\cdots a_{r}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the axioms of a bialgebra}\right) \\
&  \in1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{r}\right) \\
&  \subseteq\sum_{m=1}^{r}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{r}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{r}\right)
+\sum_{\ell=0}^{r-2}H\otimes V^{\ell}.
\end{align*}
Thus, (\ref{1402.pf.goal}) is proven in the case when $r=0$. The induction
base is hence complete.

\textit{Induction step:} Let $R\in\left\{  0,1,\ldots,s-1\right\}  $. Assume
that (\ref{1402.pf.goal}) has been proven for $r=R$. We need to prove
(\ref{1402.pf.goal}) for $r=R+1$.

We know that (\ref{1402.pf.goal}) has been proven for $r=R$. In other words,%
\begin{align}
\Delta\left(  a_{1}a_{2}\cdots a_{R}\right)   &  \in\sum_{m=1}^{R}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{R}\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)
+\sum_{\ell=0}^{R-2}H\otimes V^{\ell}. \label{1402.pf.goal.pf.indhyp}%
\end{align}


We know that $a_{R+1}\in V\subseteq\operatorname*{Prim}H=\left(  \text{the set
of all primitive elements of }H\right)  $. In other words, $a_{R+1}$ is a
primitive element of $H$. Hence, $\Delta\left(  a_{R+1}\right)  =a_{R+1}%
\otimes1_{H}+1_{H}\otimes a_{R+1}$ (by the definition of ``primitive'').

It is easy to see that%
\begin{equation}
\left(  \sum_{\ell=0}^{R-2}H\otimes V^{\ell}\right)  \cdot\left(
a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}\right)  \subseteq\sum_{\ell=0}%
^{R-1}H\otimes V^{\ell} \label{1402.pf.goal.pf.term1}%
\end{equation}
\footnote{\textit{Proof of (\ref{1402.pf.goal.pf.term1}):} Let $\ell
\in\left\{  0,1,\ldots,R-2\right\}  $. Then, $\ell\leq R-2$. Adding $1$ to
this inequality, we obtain $\ell+1\leq\left(  R-2\right)  +1=R-1$.
\par
Since $\ell\leq R-2<R-1$, we know that $H\otimes V^{\ell}$ is an addend in the
sum $\sum_{h=0}^{R-1}H\otimes V^{h}$. Hence, $H\otimes V^{\ell}\subseteq
\sum_{h=0}^{R-1}H\otimes V^{h}$.
\par
Since $\ell+1\leq R-1$, we know that $H\otimes V^{\ell+1}$ is an addend in the
sum $\sum_{h=0}^{R-1}H\otimes V^{h}$. Hence, $H\otimes V^{\ell+1}\subseteq
\sum_{h=0}^{R-1}H\otimes V^{h}$.
\par
We are now going to prove that%
\begin{equation}
\left(  H\otimes V^{\ell}\right)  \cdot\left(  a_{R+1}\otimes1_{H}%
+1_{H}\otimes a_{R+1}\right)  \subseteq\sum_{h=0}^{R-1}H\otimes V^{h}.
\label{1402.pf.goal.pf.term1.pf.1}%
\end{equation}
\par
Indeed, let $w\in\left(  H\otimes V^{\ell}\right)  \cdot\left(  a_{R+1}%
\otimes1_{H}+1_{H}\otimes a_{R+1}\right)  $. Thus, there exists an $x\in
H\otimes V^{\ell}$ such that%
\[
w=x\cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}\right)  .
\]
Consider this $x$. Since $x$ is a tensor in $H\otimes V^{\ell}$, we can write
$x$ as $x=\sum_{j=1}^{J}\lambda_{j}h_{j}\otimes y_{j}$ for some $J\in
\mathbb{N}$, some elements $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{J}$ of $k$, some elements $h_{1}$, $h_{2}$, $\ldots$, $h_{J}$ of
$H$, and some elements $y_{1}$, $y_{2}$, $\ldots$, $y_{J}$ of $V^{\ell}$.
Consider this $J$, these $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{J}%
$, these $h_{1}$, $h_{2}$, $\ldots$, $h_{J}$, and these $y_{1}$, $y_{2}$,
$\ldots$, $y_{J}$. We have%
\begin{align*}
w  &  =\underbrace{x}_{=\sum_{j=1}^{J}\lambda_{j}h_{j}\otimes y_{j}}%
\cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}\right) \\
&  =\left(  \sum_{j=1}^{J}\lambda_{j}h_{j}\otimes y_{j}\right)  \cdot\left(
a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}\right) \\
&  =\sum_{j=1}^{J}\lambda_{j}\underbrace{\left(  h_{j}\otimes y_{j}\right)
\cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}\right)  }_{=\left(
h_{j}\otimes y_{j}\right)  \cdot\left(  a_{R+1}\otimes1_{H}\right)  +\left(
h_{j}\otimes y_{j}\right)  \cdot\left(  1_{H}\otimes a_{R+1}\right)  }\\
&  =\sum_{j=1}^{J}\lambda_{j}\left(  \underbrace{\left(  h_{j}\otimes
y_{j}\right)  \cdot\left(  a_{R+1}\otimes1_{H}\right)  }_{=\left(
h_{j}a_{R+1}\right)  \otimes\left(  y_{j}1_{H}\right)  }+\underbrace{\left(
h_{j}\otimes y_{j}\right)  \cdot\left(  1_{H}\otimes a_{R+1}\right)
}_{=\left(  h_{j}1_{H}\right)  \otimes\left(  y_{j}a_{R+1}\right)  }\right) \\
&  =\sum_{j=1}^{J}\lambda_{j}\left(  \underbrace{\left(  h_{j}a_{R+1}\right)
}_{\in H}\otimes\underbrace{\left(  y_{j}1_{H}\right)  }_{=y_{j}\in V^{\ell}%
}+\underbrace{\left(  h_{j}1_{H}\right)  }_{\in H}\otimes\left(
\underbrace{y_{j}}_{\in V^{\ell}}\underbrace{a_{R+1}}_{\in V}\right)  \right)
\\
&  \in\sum_{j=1}^{J}\lambda_{j}\left(  H\otimes V^{\ell}+H\otimes
\underbrace{\left(  V^{\ell}V\right)  }_{=V^{\ell+1}}\right)  =\sum_{j=1}%
^{J}\lambda_{j}\left(  H\otimes V^{\ell}+H\otimes V^{\ell+1}\right) \\
&  \subseteq\underbrace{H\otimes V^{\ell}}_{\subseteq\sum_{h=0}^{R-1}H\otimes
V^{h}}+\underbrace{H\otimes V^{\ell+1}}_{\subseteq\sum_{h=0}^{R-1}H\otimes
V^{h}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }H\otimes V^{\ell}+H\otimes
V^{\ell+1}\text{ is a }k\text{-vector space}\right) \\
&  \subseteq\sum_{h=0}^{R-1}H\otimes V^{h}+\sum_{h=0}^{R-1}H\otimes
V^{h}\subseteq\sum_{h=0}^{R-1}H\otimes V^{h}%
\end{align*}
(since $\sum_{h=0}^{R-1}H\otimes V^{h}$ is a $k$-vector space).
\par
Now, forget that we fixed $w$. We thus have proven that every $w\in\left(
H\otimes V^{\ell}\right)  \cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes
a_{R+1}\right)  $ satisfies $w\in\sum_{h=0}^{R-1}H\otimes V^{h}$. In other
words,%
\[
\left(  H\otimes V^{\ell}\right)  \cdot\left(  a_{R+1}\otimes1_{H}%
+1_{H}\otimes a_{R+1}\right)  \subseteq\sum_{h=0}^{R-1}H\otimes V^{h}.
\]
In other words, (\ref{1402.pf.goal.pf.term1.pf.1}) is proven.
\par
Now, forget that we fixed $\ell$. We thus have shown that
(\ref{1402.pf.goal.pf.term1.pf.1}) holds for every $\ell\in\left\{
0,1,\ldots,R-2\right\}  $. Now,%
\begin{align*}
&  \left(  \sum_{\ell=0}^{R-2}H\otimes V^{\ell}\right)  \cdot\left(
a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}\right) \\
&  =\sum_{\ell=0}^{R-2}\underbrace{\left(  H\otimes V^{\ell}\right)
\cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}\right)  }%
_{\substack{\subseteq\sum_{h=0}^{R-1}H\otimes V^{h}\\\text{(by
(\ref{1402.pf.goal.pf.term1.pf.1}))}}}\\
&  \subseteq\sum_{\ell=0}^{R-2}\sum_{h=0}^{R-1}H\otimes V^{h}\subseteq
\sum_{h=0}^{R-1}H\otimes V^{h}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\sum_{h=0}^{R-1}H\otimes V^{h}\text{ is a }k\text{-vector space}\right) \\
&  =\sum_{\ell=0}^{R-1}H\otimes V^{\ell}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we renamed the summation index }h\text{ as }\ell\right)  .
\end{align*}
Thus, (\ref{1402.pf.goal.pf.term1}) is proven.}.

It is also easy to find that%
\begin{align}
&  \sum_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}%
\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right) \nonumber\\
&  =\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)  +a_{R+1}\otimes\left(
a_{1}a_{2}\cdots a_{R}\right)  . \label{1402.pf.goal.pf.term2}%
\end{align}
\footnote{\textit{Proof of (\ref{1402.pf.goal.pf.term2}):} We have%
\begin{align*}
&  \sum_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}%
\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right) \\
&  =\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +a_{R+1}\otimes\left(  \underbrace{\left(  a_{1}%
a_{2}\cdots a_{\left(  R+1\right)  -1}\right)  }_{=a_{1}a_{2}\cdots a_{R}%
}\underbrace{\left(  a_{\left(  R+1\right)  +1}a_{\left(  R+1\right)
+2}\cdots a_{R+1}\right)  }_{=\left(  \text{empty product}\right)  =1_{H}%
}\right) \\
&  =\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)  +a_{R+1}\otimes
\underbrace{\left(  \left(  a_{1}a_{2}\cdots a_{R}\right)  1_{H}\right)
}_{=a_{1}a_{2}\cdots a_{R}}\\
&  =\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)  +a_{R+1}\otimes\left(
a_{1}a_{2}\cdots a_{R}\right)  .
\end{align*}
This proves (\ref{1402.pf.goal.pf.term2}).}

Also, notice that%
\begin{equation}
\sum_{m=1}^{R}H\otimes V^{R-1}+\sum_{\ell=0}^{R-1}H\otimes V^{\ell}%
\subseteq\sum_{\ell=0}^{\left(  R+1\right)  -2}H\otimes V^{\ell}
\label{1402.pf.goal.pf.term3}%
\end{equation}
\footnote{\textit{Proof of (\ref{1402.pf.goal.pf.term3}):} If $R=0$, then
$\sum_{m=1}^{R}H\otimes V^{R-1}=\left(  \text{empty sum}\right)  =0$. Hence,
if $R=0$, then $\underbrace{\sum_{m=1}^{R}H\otimes V^{R-1}}_{=0}+\sum_{\ell
=0}^{R-1}H\otimes V^{\ell}=\sum_{\ell=0}^{R-1}H\otimes V^{\ell}=\sum_{\ell
=0}^{\left(  R+1\right)  -2}H\otimes V^{\ell}$ (since $R-1=\left(  R+1\right)
-2$). Thus, if $R=0$, then (\ref{1402.pf.goal.pf.term3}) holds. Therefore, for
the rest of this proof of (\ref{1402.pf.goal.pf.term3}), we can WLOG assume
that we don't have $R=0$. Assume this.
\par
We don't have $R=0$. Thus, we have $R\geq1$, so that $R-1\geq0$. Hence,
$R-1\in\left\{  0,1,\ldots,R-1\right\}  $. Hence, $H\otimes V^{R-1}$ is an
addend in the sum $\sum_{\ell=0}^{R-1}H\otimes V^{\ell}$. Hence, $H\otimes
V^{R-1}\subseteq\sum_{\ell=0}^{R-1}H\otimes V^{\ell}$. Now,%
\begin{align*}
\sum_{m=1}^{R}H\otimes V^{R-1}  &  \subseteq H\otimes V^{R-1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }H\otimes V^{R-1}\text{ is a
}k\text{-vector space}\right) \\
&  \subseteq\sum_{\ell=0}^{R-1}H\otimes V^{\ell},
\end{align*}
so that%
\begin{align*}
\underbrace{\sum_{m=1}^{R}H\otimes V^{R-1}}_{\subseteq\sum_{\ell=0}%
^{R-1}H\otimes V^{\ell}}+\sum_{\ell=0}^{R-1}H\otimes V^{\ell}  &
\subseteq\sum_{\ell=0}^{R-1}H\otimes V^{\ell}+\sum_{\ell=0}^{R-1}H\otimes
V^{\ell}\\
&  \subseteq\sum_{\ell=0}^{R-1}H\otimes V^{\ell}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sum_{\ell=0}^{R-1}H\otimes V^{\ell}\text{ is a }k\text{-vector
space}\right) \\
&  =\sum_{\ell=0}^{\left(  R+1\right)  -2}H\otimes V^{\ell}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }R-1=\left(  R+1\right)  -2\right)  .
\end{align*}
This proves (\ref{1402.pf.goal.pf.term3}).}.

Since $H$ is a bialgebra, the comultiplication $\Delta$ is a $k$-algebra
homomorphism (by the axioms of a bialgebra). We have%
\begin{align*}
&  \Delta\left(  \underbrace{a_{1}a_{2}\cdots a_{R+1}}_{=\left(  a_{1}%
a_{2}\cdots a_{R}\right)  a_{R+1}}\right) \\
&  =\Delta\left(  \left(  a_{1}a_{2}\cdots a_{R}\right)  a_{R+1}\right) \\
&  =\underbrace{\Delta\left(  a_{1}a_{2}\cdots a_{R}\right)  }_{\substack{\in
\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  +1_{H}\otimes\left(
a_{1}a_{2}\cdots a_{R}\right)  +\sum_{\ell=0}^{R-2}H\otimes V^{\ell
}\\\text{(by (\ref{1402.pf.goal.pf.indhyp}))}}}\cdot\underbrace{\Delta\left(
a_{R+1}\right)  }_{=a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  \in\left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  +\sum_{\ell=0}%
^{R-2}H\otimes V^{\ell}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes
a_{R+1}\right) \\
&  \subseteq\underbrace{\left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)
\right)  \right)  \cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}%
\right)  }_{\substack{=\left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)
\right)  \right)  \cdot\left(  a_{R+1}\otimes1_{H}\right)  \\+\left(
\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  \right)  \cdot\left(
1_{H}\otimes a_{R+1}\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  1_{H}\otimes\left(  a_{1}%
a_{2}\cdots a_{R}\right)  \right)  \cdot\left(  a_{R+1}\otimes1_{H}%
+1_{H}\otimes a_{R+1}\right)  }_{=\left(  1_{H}\otimes\left(  a_{1}a_{2}\cdots
a_{R}\right)  \right)  \cdot\left(  a_{R+1}\otimes1_{H}\right)  +\left(
1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  \right)  \cdot\left(
1_{H}\otimes a_{R+1}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \sum_{\ell=0}^{R-2}H\otimes
V^{\ell}\right)  \cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}\right)
}_{\subseteq\sum_{\ell=0}^{R-1}H\otimes V^{\ell}}%
\end{align*}%
\begin{align*}
&  \subseteq\underbrace{\left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)
\right)  \right)  \cdot\left(  a_{R+1}\otimes1_{H}\right)  }_{=\sum_{m=1}%
^{R}\left(  a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  \right)  \cdot\left(
a_{R+1}\otimes1_{H}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \sum_{m=1}^{R}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{R}\right)  \right)  \right)  \cdot\left(  1_{H}\otimes a_{R+1}\right)
}_{=\sum_{m=1}^{R}\left(  a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  \right)
\cdot\left(  1_{H}\otimes a_{R+1}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  1_{H}\otimes\left(  a_{1}%
a_{2}\cdots a_{R}\right)  \right)  \cdot\left(  a_{R+1}\otimes1_{H}\right)
}_{=\left(  1_{H}a_{R+1}\right)  \otimes\left(  \left(  a_{1}a_{2}\cdots
a_{R}\right)  1_{H}\right)  }+\underbrace{\left(  1_{H}\otimes\left(
a_{1}a_{2}\cdots a_{R}\right)  \right)  \cdot\left(  1_{H}\otimes
a_{R+1}\right)  }_{=\left(  1_{H}1_{H}\right)  \otimes\left(  \left(
a_{1}a_{2}\cdots a_{R}\right)  a_{R+1}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\ell=0}^{R-1}H\otimes V^{\ell}\\
&  \subseteq\sum_{m=1}^{R}\underbrace{\left(  a_{m}\otimes\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)
\right)  \right)  \cdot\left(  a_{R+1}\otimes1_{H}\right)  }_{=\left(
a_{m}a_{R+1}\right)  \otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  1_{H}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{m=1}^{R}\underbrace{\left(  a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{R}\right)  \right)  \right)  \cdot\left(  1_{H}\otimes a_{R+1}\right)
}_{=\left(  a_{m}1_{H}\right)  \otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  a_{R+1}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  1_{H}a_{R+1}\right)  }_{=a_{R+1}%
}\otimes\underbrace{\left(  \left(  a_{1}a_{2}\cdots a_{R}\right)
1_{H}\right)  }_{=a_{1}a_{2}\cdots a_{R}}+\underbrace{\left(  1_{H}%
1_{H}\right)  }_{=1_{H}}\otimes\underbrace{\left(  \left(  a_{1}a_{2}\cdots
a_{R}\right)  a_{R+1}\right)  }_{=a_{1}a_{2}\cdots a_{R}a_{R+1}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\ell=0}^{R-1}H\otimes V^{\ell}\\
&  =\sum_{m=1}^{R}\left(  a_{m}a_{R+1}\right)  \otimes\underbrace{\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{R}\right)  1_{H}\right)  }_{=\left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{m=1}^{R}\underbrace{\left(  a_{m}1_{H}\right)
}_{=a_{m}}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\underbrace{\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  a_{R+1}}_{=a_{m+1}%
a_{m+2}\cdots a_{R}a_{R+1}=a_{m+1}a_{m+2}\cdots a_{R+1}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)
+1_{H}\otimes\underbrace{\left(  a_{1}a_{2}\cdots a_{R}a_{R+1}\right)
}_{=a_{1}a_{2}\cdots a_{R+1}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\ell=0}^{R-1}H\otimes V^{\ell}%
\end{align*}%
\begin{align*}
&  =\sum_{m=1}^{R}\underbrace{\left(  a_{m}a_{R+1}\right)  }_{\in H}%
\otimes\left(  \underbrace{\left(  a_{1}a_{2}\cdots a_{m-1}\right)
}_{\substack{\in V^{m-1}\\\text{(since }a_{i}\in V\text{ for every }%
i\in\left\{  1,2,\ldots,m-1\right\}  \text{)}}}\underbrace{\left(
a_{m+1}a_{m+2}\cdots a_{R}\right)  }_{\substack{\in V^{R-m}\\\text{(since
}a_{i}\in V\text{ for every }i\in\left\{  m+1,m+2,\ldots,R\right\}  \text{)}%
}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}%
a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ +a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\ell=0}^{R-1}H\otimes V^{\ell}\\
&  \subseteq\sum_{m=1}^{R}H\otimes\left(  V^{m-1}V^{R-m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}%
a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ +a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\ell=0}^{R-1}H\otimes V^{\ell}\\
&  =\underbrace{\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)
+a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  }_{\substack{=\sum
_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)  \\\text{(by
(\ref{1402.pf.goal.pf.term2}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)
+\sum_{m=1}^{R}H\otimes\underbrace{\left(  V^{m-1}V^{R-m}\right)
}_{=V^{\left(  m-1\right)  +\left(  R-m\right)  }=V^{R-1}}+\sum_{\ell=0}%
^{R-1}H\otimes V^{\ell}\\
&  =\sum_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)
+\underbrace{\sum_{m=1}^{R}H\otimes V^{R-1}+\sum_{\ell=0}^{R-1}H\otimes
V^{\ell}}_{\substack{\subseteq\sum_{\ell=0}^{\left(  R+1\right)  -2}H\otimes
V^{\ell}\\\text{(by (\ref{1402.pf.goal.pf.term3}))}}}\\
&  \subseteq\sum_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)
+\sum_{\ell=0}^{\left(  R+1\right)  -2}H\otimes V^{\ell}.
\end{align*}
In other words, (\ref{1402.pf.goal}) holds for $r=R+1$. We have thus proven
(\ref{1402.pf.goal}) for $r=R+1$. The induction step is thus complete.

Hence, (\ref{1402.pf.goal}) is proven by induction. Now, applying
(\ref{1402.pf.goal}) to $r=s$, we obtain%
\begin{align*}
\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)   &  \in\sum_{m=1}^{s}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{s}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{s}\right)
+\sum_{\ell=0}^{s-2}H\otimes V^{\ell}.
\end{align*}
This proves Lemma~\ref{Lemma1402}.
\end{proof}

From Lemma~\ref{Lemma1402}, we easily conclude the following weaker statement:

\begin{lemma}
\label{Lemma1403} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $V$ be a
$k$-vector subspace of $\operatorname*{Prim}H$. Let $s\in\mathbb{N}$. Let
$a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$ elements of $V$. Then,%
\[
\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)  \in1_{H}\otimes\left(  a_{1}%
a_{2}\cdots a_{s}\right)  +\sum_{\ell=0}^{s-1}H\otimes V^{\ell}.
\]
\footnote{Recall that $V^{\ell}$ is defined according to
Convention~\ref{Convention100}. Hence, $V^{\ell}$ means the $\ell$-th power of
the subspace $V$ of the $k$-algebra $H$.}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1403}.]Every $m\in\left\{  1,2,\ldots,s\right\}  $
satisfies%
\begin{equation}
a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(
a_{m+1}a_{m+2}\cdots a_{s}\right)  \right)  \in\sum_{\ell=0}^{s-1}H\otimes
V^{\ell} \label{1403.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{1403.pf.1}):} Let $m\in\left\{
1,2,\ldots,s\right\}  $. Then, $1\leq m\leq s$, so that $s\geq1$ and thus
$s-1\geq0$. Hence, $s-1\in\left\{  0,1,\ldots,s-1\right\}  $. Thus, $H\otimes
V^{s-1}$ is an addend in the sum $\sum_{\ell=0}^{s-1}H\otimes V^{\ell}$.
Consequently, $H\otimes V^{s-1}\subseteq\sum_{\ell=0}^{s-1}H\otimes V^{\ell}$.
\par
Now, $\underbrace{\left(  a_{1}a_{2}\cdots a_{m-1}\right)  }_{\substack{\in
V^{m-1}\\\text{(since }a_{i}\in V\text{ for every }i\in\left\{  1,2,\ldots
,m-1\right\}  \text{)}}}\underbrace{\left(  a_{m+1}a_{m+2}\cdots a_{s}\right)
}_{\substack{\in V^{s-m}\\\text{(since }a_{i}\in V\text{ for every }%
i\in\left\{  m+1,m+2,\ldots,s\right\}  \text{)}}}\in V^{m-1}V^{s-m}=V^{\left(
m-1\right)  +\left(  s-m\right)  }=V^{s-1}$. Now,%
\[
\underbrace{a_{m}}_{\in H}\otimes\left(  \underbrace{\left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{s}\right)  }_{\in V^{s-1}%
}\right)  \in H\otimes V^{s-1}\subseteq\sum_{\ell=0}^{s-1}H\otimes V^{\ell}.
\]
This proves (\ref{1403.pf.1}).}. Hence,%
\[
\sum_{m=1}^{s}\underbrace{a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{s}\right)  \right)
}_{\substack{\in\sum_{\ell=0}^{s-1}H\otimes V^{\ell}\\\text{(by
(\ref{1403.pf.1}))}}}\in\sum_{m=1}^{s}\sum_{\ell=0}^{s-1}H\otimes V^{\ell
}\subseteq\sum_{\ell=0}^{s-1}H\otimes V^{\ell}%
\]
(since $\sum_{\ell=0}^{s-1}H\otimes V^{\ell}$ is a $k$-vector space). Now,
Lemma~\ref{Lemma1402} yields%
\begin{align*}
\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)   &  \in\underbrace{\sum_{m=1}%
^{s}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(
a_{m+1}a_{m+2}\cdots a_{s}\right)  \right)  }_{\in\sum_{\ell=0}^{s-1}H\otimes
V^{\ell}}\\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{s}\right)
+\underbrace{\sum_{\ell=0}^{s-2}H\otimes V^{\ell}}_{\substack{\subseteq
\sum_{\ell=0}^{s-1}H\otimes V^{\ell}\\\text{(since }s-2\leq s-1\text{)}}}\\
&  \subseteq\sum_{\ell=0}^{s-1}H\otimes V^{\ell}+1_{H}\otimes\left(
a_{1}a_{2}\cdots a_{s}\right)  +\sum_{\ell=0}^{s-1}H\otimes V^{\ell}\\
&  =1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{s}\right)  +\underbrace{\sum
_{\ell=0}^{s-1}H\otimes V^{\ell}+\sum_{\ell=0}^{s-1}H\otimes V^{\ell}%
}_{\substack{\subseteq\sum_{\ell=0}^{s-1}H\otimes V^{\ell}\\\text{(since }%
\sum_{\ell=0}^{s-1}H\otimes V^{\ell}\text{ is a }k\text{-vector space)}}}\\
&  \subseteq1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{s}\right)  +\sum_{\ell
=0}^{s-1}H\otimes V^{\ell}.
\end{align*}
This proves Lemma~\ref{Lemma1403}.
\end{proof}

We are now ready to prove the following result (more or less equivalent to
Theorem~\ref{Theorem1401} \textbf{(a)}):

\begin{lemma}
\label{Lemma1404} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be a
$k$-algebra. Let $r\in\mathbb{N}$. Let $f_{1}$, $f_{2}$, $\ldots$, $f_{r}$ be
$r$ maps in $\mathfrak{g}\left(  H,A\right)  $.

Every $s\in\left\{  0,1,\ldots,r-1\right\}  $ satisfies%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \left(
\operatorname*{Prim}H\right)  ^{s}\right)  =0.
\]
\footnote{Recall that $\left(  \operatorname*{Prim}H\right)  ^{s}$ is defined
according to Convention~\ref{Convention100}. Hence, $\left(
\operatorname*{Prim}H\right)  ^{s}$ means the $s$-th power of the subspace
$\operatorname*{Prim}H$ of the $k$-algebra $H$.}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1404}.]We are going to prove Lemma~\ref{Lemma1404}
by induction over $r$:

\textit{Induction base:} If $r=0$, then Lemma~\ref{Lemma1404} is vacuously
true (because if $r=0$, then there exists no $s\in\left\{  0,1,\ldots
,r-1\right\}  $). Hence, the induction base is complete.

\textit{Induction step:} Let $R\in\mathbb{N}$ be positive. Assume that
Lemma~\ref{Lemma1404} is proven for $r=R-1$. We now are going to prove that
Lemma~\ref{Lemma1404} holds for $r=R$.

Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be a $k$-algebra. Let
$f_{1}$, $f_{2}$, $\ldots$, $f_{R}$ be $R$ maps in $\mathfrak{g}\left(
H,A\right)  $. Let $s\in\left\{  0,1,\ldots,R-1\right\}  $. We are going to
show that%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  \left(
\operatorname*{Prim}H\right)  ^{s}\right)  =0.
\]


Let $g=f_{2}\ast f_{3}\ast\cdots\ast f_{R}$. Then, $g$ is a $k$-linear map
$H\rightarrow A$. Let $\ell\in\left\{  0,1,\ldots,s-1\right\}  $. Hence,
$0\leq\ell\leq s-1$. But $s\leq R-1$ (since $s\in\left\{  0,1,\ldots
,R-1\right\}  $), so that $\ell\leq\underbrace{s}_{\leq R-1}-1\leq\left(
R-1\right)  -1$. Thus, $0\leq\ell\leq\left(  R-1\right)  -1$. Hence, $\ell
\in\left\{  0,1,\ldots,\left(  R-1\right)  -1\right\}  $. Thus, we can apply
Lemma~\ref{Lemma1404} to $R-1$, $\left(  f_{2},f_{3},\ldots,f_{R}\right)  $
and $\ell$ instead of $r$, $\left(  f_{1},f_{2},\ldots,f_{r}\right)  $ and $s$
(since we assumed that Lemma~\ref{Lemma1404} is proven for $r=R-1$). As a
result, we obtain $\left(  f_{2}\ast f_{3}\ast\cdots\ast f_{R}\right)  \left(
\left(  \operatorname*{Prim}H\right)  ^{\ell}\right)  =0$. Thus,%
\[
\underbrace{g}_{=f_{2}\ast f_{3}\ast\cdots\ast f_{R}}\left(  \left(
\operatorname*{Prim}H\right)  ^{\ell}\right)  =\left(  f_{2}\ast f_{3}%
\ast\cdots\ast f_{R}\right)  \left(  \left(  \operatorname*{Prim}H\right)
^{\ell}\right)  =0.
\]


Now, forget that we fixed $\ell$. We thus have proven that%
\begin{equation}
g\left(  \left(  \operatorname*{Prim}H\right)  ^{\ell}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\left\{  0,1,\ldots,s-1\right\}
. \label{1404.pf.g.1}%
\end{equation}


We have $f_{1}\in\mathfrak{g}\left(  H,A\right)  =\left\{  f\in\mathcal{L}%
\left(  H,A\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}  $. In other
words, $f_{1}$ is an element of $\mathcal{L}\left(  H,A\right)  $ and
satisfies $f_{1}\left(  1_{H}\right)  =0$.

Now, let $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$ elements of
$\operatorname*{Prim}H$. We have $f_{1}\ast f_{2}\ast\cdots\ast f_{R}%
=f_{1}\ast\underbrace{\left(  f_{2}\ast f_{3}\ast\cdots\ast f_{R}\right)
}_{=g}=f_{1}\ast g=\mu\circ\left(  f_{1}\otimes g\right)  \circ\Delta$ (by the
definition of convolution), so that%
\begin{align*}
&  \underbrace{\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  }%
_{=\mu\circ\left(  f_{1}\otimes g\right)  \circ\Delta}\left(  a_{1}a_{2}\cdots
a_{s}\right) \\
&  =\left(  \mu\circ\left(  f_{1}\otimes g\right)  \circ\Delta\right)  \left(
a_{1}a_{2}\cdots a_{s}\right)  =\mu\left(  \left(  f_{1}\otimes g\right)
\left(  \underbrace{\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)
}_{\substack{\in1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{s}\right)  +\sum
_{\ell=0}^{s-1}H\otimes\left(  \operatorname*{Prim}H\right)  ^{\ell
}\\\text{(by Lemma~\ref{Lemma1403}, applied to }V=\operatorname*{Prim}%
H\text{)}}}\right)  \right) \\
&  \in\mu\left(  \underbrace{\left(  f_{1}\otimes g\right)  \left(
1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{s}\right)  +\sum_{\ell=0}%
^{s-1}H\otimes\left(  \operatorname*{Prim}H\right)  ^{\ell}\right)
}_{\substack{=\left(  f_{1}\otimes g\right)  \left(  1_{H}\otimes\left(
a_{1}a_{2}\cdots a_{s}\right)  \right)  +\sum_{\ell=0}^{s-1}\left(
f_{1}\otimes g\right)  \left(  H\otimes\left(  \operatorname*{Prim}H\right)
^{\ell}\right)  \\\text{(since the map }f_{1}\otimes g\text{ is }%
k\text{-linear)}}}\right) \\
&  =\mu\left(  \underbrace{\left(  f_{1}\otimes g\right)  \left(  1_{H}%
\otimes\left(  a_{1}a_{2}\cdots a_{s}\right)  \right)  }_{=f_{1}\left(
1_{H}\right)  \otimes g\left(  a_{1}a_{2}\cdots a_{s}\right)  }+\sum_{\ell
=0}^{s-1}\underbrace{\left(  f_{1}\otimes g\right)  \left(  H\otimes\left(
\operatorname*{Prim}H\right)  ^{\ell}\right)  }_{\subseteq f_{1}\left(
H\right)  \otimes g\left(  \left(  \operatorname*{Prim}H\right)  ^{\ell
}\right)  }\right) \\
&  \subseteq\mu\left(  \underbrace{f_{1}\left(  1_{H}\right)  }_{=0}\otimes
g\left(  a_{1}a_{2}\cdots a_{s}\right)  +\sum_{\ell=0}^{s-1}f_{1}\left(
H\right)  \otimes\underbrace{g\left(  \left(  \operatorname*{Prim}H\right)
^{\ell}\right)  }_{\substack{=0\\\text{(by (\ref{1404.pf.g.1}) (since }\ell
\in\left\{  0,1,\ldots,s-1\right\}  \text{))}}}\right) \\
&  =\mu\left(  \underbrace{0\otimes g\left(  a_{1}a_{2}\cdots a_{s}\right)
}_{=0}+\underbrace{\sum_{\ell=0}^{s-1}f_{1}\left(  H\right)  \otimes0}%
_{=0}\right)  =\mu\left(  0\right)  =0.
\end{align*}
Hence, $\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  a_{1}%
a_{2}\cdots a_{s}\right)  =0$.

Now, forget that we fixed $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$. We thus have
shown that whenever $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ are $s$ elements of
$\operatorname*{Prim}H$, we have%
\begin{equation}
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  a_{1}a_{2}\cdots
a_{s}\right)  =0. \label{1404.pf.almostthere}%
\end{equation}


Now, (\ref{100.Vn}) yields%
\begin{align*}
\left(  \operatorname*{Prim}H\right)  ^{s}  &  =\left\langle a_{1}a_{2}\cdots
a_{s}\ \mid\ \left(  a_{1},a_{2},\ldots,a_{s}\right)  \in\left(
\operatorname*{Prim}H\right)  ^{\times s}\right\rangle \\
&  =\left\langle \left\{  a_{1}a_{2}\cdots a_{s}\ \mid\ \left(  a_{1}%
,a_{2},\ldots,a_{s}\right)  \in\left(  \operatorname*{Prim}H\right)  ^{\times
s}\right\}  \right\rangle .
\end{align*}
Applying the map $f_{1}\ast f_{2}\ast\cdots\ast f_{R}$ to both sides of this
equality, we obtain%
\begin{align*}
&  \left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  \left(
\operatorname*{Prim}H\right)  ^{s}\right) \\
&  =\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  \left\langle
\left\{  a_{1}a_{2}\cdots a_{s}\ \mid\ \left(  a_{1},a_{2},\ldots
,a_{s}\right)  \in\left(  \operatorname*{Prim}H\right)  ^{\times s}\right\}
\right\rangle \right) \\
&  =\left\langle \underbrace{\left(  f_{1}\ast f_{2}\ast\cdots\ast
f_{R}\right)  \left(  \left\{  a_{1}a_{2}\cdots a_{s}\ \mid\ \left(
a_{1},a_{2},\ldots,a_{s}\right)  \in\left(  \operatorname*{Prim}H\right)
^{\times s}\right\}  \right)  }_{=\left\{  \left(  f_{1}\ast f_{2}\ast
\cdots\ast f_{R}\right)  \left(  a_{1}a_{2}\cdots a_{s}\right)  \ \mid
\ \left(  a_{1},a_{2},\ldots,a_{s}\right)  \in\left(  \operatorname*{Prim}%
H\right)  ^{\times s}\right\}  }\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{316.pf.3}), applied to }H\text{, }A\text{, }f_{1}\ast f_{2}%
\ast\cdots\ast f_{R}\text{ and}\\
\left\{  a_{1}a_{2}\cdots a_{s}\ \mid\ \left(  a_{1},a_{2},\ldots
,a_{s}\right)  \in\left(  \operatorname*{Prim}H\right)  ^{\times s}\right\}
\text{ instead of }M\text{, }R\text{, }\phi\text{ and }S
\end{array}
\right) \\
&  =\left\langle \left\{  \underbrace{\left(  f_{1}\ast f_{2}\ast\cdots\ast
f_{R}\right)  \left(  a_{1}a_{2}\cdots a_{s}\right)  }%
_{\substack{=0\\\text{(by (\ref{1404.pf.almostthere}))}}}\ \mid\ \left(
a_{1},a_{2},\ldots,a_{s}\right)  \in\left(  \operatorname*{Prim}H\right)
^{\times s}\right\}  \right\rangle \\
&  =\left\langle \underbrace{\left\{  0\ \mid\ \left(  a_{1},a_{2}%
,\ldots,a_{s}\right)  \in\left(  \operatorname*{Prim}H\right)  ^{\times
s}\right\}  }_{\subseteq0}\right\rangle \subseteq\left\langle 0\right\rangle
=0.
\end{align*}
In other words, $\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(
\left(  \operatorname*{Prim}H\right)  ^{s}\right)  =0$.

Now, forget that we fixed $k$, $H$, $A$, $\left(  f_{1},f_{2},\ldots
,f_{R}\right)  $ and $s$. We thus have shown that if $k$ is a field, if $H$ is
a $k$-bialgebra, if $A$ is a $k$-algebra, if $f_{1}$, $f_{2}$, $\ldots$,
$f_{R}$ are $R$ maps in $\mathfrak{g}\left(  H,A\right)  $, then every
$s\in\left\{  0,1,\ldots,R-1\right\}  $ satisfies $\left(  f_{1}\ast f_{2}%
\ast\cdots\ast f_{R}\right)  \left(  \left(  \operatorname*{Prim}H\right)
^{s}\right)  =0$. In other words, we have shown that Lemma~\ref{Lemma1404}
holds for $r=R$. This completes the induction step. The induction proof of
Lemma~\ref{Lemma1404} is thus complete.
\end{proof}

To proceed, we introduce a basic operation on permutations:

\begin{lemma}
\label{Lemma1405} As usual, let $S_{n}$ denote the $n$-th symmetric group for
every $n\in\mathbb{N}$.

Let $n$ be a positive integer.

Fix $m\in\left\{  1,2,\ldots,n\right\}  $. Let $S_{n,m}$ denote the subset
$\left\{  \sigma\in S_{n}\ \mid\ \sigma\left(  1\right)  =m\right\}  $ of
$S_{n}$.

\textbf{(a)} Define a map $\eta:\left\{  1,2,\ldots,n-1\right\}
\rightarrow\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $ by%
\begin{equation}
\left(  \eta\left(  i\right)  =\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots
,n-1\right\}  \right)  . \label{1405.eta}%
\end{equation}
This map $\eta$ is well-defined.

\textbf{(b)} Define a map $\omega:\left\{  1,2,\ldots,n\right\}
\setminus\left\{  m\right\}  \rightarrow\left\{  1,2,\ldots,n-1\right\}  $ by%
\begin{equation}
\left(  \omega\left(  j\right)  =\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{  1,2,\ldots
,n\right\}  \setminus\left\{  m\right\}  \right)  . \label{1405.omega}%
\end{equation}
This map $\omega$ is well-defined.

\textbf{(c)} The maps $\eta$ and $\omega$ are mutually inverse (where $\eta$
is defined as in Lemma~\ref{Lemma1405} \textbf{(a)}, and where $\omega$ is
defined as in Lemma~\ref{Lemma1405} \textbf{(b)}).

\textbf{(d)} For every $\sigma\in S_{n-1}$, define a map $\operatorname*{ins}%
\nolimits_{m}\sigma:\left\{  1,2,\ldots,n\right\}  \rightarrow\left\{
1,2,\ldots,n\right\}  $ by%
\begin{equation}
\left(  \left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
i\right)  =\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  i-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}i\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }i=1
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots
,n\right\}  \right)  \label{1405.insk}%
\end{equation}
(where $\eta$ is defined as in Lemma~\ref{Lemma1405} \textbf{(a)}). This map
$\operatorname*{ins}\nolimits_{m}\sigma$ is well-defined.

\textbf{(e)} Define a map $\iota_{m}:S_{n-1}\rightarrow S_{n,m}$ by%
\begin{equation}
\left(  \iota_{m}\left(  \sigma\right)  =\operatorname*{ins}\nolimits_{m}%
\sigma\ \ \ \ \ \ \ \ \ \ \text{for every }\sigma\in S_{n-1}\right)
\label{1405.iotak}%
\end{equation}
(where $\operatorname*{ins}\nolimits_{m}\sigma$ is defined as in
Lemma~\ref{Lemma1405} \textbf{(d)}). This map $\iota_{m}$ is well-defined.

\textbf{(f)} For every $\tau\in S_{n,m}$, define a map $\operatorname*{del}%
\nolimits_{m}\tau:\left\{  1,2,\ldots,n-1\right\}  \rightarrow\left\{
1,2,\ldots,n-1\right\}  $ by%
\begin{equation}
\left(  \left(  \operatorname*{del}\nolimits_{m}\tau\right)  \left(  i\right)
=\omega\left(  \tau\left(  i+1\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }i\in\left\{  1,2,\ldots,n-1\right\}  \right)  \label{1405.delk}%
\end{equation}
(where $\omega$ is defined as in Lemma~\ref{Lemma1405} \textbf{(b)}). This map
$\operatorname*{del}\nolimits_{m}\tau$ is well-defined.

\textbf{(g)} Define a map $\pi_{m}:S_{n,m}\rightarrow S_{n-1}$ by%
\begin{equation}
\left(  \pi_{m}\left(  \tau\right)  =\operatorname*{del}\nolimits_{m}%
\tau\ \ \ \ \ \ \ \ \ \ \text{for every }\tau\in S_{n,m}\right)
\label{1405.pik}%
\end{equation}
(where $\operatorname*{del}\nolimits_{m}\tau$ is defined as in
Lemma~\ref{Lemma1405} \textbf{(f)}). This map $\pi_{m}$ is well-defined.

\textbf{(h)} The maps $\iota_{m}$ and $\pi_{m}$ are mutually inverse (where
$\iota_{m}$ is defined as in Lemma~\ref{Lemma1405} \textbf{(e)}, and where
$\pi_{m}$ is defined as in Lemma~\ref{Lemma1405} \textbf{(g)}).
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1405}.]We have $1\in\left\{  1,2,\ldots,n\right\}  $
(since $n$ is positive).

\textbf{(a)} For every $i\in\left\{  1,2,\ldots,n-1\right\}  $, we have
$\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}
$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,n-1\right\}
$. We want to prove that $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} We have $i<m$.
\par
\textit{Case 2:} We have $i\geq m$.
\par
Let us first consider Case 1. In this case, we have $i<m$. Hence, $i\neq m$.
Since $i\in\left\{  1,2,\ldots,n-1\right\}  \subseteq\left\{  1,2,\ldots
,n\right\}  $ and $i\neq m$, we have $i\in\left\{  1,2,\ldots,n\right\}
\setminus\left\{  m\right\}  $. Now, $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  =i$ (since $i<m$), so that $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  =i\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $.
Thus, $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $ is
proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $i\geq m$. Hence, $i+1\geq
m+1>m$, so that $i+1\neq m$. Also, $i\in\left\{  1,2,\ldots,n-1\right\}  $, so
that $i+1\in\left\{  2,3,\ldots,n\right\}  \subseteq\left\{  1,2,\ldots
,n\right\}  $. Combining this with $i+1\neq m$, we obtain $i+1\in\left\{
1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $. Now, $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  =i+1$ (since $i\geq m$), so that $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  =i+1\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $.
Thus, $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $ is
proven in Case 2.
\par
We have thus proven $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $ in
each of the two Cases 1 and 2. Since these two Cases cover all possibilities,
this yields that $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $
always holds, qed.}. Thus, for every $i\in\left\{  1,2,\ldots,n-1\right\}  $,
the element $\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  $ is a well-defined element of $\left\{  1,2,\ldots,n\right\}
\setminus\left\{  m\right\}  $. In other words, for every $i\in\left\{
1,2,\ldots,n-1\right\}  $, the right hand side of (\ref{1405.eta}) is a
well-defined element of $\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  $. Hence, the map $\eta:\left\{  1,2,\ldots,n-1\right\}
\rightarrow\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $ is
well-defined (because the map $\eta:\left\{  1,2,\ldots,n-1\right\}
\rightarrow\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $ was
defined by (\ref{1405.eta})). Lemma~\ref{Lemma1405} \textbf{(a)} is thus proven.

\textbf{(b)} For every $j\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  $, we have $\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n-1\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,\ldots,n\right\}
\setminus\left\{  m\right\}  $. We want to prove that $\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n-1\right\}  $.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} We have $j<m$.
\par
\textit{Case 2:} We have $j\geq m$.
\par
Let us first consider Case 1. In this case, we have $j<m$. Hence, $j<m\leq n$,
so that $j\leq n-1$ (since $j$ and $n$ are integers). But $j\in\left\{
1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  \subseteq\left\{
1,2,\ldots,n\right\}  $, so that $1\leq j$. Combining $1\leq j$ and $j\leq
n-1$, we obtain $j\in\left\{  1,2,\ldots,n-1\right\}  $, so that%
\begin{align*}
\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.   &  =j\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j<m\right) \\
&  \in\left\{  1,2,\ldots,n-1\right\}  .
\end{align*}
Thus, $\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n-1\right\}  $ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $j\geq m$. Combined with
$j\neq m$ (since $j\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  $), this yields $j>m$. Thus, $j\geq m+1$ (since $j$ and $m$ are
integers), so that $j-1\geq m\geq1$. Also, $j\leq n$ (since $j\in\left\{
1,2,\ldots,n\right\}  $), so that $j-1\leq n-1$. Combining $j-1\geq1$ and
$j-1\leq n-1$, we obtain $j-1\in\left\{  1,2,\ldots,n-1\right\}  $. Now,%
\begin{align*}
\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.   &  =j-1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j\geq m\right) \\
&  \in\left\{  1,2,\ldots,n-1\right\}  .
\end{align*}
Thus, $\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n-1\right\}  $ is proven in Case 2.
\par
We now have proven $\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n-1\right\}  $ in each of the two Cases 1 and
2. Hence, $\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.  \in\left\{  1,2,\ldots,n-1\right\}  $ holds in every situation
(because the Cases 1 and 2 cover all possibilities), qed.}. In other words,
for every $j\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $,
the element $\left\{
\begin{array}
[c]{c}%
j,\ \ \ \ \ \ \ \ \ \ \text{if }j<m;\\
j-1,\ \ \ \ \ \ \ \ \ \ \text{if }j\geq m
\end{array}
\right.  $ is a well-defined element of $\left\{  1,2,\ldots,n-1\right\}  $.
In other words, for every $j\in\left\{  1,2,\ldots,n\right\}  \setminus
\left\{  m\right\}  $, the right hand side of (\ref{1405.omega}) is a
well-defined element of $\left\{  1,2,\ldots,n-1\right\}  $. In other words,
Hence, the map $\omega:\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  \rightarrow\left\{  1,2,\ldots,n-1\right\}  $ is well-defined
(because the map $\omega:\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  \rightarrow\left\{  1,2,\ldots,n-1\right\}  $ was defined by
(\ref{1405.omega})). Lemma~\ref{Lemma1405} \textbf{(b)} is thus proven.

\textbf{(c)} Clearly, the maps $\eta\circ\omega$ and $\omega\circ\eta$ are well-defined.

First, let $i\in\left\{  1,2,\ldots,n-1\right\}  $ be arbitrary. Then, we are
going to show that $\left(  \omega\circ\eta\right)  \left(  i\right)  =i$.

Clearly, $i\in\left\{  1,2,\ldots,n-1\right\}  \subseteq\left\{
1,2,\ldots,n\right\}  $. Also, from $i\in\left\{  1,2,\ldots,n-1\right\}  $,
we obtain $i+1\in\left\{  2,3,\ldots,n\right\}  \subseteq\left\{
1,2,\ldots,n\right\}  $. We must be in one of the following two cases:

\textit{Case 1:} We have $i<m$.

\textit{Case 2:} We have $i\geq m$.

Let us first consider Case 1. In this case, we have $i<m$. The definition of
$\eta$ yields $\eta\left(  i\right)  =\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  =i$ (since $i<m$). But
\begin{align*}
\left(  \omega\circ\eta\right)  \left(  i\right)   &  =\omega\left(
\underbrace{\eta\left(  i\right)  }_{=i}\right)  =\omega\left(  i\right)
=\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i-1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right)
\\
&  =i\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i<m\right)  .
\end{align*}
Thus, $\left(  \omega\circ\eta\right)  \left(  i\right)  =i$ is proven in Case 1.

Let us now consider Case 2. In this case, we have $i\geq m$. Hence, $i+1\geq
m+1>m$. But the definition of $\eta$ yields $\eta\left(  i\right)  =\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  =i+1$ (since $i\geq m$). Now,%
\begin{align*}
\left(  \omega\circ\eta\right)  \left(  i\right)   &  =\omega\left(
\underbrace{\eta\left(  i\right)  }_{=i+1}\right)  =\omega\left(  i+1\right)
=\left\{
\begin{array}
[c]{c}%
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i+1<m;\\
\left(  i+1\right)  -1,\ \ \ \ \ \ \ \ \ \ \text{if }i+1\geq m
\end{array}
\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\left(
i+1\right)  \right) \\
&  =\left(  i+1\right)  -1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i+1\geq
m\right) \\
&  =i.
\end{align*}
Hence, $\left(  \omega\circ\eta\right)  \left(  i\right)  =i$ is proven in
Case 2.

Now, we have proven that $\left(  \omega\circ\eta\right)  \left(  i\right)
=i$ in each of the two Cases 1 and 2. Thus, $\left(  \omega\circ\eta\right)
\left(  i\right)  =i$ holds in every situation (because these two Cases 1 and
2 cover all situations). Hence, $\left(  \omega\circ\eta\right)  \left(
i\right)  =i=\operatorname*{id}\left(  i\right)  $.

Now, forget that we fixed $i$. We thus have proven that for every
$i\in\left\{  1,2,\ldots,n-1\right\}  $, we have $\left(  \omega\circ
\eta\right)  \left(  i\right)  =\operatorname*{id}\left(  i\right)  $. In
other words,%
\begin{equation}
\omega\circ\eta=\operatorname*{id}. \label{1405.pf.b.1}%
\end{equation}


On the other hand, let $p\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  $ be arbitrary. We will show that $\left(  \eta\circ\omega\right)
\left(  p\right)  =p$.

Indeed, we have $p\in\left\{  1,2,\ldots,n\right\}  $ and $p\neq m$ (since
$p\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $). From
$p\in\left\{  1,2,\ldots,n\right\}  $, we obtain $1\leq p\leq n$. We must be
in one of the following two cases:

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

\textit{Case 2:} We have $p\geq m$.

Let us first consider Case 1. In this case, we have $p<m$. The definition of
$\omega\left(  p\right)  $ yields $\omega\left(  p\right)  =\left\{
\begin{array}
[c]{c}%
p,\ \ \ \ \ \ \ \ \ \ \text{if }p<m;\\
p-1,\ \ \ \ \ \ \ \ \ \ \text{if }p\geq m
\end{array}
\right.  =p$ (since $p<m$). Now,%
\begin{align*}
\left(  \eta\circ\omega\right)  \left(  p\right)   &  =\eta\left(
\underbrace{\omega\left(  p\right)  }_{=p}\right)  =\eta\left(  p\right)
=\left\{
\begin{array}
[c]{c}%
p,\ \ \ \ \ \ \ \ \ \ \text{if }p<m;\\
p+1,\ \ \ \ \ \ \ \ \ \ \text{if }p\geq m
\end{array}
\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\eta\left(
p\right)  \right) \\
&  =p\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p<m\right)  .
\end{align*}
Hence, $\left(  \eta\circ\omega\right)  \left(  p\right)  =p$ is proven in
Case 1.

Let us now consider Case 2. In this case, we have $p\geq m$. Combined with
$p\neq m$, this yields $p>m$, and therefore $p\geq m+1$ (since $p$ and $m$ are
integers). Hence, $p-1\geq m$. The definition of $\omega\left(  p\right)  $
yields $\omega\left(  p\right)  =\left\{
\begin{array}
[c]{c}%
p,\ \ \ \ \ \ \ \ \ \ \text{if }p<m;\\
p-1,\ \ \ \ \ \ \ \ \ \ \text{if }p\geq m
\end{array}
\right.  =p-1$ (since $p\geq m$). Now,%
\begin{align*}
\left(  \eta\circ\omega\right)  \left(  p\right)   &  =\eta\left(
\underbrace{\omega\left(  p\right)  }_{=p-1}\right)  =\eta\left(  p-1\right)
=\left\{
\begin{array}
[c]{c}%
p-1,\ \ \ \ \ \ \ \ \ \ \text{if }p-1<m;\\
\left(  p-1\right)  +1,\ \ \ \ \ \ \ \ \ \ \text{if }p-1\geq m
\end{array}
\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\eta\left(
p-1\right)  \right) \\
&  =\left(  p-1\right)  +1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p-1\geq
m\right) \\
&  =p.
\end{align*}
Hence, $\left(  \eta\circ\omega\right)  \left(  p\right)  =p$ is proven in
Case 2.

Now, we have proven that $\left(  \eta\circ\omega\right)  \left(  p\right)
=p$ in each of the two Cases 1 and 2. Thus, $\left(  \eta\circ\omega\right)
\left(  p\right)  =p$ holds in every situation (because these two Cases 1 and
2 cover all situations). Hence, $\left(  \eta\circ\omega\right)  \left(
p\right)  =p=\operatorname*{id}\left(  p\right)  $.

Now, forget that we fixed $p$. We thus have proven that for every
$p\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $, we have
$\left(  \eta\circ\omega\right)  \left(  p\right)  =\operatorname*{id}\left(
p\right)  $. In other words,%
\[
\eta\circ\omega=\operatorname*{id}.
\]
Combined with (\ref{1405.pf.b.1}), this yields that the maps $\eta$ and
$\omega$ are mutually inverse. This yields that $\eta$ and $\omega$ are
bijective. This proves Lemma~\ref{Lemma1405} \textbf{(c)}.

\textbf{(d)} Let $\sigma\in S_{n-1}$. For every $i\in\left\{  1,2,\ldots
,n\right\}  $, the element $\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  i-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}i\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }i=1
\end{array}
\right.  $ is a well-defined element of $\left\{  1,2,\ldots,n\right\}
$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,n\right\}  $.
We need to prove that $\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  i-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}i\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }i=1
\end{array}
\right.  $ is a well-defined element of $\left\{  1,2,\ldots,n\right\}  $.
\par
If $i=1$, then%
\[
\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  i-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}i\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }i=1
\end{array}
\right.  =m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i=1\right)
\]
is a well-defined element of $\left\{  1,2,\ldots,n\right\}  $. Hence, for the
rest of our proof (of the fact that $\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  i-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}i\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }i=1
\end{array}
\right.  $ is a well-defined element of $\left\{  1,2,\ldots,n\right\}  $), we
can WLOG assume that we don't have $i=1$. Assume this.
\par
We don't have $i=1$. Thus, $i\neq1$. Since $i\in\left\{  1,2,\ldots,n\right\}
$ and $i\neq1$, we have $i\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{
1\right\}  =\left\{  2,3,\ldots,n\right\}  $. Hence, $i\in\left\{
1,2,\ldots,n-1\right\}  $. Thus, $\sigma\left(  i-1\right)  $ is a
well-defined element of $\left\{  1,2,\ldots,n-1\right\}  $ (since $\sigma\in
S_{n-1}$), and therefore $\eta\left(  \sigma\left(  i-1\right)  \right)  $ is
a well-defined element of $\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  \subseteq\left\{  1,2,\ldots,n\right\}  $.
\par
Now,
\[
\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  i-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}i\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }i=1
\end{array}
\right.  =\eta\left(  \sigma\left(  i-1\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\neq1\right)
\]
is a well-defined element of $\left\{  1,2,\ldots,n\right\}  $, qed.}. In
other words, for every $i\in\left\{  1,2,\ldots,n\right\}  $, the right hand
side of (\ref{1405.insk}) is a well-defined element of $\left\{
1,2,\ldots,n\right\}  $. Hence, the map $\operatorname*{ins}\nolimits_{m}%
\sigma:\left\{  1,2,\ldots,n\right\}  \rightarrow\left\{  1,2,\ldots
,n\right\}  $ is well-defined (because the map $\operatorname*{ins}%
\nolimits_{m}\sigma:\left\{  1,2,\ldots,n\right\}  \rightarrow\left\{
1,2,\ldots,n\right\}  $ was defined by (\ref{1405.insk})).
Lemma~\ref{Lemma1405} \textbf{(d)} is thus proven.

\textbf{(e)} By the definition of $S_{n,m}$, we have%
\begin{equation}
S_{n,m}=\left\{  \sigma\in S_{n}\ \mid\ \sigma\left(  1\right)  =m\right\}
=\left\{  \tau\in S_{n}\ \mid\ \tau\left(  1\right)  =m\right\}
\label{1405.pf.e.Snk}%
\end{equation}
(here, we renamed the index $\sigma$ as $\tau$).

Let $\sigma\in S_{n-1}$. We are going to show that $\operatorname*{ins}%
\nolimits_{m}\sigma\in S_{n,m}$.

Consider the map $\eta$ defined in Lemma~\ref{Lemma1405} \textbf{(a)} and the
map $\omega$ defined in Lemma~\ref{Lemma1405} \textbf{(b)}. The maps $\eta$
and $\omega$ are mutually inverse (by Lemma~\ref{Lemma1405} \textbf{(c)}).
Thus, $\eta$ is bijective. Hence, $\eta$ is injective and surjective. On the
other hand, $\sigma$ is an element of $S_{n-1}$, thus a permutation of
$\left\{  1,2,\ldots,n-1\right\}  $. Hence, $\sigma$ is bijective, and thus injective.

We know that $\operatorname*{ins}\nolimits_{m}\sigma$ is a map $\left\{
1,2,\ldots,n\right\}  \rightarrow\left\{  1,2,\ldots,n\right\}  $. We now will
prove that this map $\operatorname*{ins}\nolimits_{m}\sigma$ is injective.

In fact, let $a$ and $b$ be two elements of $\left\{  1,2,\ldots,n\right\}  $
such that $\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
a\right)  =\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
b\right)  $.

We assume (for the sake of contradiction) that $a\neq b$.

We have $a\neq1$.\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then,
we don't have $a\neq1$. Hence, we have $a=1$. Now, the definition of
$\operatorname*{ins}\nolimits_{m}\sigma$ yields%
\[
\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  a\right)
=\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  a-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}a\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }a=1
\end{array}
\right.  =m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a=1\right)  ,
\]
so that $m=\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
a\right)  $. But we have $a\neq b$, so that $b\neq a=1$. The definition of
$\operatorname*{ins}\nolimits_{m}\sigma$ yields%
\begin{align*}
\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  b\right)   &
=\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  b-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}b\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }b=1
\end{array}
\right.  =\eta\left(  \sigma\left(  b-1\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }b\neq1\right) \\
&  \in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the target of }\eta\text{ is }\left\{
1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  \right)  ,
\end{align*}
so that $\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
b\right)  \neq m$. Now, we have $m=\left(  \operatorname*{ins}\nolimits_{m}%
\sigma\right)  \left(  a\right)  =\left(  \operatorname*{ins}\nolimits_{m}%
\sigma\right)  \left(  b\right)  \neq m$. This is a contradiction. This
contradiction proves that our assumption was wrong, qed.} Similarly, $b\neq1$.
Now, the definition of $\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)
\left(  a\right)  $ yields%
\[
\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  a\right)
=\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  a-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}a\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }a=1
\end{array}
\right.  =\eta\left(  \sigma\left(  a-1\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a\neq1\right)  .
\]
Similarly, $\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
b\right)  =\eta\left(  \sigma\left(  b-1\right)  \right)  $.

We have $\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
a\right)  =\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
b\right)  =\eta\left(  \sigma\left(  b-1\right)  \right)  $. Compared with
$\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  a\right)
=\eta\left(  \sigma\left(  a-1\right)  \right)  $, this yields $\eta\left(
\sigma\left(  a-1\right)  \right)  =\eta\left(  \sigma\left(  b-1\right)
\right)  $. Since $\eta$ is injective, this yields $\sigma\left(  a-1\right)
=\sigma\left(  b-1\right)  $. Hence, $a-1=b-1$ (since the map $\sigma$ is also
bijective). Thus, $a=b$, which contradicts $a\neq b$. This contradiction shows
that our assumption (that $a\neq b$) was wrong. Thus, we cannot have $a\neq
b$. We therefore have $a=b$.

Now, forget that we fixed $a$ and $b$. We thus have proven that if $a$ and $b$
are two elements of $\left\{  1,2,\ldots,n\right\}  $ such that $\left(
\operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  a\right)  =\left(
\operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  b\right)  $, then
$a=b$. In other words, the map $\operatorname*{ins}\nolimits_{m}\sigma$ is injective.

Next, let us notice that $\left(  \operatorname*{ins}\nolimits_{m}%
\sigma\right)  \left(  1\right)  $ is well-defined (since $1\in\left\{
1,2,\ldots,n\right\}  $), and the definition of $\operatorname*{ins}%
\nolimits_{m}\sigma$ yields%
\[
\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  1\right)
=\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  1-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if
}1\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }1=1
\end{array}
\right.  =m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }1=1\right)  .
\]


Next, let us show that the map $\operatorname*{ins}\nolimits_{m}\sigma$ is
surjective. In order to do so, we will show that every $j\in\left\{
1,2,\ldots,n\right\}  $ satisfies $j\in\left(  \operatorname*{ins}%
\nolimits_{m}\sigma\right)  \left(  \left\{  1,2,\ldots,n\right\}  \right)  $.

Let $j\in\left\{  1,2,\ldots,n\right\}  $. We are going to show that
$j\in\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  \left\{
1,2,\ldots,n\right\}  \right)  $.

We must be in one of the following two cases:

\textit{Case 1:} We have $j=m$.

\textit{Case 2:} We have $j\neq m$.

Let us first consider Case 1. In this case, we have $j=m$. Recall that
$\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  1\right)  =m$.
Hence, $m=\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
\underbrace{1}_{\in\left\{  1,2,\ldots,n\right\}  }\right)  \in\left(
\operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  \left\{  1,2,\ldots
,n\right\}  \right)  $. Now, $j=m\in\left(  \operatorname*{ins}\nolimits_{m}%
\sigma\right)  \left(  \left\{  1,2,\ldots,n\right\}  \right)  $. Hence,
$j\in\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  \left\{
1,2,\ldots,n\right\}  \right)  $ is proven in Case 1.

Let us now consider Case 2. In this case, we have $j\neq m$. Combining
$j\in\left\{  1,2,\ldots,n\right\}  $ and $j\neq m$, we obtain $j\in\left\{
1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  =\eta\left(  \left\{
1,2,\ldots,n-1\right\}  \right)  $ (because $\eta$ is surjective). Hence,
there exists some $i\in\left\{  1,2,\ldots,n-1\right\}  $ such that
$\eta\left(  i\right)  =j$. Consider this $i$. We have $i\in\left\{
1,2,\ldots,n-1\right\}  $, and thus the element $\sigma^{-1}\left(  i\right)
$ of $\left\{  1,2,\ldots,n-1\right\}  $ is well-defined (because $\sigma$ is
a permutation of $\left\{  1,2,\ldots,n-1\right\}  $). We have $\sigma
^{-1}\left(  i\right)  \in\left\{  1,2,\ldots,n-1\right\}  $, thus
$\sigma^{-1}\left(  i\right)  +1\in\left\{  2,3,\ldots,n\right\}
\subseteq\left\{  1,2,\ldots,n\right\}  $. Thus, $\left(  \operatorname*{ins}%
\nolimits_{m}\sigma\right)  \left(  \sigma^{-1}\left(  i\right)  +1\right)  $
is well-defined. Also, $\sigma^{-1}\left(  i\right)  +1\neq1$ (since
$\sigma^{-1}\left(  i\right)  +1\in\left\{  2,3,\ldots,n\right\}  $). The
definition of $\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
\sigma^{-1}\left(  i\right)  +1\right)  $ yields%
\begin{align*}
\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  \sigma
^{-1}\left(  i\right)  +1\right)   &  =\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  \left(  \sigma^{-1}\left(  i\right)  +1\right)
-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if }\sigma^{-1}\left(
i\right)  +1\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }\sigma^{-1}\left(  i\right)  +1=1
\end{array}
\right. \\
&  =\eta\left(  \sigma\left(  \underbrace{\left(  \sigma^{-1}\left(  i\right)
+1-1\right)  }_{=\sigma^{-1}\left(  i\right)  }\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma^{-1}\left(  i\right)
+1\neq1\right) \\
&  =\eta\left(  \underbrace{\sigma\left(  \sigma^{-1}\left(  i\right)
\right)  }_{=i}\right)  =\eta\left(  i\right)  =j.
\end{align*}
Hence,%
\[
j=\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
\underbrace{\sigma^{-1}\left(  i\right)  +1}_{\in\left\{  1,2,\ldots
,n\right\}  }\right)  \in\left(  \operatorname*{ins}\nolimits_{m}%
\sigma\right)  \left(  \left\{  1,2,\ldots,n\right\}  \right)  .
\]
Thus, $j\in\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(
\left\{  1,2,\ldots,n\right\}  \right)  $ is proven in Case 2.

We now have proven $j\in\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)
\left(  \left\{  1,2,\ldots,n\right\}  \right)  $ in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this yields that
$j\in\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  \left\{
1,2,\ldots,n\right\}  \right)  $ always holds.

Now, forget that we fixed $j$. We thus have shown that every $j\in\left\{
1,2,\ldots,n\right\}  $ satisfies $j\in\left(  \operatorname*{ins}%
\nolimits_{m}\sigma\right)  \left(  \left\{  1,2,\ldots,n\right\}  \right)  $.
In other words, $\left\{  1,2,\ldots,n\right\}  \subseteq\left(
\operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  \left\{  1,2,\ldots
,n\right\}  \right)  $. In other words, the map $\operatorname*{ins}%
\nolimits_{m}\sigma$ is surjective. Combining this with the fact that
$\operatorname*{ins}\nolimits_{m}\sigma$ is injective, we conclude that
$\operatorname*{ins}\nolimits_{m}\sigma$ is bijective. Thus,
$\operatorname*{ins}\nolimits_{m}\sigma$ is a bijective map $\left\{
1,2,\ldots,n\right\}  \rightarrow\left\{  1,2,\ldots,n\right\}  $. In other
words, $\operatorname*{ins}\nolimits_{m}\sigma$ is a permutation of $\left\{
1,2,\ldots,n\right\}  $. In other words, $\operatorname*{ins}\nolimits_{m}%
\sigma$ is an element of $S_{n}$. We thus have shown that $\operatorname*{ins}%
\nolimits_{m}\sigma\in S_{n}$.

We now know that $\operatorname*{ins}\nolimits_{m}\sigma$ is an element of
$S_{n}$ and satisfies $\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)
\left(  1\right)  =m$. In other words,%
\[
\operatorname*{ins}\nolimits_{m}\sigma\in\left\{  \tau\in S_{n}\ \mid
\ \tau\left(  1\right)  =m\right\}  =S_{n,m}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{1405.pf.e.Snk})}\right)  .
\]
Thus, $\operatorname*{ins}\nolimits_{m}\sigma$ is a well-defined element of
$S_{n,m}$.

Now, forget that we fixed $\sigma$. We thus have proven that for every
$\sigma\in S_{n-1}$, the map $\operatorname*{ins}\nolimits_{m}\sigma$ is a
well-defined element of $S_{n,m}$. In other words, for every $\sigma\in
S_{n-1}$, the right hand side of (\ref{1405.iotak}) is a well-defined element
of $S_{n,m}$. Hence, the map $\iota_{m}:S_{n-1}\rightarrow S_{n,m}$ is
well-defined (because the map $\iota_{m}:S_{n-1}\rightarrow S_{n,m}$ was
defined by (\ref{1405.iotak})). Lemma~\ref{Lemma1405} \textbf{(e)} is thus proven.

\textbf{(f)} Let $\tau\in S_{n,m}$. Then, $\tau\in S_{n,m}=\left\{  \sigma\in
S_{n}\ \mid\ \sigma\left(  1\right)  =m\right\}  $. In other words, $\tau$ is
an element of $S_{n}$ and satisfies $\tau\left(  1\right)  =m$.

We have $\tau\in S_{n}$. Thus, $\tau$ is a permutation of $\left\{
1,2,\ldots,n\right\}  $, hence a bijection. In particular, this yields that
$\tau$ is injective.

For every $i\in\left\{  1,2,\ldots,n-1\right\}  $, the element $\omega\left(
\tau\left(  i+1\right)  \right)  $ is a well-defined element of $\left\{
1,2,\ldots,n-1\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{
1,2,\ldots,n-1\right\}  $. We need to prove that $\omega\left(  \tau\left(
i+1\right)  \right)  $ is a well-defined element of $\left\{  1,2,\ldots
,n-1\right\}  $.
\par
We have $i\in\left\{  1,2,\ldots,n-1\right\}  $, so that $i+1\in\left\{
2,3,\ldots,n\right\}  $ and thus $i+1\neq1$.
\par
We have $i+1\in\left\{  2,3,\ldots,n\right\}  \subseteq\left\{  1,2,\ldots
,n\right\}  $, so that $\tau\left(  i+1\right)  $ is well-defined.
\par
If we had $\tau\left(  i+1\right)  =\tau\left(  1\right)  $, then we would
have $i+1=i$ (since $\tau$ is injective), which would contradict $i+1\neq i$.
Hence, we don't have $\tau\left(  i+1\right)  =\tau\left(  1\right)  $. Thus,
we have $\tau\left(  i+1\right)  \neq\tau\left(  1\right)  =m$. Since
$\tau\left(  i+1\right)  \in\left\{  1,2,\ldots,n\right\}  $ (because the
target of $\tau$ is $\left\{  1,2,\ldots,n\right\}  $) and $\tau\left(
i+1\right)  \neq m$, we must have $\tau\left(  i+1\right)  \in\left\{
1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $. Thus, $\omega\left(
\tau\left(  i+1\right)  \right)  $ is a well-defined element of $\left\{
1,2,\ldots,n-1\right\}  $.}. In other words, for every $i\in\left\{
1,2,\ldots,n-1\right\}  $, the right hand side of (\ref{1405.delk}) is a
well-defined element of $\left\{  1,2,\ldots,n-1\right\}  $. Hence, the map
$\operatorname*{del}\nolimits_{m}\sigma:\left\{  1,2,\ldots,n-1\right\}
\rightarrow\left\{  1,2,\ldots,n-1\right\}  $ is well-defined (because the map
$\operatorname*{del}\nolimits_{m}\sigma:\left\{  1,2,\ldots,n-1\right\}
\rightarrow\left\{  1,2,\ldots,n-1\right\}  $ was defined by (\ref{1405.delk}%
)). Lemma~\ref{Lemma1405} \textbf{(f)} is thus proven.

\textbf{(g)} Let $\tau\in S_{n,m}$. We are going to show that
$\operatorname*{del}\nolimits_{m}\tau\in S_{n-1}$.

We have $\tau\in S_{n,m}=\left\{  \sigma\in S_{n}\ \mid\ \sigma\left(
1\right)  =m\right\}  $ (by the definition of $S_{n,m}$). In other words,
$\tau$ is an element of $S_{n}$ and satisfies $\tau\left(  1\right)  =m$.

Consider the map $\eta$ defined in Lemma~\ref{Lemma1405} \textbf{(a)} and the
map $\omega$ defined in Lemma~\ref{Lemma1405} \textbf{(b)}. The maps $\eta$
and $\omega$ are mutually inverse (by Lemma~\ref{Lemma1405} \textbf{(c)}).
Thus, $\omega$ is bijective. Hence, $\omega$ is injective and surjective. On
the other hand, $\tau$ is an element of $S_{n}$, thus a permutation of
$\left\{  1,2,\ldots,n\right\}  $. Hence, $\tau$ is bijective. Thus, the
inverse $\tau^{-1}$ of $\tau$ is also bijective, and therefore injective. The
target of $\tau^{-1}$ is $\left\{  1,2,\ldots,n\right\}  $ (since $\tau
^{-1}\in S_{n}$).

We know that $\operatorname*{del}\nolimits_{m}\tau$ is a map $\left\{
1,2,\ldots,n-1\right\}  \rightarrow\left\{  1,2,\ldots,n-1\right\}  $. We now
will prove that this map $\operatorname*{del}\nolimits_{m}\tau$ is injective.

In fact, let $a$ and $b$ be two elements of $\left\{  1,2,\ldots,n-1\right\}
$ such that $\left(  \operatorname*{del}\nolimits_{m}\tau\right)  \left(
a\right)  =\left(  \operatorname*{del}\nolimits_{m}\tau\right)  \left(
b\right)  $.

The definition of $\operatorname*{del}\nolimits_{m}\sigma$ yields $\left(
\operatorname*{del}\nolimits_{m}\sigma\right)  \left(  a\right)
=\omega\left(  \tau\left(  a+1\right)  \right)  $. Hence, $\omega\left(
\tau\left(  a+1\right)  \right)  =\left(  \operatorname*{del}\nolimits_{m}%
\sigma\right)  \left(  a\right)  $.

The definition of $\operatorname*{del}\nolimits_{m}\sigma$ yields $\left(
\operatorname*{del}\nolimits_{m}\sigma\right)  \left(  b\right)
=\omega\left(  \tau\left(  b+1\right)  \right)  $. Now,%
\[
\omega\left(  \tau\left(  a+1\right)  \right)  =\left(  \operatorname*{del}%
\nolimits_{m}\sigma\right)  \left(  a\right)  =\left(  \operatorname*{del}%
\nolimits_{m}\sigma\right)  \left(  b\right)  =\omega\left(  \tau\left(
b+1\right)  \right)  .
\]
Since $\omega$ is injective, this yields $\tau\left(  a+1\right)  =\tau\left(
b+1\right)  $. Thus, $a+1=b+1$ (since $\tau$ is injective), so that $a=b$.

Now, forget that we fixed $a$ and $b$. We thus have proven that if $a$ and $b$
are two elements of $\left\{  1,2,\ldots,n-1\right\}  $ such that $\left(
\operatorname*{del}\nolimits_{m}\tau\right)  \left(  a\right)  =\left(
\operatorname*{del}\nolimits_{m}\tau\right)  \left(  b\right)  $, then $a=b$.
In other words, the map $\operatorname*{del}\nolimits_{m}\tau$ is injective.

Next, let us show that the map $\operatorname*{del}\nolimits_{m}\tau$ is
surjective. In order to do so, we will show that every $j\in\left\{
1,2,\ldots,n-1\right\}  $ satisfies $j\in\left(  \operatorname*{del}%
\nolimits_{m}\tau\right)  \left(  \left\{  1,2,\ldots,n-1\right\}  \right)  $.

Let $j\in\left\{  1,2,\ldots,n-1\right\}  $. We are going to show that
$j\in\left(  \operatorname*{del}\nolimits_{m}\tau\right)  \left(  \left\{
1,2,\ldots,n-1\right\}  \right)  $.

We have $j\in\left\{  1,2,\ldots,n-1\right\}  $. Thus, $\eta\left(  j\right)
$ is a well-defined element of $\left\{  1,2,\ldots,n\right\}  \setminus
\left\{  m\right\}  $ (since $\eta$ is a map $\left\{  1,2,\ldots,n-1\right\}
\rightarrow\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $).
Thus, $\eta\left(  j\right)  \in\left\{  1,2,\ldots,n\right\}  \setminus
\left\{  m\right\}  \subseteq\left\{  1,2,\ldots,n\right\}  $, so that
$\tau^{-1}\left(  \eta\left(  j\right)  \right)  $ is well-defined (since
$\tau^{-1}\in S_{n}$). Moreover, $\eta\left(  j\right)  \neq m$ (since
$\eta\left(  j\right)  \in\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  $). If we would have $\tau^{-1}\left(  \eta\left(  j\right)
\right)  =\tau^{-1}\left(  m\right)  $, then we would have $\eta\left(
j\right)  =m$ (since $\tau^{-1}$ is injective), which would contradict the
fact that $\eta\left(  j\right)  \neq m$. Hence, we cannot have $\tau
^{-1}\left(  \eta\left(  j\right)  \right)  =\tau^{-1}\left(  m\right)  $.
Thus, we have $\tau^{-1}\left(  \eta\left(  j\right)  \right)  \neq\tau
^{-1}\left(  m\right)  =1$ (since $\tau\left(  1\right)  =m$). But $\tau
^{-1}\left(  \eta\left(  j\right)  \right)  \in\left\{  1,2,\ldots,n\right\}
$ (since the target of $\tau^{-1}$ is $\left\{  1,2,\ldots,n\right\}  $).
Combined with $\tau^{-1}\left(  \eta\left(  j\right)  \right)  \neq1$, this
yields $\tau^{-1}\left(  \eta\left(  j\right)  \right)  \in\left\{
1,2,\ldots,n\right\}  \setminus\left\{  1\right\}  =\left\{  2,3,\ldots
,n\right\}  $. Hence, $\tau^{-1}\left(  \eta\left(  j\right)  \right)
-1\in\left\{  1,2,\ldots,n-1\right\}  $. Denote the element $\tau^{-1}\left(
\eta\left(  j\right)  \right)  -1$ of $\left\{  1,2,\ldots,n-1\right\}  $ by
$g$. Thus, $g\in\left\{  1,2,\ldots,n-1\right\}  $ and%
\[
\tau\left(  \underbrace{g}_{=\tau^{-1}\left(  \eta\left(  j\right)  \right)
-1}+1\right)  =\tau\left(  \underbrace{\tau^{-1}\left(  \eta\left(  j\right)
\right)  -1+1}_{=\tau^{-1}\left(  \eta\left(  j\right)  \right)  }\right)
=\tau\left(  \tau^{-1}\left(  \eta\left(  j\right)  \right)  \right)
=\eta\left(  j\right)  .
\]
Now, the definition of $\operatorname*{del}\nolimits_{m}\tau$ yields%
\[
\left(  \operatorname*{del}\nolimits_{m}\tau\right)  \left(  g\right)
=\omega\left(  \underbrace{\tau\left(  g+1\right)  }_{=\eta\left(  j\right)
}\right)  =\omega\left(  \eta\left(  j\right)  \right)  =\underbrace{\left(
\omega\circ\eta\right)  }_{\substack{=\operatorname*{id}\\\text{(since the
maps }\eta\text{ and }\omega\\\text{are mutually inverse)}}}\left(  j\right)
=\operatorname*{id}\left(  j\right)  =j.
\]
Thus,%
\[
j=\left(  \operatorname*{del}\nolimits_{m}\tau\right)  \left(  \underbrace{g}%
_{\in\left\{  1,2,\ldots,n-1\right\}  }\right)  \in\left(  \operatorname*{del}%
\nolimits_{m}\tau\right)  \left(  \left\{  1,2,\ldots,n-1\right\}  \right)  .
\]


Now, forget that we fixed $j$. We thus have shown that every $j\in\left\{
1,2,\ldots,n-1\right\}  $ satisfies $j\in\left(  \operatorname*{del}%
\nolimits_{m}\tau\right)  \left(  \left\{  1,2,\ldots,n-1\right\}  \right)  $.
In other words,%
\[
\left\{  1,2,\ldots,n-1\right\}  \subseteq\left(  \operatorname*{del}%
\nolimits_{m}\tau\right)  \left(  \left\{  1,2,\ldots,n-1\right\}  \right)  .
\]
In other words, the map $\operatorname*{del}\nolimits_{m}\tau$ is surjective.
Combining this with the fact that $\operatorname*{del}\nolimits_{m}\tau$ is
injective, we conclude that $\operatorname*{del}\nolimits_{m}\tau$ is
bijective. Thus, $\operatorname*{del}\nolimits_{m}\tau$ is a bijective map
$\left\{  1,2,\ldots,n-1\right\}  \rightarrow\left\{  1,2,\ldots,n-1\right\}
$. In other words, $\operatorname*{del}\nolimits_{m}\tau$ is a permutation of
$\left\{  1,2,\ldots,n-1\right\}  $. In other words, $\operatorname*{del}%
\nolimits_{m}\tau$ is an element of $S_{n-1}$. We thus have shown that
$\operatorname*{del}\nolimits_{m}\tau\in S_{n-1}$. Thus, $\operatorname*{del}%
\nolimits_{m}\tau$ is a well-defined element of $S_{n-1}$.

Now, forget that we fixed $\tau$. We thus have proven that for every $\tau\in
S_{n,m}$, the map $\operatorname*{del}\nolimits_{m}\tau$ is a well-defined
element of $S_{n-1}$. In other words, for every $\tau\in S_{n,m}$, the right
hand side of (\ref{1405.pik}) is a well-defined element of $S_{n-1}$. Hence,
the map $\pi_{m}:S_{n,m}\rightarrow S_{n-1}$ is well-defined (because the map
$\pi_{m}:S_{n,m}\rightarrow S_{n-1}$ was defined by (\ref{1405.pik})).
Lemma~\ref{Lemma1405} \textbf{(g)} is thus proven.

\textbf{(h)} Consider the map $\eta$ defined in Lemma~\ref{Lemma1405}
\textbf{(a)} and the map $\omega$ defined in Lemma~\ref{Lemma1405}
\textbf{(b)}. The maps $\eta$ and $\omega$ are mutually inverse (by
Lemma~\ref{Lemma1405} \textbf{(c)}). Clearly, the compositions $\iota_{m}%
\circ\pi_{m}$ and $\pi_{m}\circ\iota_{m}$ are well-defined.

Consider also the notation $\operatorname*{ins}\nolimits_{m}\sigma$ (for every
$\sigma\in S_{n-1}$) defined in Lemma~\ref{Lemma1405} \textbf{(d)}, and the
notation $\operatorname*{del}\nolimits_{m}\tau$ (for every $\tau\in S_{n,m}$)
defined in Lemma~\ref{Lemma1405} \textbf{(f)}.

Let us first prove that $\iota_{m}\circ\pi_{m}=\operatorname*{id}$.

Indeed, let $\tau\in S_{n,m}$ be arbitrary. We have $\tau\in S_{n,m}=\left\{
\sigma\in S_{n}\ \mid\ \sigma\left(  1\right)  =m\right\}  $ (by the
definition of $S_{n,m}$). In other words, $\tau$ is an element of $S_{n}$ and
satisfies $\tau\left(  1\right)  =m$.

Let $\tau^{\prime}=\left(  \iota_{m}\circ\pi_{m}\right)  \left(  \tau\right)
$. Notice that $\tau^{\prime}\in S_{n,m}$. We have%
\[
\tau^{\prime}=\left(  \iota_{m}\circ\pi_{m}\right)  \left(  \tau\right)
=\iota_{m}\left(  \underbrace{\pi_{m}\left(  \tau\right)  }%
_{\substack{=\operatorname*{del}\nolimits_{m}\tau\\\text{(by the definition of
}\pi_{m}\text{)}}}\right)  =\iota_{m}\left(  \operatorname*{del}%
\nolimits_{m}\tau\right)  =\operatorname*{ins}\nolimits_{m}\left(
\operatorname*{del}\nolimits_{m}\tau\right)
\]
(by the definition of $\iota_{m}$).

Let $i\in\left\{  1,2,\ldots,n\right\}  $. We are going to prove that
$\tau\left(  i\right)  =\tau^{\prime}\left(  i\right)  $.

Indeed, we must be in one of the following two cases:

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

\textit{Case 2:} We have $i\neq1$.

Let us first consider Case 1. In this case, we have $i=1$. Since $i=1$, we
have%
\begin{align*}
\tau^{\prime}\left(  i\right)   &  =\underbrace{\tau^{\prime}}%
_{=\operatorname*{ins}\nolimits_{m}\left(  \operatorname*{del}\nolimits_{m}%
\tau\right)  }\left(  1\right)  =\left(  \operatorname*{ins}\nolimits_{m}%
\left(  \operatorname*{del}\nolimits_{m}\tau\right)  \right)  \left(
1\right)  =\left\{
\begin{array}
[c]{c}%
\eta\left(  \left(  \operatorname*{del}\nolimits_{m}\tau\right)  \left(
1-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if }1\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }1=1
\end{array}
\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{ins}%
\nolimits_{m}\left(  \operatorname*{del}\nolimits_{m}\tau\right)  \right) \\
&  =m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }1=1\right) \\
&  =\tau\left(  \underbrace{1}_{=i}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\tau\left(  1\right)  =m\right) \\
&  =\tau\left(  i\right)  .
\end{align*}
Thus, $\tau\left(  i\right)  =\tau^{\prime}\left(  i\right)  $. We have thus
proven $\tau\left(  i\right)  =\tau^{\prime}\left(  i\right)  $ in Case 1.

Let us now consider Case 2. In this case, we have $i\neq1$. We have%
\begin{align*}
\underbrace{\tau^{\prime}}_{=\operatorname*{ins}\nolimits_{m}\left(
\operatorname*{del}\nolimits_{m}\tau\right)  }\left(  i\right)   &  =\left(
\operatorname*{ins}\nolimits_{m}\left(  \operatorname*{del}\nolimits_{m}%
\tau\right)  \right)  \left(  i\right)  =\left\{
\begin{array}
[c]{c}%
\eta\left(  \left(  \operatorname*{del}\nolimits_{m}\tau\right)  \left(
i-1\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ \text{if }i\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }i=1
\end{array}
\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{ins}%
\nolimits_{m}\left(  \operatorname*{del}\nolimits_{m}\tau\right)  \right) \\
&  =\eta\left(  \underbrace{\left(  \operatorname*{del}\nolimits_{m}%
\tau\right)  \left(  i-1\right)  }_{\substack{=\omega\left(  \tau\left(
\left(  i-1\right)  +1\right)  \right)  \\\text{(by the definition of
}\operatorname*{del}\nolimits_{m}\tau\text{)}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\neq1\right) \\
&  =\eta\left(  \omega\left(  \tau\left(  \underbrace{\left(  i-1\right)
+1}_{=i}\right)  \right)  \right)  =\eta\left(  \omega\left(  \tau\left(
i\right)  \right)  \right)  =\underbrace{\left(  \eta\circ\omega\right)
}_{\substack{=\operatorname*{id}\\\text{(since the maps }\eta\text{ and
}\omega\\\text{are mutually inverse)}}}\left(  \tau\left(  i\right)  \right)
\\
&  =\operatorname*{id}\left(  \tau\left(  i\right)  \right)  =\tau\left(
i\right)  .
\end{align*}
In other words, $\tau\left(  i\right)  =\tau^{\prime}\left(  i\right)  $. We
have thus proven $\tau\left(  i\right)  =\tau^{\prime}\left(  i\right)  $ in
Case 2.

We have therefore proven $\tau\left(  i\right)  =\tau^{\prime}\left(
i\right)  $ in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this shows that $\tau\left(  i\right)  =\tau^{\prime}\left(
i\right)  $ always holds.

Now, forget that we fixed $i$. We thus have shown that $\tau\left(  i\right)
=\tau^{\prime}\left(  i\right)  $ for every $i\in\left\{  1,2,\ldots
,n\right\}  $. In other words, $\tau=\tau^{\prime}$. Hence, $\tau=\tau
^{\prime}=\left(  \iota_{m}\circ\pi_{m}\right)  \left(  \tau\right)  $, so
that $\left(  \iota_{m}\circ\pi_{m}\right)  \left(  \tau\right)
=\tau=\operatorname*{id}\left(  \tau\right)  $.

Now, forget that we fixed $\tau$. We thus have proven that $\left(  \iota
_{m}\circ\pi_{m}\right)  \left(  \tau\right)  =\operatorname*{id}\left(
\tau\right)  $ for every $\tau\in S_{n,m}$. In other words, $\iota_{m}\circ
\pi_{m}=\operatorname*{id}$.

Next, we will show that $\pi_{m}\circ\iota_{m}=\operatorname*{id}$.

Let $\sigma\in S_{n-1}$. Let $\sigma^{\prime}=\left(  \pi_{m}\circ\iota
_{m}\right)  \left(  \sigma\right)  $. Notice that $\sigma^{\prime}\in
S_{n-1}$. We have%
\[
\sigma^{\prime}=\left(  \pi_{m}\circ\iota_{m}\right)  \left(  \sigma\right)
=\pi_{m}\left(  \underbrace{\iota_{m}\left(  \sigma\right)  }%
_{\substack{=\operatorname*{ins}\nolimits_{m}\sigma\\\text{(by the definition
of }\iota_{m}\text{)}}}\right)  =\pi_{m}\left(  \operatorname*{ins}%
\nolimits_{m}\sigma\right)  =\operatorname*{del}\nolimits_{m}\left(
\operatorname*{ins}\nolimits_{m}\sigma\right)
\]
(by the definition of $\pi_{m}$).

Let $j\in\left\{  1,2,\ldots,n-1\right\}  $. We are going to prove that
$\sigma\left(  j\right)  =\sigma^{\prime}\left(  j\right)  $. We have
$j\in\left\{  1,2,\ldots,n-1\right\}  $, and therefore $j+1\in\left\{
2,3,\ldots,n\right\}  $ and thus $j+1\neq1$. Now, the definition of
$\operatorname*{ins}\nolimits_{m}\sigma$ yields%
\begin{align*}
\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  j+1\right)   &
=\left\{
\begin{array}
[c]{c}%
\eta\left(  \sigma\left(  \left(  j+1\right)  -1\right)  \right)
,\ \ \ \ \ \ \ \ \ \ \text{if }j+1\neq1;\\
m,\ \ \ \ \ \ \ \ \ \ \text{if }j+1=1
\end{array}
\right. \\
&  =\eta\left(  \sigma\left(  \underbrace{\left(  j+1\right)  -1}_{=j}\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }j+1\neq1\right) \\
&  =\eta\left(  \sigma\left(  j\right)  \right)  .
\end{align*}
Now,%
\begin{align*}
\underbrace{\sigma^{\prime}}_{=\operatorname*{del}\nolimits_{m}\left(
\operatorname*{ins}\nolimits_{m}\sigma\right)  }\left(  j\right)   &  =\left(
\operatorname*{del}\nolimits_{m}\left(  \operatorname*{ins}\nolimits_{m}%
\sigma\right)  \right)  \left(  j\right)  =\omega\left(  \underbrace{\left(
\operatorname*{ins}\nolimits_{m}\sigma\right)  \left(  j+1\right)  }%
_{=\eta\left(  \sigma\left(  j\right)  \right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{del}%
\nolimits_{m}\left(  \operatorname*{ins}\nolimits_{m}\sigma\right)  \right) \\
&  =\omega\left(  \eta\left(  \sigma\left(  j\right)  \right)  \right)
=\underbrace{\left(  \omega\circ\eta\right)  }_{\substack{=\operatorname*{id}%
\\\text{(since the maps }\eta\text{ and }\omega\\\text{are mutually inverse)}%
}}\left(  \sigma\left(  j\right)  \right) \\
&  =\operatorname*{id}\left(  \sigma\left(  j\right)  \right)  =\sigma\left(
j\right)  .
\end{align*}
In other words, $\sigma\left(  j\right)  =\sigma^{\prime}\left(  j\right)  $.

Now forget that we fixed $j$. We thus have shown that every $j\in\left\{
1,2,\ldots,n-1\right\}  $ satisfies $\sigma\left(  j\right)  =\sigma^{\prime
}\left(  j\right)  $. In other words, $\sigma=\sigma^{\prime}$. Thus,
$\sigma=\sigma^{\prime}=\left(  \pi_{m}\circ\iota_{m}\right)  \left(
\sigma\right)  $, so that $\left(  \pi_{m}\circ\iota_{m}\right)  \left(
\sigma\right)  =\sigma=\operatorname*{id}\left(  \sigma\right)  $.

Now forget that we fixed $\sigma$. We thus have proven that every $\sigma\in
S_{n-1}$ satisfies $\left(  \pi_{m}\circ\iota_{m}\right)  \left(
\sigma\right)  =\operatorname*{id}\left(  \sigma\right)  $. In other words,
$\pi_{m}\circ\iota_{m}=\operatorname*{id}$. Combined with $\iota_{m}\circ
\pi_{m}=\operatorname*{id}$, this yields that the maps $\iota_{m}$ and
$\pi_{m}$ are mutually inverse. Lemma~\ref{Lemma1405} \textbf{(h)} is
therefore proven.
\end{proof}

We can now proceed to the proof of Theorem~\ref{Theorem1401}.

\begin{proof}
[Proof of Theorem~\ref{Theorem1401}.]\textbf{(a)} Let $s\in\left\{
0,1,\ldots,r-1\right\}  $. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$
elements of $\operatorname*{Prim}H$. Then, $a_{1}a_{2}\cdots a_{s}\in\left(
\operatorname*{Prim}H\right)  ^{s}$ (since $a_{i}\in\operatorname*{Prim}H$ for
every $i\in\left\{  1,2,\ldots,s\right\}  $). Hence,%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{a_{1}%
a_{2}\cdots a_{s}}_{\in\left(  \operatorname*{Prim}H\right)  ^{s}}\right)
\in\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \left(
\operatorname*{Prim}H\right)  ^{s}\right)  =0
\]
(by Lemma~\ref{Lemma1404}). Hence, $\left(  f_{1}\ast f_{2}\ast\cdots\ast
f_{r}\right)  \left(  a_{1}a_{2}\cdots a_{s}\right)  =0$. This proves
Theorem~\ref{Theorem1401} \textbf{(a)}.

\textbf{(b)} We are going to prove Theorem~\ref{Theorem1401} \textbf{(b)} by
induction over $r$:

\textit{Induction base:} If $r=0$, then Theorem~\ref{Theorem1401} \textbf{(b)}
is true\footnote{\textit{Proof.} Assume that $r=0$. Let $k$ be a field. Let
$H$ be a $k$-bialgebra. Let $A$ be a $k$-algebra. Let $f_{1}$, $f_{2}$,
$\ldots$, $f_{0}$ be $0$ maps in $\mathfrak{g}\left(  H,A\right)  $ (that is,
no maps). Let $a_{1}$, $a_{2}$, $\ldots$, $a_{0}$ be $0$ elements of
$\operatorname*{Prim}H$ (that is, no elements). Since $r=0$, we have
$S_{r}=S_{0}=\left\{  \operatorname*{id}\right\}  $. But since $r=0$, we have
$f_{1}\ast f_{2}\ast\cdots\ast f_{r}=\left(  \text{empty product with respect
to convolution}\right)  =e_{H,A}$. Also, since $r=0$, we have $a_{1}%
a_{2}\cdots a_{r}=\left(  \text{empty product}\right)  =1_{H}$. Thus,
\begin{align*}
&  \underbrace{\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  }%
_{=e_{H,A}}\left(  \underbrace{a_{1}a_{2}\cdots a_{r}}_{=1_{H}}\right) \\
&  =\underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{H}}\left(  1_{H}\right)
=\left(  \eta_{A}\circ\varepsilon_{H}\right)  \left(  1_{H}\right)  =\eta
_{A}\left(  \underbrace{\varepsilon_{H}\left(  1_{H}\right)  }%
_{\substack{=1\\\text{(since }H\text{ is a bialgebra)}}}\right)  =\eta
_{A}\left(  1\right) \\
&  =1\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\eta_{A}\right) \\
&  =1_{A}%
\end{align*}
and%
\[
\underbrace{\sum_{\sigma\in S_{r}}}_{\substack{=\sum_{\sigma\in\left\{
\operatorname*{id}\right\}  }\\\text{(since }S_{r}=\left\{  \operatorname*{id}%
\right\}  \text{)}}}\underbrace{f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{r}\left(  a_{\sigma\left(  r\right)  }\right)  }_{\substack{=\left(
\text{empty product}\right)  \\\text{(since }r=0\text{)}}}=\sum_{\sigma
\in\left\{  \operatorname*{id}\right\}  }\underbrace{\left(  \text{empty
product}\right)  }_{=1_{A}}=\sum_{\sigma\in\left\{  \operatorname*{id}%
\right\}  }1_{A}=1_{A}.
\]
Hence, $\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a_{1}%
a_{2}\cdots a_{r}\right)  =1_{A}=\sum_{\sigma\in S_{r}}f_{1}\left(
a_{\sigma\left(  1\right)  }\right)  f_{2}\left(  a_{\sigma\left(  2\right)
}\right)  \cdots f_{r}\left(  a_{\sigma\left(  r\right)  }\right)  $. Hence,
Theorem~\ref{Theorem1401} \textbf{(b)} is true for $r=0$, qed.}. Hence, the
induction base is complete.

\textit{Induction step:} Let $n\in\mathbb{N}$ be positive. Assume that
Theorem~\ref{Theorem1401} \textbf{(b)} is proven for $r=n-1$. We now are going
to prove that Theorem~\ref{Theorem1401} \textbf{(b)} holds for $r=n$.

Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be a $k$-algebra. Let
$f_{1}$, $f_{2}$, $\ldots$, $f_{n}$ be $n$ maps in $\mathfrak{g}\left(
H,A\right)  $. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{n}$ be $n$ elements of
$\operatorname*{Prim}H$. We will prove that
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}a_{2}\cdots
a_{n}\right)  =\sum_{\sigma\in S_{n}}f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  .
\]


Let $g=f_{2}\ast f_{3}\ast\cdots\ast f_{n}$. Then, $g$ is a $k$-linear map
$H\rightarrow A$. We have
\[
f_{1}\ast f_{2}\ast\cdots\ast f_{n}=f_{1}\ast\underbrace{\left(  f_{2}\ast
f_{3}\ast\cdots\ast f_{n}\right)  }_{=g}=f_{1}\ast g=\mu_{A}\circ\left(
f_{1}\otimes g\right)  \circ\Delta
\]
(by the definition of convolution). Hence,%
\begin{align}
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}a_{2}\cdots
a_{n}\right)   &  =\left(  \mu_{A}\circ\left(  f_{1}\otimes g\right)
\circ\Delta\right)  \left(  a_{1}a_{2}\cdots a_{n}\right) \nonumber\\
&  =\mu_{A}\left(  \left(  f_{1}\otimes g\right)  \left(  \Delta\left(
a_{1}a_{2}\cdots a_{n}\right)  \right)  \right)  . \label{1401.pf.b.1}%
\end{align}
We are now going to study the term $\left(  f_{1}\otimes g\right)  \left(
\Delta\left(  a_{1}a_{2}\cdots a_{n}\right)  \right)  $ on the right hand side
of this.

We have%
\begin{align}
&  \left(  f_{1}\otimes g\right)  \left(  \underbrace{\Delta\left(  a_{1}%
a_{2}\cdots a_{n}\right)  }_{\substack{\in\sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{n}\right)
+\sum_{\ell=0}^{n-2}H\otimes\left(  \operatorname*{Prim}H\right)  ^{\ell
}\\\text{(by Lemma~\ref{Lemma1402}, applied to }V=\operatorname*{Prim}H\text{
and }s=n\text{)}}}\right) \nonumber\\
&  \in\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{n}\right)
+\sum_{\ell=0}^{n-2}H\otimes\left(  \operatorname*{Prim}H\right)  ^{\ell
}\right) \nonumber\\
&  \subseteq\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{n}\right)  \right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  f_{1}\otimes g\right)  \left(  1_{H}%
\otimes\left(  a_{1}a_{2}\cdots a_{n}\right)  \right)  +\left(  f_{1}\otimes
g\right)  \left(  \sum_{\ell=0}^{n-2}H\otimes\left(  \operatorname*{Prim}%
H\right)  ^{\ell}\right)  . \label{1401.pf.b.2}%
\end{align}


But $f_{1}\in\mathfrak{g}\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}  $. In other words,
$f_{1}$ is an element of $\mathcal{L}\left(  H,A\right)  $ and satisfies
$f_{1}\left(  1_{H}\right)  =0$. Now,%
\begin{equation}
\left(  f_{1}\otimes g\right)  \left(  1_{H}\otimes\left(  a_{1}a_{2}\cdots
a_{n}\right)  \right)  =\underbrace{f_{1}\left(  1_{H}\right)  }_{=0}\otimes
g\left(  a_{1}a_{2}\cdots a_{n}\right)  =0\otimes g\left(  a_{1}a_{2}\cdots
a_{n}\right)  =0. \label{1401.pf.b.3b}%
\end{equation}


Furthermore, every $\ell\in\left\{  0,1,\ldots,\left(  n-1\right)  -1\right\}
$ satisfies
\begin{equation}
\underbrace{g}_{=f_{2}\ast f_{3}\ast\cdots\ast f_{n}}\left(  \left(
\operatorname*{Prim}H\right)  ^{\ell}\right)  =\left(  f_{2}\ast f_{3}%
\ast\cdots\ast f_{n}\right)  \left(  \left(  \operatorname*{Prim}H\right)
^{\ell}\right)  =0 \label{1401.pf.b.3c.lem}%
\end{equation}
(by Lemma~\ref{Lemma1404}, applied to $n-1$, $\left(  f_{2},f_{3},\ldots
,f_{n}\right)  $ and $\ell$ instead of $r$, $\left(  f_{1},f_{2},\ldots
,f_{r}\right)  $ and $s$). Now,
\begin{align}
&  \left(  f_{1}\otimes g\right)  \left(  \sum_{\ell=0}^{n-2}H\otimes\left(
\operatorname*{Prim}H\right)  ^{\ell}\right) \nonumber\\
&  \subseteq\sum_{\ell=0}^{n-2}\underbrace{\left(  f_{1}\otimes g\right)
\left(  H\otimes\left(  \operatorname*{Prim}H\right)  ^{\ell}\right)
}_{\subseteq f_{1}\left(  H\right)  \otimes g\left(  \left(
\operatorname*{Prim}H\right)  ^{\ell}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }f_{1}\otimes g\text{ is a }k\text{-linear map}\right) \nonumber\\
&  \subseteq\sum_{\ell=0}^{n-2}f_{1}\left(  H\right)  \otimes
\underbrace{g\left(  \left(  \operatorname*{Prim}H\right)  ^{\ell}\right)
}_{\substack{=0\\\text{(by (\ref{1401.pf.b.3c.lem}) (since}\\\ell\in\left\{
0,1,\ldots,n-2\right\}  =\left\{  0,1,\ldots,\left(  n-1\right)  -1\right\}
\text{))}}}=\sum_{\ell=0}^{n-2}\underbrace{f_{1}\left(  H\right)  \otimes
0}_{=0}=\sum_{\ell=0}^{n-2}0=0. \label{1401.pf.b.3c}%
\end{align}


Now, (\ref{1401.pf.b.2}) becomes%
\begin{align*}
&  \left(  f_{1}\otimes g\right)  \left(  \Delta\left(  a_{1}a_{2}\cdots
a_{n}\right)  \right) \\
&  \in\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  f_{1}\otimes g\right)  \left(
1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{n}\right)  \right)  }%
_{\substack{=0\\\text{(by (\ref{1401.pf.b.3b}))}}}+\underbrace{\left(
f_{1}\otimes g\right)  \left(  \sum_{\ell=0}^{n-2}H\otimes\left(
\operatorname*{Prim}H\right)  ^{\ell}\right)  }_{\substack{\subseteq
0\\\text{(by (\ref{1401.pf.b.3b}))}}}\\
&  \subseteq\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{n}\right)  \right)  \right)  +0+0\\
&  =\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  \right)  .
\end{align*}
In other words,%
\begin{align}
&  \left(  f_{1}\otimes g\right)  \left(  \Delta\left(  a_{1}a_{2}\cdots
a_{n}\right)  \right) \nonumber\\
&  =\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  \right) \nonumber\\
&  =\sum_{m=1}^{n}\underbrace{\left(  f_{1}\otimes g\right)  \left(
a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(
a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)  \right)  }_{=f_{1}\left(
a_{m}\right)  \otimes g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f_{1}\otimes g\text{ is
}k\text{-linear}\right) \nonumber\\
&  =\sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \otimes g\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{n}\right)
\right)  . \label{1401.pf.b.5}%
\end{align}


Now, (\ref{1401.pf.b.1}) becomes%
\begin{align}
&  \left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}%
a_{2}\cdots a_{n}\right) \nonumber\\
&  =\mu_{A}\left(  \underbrace{\left(  f_{1}\otimes g\right)  \left(
\Delta\left(  a_{1}a_{2}\cdots a_{n}\right)  \right)  }_{\substack{=\sum
_{m=1}^{n}f_{1}\left(  a_{m}\right)  \otimes g\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)
\\\text{(by (\ref{1401.pf.b.5}))}}}\right) \nonumber\\
&  =\mu_{A}\left(  \sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \otimes g\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  \right) \nonumber\\
&  =\sum_{m=1}^{n}\underbrace{\mu_{A}\left(  f_{1}\left(  a_{m}\right)
\otimes g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(
a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)  \right)  }_{\substack{=f_{1}%
\left(  a_{m}\right)  \cdot g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)  \\\text{(since }\mu
_{A}\text{ is the multiplication map)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mu_{A}\text{ is
}k\text{-linear}\right) \nonumber\\
&  =\sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \cdot g\left(  \left(  a_{1}%
a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{n}\right)
\right)  . \label{1401.pf.b.6}%
\end{align}


Now, we are going to show that every $m\in\left\{  1,2,\ldots,n\right\}  $
satisfies%
\begin{equation}
g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  =\sum_{\substack{\sigma\in S_{n};\\\sigma\left(
1\right)  =m}}f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  f_{3}\left(
a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  . \label{1401.pf.b.m-part.goal}%
\end{equation}


\textit{Proof of (\ref{1401.pf.b.m-part.goal}):} Let $m\in\left\{
1,2,\ldots,n\right\}  $.

We are going to use the subset $S_{n,m}$ of $S_{n}$ defined in
Lemma~\ref{Lemma1405}.

We are going to use the map $\eta:\left\{  1,2,\ldots,n-1\right\}
\rightarrow\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}  $
defined in Lemma~\ref{Lemma1405} \textbf{(a)}. (In particular, this means that
we will not use the notation $\eta$ for the unity map of a $k$-algebra in this
proof, so as to avoid conflict of notation.) We are also going to use the map
$\omega:\left\{  1,2,\ldots,n\right\}  \setminus\left\{  m\right\}
\rightarrow\left\{  1,2,\ldots,n-1\right\}  $ defined in Lemma~\ref{Lemma1405}
\textbf{(b)}. Lemma~\ref{Lemma1405} \textbf{(c)} yields that the maps $\eta$
and $\omega$ are mutually inverse. Hence, these maps $\eta$ and $\omega$ are bijections.

We will use the notion of $\operatorname*{del}\nolimits_{m}\tau$ (for every
$\tau\in S_{n,m}$) defined in Lemma~\ref{Lemma1405} \textbf{(f)}. We will also
use the map $\pi_{m}:S_{n,m}\rightarrow S_{n-1}$ defined in
Lemma~\ref{Lemma1405} \textbf{(g)}. We will furthermore use the map $\iota
_{m}$ defined in Lemma~\ref{Lemma1405} \textbf{(e)}. Lemma~\ref{Lemma1405}
\textbf{(h)} shows that the maps $\iota_{m}$ and $\pi_{m}$ are mutually
inverse. Hence, these maps $\iota_{m}$ and $\pi_{m}$ are bijections.

For every $i\in\left\{  1,2,\ldots,n-1\right\}  $, we know that $\eta\left(
i\right)  $ is a well-defined element of $\left\{  1,2,\ldots,n\right\}
\setminus\left\{  m\right\}  $ (since $\eta$ is a map $\left\{  1,2,\ldots
,n-1\right\}  \rightarrow\left\{  1,2,\ldots,n\right\}  \setminus\left\{
m\right\}  $). Hence, for every $i\in\left\{  1,2,\ldots,n-1\right\}  $, we
have $\eta\left(  i\right)  \in\left\{  1,2,\ldots,n\right\}  \setminus
\left\{  m\right\}  \subseteq\left\{  1,2,\ldots,n\right\}  $. Thus, for every
$i\in\left\{  1,2,\ldots,n-1\right\}  $, the element $a_{\eta\left(  i\right)
}$ is a well-defined element of $\operatorname*{Prim}H$ (since $a_{1}$,
$a_{2}$, $\ldots$, $a_{n}$ are elements of $\operatorname*{Prim}H$). Hence, we
can define an $\left(  n-1\right)  $-tuple $\left(  b_{1},b_{2},\ldots
,b_{n-1}\right)  \in\left(  \operatorname*{Prim}H\right)  ^{\times\left(
n-1\right)  }$ of elements of $\operatorname*{Prim}H$ by setting%
\begin{equation}
\left(  b_{i}=a_{\eta\left(  i\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every
}i\in\left\{  1,2,\ldots,n-1\right\}  \right)  . \label{1401.pf.b.m-part.b_i}%
\end{equation}
Consider this $\left(  n-1\right)  $-tuple $\left(  b_{1},b_{2},\ldots
,b_{n-1}\right)  $.

It is easy to see that%
\begin{equation}
b_{1}b_{2}\cdots b_{n-1}=\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(
a_{m+1}a_{m+2}\cdots a_{n}\right)  \label{1401.pf.b.m-part.b_prod}%
\end{equation}
\footnote{\textit{Proof of (\ref{1401.pf.b.m-part.b_prod}):} Every
$i\in\left\{  1,2,\ldots,m-1\right\}  $ satisfies%
\begin{align*}
\eta\left(  i\right)   &  =\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1405.eta})}\right) \\
&  =i\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i<m\text{ (since }i\in\left\{
1,2,\ldots,m-1\right\}  \text{)}\right)  .
\end{align*}
Hence, every $i\in\left\{  1,2,\ldots,m-1\right\}  $ satisfies
\begin{align*}
b_{i}  &  =a_{\eta\left(  i\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1401.pf.b.m-part.b_i})}\right) \\
&  =a_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta\left(  i\right)
=i\right)  .
\end{align*}
Taking the product of these equalities over all $i\in\left\{  1,2,\ldots
,m-1\right\}  $, we obtain $b_{1}b_{2}\cdots b_{m-1}=a_{1}a_{2}\cdots a_{m-1}%
$.
\par
Every $i\in\left\{  m,m+1,\ldots,n-1\right\}  $ satisfies%
\begin{align*}
\eta\left(  i\right)   &  =\left\{
\begin{array}
[c]{c}%
i,\ \ \ \ \ \ \ \ \ \ \text{if }i<m;\\
i+1,\ \ \ \ \ \ \ \ \ \ \text{if }i\geq m
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1405.eta})}\right) \\
&  =i+1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\geq m\text{ (since }%
i\in\left\{  m,m+1,\ldots,n-1\right\}  \text{)}\right)  .
\end{align*}
Hence, every $i\in\left\{  m,m+1,\ldots,n-1\right\}  $ satisfies%
\begin{align*}
b_{i}  &  =a_{\eta\left(  i\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1401.pf.b.m-part.b_i})}\right) \\
&  =a_{i+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta\left(  i\right)
=i+1\right)  .
\end{align*}
Taking the product of these equalities over all $i\in\left\{  m,m+1,\ldots
,n-1\right\}  $, we obtain $b_{m}b_{m+1}\cdots b_{n-1}=a_{m+1}a_{\left(
m+1\right)  +1}\cdots a_{\left(  n-1\right)  +1}$.
\par
Now,%
\begin{align*}
b_{1}b_{2}\cdots b_{n-1}  &  =\left(  \underbrace{b_{1}b_{2}\cdots b_{m-1}%
}_{=a_{1}a_{2}\cdots a_{m-1}}\right)  \left(  \underbrace{b_{m}b_{m+1}\cdots
b_{n-1}}_{\substack{=a_{m+1}a_{\left(  m+1\right)  +1}\cdots a_{\left(
n-1\right)  +1}\\=a_{m+1}a_{m+2}\cdots a_{n}}}\right) \\
&  =\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  .
\end{align*}
This proves (\ref{1401.pf.b.m-part.b_prod}).}.

On the other hand, we can apply Theorem~\ref{Theorem1401} \textbf{(b)} to
$n-1$, $\left(  f_{2},f_{3},\ldots,f_{n}\right)  $ and $\left(  b_{1}%
,b_{2},\ldots,b_{n-1}\right)  $ instead of $r$, $\left(  f_{1},f_{2}%
,\ldots,f_{r}\right)  $ and $\left(  a_{1},a_{2},\ldots,a_{r}\right)  $
(because we assumed that Theorem~\ref{Theorem1401} \textbf{(b)} holds for
$r=n-1$). As a result, we obtain%
\begin{align}
&  \left(  f_{2}\ast f_{3}\ast\cdots\ast f_{n}\right)  \left(  b_{1}%
b_{2}\cdots b_{n-1}\right) \nonumber\\
&  =\sum_{\sigma\in S_{n-1}}f_{2}\left(  b_{\sigma\left(  1\right)  }\right)
f_{3}\left(  b_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(
b_{\sigma\left(  n-1\right)  }\right) \nonumber\\
&  =\sum_{\sigma\in S_{n,m}}f_{2}\left(  b_{\left(  \pi_{m}\left(
\sigma\right)  \right)  \left(  1\right)  }\right)  f_{3}\left(  b_{\left(
\pi_{m}\left(  \sigma\right)  \right)  \left(  2\right)  }\right)  \cdots
f_{n}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(
n-1\right)  }\right)  \label{1401.pf.b.m-part.3}%
\end{align}
(here, we substituted $\pi_{m}\left(  \sigma\right)  $ for $\sigma$ in the
sum, because the map $\pi_{m}:S_{n,m}\rightarrow S_{n-1}$ is a bijection). But
now,%
\begin{align}
&  \underbrace{g}_{=f_{2}\ast f_{3}\ast\cdots\ast f_{n}}\left(
\underbrace{\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{n}\right)  }_{\substack{=b_{1}b_{2}\cdots b_{n-1}\\\text{(by
(\ref{1401.pf.b.m-part.b_prod}))}}}\right) \nonumber\\
&  =\left(  f_{2}\ast f_{3}\ast\cdots\ast f_{n}\right)  \left(  b_{1}%
b_{2}\cdots b_{n-1}\right) \nonumber\\
&  =\sum_{\sigma\in S_{n,m}}f_{2}\left(  b_{\left(  \pi_{m}\left(
\sigma\right)  \right)  \left(  1\right)  }\right)  f_{3}\left(  b_{\left(
\pi_{m}\left(  \sigma\right)  \right)  \left(  2\right)  }\right)  \cdots
f_{n}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(
n-1\right)  }\right)  . \label{1401.pf.b.m-part.4}%
\end{align}


Now, let $\sigma\in S_{n,m}$. Let $i\in\left\{  1,2,\ldots,n-1\right\}  $ be
arbitrary. The definition of $\pi_{m}$ yields $\pi_{m}\left(  \sigma\right)
=\operatorname*{del}\nolimits_{m}\sigma$. Hence,%
\[
\left(  \underbrace{\pi_{m}\left(  \sigma\right)  }_{=\operatorname*{del}%
\nolimits_{m}\sigma}\right)  \left(  i\right)  =\left(  \operatorname*{del}%
\nolimits_{m}\sigma\right)  \left(  i\right)  =\omega\left(  \sigma\left(
i+1\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{del}\nolimits_{m}\sigma\right)  .
\]
Thus,
\begin{align*}
b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(  i\right)  }  &
=b_{\omega\left(  \sigma\left(  i+1\right)  \right)  }=a_{\eta\left(
\omega\left(  \sigma\left(  i+1\right)  \right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{\omega\left(
\sigma\left(  i+1\right)  \right)  }\right) \\
&  =a_{\sigma\left(  i+1\right)  }%
\end{align*}
(since $\eta\left(  \omega\left(  \sigma\left(  i+1\right)  \right)  \right)
=\underbrace{\left(  \eta\circ\omega\right)  }_{\substack{=\operatorname*{id}%
\\\text{(since the maps }\eta\text{ and }\omega\\\text{are mutually inverse)}%
}}\left(  \sigma\left(  i+1\right)  \right)  =\operatorname*{id}\left(
\sigma\left(  i+1\right)  \right)  =\sigma\left(  i+1\right)  $). Hence,
$f_{i+1}\left(  \underbrace{b_{\left(  \pi_{m}\left(  \sigma\right)  \right)
\left(  i\right)  }}_{=a_{\sigma\left(  i+1\right)  }}\right)  =f_{i+1}\left(
a_{\sigma\left(  i+1\right)  }\right)  $.

Now, forget that we fixed $i$. We thus have proven the equality $f_{i+1}%
\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(  i\right)
}\right)  =f_{i+1}\left(  a_{\sigma\left(  i+1\right)  }\right)  $ for every
$i\in\left\{  1,2,\ldots,n-1\right\}  $. Taking the product of these
equalities over all $i\in\left\{  1,2,\ldots,n\right\}  $, we obtain%
\begin{align*}
&  f_{1+1}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(
1\right)  }\right)  f_{2+1}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)
\right)  \left(  2\right)  }\right)  \cdots f_{\left(  n-1\right)  +1}\left(
b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(  n-1\right)  }\right)
\\
&  =f_{1+1}\left(  a_{\sigma\left(  1+1\right)  }\right)  f_{2+1}\left(
a_{\sigma\left(  2+1\right)  }\right)  \cdots f_{\left(  n-1\right)
+1}\left(  a_{\sigma\left(  \left(  n-1\right)  +1\right)  }\right) \\
&  =f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  f_{3}\left(
a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  .
\end{align*}
Compared with%
\begin{align*}
&  f_{1+1}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(
1\right)  }\right)  f_{2+1}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)
\right)  \left(  2\right)  }\right)  \cdots f_{\left(  n-1\right)  +1}\left(
b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(  n-1\right)  }\right)
\\
&  =f_{2}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(
1\right)  }\right)  f_{3}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)
\right)  \left(  2\right)  }\right)  \cdots f_{n}\left(  b_{\left(  \pi
_{m}\left(  \sigma\right)  \right)  \left(  n-1\right)  }\right)  ,
\end{align*}
this yields%
\begin{align}
&  f_{2}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(
1\right)  }\right)  f_{3}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)
\right)  \left(  2\right)  }\right)  \cdots f_{n}\left(  b_{\left(  \pi
_{m}\left(  \sigma\right)  \right)  \left(  n-1\right)  }\right) \nonumber\\
&  =f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  f_{3}\left(
a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  . \label{1401.pf.b.m-part.8}%
\end{align}


Now, forget that we fixed $\sigma$. We thus have shown that every $\sigma\in
S_{n,m}$ satisfies (\ref{1401.pf.b.m-part.8}). Hence,%
\begin{align}
&  \sum_{\sigma\in S_{n,m}}\underbrace{f_{2}\left(  b_{\left(  \pi_{m}\left(
\sigma\right)  \right)  \left(  1\right)  }\right)  f_{3}\left(  b_{\left(
\pi_{m}\left(  \sigma\right)  \right)  \left(  2\right)  }\right)  \cdots
f_{n}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(
n-1\right)  }\right)  }_{\substack{=f_{2}\left(  b_{\sigma\left(  2\right)
}\right)  f_{3}\left(  b_{\sigma\left(  3\right)  }\right)  \cdots
f_{n}\left(  b_{\sigma\left(  n\right)  }\right)  \\\text{(by
(\ref{1401.pf.b.m-part.8}))}}}\nonumber\\
&  =\sum_{\sigma\in S_{n,m}}f_{2}\left(  a_{\sigma\left(  2\right)  }\right)
f_{3}\left(  a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(
a_{\sigma\left(  n\right)  }\right)  . \label{1401.pf.b.m-part.9}%
\end{align}


Now, let us recall that $S_{n,m}=\left\{  \sigma\in S_{n}\ \mid\ \sigma\left(
1\right)  =m\right\}  $. Hence, we can replace the summation sign
\textquotedblleft$\sum_{\sigma\in S_{n,m}}$\textquotedblright\ by a
\textquotedblleft$\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}$\textquotedblright\ in $\sum_{\sigma\in S_{n,m}}f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  f_{3}\left(  a_{\sigma\left(  3\right)
}\right)  \cdots f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  $. As a
result, we obtain%
\begin{align}
&  \sum_{\sigma\in S_{n,m}}f_{2}\left(  a_{\sigma\left(  2\right)  }\right)
f_{3}\left(  a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(
a_{\sigma\left(  n\right)  }\right) \nonumber\\
&  =\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)  =m}}f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  f_{3}\left(  a_{\sigma\left(  3\right)
}\right)  \cdots f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  .
\label{1401.pf.b.m-part.10}%
\end{align}
Now, (\ref{1401.pf.b.m-part.4}) becomes%
\begin{align*}
&  g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{n}\right)  \right) \\
&  =\sum_{\sigma\in S_{n,m}}f_{2}\left(  b_{\left(  \pi_{m}\left(
\sigma\right)  \right)  \left(  1\right)  }\right)  f_{3}\left(  b_{\left(
\pi_{m}\left(  \sigma\right)  \right)  \left(  2\right)  }\right)  \cdots
f_{n}\left(  b_{\left(  \pi_{m}\left(  \sigma\right)  \right)  \left(
n-1\right)  }\right) \\
&  =\sum_{\sigma\in S_{n,m}}f_{2}\left(  a_{\sigma\left(  2\right)  }\right)
f_{3}\left(  a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(
a_{\sigma\left(  n\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1401.pf.b.m-part.9})}\right) \\
&  =\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)  =m}}f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  f_{3}\left(  a_{\sigma\left(  3\right)
}\right)  \cdots f_{n}\left(  a_{\sigma\left(  n\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1401.pf.b.m-part.10})}\right)  .
\end{align*}
This proves (\ref{1401.pf.b.m-part.goal}).

Now, (\ref{1401.pf.b.6}) becomes%
\begin{align*}
&  \left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}%
a_{2}\cdots a_{n}\right) \\
&  =\sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \cdot\underbrace{g\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{n}\right)
\right)  }_{\substack{=\sum_{\substack{\sigma\in S_{n};\\\sigma\left(
1\right)  =m}}f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  f_{3}\left(
a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  \\\text{(by (\ref{1401.pf.b.m-part.goal}))}}}\\
&  =\sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \cdot\sum_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  =m}}f_{2}\left(  a_{\sigma\left(  2\right)
}\right)  f_{3}\left(  a_{\sigma\left(  3\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right) \\
&  =\sum_{m=1}^{n}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}f_{1}\left(  \underbrace{a_{m}}_{\substack{=a_{\sigma\left(  1\right)
}\\\text{(since }m=\sigma\left(  1\right)  \\\text{(since }\sigma\left(
1\right)  =m\text{))}}}\right)  f_{2}\left(  a_{\sigma\left(  2\right)
}\right)  f_{3}\left(  a_{\sigma\left(  3\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right) \\
&  =\sum_{m=1}^{n}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}\underbrace{f_{1}\left(  a_{\sigma\left(  1\right)  }\right)  f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  f_{3}\left(  a_{\sigma\left(  3\right)
}\right)  \cdots f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  }%
_{=f_{1}\left(  a_{\sigma\left(  1\right)  }\right)  f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  }\\
&  =\sum_{m=1}^{n}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}f_{1}\left(  a_{\sigma\left(  1\right)  }\right)  f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \sum_{\sigma\in S_{n}}f_{1}\left(  a_{\sigma\left(  1\right)  }\right)
f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(
a_{\sigma\left(  n\right)  }\right) \\
&  =\underbrace{\sum_{m\in\left\{  1,2,\ldots,n\right\}  }}_{=\sum_{m=1}^{n}%
}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)  =m}}f_{1}\left(
a_{\sigma\left(  1\right)  }\right)  f_{2}\left(  a_{\sigma\left(  2\right)
}\right)  \cdots f_{n}\left(  a_{\sigma\left(  n\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }\sigma\in S_{n}\text{
satisfies }\sigma\left(  1\right)  \in\left\{  1,2,\ldots,n\right\}  \right)
\\
&  =\sum_{m=1}^{n}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}f_{1}\left(  a_{\sigma\left(  1\right)  }\right)  f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  ,
\end{align*}
this yields%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}a_{2}\cdots
a_{n}\right)  =\sum_{\sigma\in S_{n}}f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  .
\]


Now, forget that we fixed $k$, $H$, $A$, $\left(  f_{1},f_{2},\ldots
,f_{n}\right)  $ and $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $. We thus have
shown that if $k$ is a field, if $H$ is a $k$-bialgebra, if $A$ is a
$k$-algebra, if $f_{1}$, $f_{2}$, $\ldots$, $f_{n}$ are $n$ maps in
$\mathfrak{g}\left(  H,A\right)  $, then every $n$ elements $a_{1}$, $a_{2}$,
$\ldots$, $a_{n}$ of $\operatorname*{Prim}H$ satisfy%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}a_{2}\cdots
a_{n}\right)  =\sum_{\sigma\in S_{n}}f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  .
\]
In other words, we have shown that Theorem~\ref{Theorem1401} \textbf{(b)}
holds for $r=n$. This completes the induction step. The induction proof of
Theorem~\ref{Theorem1401} \textbf{(b)} is thus complete.
\end{proof}

We can obtain some corollaries of Theorem~\ref{Theorem1401}. By setting
$f_{1}$, $f_{2}$, $\ldots$, $f_{n}$ all equal to each other, we can conclude
that the following holds:

\begin{corollary}
\label{Corollary1411} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a $k$-algebra. Let $r\in\mathbb{N}$. Let $f\in\mathfrak{g}\left(
H,A\right)  $.

\textbf{(a)} Every $s\in\left\{  0,1,\ldots,r-1\right\}  $ and every $s$
elements $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ of $\operatorname*{Prim}H$
satisfy%
\[
f^{\ast r}\left(  a_{1}a_{2}\cdots a_{s}\right)  =0.
\]


\textbf{(b)} Every $r$ elements $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ of
$\operatorname*{Prim}H$ satisfy%
\[
f^{\ast r}\left(  a_{1}a_{2}\cdots a_{r}\right)  =\sum_{\sigma\in S_{r}%
}f\left(  a_{\sigma\left(  1\right)  }\right)  f\left(  a_{\sigma\left(
2\right)  }\right)  \cdots f\left(  a_{\sigma\left(  r\right)  }\right)
\]
(where $S_{r}$ denotes the $r$-th symmetric group).

\textbf{(c)} Let $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ be $r$ elements of
$\operatorname*{Prim}H$. If the elements $f\left(  a_{1}\right)  $, $f\left(
a_{2}\right)  $, $\ldots$, $f\left(  a_{r}\right)  $ of $A$ commute pairwise,
then%
\[
f^{\ast r}\left(  a_{1}a_{2}\cdots a_{r}\right)  =r!\cdot f\left(
a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{r}\right)  .
\]

\end{corollary}

The proof of this corollary rests on what we already know along with the
following almost trivial fact:

\begin{lemma}
\label{Lemma1412} Let $k$ be a field. Let $B$ be a $k$-algebra. Let
$n\in\mathbb{N}$. Let $b_{1}$, $b_{2}$, $\ldots$, $b_{n}$ be $n$ elements of
$B$ which pairwise commute. Let $\sigma\in S_{n}$. Then,%
\[
b_{\sigma\left(  1\right)  }b_{\sigma\left(  2\right)  }\cdots b_{\sigma
\left(  n\right)  }=b_{1}b_{2}\cdots b_{n}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1412}.]Let $S=\left\{  b_{1},b_{2},\ldots
,b_{n}\right\}  $. Then, $S\subseteq B$ (since $b_{1}$, $b_{2}$, $\ldots$,
$b_{n}$ are elements of $B$). Let $A$ be the $k$-subalgebra of $B$ generated
by $S$. Then, the $k$-algebra $A$ is generated by $S$. In particular, $A$
contains $S$ as a subset. That is, $S\subseteq A$.

Moreover, any two elements of $S$ commute\footnote{\textit{Proof.} The
elements $b_{1}$, $b_{2}$, $\ldots$, $b_{n}$ are the elements of $S$ (because
$S=\left\{  b_{1},b_{2},\ldots,b_{n}\right\}  $). But we also know that the
elements $b_{1}$, $b_{2}$, $\ldots$, $b_{n}$ pairwise commute. Since the
elements $b_{1}$, $b_{2}$, $\ldots$, $b_{n}$ are the elements of $S$, this
rewrites as follows: The elements of $S$ pairwise commute. In other words, any
two elements of $S$ commute, qed.}. Thus, Proposition~\ref{Proposition47a}
yields that the $k$-algebra $A$ is commutative. Hence, products in the
$k$-algebra $A$ are well-defined without specifying the order of the factors.

Now, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies $b_{i}\in\left\{
b_{1},b_{2},\ldots,b_{n}\right\}  =S\subseteq A$. Hence, $\prod_{i\in\left\{
1,2,\ldots,n\right\}  }b_{i}$ is a well-defined product in $A$ (it is
well-defined since the $k$-algebra $A$ is commutative). Since $\sigma\in
S_{n}$, we know that $\sigma$ is a permutation of $\left\{  1,2,\ldots
,n\right\}  $. That is, $\sigma$ is a bijection $\left\{  1,2,\ldots
,n\right\}  \rightarrow\left\{  1,2,\ldots,n\right\}  $. Now,%
\begin{align*}
b_{1}b_{2}\cdots b_{n}  &  =\prod_{i\in\left\{  1,2,\ldots,n\right\}  }%
b_{i}=\prod_{i\in\left\{  1,2,\ldots,n\right\}  }b_{\sigma\left(  i\right)
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }i\text{ for }\sigma\left(  i\right)  \text{ in the
product,}\\
\text{since the map }\sigma:\left\{  1,2,\ldots,n\right\}  \rightarrow\left\{
1,2,\ldots,n\right\}  \text{ is a bijection}%
\end{array}
\right) \\
&  =b_{\sigma\left(  1\right)  }b_{\sigma\left(  2\right)  }\cdots
b_{\sigma\left(  n\right)  }.
\end{align*}
This proves Lemma~\ref{Lemma1412}.
\end{proof}

\begin{proof}
[Proof of Corollary~\ref{Corollary1411}.]\textbf{(a)} Let $s\in\left\{
0,1,\ldots,r-1\right\}  $. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$
elements of $\operatorname*{Prim}H$. Then,%
\[
\underbrace{f^{\ast r}}_{=\underbrace{f\ast f\ast\cdots\ast f}_{r\text{
times}}}\left(  a_{1}a_{2}\cdots a_{s}\right)  =\left(  \underbrace{f\ast
f\ast\cdots\ast f}_{r\text{ times}}\right)  \left(  a_{1}a_{2}\cdots
a_{s}\right)  =0
\]
(by Theorem~\ref{Theorem1401} \textbf{(a)}, applied to $\left(  f_{1}%
,f_{2},\ldots,f_{r}\right)  =\left(  \underbrace{f,f,\ldots,f}_{r\text{
times}}\right)  $). This proves Corollary~\ref{Corollary1411} \textbf{(a)}.

\textbf{(b)} Let $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ be $r$ elements of
$\operatorname*{Prim}H$. Then,%
\begin{align*}
\underbrace{f^{\ast r}}_{=\underbrace{f\ast f\ast\cdots\ast f}_{r\text{
times}}}\left(  a_{1}a_{2}\cdots a_{r}\right)   &  =\left(  \underbrace{f\ast
f\ast\cdots\ast f}_{r\text{ times}}\right)  \left(  a_{1}a_{2}\cdots
a_{r}\right) \\
&  =\sum_{\sigma\in S_{r}}f\left(  a_{\sigma\left(  1\right)  }\right)
f\left(  a_{\sigma\left(  2\right)  }\right)  \cdots f\left(  a_{\sigma\left(
r\right)  }\right)
\end{align*}
(by Theorem~\ref{Theorem1401} \textbf{(b)}, applied to $\left(  f_{1}%
,f_{2},\ldots,f_{r}\right)  =\left(  \underbrace{f,f,\ldots,f}_{r\text{
times}}\right)  $). This proves Corollary~\ref{Corollary1411} \textbf{(b)}.

\textbf{(c)} Let $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ be $r$ elements of
$\operatorname*{Prim}H$. Assume that the elements $f\left(  a_{1}\right)  $,
$f\left(  a_{2}\right)  $, $\ldots$, $f\left(  a_{r}\right)  $ of $A$ commute
pairwise. Thus, for every $\sigma\in S_{r}$, we have%
\begin{equation}
f\left(  a_{\sigma\left(  1\right)  }\right)  f\left(  a_{\sigma\left(
2\right)  }\right)  \cdots f\left(  a_{\sigma\left(  r\right)  }\right)
=f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{r}\right)
\label{1411.pf.c.1}%
\end{equation}
(by Lemma~\ref{Lemma1412}, applied to $n=r$, $B=A$ and $b_{i}=f\left(
a_{i}\right)  $). Now, Corollary~\ref{Corollary1411} \textbf{(b)} yields%
\begin{align*}
f^{\ast r}\left(  a_{1}a_{2}\cdots a_{r}\right)   &  =\sum_{\sigma\in S_{r}%
}\underbrace{f\left(  a_{\sigma\left(  1\right)  }\right)  f\left(
a_{\sigma\left(  2\right)  }\right)  \cdots f\left(  a_{\sigma\left(
r\right)  }\right)  }_{\substack{=f\left(  a_{1}\right)  f\left(
a_{2}\right)  \cdots f\left(  a_{r}\right)  \\\text{(by (\ref{1411.pf.c.1}))}%
}}\\
&  =\sum_{\sigma\in S_{r}}f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots
f\left(  a_{r}\right)  =\underbrace{\left\vert S_{r}\right\vert }_{=r!}\cdot
f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{r}\right) \\
&  =r!\cdot f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{r}\right)  .
\end{align*}
This proves Corollary~\ref{Corollary1411} \textbf{(c)}.
\end{proof}

We can apply Corollary~\ref{Corollary1411} to the Eulerian idempotent:

\begin{corollary}
\label{Corollary1413} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered bialgebra over $k$. Consider the convolution algebra
$\mathcal{L}\left(  H,H\right)  $.

\textbf{(a)} Every $s\in\left\{  0,1,\ldots,r-1\right\}  $ and every $s$
elements $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ of $\operatorname*{Prim}H$
satisfy%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  ^{\ast r}\left(
a_{1}a_{2}\cdots a_{s}\right)  =0.
\]


\textbf{(b)} Every $r$ elements $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ of
$\operatorname*{Prim}H$ satisfy%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  ^{\ast r}\left(
a_{1}a_{2}\cdots a_{r}\right)  =\sum_{\sigma\in S_{r}}a_{\sigma\left(
1\right)  }a_{\sigma\left(  2\right)  }\cdots a_{\sigma\left(  r\right)  }%
\]
(where $S_{r}$ denotes the $r$-th symmetric group).
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1413}.]We know that $\operatorname*{Log}%
F\in\mathfrak{g}\left(  H,H\right)  $ for every $F\in G\left(  H,H\right)  $
(this follows from the definition of $\operatorname*{Log}$). Applying this to
$F=\operatorname*{id}$, we obtain $\operatorname*{Log}\operatorname*{id}%
\in\mathfrak{g}\left(  H,H\right)  $. Thus, Corollary~\ref{Corollary1413}
\textbf{(a)} follows from Corollary~\ref{Corollary1411} \textbf{(a)} (applied
to $A=H$ and $f=\operatorname*{Log}\operatorname*{id}$).

\textbf{(b)} We know that $H$ is a bialgebra, and thus a unital coalgebra.
Proposition~\ref{Proposition19} \textbf{(c)} (applied to $A=H$ and
$F=\operatorname*{id}$) shows that $\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \mid_{\operatorname*{Prim}H}=\operatorname*{id}%
\mid_{\operatorname*{Prim}H}$. Thus, for every $u\in\operatorname*{Prim}H$, we
have
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(  u\right)
=\left(  \underbrace{\left(  \operatorname*{Log}\operatorname*{id}\right)
\mid_{\operatorname*{Prim}H}}_{=\operatorname*{id}\mid_{\operatorname*{Prim}%
H}}\right)  \left(  u\right)  =\left(  \operatorname*{id}\mid
_{\operatorname*{Prim}H}\right)  \left(  u\right)  =\operatorname*{id}\left(
u\right)  =u.
\]
Hence,
\begin{equation}
\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(  u\right)
=u\ \ \ \ \ \ \ \ \ \ \text{for every }u\in\operatorname*{Prim}H\text{.}
\label{1413.pf.1}%
\end{equation}


Now, let $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ be $r$ elements of
$\operatorname*{Prim}H$. Let $\sigma\in S_{r}$. Every $i\in\left\{
1,2,\ldots,r\right\}  $ satisfies $a_{\sigma\left(  i\right)  }\in
\operatorname*{Prim}H$. Hence, every $i\in\left\{  1,2,\ldots,r\right\}  $
satisfies $\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(
a_{\sigma\left(  i\right)  }\right)  =a_{\sigma\left(  i\right)  }$ (by
(\ref{1413.pf.1}), applied to $u=a_{\sigma\left(  i\right)  }$). Multiplying
these equalities over all $i\in\left\{  1,2,\ldots,r\right\}  $, we obtain%
\begin{equation}
\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(  a_{\sigma\left(
1\right)  }\right)  \cdot\left(  \operatorname*{Log}\operatorname*{id}\right)
\left(  a_{\sigma\left(  2\right)  }\right)  \cdot\cdots\cdot\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  a_{\sigma\left(
r\right)  }\right)  =a_{\sigma\left(  1\right)  }a_{\sigma\left(  2\right)
}\cdots a_{\sigma\left(  r\right)  }. \label{1413.pf.2}%
\end{equation}


Now, forget that we fixed $\sigma$. We thus have shown that (\ref{1413.pf.2})
holds for every $\sigma\in S_{r}$. Now, Corollary~\ref{Corollary1411}
\textbf{(b)} (applied to $A=H$ and $f=\operatorname*{Log}\operatorname*{id}$)
yields%
\begin{align*}
\left(  \operatorname*{Log}\operatorname*{id}\right)  ^{\ast r}\left(
a_{1}a_{2}\cdots a_{r}\right)   &  =\sum_{\sigma\in S_{r}}\underbrace{\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  a_{\sigma\left(
1\right)  }\right)  \cdot\left(  \operatorname*{Log}\operatorname*{id}\right)
\left(  a_{\sigma\left(  2\right)  }\right)  \cdot\cdots\cdot\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  a_{\sigma\left(
r\right)  }\right)  }_{\substack{=a_{\sigma\left(  1\right)  }a_{\sigma\left(
2\right)  }\cdots a_{\sigma\left(  r\right)  }\\\text{(by (\ref{1413.pf.2}))}%
}}\\
&  =\sum_{\sigma\in S_{r}}a_{\sigma\left(  1\right)  }a_{\sigma\left(
2\right)  }\cdots a_{\sigma\left(  r\right)  }.
\end{align*}
This proves Corollary~\ref{Corollary1413} \textbf{(b)}.
\end{proof}

We record a further particular case of Theorem~\ref{Theorem1401}:

\begin{corollary}
\label{Corollary1421} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a $k$-algebra. Let $r\in\mathbb{N}$. Let $f_{1}$, $f_{2}$, $\ldots$,
$f_{r}$ be $r$ maps in $\mathfrak{g}\left(  H,A\right)  $. Let $a\in
\operatorname*{Prim}H$.

\textbf{(a)} Every $s\in\left\{  0,1,\ldots,r-1\right\}  $ satisfies%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a^{s}\right)  =0.
\]


\textbf{(b)} We have%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a^{r}\right)
=r!\cdot f_{1}\left(  a\right)  f_{2}\left(  a\right)  \cdots f_{r}\left(
a\right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1421}.]\textbf{(a)} Let $s\in\left\{
0,1,\ldots,r-1\right\}  $. We have
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{a^{s}%
}_{=\underbrace{aa\cdots a}_{s\text{ times}}}\right)  =\left(  f_{1}\ast
f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{aa\cdots a}_{s\text{
times}}\right)  =0
\]
(by Theorem~\ref{Theorem1401} \textbf{(a)}, applied to $\left(  a_{1}%
,a_{2},\ldots,a_{s}\right)  =\left(  \underbrace{a,a,\ldots,a}_{s\text{
times}}\right)  $). This proves Corollary~\ref{Corollary1421} \textbf{(a)}.

\textbf{(b)} We have
\begin{align*}
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{a^{r}%
}_{=\underbrace{aa\cdots a}_{r\text{ times}}}\right)   &  =\left(  f_{1}\ast
f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{aa\cdots a}_{r\text{
times}}\right) \\
&  =\sum_{\sigma\in S_{r}}f_{1}\left(  a\right)  f_{2}\left(  a\right)  \cdots
f_{r}\left(  a\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Theorem~\ref{Theorem1401} \textbf{(b)}, applied to}\\
\left(  a_{1},a_{2},\ldots,a_{r}\right)  =\left(  \underbrace{a,a,\ldots
,a}_{r\text{ times}}\right)
\end{array}
\right) \\
&  =\underbrace{\left\vert S_{r}\right\vert }_{=r!}\cdot f_{1}\left(
a\right)  f_{2}\left(  a\right)  \cdots f_{r}\left(  a\right)  =r!\cdot
f_{1}\left(  a\right)  f_{2}\left(  a\right)  \cdots f_{r}\left(  a\right)  .
\end{align*}
This proves Corollary~\ref{Corollary1421} \textbf{(b)}.
\end{proof}

A further particular case involves only one map and only one primitive element:

\begin{corollary}
\label{Corollary1422} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a $k$-algebra. Let $r\in\mathbb{N}$. Let $f\in\mathfrak{g}\left(
H,A\right)  $. Let $a\in\operatorname*{Prim}H$.

\textbf{(a)} Every $s\in\left\{  0,1,\ldots,r-1\right\}  $ satisfies%
\[
f^{\ast r}\left(  a^{s}\right)  =0.
\]


\textbf{(b)} We have%
\[
f^{\ast r}\left(  a^{r}\right)  =r!\cdot\left(  f\left(  a\right)  \right)
^{r}.
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1422}.]\textbf{(b)} Let $s\in\left\{
0,1,\ldots,r-1\right\}  $. We have%
\[
\underbrace{f^{\ast r}}_{=\underbrace{f\ast f\ast\cdots\ast f}_{r\text{
times}}}\left(  a^{s}\right)  =\left(  \underbrace{f\ast f\ast\cdots\ast
f}_{r\text{ times}}\right)  \left(  a^{s}\right)  =0
\]
(by Corollary~\ref{Corollary1421} \textbf{(a)}, applied to $\left(
f_{1},f_{2},\ldots,f_{r}\right)  =\left(  \underbrace{f,f,\ldots,f}_{r\text{
times}}\right)  $). This proves Corollary~\ref{Corollary1422} \textbf{(a)}.

\textbf{(b)} We have
\begin{align*}
\underbrace{f^{\ast r}}_{=\underbrace{f\ast f\ast\cdots\ast f}_{r\text{
times}}}\left(  a^{r}\right)   &  =\left(  \underbrace{f\ast f\ast\cdots\ast
f}_{r\text{ times}}\right)  \left(  a^{r}\right)  =r!\cdot
\underbrace{\underbrace{f\left(  a\right)  f\left(  a\right)  \cdots f\left(
a\right)  }_{r\text{ times}}}_{=\left(  f\left(  a\right)  \right)  ^{r}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Corollary~\ref{Corollary1421} \textbf{(b)}, applied to}\\
\left(  f_{1},f_{2},\ldots,f_{r}\right)  =\left(  \underbrace{f,f,\ldots
,f}_{r\text{ times}}\right)
\end{array}
\right) \\
&  =r!\cdot\left(  f\left(  a\right)  \right)  ^{r}.
\end{align*}
This proves Corollary~\ref{Corollary1422} \textbf{(b)}.
\end{proof}

\section{\label{S33}$\left(  \varepsilon,\varepsilon\right)  $-derivations and
products}

In this section, we are going to prove some results similar to those proven in
\S \ref{S32}, although not exactly analogous (in particular, they will need
new proofs). However, before we start, let us record a really simple property
of unital coalgebras:

\begin{proposition}
\label{Proposition1500} Let $k$ be a field. Let $H$ be a unital coalgebra. Let
$x\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Then,%
\[
\Delta\left(  x\right)  \in x\otimes1_{H}+1_{H}\otimes x+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1500}.]The element $\Delta\left(
x\right)  $ is a tensor in $H\otimes H$. Hence, we can write $\Delta\left(
x\right)  $ in the form $\Delta\left(  x\right)  =\sum_{i=1}^{n}\lambda
_{i}a_{i}\otimes b_{i}$ for some $n\in\mathbb{N}$, some elements $\lambda_{1}%
$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$ of $k$, some elements $a_{1}$,
$a_{2}$, $\ldots$, $a_{n}$ of $H$, and some elements $b_{1}$, $b_{2}$,
$\ldots$, $b_{n}$ of $H$. Consider this $n$, these $\lambda_{1}$, $\lambda
_{2}$, $\ldots$, $\lambda_{n}$, these $a_{1}$, $a_{2}$, $\ldots$, $a_{n}$, and
these $b_{1}$, $b_{2}$, $\ldots$, $b_{n}$.

From $x=\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $, we obtain
$\varepsilon_{H}\left(  x\right)  =0$. Thus, $\underbrace{\varepsilon
}_{=\varepsilon_{H}}\left(  x\right)  =\varepsilon_{H}\left(  x\right)  =0$.

It is now easy to see that $\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
b_{i}\right)  a_{i}=x$\ \ \ \ \footnote{\textit{Proof.} Let
$\operatorname*{kan}:H\rightarrow H\otimes k$ be the canonical isomorphism
which sends every $y\in H$ to $y\otimes1$. Then, by the axioms of a coalgebra,
we have $\left(  \operatorname*{id}\otimes\varepsilon_{H}\right)  \circ
\Delta_{H}=\operatorname*{kan}$ (since $H$ is a coalgebra). Now,%
\begin{align*}
\left(  \left(  \operatorname*{id}\otimes\varepsilon_{H}\right)  \circ
\Delta_{H}\right)  \left(  x\right)   &  =\left(  \operatorname*{id}%
\otimes\varepsilon_{H}\right)  \left(  \underbrace{\Delta_{H}\left(  x\right)
}_{=\Delta\left(  x\right)  =\sum_{i=1}^{n}\lambda_{i}a_{i}\otimes b_{i}%
}\right)  =\left(  \operatorname*{id}\otimes\varepsilon_{H}\right)  \left(
\sum_{i=1}^{n}\lambda_{i}a_{i}\otimes b_{i}\right) \\
&  =\sum_{i=1}^{n}\lambda_{i}\underbrace{\left(  \operatorname*{id}%
\otimes\varepsilon_{H}\right)  \left(  a_{i}\otimes b_{i}\right)
}_{=\operatorname*{id}\left(  a_{i}\right)  \otimes\varepsilon_{H}\left(
b_{i}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
\operatorname*{id}\otimes\varepsilon_{H}\text{ is }k\text{-linear}\right) \\
&  =\sum_{i=1}^{n}\lambda_{i}\underbrace{\operatorname*{id}\left(
a_{i}\right)  }_{=a_{i}}\otimes\underbrace{\varepsilon_{H}\left(
b_{i}\right)  }_{=\varepsilon\left(  b_{i}\right)  =\varepsilon\left(
b_{i}\right)  1}=\sum_{i=1}^{n}\lambda_{i}\underbrace{a_{i}\otimes
\varepsilon\left(  b_{i}\right)  1}_{=\varepsilon\left(  b_{i}\right)
a_{i}\otimes1}\\
&  =\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  b_{i}\right)  a_{i}%
\otimes1=\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  b_{i}\right)
a_{i}\right)  \otimes1=\operatorname*{kan}\left(  \sum_{i=1}^{n}\lambda
_{i}\varepsilon\left(  b_{i}\right)  a_{i}\right)
\end{align*}
(since $\operatorname*{kan}\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
b_{i}\right)  a_{i}\right)  =\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon
\left(  b_{i}\right)  a_{i}\right)  \otimes1$ (by the definition of
$\operatorname*{kan}$)). Hence,%
\[
\operatorname*{kan}\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
b_{i}\right)  a_{i}\right)  =\left(  \underbrace{\left(  \operatorname*{id}%
\otimes\varepsilon_{H}\right)  \circ\Delta_{H}}_{=\operatorname*{kan}}\right)
\left(  x\right)  =\operatorname*{kan}\left(  x\right)  .
\]
Since the map $\operatorname*{kan}$ is injective (because $\operatorname*{kan}%
$ is an isomorphism), this yields $\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
b_{i}\right)  a_{i}=x$, qed.} and $\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
a_{i}\right)  b_{i}=x$\ \ \ \ \footnote{\textit{Proof.} Let
$\operatorname*{kan}:H\rightarrow k\otimes H$ be the canonical isomorphism
which sends every $y\in H$ to $1\otimes y$. Then, by the axioms of a
coalgebra, we have $\left(  \varepsilon_{H}\otimes\operatorname*{id}\right)
\circ\Delta_{H}=\operatorname*{kan}$ (since $H$ is a coalgebra). Now,%
\begin{align*}
\left(  \left(  \varepsilon_{H}\otimes\operatorname*{id}\right)  \circ
\Delta_{H}\right)  \left(  x\right)   &  =\left(  \varepsilon_{H}%
\otimes\operatorname*{id}\right)  \left(  \underbrace{\Delta_{H}\left(
x\right)  }_{=\Delta\left(  x\right)  =\sum_{i=1}^{n}\lambda_{i}a_{i}\otimes
b_{i}}\right)  \left(  \varepsilon_{H}\otimes\operatorname*{id}\right)
\left(  \sum_{i=1}^{n}\lambda_{i}a_{i}\otimes b_{i}\right) \\
&  =\sum_{i=1}^{n}\lambda_{i}\underbrace{\left(  \varepsilon_{H}%
\otimes\operatorname*{id}\right)  \left(  a_{i}\otimes b_{i}\right)
}_{=\operatorname*{id}\left(  a_{i}\right)  \otimes\varepsilon_{H}\left(
b_{i}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
\varepsilon_{H}\otimes\operatorname*{id}\text{ is }k\text{-linear}\right) \\
&  =\sum_{i=1}^{n}\lambda_{i}\underbrace{\varepsilon_{H}\left(  a_{i}\right)
}_{=\varepsilon\left(  a_{i}\right)  =\varepsilon\left(  a_{i}\right)
1}\otimes\underbrace{\operatorname*{id}\left(  b_{i}\right)  }_{=b_{i}}%
=\sum_{i=1}^{n}\underbrace{\lambda_{i}\varepsilon\left(  a_{i}\right)
1\otimes b_{i}}_{=1\otimes\lambda_{i}\varepsilon\left(  a_{i}\right)  b_{i}}\\
&  =\sum_{i=1}^{n}1\otimes\lambda_{i}\varepsilon\left(  a_{i}\right)
b_{i}=1\otimes\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
a_{i}\right)  b_{i}\right)  =\operatorname*{kan}\left(  \sum_{i=1}^{n}%
\lambda_{i}\varepsilon\left(  a_{i}\right)  b_{i}\right)
\end{align*}
(since $\operatorname*{kan}\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
a_{i}\right)  b_{i}\right)  =1\otimes\left(  \sum_{i=1}^{n}\lambda
_{i}\varepsilon\left(  a_{i}\right)  b_{i}\right)  $ (by the definition of
$\operatorname*{kan}$)). Hence,%
\[
\operatorname*{kan}\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
a_{i}\right)  b_{i}\right)  =\left(  \underbrace{\left(  \varepsilon
_{H}\otimes\operatorname*{id}\right)  \circ\Delta_{H}}_{=\operatorname*{kan}%
}\right)  \left(  x\right)  =\operatorname*{kan}\left(  x\right)  .
\]
Since the map $\operatorname*{kan}$ is injective (because $\operatorname*{kan}%
$ is an isomorphism), this yields $\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
a_{i}\right)  b_{i}=x$, qed.}. Also, $\sum_{i=1}^{n}\lambda_{i}\varepsilon
\left(  a_{i}\right)  \varepsilon\left(  b_{i}\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} We know that $\sum_{i=1}^{n}\lambda
_{i}\varepsilon\left(  a_{i}\right)  b_{i}=x$. Applying the map $\varepsilon$
to this equality, we obtain $\varepsilon\left(  \sum_{i=1}^{n}\lambda
_{i}\varepsilon\left(  a_{i}\right)  b_{i}\right)  =\varepsilon\left(
x\right)  =0$. Hence,%
\[
0=\varepsilon\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)
b_{i}\right)  =\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)
\varepsilon\left(  b_{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the
map }\varepsilon\text{ is }k\text{-linear}\right)  ,
\]
qed.}.

By the axioms of a unital coalgebra, we have $\varepsilon\left(  1_{H}\right)
=1$ (since $H$ is a unital coalgebra).

For every $i\in\left\{  1,2,\ldots,n\right\}  $, define an element
$a_{i}^{\prime}$ of $H$ by $a_{i}^{\prime}=a_{i}-\varepsilon\left(
a_{i}\right)  1_{H}$. Then, for every $i\in\left\{  1,2,\ldots,n\right\}  $,
we have $a_{i}^{\prime}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,n\right\}  $.
Applying the map $\varepsilon$ to the equality $a_{i}^{\prime}=a_{i}%
-\varepsilon\left(  a_{i}\right)  1_{H}$, we obtain%
\begin{align*}
\varepsilon\left(  a_{i}^{\prime}\right)   &  =\varepsilon\left(
a_{i}-\varepsilon\left(  a_{i}\right)  1_{H}\right)  =\varepsilon\left(
a_{i}\right)  -\varepsilon\left(  a_{i}\right)  \underbrace{\varepsilon\left(
1_{H}\right)  }_{=1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\varepsilon\text{ is }k\text{-linear}\right) \\
&  =\varepsilon\left(  a_{i}\right)  -\varepsilon\left(  a_{i}\right)  =0,
\end{align*}
so that $a_{i}^{\prime}\in\operatorname*{Ker}\underbrace{\varepsilon
}_{=\varepsilon_{H}}=\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $,
qed.}.

For every $i\in\left\{  1,2,\ldots,n\right\}  $, define an element
$b_{i}^{\prime}$ of $H$ by $b_{i}^{\prime}=b_{i}-\varepsilon\left(
b_{i}\right)  1_{H}$. Then, for every $i\in\left\{  1,2,\ldots,n\right\}  $,
we have $b_{i}^{\prime}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,n\right\}  $.
Applying the map $\varepsilon$ to the equality $b_{i}^{\prime}=b_{i}%
-\varepsilon\left(  b_{i}\right)  1_{H}$, we obtain%
\begin{align*}
\varepsilon\left(  b_{i}^{\prime}\right)   &  =\varepsilon\left(
b_{i}-\varepsilon\left(  b_{i}\right)  1_{H}\right)  =\varepsilon\left(
b_{i}\right)  -\varepsilon\left(  b_{i}\right)  \underbrace{\varepsilon\left(
1_{H}\right)  }_{=1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\varepsilon\text{ is }k\text{-linear}\right) \\
&  =\varepsilon\left(  b_{i}\right)  -\varepsilon\left(  b_{i}\right)  =0,
\end{align*}
so that $b_{i}^{\prime}\in\operatorname*{Ker}\underbrace{\varepsilon
}_{=\varepsilon_{H}}=\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $,
qed.}. Also, $\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)
b_{i}^{\prime}=x$\ \ \ \ \footnote{\textit{Proof.} We have%
\begin{align*}
\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)  \underbrace{b_{i}%
^{\prime}}_{=b_{i}-\varepsilon\left(  b_{i}\right)  1_{H}}  &  =\sum_{i=1}%
^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)  \left(  b_{i}-\varepsilon
\left(  b_{i}\right)  1_{H}\right)  =\underbrace{\sum_{i=1}^{n}\lambda
_{i}\varepsilon\left(  a_{i}\right)  b_{i}}_{=x}-\underbrace{\sum_{i=1}%
^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)  \varepsilon\left(
b_{i}\right)  1_{H}}_{=\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
a_{i}\right)  \varepsilon\left(  b_{i}\right)  \right)  1_{H}}\\
&  =x-\underbrace{\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
a_{i}\right)  \varepsilon\left(  b_{i}\right)  \right)  }_{=0}1_{H}%
=x-\underbrace{0\cdot1_{H}}_{=0}=x,
\end{align*}
qed.}.

Now,%
\begin{align*}
\Delta\left(  x\right)   &  =\sum_{i=1}^{n}\lambda_{i}a_{i}\otimes
\underbrace{b_{i}}_{\substack{=b_{i}^{\prime}+\varepsilon\left(  b_{i}\right)
1_{H}\\\text{(since }b_{i}^{\prime}=b_{i}-\varepsilon\left(  b_{i}\right)
1_{H}\text{)}}}=\sum_{i=1}^{n}\lambda_{i}\underbrace{a_{i}\otimes\left(
b_{i}^{\prime}+\varepsilon\left(  b_{i}\right)  1_{H}\right)  }_{=a_{i}\otimes
b_{i}^{\prime}+\varepsilon\left(  b_{i}\right)  a_{i}\otimes1_{H}}\\
&  =\sum_{i=1}^{n}\lambda_{i}\left(  a_{i}\otimes b_{i}^{\prime}%
+\varepsilon\left(  b_{i}\right)  a_{i}\otimes1_{H}\right) \\
&  =\sum_{i=1}^{n}\lambda_{i}\underbrace{a_{i}}_{\substack{=a_{i}^{\prime
}+\varepsilon\left(  a_{i}\right)  1_{H}\\\text{(since }a_{i}^{\prime}%
=a_{i}-\varepsilon\left(  a_{i}\right)  1_{H}\text{)}}}\otimes b_{i}^{\prime
}+\underbrace{\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  b_{i}\right)
a_{i}\otimes1_{H}}_{=\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon\left(
b_{i}\right)  a_{i}\right)  \otimes1_{H}}\\
&  =\sum_{i=1}^{n}\lambda_{i}\underbrace{\left(  a_{i}^{\prime}+\varepsilon
\left(  a_{i}\right)  1_{H}\right)  \otimes b_{i}^{\prime}}_{=a_{i}^{\prime
}\otimes b_{i}^{\prime}+\varepsilon\left(  a_{i}\right)  1_{H}\otimes
b_{i}^{\prime}}+\underbrace{\left(  \sum_{i=1}^{n}\lambda_{i}\varepsilon
\left(  b_{i}\right)  a_{i}\right)  }_{=x}\otimes1_{H}\\
&  =\underbrace{\sum_{i=1}^{n}\lambda_{i}\left(  a_{i}^{\prime}\otimes
b_{i}^{\prime}+\varepsilon\left(  a_{i}\right)  1_{H}\otimes b_{i}^{\prime
}\right)  }_{=\sum_{i=1}^{n}\lambda_{i}a_{i}^{\prime}\otimes b_{i}^{\prime
}+\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)  1_{H}\otimes
b_{i}^{\prime}}+x\otimes1_{H}\\
&  =\sum_{i=1}^{n}\lambda_{i}\underbrace{a_{i}^{\prime}}_{\in
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  }\otimes\underbrace{b_{i}%
^{\prime}}_{\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)
}+\underbrace{\sum_{i=1}^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)
1_{H}\otimes b_{i}^{\prime}}_{=1_{H}\otimes\sum_{i=1}^{n}\lambda
_{i}\varepsilon\left(  a_{i}\right)  b_{i}^{\prime}}+x\otimes1_{H}\\
&  \in\underbrace{\sum_{i=1}^{n}\lambda_{i}\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  }_{\substack{\subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \\\text{(since
}\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\text{ is}\\\text{a }k\text{-vector space)}}}+1_{H}\otimes\underbrace{\sum
_{i=1}^{n}\lambda_{i}\varepsilon\left(  a_{i}\right)  b_{i}^{\prime}}%
_{=x}+x\otimes1_{H}\\
&  \subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  +1_{H}\otimes x+x\otimes1_{H}\\
&  =x\otimes1_{H}+1_{H}\otimes x+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  .
\end{align*}
This proves Proposition~\ref{Proposition1500}.
\end{proof}

The following theorem is similar to Theorem~\ref{Theorem1401}:

\begin{theorem}
\label{Theorem1501} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be
a $k$-algebra. Let $r\in\mathbb{N}$. Let $f_{1}$, $f_{2}$, $\ldots$, $f_{r}$
be $r$ elements of $\mathcal{L}\left(  H,A\right)  $ such that $f_{1}$,
$f_{2}$, $\ldots$, $f_{r}$ are $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivations.

\textbf{(a)} Every $s\in\mathbb{N}$ and every $s$ elements $a_{1}$, $a_{2}$,
$\ldots$, $a_{s}$ of $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $
such that $s>r$ satisfy%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a_{1}a_{2}\cdots
a_{s}\right)  =0.
\]


\textbf{(b)} Every $r$ elements $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ satisfy%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a_{1}a_{2}\cdots
a_{r}\right)  =\sum_{\sigma\in S_{r}}f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{r}\left(  a_{\sigma\left(  r\right)  }\right)
\]
(where $S_{r}$ denotes the $r$-th symmetric group).
\end{theorem}

We prepare for the proof of this theorem by showing a fact resembling of
Lemma~\ref{Lemma1402}:

\begin{lemma}
\label{Lemma1502} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let
$s\in\mathbb{N}$. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Then,%
\begin{align*}
\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)   &  \in\sum_{m=1}^{s}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{s}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s}%
\end{align*}
\footnote{Recall that $\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{\ell}$ is defined according to
Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{\ell}$ means the $\ell$-th power of the
subspace $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ of the
$k$-algebra $H$ for every $\ell\in\mathbb{N}$.}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1502}.]Let us first notice that $\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $ is the kernel of $\varepsilon_{H}$. Thus,
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ is the kernel of an
algebra homomorphism (since $\varepsilon_{H}$ is an algebra homomorphism).
Thus, $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ is an ideal of $H$
(since a kernel of an algebra homomorphism is always an ideal).

We will prove that every $r\in\left\{  0,1,\ldots,s\right\}  $ satisfies%
\begin{align}
\Delta\left(  a_{1}a_{2}\cdots a_{r}\right)   &  \in\sum_{m=1}^{r}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{r}\right)  \right)  +1_{H}\otimes\left(  a_{1}a_{2}\cdots
a_{r}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{r}. \label{1502.pf.goal}%
\end{align}


\textit{Proof of (\ref{1502.pf.goal}):} We will prove (\ref{1502.pf.goal}) by
induction over $r$:

\textit{Induction base:} Let $r=0$. Then, $a_{1}a_{2}\cdots a_{r}=a_{1}%
a_{2}\cdots a_{0}=\left(  \text{empty product}\right)  =1_{H}$. Also, from
$r=0$, we obtain $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{r}=\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{0}=k\cdot1_{H}\ni1_{H}$, so that $1_{H}\in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{r}$. Now,%
\begin{align*}
\Delta\left(  \underbrace{a_{1}a_{2}\cdots a_{r}}_{=1_{H}}\right)   &
=\Delta\left(  1_{H}\right)  =1_{H}\otimes\underbrace{1_{H}}_{=a_{1}%
a_{2}\cdots a_{r}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the axioms of a
bialgebra}\right) \\
&  =1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{r}\right) \\
&  \subseteq\sum_{m=1}^{r}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{r}\right)  \right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{r}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{r}.
\end{align*}
Thus, (\ref{1502.pf.goal}) is proven in the case when $r=0$. The induction
base is hence complete.

\textit{Induction step:} Let $R\in\left\{  0,1,\ldots,s-1\right\}  $. Assume
that (\ref{1502.pf.goal}) has been proven for $r=R$. We need to prove
(\ref{1502.pf.goal}) for $r=R+1$.

We know that (\ref{1502.pf.goal}) has been proven for $r=R$. In other words,%
\begin{align}
\Delta\left(  a_{1}a_{2}\cdots a_{R}\right)   &  \in\sum_{m=1}^{R}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{R}\right)  \right)  +1_{H}\otimes\left(  a_{1}a_{2}\cdots
a_{R}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{R}. \label{1502.pf.goal.pf.indhyp}%
\end{align}


Define a $k$-vector subspace $\mathfrak{R}$ of $H\otimes H$ by%
\begin{equation}
\mathfrak{R}=\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{R}. \label{1502.pf.goal.pf.R}%
\end{equation}
Then, (\ref{1502.pf.goal.pf.indhyp}) becomes%
\begin{align}
\Delta\left(  a_{1}a_{2}\cdots a_{R}\right)   &  \in\sum_{m=1}^{R}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{R}\right)  \right)  +1_{H}\otimes\left(  a_{1}a_{2}\cdots
a_{R}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{R}}_{=\mathfrak{R}}\nonumber\\
&  =\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  +1_{H}\otimes\left(
a_{1}a_{2}\cdots a_{R}\right)  +\mathfrak{R}. \label{1502.pf.goal.pf.indhyp.R}%
\end{align}


We have $a_{R+1}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $.
Hence, Proposition~\ref{Proposition1500} (applied to $x=a_{R+1}$) yields%
\begin{align}
\Delta\left(  a_{R+1}\right)   &  \in\underbrace{a_{R+1}}_{\in
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  }\otimes\underbrace{1_{H}%
}_{\in H}+1_{H}\otimes a_{R+1}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\underbrace{\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  }_{\subseteq
H}\label{1502.pf.goal.pf.newterm.1}\\
&  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\otimes H+1_{H}\otimes a_{R+1}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes H\nonumber\\
&  =\underbrace{\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \otimes H+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \otimes H}_{\substack{\subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes H\\\text{(since }\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes H\text{ is
a }k\text{-vector space)}}}+1_{H}\otimes a_{R+1}\nonumber\\
&  \subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \otimes H+1_{H}\otimes a_{R+1}. \label{1502.pf.goal.pf.newterm.4}%
\end{align}


Now, let us define a $k$-vector subspace $\mathfrak{R}^{\prime}$ of $H\otimes
H$ by%
\begin{equation}
\mathfrak{R}^{\prime}=\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{R+1}. \label{1502.pf.goal.pf.R'}%
\end{equation}


It is easy to see that%
\begin{align}
&  \left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  \right)
\cdot\Delta\left(  a_{R+1}\right) \nonumber\\
&  \in\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)
+\mathfrak{R}^{\prime}. \label{1502.pf.goal.pf.term1a}%
\end{align}
\footnote{\textit{Proof of (\ref{1502.pf.goal.pf.term1a}):} Thus,%
\begin{align*}
&  \left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  \right)
\cdot\underbrace{\Delta\left(  a_{R+1}\right)  }_{\substack{\in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes
H+1_{H}\otimes a_{R+1}\\\text{(by (\ref{1502.pf.goal.pf.newterm.4}))}}}\\
&  \in\left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes H+1_{H}\otimes a_{R+1}\right) \\
&  \subseteq\left(  \sum_{m=1}^{R}\underbrace{a_{m}}_{\in\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  }\otimes\left(  \underbrace{\left(  a_{1}%
a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  }_{\in
H}\right)  \right)  \cdot\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes H\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \sum_{m=1}^{R}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{R}\right)  \right)  \right)  \cdot\left(  1_{H}\otimes a_{R+1}\right)
}_{=\sum_{m=1}^{R}\left(  a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  \right)
\cdot\left(  1_{H}\otimes a_{R+1}\right)  }\\
&  \subseteq\underbrace{\left(  \sum_{m=1}^{R}\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \otimes H\right)  }%
_{\substack{\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \otimes H\\\text{(since }\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \otimes H\\\text{is a }k\text{-vector
space)}}}\cdot\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \otimes H\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{m=1}^{R}\underbrace{\left(  a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{R}\right)  \right)  \right)  \cdot\left(  1_{H}\otimes a_{R+1}\right)
}_{=\left(  a_{m}1_{H}\right)  \otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  a_{R+1}\right)  }%
\end{align*}%
\begin{align*}
&  \subseteq\underbrace{\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes H\right)  \cdot\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes H\right)
}_{\substack{=\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \cdot\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \right)  \otimes\left(  H\cdot H\right)  \\\text{(by
Lemma~\ref{Lemma1315}, applied to }H\text{, }H\text{, }\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \text{, }\operatorname*{Ker}\left(
\varepsilon_{H}\right)  \text{, }H\text{ and }H\\\text{instead of }U\text{,
}V\text{, }A\text{, }C\text{, }B\text{ and }D\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{m=1}^{R}\underbrace{\left(  a_{m}1_{H}\right)
}_{=a_{m}}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\underbrace{\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  a_{R+1}}_{=a_{m+1}%
a_{m+2}\cdots a_{R+1}}\right) \\
&  =\underbrace{\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \cdot\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \right)  }_{=\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}}\otimes\underbrace{\left(  H\cdot
H\right)  }_{=H}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}%
a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)
\right) \\
&  =\underbrace{\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}\otimes H}_{\substack{\subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{R+1}\\=\mathfrak{R}^{\prime
}\\\text{(by (\ref{1502.pf.goal.pf.R'}))}}}+\sum_{m=1}^{R}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{R+1}\right)  \right) \\
&  =\mathfrak{R}^{\prime}+\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}%
a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)
\right) \\
&  =\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)  +\mathfrak{R}^{\prime}.
\end{align*}
This proves (\ref{1502.pf.goal.pf.term1a}).} Also,%
\begin{align}
&  \left(  1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  \right)
\cdot\Delta\left(  a_{R+1}\right) \nonumber\\
&  \in a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  +1_{H}%
\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)  +\mathfrak{R}^{\prime}.
\label{1502.pf.goal.pf.term1b}%
\end{align}
\footnote{\textit{Proof of (\ref{1502.pf.goal.pf.term1b}):} We have
\begin{align*}
&  \left(  1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  \right)
\cdot\underbrace{\Delta\left(  a_{R+1}\right)  }_{\substack{\in a_{R+1}%
\otimes1_{H}+1_{H}\otimes a_{R+1}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \\\text{(by (\ref{1502.pf.goal.pf.newterm.1}%
))}}}\\
&  \in\left(  1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  \right)
\cdot\left(  a_{R+1}\otimes1_{H}+1_{H}\otimes a_{R+1}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \right) \\
&  \subseteq\underbrace{\left(  1_{H}\otimes\left(  a_{1}a_{2}\cdots
a_{R}\right)  \right)  \cdot\left(  a_{R+1}\otimes1_{H}\right)  }_{=\left(
1_{H}a_{R+1}\right)  \otimes\left(  \left(  a_{1}a_{2}\cdots a_{R}\right)
1_{H}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  1_{H}\otimes\left(  a_{1}%
a_{2}\cdots a_{R}\right)  \right)  \cdot\left(  1_{H}\otimes a_{R+1}\right)
}_{=\left(  1_{H}1_{H}\right)  \otimes\left(  \left(  a_{1}a_{2}\cdots
a_{R}\right)  a_{R+1}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \underbrace{1_{H}}_{\in H}\otimes
\underbrace{\left(  a_{1}a_{2}\cdots a_{R}\right)  }_{\substack{\in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R}\\\text{(since
}a_{i}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \text{
for}\\\text{every }i\in\left\{  1,2,\ldots,R\right\}  \text{)}}}\right)
\cdot\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \right) \\
&  \subseteq\underbrace{\left(  1_{H}a_{R+1}\right)  }_{=a_{R+1}}%
\otimes\underbrace{\left(  \left(  a_{1}a_{2}\cdots a_{R}\right)
1_{H}\right)  }_{=a_{1}a_{2}\cdots a_{R}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  1_{H}1_{H}\right)  }_{=1_{H}%
}\otimes\underbrace{\left(  \left(  a_{1}a_{2}\cdots a_{R}\right)
a_{R+1}\right)  }_{=a_{1}a_{2}\cdots a_{R+1}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R}\right)
\cdot\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \right)  }_{\substack{=\left(  H\cdot\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \right)  \otimes\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R}\cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \right)
\\\text{(by Lemma~\ref{Lemma1315}, applied to }H\text{, }H\text{, }H\text{,
}\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \text{, }\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R}\text{ and
}\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \\\text{instead of
}U\text{, }V\text{, }A\text{, }C\text{, }B\text{ and }D\text{)}}}
\end{align*}%
\begin{align*}
&  =a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  H\cdot\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \right)  \otimes\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R}\cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \right) \\
&  =a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  +1_{H}\otimes\left(
a_{1}a_{2}\cdots a_{R+1}\right)  +\underbrace{\left(  H\cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \right)
}_{\substack{\subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)
\\\text{(since }\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \text{ is
an ideal of }H\text{)}}}\otimes\underbrace{\left(  \left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{R}\cdot\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \right)  }_{=\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R+1}}\\
&  \subseteq a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)  +\underbrace{\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R+1}%
}_{\substack{\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{R+1}\\=\mathfrak{R}^{\prime}\\\text{(by
(\ref{1502.pf.goal.pf.R'}))}}}\\
&  \subseteq a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)  +\mathfrak{R}^{\prime},
\end{align*}
so that (\ref{1502.pf.goal.pf.term1b}) is proven.} Furthermore,%
\begin{equation}
\mathfrak{R}\cdot\Delta\left(  a_{R+1}\right)  \subseteq\mathfrak{R}^{\prime}.
\label{1502.pf.goal.pf.term1c}%
\end{equation}
\footnote{\textit{Proof of (\ref{1502.pf.goal.pf.term1c}):} We have
\[
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\cdot
H=\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\cdot\underbrace{\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \cdot H}_{\substack{\subseteq\operatorname*{Ker}\left(
\varepsilon_{H}\right)  \\\text{(since }\operatorname*{Ker}\left(
\varepsilon_{H}\right)  \text{ is}\\\text{an ideal of }H\text{)}}%
}\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
=\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}.
\]
\par
We have $a_{R+1}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ (since
$a_{i}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ for every
$i\in\left\{  1,2,\ldots,s\right\}  $). Now,%
\begin{align}
&  \underbrace{\mathfrak{R}}_{\substack{=\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{R}\\\text{(by
(\ref{1502.pf.goal.pf.R}))}}}\cdot\left(  \underbrace{1_{H}}_{\in H}%
\otimes\underbrace{a_{R+1}}_{\in\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  }\right) \nonumber\\
&  \subseteq\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{R}\right)  \cdot\left(  H\otimes\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right) \nonumber\\
&  \subseteq\underbrace{\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\otimes H\right)  \cdot\left(
H\otimes\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
}_{\substack{=\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\cdot H\right)  \otimes\left(  H\cdot
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \\\text{(by
Lemma~\ref{Lemma1315}, applied to } H\text{, }H\text{, }\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\text{,
}H\text{, }H\text{ and }\operatorname*{Ker}\left(  \varepsilon_{H}\right)
\\\text{instead of }U\text{, }V\text{, }A\text{, }C\text{, }B\text{ and
}D\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{R}\right)  \cdot\left(  H\otimes
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  }%
_{\substack{=\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \cdot H\right)  \otimes\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R}%
\cdot\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \\\text{(by
Lemma~\ref{Lemma1315}, applied to }H\text{, }H\text{, }\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \text{, }H\text{, }\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{R}\text{ and }\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \\\text{instead of }U\text{, }V\text{,
}A\text{, }C\text{, }B\text{ and }D\text{)}}}\nonumber\\
&  =\underbrace{\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\cdot H\right)  }_{\subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}}%
\otimes\underbrace{\left(  H\cdot\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  }_{\subseteq H}+\underbrace{\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot H\right)
}_{\substack{\subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)
\\\text{(since }\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \text{ is
an}\\\text{ideal of }H\text{)}}}\otimes\underbrace{\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R}%
\cdot\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  }_{=\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R+1}}\nonumber\\
&  \subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{R+1}=\mathfrak{R}^{\prime}
\label{1502.pf.goal.pf.term-triv}%
\end{align}
(by (\ref{1502.pf.goal.pf.R'})).
\par
We have $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}=\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\cdot\underbrace{\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  }_{\subseteq H}\subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \cdot H\subseteq\operatorname*{Ker}\left(
\varepsilon_{H}\right)  $ (since $\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  $ is an ideal of $H$). Now, the definition of $\mathfrak{R}$
yields%
\begin{align*}
\mathfrak{R}  &  =\underbrace{\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}}_{=\operatorname*{Ker}\left(
\varepsilon_{H}\right)  }\otimes H+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\underbrace{\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{R}}_{\subseteq
H}\\
&  \subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \otimes H+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \otimes H\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \otimes H
\end{align*}
(since $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\otimes H$ is a $k$-vector space). Thus,%
\begin{align*}
&  \mathfrak{R}\cdot\underbrace{\Delta\left(  a_{R+1}\right)  }_{\substack{\in
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes
H+1_{H}\otimes a_{R+1}\\\text{(by (\ref{1502.pf.goal.pf.newterm.4}))}}}\\
&  \subseteq\mathfrak{R}\cdot\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes H+1_{H}\otimes a_{R+1}\right) \\
&  \subseteq\underbrace{\mathfrak{R}}_{\subseteq\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  \otimes H}\cdot\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes H\right)
+\underbrace{\mathfrak{R}\cdot\left(  1_{H}\otimes a_{R+1}\right)
}_{\substack{\subseteq\mathfrak{R}^{\prime}\\\text{(by
(\ref{1502.pf.goal.pf.term-triv}))}}}\\
&  \subseteq\underbrace{\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes H\right)  \cdot\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \otimes H\right)
}_{\substack{=\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \cdot\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \right)  \otimes\left(  H\cdot H\right)  \\\text{(by
Lemma~\ref{Lemma1315}, applied to }H\text{, }H\text{, }\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \text{, }\operatorname*{Ker}\left(
\varepsilon_{H}\right)  \text{, }H\text{ and }H\\\text{instead of }U\text{,
}V\text{, }A\text{, }C\text{, }B\text{ and }D\text{)}}}+\mathfrak{R}^{\prime
}\\
&  =\underbrace{\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \cdot\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  \right)  }_{=\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}}\otimes\underbrace{\left(  H\cdot
H\right)  }_{\subseteq H}+\mathfrak{R}^{\prime}\\
&  \subseteq\underbrace{\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H}_{\substack{\subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\otimes
H+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{R+1}=\mathfrak{R}^{\prime}\\\text{(by (\ref{1502.pf.goal.pf.R'}))}%
}}+\mathfrak{R}^{\prime}\subseteq\mathfrak{R}^{\prime}+\mathfrak{R}^{\prime
}\subseteq\mathfrak{R}^{\prime}%
\end{align*}
(since $\mathfrak{R}^{\prime}$ is a $k$-vector space). This proves
(\ref{1502.pf.goal.pf.term1c}).}

It is also easy to find that%
\begin{align}
&  \sum_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}%
\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right) \nonumber\\
&  =\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)  +a_{R+1}\otimes\left(
a_{1}a_{2}\cdots a_{R}\right)  . \label{1502.pf.goal.pf.term2}%
\end{align}
\footnote{The proof of this is identical to the proof of
(\ref{1402.pf.goal.pf.term2}).}

Since $H$ is a bialgebra, the comultiplication $\Delta$ is a $k$-algebra
homomorphism (by the axioms of a bialgebra). We have%
\begin{align*}
&  \Delta\left(  \underbrace{a_{1}a_{2}\cdots a_{R+1}}_{=\left(  a_{1}%
a_{2}\cdots a_{R}\right)  a_{R+1}}\right) \\
&  =\Delta\left(  \left(  a_{1}a_{2}\cdots a_{R}\right)  a_{R+1}\right) \\
&  =\underbrace{\Delta\left(  a_{1}a_{2}\cdots a_{R}\right)  }_{\substack{\in
\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)  +1_{H}\otimes\left(
a_{1}a_{2}\cdots a_{R}\right)  +\mathfrak{R}\\\text{(by
(\ref{1502.pf.goal.pf.indhyp.R}))}}}\cdot\Delta\left(  a_{R+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  \in\left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)  \right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  +\mathfrak{R}\right)
\cdot\Delta\left(  a_{R+1}\right) \\
&  \subseteq\underbrace{\left(  \sum_{m=1}^{R}a_{m}\otimes\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R}\right)
\right)  \right)  \cdot\Delta\left(  a_{R+1}\right)  }_{\substack{\in
\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)  +\mathfrak{R}^{\prime
}\\\text{(by (\ref{1502.pf.goal.pf.term1a}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  1_{H}\otimes\left(  a_{1}%
a_{2}\cdots a_{R}\right)  \right)  \cdot\Delta\left(  a_{R+1}\right)
}_{\substack{=a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)  +\mathfrak{R}^{\prime
}\\\text{(by (\ref{1502.pf.goal.pf.term1b}))}}}+\underbrace{\mathfrak{R}%
\cdot\Delta\left(  a_{R+1}\right)  }_{\substack{\subseteq\mathfrak{R}^{\prime
}\\\text{(by (\ref{1502.pf.goal.pf.term1c}))}}}\\
&  \subseteq\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)
+\mathfrak{R}^{\prime}\\
&  \ \ \ \ \ \ \ \ \ \ +a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)  +\mathfrak{R}^{\prime
}+\mathfrak{R}^{\prime}%
\end{align*}%
\begin{align*}
&  =\underbrace{\sum_{m=1}^{R}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)
+a_{R+1}\otimes\left(  a_{1}a_{2}\cdots a_{R}\right)  }_{\substack{=\sum
_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)  \\\text{(by
(\ref{1502.pf.goal.pf.term2}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right)
+\underbrace{\mathfrak{R}^{\prime}+\mathfrak{R}^{\prime}+\mathfrak{R}^{\prime
}}_{\substack{\subseteq\mathfrak{R}^{\prime}\\\text{(since }\mathfrak{R}%
^{\prime}\text{ is a }k\text{-vector space)}}}\\
&  \subseteq\sum_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\mathfrak{R}^{\prime}}_{\substack{=\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\otimes
H+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{R+1}\\\text{(by the definition of }\mathfrak{R}^{\prime}\text{)}}}\\
&  =\sum_{m=1}^{R+1}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{R+1}\right)  \right)
+1_{H}\otimes\left(  a_{1}a_{2}\cdots a_{R+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  \otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{R+1}.
\end{align*}
In other words, (\ref{1502.pf.goal}) holds for $r=R+1$. We have thus proven
(\ref{1502.pf.goal}) for $r=R+1$. The induction step is thus complete.

Hence, (\ref{1502.pf.goal}) is proven by induction. Now, applying
(\ref{1502.pf.goal}) to $r=s$, we obtain%
\begin{align*}
\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)   &  \in\sum_{m=1}^{s}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{s}\right)  \right)  +\underbrace{1_{H}}_{\in H}%
\otimes\underbrace{\left(  a_{1}a_{2}\cdots a_{s}\right)  }_{\substack{\in
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s}\\\text{(since }a_{i}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)
\text{ for}\\\text{every }i\in\left\{  1,2,\ldots,s\right\}  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+\underbrace{\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  }_{\subseteq H}\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\\
&  \subseteq\sum_{m=1}^{s}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{s}\right)  \right)
+H\otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s}\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s}\\
&  =\sum_{m=1}^{s}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{s}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+\underbrace{H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}%
+H\otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s}}_{\substack{\subseteq H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s}\\\text{(since }H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\text{ is a
}k\text{-vector space)}}}\\
&  \subseteq\sum_{m=1}^{s}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{s}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s}.
\end{align*}
This proves Lemma~\ref{Lemma1502}.
\end{proof}

From Lemma~\ref{Lemma1502}, we easily conclude the following weaker statement:

\begin{lemma}
\label{Lemma1503} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let
$s\in\mathbb{N}$ be positive. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$
elements of $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Then,%
\[
\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)  \in H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s-1}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\otimes
H+H\otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s}.
\]
\footnote{Recall that $\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{\ell}$ is defined according to
Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{\ell}$ means the $\ell$-th power of the
subspace $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ of the
$k$-algebra $H$ for every $\ell\in\mathbb{N}$.}
\end{lemma}

(Note that we couldn't remove the $H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s-1}$ term from the right hand side in
Lemma~\ref{Lemma1503}; in fact, if $s=1$, then we don't have $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}%
\subseteq\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s-1}$, because in this case we have $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s-1}=\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{0}=k\cdot1_{H}\not \supseteq
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $.)

\begin{proof}
[Proof of Lemma~\ref{Lemma1503}.]Every $m\in\left\{  1,2,\ldots,s\right\}  $
satisfies%
\begin{equation}
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{s}\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{s-1} \label{1503.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{1503.pf.1}):} Let $m\in\left\{
1,2,\ldots,s\right\}  $. Then, $a_{1}a_{2}\cdots a_{m-1}\in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{m-1}$ (since
$a_{i}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ for all
$i\in\left\{  1,2,\ldots,m-1\right\}  $) and $a_{m+1}a_{m+2}\cdots a_{s}%
\in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s-m}$
(since $a_{i}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ for all
$i\in\left\{  m+1,m+2,\ldots,s\right\}  $). Now,%
\begin{align*}
\underbrace{\left(  a_{1}a_{2}\cdots a_{m-1}\right)  }_{\in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{m-1}%
}\underbrace{\left(  a_{m+1}a_{m+2}\cdots a_{s}\right)  }_{\in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s-m}}  &
\in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{m-1}\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s-m}\\
&  =\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{\left(  m-1\right)  +\left(  s-m\right)  }=\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{s-1}.
\end{align*}
This proves (\ref{1503.pf.1}).}. Hence,%
\begin{align}
\sum_{m=1}^{s}\underbrace{a_{m}}_{\in H}\otimes\underbrace{\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{s}\right)
\right)  }_{\substack{\in\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{s-1}\\\text{(by (\ref{1503.pf.1}))}}}  &  \in
\sum_{m=1}^{s}H\otimes\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{s-1}\nonumber\\
&  \subseteq H\otimes\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{s-1} \label{1503.pf.2}%
\end{align}
(since $H\otimes\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{s-1}$ is a $k$-vector space).

Now, Lemma~\ref{Lemma1502} yields%
\begin{align*}
\Delta\left(  a_{1}a_{2}\cdots a_{s}\right)   &  \in\underbrace{\sum_{m=1}%
^{s}a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(
a_{m+1}a_{m+2}\cdots a_{s}\right)  \right)  }_{\in H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s-1}}\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s}\\
&  \subseteq H\otimes\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{s-1}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s}.
\end{align*}
This proves Lemma~\ref{Lemma1503}.
\end{proof}

We are now ready to prove the following result (more or less equivalent to
Theorem~\ref{Theorem1501} \textbf{(a)}):

\begin{lemma}
\label{Lemma1504} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be a
$k$-algebra. Let $r\in\mathbb{N}$. Let $f_{1}$, $f_{2}$, $\ldots$, $f_{r}$ be
$r$ elements of $\mathcal{L}\left(  H,A\right)  $ such that $f_{1}$, $f_{2}$,
$\ldots$, $f_{r}$ are $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivations.

Every $s\in\mathbb{N}$ such that $s>r$ satisfies%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\right)  =0.
\]
\footnote{Recall that $\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{\ell}$ is defined according to
Convention~\ref{Convention100}. Hence, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{\ell}$ means the $\ell$-th power of the
subspace $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ of the
$k$-algebra $H$ for every $\ell\in\mathbb{N}$.}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1504}.]We are going to prove Lemma~\ref{Lemma1504}
by induction over $r$:

\textit{Induction base:} If $r=0$, then Lemma~\ref{Lemma1504} is
true\footnote{\textit{Proof.} Let $r=0$. Let $k$ be a field. Let $H$ be a
$k$-bialgebra. Let $A$ be a $k$-algebra. Let $f_{1}$, $f_{2}$, $\ldots$,
$f_{0}$ be $0$ elements of $\mathcal{L}\left(  H,A\right)  $ (that is, no
elements) such that $f_{1}$, $f_{2}$, $\ldots$, $f_{0}$ are $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivations. Let $s\in\mathbb{N}$
such that $s>r$. Recall that $\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  $ is an ideal of $H$ (this is proven as in the proof of
Lemma~\ref{Lemma1502}). We have $s>r=0$, so that $\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{s}=\underbrace{\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s-1}}_{\subseteq
H}\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\subseteq H\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  \subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ (since
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ is an ideal of $H$). On
the other hand, $r=0$, so that%
\begin{align*}
f_{1}\ast f_{2}\ast\cdots\ast f_{r}  &  =f_{1}\ast f_{2}\ast\cdots\ast
f_{0}=\left(  \text{empty product in }\mathcal{L}\left(  H,A\right)  \right)
\\
&  =e_{H,A}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the unity of the
}k\text{-algebra }\mathcal{L}\left(  H,A\right)  \text{ is }e_{H,A}\right) \\
&  =\eta_{A}\circ\varepsilon_{H}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\varepsilon_{H}\right)  ,
\end{align*}
so that%
\begin{align*}
\left(  \underbrace{f_{1}\ast f_{2}\ast\cdots\ast f_{r}}_{=\eta_{A}%
\circ\varepsilon_{H}}\right)  \left(  \underbrace{\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{s}}_{\subseteq\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  }\right)   &  \subseteq\left(  \eta_{A}%
\circ\varepsilon_{H}\right)  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right) \\
&  =\eta_{A}\left(  \underbrace{\varepsilon_{H}\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  }_{\substack{=0\\\text{(by the
definition of a kernel)}}}\right)  =\eta_{A}\left(  0\right)  =0
\end{align*}
(since the map $\eta_{A}$ is $k$-linear). Hence, $\left(  f_{1}\ast f_{2}%
\ast\cdots\ast f_{r}\right)  \left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{s}\right)  =0$. Thus, Lemma~\ref{Lemma1504}
is proven under the assumption that $r=0$, qed.}. Hence, the induction base is complete.

\textit{Induction step:} Let $R\in\mathbb{N}$ be positive. Assume that
Lemma~\ref{Lemma1504} is proven for $r=R-1$. We now are going to prove that
Lemma~\ref{Lemma1504} holds for $r=R$.

Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be a $k$-algebra. Let
$f_{1}$, $f_{2}$, $\ldots$, $f_{R}$ be $R$ elements of $\mathcal{L}\left(
H,A\right)  $ such that $f_{1}$, $f_{2}$, $\ldots$, $f_{R}$ are $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivations. Let $s$ be an element
of $\mathbb{N}$ such that $s>R$. We are going to show that%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\right)  =0.
\]


We have $s>R\geq0$, and thus $s$ is positive.

Let $g=f_{2}\ast f_{3}\ast\cdots\ast f_{R}$. Then, $g$ is a $k$-linear map
$H\rightarrow A$. Let $\ell\in\mathbb{N}$ be such that $\ell>R-1$. Thus, we
can apply Lemma~\ref{Lemma1504} to $R-1$, $\left(  f_{2},f_{3},\ldots
,f_{R}\right)  $ and $\ell$ instead of $r$, $\left(  f_{1},f_{2},\ldots
,f_{r}\right)  $ and $s$ (since we assumed that Lemma~\ref{Lemma1504} is
proven for $r=R-1$). As a result, we obtain $\left(  f_{2}\ast f_{3}\ast
\cdots\ast f_{R}\right)  \left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{\ell}\right)  =0$. Thus,%
\[
\underbrace{g}_{=f_{2}\ast f_{3}\ast\cdots\ast f_{R}}\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{\ell}\right)
=\left(  f_{2}\ast f_{3}\ast\cdots\ast f_{R}\right)  \left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{\ell}\right)
=0.
\]


Now, forget that we fixed $\ell$. We thus have proven that%
\begin{equation}
g\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{\ell}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{N}\text{
such that }\ell\geq R-1. \label{1504.pf.g.1}%
\end{equation}


Applying (\ref{1504.pf.g.1}) to $\ell=s-1$, we obtain%
\begin{equation}
g\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s-1}\right)  =0 \label{1504.pf.g.2}%
\end{equation}
(since $\underbrace{s}_{>R}-1>R-1$). But applying (\ref{1504.pf.g.1}) to
$\ell=s$, we obtain%
\begin{equation}
g\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s}\right)  =0 \label{1504.pf.g.3}%
\end{equation}
(since $s>R>R-1$).

We know that $f_{1}$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation. But the axioms of a $k$-bialgebra yield that $\varepsilon_{H}$
is a $k$-algebra homomorphism (since $H$ is a $k$-bialgebra). Hence,
Theorem~\ref{Theorem115} (applied to $f=f_{1}$) yields that $f_{1}$ is an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation if and only if
$f_{1}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0$. Thus, $f_{1}\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot
1_{H}\right)  =0$ (since we know that $f_{1}$ is an $\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivation). Hence,
\[
f_{1}\left(  \underbrace{\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}}_{\subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}}\right)  \subseteq
f_{1}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0,
\]
so that $f_{1}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)  =0$. Moreover,
\[
f_{1}\left(  \underbrace{1_{H}}_{\in k\cdot1_{H}\subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}%
}\right)  \in f_{1}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0,
\]
so that $f_{1}\left(  1_{H}\right)  =0$.

Now, let $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. We are going to prove
that $\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  a_{1}%
a_{2}\cdots a_{s}\right)  =0$. First, we have%
\begin{align*}
&  \left(  f_{1}\otimes g\right)  \left(  \underbrace{\Delta\left(  a_{1}%
a_{2}\cdots a_{s}\right)  }_{\substack{\in H\otimes\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{s-1}+\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}\otimes H+H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\\\text{(by
Lemma~\ref{Lemma1503})}}}\right) \\
&  \in\underbrace{\left(  f_{1}\otimes g\right)  \left(  H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s-1}\right)
}_{\subseteq f_{1}\left(  H\right)  \otimes g\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s-1}\right)
}+\underbrace{\left(  f_{1}\otimes g\right)  \left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\otimes
H\right)  }_{\subseteq f_{1}\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\right)  \otimes g\left(  H\right)
}+\underbrace{\left(  f_{1}\otimes g\right)  \left(  H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\right)
}_{\subseteq f_{1}\left(  H\right)  \otimes g\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\right)  }\\
&  \subseteq f_{1}\left(  H\right)  \otimes\underbrace{g\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s-1}\right)
}_{\substack{=0\\\text{(by (\ref{1504.pf.g.2}))}}}+\underbrace{f_{1}\left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\right)  }_{=0}\otimes g\left(  H\right)  +f_{1}\left(  H\right)
\otimes\underbrace{g\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{s}\right)  }_{\substack{=0\\\text{(by
(\ref{1504.pf.g.3}))}}}\\
&  =\underbrace{f_{1}\left(  H\right)  \otimes0}_{=0}+\underbrace{0\otimes
g\left(  H\right)  }_{=0}+\underbrace{f_{1}\left(  H\right)  \otimes0}_{=0}=0.
\end{align*}
Thus, $\left(  f_{1}\otimes g\right)  \left(  \Delta\left(  a_{1}a_{2}\cdots
a_{s}\right)  \right)  =0$. Now, $f_{1}\ast f_{2}\ast\cdots\ast f_{R}%
=f_{1}\ast\underbrace{\left(  f_{2}\ast f_{3}\ast\cdots\ast f_{R}\right)
}_{=g}=f_{1}\ast g=\mu\circ\left(  f_{1}\otimes g\right)  \circ\Delta$ (by the
definition of convolution), so that%
\begin{align*}
&  \underbrace{\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  }%
_{=\mu\circ\left(  f_{1}\otimes g\right)  \circ\Delta}\left(  a_{1}a_{2}\cdots
a_{s}\right) \\
&  =\left(  \mu\circ\left(  f_{1}\otimes g\right)  \circ\Delta\right)  \left(
a_{1}a_{2}\cdots a_{s}\right)  =\mu\left(  \underbrace{\left(  f_{1}\otimes
g\right)  \left(  \Delta\left(  a_{1}a_{2}\cdots a_{s}\right)  \right)  }%
_{=0}\right)  =\mu\left(  0\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mu\text{ is
}k\text{-linear}\right)  .
\end{align*}


Now, forget that we fixed $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$. We thus have
shown that whenever $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ are $s$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $, we have%
\begin{equation}
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  a_{1}a_{2}\cdots
a_{s}\right)  =0. \label{1504.pf.almostthere}%
\end{equation}


Now, (\ref{100.Vn}) yields%
\begin{align*}
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}  &
=\left\langle a_{1}a_{2}\cdots a_{s}\ \mid\ \left(  a_{1},a_{2},\ldots
,a_{s}\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{\times s}\right\rangle \\
&  =\left\langle \left\{  a_{1}a_{2}\cdots a_{s}\ \mid\ \left(  a_{1}%
,a_{2},\ldots,a_{s}\right)  \in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{\times s}\right\}  \right\rangle .
\end{align*}
Applying the map $f_{1}\ast f_{2}\ast\cdots\ast f_{R}$ to both sides of this
equality, we obtain%
\begin{align*}
&  \left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\right) \\
&  =\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  \left\langle
\left\{  a_{1}a_{2}\cdots a_{s}\ \mid\ \left(  a_{1},a_{2},\ldots
,a_{s}\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{\times s}\right\}  \right\rangle \right) \\
&  =\left\langle \underbrace{\left(  f_{1}\ast f_{2}\ast\cdots\ast
f_{R}\right)  \left(  \left\{  a_{1}a_{2}\cdots a_{s}\ \mid\ \left(
a_{1},a_{2},\ldots,a_{s}\right)  \in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{\times s}\right\}  \right)  }_{=\left\{
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  a_{1}a_{2}\cdots
a_{s}\right)  \ \mid\ \left(  a_{1},a_{2},\ldots,a_{s}\right)  \in\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{\times
s}\right\}  }\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{316.pf.3}), applied to }H\text{, }A\text{, }f_{1}\ast f_{2}%
\ast\cdots\ast f_{R}\text{ and}\\
\left\{  a_{1}a_{2}\cdots a_{s}\ \mid\ \left(  a_{1},a_{2},\ldots
,a_{s}\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{\times s}\right\}  \text{ instead of }M\text{, }R\text{, }%
\phi\text{ and }S
\end{array}
\right) \\
&  =\left\langle \left\{  \underbrace{\left(  f_{1}\ast f_{2}\ast\cdots\ast
f_{R}\right)  \left(  a_{1}a_{2}\cdots a_{s}\right)  }%
_{\substack{=0\\\text{(by (\ref{1504.pf.almostthere}))}}}\ \mid\ \left(
a_{1},a_{2},\ldots,a_{s}\right)  \in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{\times s}\right\}  \right\rangle \\
&  =\left\langle \underbrace{\left\{  0\ \mid\ \left(  a_{1},a_{2}%
,\ldots,a_{s}\right)  \in\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{\times s}\right\}  }_{\subseteq0}\right\rangle
\subseteq\left\langle 0\right\rangle =0.
\end{align*}
In other words, $\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s}\right)  =0$.

Now, forget that we fixed $k$, $H$, $A$, $\left(  f_{1},f_{2},\ldots
,f_{R}\right)  $ and $s$. We thus have shown that if $k$ is a field, if $H$ is
a $k$-bialgebra, if $A$ is a $k$-algebra, if $f_{1}$, $f_{2}$, $\ldots$,
$f_{R}$ be $R$ elements of $\mathcal{L}\left(  H,A\right)  $ such that $f_{1}%
$, $f_{2}$, $\ldots$, $f_{R}$ are $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivations, then every $s\in\mathbb{N}$ satisfying $s>R$
satisfies $\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{R}\right)  \left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{s}\right)  =0$.
In other words, we have shown that Lemma~\ref{Lemma1504} holds for $r=R$. This
completes the induction step. The induction proof of Lemma~\ref{Lemma1504} is
thus complete.
\end{proof}

We can now proceed to the proof of Theorem~\ref{Theorem1501}.

\begin{proof}
[Proof of Theorem~\ref{Theorem1501}.]\textbf{(a)} Let $s\in\mathbb{N}$ be such
that $s>r$. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Then, $a_{1}a_{2}\cdots
a_{s}\in\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{s}$ (since $a_{i}\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ for
every $i\in\left\{  1,2,\ldots,s\right\}  $). Hence,%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{a_{1}%
a_{2}\cdots a_{s}}_{\in\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{s}}\right)  \in\left(  f_{1}\ast f_{2}\ast\cdots\ast
f_{r}\right)  \left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{s}\right)  =0
\]
(by Lemma~\ref{Lemma1504}). Hence, $\left(  f_{1}\ast f_{2}\ast\cdots\ast
f_{r}\right)  \left(  a_{1}a_{2}\cdots a_{s}\right)  =0$. This proves
Theorem~\ref{Theorem1501} \textbf{(a)}.

\textbf{(b)} We are going to prove Theorem~\ref{Theorem1501} \textbf{(b)} by
induction over $r$:

\textit{Induction base:} If $r=0$, then Theorem~\ref{Theorem1501} \textbf{(b)}
is true\footnote{\textit{Proof.} Assume that $r=0$. Let $k$ be a field. Let
$H$ be a $k$-bialgebra. Let $A$ be a $k$-algebra. Let $f_{1}$, $f_{2}$,
$\ldots$, $f_{0}$ be $0$ elements of $\mathcal{L}\left(  H,A\right)  $ (that
is, no elements) such that $f_{1}$, $f_{2}$, $\ldots$, $f_{0}$ are $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivations. Let $a_{1}$, $a_{2}$,
$\ldots$, $a_{0}$ be $0$ elements of $\operatorname*{Ker}\left(
\varepsilon_{H}\right)  $ (that is, no elements). Since $r=0$, we have
$S_{r}=S_{0}=\left\{  \operatorname*{id}\right\}  $. But since $r=0$, we have
$f_{1}\ast f_{2}\ast\cdots\ast f_{r}=\left(  \text{empty product with respect
to convolution}\right)  =e_{H,A}$. Also, since $r=0$, we have $a_{1}%
a_{2}\cdots a_{r}=\left(  \text{empty product}\right)  =1_{H}$. Thus,
\begin{align*}
&  \underbrace{\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  }%
_{=e_{H,A}}\left(  \underbrace{a_{1}a_{2}\cdots a_{r}}_{=1_{H}}\right) \\
&  =\underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{H}}\left(  1_{H}\right)
=\left(  \eta_{A}\circ\varepsilon_{H}\right)  \left(  1_{H}\right)  =\eta
_{A}\left(  \underbrace{\varepsilon_{H}\left(  1_{H}\right)  }%
_{\substack{=1\\\text{(since }H\text{ is a bialgebra)}}}\right)  =\eta
_{A}\left(  1\right) \\
&  =1\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\eta_{A}\right) \\
&  =1_{A}%
\end{align*}
and%
\[
\underbrace{\sum_{\sigma\in S_{r}}}_{\substack{=\sum_{\sigma\in\left\{
\operatorname*{id}\right\}  }\\\text{(since }S_{r}=\left\{  \operatorname*{id}%
\right\}  \text{)}}}\underbrace{f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{r}\left(  a_{\sigma\left(  r\right)  }\right)  }_{\substack{=\left(
\text{empty product}\right)  \\\text{(since }r=0\text{)}}}=\sum_{\sigma
\in\left\{  \operatorname*{id}\right\}  }\underbrace{\left(  \text{empty
product}\right)  }_{=1_{A}}=\sum_{\sigma\in\left\{  \operatorname*{id}%
\right\}  }1_{A}=1_{A}.
\]
Hence, $\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a_{1}%
a_{2}\cdots a_{r}\right)  =1_{A}=\sum_{\sigma\in S_{r}}f_{1}\left(
a_{\sigma\left(  1\right)  }\right)  f_{2}\left(  a_{\sigma\left(  2\right)
}\right)  \cdots f_{r}\left(  a_{\sigma\left(  r\right)  }\right)  $. Hence,
Theorem~\ref{Theorem1501} \textbf{(b)} is true for $r=0$, qed.}. Hence, the
induction base is complete.

\textit{Induction step:} Let $n\in\mathbb{N}$ be positive. Assume that
Theorem~\ref{Theorem1501} \textbf{(b)} is proven for $r=n-1$. We now are going
to prove that Theorem~\ref{Theorem1501} \textbf{(b)} holds for $r=n$.

Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$ be a $k$-algebra. Let
$f_{1}$, $f_{2}$, $\ldots$, $f_{n}$ be $n$ elements of $\mathcal{L}\left(
H,A\right)  $ such that $f_{1}$, $f_{2}$, $\ldots$, $f_{n}$ are $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivations. Let $a_{1}$, $a_{2}$,
$\ldots$, $a_{n}$ be $n$ elements of $\operatorname*{Ker}\left(
\varepsilon_{H}\right)  $. We will prove that
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}a_{2}\cdots
a_{n}\right)  =\sum_{\sigma\in S_{n}}f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  .
\]


Let $g=f_{2}\ast f_{3}\ast\cdots\ast f_{n}$. Then, $g$ is a $k$-linear map
$H\rightarrow A$. We have
\[
f_{1}\ast f_{2}\ast\cdots\ast f_{n}=f_{1}\ast\underbrace{\left(  f_{2}\ast
f_{3}\ast\cdots\ast f_{n}\right)  }_{=g}=f_{1}\ast g=\mu_{A}\circ\left(
f_{1}\otimes g\right)  \circ\Delta
\]
(by the definition of convolution). Hence,%
\begin{align}
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}a_{2}\cdots
a_{n}\right)   &  =\left(  \mu_{A}\circ\left(  f_{1}\otimes g\right)
\circ\Delta\right)  \left(  a_{1}a_{2}\cdots a_{n}\right) \nonumber\\
&  =\mu_{A}\left(  \left(  f_{1}\otimes g\right)  \left(  \Delta\left(
a_{1}a_{2}\cdots a_{n}\right)  \right)  \right)  . \label{1501.pf.b.1}%
\end{align}
We are now going to study the term $\left(  f_{1}\otimes g\right)  \left(
\Delta\left(  a_{1}a_{2}\cdots a_{n}\right)  \right)  $ on the right hand side
of this.

We have%
\begin{align}
&  \left(  f_{1}\otimes g\right)  \left(  \underbrace{\Delta\left(  a_{1}%
a_{2}\cdots a_{n}\right)  }_{\substack{\in\sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{n}\\\text{(by Lemma~\ref{Lemma1502},
applied to }s=n\text{)}}}\right) \nonumber\\
&  \in\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  +\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\otimes H+H\otimes\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{n}\right) \nonumber\\
&  \subseteq\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{n}\right)  \right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  f_{1}\otimes g\right)  \left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\otimes
H\right)  +\left(  f_{1}\otimes g\right)  \left(  H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{n}\right)  .
\label{1501.pf.b.2}%
\end{align}


But $f_{1}$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation. From this, it is easy to conclude that $f_{1}\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\right)
=0$\ \ \ \ \footnote{\textit{Proof.} The axioms of a $k$-bialgebra yield that
$\varepsilon_{H}$ is a $k$-algebra homomorphism (since $H$ is a $k$%
-bialgebra). Hence, Theorem~\ref{Theorem115} (applied to $f=f_{1}$) yields
that $f_{1}$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation if and only if $f_{1}\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$. Thus,
$f_{1}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0$ (since we know that $f_{1}$ is an
$\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation). Hence,
\[
f_{1}\left(  \underbrace{\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}}_{\subseteq\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}}\right)  \subseteq
f_{1}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0,
\]
so that $f_{1}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)  =0$, qed.}. Now,%
\begin{equation}
\left(  f_{1}\otimes g\right)  \left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}\otimes H\right)  \subseteq
\underbrace{f_{1}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}\right)  }_{=0}\otimes g\left(  H\right)  =0\otimes
g\left(  H\right)  =0. \label{1501.pf.b.3b}%
\end{equation}


Furthermore, $n>n-1$. Hence, Lemma~\ref{Lemma1504} (applied to $n-1$, $\left(
f_{2},f_{3},\ldots,f_{n}\right)  $ and $n$ instead of $r$, $\left(
f_{1},f_{2},\ldots,f_{r}\right)  $ and $s$) yields $\left(  f_{2}\ast
f_{3}\ast\cdots\ast f_{n}\right)  \left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{n}\right)  =0$. Thus,
\begin{equation}
\underbrace{g}_{=f_{2}\ast f_{3}\ast\cdots\ast f_{n}}\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{n}\right)
=\left(  f_{2}\ast f_{3}\ast\cdots\ast f_{n}\right)  \left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{n}\right)  =0.
\label{1501.pf.b.3c.lem}%
\end{equation}
Now,%
\begin{equation}
\left(  f_{1}\otimes g\right)  \left(  H\otimes\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{n}\right)  \subseteq f_{1}\left(
H\right)  \otimes\underbrace{g\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{n}\right)  }_{=0}=f_{1}\left(  H\right)
\otimes0=0. \label{1501.pf.b.3c}%
\end{equation}


Now, (\ref{1501.pf.b.2}) becomes%
\begin{align*}
&  \left(  f_{1}\otimes g\right)  \left(  \Delta\left(  a_{1}a_{2}\cdots
a_{n}\right)  \right) \\
&  \in\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  f_{1}\otimes g\right)  \left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}\otimes H\right)  }_{\substack{\subseteq0\\\text{(by (\ref{1501.pf.b.3b}%
))}}}+\underbrace{\left(  f_{1}\otimes g\right)  \left(  H\otimes\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{n}\right)
}_{\substack{\subseteq0\\\text{(by (\ref{1501.pf.b.3b}))}}}\\
&  \subseteq\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}%
\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}%
a_{m+2}\cdots a_{n}\right)  \right)  \right)  +0+0\\
&  =\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  \right)  .
\end{align*}
In other words,%
\begin{align}
&  \left(  f_{1}\otimes g\right)  \left(  \Delta\left(  a_{1}a_{2}\cdots
a_{n}\right)  \right) \nonumber\\
&  =\left(  f_{1}\otimes g\right)  \left(  \sum_{m=1}^{n}a_{m}\otimes\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  \right) \nonumber\\
&  =\sum_{m=1}^{n}\underbrace{\left(  f_{1}\otimes g\right)  \left(
a_{m}\otimes\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(
a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)  \right)  }_{=f_{1}\left(
a_{m}\right)  \otimes g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f_{1}\otimes g\text{ is
}k\text{-linear}\right) \nonumber\\
&  =\sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \otimes g\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{n}\right)
\right)  . \label{1501.pf.b.5}%
\end{align}


Now, (\ref{1501.pf.b.1}) becomes%
\begin{align}
&  \left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}%
a_{2}\cdots a_{n}\right) \nonumber\\
&  =\mu_{A}\left(  \underbrace{\left(  f_{1}\otimes g\right)  \left(
\Delta\left(  a_{1}a_{2}\cdots a_{n}\right)  \right)  }_{\substack{=\sum
_{m=1}^{n}f_{1}\left(  a_{m}\right)  \otimes g\left(  \left(  a_{1}a_{2}\cdots
a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)
\\\text{(by (\ref{1501.pf.b.5}))}}}\right) \nonumber\\
&  =\mu_{A}\left(  \sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \otimes g\left(
\left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  \right) \nonumber\\
&  =\sum_{m=1}^{n}\underbrace{\mu_{A}\left(  f_{1}\left(  a_{m}\right)
\otimes g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(
a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)  \right)  }_{\substack{=f_{1}%
\left(  a_{m}\right)  \cdot g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)
\left(  a_{m+1}a_{m+2}\cdots a_{n}\right)  \right)  \\\text{(since }\mu
_{A}\text{ is the multiplication map)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mu_{A}\text{ is
}k\text{-linear}\right) \nonumber\\
&  =\sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \cdot g\left(  \left(  a_{1}%
a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{n}\right)
\right)  . \label{1501.pf.b.6}%
\end{align}


Now, we are going to show that every $m\in\left\{  1,2,\ldots,n\right\}  $
satisfies%
\begin{equation}
g\left(  \left(  a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots
a_{n}\right)  \right)  =\sum_{\substack{\sigma\in S_{n};\\\sigma\left(
1\right)  =m}}f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  f_{3}\left(
a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  . \label{1501.pf.b.m-part.goal}%
\end{equation}


The proof of (\ref{1501.pf.b.m-part.goal}) proceeds exactly as the proof of
(\ref{1401.pf.b.m-part.goal}), with the only difference that:

-- every appearance of \textquotedblleft$\operatorname*{Prim}H$%
\textquotedblright\ has to be replaced by \textquotedblleft%
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $\textquotedblright;

-- every reference to Theorem~\ref{Theorem1401} has to be replaced by a
reference to Theorem~\ref{Theorem1501}.

Therefore, (\ref{1501.pf.b.m-part.goal}) is proven.

Now, (\ref{1501.pf.b.6}) becomes%
\begin{align*}
&  \left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}%
a_{2}\cdots a_{n}\right) \\
&  =\sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \cdot\underbrace{g\left(  \left(
a_{1}a_{2}\cdots a_{m-1}\right)  \left(  a_{m+1}a_{m+2}\cdots a_{n}\right)
\right)  }_{\substack{=\sum_{\substack{\sigma\in S_{n};\\\sigma\left(
1\right)  =m}}f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  f_{3}\left(
a_{\sigma\left(  3\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  \\\text{(by (\ref{1501.pf.b.m-part.goal}))}}}\\
&  =\sum_{m=1}^{n}f_{1}\left(  a_{m}\right)  \cdot\sum_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  =m}}f_{2}\left(  a_{\sigma\left(  2\right)
}\right)  f_{3}\left(  a_{\sigma\left(  3\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right) \\
&  =\sum_{m=1}^{n}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}f_{1}\left(  \underbrace{a_{m}}_{\substack{=a_{\sigma\left(  1\right)
}\\\text{(since }m=\sigma\left(  1\right)  \\\text{(since }\sigma\left(
1\right)  =m\text{))}}}\right)  f_{2}\left(  a_{\sigma\left(  2\right)
}\right)  f_{3}\left(  a_{\sigma\left(  3\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right) \\
&  =\sum_{m=1}^{n}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}\underbrace{f_{1}\left(  a_{\sigma\left(  1\right)  }\right)  f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  f_{3}\left(  a_{\sigma\left(  3\right)
}\right)  \cdots f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  }%
_{=f_{1}\left(  a_{\sigma\left(  1\right)  }\right)  f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  }\\
&  =\sum_{m=1}^{n}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}f_{1}\left(  a_{\sigma\left(  1\right)  }\right)  f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \sum_{\sigma\in S_{n}}f_{1}\left(  a_{\sigma\left(  1\right)  }\right)
f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(
a_{\sigma\left(  n\right)  }\right) \\
&  =\underbrace{\sum_{m\in\left\{  1,2,\ldots,n\right\}  }}_{=\sum_{m=1}^{n}%
}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)  =m}}f_{1}\left(
a_{\sigma\left(  1\right)  }\right)  f_{2}\left(  a_{\sigma\left(  2\right)
}\right)  \cdots f_{n}\left(  a_{\sigma\left(  n\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }\sigma\in S_{n}\text{
satisfies }\sigma\left(  1\right)  \in\left\{  1,2,\ldots,n\right\}  \right)
\\
&  =\sum_{m=1}^{n}\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)
=m}}f_{1}\left(  a_{\sigma\left(  1\right)  }\right)  f_{2}\left(
a_{\sigma\left(  2\right)  }\right)  \cdots f_{n}\left(  a_{\sigma\left(
n\right)  }\right)  ,
\end{align*}
this yields%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}a_{2}\cdots
a_{n}\right)  =\sum_{\sigma\in S_{n}}f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  .
\]


Now, forget that we fixed $H$, $A$, $\left(  f_{1},f_{2},\ldots,f_{n}\right)
$ and $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $. We thus have shown that if
$H$ is a $k$-bialgebra, if $A$ is a $k$-algebra, if $f_{1}$, $f_{2}$, $\ldots
$, $f_{n}$ are $n$ elements of $\mathcal{L}\left(  H,A\right)  $ such that
$f_{1}$, $f_{2}$, $\ldots$, $f_{n}$ are $\left(  \varepsilon_{H}%
,\varepsilon_{H}\right)  $-derivations, then every $n$ elements $a_{1}$,
$a_{2}$, $\ldots$, $a_{n}$ of $\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  $ satisfy%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{n}\right)  \left(  a_{1}a_{2}\cdots
a_{n}\right)  =\sum_{\sigma\in S_{n}}f_{1}\left(  a_{\sigma\left(  1\right)
}\right)  f_{2}\left(  a_{\sigma\left(  2\right)  }\right)  \cdots
f_{n}\left(  a_{\sigma\left(  n\right)  }\right)  .
\]
In other words, we have shown that Theorem~\ref{Theorem1501} \textbf{(b)}
holds for $r=n$. This completes the induction step. The induction proof of
Theorem~\ref{Theorem1501} \textbf{(b)} is thus complete.
\end{proof}

We can obtain some corollaries of Theorem~\ref{Theorem1501}. By setting
$f_{1}$, $f_{2}$, $\ldots$, $f_{n}$ all equal to each other, we can conclude
that the following holds:

\begin{corollary}
\label{Corollary1511} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a $k$-algebra. Let $r\in\mathbb{N}$. Let $f\in\mathcal{L}\left(
H,A\right)  $ be such that $f$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation.

\textbf{(a)} Every $s\in\mathbb{N}$ and every $s$ elements $a_{1}$, $a_{2}$,
$\ldots$, $a_{s}$ of $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $
such that $s>r$ satisfy%
\[
f^{\ast r}\left(  a_{1}a_{2}\cdots a_{s}\right)  =0.
\]


\textbf{(b)} Every $r$ elements $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ satisfy%
\[
f^{\ast r}\left(  a_{1}a_{2}\cdots a_{r}\right)  =\sum_{\sigma\in S_{r}%
}f\left(  a_{\sigma\left(  1\right)  }\right)  f\left(  a_{\sigma\left(
2\right)  }\right)  \cdots f\left(  a_{\sigma\left(  r\right)  }\right)
\]
(where $S_{r}$ denotes the $r$-th symmetric group).

\textbf{(c)} Let $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ be $r$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. If the elements
$f\left(  a_{1}\right)  $, $f\left(  a_{2}\right)  $, $\ldots$, $f\left(
a_{r}\right)  $ of $A$ commute pairwise, then%
\[
f^{\ast r}\left(  a_{1}a_{2}\cdots a_{r}\right)  =r!\cdot f\left(
a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{r}\right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1511}.]\textbf{(a)} Let $s\in\mathbb{N}$ be
such that $s>r$. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ be $s$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Then,%
\[
\underbrace{f^{\ast r}}_{=\underbrace{f\ast f\ast\cdots\ast f}_{r\text{
times}}}\left(  a_{1}a_{2}\cdots a_{s}\right)  =\left(  \underbrace{f\ast
f\ast\cdots\ast f}_{r\text{ times}}\right)  \left(  a_{1}a_{2}\cdots
a_{s}\right)  =0
\]
(by Theorem~\ref{Theorem1501} \textbf{(a)}, applied to $\left(  f_{1}%
,f_{2},\ldots,f_{r}\right)  =\left(  \underbrace{f,f,\ldots,f}_{r\text{
times}}\right)  $). This proves Corollary~\ref{Corollary1511} \textbf{(a)}.

\textbf{(b)} Let $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ be $r$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Then,%
\begin{align*}
\underbrace{f^{\ast r}}_{=\underbrace{f\ast f\ast\cdots\ast f}_{r\text{
times}}}\left(  a_{1}a_{2}\cdots a_{r}\right)   &  =\left(  \underbrace{f\ast
f\ast\cdots\ast f}_{r\text{ times}}\right)  \left(  a_{1}a_{2}\cdots
a_{r}\right) \\
&  =\sum_{\sigma\in S_{r}}f\left(  a_{\sigma\left(  1\right)  }\right)
f\left(  a_{\sigma\left(  2\right)  }\right)  \cdots f\left(  a_{\sigma\left(
r\right)  }\right)
\end{align*}
(by Theorem~\ref{Theorem1501} \textbf{(b)}, applied to $\left(  f_{1}%
,f_{2},\ldots,f_{r}\right)  =\left(  \underbrace{f,f,\ldots,f}_{r\text{
times}}\right)  $). This proves Corollary~\ref{Corollary1511} \textbf{(b)}.

\textbf{(c)} Let $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ be $r$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Assume that the
elements $f\left(  a_{1}\right)  $, $f\left(  a_{2}\right)  $, $\ldots$,
$f\left(  a_{r}\right)  $ of $A$ commute pairwise. Thus, for every $\sigma\in
S_{n}$, we have%
\begin{equation}
f\left(  a_{\sigma\left(  1\right)  }\right)  f\left(  a_{\sigma\left(
2\right)  }\right)  \cdots f\left(  a_{\sigma\left(  r\right)  }\right)
=f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{r}\right)
\label{1511.pf.c.1}%
\end{equation}
(by Lemma~\ref{Lemma1412}, applied to $n=r$, $B=A$ and $b_{i}=f\left(
a_{i}\right)  $). Now, Corollary~\ref{Corollary1511} \textbf{(b)} yields%
\begin{align*}
f^{\ast r}\left(  a_{1}a_{2}\cdots a_{r}\right)   &  =\sum_{\sigma\in S_{r}%
}\underbrace{f\left(  a_{\sigma\left(  1\right)  }\right)  f\left(
a_{\sigma\left(  2\right)  }\right)  \cdots f\left(  a_{\sigma\left(
r\right)  }\right)  }_{\substack{=f\left(  a_{1}\right)  f\left(
a_{2}\right)  \cdots f\left(  a_{r}\right)  \\\text{(by (\ref{1511.pf.c.1}))}%
}}\\
&  =\sum_{\sigma\in S_{r}}f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots
f\left(  a_{r}\right)  =\underbrace{\left\vert S_{r}\right\vert }_{=r!}\cdot
f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{r}\right) \\
&  =r!\cdot f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{r}\right)  .
\end{align*}
This proves Corollary~\ref{Corollary1511} \textbf{(c)}.
\end{proof}

We can apply Corollary~\ref{Corollary1511} to the Eulerian idempotent:

\begin{corollary}
\label{Corollary1513} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered commutative bialgebra over $k$. Consider the convolution
algebra $\mathcal{L}\left(  H,H\right)  $.

\textbf{(a)} Every $s\in\mathbb{N}$ and every $s$ elements $a_{1}$, $a_{2}$,
$\ldots$, $a_{s}$ of $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $
such that $s>r$ satisfy%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  ^{\ast r}\left(
a_{1}a_{2}\cdots a_{s}\right)  =0.
\]


\textbf{(b)} Every $r$ elements $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ satisfy%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  ^{\ast r}\left(
a_{1}a_{2}\cdots a_{r}\right)  =r!\cdot\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  a_{1}\right)  \cdot\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  a_{2}\right)
\cdot\cdots\cdot\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(
a_{r}\right)  .
\]

\end{corollary}

Notice that Corollary~\ref{Corollary1513} generalizes an observation made in
Example 3.18 of \cite{8}.

\begin{proof}
[Proof of Corollary~\ref{Corollary1513}.]Let $f=\operatorname*{Log}%
\operatorname*{id}$. Then, $f$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation (this was shown in the proof of
Theorem~\ref{Theorem120}). Hence, Corollary~\ref{Corollary1511} \textbf{(a)}
(applied to $A=H$) yields that every $s\in\mathbb{N}$ and every $s$ elements
$a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ of $\operatorname*{Ker}\left(
\varepsilon_{H}\right)  $ such that $s>r$ satisfy $f^{\ast r}\left(
a_{1}a_{2}\cdots a_{s}\right)  =0$. Since $f=\operatorname*{Log}%
\operatorname*{id}$, this rewrites as follows: Every $s\in\mathbb{N}$ and
every $s$ elements $a_{1}$, $a_{2}$, $\ldots$, $a_{s}$ of $\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $ such that $s>r$ satisfy%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  ^{\ast r}\left(
a_{1}a_{2}\cdots a_{s}\right)  =0.
\]
This proves Corollary~\ref{Corollary1513} \textbf{(a)}.

\textbf{(b)} Let $a_{1}$, $a_{2}$, $\ldots$, $a_{r}$ be $r$ elements of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Then, the elements
$f\left(  a_{1}\right)  $, $f\left(  a_{2}\right)  $, $\ldots$, $f\left(
a_{r}\right)  $ of $H$ commute pairwise (since $H$ is commutative). Therefore,
Corollary~\ref{Corollary1511} \textbf{(b)} (applied to $A=H$) yields that
$f^{\ast r}\left(  a_{1}a_{2}\cdots a_{r}\right)  =r!\cdot f\left(
a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{r}\right)  $. Since
$f=\operatorname*{Log}\operatorname*{id}$, this rewrites as%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  ^{\ast r}\left(
a_{1}a_{2}\cdots a_{r}\right)  =r!\cdot\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  a_{1}\right)  \cdot\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  a_{2}\right)
\cdot\cdots\cdot\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(
a_{r}\right)  .
\]
This proves Corollary~\ref{Corollary1513} \textbf{(b)}.
\end{proof}

We record a further particular case of Theorem~\ref{Theorem1501}:

\begin{corollary}
\label{Corollary1521} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a $k$-algebra. Let $r\in\mathbb{N}$. Let $f_{1}$, $f_{2}$, $\ldots$,
$f_{r}$ be $r$ elements of $\mathcal{L}\left(  H,A\right)  $ such that $f_{1}%
$, $f_{2}$, $\ldots$, $f_{r}$ are $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivations. Let $a\in\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  $.

\textbf{(a)} Every $s\in\mathbb{N}$ such that $s>r$ satisfies%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a^{s}\right)  =0.
\]


\textbf{(b)} We have%
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  a^{r}\right)
=r!\cdot f_{1}\left(  a\right)  f_{2}\left(  a\right)  \cdots f_{r}\left(
a\right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1521}.]\textbf{(a)} Let $s\in\mathbb{N}$ be
such that $s>r$. We have
\[
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{a^{s}%
}_{=\underbrace{aa\cdots a}_{s\text{ times}}}\right)  =\left(  f_{1}\ast
f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{aa\cdots a}_{s\text{
times}}\right)  =0
\]
(by Theorem~\ref{Theorem1501} \textbf{(a)}, applied to $\left(  a_{1}%
,a_{2},\ldots,a_{s}\right)  =\left(  \underbrace{a,a,\ldots,a}_{s\text{
times}}\right)  $). This proves Corollary~\ref{Corollary1521} \textbf{(a)}.

\textbf{(b)} We have
\begin{align*}
\left(  f_{1}\ast f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{a^{r}%
}_{=\underbrace{aa\cdots a}_{r\text{ times}}}\right)   &  =\left(  f_{1}\ast
f_{2}\ast\cdots\ast f_{r}\right)  \left(  \underbrace{aa\cdots a}_{r\text{
times}}\right) \\
&  =\sum_{\sigma\in S_{r}}f_{1}\left(  a\right)  f_{2}\left(  a\right)  \cdots
f_{r}\left(  a\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Theorem~\ref{Theorem1501} \textbf{(b)}, applied to}\\
\left(  a_{1},a_{2},\ldots,a_{r}\right)  =\left(  \underbrace{a,a,\ldots
,a}_{r\text{ times}}\right)
\end{array}
\right) \\
&  =\underbrace{\left\vert S_{r}\right\vert }_{=r!}\cdot f_{1}\left(
a\right)  f_{2}\left(  a\right)  \cdots f_{r}\left(  a\right)  =r!\cdot
f_{1}\left(  a\right)  f_{2}\left(  a\right)  \cdots f_{r}\left(  a\right)  .
\end{align*}
This proves Corollary~\ref{Corollary1521} \textbf{(b)}.
\end{proof}

A further particular case involves only one map and only one element of
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $:

\begin{corollary}
\label{Corollary1522} Let $k$ be a field. Let $H$ be a $k$-bialgebra. Let $A$
be a $k$-algebra. Let $r\in\mathbb{N}$. Let $f\in\mathcal{L}\left(
H,A\right)  $ be such that $f$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation. Let $a\in\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  $.

\textbf{(a)} Every $s\in\mathbb{N}$ such that $s>r$ satisfies%
\[
f^{\ast r}\left(  a^{s}\right)  =0.
\]


\textbf{(b)} We have%
\[
f^{\ast r}\left(  a^{r}\right)  =r!\cdot\left(  f\left(  a\right)  \right)
^{r}.
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1522}.]\textbf{(b)} Let $s\in\mathbb{N}$ be
such that $s>r$. We have%
\[
\underbrace{f^{\ast r}}_{=\underbrace{f\ast f\ast\cdots\ast f}_{r\text{
times}}}\left(  a^{s}\right)  =\left(  \underbrace{f\ast f\ast\cdots\ast
f}_{r\text{ times}}\right)  \left(  a^{s}\right)  =0
\]
(by Corollary~\ref{Corollary1521} \textbf{(a)}, applied to $\left(
f_{1},f_{2},\ldots,f_{r}\right)  =\left(  \underbrace{f,f,\ldots,f}_{r\text{
times}}\right)  $). This proves Corollary~\ref{Corollary1522} \textbf{(a)}.

\textbf{(b)} We have
\begin{align*}
\underbrace{f^{\ast r}}_{=\underbrace{f\ast f\ast\cdots\ast f}_{r\text{
times}}}\left(  a^{r}\right)   &  =\left(  \underbrace{f\ast f\ast\cdots\ast
f}_{r\text{ times}}\right)  \left(  a^{r}\right)  =r!\cdot
\underbrace{\underbrace{f\left(  a\right)  f\left(  a\right)  \cdots f\left(
a\right)  }_{r\text{ times}}}_{=\left(  f\left(  a\right)  \right)  ^{r}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Corollary~\ref{Corollary1521} \textbf{(b)}, applied to}\\
\left(  f_{1},f_{2},\ldots,f_{r}\right)  =\left(  \underbrace{f,f,\ldots
,f}_{r\text{ times}}\right)
\end{array}
\right) \\
&  =r!\cdot\left(  f\left(  a\right)  \right)  ^{r}.
\end{align*}
This proves Corollary~\ref{Corollary1522} \textbf{(b)}.
\end{proof}

\section{\label{S60}An invertibility criterion for coalgebra homomorphisms}

We next prove a fact which comes handy when one desires to show that certain
coalgebra maps are invertible:

\begin{theorem}
\label{Theorem1601} Let $k$ be a field. Let $C$ be a connected filtered
$k$-coalgebra. As we know, $C$ thus becomes a unital coalgebra. Let
$f:C\rightarrow C$ be a $k$-coalgebra homomorphism satisfying $f\left(
1_{C}\right)  =1_{C}$. Assume that%
\begin{equation}
f\left(  x\right)  =x\ \ \ \ \ \ \ \ \ \ \text{for every primitive element
}x\text{ of }C. \label{1601.prim.act}%
\end{equation}


\textbf{(a)} We have $\left(  \operatorname{id}_{C}-f\right)  ^{n}\left(
C_{\leq n}\right)  =0$ for every integer $n\geq1$.

\textbf{(b)} The map $f$ is a $k$-coalgebra isomorphism.
\end{theorem}

Note that $f$ is not required to satisfy $f\left(  C_{\leq n}\right)
\subseteq C_{\leq n}$ in Theorem~\ref{Theorem1601}.

Before we prove Theorem~\ref{Theorem1601}, we are going to show some lemmas.
First, here is a very basic invertibility criterion for linear maps:

\begin{lemma}
\label{Lemma1602} Let $k$ be a field. Let $V$ be a $k$-vector space. Let
$f:V\rightarrow V$ be a $k$-linear map. Let $\left(  V_{1},V_{2},V_{3}%
,\ldots\right)  $ be a sequence of $k$-vector subspaces of $V$ such that
$V=\bigcup_{n\geq1}V_{n}$. Assume that $\left(  \operatorname{id}%
_{V}-f\right)  ^{n}\left(  V_{n}\right)  =0$ for every integer $n\geq1$. Then,
the map $f$ is invertible.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1602}.]Clearly, the elements $f$ and
$\operatorname{id}_{V}$ of $\operatorname*{End}V$ commute (since
$f\circ\operatorname{id}_{V}=f=\operatorname{id}_{V}\circ f$). Let
$\mathfrak{H}$ be the $k$-subalgebra of $\operatorname*{End}V$ generated by
the elements $f$ and $\operatorname{id}_{V}$.\ \ \ \ \footnote{Of course, this
$\mathfrak{H}$ is also the $k$-subalgebra of $\operatorname*{End}V$ generated
by the element $f$ alone (because the element $\operatorname{id}_{V}$ is the
unity of the $k$-algebra $\operatorname*{End}V$, and thus lies in any
$k$-subalgebra of $\operatorname*{End}V$). But we will not need this.} Then,
the $k$-algebra $\mathfrak{H}$ is commutative (by Corollary~\ref{Corollary47b}%
, applied to $A=\operatorname*{End}V$ and $g=\operatorname{id}_{V}$).
Therefore, we can apply the binomial theorem in this $k$-algebra
$\mathfrak{H}$. Since $f$ and $\operatorname{id}_{V}$ both belong to
$\mathfrak{H}$ (because $\mathfrak{H}$ is generated by $f$ and
$\operatorname{id}_{V}$), the binomial theorem thus yields%
\begin{align}
\left(  \operatorname{id}_{V}-f\right)  ^{n}  &  =\sum_{i=0}^{n}\left(
-1\right)  ^{n-i}\dbinom{n}{i}\underbrace{\operatorname{id}_{V}^{i}%
}_{=\operatorname{id}_{V}}\circ f^{n-i}=\sum_{i=0}^{n}\left(  -1\right)
^{n-i}\dbinom{n}{i}\underbrace{\operatorname{id}_{V}\circ f^{n-i}}_{=f^{n-i}%
}\nonumber\\
&  =\sum_{i=0}^{n}\left(  -1\right)  ^{n-i}\dbinom{n}{i}f^{n-i}%
=\underbrace{\left(  -1\right)  ^{n-n}}_{=\left(  -1\right)  ^{0}%
=1}\underbrace{\dbinom{n}{n}}_{=1}\underbrace{f^{n-n}}_{=f^{0}%
=\operatorname{id}_{V}}+\sum_{i=0}^{n-1}\left(  -1\right)  ^{n-i}\dbinom{n}%
{i}\underbrace{f^{n-i}}_{=f^{\left(  n-i-1\right)  +1}}\nonumber\\
&  =\operatorname{id}_{V}+\sum_{i=0}^{n-1}\left(  -1\right)  ^{n-i}\dbinom
{n}{i}\underbrace{f^{\left(  n-i-1\right)  +1}}_{\substack{=f\circ
f^{n-i-1}\\\text{(since }n-i-1\geq0\text{ (because }i\leq n-1\text{))}%
}}\label{1602.pf.1}\\
&  =\operatorname{id}_{V}+\underbrace{\sum_{i=0}^{n-1}\left(  -1\right)
^{n-i}\dbinom{n}{i}f\circ f^{n-i-1}}_{\substack{=f\circ\left(  \sum
_{i=0}^{n-1}\left(  -1\right)  ^{n-i}\dbinom{n}{i}f^{n-i-1}\right)
\\\text{(since composition of }k\text{-linear maps is }k\text{-bilinear)}%
}}\nonumber\\
&  =\operatorname{id}_{V}+f\circ\left(  \sum_{i=0}^{n-1}\left(  -1\right)
^{n-i}\dbinom{n}{i}f^{n-i-1}\right)  \label{1602.pf.2}%
\end{align}
for each $n\in\mathbb{N}$.

Let us now show that the map $f$ is surjective.

Indeed, let $v\in V$ be arbitrary. Then, $v\in V=\bigcup_{n\geq1}V_{n}$.
Hence, there exists an integer $n\geq1$ such that $v\in V_{n}$. Consider this
$n$.

Since $v\in V_{n}$, we have $\left(  \operatorname{id}_{V}-f\right)
^{n}\left(  \underbrace{v}_{\in V_{n}}\right)  \in\left(  \operatorname{id}%
_{V}-f\right)  ^{n}\left(  V_{n}\right)  =0$, and thus $\left(
\operatorname{id}_{V}-f\right)  ^{n}\left(  v\right)  =0$.

Now, recall that $\left(  \operatorname{id}_{V}-f\right)  ^{n}\left(
v\right)  =0$, so that%
\begin{align*}
0  &  =\underbrace{\left(  \operatorname{id}_{V}-f\right)  ^{n}}%
_{\substack{=\operatorname{id}_{V}+f\circ\left(  \sum_{i=0}^{n-1}\left(
-1\right)  ^{n-i}\dbinom{n}{i}f^{n-i-1}\right)  \\\text{(by (\ref{1602.pf.2}%
))}}}\left(  v\right)  =\left(  \operatorname{id}_{V}+f\circ\left(  \sum
_{i=0}^{n-1}\left(  -1\right)  ^{n-i}\dbinom{n}{i}f^{n-i-1}\right)  \right)
\left(  v\right) \\
&  =\underbrace{\operatorname{id}_{V}\left(  v\right)  }_{=v}+\left(
f\circ\left(  \sum_{i=0}^{n-1}\left(  -1\right)  ^{n-i}\dbinom{n}{i}%
f^{n-i-1}\right)  \right)  \left(  v\right) \\
&  =v+\left(  f\circ\left(  \sum_{i=0}^{n-1}\left(  -1\right)  ^{n-i}%
\dbinom{n}{i}f^{n-i-1}\right)  \right)  \left(  v\right)  .
\end{align*}
Subtracting $v$ from this equality, we obtain%
\begin{align*}
-v  &  =\left(  f\circ\left(  \sum_{i=0}^{n-1}\left(  -1\right)  ^{n-i}%
\dbinom{n}{i}f^{n-i-1}\right)  \right)  \left(  v\right) \\
&  =f\left(  \underbrace{\left(  \sum_{i=0}^{n-1}\left(  -1\right)
^{n-i}\dbinom{n}{i}f^{n-i-1}\right)  \left(  v\right)  }_{\in V}\right)  \in
f\left(  V\right)  .
\end{align*}
Thus, $v=-\underbrace{\left(  -v\right)  }_{\in f\left(  V\right)  }%
\in-f\left(  V\right)  \subseteq f\left(  V\right)  $ (since $f\left(
V\right)  $ is a $k$-vector space).

Now, forget that we fixed $v$. We thus have proven that every $v\in V$
satisfies $v\in f\left(  V\right)  $. In other words, $V\subseteq f\left(
V\right)  $. In other words, the map $f$ is surjective.

Now, let $v\in\operatorname*{Ker}f$ be arbitrary. Then, $f\left(  v\right)
=0$. But on the other hand, $v\in\operatorname*{Ker}f\subseteq V=\bigcup
_{n\geq1}V_{n}$. Hence, there exists an integer $n\geq1$ such that $v\in
V_{n}$. Consider this $n$.

Since $v\in V_{n}$, we have $\left(  \operatorname{id}_{V}-f\right)
^{n}\left(  \underbrace{v}_{\in V_{n}}\right)  \in\left(  \operatorname{id}%
_{V}-f\right)  ^{n}\left(  V_{n}\right)  =0$, and thus $\left(
\operatorname{id}_{V}-f\right)  ^{n}\left(  v\right)  =0$. Thus,
\begin{align*}
0  &  =\underbrace{\left(  \operatorname{id}_{V}-f\right)  ^{n}}%
_{\substack{=\operatorname{id}_{V}+\sum_{i=0}^{n-1}\left(  -1\right)
^{n-i}\dbinom{n}{i}f^{\left(  n-i-1\right)  +1}\\\text{(by (\ref{1602.pf.1}%
))}}}\left(  v\right) \\
&  =\left(  \operatorname{id}_{V}+\sum_{i=0}^{n-1}\left(  -1\right)
^{n-i}\dbinom{n}{i}f^{\left(  n-i-1\right)  +1}\right)  \left(  v\right) \\
&  =\underbrace{\operatorname{id}_{V}\left(  v\right)  }_{=v}+\sum_{i=0}%
^{n-1}\left(  -1\right)  ^{n-i}\dbinom{n}{i}\underbrace{f^{\left(
n-i-1\right)  +1}}_{\substack{=f^{n-i-1}\circ f\\\text{(since }n-i-1\geq
0\text{ (because }i\leq n-1\text{))}}}\left(  v\right) \\
&  =v+\sum_{i=0}^{n-1}\left(  -1\right)  ^{n-i}\dbinom{n}{i}%
\underbrace{\left(  f^{n-i-1}\circ f\right)  \left(  v\right)  }%
_{=f^{n-i-1}\left(  f\left(  v\right)  \right)  }\\
&  =v+\sum_{i=0}^{n-1}\left(  -1\right)  ^{n-i}\dbinom{n}{i}f^{n-i-1}\left(
\underbrace{f\left(  v\right)  }_{=0}\right)  =v+\sum_{i=0}^{n-1}\left(
-1\right)  ^{n-i}\dbinom{n}{i}\underbrace{f^{n-i-1}\left(  0\right)
}_{\substack{=0\\\text{(since }f^{n-i-1}\text{ is }k\text{-linear)}}}\\
&  =v+\underbrace{\sum_{i=0}^{n-1}\left(  -1\right)  ^{n-i}\dbinom{n}{i}%
0}_{=0}=v,
\end{align*}
so that $v=0$.

Now, forget that we fixed $v$. We thus have proven that every $v\in
\operatorname*{Ker}f$ satisfies $v=0$. In other words, $\operatorname*{Ker}%
f=0$. Thus, the $k$-linear map $f$ is injective. Combined with the fact that
$f$ is surjective, this yields that $f$ is bijective. Hence, $f$ is
invertible. This proves Lemma~\ref{Lemma1602}.
\end{proof}

Let us show a simple property of connected filtered $k$-coalgebras:

\begin{proposition}
\label{Proposition1603} Let $k$ be a field. Let $C$ be a connected filtered
$k$-coalgebra. As we know, $C$ thus becomes a unital coalgebra. Let $x\in
C_{\leq1}$. Then, $x-\varepsilon\left(  x\right)  \cdot1_{C}$ is a primitive
element of $C$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1603}.]By the axioms of a unital
coalgebra, we have $\varepsilon\left(  1_{C}\right)  =1$ (since $C$ is a
unital coalgebra).

Let $y=x-\varepsilon\left(  x\right)  \cdot1_{C}$. We have%
\begin{align*}
\varepsilon\left(  \underbrace{y}_{=x-\varepsilon\left(  x\right)  \cdot1_{C}%
}\right)   &  =\varepsilon\left(  x-\varepsilon\left(  x\right)  \cdot
1_{C}\right)  =\varepsilon\left(  x\right)  -\varepsilon\left(  x\right)
\cdot\underbrace{\varepsilon\left(  1_{C}\right)  }_{=1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\varepsilon\text{ is
}k\text{-linear}\right) \\
&  =\varepsilon\left(  x\right)  -\varepsilon\left(  x\right)  =0.
\end{align*}
Thus, $y\in\operatorname*{Ker}\varepsilon$. Combined with $y\in C_{\leq1}$,
this yields $y\in C_{\leq1}\cap\operatorname*{Ker}\varepsilon$. Thus,
Proposition~\ref{Proposition276} (applied to $1$ and $y$ instead of $\ell$ and
$x$) yields%
\[
\Delta_{C}\left(  y\right)  \in y\otimes1_{C}+1_{C}\otimes y+\underbrace{\sum
_{u=1}^{1-1}C_{\leq u}\otimes C_{\leq1-u}}_{=\left(  \text{empty sum}\right)
=0}=y\otimes1_{C}+1_{C}\otimes y.
\]
In other words, $\Delta_{C}\left(  y\right)  =y\otimes1_{C}+1_{C}\otimes y$.
In other words, $y$ is primitive. In other words, $x-\varepsilon\left(
x\right)  \cdot1_{C}$ is primitive (since $y=x-\varepsilon\left(  x\right)
\cdot1_{C}$). This proves Proposition~\ref{Proposition1603}.
\end{proof}

Next, we state two lemmas from linear algebra:

\begin{lemma}
\label{Lemma1604} Let $k$ be a field. Let $V$ and $V^{\prime}$ be two
$k$-vector spaces. Let $\beta\in\operatorname*{End}V$ and $\beta^{\prime}%
\in\operatorname*{End}\left(  V^{\prime}\right)  $. Then,
\begin{equation}
\beta^{i}\otimes\left(  \beta^{\prime}\right)  ^{i}=\left(  \beta\otimes
\beta^{\prime}\right)  ^{i} \label{1604}%
\end{equation}
for every $i\in\mathbb{N}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1604}.]We shall prove (\ref{1604}) by induction over
$i$:

\textit{Induction base:} If $i=0$, then%
\begin{align*}
\beta^{i}\otimes\left(  \beta^{\prime}\right)  ^{i}  &  =\underbrace{\beta
^{0}}_{=\operatorname{id}_{V}}\otimes\underbrace{\left(  \beta^{\prime
}\right)  ^{0}}_{=\operatorname{id}_{V^{\prime}}}=\operatorname{id}_{V}%
\otimes\operatorname{id}_{V^{\prime}}=\operatorname{id}_{V\otimes V^{\prime}%
}=\left(  \beta\otimes\beta^{\prime}\right)  ^{0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \beta\otimes\beta^{\prime
}\right)  ^{0}=\operatorname{id}_{V\otimes V^{\prime}}\right) \\
&  =\left(  \beta\otimes\beta^{\prime}\right)  ^{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }0=i\text{ (because }i=0\text{)}\right)  .
\end{align*}
Thus, (\ref{1604}) is proven for $i=0$. This completes the induction base.

\textit{Induction step:} Let $K\in\mathbb{N}$. Assume that (\ref{1604}) holds
for $i=K$. We need to show that (\ref{1604}) holds for $i=K+1$.

We know that (\ref{1604}) holds for $i=K$. In other words, $\beta^{K}%
\otimes\left(  \beta^{\prime}\right)  ^{K}=\left(  \beta\otimes\beta^{\prime
}\right)  ^{K}$. Now,%
\[
\underbrace{\beta^{K+1}}_{=\beta^{K}\circ\beta}\otimes\underbrace{\left(
\beta^{\prime}\right)  ^{K+1}}_{=\left(  \beta^{\prime}\right)  ^{K}\circ
\beta^{\prime}}=\left(  \beta^{K}\circ\beta\right)  \otimes\left(  \left(
\beta^{\prime}\right)  ^{K}\circ\beta^{\prime}\right)  =\left(  \beta
^{K}\otimes\left(  \beta^{\prime}\right)  ^{K}\right)  \circ\left(
\beta\otimes\beta^{\prime}\right)
\]
(by (\ref{(X)(X)}), applied to $V$, $V$, $V$, $V^{\prime}$, $V^{\prime}$,
$V^{\prime}$, $\beta$, $\beta^{K}$, $\beta^{\prime}$ and $\left(
\beta^{\prime}\right)  ^{K}$ instead of $U$, $V$, $W$, $U^{\prime}$,
$V^{\prime}$, $W^{\prime}$, $\alpha$, $\beta$, $\alpha^{\prime}$ and
$\beta^{\prime}$). Thus,%
\[
\beta^{K+1}\otimes\left(  \beta^{\prime}\right)  ^{K+1}=\underbrace{\left(
\beta^{K}\otimes\left(  \beta^{\prime}\right)  ^{K}\right)  }_{=\left(
\beta\otimes\beta^{\prime}\right)  ^{K}}\circ\left(  \beta\otimes\beta
^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)  ^{K}\circ\left(
\beta\otimes\beta^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)
^{K+1}.
\]
In other words, (\ref{1604}) holds for $i=K+1$. This completes the induction
step. Thus, (\ref{1604}) is proven by induction. Hence, Lemma~\ref{Lemma1604}
is proven.
\end{proof}

\begin{lemma}
\label{Lemma1605} Let $k$ be a field. Let $V$ be a $k$-vector space. Let
$f\in\operatorname*{End}V$. Let $n\in\mathbb{N}$. Then,%
\[
\left(  \operatorname{id}_{V}\otimes\operatorname{id}_{V}-f\otimes f\right)
^{n}=\sum_{i=0}^{n}\dbinom{n}{i}\left(  \operatorname{id}_{V}-f\right)
^{n-i}\otimes\left(  f^{n-i}\circ\left(  \operatorname{id}_{V}-f\right)
^{i}\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1605}.]First of all, it is easy to see that
\begin{equation}
\left(  \operatorname{id}_{V}-f\right)  ^{i}\circ f^{n-i}=f^{n-i}\circ\left(
\operatorname{id}_{V}-f\right)  ^{i}\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  0,1,\ldots,n\right\}  . \label{1605.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{1605.pf.1}):} Let $i\in\left\{
0,1,\ldots,n\right\}  $. The elements $f$ and $\operatorname{id}_{V}$ of
$\operatorname*{End}V$ clearly commute (since $f\circ\operatorname{id}%
_{V}=f=\operatorname{id}_{V}\circ f$). Let $\mathfrak{G}$ be the
$k$-subalgebra of $\operatorname*{End}V$ generated by the elements $f$ and
$\operatorname{id}_{V}$. Then, the $k$-algebra $\mathfrak{G}$ is commutative
(by Corollary~\ref{Corollary47b}, applied to $A=\operatorname*{End}V$,
$g=\operatorname{id}_{V}$ and $\mathfrak{H}=\mathfrak{G}$). But the elements
$f$ and $\operatorname{id}_{V}$ belong to $\mathfrak{G}$ (because the
$k$-algebra $\mathfrak{G}$ is generated by $f$ and $\operatorname{id}_{V}$).
Thus, $f^{n-i}$ and $\left(  \operatorname{id}_{V}-f\right)  ^{i}$ belong to
$\mathfrak{G}$. As a consequence, $f^{n-i}\circ\left(  \operatorname{id}%
_{V}-f\right)  ^{i}=\left(  \operatorname{id}_{V}-f\right)  ^{i}\circ f^{n-i}$
in $\mathfrak{G}$ (since $\mathfrak{G}$ is commutative). This proves
(\ref{1605.pf.1}).}

Also,
\begin{align*}
\left(  \operatorname{id}_{V}-f\right)  \circ f  &
=\underbrace{\operatorname{id}_{V}\circ f}_{=f=f\circ\operatorname{id}_{V}%
}-f\circ f\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of
}k\text{-linear maps is }k\text{-bilinear}\right) \\
&  =f\circ\operatorname{id}_{V}-f\circ f\\
&  =f\circ\left(  \operatorname{id}_{V}-f\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since composition of }k\text{-linear maps is }k\text{-bilinear}\right)
.
\end{align*}


Now, (\ref{(X)(X)}) (applied to $V$, $V$, $V$, $V$, $V$, $V$,
$\operatorname{id}_{V}-f$, $\operatorname{id}_{V}$, $f$ and $\operatorname{id}%
_{V}-f$ instead of $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$,
$\alpha$, $\beta$, $\alpha^{\prime}$ and $\beta^{\prime}$) yields
\[
\left(  \operatorname{id}_{V}\circ\left(  \operatorname{id}_{V}-f\right)
\right)  \otimes\left(  \left(  \operatorname{id}_{V}-f\right)  \circ
f\right)  =\left(  \operatorname{id}_{V}\otimes\left(  \operatorname{id}%
_{V}-f\right)  \right)  \circ\left(  \left(  \operatorname{id}_{V}-f\right)
\otimes f\right)  ,
\]
so that%
\begin{align*}
&  \left(  \operatorname{id}_{V}\otimes\left(  \operatorname{id}_{V}-f\right)
\right)  \circ\left(  \left(  \operatorname{id}_{V}-f\right)  \otimes f\right)
\\
&  =\left(  \underbrace{\operatorname{id}_{V}\circ\left(  \operatorname{id}%
_{V}-f\right)  }_{=\operatorname{id}_{V}-f=\left(  \operatorname{id}%
_{V}-f\right)  \circ\operatorname{id}_{V}}\right)  \otimes\left(
\underbrace{\left(  \operatorname{id}_{V}-f\right)  \circ f}_{=f\circ\left(
\operatorname{id}_{V}-f\right)  }\right) \\
&  =\left(  \left(  \operatorname{id}_{V}-f\right)  \circ\operatorname{id}%
_{V}\right)  \otimes\left(  f\circ\left(  \operatorname{id}_{V}-f\right)
\right) \\
&  =\left(  \left(  \operatorname{id}_{V}-f\right)  \otimes f\right)
\circ\left(  \operatorname{id}_{V}\otimes\left(  \operatorname{id}%
_{V}-f\right)  \right)
\end{align*}
(by (\ref{(X)(X)}), applied to $V$, $V$, $V$, $V$, $V$, $V$,
$\operatorname{id}_{V}-f$, $\operatorname{id}_{V}$, $\operatorname{id}_{V}-f$
and $f$ instead of $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$,
$\alpha$, $\beta$, $\alpha^{\prime}$ and $\beta^{\prime}$). In other words,
the elements $\operatorname{id}_{V}\otimes\left(  \operatorname{id}%
_{V}-f\right)  $ and $\left(  \operatorname{id}_{V}-f\right)  \otimes f$ of
$\operatorname*{End}\left(  V\otimes V\right)  $ commute.

Now, let $\mathfrak{H}$ be the the $k$-subalgebra of $\operatorname*{End}%
\left(  V\otimes V\right)  $ generated by the elements $\operatorname{id}%
_{V}\otimes\left(  \operatorname{id}_{V}-f\right)  $ and $\left(
\operatorname{id}_{V}-f\right)  \otimes f$. Then, the $k$-algebra
$\mathfrak{H}$ is commutative (by Corollary~\ref{Corollary47b}, applied to
$\operatorname*{End}\left(  V\otimes V\right)  $, $\operatorname{id}%
_{V}\otimes\left(  \operatorname{id}_{V}-f\right)  $ and $\left(
\operatorname{id}_{V}-f\right)  \otimes f$ instead of $A$, $f$ and $g$).
Therefore, we can apply the binomial theorem in this $k$-algebra
$\mathfrak{H}$. Since $\operatorname{id}_{V}\otimes\left(  \operatorname{id}%
_{V}-f\right)  $ and $\left(  \operatorname{id}_{V}-f\right)  \otimes f$ both
belong to $\mathfrak{H}$ (because $\mathfrak{H}$ is generated by
$\operatorname{id}_{V}\otimes\left(  \operatorname{id}_{V}-f\right)  $ and
$\left(  \operatorname{id}_{V}-f\right)  \otimes f$), the binomial theorem
thus yields%
\begin{align*}
&  \left(  \operatorname{id}_{V}\otimes\left(  \operatorname{id}_{V}-f\right)
+\left(  \operatorname{id}_{V}-f\right)  \otimes f\right)  ^{n}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  \operatorname{id}%
_{V}\otimes\left(  \operatorname{id}_{V}-f\right)  \right)  ^{i}%
}_{\substack{=\operatorname{id}_{V}^{i}\otimes\left(  \operatorname{id}%
_{V}-f\right)  ^{i}\\\text{(because Lemma~\ref{Lemma1604}}\\\text{(applied to
}V\text{, }\operatorname{id}_{V}\text{ and }\operatorname{id}_{V}%
-f\\\text{instead of }V^{\prime}\text{, }\beta\text{ and }\beta^{\prime
}\text{) yields}\\\operatorname{id}_{V}^{i}\otimes\left(  \operatorname{id}%
_{V}-f\right)  ^{i}=\left(  \operatorname{id}_{V}\otimes\left(
\operatorname{id}_{V}-f\right)  \right)  ^{i}\text{)}}}\circ
\underbrace{\left(  \left(  \operatorname{id}_{V}-f\right)  \otimes f\right)
^{n-i}}_{\substack{=\left(  \operatorname{id}_{V}-f\right)  ^{n-i}\otimes
f^{n-i}\\\text{(because Lemma~\ref{Lemma1604}}\\\text{(applied to }V\text{,
}\operatorname{id}_{V}-f\text{, }f\text{ and }n-i\\\text{instead of }%
V^{\prime}\text{, }\beta\text{, }\beta^{\prime}\text{ and }i\text{)
yields}\\\left(  \operatorname{id}_{V}-f\right)  ^{n-i}\otimes f^{n-i}=\left(
\left(  \operatorname{id}_{V}-f\right)  \otimes f\right)  ^{n-i}\text{)}}}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  \operatorname{id}_{V}%
^{i}\otimes\left(  \operatorname{id}_{V}-f\right)  ^{i}\right)  \circ\left(
\left(  \operatorname{id}_{V}-f\right)  ^{n-i}\otimes f^{n-i}\right)
}_{\substack{=\left(  \operatorname{id}_{V}^{i}\circ\left(  \operatorname{id}%
_{V}-f\right)  ^{n-i}\right)  \otimes\left(  \left(  \operatorname{id}%
_{V}-f\right)  ^{i}\circ f^{n-i}\right)  \\\text{(since (\ref{(X)(X)})
(applied to }V\text{, }V\text{, }V\text{, }V\text{, }V\text{, }V\text{,}%
\\\left(  \operatorname{id}_{V}-f\right)  ^{n-i}\text{, }\operatorname{id}%
_{V}^{i}\text{, }f^{n-i}\text{ and }\left(  \operatorname{id}_{V}-f\right)
^{i}\text{ instead of}\\U\text{, }V\text{, }W\text{, }U^{\prime}\text{,
}V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }%
\alpha^{\prime}\text{ and }\beta^{\prime}\text{) yields}\\\left(
\operatorname{id}_{V}^{i}\circ\left(  \operatorname{id}_{V}-f\right)
^{n-i}\right)  \otimes\left(  \left(  \operatorname{id}_{V}-f\right)
^{i}\circ f^{n-i}\right)  \\=\left(  \operatorname{id}_{V}^{i}\otimes\left(
\operatorname{id}_{V}-f\right)  ^{i}\right)  \circ\left(  \left(
\operatorname{id}_{V}-f\right)  ^{n-i}\otimes f^{n-i}\right)  \text{)}}}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\left(  \underbrace{\operatorname{id}_{V}^{i}%
}_{=\operatorname{id}_{V}}\circ\left(  \operatorname{id}_{V}-f\right)
^{n-i}\right)  \otimes\left(  \underbrace{\left(  \operatorname{id}%
_{V}-f\right)  ^{i}\circ f^{n-i}}_{\substack{=f^{n-i}\circ\left(
\operatorname{id}_{V}-f\right)  ^{i}\\\text{(by (\ref{1605.pf.1}))}}}\right)
\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  \operatorname{id}_{V}%
\circ\left(  \operatorname{id}_{V}-f\right)  ^{n-i}\right)  }_{=\left(
\operatorname{id}_{V}-f\right)  ^{n-i}}\otimes\left(  f^{n-i}\circ\left(
\operatorname{id}_{V}-f\right)  ^{i}\right) \\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\left(  \operatorname{id}_{V}-f\right)
^{n-i}\otimes\left(  f^{n-i}\circ\left(  \operatorname{id}_{V}-f\right)
^{i}\right)  .
\end{align*}
This proves Lemma~\ref{Lemma1605}.
\end{proof}

Here comes another little lemma about coalgebras:

\begin{lemma}
\label{Lemma1606} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let
$f:C\rightarrow C$ and $g:C\rightarrow C$ be two $k$-coalgebra homomorphisms.
Then,%
\begin{equation}
\left(  f\otimes f-g\otimes g\right)  ^{n}\circ\Delta_{C}=\Delta_{C}%
\circ\left(  f-g\right)  ^{n} \label{1606}%
\end{equation}
for every $n\in\mathbb{N}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1606}.]We will prove Lemma~\ref{Lemma1606} by
induction over $n$:

\textit{Induction base:} If $n=0$, then (\ref{1606})
holds.\footnote{\textit{Proof.} Assume that $n=0$. Then,
\begin{align*}
\left(  f\otimes f-g\otimes g\right)  ^{n}\circ\Delta_{C}  &
=\underbrace{\left(  f\otimes f-g\otimes g\right)  ^{0}}_{=\operatorname{id}%
_{C\otimes C}}\circ\Delta_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}n=0\right) \\
&  =\Delta_{C}.
\end{align*}
Compared with%
\begin{align*}
\Delta_{C}\circ\left(  f-g\right)  ^{n}  &  =\Delta_{C}\circ
\underbrace{\left(  f-g\right)  ^{0}}_{=\operatorname{id}_{C}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n=0\right) \\
&  =\Delta_{C},
\end{align*}
this yields $\left(  f\otimes f-g\otimes g\right)  ^{n}\circ\Delta_{C}%
=\Delta_{C}\circ\left(  f-g\right)  ^{n}$. Thus, (\ref{1606}) holds if $n=0$,
qed.} This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that (\ref{1606}) holds
for $n=N$. We now must show that (\ref{1606}) also holds for $n=N+1$.

We know that (\ref{1606}) holds for $n=N$. In other words, we have%
\[
\left(  f\otimes f-g\otimes g\right)  ^{N}\circ\Delta_{C}=\Delta_{C}%
\circ\left(  f-g\right)  ^{N}.
\]


Since $f$ is a $k$-coalgebra homomorphism, we have $\Delta_{C}\circ f=\left(
f\otimes f\right)  \circ\Delta_{C}$ and $\varepsilon_{C}\circ f=\varepsilon
_{C}$. Since $g$ is a $k$-coalgebra homomorphism, we have $\Delta_{C}\circ
g=\left(  g\otimes g\right)  \circ\Delta_{C}$ and $\varepsilon_{C}\circ
g=\varepsilon_{C}$.

Since composition of $k$-linear maps is $k$-bilinear, we have
\begin{align*}
\Delta_{C}\circ\left(  f-g\right)   &  =\underbrace{\Delta_{C}\circ
f}_{=\left(  f\otimes f\right)  \circ\Delta_{C}}-\underbrace{\Delta_{C}\circ
g}_{=\left(  g\otimes g\right)  \circ\Delta_{C}}=\left(  f\otimes f\right)
\circ\Delta_{C}-\left(  g\otimes g\right)  \circ\Delta_{C}\\
&  =\left(  f\otimes f-g\otimes g\right)  \circ\Delta_{C}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps is
}k\text{-bilinear}\right)  .
\end{align*}
Thus,%
\[
\left(  f\otimes f-g\otimes g\right)  \circ\Delta_{C}=\Delta_{C}\circ\left(
f-g\right)  .
\]
But now,%
\begin{align*}
&  \underbrace{\left(  f\otimes f-g\otimes g\right)  ^{N+1}}_{=\left(
f\otimes f-g\otimes g\right)  \circ\left(  f\otimes f-g\otimes g\right)  ^{N}%
}\circ\Delta_{C}\\
&  =\left(  f\otimes f-g\otimes g\right)  \circ\underbrace{\left(  f\otimes
f-g\otimes g\right)  ^{N}\circ\Delta_{C}}_{=\Delta_{C}\circ\left(  f-g\right)
^{N}}\\
&  =\underbrace{\left(  f\otimes f-g\otimes g\right)  \circ\Delta_{C}%
}_{=\Delta_{C}\circ\left(  f-g\right)  }\circ\left(  f-g\right)  ^{N}%
=\Delta_{C}\circ\underbrace{\left(  f-g\right)  \circ\left(  f-g\right)  ^{N}%
}_{=\left(  f-g\right)  ^{N+1}}=\Delta_{C}\circ\left(  f-g\right)  ^{N+1}.
\end{align*}
In other words, (\ref{1606}) holds for $n=N+1$. This completes the induction
step. The induction proof of (\ref{1606}) is thus complete. Hence,
Lemma~\ref{Lemma1606} is proven.
\end{proof}

We will now finally prove Theorem~\ref{Theorem1601}.

\begin{proof}
[Proof of Theorem~\ref{Theorem1601}.]\textbf{(a)} We need to prove that%
\begin{equation}
\left(  \operatorname{id}_{C}-f\right)  ^{n}\left(  C_{\leq n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every integer }n\geq1. \label{1601.a.pf.1}%
\end{equation}


\textit{Proof of (\ref{1601.a.pf.1}):} We will prove (\ref{1601.a.pf.1}) by
strong induction over $n$:

\textit{Induction step:}\footnote{A strong induction needs no induction base.}
Let $N$ be an integer $\geq1$. We assume that (\ref{1601.a.pf.1}) holds
whenever $n<N$. We now need to prove that (\ref{1601.a.pf.1}) holds for $n=N$.

We have assumed that (\ref{1601.a.pf.1}) holds whenever $n<N$. In other words,%
\begin{equation}
\left(  \operatorname{id}_{C}-f\right)  ^{n}\left(  C_{\leq n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every integer }n\geq1\text{ satisfying }n<N.
\label{1601.a.pf.2}%
\end{equation}


Now, let $x\in C_{\leq N}$ be arbitrary. Let $y=x-\varepsilon\left(  x\right)
\cdot1_{C}$. Then, it is easy to see that $y\in C_{\leq N}\cap
\operatorname*{Ker}\varepsilon$\ \ \ \ \footnote{\textit{Proof.} By the axioms
of a connected filtered $k$-coalgebra, we have $1_{C}\in C_{\leq0}$ (since $C$
is a connected filtered $k$-coalgebra). But $C_{\leq0}\subseteq C_{\leq
1}\subseteq C_{\leq2}\subseteq\cdots$ (since $C$ is a filtered $k$-coalgebra),
hence $C_{\leq0}\subseteq C_{\leq N}$. Thus, $1_{C}\in C_{\leq0}\subseteq
C_{\leq N}$. Now, $y=\underbrace{x}_{\in C_{\leq N}}-\varepsilon\left(
x\right)  \cdot\underbrace{1_{C}}_{\in C_{\leq N}}\in C_{\leq N}%
-\varepsilon\left(  x\right)  \cdot C_{\leq N}\subseteq C_{\leq N}$ (since
$C_{\leq N}$ is a $k$-vector space).
\par
By the axioms of a unital coalgebra, we have $\varepsilon\left(  1_{C}\right)
=1$ (since $C$ is a unital coalgebra).
\par
Also,%
\begin{align*}
\varepsilon\left(  \underbrace{y}_{=x-\varepsilon\left(  x\right)  \cdot1_{C}%
}\right)   &  =\varepsilon\left(  x-\varepsilon\left(  x\right)  \cdot
1_{C}\right)  =\varepsilon\left(  x\right)  -\varepsilon\left(  x\right)
\cdot\underbrace{\varepsilon\left(  1_{C}\right)  }_{=1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\varepsilon\text{ is
}k\text{-linear}\right) \\
&  =\varepsilon\left(  x\right)  -\varepsilon\left(  x\right)  =0.
\end{align*}
Thus, $y\in\operatorname*{Ker}\varepsilon$. Combined with $y\in C_{\leq N}$,
this yields $y\in C_{\leq N}\cap\operatorname*{Ker}\varepsilon$, qed.}. Hence,
Proposition~\ref{Proposition276} (applied to $N$ and $y$ instead of $\ell$ and
$x$) yields%
\begin{equation}
\Delta_{C}\left(  y\right)  \in y\otimes1_{C}+1_{C}\otimes y+\sum_{u=1}%
^{N-1}C_{\leq u}\otimes C_{\leq N-u}. \label{1601.a.pf.3}%
\end{equation}


But Lemma~\ref{Lemma1606} (applied to $\operatorname{id}_{C}$, $f$ and $N-1$
instead of $f$, $g$ and $n$) yields%
\begin{equation}
\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes f\right)
^{N-1}\circ\Delta_{C}=\Delta_{C}\circ\left(  \operatorname{id}_{C}-f\right)
^{N-1}. \label{1601.a.pf.4}%
\end{equation}


Now, define an element $z\in C$ by $z=\left(  \operatorname{id}_{C}-f\right)
^{N-1}\left(  y\right)  $. Then,%
\begin{align}
\Delta_{C}\left(  \underbrace{z}_{=\left(  \operatorname{id}_{C}-f\right)
^{N-1}\left(  y\right)  }\right)   &  =\Delta_{C}\left(  \left(
\operatorname{id}_{C}-f\right)  ^{N-1}\left(  y\right)  \right)  =\left(
\underbrace{\Delta_{C}\circ\left(  \operatorname{id}_{C}-f\right)  ^{N-1}%
}_{\substack{=\left(  \operatorname{id}_{C}\otimes\operatorname{id}%
_{C}-f\otimes f\right)  ^{N-1}\circ\Delta_{C}\\\text{(by (\ref{1601.a.pf.4}%
))}}}\right)  \left(  y\right) \nonumber\\
&  =\left(  \left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes
f\right)  ^{N-1}\circ\Delta_{C}\right)  \left(  y\right) \nonumber\\
&  =\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes
f\right)  ^{N-1}\left(  \underbrace{\Delta_{C}\left(  y\right)  }%
_{\substack{\in y\otimes1_{C}+1_{C}\otimes y+\sum_{u=1}^{N-1}C_{\leq u}\otimes
C_{\leq N-u}\\\text{(by (\ref{1601.a.pf.3}))}}}\right) \nonumber\\
&  \in\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes
f\right)  ^{N-1}\left(  y\otimes1_{C}+1_{C}\otimes y+\sum_{u=1}^{N-1}C_{\leq
u}\otimes C_{\leq N-u}\right) \nonumber\\
&  \subseteq\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes
f\right)  ^{N-1}\left(  y\otimes1_{C}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname{id}_{C}\otimes\operatorname{id}%
_{C}-f\otimes f\right)  ^{N-1}\left(  1_{C}\otimes y\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname{id}_{C}\otimes\operatorname{id}%
_{C}-f\otimes f\right)  ^{N-1}\left(  \sum_{u=1}^{N-1}C_{\leq u}\otimes
C_{\leq N-u}\right)  \label{1601.a.pf.5}%
\end{align}
(since the map $\left(  \operatorname{id}_{C}\otimes\operatorname{id}%
_{C}-f\otimes f\right)  ^{N-1}$ is $k$-linear).

Recall that $N\geq1$, hence $N-1\geq0$. Thus, $N-1\in\mathbb{N}$. But
Lemma~\ref{Lemma1605} (applied to $V=C$ and $n=N-1$) yields%
\begin{align}
&  \left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes
f\right)  ^{N-1}\nonumber\\
&  =\sum_{i=0}^{N-1}\dbinom{N-1}{i}\left(  \operatorname{id}_{C}-f\right)
^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(  \operatorname{id}_{C}-f\right)
^{i}\right)  . \label{1601.a.pf.10}%
\end{align}
Also, we have%
\begin{equation}
\left(  \operatorname{id}_{C}-f\right)  ^{\ell}\left(  1_{C}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every integer }\ell\geq1. \label{1601.a.pf.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{1601.a.pf.9}):} Let $\ell$ be an integer
$\geq1$. Then, $\left(  \operatorname{id}_{C}-f\right)  ^{\ell}=\left(
\operatorname{id}_{C}-f\right)  ^{\ell-1}\circ\left(  \operatorname{id}%
_{C}-f\right)  $. But $f\left(  1_{C}\right)  =1_{C}$, so that $\left(
\operatorname{id}_{C}-f\right)  \left(  1_{C}\right)
=\underbrace{\operatorname{id}_{C}\left(  1_{C}\right)  }_{=1_{C}%
}-\underbrace{f\left(  1_{C}\right)  }_{=1_{C}}=1_{C}-1_{C}=0$ and
\begin{align*}
\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{\ell}}_{=\left(
\operatorname{id}_{C}-f\right)  ^{\ell-1}\circ\left(  \operatorname{id}%
_{C}-f\right)  }\left(  1_{C}\right)   &  =\left(  \left(  \operatorname{id}%
_{C}-f\right)  ^{\ell-1}\circ\left(  \operatorname{id}_{C}-f\right)  \right)
\left(  1_{C}\right) \\
&  =\left(  \operatorname{id}_{C}-f\right)  ^{\ell-1}\left(
\underbrace{\left(  \operatorname{id}_{C}-f\right)  \left(  1_{C}\right)
}_{=0}\right)  =\left(  \operatorname{id}_{C}-f\right)  ^{\ell-1}\left(
0\right)  =0
\end{align*}
(since the map $\left(  \operatorname{id}_{C}-f\right)  ^{\ell-1}$ is
$k$-linear). This proves (\ref{1601.a.pf.9}).} Moreover,%
\begin{equation}
f^{\ell}\left(  1_{C}\right)  =1_{C}\ \ \ \ \ \ \ \ \ \ \text{for every }%
\ell\in\mathbb{N}. \label{1601.a.pf.8}%
\end{equation}
\footnote{\textit{Proof of (\ref{1601.a.pf.8}):} We will prove
(\ref{1601.a.pf.8}) by induction over $\ell$:
\par
\textit{Induction base:} For $\ell=0$, we have $f^{\ell}\left(  1_{C}\right)
=\underbrace{f^{0}}_{=\operatorname{id}_{C}}\left(  1_{C}\right)
=\operatorname{id}_{C}\left(  1_{C}\right)  =1_{C}$. Hence, (\ref{1601.a.pf.8}%
) is proven for $\ell=0$. This completes the induction base.
\par
\textit{Induction step:} Let $L\in\mathbb{N}$. Assume that (\ref{1601.a.pf.8})
holds for $\ell=L$. We need to prove that (\ref{1601.a.pf.8}) holds for
$\ell=L+1$.
\par
We know that (\ref{1601.a.pf.8}) holds for $\ell=L$. In other words,
$f^{L}\left(  1_{C}\right)  =1_{C}$. Now, $\underbrace{f^{L+1}}_{=f\circ
f^{L}}\left(  1_{C}\right)  =\left(  f\circ f^{L}\right)  \left(
1_{C}\right)  =f\left(  \underbrace{f^{L}\left(  1_{C}\right)  }_{=1_{C}%
}\right)  =f\left(  1_{C}\right)  =1_{C}$. In other words, (\ref{1601.a.pf.8})
holds for $\ell=L+1$. This completes the induction step. The induction proof
of (\ref{1601.a.pf.8}) is thus complete.}

Now,%
\begin{equation}
\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes f\right)
^{N-1}\left(  y\otimes1_{C}\right)  =z\otimes1_{C} \label{1601.a.pf.11}%
\end{equation}
\footnote{\textit{Proof of (\ref{1601.a.pf.11}):} Every $i\in\left\{
1,2,\ldots,N-1\right\}  $ satisfies $i\geq1$. Hence, every $i\in\left\{
1,2,\ldots,N-1\right\}  $ satisfies $\left(  \operatorname{id}_{C}-f\right)
^{i}\left(  1_{C}\right)  =0$ (by (\ref{1601.a.pf.9}) (applied to $\ell=i$)).
Thus, every $i\in\left\{  1,2,\ldots,N-1\right\}  $ satisfies%
\begin{equation}
\left(  f^{N-1-i}\circ\left(  \operatorname{id}_{C}-f\right)  ^{i}\right)
\left(  1_{C}\right)  =f^{N-1-i}\left(  \underbrace{\left(  \operatorname{id}%
_{C}-f\right)  ^{i}\left(  1_{C}\right)  }_{\substack{=0}}\right)
=f^{N-1-i}\left(  0\right)  =0 \label{1601.a.pf.11.pf.1}%
\end{equation}
(since $f^{N-1-i}$ is $k$-linear). Now,%
\begin{align*}
&  \underbrace{\left(  \operatorname{id}_{C}\otimes\operatorname{id}%
_{C}-f\otimes f\right)  ^{N-1}}_{\substack{=\sum_{i=0}^{N-1}\dbinom{N-1}%
{i}\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\otimes\left(
f^{N-1-i}\circ\left(  \operatorname{id}_{C}-f\right)  ^{i}\right)  \\\text{(by
(\ref{1601.a.pf.10}))}}}\left(  y\otimes1_{C}\right) \\
&  =\left(  \sum_{i=0}^{N-1}\dbinom{N-1}{i}\left(  \operatorname{id}%
_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{i}\right)  \right)  \left(  y\otimes1_{C}\right) \\
&  =\sum_{i=0}^{N-1}\dbinom{N-1}{i}\underbrace{\left(  \left(
\operatorname{id}_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \right)  \left(  y\otimes
1_{C}\right)  }_{=\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\left(
y\right)  \otimes\left(  f^{N-1-i}\circ\left(  \operatorname{id}_{C}-f\right)
^{i}\right)  \left(  1_{C}\right)  }\\
&  =\sum_{i=0}^{N-1}\dbinom{N-1}{i}\left(  \operatorname{id}_{C}-f\right)
^{N-1-i}\left(  y\right)  \otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \left(  1_{C}\right) \\
&  =\sum_{i=1}^{N-1}\dbinom{N-1}{i}\left(  \operatorname{id}_{C}-f\right)
^{N-1-i}\left(  y\right)  \otimes\underbrace{\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \left(  1_{C}\right)
}_{\substack{=0\\\text{(by (\ref{1601.a.pf.11.pf.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\dbinom{N-1}{0}}_{=1}\underbrace{\left(
\operatorname{id}_{C}-f\right)  ^{N-1-0}}_{=\left(  \operatorname{id}%
_{C}-f\right)  ^{N-1}}\left(  y\right)  \otimes\left(  \underbrace{f^{N-1-0}%
}_{=f^{N-1}}\circ\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{0}%
}_{=\operatorname{id}_{C}}\right)  \left(  1_{C}\right) \\
&  =\sum_{i=1}^{N-1}\dbinom{N-1}{i}\underbrace{\left(  \operatorname{id}%
_{C}-f\right)  ^{N-1-i}\left(  y\right)  \otimes0}_{=0}+\underbrace{1\left(
\operatorname{id}_{C}-f\right)  ^{N-1}\left(  y\right)  }_{\substack{=\left(
\operatorname{id}_{C}-f\right)  ^{N-1}\left(  y\right)  =z\\\text{(since
}z=\left(  \operatorname{id}_{C}-f\right)  ^{N-1}\left(  y\right)  \text{)}%
}}\otimes\underbrace{\left(  f^{N-1}\circ\operatorname{id}_{C}\right)
}_{=f^{N-1}}\left(  1_{C}\right) \\
&  =\underbrace{\sum_{i=1}^{N-1}\dbinom{N-1}{i}0}_{=0}+z\otimes f^{N-1}\left(
1_{C}\right)  =z\otimes\underbrace{f^{N-1}\left(  1_{C}\right)  }%
_{\substack{=1_{C}\\\text{(by (\ref{1601.a.pf.8}), applied to }\ell
=N-1\text{)}}}=z\otimes1_{C}.
\end{align*}
This proves (\ref{1601.a.pf.11}).} and%
\begin{equation}
\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes f\right)
^{N-1}\left(  1_{C}\otimes y\right)  =1_{C}\otimes z \label{1601.a.pf.12}%
\end{equation}
\footnote{\textit{Proof of (\ref{1601.a.pf.12}):} Every $i\in\left\{
0,1,\ldots,N-2\right\}  $ satisfies $N-1-i\geq1$ (since every $i\in\left\{
0,1,\ldots,N-2\right\}  $ satisfies $i\leq N-2$ and thus $N-1-\underbrace{i}%
_{\leq N-2}\geq N-1-\left(  N-2\right)  =1$). Hence, every $i\in\left\{
0,1,\ldots,N-2\right\}  $ satisfies
\begin{equation}
\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\left(  1_{C}\right)  =0
\label{1601.a.pf.12.pf.1}%
\end{equation}
(by (\ref{1601.a.pf.9}) (applied to $\ell=N-1-i$)). Now,
\begin{align*}
&  \underbrace{\left(  \operatorname{id}_{C}\otimes\operatorname{id}%
_{C}-f\otimes f\right)  ^{N-1}}_{\substack{=\sum_{i=0}^{N-1}\dbinom{N-1}%
{i}\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\otimes\left(
f^{N-1-i}\circ\left(  \operatorname{id}_{C}-f\right)  ^{i}\right)  \\\text{(by
(\ref{1601.a.pf.10}))}}}\left(  1_{C}\otimes y\right) \\
&  =\left(  \sum_{i=0}^{N-1}\dbinom{N-1}{i}\left(  \operatorname{id}%
_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{i}\right)  \right)  \left(  1_{C}\otimes y\right) \\
&  =\sum_{i=0}^{N-1}\dbinom{N-1}{i}\underbrace{\left(  \left(
\operatorname{id}_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \right)  \left(  1_{C}\otimes
y\right)  }_{=\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\left(
1_{C}\right)  \otimes\left(  f^{N-1-i}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{i}\right)  \left(  y\right)  }\\
&  =\sum_{i=0}^{N-1}\dbinom{N-1}{i}\left(  \operatorname{id}_{C}-f\right)
^{N-1-i}\left(  1_{C}\right)  \otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \left(  y\right) \\
&  =\sum_{i=0}^{N-2}\dbinom{N-1}{i}\underbrace{\left(  \operatorname{id}%
_{C}-f\right)  ^{N-1-i}\left(  1_{C}\right)  }_{\substack{=0\\\text{(by
(\ref{1601.a.pf.12.pf.1}))}}}\otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \left(  y\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\dbinom{N-1}{N-1}}_{=1}\underbrace{\left(
\operatorname{id}_{C}-f\right)  ^{N-1-\left(  N-1\right)  }}_{=\left(
\operatorname{id}_{C}-f\right)  ^{0}=\operatorname{id}_{C}}\left(
1_{C}\right)  \otimes\left(  \underbrace{f^{N-1-\left(  N-1\right)  }}%
_{=f^{0}=\operatorname{id}_{C}}\circ\left(  \operatorname{id}_{C}-f\right)
^{N-1}\right)  \left(  y\right) \\
&  =\sum_{i=0}^{N-2}\dbinom{N-1}{i}\underbrace{0\otimes\left(  f^{N-1-i}%
\circ\left(  \operatorname{id}_{C}-f\right)  ^{i}\right)  \left(  y\right)
}_{=0}+\underbrace{1\operatorname{id}_{C}\left(  1_{C}\right)  }%
_{=\operatorname{id}_{C}\left(  1_{C}\right)  =1_{C}}\otimes
\underbrace{\left(  \operatorname{id}_{C}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{N-1}\right)  }_{=\left(  \operatorname{id}_{C}-f\right)
^{N-1}}\left(  y\right) \\
&  =\underbrace{\sum_{i=0}^{N-2}\dbinom{N-1}{i}0}_{=0}+1_{C}\otimes\left(
\operatorname{id}_{C}-f\right)  ^{N-1}\left(  y\right)  =1_{C}\otimes
\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{N-1}\left(  y\right)
}_{\substack{=z\\\text{(since }z=\left(  \operatorname{id}_{C}-f\right)
^{N-1}\left(  y\right)  \text{)}}}=1_{C}\otimes z.
\end{align*}
This proves (\ref{1601.a.pf.12}).}. Furthermore, every $u\in\left\{
1,2,\ldots,N-1\right\}  $ and every $i\in\left\{  0,1,\ldots,N-1\right\}  $
satisfy%
\begin{equation}
\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\left(  C_{\leq u}\right)
\otimes f^{N-1-i}\left(  \left(  \operatorname{id}_{C}-f\right)  ^{i}\left(
C_{\leq N-u}\right)  \right)  =0. \label{1601.a.pf.13o}%
\end{equation}
\footnote{\textit{Proof of (\ref{1601.a.pf.13o}):} Let $u\in\left\{
1,2,\ldots,N-1\right\}  $ and $i\in\left\{  0,1,\ldots,N-1\right\}  $. Notice
that $1\leq u\leq N-1$ (since $u\in\left\{  1,2,\ldots,N-1\right\}  $), hence
$u\geq1$ and $u\leq N-1<N$. Hence, (\ref{1601.a.pf.2}) (applied to $n=u$)
yields $\left(  \operatorname{id}_{C}-f\right)  ^{u}\left(  C_{\leq u}\right)
=0$. On the other hand, $N-\underbrace{u}_{\leq N-1}\geq N-\left(  N-1\right)
=1$ and $N-\underbrace{u}_{\geq1}\leq N-1<N$; therefore, (\ref{1601.a.pf.2})
(applied to $n=N-u$) yields $\left(  \operatorname{id}_{C}-f\right)
^{N-u}\left(  C_{\leq N-u}\right)  =0$.
\par
We want to prove (\ref{1601.a.pf.13o}). We must be in one of the following two
cases:
\par
\textit{Case 1:} We have $u\leq N-1-i$.
\par
\textit{Case 2:} We have $u>N-1-i$.
\par
Let us first consider Case 1. In this case, we have $u\leq N-1-i$. Thus,
$N-1-i\geq u$, so that $\left(  N-1-i\right)  -u$ is a nonnegative integer.
Denote this nonnegative integer by $\gamma$. Thus, $\left(  N-1-i\right)
-u=\gamma$, so that $N-1-i=\gamma+u$ and therefore%
\[
\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}=\left(  \operatorname{id}%
_{C}-f\right)  ^{\gamma+u}=\left(  \operatorname{id}_{C}-f\right)  ^{\gamma
}\circ\left(  \operatorname{id}_{C}-f\right)  ^{u},
\]
and thus%
\begin{align*}
\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}}_{=\left(
\operatorname{id}_{C}-f\right)  ^{\gamma}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{u}}\left(  C_{\leq u}\right)   &  =\left(  \left(
\operatorname{id}_{C}-f\right)  ^{\gamma}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{u}\right)  \left(  C_{\leq u}\right) \\
&  =\left(  \operatorname{id}_{C}-f\right)  ^{\gamma}\left(
\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{u}\left(  C_{\leq
u}\right)  }_{=0}\right)  =\left(  \operatorname{id}_{C}-f\right)  ^{\gamma
}\left(  0\right)  =0
\end{align*}
(since the map $\left(  \operatorname{id}_{C}-f\right)  ^{\gamma}$ is
$k$-linear). Hence,%
\[
\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\left(  C_{\leq
u}\right)  }_{=0}\otimes f^{N-1-i}\left(  \left(  \operatorname{id}%
_{C}-f\right)  ^{i}\left(  C_{\leq N-u}\right)  \right)  =0\otimes
f^{N-1-i}\left(  \left(  \operatorname{id}_{C}-f\right)  ^{i}\left(  C_{\leq
N-u}\right)  \right)  =0.
\]
Therefore, (\ref{1601.a.pf.13o}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $u>N-1-i$. Since $u$ and
$N-1-i$ are integers, this yields $u\geq\left(  N-1-i\right)  +1=N-i$. Thus,
$u+i\geq N$, so that $i\geq N-u$. In other words, $i-\left(  N-u\right)  $ is
a nonnegative integer. Denote this nonnegative integer by $\gamma$. Thus,
$i-\left(  N-u\right)  =\gamma$, so that $i=\gamma+\left(  N-u\right)  $ and
therefore%
\[
\left(  \operatorname{id}_{C}-f\right)  ^{i}=\left(  \operatorname{id}%
_{C}-f\right)  ^{\gamma+\left(  N-u\right)  }=\left(  \operatorname{id}%
_{C}-f\right)  ^{\gamma}\circ\left(  \operatorname{id}_{C}-f\right)  ^{N-u},
\]
and thus%
\begin{align*}
\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{i}}_{=\left(
\operatorname{id}_{C}-f\right)  ^{\gamma}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{N-u}}\left(  C_{\leq N-u}\right)   &  =\left(  \left(
\operatorname{id}_{C}-f\right)  ^{\gamma}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{N-u}\right)  \left(  C_{\leq N-u}\right) \\
&  =\left(  \operatorname{id}_{C}-f\right)  ^{\gamma}\left(
\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{N-u}\left(  C_{\leq
N-u}\right)  }_{=0}\right)  =\left(  \operatorname{id}_{C}-f\right)  ^{\gamma
}\left(  0\right)  =0
\end{align*}
(since the map $\left(  \operatorname{id}_{C}-f\right)  ^{\gamma}$ is
$k$-linear). Thus,%
\begin{align*}
&  \left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\left(  C_{\leq u}\right)
\otimes f^{N-1-i}\left(  \underbrace{\left(  \operatorname{id}_{C}-f\right)
^{i}\left(  C_{\leq N-u}\right)  }_{=0}\right) \\
&  =\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\left(  C_{\leq u}\right)
\otimes\underbrace{f^{N-1-i}\left(  0\right)  }_{\substack{=0\\\text{(since
}f^{N-1-i}\text{ is }k\text{-linear)}}}=\left(  \operatorname{id}%
_{C}-f\right)  ^{N-1-i}\left(  C_{\leq u}\right)  \otimes0=0.
\end{align*}
Thus, (\ref{1601.a.pf.13o}) is proven in Case 2.
\par
We have now proven (\ref{1601.a.pf.13o}) in both Cases 1 and 2. Since these
two Cases cover all possibilities, this yields that (\ref{1601.a.pf.13o})
always holds. The proof of (\ref{1601.a.pf.13o}) is thus complete.}. Every
$u\in\left\{  1,2,\ldots,N-1\right\}  $ satisfies%
\begin{equation}
\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes f\right)
^{N-1}\left(  C_{\leq u}\otimes C_{\leq N-u}\right)  =0. \label{1601.a.pf.13a}%
\end{equation}
\footnote{\textit{Proof of (\ref{1601.a.pf.13a}):} Let $u\in\left\{
1,2,\ldots,N-1\right\}  $. Then, $1\leq u\leq N-1$. Now,
\begin{align*}
&  \underbrace{\left(  \operatorname{id}_{C}\otimes\operatorname{id}%
_{C}-f\otimes f\right)  ^{N-1}}_{\substack{=\sum_{i=0}^{N-1}\dbinom{N-1}%
{i}\left(  \operatorname{id}_{C}-f\right)  ^{N-1-i}\otimes\left(
f^{N-1-i}\circ\left(  \operatorname{id}_{C}-f\right)  ^{i}\right)  \\\text{(by
(\ref{1601.a.pf.10}))}}}\left(  C_{\leq u}\otimes C_{\leq N-u}\right) \\
&  =\left(  \sum_{i=0}^{N-1}\dbinom{N-1}{i}\left(  \operatorname{id}%
_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{i}\right)  \right)  \left(  C_{\leq u}\otimes C_{\leq
N-u}\right) \\
&  \subseteq\sum_{i=0}^{N-1}\underbrace{\dbinom{N-1}{i}\left(  \left(
\operatorname{id}_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \right)  \left(  C_{\leq
u}\otimes C_{\leq N-u}\right)  }_{\substack{\subseteq\left(  \left(
\operatorname{id}_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \right)  \left(  C_{\leq
u}\otimes C_{\leq N-u}\right)  \\\text{(since }\left(  \left(
\operatorname{id}_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \right)  \left(  C_{\leq
u}\otimes C_{\leq N-u}\right)  \\\text{is a }k\text{-vector space)}}}\\
&  \subseteq\sum_{i=0}^{N-1}\underbrace{\left(  \left(  \operatorname{id}%
_{C}-f\right)  ^{N-1-i}\otimes\left(  f^{N-1-i}\circ\left(  \operatorname{id}%
_{C}-f\right)  ^{i}\right)  \right)  \left(  C_{\leq u}\otimes C_{\leq
N-u}\right)  }_{\subseteq\left(  \operatorname{id}_{C}-f\right)
^{N-1-i}\left(  C_{\leq u}\right)  \otimes\left(  f^{N-1-i}\circ\left(
\operatorname{id}_{C}-f\right)  ^{i}\right)  \left(  C_{\leq N-u}\right)  }\\
&  \subseteq\sum_{i=0}^{N-1}\left(  \operatorname{id}_{C}-f\right)
^{N-1-i}\left(  C_{\leq u}\right)  \otimes\underbrace{\left(  f^{N-1-i}%
\circ\left(  \operatorname{id}_{C}-f\right)  ^{i}\right)  \left(  C_{\leq
N-u}\right)  }_{=f^{N-1-i}\left(  \left(  \operatorname{id}_{C}-f\right)
^{i}\left(  C_{\leq N-u}\right)  \right)  }\\
&  =\sum_{i=0}^{N-1}\underbrace{\left(  \operatorname{id}_{C}-f\right)
^{N-1-i}\left(  C_{\leq u}\right)  \otimes f^{N-1-i}\left(  \left(
\operatorname{id}_{C}-f\right)  ^{i}\left(  C_{\leq N-u}\right)  \right)
}_{\substack{=0\\\text{(by (\ref{1601.a.pf.13o}))}}}\\
&  =\sum_{i=0}^{N-1}0=0.
\end{align*}
In other words, $\left(  \operatorname{id}_{C}\otimes\operatorname{id}%
_{C}-f\otimes f\right)  ^{N-1}\left(  C_{\leq u}\otimes C_{\leq N-u}\right)
=0$. This proves (\ref{1601.a.pf.13a}).} Thus,%
\begin{equation}
\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes f\right)
^{N-1}\left(  \sum_{u=1}^{N-1}C_{\leq u}\otimes C_{\leq N-u}\right)  =0.
\label{1601.a.pf.13b}%
\end{equation}
\footnote{\textit{Proof of (\ref{1601.a.pf.13b}):} The map $\left(
\operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes f\right)  ^{N-1}$
is $k$-linear. Thus,%
\begin{align*}
&  \left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes
f\right)  ^{N-1}\left(  \sum_{u=1}^{N-1}C_{\leq u}\otimes C_{\leq N-u}\right)
\\
&  \subseteq\sum_{u=1}^{N-1}\underbrace{\left(  \operatorname{id}_{C}%
\otimes\operatorname{id}_{C}-f\otimes f\right)  ^{N-1}\left(  C_{\leq
u}\otimes C_{\leq N-u}\right)  }_{\substack{=0\\\text{(by (\ref{1601.a.pf.13b}%
))}}}=\sum_{u=1}^{N-1}0=0.
\end{align*}
Therefore, $\left(  \operatorname{id}_{C}\otimes\operatorname{id}_{C}-f\otimes
f\right)  ^{N-1}\left(  \sum_{u=1}^{N-1}C_{\leq u}\otimes C_{\leq N-u}\right)
=0$. This proves (\ref{1601.a.pf.13b}).} Now, (\ref{1601.a.pf.5}) becomes%
\begin{align*}
\Delta_{C}\left(  z\right)   &  \in\underbrace{\left(  \operatorname{id}%
_{C}\otimes\operatorname{id}_{C}-f\otimes f\right)  ^{N-1}\left(
y\otimes1_{C}\right)  }_{\substack{=z\otimes1_{C}\\\text{(by
(\ref{1601.a.pf.11}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \operatorname{id}_{C}%
\otimes\operatorname{id}_{C}-f\otimes f\right)  ^{N-1}\left(  1_{C}\otimes
y\right)  }_{\substack{=1_{C}\otimes z\\\text{(by (\ref{1601.a.pf.12}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \operatorname{id}_{C}%
\otimes\operatorname{id}_{C}-f\otimes f\right)  ^{N-1}\left(  \sum_{u=1}%
^{N-1}C_{\leq u}\otimes C_{\leq N-u}\right)  }_{\substack{=0\\\text{(by
(\ref{1601.a.pf.13b}))}}}\\
&  =z\otimes1_{C}+1_{C}\otimes z+0=z\otimes1_{C}+1_{C}\otimes z.
\end{align*}
In other words, $\Delta_{C}\left(  z\right)  =z\otimes1_{C}+1_{C}\otimes z$.
In other words, the element $z$ of $C$ is primitive (because the definition of
\textquotedblleft primitive\textquotedblright\ yields that the element $z$ of
$C$ is primitive if and only if $\Delta_{C}\left(  z\right)  =z\otimes
1_{C}+1_{C}\otimes z$). Hence, (\ref{1601.prim.act}) (applied to $z$ instead
of $x$) yields%
\begin{equation}
f\left(  z\right)  =z. \label{1601.a.pf.almost}%
\end{equation}


Now,%
\begin{align*}
\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{N}}_{=\left(
\operatorname{id}_{C}-f\right)  \circ\left(  \operatorname{id}_{C}-f\right)
^{N-1}}\left(  y\right)   &  =\left(  \left(  \operatorname{id}_{C}-f\right)
\circ\left(  \operatorname{id}_{C}-f\right)  ^{N-1}\right)  \left(  y\right)
=\left(  \operatorname{id}_{C}-f\right)  \left(  \underbrace{\left(
\operatorname{id}_{C}-f\right)  ^{N-1}\left(  y\right)  }%
_{\substack{=z\\\text{(since }z=\left(  \operatorname{id}_{C}-f\right)
^{N-1}\left(  y\right)  \text{)}}}\right) \\
&  =\left(  \operatorname{id}_{C}-f\right)  \left(  z\right)
=\underbrace{\operatorname{id}_{C}\left(  z\right)  }_{=z}%
-\underbrace{f\left(  z\right)  }_{\substack{=z\\\text{(by
(\ref{1601.a.pf.almost}))}}}=z-z=0.
\end{align*}


But recall that $y=x-\varepsilon\left(  x\right)  \cdot1_{C}$, so that
$x=y+\varepsilon\left(  x\right)  \cdot1_{C}$. Applying the map $\left(
\operatorname{id}_{C}-f\right)  ^{N}$ to both sides of this equality, we
obtain
\begin{align*}
&  \left(  \operatorname{id}_{C}-f\right)  ^{N}\left(  x\right) \\
&  =\left(  \operatorname{id}_{C}-f\right)  ^{N}\left(  y+\varepsilon\left(
x\right)  \cdot1_{C}\right) \\
&  =\underbrace{\left(  \operatorname{id}_{C}-f\right)  ^{N}\left(  y\right)
}_{=0}+\varepsilon\left(  x\right)  \cdot\underbrace{\left(  \operatorname{id}%
_{C}-f\right)  ^{N}\left(  1_{C}\right)  }_{\substack{=0\\\text{(by
(\ref{1601.a.pf.9}), applied to }\ell=N\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\left(  \operatorname{id}%
_{C}-f\right)  ^{N}\text{ is }k\text{-linear}\right) \\
&  =0+\varepsilon\left(  x\right)  \cdot0=0.
\end{align*}
In other words, $x\in\operatorname*{Ker}\left(  \left(  \operatorname{id}%
_{C}-f\right)  ^{N}\right)  $.

Now, forget that we fixed $x$. We thus have proven that every $x\in C_{\leq
N}$ satisfies $x\in\operatorname*{Ker}\left(  \left(  \operatorname{id}%
_{C}-f\right)  ^{N}\right)  $. In other words, $C_{\leq N}\subseteq
\operatorname*{Ker}\left(  \left(  \operatorname{id}_{C}-f\right)
^{N}\right)  $. Hence,%
\[
\left(  \operatorname{id}_{C}-f\right)  ^{N}\left(  \underbrace{C_{\leq N}%
}_{\subseteq\operatorname*{Ker}\left(  \left(  \operatorname{id}_{C}-f\right)
^{N}\right)  }\right)  \subseteq\left(  \operatorname{id}_{C}-f\right)
^{N}\left(  \operatorname*{Ker}\left(  \left(  \operatorname{id}_{C}-f\right)
^{N}\right)  \right)  =0.
\]
In other words, (\ref{1601.a.pf.1}) holds for $n=N$. This completes the
induction step. Thus, the induction proof of (\ref{1601.a.pf.1}) is complete.

Now, we have proven that (\ref{1601.a.pf.1}) holds for every integer $n\geq1$.
In other words, Theorem~\ref{Theorem1601} \textbf{(a)} is proven.

\textbf{(b)} We have $C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}%
\subseteq\cdots$ (since $C$ is filtered), so that $C_{\leq0}\subseteq
C_{\leq1}\subseteq\bigcup_{n\geq1}C_{\leq n}$ (because $C_{\leq1}$ is a term
in the union $\bigcup_{n\geq1}C_{\leq n}$). Now, since $C$ is filtered, we
have%
\[
C=\bigcup_{n\geq0}C_{\leq n}=\underbrace{C_{\leq0}}_{\subseteq\bigcup_{n\geq
1}C_{\leq n}}\cup\left(  \bigcup_{n\geq1}C_{\leq n}\right)  \subseteq\left(
\bigcup_{n\geq1}C_{\leq n}\right)  \cup\left(  \bigcup_{n\geq1}C_{\leq
n}\right)  =\bigcup_{n\geq1}C_{\leq n}.
\]
Combined with $\bigcup_{n\geq1}C_{\leq n}\subseteq C$ (this is obvious), this
yields $C=\bigcup_{n\geq1}C_{\leq n}$. Moreover, Theorem~\ref{Theorem1601}
\textbf{(a)} yields that $\left(  \operatorname{id}_{C}-f\right)  ^{n}\left(
C_{\leq n}\right)  =0$ for every integer $n\geq1$. Hence,
Lemma~\ref{Lemma1602} (applied to $V=C$ and $V_{i}=C_{\leq i}$) yields that
the map $f$ is invertible. Thus, $f$ is an invertible $k$-coalgebra
homomorphism. Therefore, Proposition~\ref{Proposition1317} (applied to $D=C$)
yields that $f$ is a $k$-coalgebra isomorphism. This proves
Theorem~\ref{Theorem1601} \textbf{(b)}.
\end{proof}

As a consequence of Theorem~\ref{Theorem1601}, we can obtain the curious fact
that the antipode of a connected filtered $k$-Hopf algebra is invertible, and
somewhat more:

\begin{theorem}
\label{Theorem1621} Let $k$ be a field. Let $H$ be a connected filtered
$k$-Hopf algebra. Let $S$ be the antipode of $H$.

\textbf{(a)} We have $\left(  \operatorname{id}_{H}-S^{2}\right)  ^{n}\left(
H_{\leq n}\right)  =0$ for every integer $n\geq1$.

\textbf{(b)} The map $S$ is invertible.
\end{theorem}

We notice that Theorem~\ref{Theorem1621} generalizes a result of Aguiar and
Lauve (\cite[Corollary 5]{AguLau14})\footnote{Here are the details:
\par
If $k$ is a field, and $H$ is a connected filtered $k$-Hopf algebra, then
Theorem~\ref{Theorem1621} \textbf{(a)} shows that the only possible eigenvalue
of $S^{2}$ on $H_{\leq n}$ is $1$, whence the only possible eigenvalues of $S$
on $H_{\leq n}$ are $1$ and $-1$. This readily yields \cite[Corollary
5]{AguLau14}, but is itself a more general result. (Note also that the paper
\cite{AguLau14} works entirely over a field of characteristic $0$, and does
use this assumption in the proof; in contrast, what we are doing makes sense
over any commutative ring $k$.)}.

Before we prove Theorem~\ref{Theorem1621}, let us show a lemma concerning
arbitrary Hopf algebras:

\begin{lemma}
\label{Lemma1621b} Let $k$ be a field. Let $H$ be a $k$-Hopf algebra. Let $S$
be the antipode of $H$.

\textbf{(a)} The map $S^{2}:H\rightarrow H$ is a $k$-coalgebra homomorphism.

\textbf{(b)} We have $S^{2}\left(  1_{H}\right)  =1_{H}$.

\textbf{(c)} We have $S^{2}\left(  x\right)  =x$ for every primitive element
$x$ of $H$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1621b}.]\textbf{(a)} Recall that the coopposite
coalgebra $H^{\operatorname*{cop}}$ of $H$ is defined as the $k$-coalgebra
$\left(  H,\tau_{H,H}\circ\Delta_{H},\varepsilon_{H}\right)  $. Thus,
$\Delta_{H^{\operatorname*{cop}}}=\tau_{H,H}\circ\Delta_{H}$ and
$\varepsilon_{H^{\operatorname*{cop}}}=\varepsilon_{H}$.

Proposition~\ref{Proposition553} yields that the antipode of $H$ is a
$k$-coalgebra homomorphism $H^{\operatorname*{cop}}\rightarrow H$. In other
words, $S$ is a $k$-coalgebra homomorphism $H^{\operatorname*{cop}}\rightarrow
H$ (since $S$ is the antipode of $H$). Thus, $\Delta_{H}\circ S=\left(
S\otimes S\right)  \circ\Delta_{H^{\operatorname*{cop}}}$ and $\varepsilon
_{H}\circ S=\varepsilon_{H^{\operatorname*{cop}}}$.

Clearly, (\ref{(X)(X)}) (applied to $U=H$, $V=H$, $W=H$, $U^{\prime}=H$,
$V^{\prime}=H$, $W^{\prime}=H$, $\alpha=S$, $\beta=S$, $\alpha^{\prime}=S$ and
$\beta^{\prime}=S$) yields $\left(  S\circ S\right)  \otimes\left(  S\circ
S\right)  =\left(  S\otimes S\right)  \circ\left(  S\otimes S\right)  $.
Hence,%
\[
\left(  S\otimes S\right)  \circ\left(  S\otimes S\right)
=\underbrace{\left(  S\circ S\right)  }_{=S^{2}}\otimes\underbrace{\left(
S\circ S\right)  }_{=S^{2}}=S^{2}\otimes S^{2}.
\]


A very basic property of the flip maps says that if $V$ and $W$ are two
$k$-vector spaces, then $\tau_{W,V}\circ\tau_{V,W}=\operatorname{id}_{V\otimes
W}$. Applying this to $V=H$ and $W=H$, we obtain $\tau_{H,H}\circ\tau
_{H,H}=\operatorname{id}_{H\otimes H}$. Now,%
\begin{align*}
\Delta_{H}\circ\underbrace{S^{2}}_{=S\circ S}  &  =\underbrace{\Delta_{H}\circ
S}_{=\left(  S\otimes S\right)  \circ\Delta_{H^{\operatorname*{cop}}}}\circ
S=\left(  S\otimes S\right)  \circ\underbrace{\Delta_{H^{\operatorname*{cop}}%
}}_{=\tau_{H,H}\circ\Delta_{H}}\circ S\\
&  =\left(  S\otimes S\right)  \circ\tau_{H,H}\circ\underbrace{\Delta_{H}\circ
S}_{=\left(  S\otimes S\right)  \circ\Delta_{H^{\operatorname*{cop}}}}=\left(
S\otimes S\right)  \circ\tau_{H,H}\circ\left(  S\otimes S\right)
\circ\underbrace{\Delta_{H^{\operatorname*{cop}}}}_{=\tau_{H,H}\circ\Delta
_{H}}\\
&  =\left(  S\otimes S\right)  \circ\tau_{H,H}\circ\underbrace{\left(
S\otimes S\right)  \circ\tau_{H,H}}_{\substack{=\tau_{H,H}\circ\left(
S\otimes S\right)  \\\text{(by Proposition~\ref{Proposition35} \textbf{(a)}%
,}\\\text{applied to }V=H\text{, }W=H\text{, }V^{\prime}=H\text{, }W^{\prime
}=H\text{,}\\f=S\text{ and }g=S\text{)}}}\circ\Delta_{H}\\
&  =\left(  S\otimes S\right)  \circ\underbrace{\tau_{H,H}\circ\tau_{H,H}%
}_{=\operatorname{id}_{H\otimes H}}\circ\left(  S\otimes S\right)  \circ
\Delta_{H}\\
&  =\underbrace{\left(  S\otimes S\right)  \circ\left(  S\otimes S\right)
}_{=S^{2}\otimes S^{2}}\circ\Delta_{H}=\left(  S^{2}\otimes S^{2}\right)
\circ\Delta_{H}.
\end{align*}
Combined with%
\[
\varepsilon_{H}\circ\underbrace{S^{2}}_{=S\circ S}=\underbrace{\varepsilon
_{H}\circ S}_{=\varepsilon_{H^{\operatorname*{cop}}}=\varepsilon_{H}}\circ
S=\varepsilon_{H}\circ S=\varepsilon_{H^{\operatorname*{cop}}}=\varepsilon
_{H},
\]
this yields that $S^{2}:H\rightarrow H$ is a $k$-coalgebra homomorphism. This
proves Lemma~\ref{Lemma1621b} \textbf{(a)}.

\textbf{(b)} Just as in the proof of Corollary~\ref{Corollary813}, we can show
that $S\left(  1_{H}\right)  =1_{H}$. Hence, $\underbrace{S^{2}}_{=S\circ
S}\left(  1_{H}\right)  =\left(  S\circ S\right)  \left(  1_{H}\right)
=S\left(  \underbrace{S\left(  1_{H}\right)  }_{=1_{H}}\right)  =S\left(
1_{H}\right)  =1_{H}$. This proves Lemma~\ref{Lemma1621b} \textbf{(b)}.

\textbf{(c)} Let $x$ be a primitive element of $H$. Then, $x$ belongs to the
set of all primitive elements of $H$. In other words, $x$ belongs to
$\operatorname*{Prim}H$ (since $\operatorname*{Prim}H$ is the set of all
primitive elements of $H$). In other words, $x\in\operatorname*{Prim}H$.
Hence, Proposition~\ref{Proposition843} yields $S\left(  x\right)  =-x$. Thus,%
\begin{align*}
\underbrace{S^{2}}_{=S\circ S}\left(  x\right)   &  =\left(  S\circ S\right)
\left(  x\right)  =S\left(  \underbrace{S\left(  x\right)  }_{=-x}\right)
=S\left(  -x\right)  =-\underbrace{S\left(  x\right)  }_{=-x}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }S\text{ is }k\text{-linear}%
\right) \\
&  =-\left(  -x\right)  =x.
\end{align*}
This proves Lemma~\ref{Lemma1621b} \textbf{(c)}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{Theorem1621}.]Lemma~\ref{Lemma1621b} \textbf{(a)} shows
that the map $S^{2}:H\rightarrow H$ is a $k$-coalgebra homomorphism.

Lemma~\ref{Lemma1621b} \textbf{(b)} yields $S^{2}\left(  1_{H}\right)  =1_{H}$.

Moreover, Lemma~\ref{Lemma1621b} \textbf{(c)} shows that%
\[
S^{2}\left(  x\right)  =x\ \ \ \ \ \ \ \ \ \ \text{for every primitive element
}x\text{ of }H.
\]
Hence, we can apply Theorem~\ref{Theorem1601} to $C=H$ and $f=S^{2}$.

\textbf{(a)} Theorem~\ref{Theorem1601} \textbf{(a)} (applied to $C=H$ and
$f=S^{2}$) yields that we have $\left(  \operatorname{id}_{H}-S^{2}\right)
^{n}\left(  H_{\leq n}\right)  =0$ for every integer $n\geq1$. This proves
Theorem~\ref{Theorem1621} \textbf{(a)}.

\textbf{(b)} Theorem~\ref{Theorem1601} \textbf{(b)} (applied to $C=H$ and
$f=S^{2}$) yields that the map $S^{2}$ is a $k$-coalgebra isomorphism. In
particular, the map $S^{2}$ is invertible. In other words, there exists a map
$T:H\rightarrow H$ satisfying $S^{2}\circ T=\operatorname{id}_{H}$ and $T\circ
S^{2}=\operatorname{id}_{H}$. Consider this map $T$. We have%
\[
S\circ\left(  S\circ T\right)  =\underbrace{\left(  S\circ S\right)  }%
_{=S^{2}}\circ T=S^{2}\circ T=\operatorname{id}_{H};
\]
therefore, the map $S$ is right-invertible. Also,%
\[
\left(  T\circ S\right)  \circ S=T\circ\underbrace{\left(  S\circ S\right)
}_{=S^{2}}=T\circ S^{2}=\operatorname{id}_{H};
\]
therefore, the map $S$ is left-invertible. Any map which is both
left-invertible and right-invertible must be invertible. Applying this to the
map $S$, we conclude that the map $S$ is invertible (since we know that the
map $S$ is both left-invertible and right-invertible). This proves
Theorem~\ref{Theorem1621} \textbf{(b)}.
\end{proof}

\section{\label{S61}Leray's theorem for the Eulerian idempotent}

Our next goal is to prove (one form of) Leray's theorem. We first introduce
some notation:

\begin{definition}
\label{Definition1701} Let $V$ be any $k$-vector space. We use the notation
$\operatorname*{Sym}V$ for the symmetric algebra of $V$. For every
$n\in\mathbb{N}$, we denote by $\operatorname*{Sym}\nolimits^{n}V$ the $n$-th
symmetric power of $V$. Thus, $\operatorname*{Sym}V=\bigoplus_{n\in\mathbb{N}%
}\operatorname*{Sym}\nolimits^{n}V$. Thus, $\operatorname*{Sym}\nolimits^{1}%
V\subseteq\operatorname*{Sym}V$.

There is a canonical injection $V\rightarrow\operatorname*{Sym}V$ of vector
spaces (obtained by composing the canonical isomorphism $V\rightarrow
\operatorname*{Sym}\nolimits^{1}V$ with the canonical inclusion
$\operatorname*{Sym}\nolimits^{1}V\rightarrow\operatorname*{Sym}V$). We denote
this injection by $\operatorname*{syminc}\nolimits_{V}$. We will often
identify $V$ with a $k$-vector subspace of $\operatorname*{Sym}V$ along this
injection (whenever this does not cause misunderstandings).

The universal property of $\operatorname*{Sym}V$ says that if $A$ is any
commutative $k$-algebra and if $\varphi:V\rightarrow A$ is any $k$-linear map,
then there exists a unique $k$-algebra homomorphism $\Phi:\operatorname*{Sym}%
V\rightarrow A$ satisfying $\Phi\circ\operatorname*{syminc}\nolimits_{V}%
=\varphi$. This homomorphism $\Phi$ will be denoted by
$\operatorname*{symlift}\varphi$. It can be explicitly computed by the formula%
\begin{align}
\left(  \operatorname*{symlift}\varphi\right)  \left(  v_{1}v_{2}\cdots
v_{n}\right)   &  =\varphi\left(  v_{1}\right)  \varphi\left(  v_{2}\right)
\cdots\varphi\left(  v_{n}\right) \label{1701.symlift.varphi}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for any }n\in\mathbb{N}\text{ and }\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}.\nonumber
\end{align}
(Here, $v_{1}v_{2}\cdots v_{n}$ denotes the projection of the tensor
$v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\in V^{\otimes n}$ onto the
$n$-th symmetric power $\operatorname*{Sym}\nolimits^{n}V$, or, equivalently,
the product of the elements $v_{1}$, $v_{2}$, $\ldots$, $v_{n}$ of
$\operatorname*{Sym}V$.)
\end{definition}

\begin{theorem}
\label{Theorem1702} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered commutative bialgebra over $k$. Consider the convolution
algebra $\mathcal{L}\left(  H,H\right)  $. Let $\mathfrak{e}$ denote the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $.

\textbf{(a)} The map $\mathfrak{e}$ is a projection.

\textbf{(b)} We have $\operatorname*{Ker}\mathfrak{e}=\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}%
$.\ \ \ \ \footnote{Recall that $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ is to be understood according to
Convention~\ref{Convention100}. Thus, $\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$ means the subspace $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  \cdot\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of $H$.}

\textbf{(c)} The map $\mathfrak{e}$ is an $\left(  \varepsilon_{H}%
,\varepsilon_{H}\right)  $-derivation.

\textbf{(d)} Let $\mathfrak{j}$ denote the inclusion map $\mathfrak{e}\left(
H\right)  \rightarrow H$. Define a map $\mathfrak{e}^{\prime}:H\rightarrow
\mathfrak{e}\left(  H\right)  $ by%
\[
\left(  \mathfrak{e}^{\prime}\left(  h\right)  =\mathfrak{e}\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H\right)  .
\]
\footnote{This is well-defined, since $\mathfrak{e}\left(  h\right)
\in\mathfrak{e}\left(  H\right)  $ for every $h\in H$.} This map
$\mathfrak{e}^{\prime}$ is clearly $k$-linear\footnote{In fact, this map
$\mathfrak{e}^{\prime}$ is obtained from the $k$-linear map $\mathfrak{e}$ by
changing the target to $\mathfrak{e}\left(  H\right)  $.}. Define a $k$-linear
map $\mathfrak{q}:H\rightarrow\operatorname*{Sym}\left(  \mathfrak{e}\left(
H\right)  \right)  $ by $\mathfrak{q}=\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }\circ\mathfrak{e}^{\prime}$. Then,
the two $k$-linear maps $\operatorname*{symlift}\mathfrak{j}%
:\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  \rightarrow
H$ and $e^{\ast\mathfrak{q}}:H\rightarrow\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  $ are well-defined $k$-algebra
homomorphisms and mutually inverse.

\textbf{(e)} We have $\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)
\right)  \cong H$ as $k$-algebras.
\end{theorem}

Much of Theorem~\ref{Theorem1702} has already been proven (e.g., parts
\textbf{(a)} and \textbf{(b)} follow readily from Theorem~\ref{Theorem101}).
It is part \textbf{(d)}, and its consequence part \textbf{(e)}, which are the
most important for us. Theorem~\ref{Theorem1702} \textbf{(e)} is more or less
a generalization of Theorem 3.8.3 in \cite{9}\footnote{It is not exactly a
generalization of Theorem 3.8.3 in \cite{9}, since \cite{9} works with graded
$k$-algebras, and consequently assumes $H$ to be graded and claims
$\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $ to be
isomorphic to $H$ as \textbf{graded} $k$-algebra. But this can also be easily
derived from our Theorem~\ref{Theorem1702} \textbf{(d)}, since one can easily
show that $\mathfrak{e}\left(  H\right)  $ is a homogeneous $k$-vector
subspace of $H$ and that the $k$-linear map $\operatorname*{symlift}%
\mathfrak{j}$ is graded (with the appropriate grading on $\operatorname*{Sym}%
\left(  \mathfrak{e}\left(  H\right)  \right)  $).}.

Before we step to proving Theorem~\ref{Theorem1702}, let us show a general
fact about convolution:\footnote{I have now realized that
Proposition~\ref{Proposition1711} is just a repetition of parts \textbf{(a)},
\textbf{(b)}, \textbf{(c)} and \textbf{(d)} of
Proposition~\ref{Proposition1102} (with $\psi$ renamed as $p$).}

\begin{proposition}
\label{Proposition1711} Let $k$ be a field. Let $C$ be a $k$-coalgebra. Let
$A$ and $B$ be two $k$-algebras. Let $p:A\rightarrow B$ be a $k$-algebra homomorphism.

\textbf{(a)} We have $p\circ e_{C,A}=e_{C,B}$. (Here, the maps $e_{C,A}%
:C\rightarrow A$ and $e_{C,B}:C\rightarrow B$ are defined in the same way as
the map $e_{H,A}:H\rightarrow A$ in Definition~\ref{def.eHA}.)

\textbf{(b)} For every $f\in\mathcal{L}\left(  C,A\right)  $ and
$g\in\mathcal{L}\left(  C,A\right)  $, we have $p\circ\left(  f\ast g\right)
=\left(  p\circ f\right)  \ast\left(  p\circ g\right)  $.

\textbf{(c)} For every $f\in\mathcal{L}\left(  C,A\right)  $ and
$n\in\mathbb{N}$, we have $p\circ\left(  f^{\ast n}\right)  =\left(  p\circ
f\right)  ^{\ast n}$.

\textbf{(d)} Assume that the field $k$ has characteristic $0$. Assume also
that $C$ is a connected filtered $k$-coalgebra. Let $f\in\mathfrak{g}\left(
C,A\right)  $. (See Definition~\ref{Definition8} for the meaning of
$\mathfrak{g}\left(  C,A\right)  $.) Then, $p\circ f\in\mathfrak{g}\left(
C,B\right)  $ and $p\circ e^{\ast f}=e^{\ast\left(  p\circ f\right)  }$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1711}.]We know that $p$ is a $k$-algebra
homomorphism if and only if it satisfies $p\circ\eta_{A}=\eta_{B}$ and
$p\circ\mu_{A}=\mu_{B}\circ\left(  p\otimes p\right)  $ (due to the definition
of $k$-algebra homomorphisms using arrows). Thus, $p$ satisfies $p\circ
\eta_{A}=\eta_{B}$ and $p\circ\mu_{A}=\mu_{B}\circ\left(  p\otimes p\right)  $
(since we know that $p$ is a $k$-algebra homomorphism).

By the definition of $\mathfrak{g}\left(  C,A\right)  $, we have%
\[
\mathfrak{g}\left(  C,A\right)  =\left\{  f\in\mathcal{L}\left(  C,A\right)
\ \mid\ f\left(  1_{C}\right)  =0\right\}  =\left\{  h\in\mathcal{L}\left(
C,A\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}
\]
(here we renamed $f$ as $h$). By the definition of $\mathfrak{g}\left(
C,B\right)  $, we have%
\[
\mathfrak{g}\left(  C,B\right)  =\left\{  f\in\mathcal{L}\left(  C,B\right)
\ \mid\ f\left(  1_{C}\right)  =0\right\}  =\left\{  h\in\mathcal{L}\left(
C,B\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}
\]
(here we renamed $f$ as $h$).

\textbf{(a)} The definition of $e_{C,A}$ yields $e_{C,A}=\eta_{A}%
\circ\varepsilon_{C}$. The definition of $e_{C,B}$ yields $e_{C,B}=\eta
_{B}\circ\varepsilon_{C}$. Now, $p\circ\underbrace{e_{C,A}}_{=\eta_{A}%
\circ\varepsilon_{C}}=\underbrace{p\circ\eta_{A}}_{=\eta_{B}}\circ
\varepsilon_{C}=\eta_{B}\circ\varepsilon_{C}=e_{C,B}$. This proves
Proposition~\ref{Proposition1711} \textbf{(a)}.

\textbf{(b)} Let $f\in\mathcal{L}\left(  C,A\right)  $ and $g\in
\mathcal{L}\left(  C,A\right)  $. The definition of convolution yields%
\begin{align*}
\left(  p\circ f\right)  \ast\left(  p\circ g\right)   &  =\mu_{B}%
\circ\underbrace{\left(  \left(  p\circ f\right)  \otimes\left(  p\circ
g\right)  \right)  }_{\substack{=\left(  p\otimes p\right)  \circ\left(
f\otimes g\right)  \\\text{(by (\ref{(X)(X)}), applied to }C\text{, }A\text{,
}B\text{, }C\text{, }A\text{, }B\text{, }f\text{, }p\text{, }g\text{ and
}p\\\text{instead of }U\text{, }V\text{, }W\text{, }U^{\prime}\text{,
}V^{\prime}\text{, }W^{\prime}\text{, }\alpha\text{, }\beta\text{, }%
\alpha^{\prime}\text{ and }\beta\text{')}}}\circ\Delta_{C}\\
&  =\underbrace{\mu_{B}\circ\left(  p\otimes p\right)  }_{\substack{=p\circ
\mu_{A}\\\text{(since }p\circ\mu_{A}=\mu_{B}\circ\left(  p\otimes p\right)
\text{)}}}\circ\left(  f\otimes g\right)  \circ\Delta_{C}=p\circ
\underbrace{\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
}_{\substack{=f\ast g\\\text{(since }f\ast g=\mu_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}\\\text{(by the definition of convolution))}}}\\
&  =p\circ\left(  f\ast g\right)  .
\end{align*}
This proves Proposition~\ref{Proposition1711} \textbf{(b)}.

\textbf{(c)} Let $f\in\mathcal{L}\left(  C,A\right)  $. We need to show that%
\begin{equation}
p\circ\left(  f^{\ast n}\right)  =\left(  p\circ f\right)  ^{\ast
n}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{.}
\label{1711.pf.c.goal}%
\end{equation}


\textit{Proof of (\ref{1711.pf.c.goal}):} We shall prove (\ref{1711.pf.c.goal}%
) by induction over $n$:

\textit{Induction base:} We have $p\circ\underbrace{\left(  f^{\ast0}\right)
}_{=e_{C,A}}=p\circ e_{C,A}=e_{C,B}$ (according to
Proposition~\ref{Proposition1711} \textbf{(a)}) and $\left(  p\circ f\right)
^{\ast0}=e_{C,B}$. Hence, $p\circ\left(  f^{\ast0}\right)  =e_{C,B}=\left(
p\circ f\right)  ^{\ast0}$. In other words, (\ref{1711.pf.c.goal}) holds for
$n=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
(\ref{1711.pf.c.goal}) holds for $n=N$. We now need to show that
(\ref{1711.pf.c.goal}) holds for $n=N+1$.

We know that (\ref{1711.pf.c.goal}) holds for $n=N$. In other words, we have
$p\circ\left(  f^{\ast N}\right)  =\left(  p\circ f\right)  ^{\ast N}$. Now,%
\begin{align*}
p\circ\underbrace{\left(  f^{\ast\left(  N+1\right)  }\right)  }_{=f^{\ast
N}\circ f}  &  =p\circ\left(  f^{\ast N}\circ f\right)  =\underbrace{\left(
p\circ f^{\ast N}\right)  }_{=\left(  p\circ f\right)  ^{\ast N}}\ast\left(
p\circ f\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition~\ref{Proposition1711}
\textbf{(b)}, applied to }f^{\ast N}\text{ and }f\text{ instead of }f\text{
and }g\right) \\
&  =\left(  p\circ f\right)  ^{\ast N}\ast\left(  p\circ f\right)  =\left(
p\circ f\right)  ^{\ast\left(  N+1\right)  }.
\end{align*}
In other words, (\ref{1711.pf.c.goal}) holds for $n=N+1$. This completes the
induction step.

Thus, the induction proof of (\ref{1711.pf.c.goal}) is complete. In other
words, Proposition~\ref{Proposition1711} \textbf{(c)} is proven.

\textbf{(d)} We have $f\in\mathfrak{g}\left(  C,A\right)  =\left\{
h\in\mathcal{L}\left(  C,A\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}
$. In other words, $f$ is an element of $\mathcal{L}\left(  C,A\right)  $ and
satisfies $f\left(  1_{C}\right)  =0$. Now, $\left(  p\circ f\right)  \left(
1_{C}\right)  =p\left(  \underbrace{f\left(  1_{C}\right)  }_{=0}\right)
=p\left(  0\right)  =0$ (since the map $p$ is $k$-linear). Thus, $p\circ f$ is
an element of $\mathcal{L}\left(  C,B\right)  $ and satisfies $\left(  p\circ
f\right)  \left(  1_{C}\right)  =0$. In other words, $p\circ f\in\left\{
h\in\mathcal{L}\left(  C,B\right)  \ \mid\ h\left(  1_{C}\right)  =0\right\}
=\mathfrak{g}\left(  C,B\right)  $ (since $\mathfrak{g}\left(  C,B\right)
=\left\{  h\in\mathcal{L}\left(  C,B\right)  \ \mid\ h\left(  1_{C}\right)
=0\right\}  $). Thus, $e^{\ast\left(  p\circ f\right)  }$ is a well-defined
$k$-linear map $C\rightarrow B$.

Now, let $x\in C$. Then, the definition of $e^{\ast\left(  p\circ f\right)
}\left(  x\right)  $ yields $e^{\ast\left(  p\circ f\right)  }\left(
x\right)  =\sum_{i\geq0}\dfrac{\left(  p\circ f\right)  ^{\ast i}\left(
x\right)  }{i!}$. On the other hand, the definition of $e^{\ast f}\left(
x\right)  $ yields
\begin{equation}
e^{\ast f}\left(  x\right)  =\sum_{i\geq0}\dfrac{f^{\ast i}\left(  x\right)
}{i!}=\sum_{i\geq0}\dfrac{1}{i!}f^{\ast i}\left(  x\right)  .
\label{1711.pf.d.1}%
\end{equation}
Notice that the sum $\sum_{i\geq0}\dfrac{1}{i!}f^{\ast i}\left(  x\right)  $
converges with respect to the discrete topology.

Applying the map $p$ to both sides of the equality (\ref{1711.pf.d.1}), we
obtain%
\begin{align*}
p\left(  e^{\ast f}\left(  x\right)  \right)   &  =p\left(  \sum_{i\geq
0}\dfrac{1}{i!}f^{\ast i}\left(  x\right)  \right)  =\sum_{i\geq0}\dfrac
{1}{i!}\underbrace{p\left(  f^{\ast i}\left(  x\right)  \right)  }_{=\left(
p\circ\left(  f^{\ast i}\right)  \right)  \left(  x\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }p\text{ is }k\text{-linear, and since the}\\
\text{sum }\sum_{i\geq0}\dfrac{1}{i!}f^{\ast i}\left(  x\right)  \text{
converges}\\
\text{with respect to the discrete topology}%
\end{array}
\right) \\
&  =\sum_{i\geq0}\dfrac{1}{i!}\underbrace{\left(  p\circ\left(  f^{\ast
i}\right)  \right)  }_{\substack{=\left(  p\circ f\right)  ^{\ast
i}\\\text{(by Proposition~\ref{Proposition1711} \textbf{(c)},}\\\text{applied
to }n=i\text{)}}}\left(  x\right)  =\sum_{i\geq0}\dfrac{1}{i!}\left(  p\circ
f\right)  ^{\ast i}\left(  x\right)  =\sum_{i\geq0}\dfrac{\left(  p\circ
f\right)  ^{\ast i}\left(  x\right)  }{i!}\\
&  =e^{\ast\left(  p\circ f\right)  }\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }e^{\ast\left(  p\circ f\right)
}\left(  x\right)  =\sum_{i\geq0}\dfrac{\left(  p\circ f\right)  ^{\ast
i}\left(  x\right)  }{i!}\right)  .
\end{align*}
Thus, $e^{\ast\left(  p\circ f\right)  }\left(  x\right)  =p\left(  e^{\ast
f}\left(  x\right)  \right)  =\left(  p\circ e^{\ast f}\right)  \left(
x\right)  $.

Now, let us forget that we fixed $x$. We thus have shown that $e^{\ast\left(
p\circ f\right)  }\left(  x\right)  =\left(  p\circ e^{\ast f}\right)  \left(
x\right)  $ for every $x\in C$. In other words, $e^{\ast\left(  p\circ
f\right)  }=p\circ e^{\ast f}$. This proves Proposition~\ref{Proposition1711}
\textbf{(d)}.
\end{proof}

The following proposition neatly complements Proposition~\ref{Proposition1711}%
:\footnote{I have now realized that Proposition~\ref{Proposition1712} is just
a repetition of Proposition~\ref{Proposition1102} \textbf{(e)} (with $\psi$
renamed as $p$).}

\begin{proposition}
\label{Proposition1712} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered $k$-coalgebra. Let $A$ and $B$ be two $k$-algebras. Let
$p:A\rightarrow B$ be a $k$-algebra homomorphism. Let $F\in G\left(
C,A\right)  $. (See Definition~\ref{Definition8} for the meaning of $G\left(
C,A\right)  $.) Then, $p\circ F\in G\left(  C,B\right)  $ and $p\circ
\operatorname*{Log}F=\operatorname*{Log}\left(  p\circ F\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1712}.]By the definition of $G\left(
C,A\right)  $, we have%
\[
G\left(  C,A\right)  =\left\{  f\in\mathcal{L}\left(  C,A\right)
\ \mid\ f\left(  1_{C}\right)  =1_{A}\right\}  .
\]
By the definition of $G\left(  C,B\right)  $, we have%
\[
G\left(  C,B\right)  =\left\{  f\in\mathcal{L}\left(  C,B\right)
\ \mid\ f\left(  1_{C}\right)  =1_{B}\right\}  .
\]


We have $F\in G\left(  C,A\right)  =\left\{  f\in\mathcal{L}\left(
C,A\right)  \ \mid\ f\left(  1_{C}\right)  =1_{A}\right\}  $. In other words,
$F$ is an element $f\in\mathcal{L}\left(  C,A\right)  $ satisfying $f\left(
1_{C}\right)  =1_{A}$. In other words, $F$ is an element of $\mathcal{L}%
\left(  C,A\right)  $ and satisfies $F\left(  1_{C}\right)  =1_{A}$.

But $p$ is a $k$-algebra homomorphism, and therefore satisfies $p\left(
1_{A}\right)  =1_{B}$. Now, $\left(  p\circ F\right)  \left(  1_{C}\right)
=p\left(  \underbrace{F\left(  1_{C}\right)  }_{=1_{A}}\right)  =p\left(
1_{A}\right)  =1_{B}$. Hence, $p\circ F$ is an element of $\mathcal{L}\left(
C,B\right)  $ and satisfies $\left(  p\circ F\right)  \left(  1_{C}\right)
=1_{B}$. In other words, $p\circ F$ is an element $f\in\mathcal{L}\left(
C,B\right)  $ satisfying $f\left(  1_{C}\right)  =1_{B}$. In other words,%
\[
p\circ F\in\left\{  f\in\mathcal{L}\left(  C,B\right)  \ \mid\ f\left(
1_{C}\right)  =1_{B}\right\}  =G\left(  C,B\right)
\]
(since $G\left(  C,B\right)  =\left\{  f\in\mathcal{L}\left(  C,B\right)
\ \mid\ f\left(  1_{C}\right)  =1_{B}\right\}  $). Thus, $\operatorname*{Log}%
\left(  p\circ F\right)  $ is a well-defined $k$-linear map $C\rightarrow B$.

Consider the maps $e_{C,A}:C\rightarrow A$ and $e_{C,B}:C\rightarrow B$ that
are defined in the same way as the map $e_{H,A}:H\rightarrow A$ in
Definition~\ref{def.eHA}.

The definition of $\operatorname*{Log}\left(  p\circ F\right)  $ yields
$\operatorname*{Log}\left(  p\circ F\right)  =\operatorname{Log}_{1}\left(
p\circ F-e_{C,B}\right)  $. Thus, $\operatorname{Log}_{1}\left(  p\circ
F-e_{C,B}\right)  $ is well-defined; hence, $p\circ F-e_{C,B}\in
\mathfrak{g}\left(  C,B\right)  $ (since $\operatorname{Log}_{1}f$ is
well-defined for an $f\in\mathcal{L}\left(  C,B\right)  $ only when
$f\in\mathfrak{g}\left(  C,B\right)  $).

The definition of $\operatorname*{Log}F$ yields $\operatorname*{Log}%
F=\operatorname{Log}_{1}\left(  F-e_{C,A}\right)  $. Thus, $\operatorname{Log}%
_{1}\left(  F-e_{C,A}\right)  $ is well-defined; hence, $F-e_{C,A}%
\in\mathfrak{g}\left(  C,A\right)  $ (since $\operatorname{Log}_{1}f$ is
well-defined for an $f\in\mathcal{L}\left(  C,A\right)  $ only when
$f\in\mathfrak{g}\left(  C,A\right)  $).

Define an $f\in\mathcal{L}\left(  C,A\right)  $ by $f=F-e_{C,A}$. Then,
$f=F-e_{C,A}\in\mathfrak{g}\left(  C,A\right)  $. Also,%
\begin{align*}
p\circ\underbrace{f}_{=F-e_{C,A}}  &  =p\circ\left(  F-e_{C,A}\right)  =p\circ
F-\underbrace{p\circ e_{C,A}}_{\substack{=e_{C,B}\\\text{(by
Proposition~\ref{Proposition1711} \textbf{(a)})}}}=p\circ F-e_{C,B}\\
&  \in\mathfrak{g}\left(  C,B\right)  .
\end{align*}


Now let $x\in C$. Then, the definition of $\operatorname{Log}_{1}f$ yields
\begin{equation}
\left(  \operatorname{Log}_{1}f\right)  \left(  x\right)  =\sum_{i\geq1}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\in\mathfrak{g}\left(  C,A\right)
\right)  . \label{1712.pf.3}%
\end{equation}
In particular, the sum $\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}%
{i}f^{\ast i}\left(  x\right)  $ converges with respect to the discrete
topology. Applying the map $p$ to both sides of the equality (\ref{1712.pf.3}%
), we find%
\begin{equation}
p\left(  \left(  \operatorname{Log}_{1}f\right)  \left(  x\right)  \right)
=p\left(  \sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
x\right)  \right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}p\left(
f^{\ast i}\left(  x\right)  \right)  \label{1712.pf.4}%
\end{equation}
(since the map $p$ is $k$-linear, and since the sum $\sum_{i\geq1}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  x\right)  $ converges
with respect to the discrete topology). Also, the definition of
$\operatorname{Log}_{1}\left(  p\circ f\right)  $ yields%
\begin{align*}
\left(  \operatorname{Log}_{1}\left(  p\circ f\right)  \right)  \left(
x\right)   &  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}%
\underbrace{\left(  p\circ f\right)  ^{\ast i}}_{\substack{=p\circ\left(
f^{\ast i}\right)  \\\text{(since }p\circ\left(  f^{\ast i}\right)  =\left(
p\circ f\right)  ^{\ast i}\\\text{(by Proposition~\ref{Proposition1711}
\textbf{(c)},}\\\text{applied to }n=i\text{))}}}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p\circ f\in\mathfrak{g}\left(
C,B\right)  \right) \\
&  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{\left(
p\circ\left(  f^{\ast i}\right)  \right)  \left(  x\right)  }_{=p\left(
f^{\ast i}\left(  x\right)  \right)  }=\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}p\left(  f^{\ast i}\left(  x\right)  \right) \\
&  =p\left(  \left(  \operatorname{Log}_{1}f\right)  \left(  x\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1712.pf.4})}\right) \\
&  =\left(  p\circ\left(  \operatorname{Log}_{1}f\right)  \right)  \left(
x\right)  .
\end{align*}


Now, forget that we fixed $x$. We thus have shown that $\left(
\operatorname{Log}_{1}\left(  p\circ f\right)  \right)  \left(  x\right)
=\left(  p\circ\left(  \operatorname{Log}_{1}f\right)  \right)  \left(
x\right)  $ for each $x\in C$. In other words, $\operatorname{Log}_{1}\left(
p\circ f\right)  =p\circ\left(  \operatorname{Log}_{1}f\right)  $. Hence,
$p\circ\left(  \operatorname{Log}_{1}f\right)  =\operatorname{Log}_{1}\left(
p\circ f\right)  $. But
\begin{align*}
p\circ\left(  \underbrace{\operatorname*{Log}F}_{=\operatorname{Log}%
_{1}\left(  F-e_{C,A}\right)  }\right)   &  =p\circ\left(  \operatorname{Log}%
_{1}\left(  \underbrace{F-e_{C,A}}_{=f}\right)  \right)  =p\circ\left(
\operatorname{Log}_{1}f\right)  =\operatorname{Log}_{1}\left(
\underbrace{p\circ f}_{=p\circ F-e_{C,B}}\right) \\
&  =\operatorname{Log}_{1}\left(  p\circ F-e_{C,B}\right)
=\operatorname*{Log}\left(  p\circ F\right)
\end{align*}
(since we have shown that $\operatorname*{Log}\left(  p\circ F\right)
=\operatorname{Log}_{1}\left(  p\circ F-e_{C,B}\right)  $). This completes the
proof of Proposition~\ref{Proposition1712} (since $p\circ F\in G\left(
C,B\right)  $ has already been shown).
\end{proof}

The next proposition can be regarded as an addendum to
Proposition~\ref{Proposition275}:

\begin{proposition}
\label{Proposition1713} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered $k$-coalgebra. Let $A$ be a filtered $k$-algebra.

Let $F\in G\left(  C,A\right)  $ be a $k$-linear map respecting the
filtration. Then, the $k$-linear map $\operatorname*{Log}F$ also respects the filtration.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1713}.]We know that the map $F$ respects
the filtration. In other words,%
\begin{equation}
F\left(  C_{\leq n}\right)  \subseteq A_{\leq n}\ \ \ \ \ \ \ \ \ \ \text{for
each }n\in\mathbb{N}. \label{1713.pf.F-resp}%
\end{equation}


On the other hand, Proposition~\ref{Proposition275} \textbf{(a)} shows that
the map $e_{C,A}:C\rightarrow A$ respects the filtration. In other words,%
\begin{equation}
e_{C,A}\left(  C_{\leq n}\right)  \subseteq A_{\leq n}%
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}. \label{1713.pf.eCA-resp}%
\end{equation}


The definition of $\operatorname*{Log}F$ yields $\operatorname*{Log}%
F=\operatorname{Log}_{1}\left(  F-e_{C,A}\right)  $. In particular,
$\operatorname{Log}_{1}\left(  F-e_{C,A}\right)  $ is well-defined; thus,
$F-e_{C,A}\in\mathfrak{g}\left(  C,A\right)  $ (since $\operatorname{Log}%
_{1}f$ is well-defined for an $f\in\mathcal{L}\left(  C,A\right)  $ only when
$f\in\mathfrak{g}\left(  C,A\right)  $). Thus, we can define a $k$-linear map
$f\in\mathfrak{g}\left(  C,A\right)  $ by $f=F-e_{C,A}$. Consider this $f$. We
have $\operatorname*{Log}F=\operatorname{Log}_{1}\underbrace{\left(
F-e_{C,A}\right)  }_{=f}=\operatorname{Log}_{1}f$.

We have $f\left(  C_{\leq n}\right)  \subseteq A_{\leq n}$ for each
$n\in\mathbb{N}$\ \ \ \ \footnote{\textit{Proof.} Let $n\in\mathbb{N}$. Then,%
\begin{align*}
\underbrace{f}_{=F-e_{C,A}}\left(  C_{\leq n}\right)   &  =\left(
F-e_{C,A}\right)  \left(  C_{\leq n}\right)  \subseteq\underbrace{F\left(
C_{\leq n}\right)  }_{\substack{\subseteq A_{\leq n}\\\text{(by
(\ref{1713.pf.F-resp}))}}}-\underbrace{e_{C,A}\left(  C_{\leq n}\right)
}_{\substack{\subseteq A_{\leq n}\\\text{(by (\ref{1713.pf.eCA-resp}))}%
}}\subseteq A_{\leq n}-A_{\leq n}\\
&  \subseteq A_{\leq n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A_{\leq
n}\text{ is a }k\text{-vector space}\right)  .
\end{align*}
Qed.}. In other words, the map $f:C\rightarrow A$ respects the filtration.

Now, fix $i\in\mathbb{N}$. Thus, $i$ is a nonnegative integer. Recall that the
map $f:C\rightarrow A$ respects the filtration. Hence,
Proposition~\ref{Proposition275} \textbf{(c)} (applied to $n=i$) shows that
$f^{\ast i}$ also respects the filtration. In other words,
\begin{equation}
f^{\ast i}\left(  C_{\leq n}\right)  \subseteq A_{\leq n}%
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}. \label{1713.pf.fi-resp}%
\end{equation}


Now, forget that we fixed $i$. We thus have proven (\ref{1713.pf.fi-resp}) for
each $i\in\mathbb{N}$.

Fix $n\in\mathbb{N}$. Let $x\in C_{\leq n}$. Then, each integer $i>n$
satisfies%
\begin{equation}
f^{\ast i}\left(  x\right)  =0 \label{1713.pf.fix=0}%
\end{equation}
\footnote{\textit{Proof of (\ref{1713.pf.fix=0}):} Let $i>n$ be an integer.
Then, Remark~\ref{Remark9} (applied to $H=C$) yields $f^{\ast i}\left(
C_{\leq n}\right)  =0$ (since $i>n$). But from $x\in C_{\leq n}$, we obtain
$f^{\ast i}\left(  x\right)  \in f^{\ast i}\left(  C_{\leq n}\right)  =0$,
thus $f^{\ast i}\left(  x\right)  =0$. This proves (\ref{1713.pf.fix=0}).}.
Now,%
\begin{align*}
\underbrace{\left(  \operatorname*{Log}F\right)  }_{=\operatorname{Log}_{1}%
f}\left(  x\right)   &  =\left(  \operatorname{Log}_{1}f\right)  \left(
x\right)  =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{log})}\right) \\
&  =\underbrace{\sum_{\substack{i\geq1;\\i\leq n}}}_{=\sum_{i=1}^{n}}%
\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(  \underbrace{x}_{\in
C_{\leq n}}\right)  +\sum_{\substack{i\geq1;\\i>n}}\dfrac{\left(  -1\right)
^{i-1}}{i}\underbrace{f^{\ast i}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{1713.pf.fix=0}))}}}\\
&  \in\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}f^{\ast i}\left(
C_{\leq n}\right)  +\underbrace{\sum_{\substack{i\geq1;\\i>n}}\dfrac{\left(
-1\right)  ^{i-1}}{i}0}_{=0}=\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}%
}{i}\underbrace{f^{\ast i}\left(  C_{\leq n}\right)  }_{\substack{\subseteq
A_{\leq n}\\\text{(by (\ref{1713.pf.fi-resp}))}}}\\
&  \subseteq\sum_{i=1}^{n}\dfrac{\left(  -1\right)  ^{i-1}}{i}A_{\leq
n}\subseteq A_{\leq n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A_{\leq
n}\text{ is a }k\text{-vector space}\right)  .
\end{align*}


Now, forget that we fixed $x$. We thus have proven that $\left(
\operatorname*{Log}F\right)  \left(  x\right)  \in A_{\leq n}$ for each $x\in
C_{\leq n}$. In other words, $\left(  \operatorname*{Log}F\right)  \left(
C_{\leq n}\right)  \subseteq A_{\leq n}$.

Now, forget that we fixed $n$. We thus have shown that $\left(
\operatorname*{Log}F\right)  \left(  C_{\leq n}\right)  \subseteq A_{\leq n}$
for each $n\in\mathbb{N}$. In other words, the map $\operatorname*{Log}F$
respects the filtration. This proves Proposition~\ref{Proposition1713}.
\end{proof}

\begin{proposition}
\label{Proposition1714} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered commutative bialgebra over $k$. Consider the convolution
algebra $\mathcal{L}\left(  H,H\right)  $. Let $\mathfrak{e}$ denote the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $.

\textbf{(a)} We have $\operatorname*{id}\nolimits^{\ast n}\circ\mathfrak{e}%
=n\mathfrak{e}$ for each $n\in\mathbb{N}$.

\textbf{(b)} We have $\mathfrak{e}^{\ast n}\circ\mathfrak{e}=\delta
_{n,1}\mathfrak{e}$ for each $n\in\mathbb{N}$. Here, we are using the
\textit{Kronecker delta notation} (i.e., whenever $u$ and $v$ are two objects,
we set $\delta_{u,v}=%
\begin{cases}
1, & \text{if }u=v;\\
0, & \text{if }u\neq v
\end{cases}
$).
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1714}.]Recall that $\operatorname*{Log}%
F\in\mathfrak{g}\left(  H,H\right)  $ for each $F\in G\left(  H,H\right)  $
(by the definition of $\operatorname*{Log}F$). Applying this to
$F=\operatorname*{id}$, we obtain $\operatorname*{Log}\operatorname*{id}%
\in\mathfrak{g}\left(  H,H\right)  $. Thus, $\mathfrak{e}=\operatorname*{Log}%
\operatorname*{id}\in\mathfrak{g}\left(  H,H\right)  $. Hence, $\mathfrak{e}%
\left(  1_{H}\right)  =0$\ \ \ \ \footnote{\textit{Proof.} We have
$\mathfrak{e}\in\mathfrak{g}\left(  H,H\right)  =\left\{  f\in\mathcal{L}%
\left(  H,H\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}  $ (by the
definition of $\mathfrak{g}\left(  H,H\right)  $). In other words,
$\mathfrak{e}$ is an element $f\in\mathcal{L}\left(  H,H\right)  $ satisfying
$f\left(  1_{H}\right)  =0$. In other words, $\mathfrak{e}$ is an element of
$\mathcal{L}\left(  H,H\right)  $ and satisfies $\mathfrak{e}\left(
1_{H}\right)  =0$. Qed.}.

Proposition~\ref{Proposition13} \textbf{(b)} (applied to $A=H$ and
$F=\operatorname*{id}$) yields $e^{\ast\left(  \operatorname*{Log}%
\operatorname*{id}\right)  }=\operatorname*{id}$. Since $\mathfrak{e}%
=\operatorname*{Log}\operatorname*{id}$, this rewrites as $e^{\ast
\mathfrak{e}}=\operatorname*{id}$.

\textbf{(a)} Let $n\in\mathbb{N}$. Clearly, $\operatorname*{id}:H\rightarrow
H$ is a $k$-algebra homomorphism. Hence, Corollary~\ref{Corollary136} (applied
to $A=H$ and $f=\operatorname*{id}$) shows that $\operatorname*{id}%
\nolimits^{\ast n}:H\rightarrow H$ is a $k$-algebra homomorphism.

Also, $\operatorname*{id}:H\rightarrow H$ is an element of $G\left(
H,H\right)  $.

But $H$ is a connected filtered $k$-coalgebra (since $H$ is a connected
filtered bialgebra over $k$). Hence, Proposition~\ref{Proposition1712}
(applied to $C=H$, $A=H$, $B=H$, $p=\operatorname*{id}\nolimits^{\ast n}$ and
$F=\operatorname*{id}$) shows that $\operatorname*{id}\nolimits^{\ast n}%
\circ\operatorname*{id}\in G\left(  H,H\right)  $ and $\operatorname*{id}%
\nolimits^{\ast n}\circ\operatorname*{Log}\operatorname*{id}%
=\operatorname*{Log}\left(  \operatorname*{id}\nolimits^{\ast n}%
\circ\operatorname*{id}\right)  $. Thus,%
\begin{equation}
\operatorname*{id}\nolimits^{\ast n}\circ\underbrace{\mathfrak{e}%
}_{=\operatorname*{Log}\operatorname*{id}}=\operatorname*{id}\nolimits^{\ast
n}\circ\operatorname*{Log}\operatorname*{id}=\operatorname*{Log}\left(
\underbrace{\operatorname*{id}\nolimits^{\ast n}\circ\operatorname*{id}%
}_{=\operatorname*{id}\nolimits^{\ast n}}\right)  =\operatorname*{Log}\left(
\operatorname*{id}\nolimits^{\ast n}\right)  . \label{1714.pf.a.1}%
\end{equation}


On the other hand, $n\underbrace{\mathfrak{e}}_{\in\mathfrak{g}\left(
H,H\right)  }\in n\mathfrak{g}\left(  H,H\right)  \subseteq\mathfrak{g}\left(
H,H\right)  $ (since $\mathfrak{g}\left(  H,H\right)  $ is a $k$-vector space).

Recall that $\mathfrak{e}\in\mathfrak{g}\left(  H,H\right)  $. Thus,
Corollary~\ref{Corollary48} (applied to $C=H$ and $f=\mathfrak{e}$) yields
$e^{\ast\left(  n\mathfrak{e}\right)  }=\left(  e^{\ast\mathfrak{e}}\right)
^{\ast n}=\operatorname*{id}\nolimits^{\ast n}$ (since $e^{\ast\mathfrak{e}%
}=\operatorname*{id}$). Thus, $\operatorname*{Log}\left(  \underbrace{e^{\ast
\left(  n\mathfrak{e}\right)  }}_{=\operatorname*{id}\nolimits^{\ast n}%
}\right)  =\operatorname*{Log}\left(  \operatorname*{id}\nolimits^{\ast
n}\right)  =\operatorname*{id}\nolimits^{\ast n}\circ\mathfrak{e}$ (by
(\ref{1714.pf.a.1})). But Proposition~\ref{Proposition13} \textbf{(a)}
(applied to $A=H$ and $f=n\mathfrak{e}$) yields $\operatorname*{Log}\left(
e^{\ast\left(  n\mathfrak{e}\right)  }\right)  =n\mathfrak{e}$ (since
$n\mathfrak{e}\in\mathfrak{g}\left(  H,H\right)  $). Comparing this with
$\operatorname*{Log}\left(  e^{\ast\left(  n\mathfrak{e}\right)  }\right)
=\operatorname*{id}\nolimits^{\ast n}\circ\mathfrak{e}$, we obtain
$\operatorname*{id}\nolimits^{\ast n}\circ\mathfrak{e}=n\mathfrak{e}$. This
proves Proposition~\ref{Proposition1714} \textbf{(a)}.

\textbf{(b)} For each $n\in\mathbb{N}$, we define a $k$-linear map
$g_{n}:H\rightarrow H$ by $g_{n}=\mathfrak{e}^{\ast n}\circ\mathfrak{e}%
-\delta_{n,1}\mathfrak{e}$. Then, $g_{n}\left(  H_{\leq n-1}\right)  =0$ for
each $n\in\mathbb{N}$\ \ \ \ \footnote{\textit{Proof.} Let $n\in\mathbb{N}$.
We must show that $g_{n}\left(  H_{\leq n-1}\right)  =0$.
\par
We have $n\in\mathbb{N}$. Hence, we are in one of the following three cases:
\par
\textit{Case 1:} We have $n=0$.
\par
\textit{Case 2:} We have $n=1$.
\par
\textit{Case 3:} We have $n>1$.
\par
Let us first consider Case 1. In this case, we have $n=0$. Thus, $H_{\leq
n-1}=H_{\leq0-1}=H_{\leq-1}=0$. Hence, $g_{n}\left(  \underbrace{H_{\leq n-1}%
}_{=0}\right)  =g_{n}\left(  0\right)  =0$ (since the map $g_{n}$ is
$k$-linear). Thus, $g_{n}\left(  H_{\leq n-1}\right)  =0$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $n=1$. Thus, $H_{\leq
n-1}=H_{\leq1-1}=H_{\leq0}$.
\par
But the filtered $k$-bialgebra $H$ is connected if and only if $H_{\leq
0}=k\cdot1_{H}$ (by Remark~\ref{Remark7m}). Thus, $H_{\leq0}=k\cdot1_{H}$
(since the $k$-bialgebra $H$ is connected). Hence, $H_{\leq n-1}=H_{\leq
0}=k\cdot1_{H}$.
\par
Now,
\begin{align*}
\underbrace{g_{n}}_{=\mathfrak{e}^{\ast n}\circ\mathfrak{e}-\delta_{n,1}%
\circ\mathfrak{e}}\left(  1_{H}\right)   &  =\left(  \mathfrak{e}^{\ast
n}\circ\mathfrak{e}-\delta_{n,1}\circ\mathfrak{e}\right)  \left(
1_{H}\right)  =\underbrace{\left(  \mathfrak{e}^{\ast n}\circ\mathfrak{e}%
\right)  \left(  1_{H}\right)  }_{=\mathfrak{e}^{\ast n}\left(  \mathfrak{e}%
\left(  1_{H}\right)  \right)  }-\delta_{n,1}\underbrace{\mathfrak{e}\left(
1_{H}\right)  }_{=0}\\
&  =\mathfrak{e}^{\ast n}\left(  \mathfrak{e}\left(  1_{H}\right)  \right)
-\underbrace{\delta_{n,1}0}_{=0}=\mathfrak{e}^{\ast n}\left(
\underbrace{\mathfrak{e}\left(  1_{H}\right)  }_{=0}\right)  =\mathfrak{e}%
^{\ast n}\left(  0\right)  =0
\end{align*}
(since the map $\mathfrak{e}^{\ast n}$ is $k$-linear). Finally,%
\begin{align*}
g_{n}\left(  \underbrace{H_{\leq n-1}}_{=k\cdot1_{H}}\right)   &
=g_{n}\left(  k\cdot1_{H}\right)  =k\cdot\underbrace{g_{n}\left(
1_{H}\right)  }_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
g_{n}\text{ is }k\text{-linear}\right) \\
&  =0.
\end{align*}
Thus, $g_{n}\left(  H_{\leq n-1}\right)  =0$ is proven in Case 2.
\par
Let us finally consider Case 3. In this case, we have $n>1$. Thus, $n\geq2$
(since $n\in\mathbb{N}$), so that $n-1\geq1$ and thus $n-1\in\mathbb{N}$. Now,
Remark~\ref{Remark9} (applied to $H$, $n$, $n-1$ and $\mathfrak{e}$ instead of
$A$, $i$, $n$ and $f$) shows that $\mathfrak{e}^{\ast n}\left(  H_{\leq
n-1}\right)  =0$ (since $\mathfrak{e}\in\mathfrak{g}\left(  H,H\right)  $ and
$n>n-1$). Also, $n\neq1$ (since $n>1$) and thus $\delta_{n,1}=0$.
\par
Furthermore, the map $\operatorname*{id}\in G\left(  H,H\right)  $ is a
$k$-linear map respecting the filtration. Hence,
Proposition~\ref{Proposition1713} (applied to $C=H$, $A=H$ and
$F=\operatorname*{id}$) shows that the $k$-linear map $\operatorname*{Log}%
\operatorname*{id}$ also respects the filtration. In other words, the
$k$-linear map $\mathfrak{e}$ respects the filtration (since $\mathfrak{e}%
=\operatorname*{Log}\operatorname*{id}$). In other words, $\mathfrak{e}\left(
H_{\leq m}\right)  \subseteq H_{\leq m}$ for each $m\in\mathbb{N}$. Applying
this to $m=n-1$, we find $\mathfrak{e}\left(  H_{\leq n-1}\right)  \subseteq
H_{\leq n-1}$.
\par
Now,%
\begin{align*}
\underbrace{g_{n}}_{=\mathfrak{e}^{\ast n}\circ\mathfrak{e}-\delta_{n,1}%
\circ\mathfrak{e}}\left(  H_{\leq n-1}\right)   &  =\left(  \mathfrak{e}^{\ast
n}\circ\mathfrak{e}-\delta_{n,1}\circ\mathfrak{e}\right)  \left(  H_{\leq
n-1}\right)  \subseteq\underbrace{\left(  \mathfrak{e}^{\ast n}\circ
\mathfrak{e}\right)  \left(  H_{\leq n-1}\right)  }_{=\mathfrak{e}^{\ast
n}\left(  \mathfrak{e}\left(  H_{\leq n-1}\right)  \right)  }%
-\underbrace{\left(  \delta_{n,1}\circ\mathfrak{e}\right)  \left(  H_{\leq
n-1}\right)  }_{=\delta_{n,1}\mathfrak{e}\left(  H_{\leq n-1}\right)  }\\
&  =\mathfrak{e}^{\ast n}\left(  \underbrace{\mathfrak{e}\left(  H_{\leq
n-1}\right)  }_{\subseteq H_{\leq n-1}}\right)  -\underbrace{\delta_{n,1}%
}_{=0}\mathfrak{e}\left(  H_{\leq n-1}\right)  \subseteq
\underbrace{\mathfrak{e}^{\ast n}\left(  H_{\leq n-1}\right)  }_{=0}%
-\underbrace{0\mathfrak{e}\left(  H_{\leq n-1}\right)  }_{=0}=0-0=0.
\end{align*}
Thus, $g_{n}\left(  H_{\leq n-1}\right)  =0$ is proven in Case 3.
\par
We have now proven $g_{n}\left(  H_{\leq n-1}\right)  =0$ in all three Cases
1, 2 and 3. Since these three Cases cover all possibilities, we therefore
conclude that $g_{n}\left(  H_{\leq n-1}\right)  =0$ always holds. Qed.}.

For each $n\in\mathbb{N}$, define a $k$-linear map $h_{n}:H\rightarrow H$ by
$h_{n}=\dfrac{1}{n!}g_{n}$. Then, $h_{n}\left(  H_{\leq n-1}\right)  =0$ for
each $n\in\mathbb{N}$\ \ \ \ \footnote{\textit{Proof.} We have previously
shown that $g_{n}\left(  H_{\leq n-1}\right)  =0$ for each $n\in\mathbb{N}$.
Now, for each $n\in\mathbb{N}$, we have%
\[
\underbrace{h_{n}}_{=\dfrac{1}{n!}g_{n}}\left(  H_{\leq n-1}\right)  =\left(
\dfrac{1}{n!}g_{n}\right)  \left(  H_{\leq n-1}\right)  =\dfrac{1}%
{n!}\underbrace{g_{n}\left(  H_{\leq n-1}\right)  }_{=0}=0.
\]
Qed.}.

For each $i\in\mathbb{N}$, we have%
\begin{align}
h_{i}  &  =\dfrac{1}{i!}\underbrace{g_{i}}_{\substack{=\mathfrak{e}^{\ast
i}\circ\mathfrak{e}-\delta_{i,1}\mathfrak{e}\\\text{(by the definition of
}g_{i}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
h_{i}\right) \nonumber\\
&  =\dfrac{1}{i!}\left(  \mathfrak{e}^{\ast i}\circ\mathfrak{e}-\delta
_{i,1}\mathfrak{e}\right)  . \label{1714.pf.b.hi=}%
\end{align}


Now, let $t\in\mathbb{N}$. Recall that $\mathfrak{e}\in\mathfrak{g}\left(
H,H\right)  $. Thus, Corollary~\ref{Corollary48} (applied to $C=H$, $n=t$ and
$f=\mathfrak{e}$) yields $e^{\ast\left(  t\mathfrak{e}\right)  }=\left(
e^{\ast\mathfrak{e}}\right)  ^{\ast t}=\operatorname*{id}\nolimits^{\ast t}$
(since $e^{\ast\mathfrak{e}}=\operatorname*{id}$).

On the other hand, $t\underbrace{\mathfrak{e}}_{\in\mathfrak{g}\left(
H,H\right)  }\in t\mathfrak{g}\left(  H,H\right)  \subseteq\mathfrak{g}\left(
H,H\right)  $ (since $\mathfrak{g}\left(  H,H\right)  $ is a $k$-vector space).

Let $x\in H$. Set $y=\mathfrak{e}\left(  x\right)  $.
Proposition~\ref{Proposition1714} \textbf{(a)} (applied to $n=t$) yields
$\operatorname*{id}\nolimits^{\ast t}\circ\mathfrak{e}=t\mathfrak{e}$. Thus,%
\[
\left(  \operatorname*{id}\nolimits^{\ast t}\circ\mathfrak{e}\right)  \left(
x\right)  =\left(  t\mathfrak{e}\right)  \left(  x\right)
=t\underbrace{\mathfrak{e}\left(  x\right)  }_{=y}=ty,
\]
so that%
\begin{align}
ty  &  =\left(  \operatorname*{id}\nolimits^{\ast t}\circ\mathfrak{e}\right)
\left(  x\right)  =\underbrace{\operatorname*{id}\nolimits^{\ast t}}%
_{=e^{\ast\left(  t\mathfrak{e}\right)  }}\left(  \underbrace{\mathfrak{e}%
\left(  x\right)  }_{=y}\right)  =e^{\ast\left(  t\mathfrak{e}\right)
}\left(  y\right) \nonumber\\
&  =\sum_{i\geq0}\dfrac{\left(  t\mathfrak{e}\right)  ^{\ast i}\left(
y\right)  }{i!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{exp}), applied to
}t\mathfrak{e}\text{ and }y\text{ instead of }f\text{ and }x\right)
\nonumber\\
&  =\sum_{i\geq0}\dfrac{1}{i!}\underbrace{\left(  t\mathfrak{e}\right)  ^{\ast
i}}_{=t^{i}\mathfrak{e}^{\ast i}}\left(  y\right)  =\sum_{i\geq0}\dfrac{1}%
{i!}\underbrace{\left(  t^{i}\mathfrak{e}^{\ast i}\right)  \left(  y\right)
}_{=t^{i}\mathfrak{e}^{\ast i}\left(  y\right)  }=\sum_{i\geq0}\dfrac{1}%
{i!}t^{i}\mathfrak{e}^{\ast i}\left(  \underbrace{y}_{=\mathfrak{e}\left(
x\right)  }\right) \nonumber\\
&  =\sum_{i\geq0}\dfrac{1}{i!}t^{i}\underbrace{\mathfrak{e}^{\ast i}\left(
\mathfrak{e}\left(  x\right)  \right)  }_{=\left(  \mathfrak{e}^{\ast i}%
\circ\mathfrak{e}\right)  \left(  x\right)  }=\sum_{i\geq0}\dfrac{1}{i!}%
t^{i}\left(  \mathfrak{e}^{\ast i}\circ\mathfrak{e}\right)  \left(  x\right)
. \label{1714.pf.b.4}%
\end{align}
On the other hand, all but finitely many $i\in\mathbb{N}$ satisfy $\dfrac
{1}{i!}t^{i}\delta_{i,1}\mathfrak{e}\left(  x\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} Each $i\in\mathbb{N}$ satisfying $i\neq1$
satisfies $\dfrac{1}{i!}t^{i}\underbrace{\delta_{i,1}}%
_{\substack{=0\\\text{(since }i\neq1\text{)}}}\mathfrak{e}\left(  x\right)
=0$. Hence, all but finitely many $i\in\mathbb{N}$ satisfy $\dfrac{1}{i!}%
t^{i}\delta_{i,1}\mathfrak{e}\left(  x\right)  =0$ (since all but finitely
many $i\in\mathbb{N}$ satisfy $i\neq1$).}. Hence, the infinite sum
$\sum_{i\geq0}\dfrac{1}{i!}t^{i}\delta_{i,1}\mathfrak{e}\left(  x\right)  $
converges. This infinite sum rewrites as follows:%
\begin{align*}
\sum_{i\geq0}\dfrac{1}{i!}t^{i}\delta_{i,1}\mathfrak{e}\left(  x\right)   &
=\sum_{\substack{i\geq0;\\i\neq1}}\dfrac{1}{i!}t^{i}\underbrace{\delta_{i,1}%
}_{\substack{=0\\\text{(since }i\neq1\text{)}}}\mathfrak{e}\left(  x\right)
+\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}=1}\underbrace{t^{1}}_{=t}%
\underbrace{\delta_{1,1}}_{\substack{=1\\\text{(since }1=1\text{)}%
}}\underbrace{\mathfrak{e}\left(  x\right)  }_{=y}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=1\text{ from the sum}\right) \\
&  =\underbrace{\sum_{\substack{i\geq0;\\i\neq1}}\dfrac{1}{i!}t^{i}%
0\mathfrak{e}\left(  x\right)  }_{=0}+ty=ty\\
&  =\sum_{i\geq0}\dfrac{1}{i!}t^{i}\left(  \mathfrak{e}^{\ast i}%
\circ\mathfrak{e}\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{1714.pf.b.4})}\right)  .
\end{align*}
Hence,%
\begin{align*}
0  &  =\sum_{i\geq0}\dfrac{1}{i!}t^{i}\left(  \mathfrak{e}^{\ast i}%
\circ\mathfrak{e}\right)  \left(  x\right)  -\sum_{i\geq0}\dfrac{1}{i!}%
t^{i}\delta_{i,1}\mathfrak{e}\left(  x\right)  =\sum_{i\geq0}%
\underbrace{\dfrac{1}{i!}t^{i}}_{=t^{i}\cdot\dfrac{1}{i!}}\underbrace{\left(
\left(  \mathfrak{e}^{\ast i}\circ\mathfrak{e}\right)  \left(  x\right)
-\delta_{i,1}\mathfrak{e}\left(  x\right)  \right)  }_{=\left(  \mathfrak{e}%
^{\ast i}\circ\mathfrak{e}-\delta_{i,1}\mathfrak{e}\right)  \left(  x\right)
}\\
&  =\sum_{i\geq0}t^{i}\cdot\underbrace{\dfrac{1}{i!}\left(  \mathfrak{e}^{\ast
i}\circ\mathfrak{e}-\delta_{i,1}\mathfrak{e}\right)  \left(  x\right)
}_{=\left(  \dfrac{1}{i!}\left(  \mathfrak{e}^{\ast i}\circ\mathfrak{e}%
-\delta_{i,1}\mathfrak{e}\right)  \right)  \left(  x\right)  }=\sum_{i\geq
0}t^{i}\cdot\underbrace{\left(  \dfrac{1}{i!}\left(  \mathfrak{e}^{\ast
i}\circ\mathfrak{e}-\delta_{i,1}\mathfrak{e}\right)  \right)  }%
_{\substack{=h_{i}\\\text{(by (\ref{1714.pf.b.hi=}))}}}\left(  x\right) \\
&  =\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  .
\end{align*}
In other words, $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)  =0$.

Now, forget that we fixed $x$ and $t$. We thus have shown that every $x\in H$
and every $t\in\mathbb{N}$ satisfy $\sum_{i\geq0}t^{i}h_{i}\left(  x\right)
=0$. Hence, Proposition~\ref{Proposition49} \textbf{(b)} shows that%
\begin{equation}
h_{n}=0\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}
\label{1714.pf.b.at}%
\end{equation}
(since we already have proven that $h_{n}\left(  H_{\leq n-1}\right)  =0$ for
each $n\in\mathbb{N}$).

Now, fix $n\in\mathbb{N}$. From (\ref{1714.pf.b.at}), we obtain $h_{n}=0$. But
the definition of $h_{n}$ yields $h_{n}=\dfrac{1}{n!}g_{n}$, so that
$g_{n}=n!\underbrace{h_{n}}_{=0}=0$. Furthermore, the definition of $g_{n}$
yields $g_{n}=\mathfrak{e}^{\ast n}\circ\mathfrak{e}-\delta_{n,1}\mathfrak{e}%
$, so that $\mathfrak{e}^{\ast n}\circ\mathfrak{e}-\delta_{n,1}\mathfrak{e}%
=g_{n}=0$, and thus $\mathfrak{e}^{\ast n}\circ\mathfrak{e}=\delta
_{n,1}\mathfrak{e}$. This proves Proposition~\ref{Proposition1714}
\textbf{(b)}.
\end{proof}

\begin{proposition}
\label{Proposition1715} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered bialgebra over $k$. Consider the convolution algebra
$\mathcal{L}\left(  H,H\right)  $. Let $\mathfrak{e}$ denote the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $.
Then, the image $\mathfrak{e}\left(  H\right)  $ generates the $k$-algebra $H$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1715}.]Let $\mathfrak{E}$ be the
$k$-vector subspace $\mathfrak{e}\left(  H\right)  $ of $H$.

Recall Convention~\ref{Convention230}. For every $\ell\in\mathbb{N}$,
Proposition~\ref{Proposition272a} yields%
\[
H_{\leq\ell}\subseteq\sum_{i=0}^{\ell}\underbrace{\mathfrak{E}^{i}}%
_{\subseteq\operatorname{AlgGen}_{k}\mathfrak{E}}\subseteq\sum_{i=0}^{\ell
}\operatorname{AlgGen}_{k}\mathfrak{E}\subseteq\operatorname{AlgGen}%
_{k}\mathfrak{E}%
\]
(since $\operatorname{AlgGen}_{k}\mathfrak{E}$ is a $k$-vector space). Since
$H$ is filtered, we have%
\[
H=\bigcup_{\ell\in\mathbb{N}}\underbrace{H_{\leq\ell}}_{\subseteq
\operatorname{AlgGen}_{k}\mathfrak{E}}\subseteq\bigcup_{\ell\in\mathbb{N}%
}\operatorname{AlgGen}_{k}\mathfrak{E}=\operatorname{AlgGen}_{k}\mathfrak{E}.
\]
Combined with $\operatorname{AlgGen}_{k}\mathfrak{E}\subseteq H$ (which is
trivial), this yields $H=\operatorname{AlgGen}_{k}\mathfrak{E}$. In other
words, $H$ is the $k$-subalgebra of $H$ generated by $\mathfrak{E}$ (since
$\operatorname{AlgGen}_{k}\mathfrak{E}$ is the $k$-subalgebra of $H$ generated
by $\mathfrak{E}$). In other words, the subset $\mathfrak{E}$ of $H$ generates
the $k$-algebra $H$. Since $\mathfrak{E}=\mathfrak{e}\left(  H\right)  $, this
rewrites as follows: The subset $\mathfrak{e}\left(  H\right)  $ of $H$
generates the $k$-algebra $H$. This proves Proposition~\ref{Proposition1715}.
\end{proof}

\begin{lemma}
\label{Lemma1721} Let $k$ be a field. Let $V$ be any $k$-vector space. Recall
the notations introduced in Definition~\ref{Definition1701}.

Let $A$ be a commutative $k$-algebra. Let $\varphi:V\rightarrow A$ be a
$k$-linear map.

\textbf{(a)} If the image $\varphi\left(  V\right)  $ generates the
$k$-algebra $A$, then the $k$-algebra homomorphism $\operatorname*{symlift}%
\varphi:\operatorname*{Sym}V\rightarrow A$ is surjective.

\textbf{(b)} Let $\rho:A\rightarrow\operatorname*{Sym}V$ be a $k$-algebra
homomorphism such that $\rho\circ\varphi=\operatorname*{syminc}\nolimits_{V}$.
Then, $\rho\circ\left(  \operatorname*{symlift}\varphi\right)
=\operatorname{id}_{\operatorname*{Sym}V}$.

\textbf{(c)} Let $\rho:A\rightarrow\operatorname*{Sym}V$ be a $k$-algebra
homomorphism such that $\rho\circ\varphi=\operatorname*{syminc}\nolimits_{V}$.
Assume furthermore that the image $\varphi\left(  V\right)  $ generates the
$k$-algebra $A$. Then, the two $k$-algebra homomorphisms
$\operatorname*{symlift}\varphi:\operatorname*{Sym}V\rightarrow A$ and
$\rho:A\rightarrow\operatorname*{Sym}V$ are mutually inverse.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1721}.]Recall that $\operatorname*{symlift}\varphi$
is the unique $k$-algebra homomorphism $\Phi:\operatorname*{Sym}V\rightarrow
A$ satisfying $\Phi\circ\operatorname*{syminc}\nolimits_{V}=\varphi$. Hence,
$\operatorname*{symlift}\varphi$ is a $k$-algebra homomorphism
$\operatorname*{Sym}V\rightarrow A$ and satisfies $\left(
\operatorname*{symlift}\varphi\right)  \circ\operatorname*{syminc}%
\nolimits_{V}=\varphi$.

\textbf{(a)} Assume that the image $\varphi\left(  V\right)  $ generates the
$k$-algebra $A$. In other words, the $k$-subalgebra of $A$ generated by
$\varphi\left(  V\right)  $ is $A$.

Let $S$ denote the image $\varphi\left(  V\right)  $.

Recall that the $k$-subalgebra of $A$ generated by $\varphi\left(  V\right)  $
is $A$. In other words, the $k$-subalgebra of $A$ generated by $S$ is $A$
(since $S=\varphi\left(  V\right)  $). In other words, $A$ is the
$k$-subalgebra of $A$ generated by $S$. In other words, $A$ is the smallest
$k$-subalgebra of $A$ which contains $S$ as a subset (because the
$k$-subalgebra of $A$ generated by $S$ is the smallest $k$-subalgebra of $A$
which contains $S$ as a subset). Hence,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{whenever }U\text{ is a }k\text{-subalgebra of }A\text{ which contains
}S\text{ as}\\
\text{a subset, we must necessarily have }A\subseteq U
\end{array}
\right)  . \label{1721.pf.a.AinU}%
\end{equation}


But the map $\operatorname*{symlift}\varphi:\operatorname*{Sym}V\rightarrow A$
is a $k$-algebra homomorphism. Thus, its image $\left(
\operatorname*{symlift}\varphi\right)  \left(  \operatorname*{Sym}V\right)  $
is a $k$-subalgebra of $A$. Furthermore, this $k$-subalgebra $\left(
\operatorname*{symlift}\varphi\right)  \left(  \operatorname*{Sym}V\right)  $
contains $S$ as a subset\footnote{\textit{Proof.} We have%
\begin{align*}
S  &  =\underbrace{\varphi}_{=\left(  \operatorname*{symlift}\varphi\right)
\circ\operatorname*{syminc}\nolimits_{V}}\left(  V\right)  =\left(  \left(
\operatorname*{symlift}\varphi\right)  \circ\operatorname*{syminc}%
\nolimits_{V}\right)  \left(  V\right)  =\left(  \operatorname*{symlift}%
\varphi\right)  \left(  \underbrace{\operatorname*{syminc}\nolimits_{V}\left(
V\right)  }_{\subseteq\operatorname*{Sym}V}\right) \\
&  \subseteq\left(  \operatorname*{symlift}\varphi\right)  \left(
\operatorname*{Sym}V\right)  .
\end{align*}
In other words, $\left(  \operatorname*{symlift}\varphi\right)  \left(
\operatorname*{Sym}V\right)  $ contains $S$ as a subset.}. Hence,
(\ref{1721.pf.a.AinU}) (applied to $U=\left(  \operatorname*{symlift}%
\varphi\right)  \left(  \operatorname*{Sym}V\right)  $) shows that
$A\subseteq\left(  \operatorname*{symlift}\varphi\right)  \left(
\operatorname*{Sym}V\right)  $. Combined with $\left(  \operatorname*{symlift}%
\varphi\right)  \left(  \operatorname*{Sym}V\right)  \subseteq A$ (which is
obvious), this yields $A=\left(  \operatorname*{symlift}\varphi\right)
\left(  \operatorname*{Sym}V\right)  $. In other words, the map
$\operatorname*{symlift}\varphi$ is surjective. This proves
Lemma~\ref{Lemma1721} \textbf{(a)}.

\textbf{(b)} The $k$-algebra $\operatorname*{Sym}V$ is commutative, and the
map $\operatorname*{syminc}\nolimits_{V}:V\rightarrow\operatorname*{Sym}V$ is
a $k$-linear map. Hence, a $k$-algebra homomorphism $\operatorname*{symlift}%
\left(  \operatorname*{syminc}\nolimits_{V}\right)  :\operatorname*{Sym}%
V\rightarrow\operatorname*{Sym}V$ is well-defined. The definition of this
homomorphism shows that $\operatorname*{symlift}\left(  \operatorname*{syminc}%
\nolimits_{V}\right)  $ is the unique $k$-algebra homomorphism $\Phi
:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$ satisfying $\Phi
\circ\operatorname*{syminc}\nolimits_{V}=\operatorname*{syminc}\nolimits_{V}$.
In particular, this shows that there exists \textbf{at most one} $k$-algebra
homomorphism $\Phi:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$
satisfying $\Phi\circ\operatorname*{syminc}\nolimits_{V}%
=\operatorname*{syminc}\nolimits_{V}$ (namely, the homomorphism
$\operatorname*{symlift}\left(  \operatorname*{syminc}\nolimits_{V}\right)
$). In other words, if $\Phi_{1}$ and $\Phi_{2}$ are two $k$-algebra
homomorphisms $\Phi:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$
satisfying $\Phi\circ\operatorname*{syminc}\nolimits_{V}%
=\operatorname*{syminc}\nolimits_{V}$, then%
\begin{equation}
\Phi_{1}=\Phi_{2}. \label{1721.pf.b.eq}%
\end{equation}


Now, $\operatorname{id}_{\operatorname*{Sym}V}$ is a $k$-algebra homomorphism
$\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$ satisfying
$\operatorname{id}_{\operatorname*{Sym}V}\circ\operatorname*{syminc}%
\nolimits_{V}=\operatorname*{syminc}\nolimits_{V}$. In other words,
$\operatorname{id}_{\operatorname*{Sym}V}$ is a $k$-algebra homomorphism
$\Phi:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$ satisfying
$\Phi\circ\operatorname*{syminc}\nolimits_{V}=\operatorname*{syminc}%
\nolimits_{V}$.

On the other hand, $\rho$ and $\operatorname*{symlift}\varphi$ are $k$-algebra
homomorphisms. Thus, their composition $\rho\circ\left(
\operatorname*{symlift}\varphi\right)  $ is also a $k$-algebra homomorphism
(since the composition of two $k$-algebra homomorphisms is always a
$k$-algebra homomorphism). Hence, $\rho\circ\left(  \operatorname*{symlift}%
\varphi\right)  $ is a $k$-algebra homomorphism $\operatorname*{Sym}%
V\rightarrow\operatorname*{Sym}V$ satisfying
\[
\left(  \rho\circ\left(  \operatorname*{symlift}\varphi\right)  \right)
\circ\operatorname*{syminc}\nolimits_{V}=\rho\circ\underbrace{\left(
\operatorname*{symlift}\varphi\right)  \circ\operatorname*{syminc}%
\nolimits_{V}}_{=\varphi}=\rho\circ\varphi=\operatorname*{syminc}%
\nolimits_{V}.
\]
In other words, $\rho\circ\left(  \operatorname*{symlift}\varphi\right)  $ is
a $k$-algebra homomorphism $\Phi:\operatorname*{Sym}V\rightarrow
\operatorname*{Sym}V$ satisfying $\Phi\circ\operatorname*{syminc}%
\nolimits_{V}=\operatorname*{syminc}\nolimits_{V}$.

Thus, (\ref{1721.pf.b.eq}) (applied to $\Phi_{1}=\rho\circ\left(
\operatorname*{symlift}\varphi\right)  $ and $\Phi_{2}=\operatorname{id}%
_{\operatorname*{Sym}V}$) shows that $\rho\circ\left(  \operatorname*{symlift}%
\varphi\right)  =\operatorname{id}_{\operatorname*{Sym}V}$. This proves
Lemma~\ref{Lemma1721} \textbf{(b)}.

\textbf{(c)} Lemma~\ref{Lemma1721} \textbf{(a)} shows that the $k$-algebra
homomorphism $\operatorname*{symlift}\varphi:\operatorname*{Sym}V\rightarrow
A$ is surjective. Lemma~\ref{Lemma1721} \textbf{(b)} shows that $\rho
\circ\left(  \operatorname*{symlift}\varphi\right)  =\operatorname{id}%
_{\operatorname*{Sym}V}$. Hence, $\left(  \operatorname*{symlift}%
\varphi\right)  \circ\rho=\operatorname{id}_{A}$%
\ \ \ \ \footnote{\textit{Proof.} Let $a\in A$. Then, $a\in A=\left(
\operatorname*{symlift}\varphi\right)  \left(  \operatorname*{Sym}V\right)  $
(since the homomorphism $\operatorname*{symlift}\varphi$ is surjective). In
other words, there exists some $b\in\operatorname*{Sym}V$ such that $a=\left(
\operatorname*{symlift}\varphi\right)  \left(  b\right)  $. Consider this $b$.
Now, $\underbrace{\left(  \rho\circ\left(  \operatorname*{symlift}%
\varphi\right)  \right)  }_{=\operatorname{id}_{\operatorname*{Sym}V}}\left(
b\right)  =\operatorname{id}_{\operatorname*{Sym}V}\left(  b\right)  =b$, so
that $b=\left(  \rho\circ\left(  \operatorname*{symlift}\varphi\right)
\right)  \left(  b\right)  =\rho\left(  \underbrace{\left(
\operatorname*{symlift}\varphi\right)  \left(  b\right)  }_{=a}\right)
=\rho\left(  a\right)  $. Hence, $\rho\left(  a\right)  =b$.
\par
Now,%
\[
\left(  \left(  \operatorname*{symlift}\varphi\right)  \circ\rho\right)
\left(  a\right)  =\left(  \operatorname*{symlift}\varphi\right)  \left(
\underbrace{\rho\left(  a\right)  }_{=b}\right)  =\left(
\operatorname*{symlift}\varphi\right)  \left(  b\right)  =a=\operatorname{id}%
_{A}\left(  a\right)  .
\]
\par
Now, forget that we fixed $a$. We thus have shown that $\left(  \left(
\operatorname*{symlift}\varphi\right)  \circ\rho\right)  \left(  a\right)
=\operatorname{id}_{A}\left(  a\right)  $ for each $a\in A$. In other words,
$\left(  \operatorname*{symlift}\varphi\right)  \circ\rho=\operatorname{id}%
_{A}$.}. Combining this with $\rho\circ\left(  \operatorname*{symlift}%
\varphi\right)  =\operatorname{id}_{\operatorname*{Sym}V}$, we conclude that
the maps $\operatorname*{symlift}\varphi:\operatorname*{Sym}V\rightarrow A$
and $\rho:A\rightarrow\operatorname*{Sym}V$ are mutually inverse. This proves
Lemma~\ref{Lemma1721} \textbf{(c)}.
\end{proof}

The following lemma just recapitulates some trivial observations:

\begin{lemma}
\label{Lemma1730} Let $k$ be a field. Let $A$ be a $k$-algebra. Let $C$ be a
connected filtered $k$-coalgebra. Let $h\in\mathfrak{g}\left(  C,A\right)  $.

\textbf{(a)} We have $h\left(  1_{C}\right)  =0$.

\textbf{(b)} We have $h\left(  C_{\leq0}\right)  =0$.

\textbf{(c)} Assume that the field $k$ has characteristic $0$. Then, $e^{\ast
h}\left(  1_{C}\right)  =1_{A}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1730}.]We have $h\in\mathfrak{g}\left(  C,A\right)
=\left\{  f\in\mathcal{L}\left(  C,A\right)  \ \mid\ f\left(  1_{C}\right)
=0\right\}  $ (by the definition of $\mathfrak{g}\left(  C,A\right)  $). In
other words, $h$ is an element $f\in\mathcal{L}\left(  C,A\right)  $
satisfying $f\left(  1_{C}\right)  =0$. In other words, $h$ is an element of
$\mathcal{L}\left(  C,A\right)  $ and satisfies $h\left(  1_{C}\right)  =0$.
Thus, Lemma~\ref{Lemma1730} \textbf{(a)} is proven.

\textbf{(b)} Remark~\ref{Remark7l} gives $C_{\leq0}=k\cdot1_{C}$. Thus,%
\begin{align*}
h\left(  \underbrace{C_{\leq0}}_{=k\cdot1_{C}}\right)   &  =h\left(
k\cdot1_{C}\right)  =k\cdot\underbrace{h\left(  1_{C}\right)  }_{=0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }h\text{ is }k\text{-linear}\right) \\
&  =k\cdot0=0.
\end{align*}
This proves Lemma~\ref{Lemma1730} \textbf{(b)}.

\textbf{(c)} In Definition~\ref{Definition10}, it was shown that if $H$ is a
connected filtered $k$-coalgebra, and if $f\in\mathfrak{g}\left(  H,A\right)
$, then $e^{\ast f}\in G\left(  H,A\right)  $. Applying this to $H=C$ and
$f=h$, we obtain $e^{\ast h}\in G\left(  C,A\right)  =\left\{  f\in
\mathcal{L}\left(  C,A\right)  \ \mid\ f\left(  1_{C}\right)  =1_{A}\right\}
$ (by the definition of $G\left(  C,A\right)  $). In other words, $e^{\ast h}$
is an element $f\in\mathcal{L}\left(  C,A\right)  $ satisfying $f\left(
1_{C}\right)  =1_{A}$. In other words, $e^{\ast h}$ is an element of
$\mathcal{L}\left(  C,A\right)  $ and satisfies $e^{\ast h}\left(
1_{C}\right)  =1_{A}$. Thus, Lemma~\ref{Lemma1730} \textbf{(c)} is proven.
\end{proof}

Next, we state a lemma which is somewhat similar to Lemma~\ref{Lemma240}:

\begin{lemma}
\label{Lemma1731} Let $k$ be a field. Let $A$ be a $k$-algebra. Let $V$ be a
filtered $k$-vector space. Let $\alpha:V\rightarrow A$ and $\beta:V\rightarrow
A$ be two $k$-linear maps. Assume that each $i\in\mathbb{N}$ satisfies%
\begin{equation}
\left(  \alpha-\beta\right)  \left(  V_{\leq i}\right)  \subseteq
\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq i-1}\right)  \right)  .
\label{1731.eq.ass}%
\end{equation}
Then:

\textbf{(a)} For each $n\in\left\{  -1,0,1,\ldots\right\}  $, we have
$\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq n}\right)  \right)
=\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq n}\right)  \right)  $.

\textbf{(b)} We have $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V\right)
\right)  =\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)  \right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1731}.]\textbf{(a)} Let us prove
Lemma~\ref{Lemma1731} \textbf{(a)} by induction over $n$:

\textit{Induction base:} We have $\alpha\left(  V_{\leq-1}\right)
=\beta\left(  V_{\leq-1}\right)  $\ \ \ \ \footnote{\textit{Proof.} We have
$V_{\leq-1}=0$ and thus $\alpha\left(  V_{\leq-1}\right)  =\alpha\left(
0\right)  =0$ (since the map $\alpha$ is $k$-linear). The same argument
(applied to $\beta$ instead of $\alpha$) shows that $\beta\left(  V_{\leq
-1}\right)  =0$. Thus, $\alpha\left(  V_{\leq-1}\right)  =0=\beta\left(
V_{\leq-1}\right)  $.} and thus $\operatorname{AlgGen}_{k}\left(
\underbrace{\alpha\left(  V_{\leq-1}\right)  }_{=\beta\left(  V_{\leq
-1}\right)  }\right)  =\operatorname{AlgGen}_{k}\left(  \beta\left(
V_{\leq-1}\right)  \right)  $. In other words, Lemma~\ref{Lemma1731}
\textbf{(a)} holds for $n=-1$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
Lemma~\ref{Lemma1731} \textbf{(a)} holds for $n=N-1$. We now must prove that
Lemma~\ref{Lemma1731} \textbf{(a)} also holds for $n=N$.

The assumption (\ref{1731.eq.ass}) (applied to $n=N$) yields
\begin{equation}
\left(  \alpha-\beta\right)  \left(  V_{\leq N}\right)  \subseteq
\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N-1}\right)  \right)  .
\label{1731.pf.1}%
\end{equation}


Since Lemma~\ref{Lemma1731} \textbf{(a)} holds for $n=N-1$, we have
$\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq N-1}\right)  \right)
=\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N-1}\right)  \right)
$. Thus, (\ref{1731.pf.1}) becomes
\[
\left(  \alpha-\beta\right)  \left(  V_{\leq N}\right)  \subseteq
\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N-1}\right)  \right)
=\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq N-1}\right)  \right)
.
\]
Hence,%
\begin{align}
\underbrace{\left(  \beta-\alpha\right)  }_{=-\left(  \alpha-\beta\right)
}\left(  V_{\leq N}\right)   &  =\left(  -\left(  \alpha-\beta\right)
\right)  \left(  V_{\leq N}\right)  =-\underbrace{\left(  \alpha-\beta\right)
\left(  V_{\leq N}\right)  }_{\subseteq\operatorname{AlgGen}_{k}\left(
\alpha\left(  V_{\leq N-1}\right)  \right)  }\nonumber\\
&  \subseteq-\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq
N-1}\right)  \right)  \subseteq\operatorname{AlgGen}_{k}\left(  \alpha\left(
V_{\leq N-1}\right)  \right)  \label{1731.pf.2}%
\end{align}
(since $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq N-1}\right)
\right)  $ is a $k$-vector space).

We have $V_{\leq N-1}\subseteq V_{\leq N}$\ \ \ \ \footnote{\textit{Proof.} If
$N=0$, then $V_{\leq N-1}\subseteq V_{\leq N}$ holds (because if $N=0$, then
$V_{\leq N-1}=V_{\leq0-1}=V_{\leq-1}=0\subseteq V_{\leq N}$). 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 $N\geq1$ (since
$N\in\mathbb{N}$). Hence, $N-1\in\mathbb{N}$.
\par
But since $V$ is a filtered $k$-algebra, we have $V_{\leq0}\subseteq V_{\leq
1}\subseteq V_{\leq2}\subseteq\cdots$. Hence, $V_{\leq N-1}\subseteq V_{\leq
N}$ (since $N-1\in\mathbb{N}$). Qed.}.

Applying (\ref{1731.pf.triv-in}) to $S=\beta\left(  V_{\leq N}\right)  $, we
obtain $\beta\left(  V_{\leq N}\right)  \subseteq\operatorname{AlgGen}%
_{k}\left(  \beta\left(  V_{\leq N}\right)  \right)  $.

Now, define a $k$-linear map $\xi:V\rightarrow A$ by $\xi=\alpha-\beta$. Then,%
\begin{align*}
\underbrace{\xi}_{=\alpha-\beta}\left(  V_{\leq N}\right)   &  =\left(
\alpha-\beta\right)  \left(  V_{\leq N}\right)  \subseteq\operatorname{AlgGen}%
_{k}\left(  \beta\left(  \underbrace{V_{\leq N-1}}_{\subseteq V_{\leq N}%
}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1731.pf.1}%
)}\right) \\
&  =\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N}\right)  \right)
.
\end{align*}


But $\alpha=\xi+\beta$ (since $\xi=\alpha-\beta$). Now,%
\begin{align*}
\underbrace{\alpha}_{=\xi+\beta}\left(  V_{\leq N}\right)   &  =\left(
\xi+\beta\right)  \left(  V_{\leq N}\right)  \subseteq\underbrace{\xi\left(
V_{\leq N}\right)  }_{\subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(
V_{\leq N}\right)  \right)  }+\underbrace{\beta\left(  V_{\leq N}\right)
}_{\subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N}\right)
\right)  }\\
&  \subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N}\right)
\right)  +\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N}\right)
\right)  \subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq
N}\right)  \right)
\end{align*}
(since $\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N}\right)
\right)  $ is a $k$-vector space). In other words, $\operatorname{AlgGen}%
_{k}\left(  \beta\left(  V_{\leq N}\right)  \right)  $ contains $\alpha\left(
V_{\leq N}\right)  $ as a subset. Hence, $\operatorname{AlgGen}_{k}\left(
\beta\left(  V_{\leq N}\right)  \right)  $ is a $k$-subalgebra of $A$
containing $\alpha\left(  V_{\leq N}\right)  $ as a subset\footnote{since
$\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N}\right)  \right)  $
is clearly a $k$-subalgebra of $A$}.

But we know that $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq
N}\right)  \right)  $ is the smallest $k$-subalgebra of $A$ containing
$\alpha\left(  V_{\leq N}\right)  $ as a subset. This means that whenever $U$
is a $k$-subalgebra of $A$ containing $\alpha\left(  V_{\leq N}\right)  $ as a
subset, we must necessarily have $\operatorname{AlgGen}_{k}\left(
\alpha\left(  V_{\leq N}\right)  \right)  \subseteq U$. Applied to
$U=\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq N}\right)  \right)
$, this yields that $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq
N}\right)  \right)  \subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(
V_{\leq N}\right)  \right)  $.

So we have proven $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq
N}\right)  \right)  \subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(
V_{\leq N}\right)  \right)  $. But the same argument, with the maps $\alpha$
and $\beta$ interchanged (and using the equality (\ref{1731.pf.2}) instead of
(\ref{1731.pf.1})), shows that $\operatorname{AlgGen}_{k}\left(  \beta\left(
V_{\leq N}\right)  \right)  \subseteq\operatorname{AlgGen}_{k}\left(
\alpha\left(  V_{\leq N}\right)  \right)  $.

Combining $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V_{\leq N}\right)
\right)  \subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(  V_{\leq
N}\right)  \right)  $ and $\operatorname{AlgGen}_{k}\left(  \beta\left(
V_{\leq N}\right)  \right)  \subseteq\operatorname{AlgGen}_{k}\left(
\alpha\left(  V_{\leq N}\right)  \right)  $, we obtain $\operatorname{AlgGen}%
_{k}\left(  \alpha\left(  V_{\leq N}\right)  \right)  =\operatorname{AlgGen}%
_{k}\left(  \beta\left(  V_{\leq N}\right)  \right)  $. In other words,
Lemma~\ref{Lemma1731} \textbf{(a)} holds for $n=N$. This completes the
induction step.

Thus, the induction proof of Lemma~\ref{Lemma1731} \textbf{(a)} is complete.

\textbf{(b)} For every $n\in\mathbb{N}$, we have%
\begin{align}
\alpha\left(  V_{\leq n}\right)   &  \subseteq\operatorname{AlgGen}_{k}\left(
\alpha\left(  V_{\leq n}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1731.pf.triv-in}), applied to }S=\alpha\left(  V_{\leq n}\right)
\right) \nonumber\\
&  =\operatorname{AlgGen}_{k}\left(  \beta\left(  \underbrace{V_{\leq n}%
}_{\subseteq V}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Lemma~\ref{Lemma1731} \textbf{(a)}}\right) \nonumber\\
&  \subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)  \right)  .
\label{1731.pf.b.1}%
\end{align}
But $V$ is a filtered $k$-vector space. Hence, $V=\bigcup_{n\in\mathbb{N}%
}V_{\leq n}$. Applying the map $\alpha$ to this equality, we obtain%
\begin{align*}
\alpha\left(  V\right)   &  =\alpha\left(  \bigcup_{n\in\mathbb{N}}V_{\leq
n}\right)  =\bigcup_{n\in\mathbb{N}}\underbrace{\alpha\left(  V_{\leq
n}\right)  }_{\substack{\subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(
V\right)  \right)  \\\text{(by (\ref{1731.pf.b.1}))}}}\\
&  \subseteq\bigcup_{n\in\mathbb{N}}\operatorname{AlgGen}_{k}\left(
\beta\left(  V\right)  \right)  =\operatorname{AlgGen}_{k}\left(  \beta\left(
V\right)  \right)  .
\end{align*}
In other words, $\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)
\right)  $ contains $\alpha\left(  V\right)  $ as a subset. Hence,
$\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)  \right)  $ is a
$k$-subalgebra of $A$ containing $\alpha\left(  V\right)  $ as a
subset\footnote{since $\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)
\right)  $ is clearly a $k$-subalgebra of $A$}.

But we know that $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V\right)
\right)  $ is the smallest $k$-subalgebra of $A$ containing $\alpha\left(
V\right)  $ as a subset. This means that whenever $U$ is a $k$-subalgebra of
$A$ containing $\alpha\left(  V\right)  $ as a subset, we must necessarily
have $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V\right)  \right)
\subseteq U$. Applied to $U=\operatorname{AlgGen}_{k}\left(  \beta\left(
V\right)  \right)  $, this yields that $\operatorname{AlgGen}_{k}\left(
\alpha\left(  V\right)  \right)  \subseteq\operatorname{AlgGen}_{k}\left(
\beta\left(  V\right)  \right)  $.

So we have proven $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V\right)
\right)  \subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)
\right)  $. But the same argument, with the maps $\alpha$ and $\beta$
interchanged, shows that $\operatorname{AlgGen}_{k}\left(  \beta\left(
V\right)  \right)  \subseteq\operatorname{AlgGen}_{k}\left(  \alpha\left(
V\right)  \right)  $.

Combining $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V\right)  \right)
\subseteq\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)  \right)  $
and $\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)  \right)
\subseteq\operatorname{AlgGen}_{k}\left(  \alpha\left(  V\right)  \right)  $,
we obtain $\operatorname{AlgGen}_{k}\left(  \alpha\left(  V\right)  \right)
=\operatorname{AlgGen}_{k}\left(  \beta\left(  V\right)  \right)  $. This
proves Lemma~\ref{Lemma1731} \textbf{(b)}.
\end{proof}

\begin{proposition}
\label{Proposition1732} Let $k$ be a field. Let $C$ be a connected filtered
$k$-coalgebra. Let $A$ be a $k$-algebra. Let $f\in\mathfrak{g}\left(
C,A\right)  $. Then:

\textbf{(a)} For each $n\in\mathbb{N}$ and $i\in\mathbb{N}$, we have%
\[
f^{\ast i}\left(  C_{\leq n}\right)  \subseteq\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n}\right)  \right)  .
\]


\textbf{(b)} For each $n\in\mathbb{N}$ and $i\in\mathbb{N}$ satisfying
$i\geq2$, we have%
\[
f^{\ast i}\left(  C_{\leq n}\right)  \subseteq\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n-1}\right)  \right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1732}.]\textbf{(a)} We shall prove
Proposition~\ref{Proposition1732} \textbf{(a)} by induction on $i$:

\textit{Induction base:} For each $n\in\mathbb{N}$, we have $f^{\ast0}\left(
C_{\leq n}\right)  \subseteq\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq
n}\right)  \right)  $\ \ \ \ \footnote{\textit{Proof.} Let $n\in\mathbb{N}$.
The set $\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n}\right)  \right)
$ is a $k$-subalgebra of $A$, and thus contains the element $1_{A}$. In other
words, $1_{A}\in\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n}\right)
\right)  $.
\par
But $f^{\ast0}=e_{C,A}=\eta_{A}\circ\varepsilon_{C}$ (by the definition of
$e_{C,A}$). Hence, each $x\in C_{\leq n}$ satisfies
\begin{align*}
\underbrace{f^{\ast0}}_{=\eta_{A}\circ\varepsilon_{C}}\left(  x\right)   &
=\left(  \eta_{A}\circ\varepsilon_{C}\right)  \left(  x\right)  =\eta
_{A}\left(  \varepsilon_{C}\left(  x\right)  \right)  =\varepsilon_{C}\left(
x\right)  \cdot\underbrace{1_{A}}_{\in\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n}\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\eta_{A}\right) \\
&  \in\varepsilon_{C}\left(  x\right)  \cdot\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n}\right)  \right)  \subseteq\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n}\right)  \right)
\end{align*}
(since $\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n}\right)  \right)
$ is a $k$-vector space). In other words, $f^{\ast0}\left(  C_{\leq n}\right)
\subseteq\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n}\right)  \right)
$. Qed.}. In other words, Proposition~\ref{Proposition1732} \textbf{(a)} holds
for $i=0$. This completes the induction base.

\textit{Induction step:} Let $j$ be a positive integer. Assume that
Proposition~\ref{Proposition1732} \textbf{(a)} holds for $i=j-1$. We must then
show that Proposition~\ref{Proposition1732} \textbf{(a)} holds for $i=j$.

We have assumed that Proposition~\ref{Proposition1732} \textbf{(a)} holds for
$i=j-1$. In other words, whenever $k$, $C$, $A$ and $f$ are as in
Proposition~\ref{Proposition1732} \textbf{(a)}, we have%
\begin{equation}
f^{\ast\left(  j-1\right)  }\left(  C_{\leq n}\right)  \subseteq
\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n}\right)  \right)
\label{1732.pf.a.indhyp}%
\end{equation}
for each $n\in\mathbb{N}$.

Now, we must prove that Proposition~\ref{Proposition1732} \textbf{(a)} holds
for $i=j$. So let $k$, $C$, $A$ and $f$ be as in
Proposition~\ref{Proposition1732} \textbf{(a)}. Let $n\in\mathbb{N}$.

Let $\mathfrak{A}$ be the $k$-subalgebra $\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n}\right)  \right)  $ of $A$. Thus, $\mathfrak{A}%
=\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n}\right)  \right)  $.

Each $u\in\left\{  0,1,\ldots,n\right\}  $ satisfies%
\begin{equation}
f\left(  C_{\leq u}\right)  \subseteq\mathfrak{A} \label{1732.pf.a.3a}%
\end{equation}
\footnote{\textit{Proof of (\ref{1732.pf.a.3a}):} Let $u\in\left\{
0,1,\ldots,n\right\}  $. Then, $u\in\mathbb{N}$ and $u\leq n$.
\par
Applying (\ref{1731.pf.triv-in}) to $S=f\left(  C_{\leq n}\right)  $, we
obtain $f\left(  C_{\leq n}\right)  \subseteq\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n}\right)  \right)  =\mathfrak{A}$.
\par
Since $C$ is filtered, we have $C_{\leq0}\subseteq C_{\leq1}\subseteq
C_{\leq2}\subseteq\cdots$. In other words, $C_{\leq x}\subseteq C_{\leq y}$
whenever $x$ and $y$ are two elements of $\mathbb{N}$ satisfying $x\leq y$.
Applying this to $x=u$ and $y=n$, we obtain $C_{\leq u}\subseteq C_{\leq n}$
(since $u\leq n$). Thus, $f\left(  \underbrace{C_{\leq u}}_{\subseteq C_{\leq
n}}\right)  \subseteq f\left(  C_{\leq n}\right)  \subseteq\mathfrak{A}$. This
proves (\ref{1732.pf.a.3a}).} and%
\begin{equation}
f^{\ast\left(  j-1\right)  }\left(  C_{\leq n-u}\right)  \subseteq\mathfrak{A}
\label{1732.pf.a.3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{1732.pf.a.3b}):} Let $u\in\left\{
0,1,\ldots,n\right\}  $. Then, $n-u\in\left\{  0,1,\ldots,n\right\}
\subseteq\mathbb{N}$. But $u\in\left\{  0,1,\ldots,n\right\}  $ shows that
$u\geq0$ and thus $n-\underbrace{u}_{\geq0}\leq n$.
\par
Since $C$ is filtered, we have $C_{\leq0}\subseteq C_{\leq1}\subseteq
C_{\leq2}\subseteq\cdots$. In other words, $C_{\leq x}\subseteq C_{\leq y}$
whenever $x$ and $y$ are two elements of $\mathbb{N}$ satisfying $x\leq y$.
Applying this to $x=n-u$ and $y=n$, we obtain $C_{\leq n-u}\subseteq C_{\leq
n}$ (since $n-u\leq n$). Thus,
\begin{align*}
f^{\ast\left(  j-1\right)  }\left(  \underbrace{C_{\leq n-u}}_{\subseteq
C_{\leq n}}\right)   &  \subseteq f^{\ast\left(  j-1\right)  }\left(  C_{\leq
n}\right)  \subseteq\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq
n}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1732.pf.a.indhyp})}\right) \\
&  =\mathfrak{A}.
\end{align*}
This proves (\ref{1732.pf.a.3b}).}.

Since $j$ is a positive integer, we have $f^{\ast j}=f\ast f^{\ast\left(
j-1\right)  }=\mu_{A}\circ\left(  f\otimes f^{\ast\left(  j-1\right)
}\right)  \circ\Delta_{C}$ (by the definition of convolution). Hence,%
\begin{align*}
&  \underbrace{f^{\ast j}}_{=\mu_{A}\circ\left(  f\otimes f^{\ast\left(
j-1\right)  }\right)  \circ\Delta_{C}}\left(  C_{\leq n}\right) \\
&  =\left(  \mu_{A}\circ\left(  f\otimes f^{\ast\left(  j-1\right)  }\right)
\circ\Delta_{C}\right)  \left(  C_{\leq n}\right) \\
&  =\mu_{A}\left(  \left(  f\otimes f^{\ast\left(  j-1\right)  }\right)
\left(  \underbrace{\Delta_{C}\left(  C_{\leq n}\right)  }%
_{\substack{\subseteq\sum_{u=0}^{n}C_{\leq u}\otimes C_{\leq n-u}%
\\\text{(since }C\text{ is a filtered }k\text{-coalgebra)}}}\right)  \right)
\\
&  \subseteq\mu_{A}\left(  \underbrace{\left(  f\otimes f^{\ast\left(
j-1\right)  }\right)  \left(  \sum_{u=0}^{n}C_{\leq u}\otimes C_{\leq
n-u}\right)  }_{\substack{\subseteq\sum_{u=0}^{n}\left(  f\otimes
f^{\ast\left(  j-1\right)  }\right)  \left(  C_{\leq u}\otimes C_{\leq
n-u}\right)  \\\text{(since the map }f\otimes f^{\ast\left(  j-1\right)
}\text{ is }k\text{-linear)}}}\right) \\
&  \subseteq\mu_{A}\left(  \sum_{u=0}^{n}\underbrace{\left(  f\otimes
f^{\ast\left(  j-1\right)  }\right)  \left(  C_{\leq u}\otimes C_{\leq
n-u}\right)  }_{\subseteq f\left(  C_{\leq u}\right)  \otimes f^{\ast\left(
j-1\right)  }\left(  C_{\leq n-u}\right)  }\right)  \subseteq\mu_{A}\left(
\sum_{u=0}^{n}\underbrace{f\left(  C_{\leq u}\right)  }_{\substack{\subseteq
\mathfrak{A}\\\text{(by (\ref{1732.pf.a.3a}))}}}\otimes\underbrace{f^{\ast
\left(  j-1\right)  }\left(  C_{\leq n-u}\right)  }_{\substack{\subseteq
\mathfrak{A}\\\text{(by (\ref{1732.pf.a.3b}))}}}\right) \\
&  \subseteq\mu_{A}\left(  \underbrace{\sum_{u=0}^{n}\mathfrak{A}%
\otimes\mathfrak{A}}_{\substack{\subseteq\mathfrak{A}\otimes\mathfrak{A}%
\\\text{(since }\mathfrak{A}\otimes\mathfrak{A}\text{ is a }k\text{-vector
space)}}}\right)  \subseteq\mu_{A}\left(  \mathfrak{A}\otimes\mathfrak{A}%
\right) \\
&  =\mathfrak{AA}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{104.UV}), applied
to }U=\mathfrak{A}\text{ and }V=\mathfrak{A}\right) \\
&  \subseteq\mathfrak{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mathfrak{A}\text{ is a }k\text{-algebra}\right) \\
&  =\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n}\right)  \right)  .
\end{align*}


Now, forget that we fixed $n$. We thus have proven that $f^{\ast j}\left(
C_{\leq n}\right)  \subseteq\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq
n}\right)  \right)  $ for each $n\in\mathbb{N}$.

Now, forget that we fixed $k$, $C$, $A$ and $f$. We thus have shown that if
$k$, $C$, $A$ and $f$ are as in Proposition~\ref{Proposition1732}
\textbf{(a)}, then we have $f^{\ast j}\left(  C_{\leq n}\right)
\subseteq\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n}\right)  \right)
$ for each $n\in\mathbb{N}$. In other words, Proposition~\ref{Proposition1732}
\textbf{(a)} holds for $i=j$. This completes the induction step. Thus, the
induction proof of Proposition~\ref{Proposition1732} \textbf{(a)} is complete.

\textbf{(b)} Let $n\in\mathbb{N}$ and $i\in\mathbb{N}$ be such that $i\geq2$.

From $i\geq2\geq1$, we conclude that $i$ is positive. Hence,
Lemma~\ref{Lemma277} (applied to $C$ and $f$ instead of $H$ and $h$) yields
$f^{\ast i}\in\mathfrak{g}\left(  C,A\right)  $. Hence, Lemma~\ref{Lemma1730}
\textbf{(b)} (applied to $f^{\ast i}$ instead of $h$) yields%
\[
f^{\ast i}\left(  C_{\leq0}\right)  =0\subseteq\operatorname{AlgGen}%
_{k}\left(  f\left(  C_{\leq0-1}\right)  \right)
\]
(since $\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq0-1}\right)  \right)
$ is a $k$-subalgebra of $A$). In other words,
Proposition~\ref{Proposition1732} \textbf{(b)} holds if $n=0$. Thus, 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 $n=0$). Thus, $n$ is a positive integer
(since $n\in\mathbb{N}$). Hence, $n-1\in\mathbb{N}$. Also, $i-1\in\mathbb{N}$
(since $i$ is a positive integer).

Let $\mathfrak{A}$ be the $k$-subalgebra $\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n-1}\right)  \right)  $ of $A$. Thus, $\mathfrak{A}%
=\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n-1}\right)  \right)  $.

For each $u\in\left\{  1,2,\ldots,n-1\right\}  $, we have%
\begin{equation}
f\left(  C_{\leq u}\right)  \subseteq\mathfrak{A} \label{1732.pf.b.3a}%
\end{equation}
\footnote{\textit{Proof of (\ref{1732.pf.b.3a}):} Let $u\in\left\{
1,2,\ldots,n-1\right\}  $. Then, $1\leq u\leq n-1$.
\par
Applying (\ref{1731.pf.triv-in}) to $S=f\left(  C_{\leq n-1}\right)  $, we
obtain $f\left(  C_{\leq n-1}\right)  \subseteq\operatorname{AlgGen}%
_{k}\left(  f\left(  C_{\leq n-1}\right)  \right)  =\mathfrak{A}$.
\par
Since $C$ is filtered, we have $C_{\leq0}\subseteq C_{\leq1}\subseteq
C_{\leq2}\subseteq\cdots$. In other words, $C_{\leq x}\subseteq C_{\leq y}$
whenever $x$ and $y$ are two elements of $\mathbb{N}$ satisfying $x\leq y$.
Applying this to $x=u$ and $y=n-1$, we obtain $C_{\leq u}\subseteq C_{\leq
n-1}$ (since $u\leq n-1$). Thus, $f\left(  \underbrace{C_{\leq u}}_{\subseteq
C_{\leq n-1}}\right)  \subseteq f\left(  C_{\leq n-1}\right)  \subseteq
\mathfrak{A}$. This proves (\ref{1732.pf.b.3a}).} and%
\begin{equation}
f^{\ast\left(  i-1\right)  }\left(  C_{\leq n-u}\right)  \subseteq\mathfrak{A}
\label{1732.pf.b.3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{1732.pf.b.3b}):} Let $u\in\left\{
1,2,\ldots,n-1\right\}  $. Then, $n-u\in\left\{  1,2,\ldots,n-1\right\}
\subseteq\mathbb{N}$. Also, from $u\in\left\{  1,2,\ldots,n-1\right\}  $, we
obtain $1\leq u\leq n-1$. Thus, $n-\underbrace{u}_{\geq1}\leq n-1$.
\par
Since $C$ is filtered, we have $C_{\leq0}\subseteq C_{\leq1}\subseteq
C_{\leq2}\subseteq\cdots$. In other words, $C_{\leq x}\subseteq C_{\leq y}$
whenever $x$ and $y$ are two elements of $\mathbb{N}$ satisfying $x\leq y$.
Applying this to $x=n-u$ and $y=n-1$, we obtain $C_{\leq n-u}\subseteq C_{\leq
n-1}$ (since $n-u\leq n-1$). Thus,
\begin{align*}
f^{\ast\left(  i-1\right)  }\left(  \underbrace{C_{\leq n-u}}_{\subseteq
C_{\leq n-1}}\right)   &  \subseteq f^{\ast\left(  i-1\right)  }\left(
C_{\leq n-1}\right)  \subseteq\operatorname{AlgGen}_{k}\left(  f\left(
C_{\leq n-1}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition~\ref{Proposition1732}
\textbf{(a)}, applied to }n-1\text{ and }i-1\text{ instead of }n\text{ and
}i\right) \\
&  =\mathfrak{A}.
\end{align*}
This proves (\ref{1732.pf.b.3b}).}.

From $i\geq2$, we obtain $i-1\geq1$. Thus, $i-1$ is a positive element of
$\mathbb{N}$. Hence, Lemma~\ref{Lemma277} (applied to $C$, $f$ and $i-1$
instead of $H$, $h$ and $i$) yields $f^{\ast\left(  i-1\right)  }%
\in\mathfrak{g}\left(  C,A\right)  $. Thus, Proposition~\ref{Proposition274a}
(applied to $g=f^{\ast\left(  i-1\right)  }$ and $\ell=n$) yields%
\[
\left(  f\ast f^{\ast\left(  i-1\right)  }\right)  \left(  C_{\leq n}\right)
\subseteq\sum_{u=1}^{n-1}\underbrace{f\left(  C_{\leq u}\right)
}_{\substack{\subseteq\mathfrak{A}\\\text{(by (\ref{1732.pf.b.3a}))}%
}}\underbrace{f^{\ast\left(  i-1\right)  }\left(  C_{\leq n-u}\right)
}_{\substack{\subseteq\mathfrak{A}\\\text{(by (\ref{1732.pf.b.3b}))}%
}}\subseteq\sum_{u=1}^{n-1}\mathfrak{AA}\subseteq\mathfrak{A}%
\]
(since $\mathfrak{A}$ is a $k$-subalgebra of $A$). Now,%
\[
\underbrace{f^{\ast i}}_{=f\ast f^{\ast\left(  i-1\right)  }}\left(  C_{\leq
n}\right)  =\left(  f\ast f^{\ast\left(  i-1\right)  }\right)  \left(  C_{\leq
n}\right)  \subseteq\mathfrak{A}=\operatorname{AlgGen}_{k}\left(  f\left(
C_{\leq n-1}\right)  \right)  .
\]
This proves Proposition~\ref{Proposition1732} \textbf{(b)}.
\end{proof}

\begin{proposition}
\label{Proposition1733} Let $k$ be a field of characteristic $0$. Let $C$ be a
connected filtered $k$-coalgebra. Let $A$ be a $k$-algebra. Let $f\in
\mathfrak{g}\left(  C,A\right)  $. Then:

\textbf{(a)} Each $i\in\mathbb{N}$ satisfies%
\[
\left(  e^{\ast f}-f\right)  \left(  C_{\leq i}\right)  \subseteq
\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq i-1}\right)  \right)  .
\]


\textbf{(b)} We have $\operatorname{AlgGen}_{k}\left(  e^{\ast f}\left(
C\right)  \right)  =\operatorname{AlgGen}_{k}\left(  f\left(  C\right)
\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1733}.]\textbf{(a)} Let $n$ be a
positive integer. We shall prove that%
\begin{equation}
\left(  e^{\ast f}-f\right)  \left(  C_{\leq n}\right)  \subseteq
\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n-1}\right)  \right)  .
\label{1733.pf.a.semigoal}%
\end{equation}


Indeed, let $x\in C_{\leq n}$ be arbitrary. Then,
\begin{equation}
f^{\ast i}\left(  x\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every integer }i>n
\label{1733.pf.a.fix=0}%
\end{equation}
\footnote{\textit{Proof of (\ref{1733.pf.a.fix=0}):} Let $i>n$ be an integer.
Thus, $i>n\geq0$, so that $i\in\mathbb{N}$. Now, Remark~\ref{Remark9} (applied
to $H=C$) yields $f^{\ast i}\left(  C_{\leq n}\right)  =0$ (since $i>n$). But
$x\in C_{\leq n}$ yields $f^{\ast i}\left(  x\right)  \in f^{\ast i}\left(
C_{\leq n}\right)  =0$, thus $f^{\ast i}\left(  x\right)  =0$. This proves
(\ref{1733.pf.a.fix=0}).}.

Let $\mathfrak{A}$ denote the $k$-subalgebra $\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n-1}\right)  \right)  $ of $A$. Then, $1_{A}\in\mathfrak{A}$
(since $\mathfrak{A}$ is a $k$-subalgebra of $A$).

We have%
\begin{equation}
f^{\ast i}\left(  x\right)  \in\mathfrak{A}\ \ \ \ \ \ \ \ \ \ \text{for each
integer }i\geq2 \label{1733.pf.a.fix2}%
\end{equation}
\footnote{\textit{Proof of (\ref{1733.pf.a.fix2}):} Let $i\geq2$ be an
integer. Thus, $i\in\mathbb{N}$. Now, Proposition~\ref{Proposition1732}
\textbf{(b)} yields $f^{\ast i}\left(  C_{\leq n}\right)  \subseteq
\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq n-1}\right)  \right)
=\mathfrak{A}$ (since $\mathfrak{A}=\operatorname{AlgGen}_{k}\left(  f\left(
C_{\leq n-1}\right)  \right)  $). But $f^{\ast i}\left(  \underbrace{x}_{\in
C_{\leq n}}\right)  \in f^{\ast i}\left(  C_{\leq n}\right)  \subseteq
\mathfrak{A}$. This proves (\ref{1733.pf.a.fix2}).}.

On the other hand, $n\geq1$ (since $n$ is a positive integer). Furthermore,
$f^{\ast0}=e_{C,A}=\eta_{A}\circ\varepsilon_{C}$ (by the definition of
$e_{C,A}$) and thus
\begin{align*}
\underbrace{f^{\ast0}}_{=\eta_{A}\circ\varepsilon_{C}}\left(  x\right)   &
=\left(  \eta_{A}\circ\varepsilon_{C}\right)  \left(  x\right)  =\eta
_{A}\left(  \varepsilon_{C}\left(  x\right)  \right)  =\varepsilon_{C}\left(
x\right)  \cdot\underbrace{1_{A}}_{\in\mathfrak{A}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\eta_{A}\right) \\
&  \in\varepsilon_{C}\left(  x\right)  \cdot\mathfrak{A}\subseteq\mathfrak{A}%
\end{align*}
(since $\mathfrak{A}$ is a $k$-vector space).

But the definition of $e^{\ast f}$ yields%
\begin{align*}
e^{\ast f}\left(  x\right)   &  =\sum_{i\geq0}\underbrace{\dfrac{f^{\ast
i}\left(  x\right)  }{i!}}_{=\dfrac{1}{i!}f^{\ast i}\left(  x\right)  }%
=\sum_{i\geq0}\dfrac{1}{i!}f^{\ast i}\left(  x\right) \\
&  =\underbrace{\sum_{\substack{i\geq0;\\i\leq n}}}_{=\sum_{i=0}^{n}}\dfrac
{1}{i!}f^{\ast i}\left(  x\right)  +\sum_{\substack{i\geq0;\\i>n}}\dfrac
{1}{i!}\underbrace{f^{\ast i}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{1733.pf.a.fix=0}))}}}\\
&  =\sum_{i=0}^{n}\dfrac{1}{i!}f^{\ast i}\left(  x\right)  +\underbrace{\sum
_{\substack{i\geq0;\\i>n}}\dfrac{1}{i!}0}_{=0}=\sum_{i=0}^{n}\dfrac{1}%
{i!}f^{\ast i}\left(  x\right) \\
&  =\underbrace{\dfrac{1}{0!}}_{=\dfrac{1}{1}=1}f^{\ast0}\left(  x\right)
+\underbrace{\sum_{i=1}^{n}\dfrac{1}{i!}f^{\ast i}\left(  x\right)
}_{\substack{=\dfrac{1}{1!}f^{\ast1}\left(  x\right)  +\sum_{i=2}^{n}\dfrac
{1}{i!}f^{\ast i}\left(  x\right)  \\\text{(here, we have split off the
addend}\\\text{for }i=1\text{ from the sum (since }n\geq1\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=0\\
\text{from the sum (because }n\geq1\geq0\text{)}%
\end{array}
\right) \\
&  =f^{\ast0}\left(  x\right)  +\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}%
=1}\underbrace{f^{\ast1}}_{=f}\left(  x\right)  +\sum_{i=2}^{n}\dfrac{1}%
{i!}f^{\ast i}\left(  x\right) \\
&  =f^{\ast0}\left(  x\right)  +f\left(  x\right)  +\sum_{i=2}^{n}\dfrac
{1}{i!}f^{\ast i}\left(  x\right)  .
\end{align*}
Subtracting $f\left(  x\right)  $ from both sides of this equality, we obtain%
\begin{align*}
e^{\ast f}\left(  x\right)  -f\left(  x\right)   &  =\underbrace{f^{\ast
0}\left(  x\right)  }_{\in\mathfrak{A}}+\sum_{i=2}^{n}\dfrac{1}{i!}%
\underbrace{f^{\ast i}\left(  x\right)  }_{\substack{\in\mathfrak{A}%
\\\text{(by (\ref{1733.pf.a.fix2}))}}}\in\mathfrak{A}+\underbrace{\sum
_{i=2}^{n}\dfrac{1}{i!}\mathfrak{A}}_{\substack{\subseteq\mathfrak{A}%
\\\text{(since }\mathfrak{A}\text{ is a }k\text{-vector space)}}}\\
&  \subseteq\mathfrak{A}+\mathfrak{A}\subseteq\mathfrak{A}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{A}\text{ is a
}k\text{-vector space}\right)  .
\end{align*}


Now, $\left(  e^{\ast f}-f\right)  \left(  x\right)  =e^{\ast f}\left(
x\right)  -f\left(  x\right)  \in\mathfrak{A}$.

Now, let us forget that we fixed $x$. We thus have proven that $\left(
e^{\ast f}-f\right)  \left(  x\right)  \in\mathfrak{A}$ for each $x\in C_{\leq
n}$. In other words, $\left(  e^{\ast f}-f\right)  \left(  C_{\leq n}\right)
\subseteq\mathfrak{A}$. This rewrites as $\left(  e^{\ast f}-f\right)  \left(
C_{\leq n}\right)  \subseteq\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq
n-1}\right)  \right)  $ (since $\mathfrak{A}=\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq n-1}\right)  \right)  $). Thus, (\ref{1733.pf.a.semigoal}) is proven.

Now, let us forget that we fixed $n$. We thus have proven that
(\ref{1733.pf.a.semigoal}) holds for each positive integer $n$.

Now, let $i\in\mathbb{N}$. We must prove that $\left(  e^{\ast f}-f\right)
\left(  C_{\leq i}\right)  \subseteq\operatorname{AlgGen}_{k}\left(  f\left(
C_{\leq i-1}\right)  \right)  $. If $i$ is a positive integer, then this
follows immediately from (\ref{1733.pf.a.semigoal}) (applied to $n=i$). Hence,
for the rest of this proof, we can WLOG assume that $i$ is not a positive
integer. Assume this.

We have $i\notin\left\{  1,2,3,\ldots\right\}  $ (since $i$ is not a positive
integer). Combining this with $i\in\mathbb{N}$, we obtain $i\in\mathbb{N}%
\setminus\left\{  1,2,3,\ldots\right\}  =\left\{  0\right\}  $. In other
words, $i=0$. Hence, $C_{\leq i}=C_{\leq0}=k\cdot1_{C}$ (by
Remark~\ref{Remark7l}).

But%
\[
\left(  e^{\ast f}-f\right)  \left(  1_{C}\right)  =\underbrace{e^{\ast
f}\left(  1_{C}\right)  }_{\substack{=1_{A}\\\text{(by Lemma~\ref{Lemma1730}
\textbf{(c)},}\\\text{applied to }h=f\text{)}}}-\underbrace{f\left(
1_{C}\right)  }_{\substack{=0\\\text{(by Lemma~\ref{Lemma1730} \textbf{(a)}%
,}\\\text{applied to }h=f\text{)}}}=1_{A}-0=1_{A}.
\]
Now,%
\begin{align*}
\left(  e^{\ast f}-f\right)  \left(  \underbrace{C_{\leq i}}_{=k\cdot1_{C}%
}\right)   &  =\left(  e^{\ast f}-f\right)  \left(  k\cdot1_{C}\right) \\
&  =k\cdot\underbrace{\left(  e^{\ast f}-f\right)  \left(  1_{C}\right)
}_{=1_{A}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }e^{\ast f}-f\text{
is }k\text{-linear}\right) \\
&  =k\cdot\underbrace{1_{A}}_{\substack{\in\operatorname{AlgGen}_{k}\left(
f\left(  C_{\leq i-1}\right)  \right)  \\\text{(since }\operatorname{AlgGen}%
_{k}\left(  f\left(  C_{\leq i-1}\right)  \right)  \\\text{is a }%
k\text{-subalgebra of }A\text{)}}}\subseteq k\cdot\operatorname{AlgGen}%
_{k}\left(  f\left(  C_{\leq i-1}\right)  \right) \\
&  \subseteq\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq i-1}\right)
\right)
\end{align*}
(since $\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq i-1}\right)
\right)  $ is a $k$-vector subspace of $A$). Thus, $\left(  e^{\ast
f}-f\right)  \left(  C_{\leq i}\right)  \subseteq\operatorname{AlgGen}%
_{k}\left(  f\left(  C_{\leq i-1}\right)  \right)  $ is proven. The proof of
Proposition~\ref{Proposition1733} \textbf{(a)} is thus complete.

\textbf{(b)} Proposition~\ref{Proposition1733} \textbf{(a)} says that each
$i\in\mathbb{N}$ satisfies $\left(  e^{\ast f}-f\right)  \left(  C_{\leq
i}\right)  \subseteq\operatorname{AlgGen}_{k}\left(  f\left(  C_{\leq
i-1}\right)  \right)  $. Hence, Lemma~\ref{Lemma1731} \textbf{(b)} (applied to
$V=C$, $\alpha=e^{\ast f}$ and $\beta=f$) shows that $\operatorname{AlgGen}%
_{k}\left(  e^{\ast f}\left(  C\right)  \right)  =\operatorname{AlgGen}%
_{k}\left(  f\left(  C\right)  \right)  $. This proves
Proposition~\ref{Proposition1733} \textbf{(b)}.
\end{proof}

\begin{corollary}
\label{Corollary1734} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $A$ be a $k$-algebra. Let
$f\in\mathfrak{g}\left(  H,A\right)  $ be an $\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivation. Assume that the subset $f\left(
H\right)  $ of $A$ generates the $k$-algebra $A$. Then, the map $e^{\ast
f}:H\rightarrow A$ is a surjective $k$-algebra homomorphism.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1734}.]Lemma~\ref{Lemma121} yields that
$e^{\ast f}$ is a $k$-algebra homomorphism. It remains to show that this
homomorphism $e^{\ast f}$ is surjective.

The subset $f\left(  H\right)  $ of $A$ generates the $k$-algebra $A$. In
other words, the $k$-subalgebra of $A$ generated by $f\left(  H\right)  $ is
$A$. In other words, $\operatorname{AlgGen}_{k}\left(  f\left(  H\right)
\right)  $ is $A$ (since $\operatorname{AlgGen}_{k}\left(  f\left(  H\right)
\right)  $ is the $k$-subalgebra of $A$ generated by $f\left(  H\right)  $).
In other words, $\operatorname{AlgGen}_{k}\left(  f\left(  H\right)  \right)
=A$.

Proposition~\ref{Proposition1733} (applied to $C=H$) yields
$\operatorname{AlgGen}_{k}\left(  e^{\ast f}\left(  H\right)  \right)
=\operatorname{AlgGen}_{k}\left(  f\left(  H\right)  \right)  =A$.

The image $e^{\ast f}\left(  H\right)  $ is a $k$-subalgebra of $A$ (since
$e^{\ast f}:H\rightarrow A$ is a $k$-algebra homomorphism).

Let $S$ denote the image $e^{\ast f}\left(  H\right)  $. Then, $S=e^{\ast
f}\left(  H\right)  $. Hence, $\operatorname{AlgGen}_{k}\underbrace{S}%
_{=e^{\ast f}\left(  H\right)  }=\operatorname{AlgGen}_{k}\left(  e^{\ast
f}\left(  H\right)  \right)  =A$.

But $S=e^{\ast f}\left(  H\right)  $. Hence, $S$ is a $k$-subalgebra of $A$
(since $e^{\ast f}\left(  H\right)  $ is a $k$-subalgebra of $A$).

The $k$-subalgebra of $A$ generated by $S$ is $\operatorname{AlgGen}_{k}S=A$.
In other words, $A$ is the $k$-subalgebra of $A$ generated by $S$. In other
words, $A$ is the smallest $k$-subalgebra of $A$ which contains $S$ as a
subset (because the $k$-subalgebra of $A$ generated by $S$ is the smallest
$k$-subalgebra of $A$ which contains $S$ as a subset). Hence,%
\[
\left(
\begin{array}
[c]{c}%
\text{whenever }U\text{ is a }k\text{-subalgebra of }A\text{ which contains
}S\text{ as}\\
\text{a subset, we must necessarily have }A\subseteq U
\end{array}
\right)  .
\]
Applying this to $U=S$, we conclude that $A\subseteq S$ (since $S$ is a
$k$-subalgebra of $A$ that contains $S$ as a subset). Combined with
$S\subseteq A$ (which is obvious), this yields $A=S$. Thus, $A=S=e^{\ast
f}\left(  H\right)  $. In other words, the map $e^{\ast f}$ is surjective.
This completes the proof of Corollary~\ref{Corollary1734}.
\end{proof}

\begin{proposition}
\label{Proposition1735} Let $k$ be a field. Let $H$ be a connected filtered
$k$-coalgebra and, at the same time, a $k$-bialgebra with the same underlying
$k$-coalgebra structure. Let $A$ be a $k$-algebra. Let $h:H\rightarrow A$ be
an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation. Then,
$h\in\mathfrak{g}\left(  H,A\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1735}.]Theorem~\ref{Theorem115} (applied
to $f=h$) shows that $h$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation if and only if $h\left(  \left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$. Thus,
$h\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  =0$ (since $h$ is an $\left(  \varepsilon
_{H},\varepsilon_{H}\right)  $-derivation).

But $1_{H}=\underbrace{1}_{\in k}\cdot1_{H}\in k\cdot1_{H}\subseteq\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}$.
Hence,%
\[
h\left(  \underbrace{1_{H}}_{\in\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}}\right)  \subseteq h\left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  =0.
\]
Thus, $h\left(  1_{H}\right)  =0$.

Recall that $\mathfrak{g}\left(  H,A\right)  =\left\{  f\in\mathcal{L}\left(
H,A\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}  $ (by the definition of
$\mathfrak{g}\left(  H,A\right)  $). Now, $h$ is an element $f\in
\mathcal{L}\left(  H,A\right)  $ satisfying $f\left(  1_{H}\right)  =0$ (since
$h\in\mathcal{L}\left(  H,A\right)  $ and $h\left(  1_{H}\right)  =0$). In
other words, $h\in\left\{  f\in\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(
1_{H}\right)  =0\right\}  $. This rewrites as $h\in\mathfrak{g}\left(
H,A\right)  $ (since $\mathfrak{g}\left(  H,A\right)  =\left\{  f\in
\mathcal{L}\left(  H,A\right)  \ \mid\ f\left(  1_{H}\right)  =0\right\}  $).
This proves Proposition~\ref{Proposition1735}.
\end{proof}

Finally, having collected all the pieces of the puzzle, we can prove
Theorem~\ref{Theorem1702}:

\begin{proof}
[Proof of Theorem~\ref{Theorem1702}.]Theorem~\ref{Theorem101} shows that the
map $\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $
is a projection such that $\operatorname*{Ker}\left(  \operatorname*{Log}%
\operatorname*{id}\right)  =\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}+k\cdot1_{H}$.\ \ \ \ \footnote{Recall that $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ is to be
understood according to Convention~\ref{Convention100}. Thus, $\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$ means the
subspace $\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
\cdot\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  $ of
$H$.}

But $\mathfrak{e}=\operatorname*{Log}\operatorname*{id}$ (by the definition of
$\mathfrak{e}$). Hence, the map $\mathfrak{e}$ is a projection (since the map
$\operatorname*{Log}\operatorname*{id}$ is a projection). This proves
Theorem~\ref{Theorem1702} \textbf{(a)}.

From $\mathfrak{e}=\operatorname*{Log}\operatorname*{id}$, we obtain
$\operatorname*{Ker}\mathfrak{e}=\operatorname*{Ker}\left(
\operatorname*{Log}\operatorname*{id}\right)  =\left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}$. This proves
Theorem~\ref{Theorem1702} \textbf{(b)}.

We have $\mathfrak{e}=\operatorname*{Log}\operatorname*{id}\in\mathfrak{g}%
\left(  H,H\right)  $ (because $\operatorname*{Log}F\in\mathfrak{g}\left(
H,H\right)  $ for every $F\in G\left(  H,H\right)  $). Besides, $\mathfrak{e}%
=\operatorname*{Log}\operatorname*{id}$ yields $e^{\ast\mathfrak{e}}%
=e^{\ast\left(  \operatorname*{Log}\operatorname*{id}\right)  }%
=\operatorname*{id}$ (by Proposition~\ref{Proposition13} \textbf{(b)}, applied
to $F=\operatorname*{id}$ and $A=H$). Hence, $e^{\ast\mathfrak{e}}$ is a
$k$-algebra homomorphism (since $\operatorname*{id}$ is a $k$-algebra
homomorphism). By Lemma~\ref{Lemma122} (applied to $A=H$ and $f=\mathfrak{e}%
$), this yields that $\mathfrak{e}$ is an $\left(  \varepsilon_{H}%
,\varepsilon_{H}\right)  $-derivation. This proves Theorem~\ref{Theorem1702}
\textbf{(c)}.

Definition~\ref{Definition109} (applied to $A=H$ and $f=\mathfrak{e}$) shows
that the map $\mathfrak{e}$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation if and only if every $\left(  a,b\right)  \in
H\times H$ satisfies $\mathfrak{e}\left(  ab\right)  =\mathfrak{e}\left(
a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)
\mathfrak{e}\left(  b\right)  $. Thus, every $\left(  a,b\right)  \in H\times
H$ satisfies%
\begin{equation}
\mathfrak{e}\left(  ab\right)  =\mathfrak{e}\left(  a\right)  \varepsilon
_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)  \mathfrak{e}\left(
b\right)  \label{1702.pf.der-1}%
\end{equation}
(since the map $\mathfrak{e}$ is an $\left(  \varepsilon_{H},\varepsilon
_{H}\right)  $-derivation). Hence, every $\left(  a,b\right)  \in H\times H$
satisfies%
\begin{equation}
\mathfrak{q}\left(  ab\right)  =\mathfrak{q}\left(  a\right)  \varepsilon
_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)  \mathfrak{q}\left(
b\right)  \label{1702.pf.der-2}%
\end{equation}
\footnote{\textit{Proof of (\ref{1702.pf.der-2}):} Let $\left(  a,b\right)
\in H\times H$. The definition of the map $\mathfrak{e}^{\prime}$ yields
$\mathfrak{e}^{\prime}\left(  ab\right)  =\mathfrak{e}\left(  ab\right)  $ and
$\mathfrak{e}^{\prime}\left(  a\right)  =\mathfrak{e}\left(  a\right)  $ and
$\mathfrak{e}^{\prime}\left(  b\right)  =\mathfrak{e}\left(  b\right)  $. But
recall that $\mathfrak{q}=\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(
H\right)  }\circ\mathfrak{e}^{\prime}$. Hence,%
\begin{equation}
\underbrace{\mathfrak{q}}_{=\operatorname*{syminc}\nolimits_{\mathfrak{e}%
\left(  H\right)  }\circ\mathfrak{e}^{\prime}}\left(  a\right)  =\left(
\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }%
\circ\mathfrak{e}^{\prime}\right)  \left(  a\right)  =\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }\left(  \underbrace{\mathfrak{e}%
^{\prime}\left(  a\right)  }_{=\mathfrak{e}\left(  a\right)  }\right)
=\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }\left(
\mathfrak{e}\left(  a\right)  \right)  . \label{1702.pf.der-2.pf.1}%
\end{equation}
The same argument (applied to $b$ instead of $a$) shows that%
\begin{equation}
\mathfrak{q}\left(  b\right)  =\operatorname*{syminc}\nolimits_{\mathfrak{e}%
\left(  H\right)  }\left(  \mathfrak{e}\left(  b\right)  \right)  .
\label{1702.pf.der-2.pf.2}%
\end{equation}
Finally,%
\begin{align*}
\underbrace{\mathfrak{q}}_{=\operatorname*{syminc}\nolimits_{\mathfrak{e}%
\left(  H\right)  }\circ\mathfrak{e}^{\prime}}\left(  ab\right)   &  =\left(
\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }%
\circ\mathfrak{e}^{\prime}\right)  \left(  ab\right)  =\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }\left(  \underbrace{\mathfrak{e}%
^{\prime}\left(  ab\right)  }_{=\mathfrak{e}\left(  ab\right)  }\right)
=\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }\left(
\underbrace{\mathfrak{e}\left(  ab\right)  }_{\substack{=\mathfrak{e}\left(
a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)
\mathfrak{e}\left(  b\right)  \\\text{(by (\ref{1702.pf.der-1}))}}}\right) \\
&  =\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }\left(
\mathfrak{e}\left(  a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon
_{H}\left(  a\right)  \mathfrak{e}\left(  b\right)  \right) \\
&  =\underbrace{\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)
}\left(  \mathfrak{e}\left(  a\right)  \right)  }_{\substack{=\mathfrak{q}%
\left(  a\right)  \\\text{(by (\ref{1702.pf.der-2.pf.1}))}}}\varepsilon
_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)
\underbrace{\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)
}\left(  \mathfrak{e}\left(  b\right)  \right)  }_{\substack{=\mathfrak{q}%
\left(  b\right)  \\\text{(by (\ref{1702.pf.der-2.pf.2}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }\text{ is }k\text{-linear}\right) \\
&  =\mathfrak{q}\left(  a\right)  \varepsilon_{H}\left(  b\right)
+\varepsilon_{H}\left(  a\right)  \mathfrak{q}\left(  b\right)  .
\end{align*}
This proves (\ref{1702.pf.der-2}).}.

Definition~\ref{Definition109} (applied to $A=\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  $ and $f=\mathfrak{q}$) shows that the
map $\mathfrak{q}$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation if and only if every $\left(  a,b\right)  \in H\times H$
satisfies $\mathfrak{q}\left(  ab\right)  =\mathfrak{q}\left(  a\right)
\varepsilon_{H}\left(  b\right)  +\varepsilon_{H}\left(  a\right)
\mathfrak{q}\left(  b\right)  $. Thus, the map $\mathfrak{q}$ is an $\left(
\varepsilon_{H},\varepsilon_{H}\right)  $-derivation (since every $\left(
a,b\right)  \in H\times H$ satisfies $\mathfrak{q}\left(  ab\right)
=\mathfrak{q}\left(  a\right)  \varepsilon_{H}\left(  b\right)  +\varepsilon
_{H}\left(  a\right)  \mathfrak{q}\left(  b\right)  $). Therefore,
Proposition~\ref{Proposition1735} (applied to $A=\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  $ and $h=\mathfrak{q}$) yields
$\mathfrak{q}\in\mathfrak{g}\left(  H,\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  \right)  $. Hence, the map
$e^{\ast\mathfrak{q}}:H\rightarrow\operatorname*{Sym}\left(  \mathfrak{e}%
\left(  H\right)  \right)  $ is well-defined.

Furthermore, the subset $\mathfrak{q}\left(  H\right)  $ of
$\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $ generates
the $k$-algebra $\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)
\right)  $\ \ \ \ \footnote{\textit{Proof.} We have
\[
\mathfrak{e}^{\prime}\left(  H\right)  =\left\{  \underbrace{\mathfrak{e}%
^{\prime}\left(  h\right)  }_{\substack{=\mathfrak{e}\left(  h\right)
\\\text{(by the definition of }\mathfrak{e}^{\prime}\text{)}}}\ \mid\ h\in
H\right\}  =\left\{  \mathfrak{e}\left(  h\right)  \ \mid\ h\in H\right\}
=\mathfrak{e}\left(  H\right)  .
\]
\par
On the other hand, the definition of $\mathfrak{q}$ yields $\mathfrak{q}%
=\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }%
\circ\mathfrak{e}^{\prime}$. Hence,%
\begin{align*}
\underbrace{\mathfrak{q}}_{=\operatorname*{syminc}\nolimits_{\mathfrak{e}%
\left(  H\right)  }\circ\mathfrak{e}^{\prime}}\left(  H\right)   &  =\left(
\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }%
\circ\mathfrak{e}^{\prime}\right)  \left(  H\right)  =\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }\left(  \mathfrak{e}^{\prime}\left(
H\right)  \right)  =\mathfrak{e}^{\prime}\left(  H\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }\text{ is merely an inclusion
map}\right) \\
&  =\mathfrak{e}\left(  H\right)  .
\end{align*}
\par
It is well-known that if $V$ is a $k$-vector space, then the subset $V$ of
$\operatorname*{Sym}V$ generates the $k$-algebra $\operatorname*{Sym}V$.
Applying this to $V=\mathfrak{e}\left(  H\right)  $, we conclude that the
subset $\mathfrak{e}\left(  H\right)  $ of $\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  $ generates the $k$-algebra
$\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $. In other
words, the subset $\mathfrak{q}\left(  H\right)  $ of $\operatorname*{Sym}%
\left(  \mathfrak{e}\left(  H\right)  \right)  $ generates the $k$-algebra
$\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $ (since
$\mathfrak{q}\left(  H\right)  =\mathfrak{e}\left(  H\right)  $).}. Hence,
Corollary~\ref{Corollary1734} (applied to $A=\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  $ and $f=\mathfrak{q}$) shows that the
map $e^{\ast\mathfrak{q}}:H\rightarrow\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  $ is a surjective $k$-algebra homomorphism.

On the other hand, $\operatorname*{symlift}\mathfrak{j}$ is the unique
$k$-algebra homomorphism $\Phi:\operatorname*{Sym}\left(  \mathfrak{e}\left(
H\right)  \right)  \rightarrow H$ satisfying $\Phi\circ\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }=\mathfrak{j}$ (by the definition of
$\operatorname*{symlift}\mathfrak{j}$). Thus, $\operatorname*{symlift}%
\mathfrak{j}$ is a $k$-algebra homomorphism $\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  \rightarrow H$ and satisfies $\left(
\operatorname*{symlift}\mathfrak{j}\right)  \circ\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }=\mathfrak{j}$. Therefore, $\left(
\operatorname*{symlift}\mathfrak{j}\right)  \circ\mathfrak{q}=\mathfrak{e}%
$\ \ \ \ \footnote{\textit{Proof.} Each $h\in H$ satisfies%
\begin{align*}
\left(  \mathfrak{j}\circ\mathfrak{e}^{\prime}\right)  \left(  h\right)   &
=\mathfrak{j}\left(  \mathfrak{e}^{\prime}\left(  h\right)  \right)
=\mathfrak{e}^{\prime}\left(  h\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mathfrak{j}\text{ is just an inclusion map}\right) \\
&  =\mathfrak{e}\left(  h\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mathfrak{e}\right)  .
\end{align*}
Thus, $\mathfrak{j}\circ\mathfrak{e}^{\prime}=\mathfrak{e}$ (since both
$\mathfrak{j}\circ\mathfrak{e}^{\prime}$ and $\mathfrak{e}$ are maps from $H$
to $H$). Now,%
\[
\left(  \operatorname*{symlift}\mathfrak{j}\right)  \circ
\underbrace{\mathfrak{q}}_{=\operatorname*{syminc}\nolimits_{\mathfrak{e}%
\left(  H\right)  }\circ\mathfrak{e}^{\prime}}=\underbrace{\left(
\operatorname*{symlift}\mathfrak{j}\right)  \circ\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }}_{=\mathfrak{j}}\circ
\mathfrak{e}^{\prime}=\mathfrak{j}\circ\mathfrak{e}^{\prime}=\mathfrak{e}.
\]
}.

Now, Proposition~\ref{Proposition1711} \textbf{(d)} (applied to $C=H$,
$A=\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $, $B=H$,
$p=\operatorname*{symlift}\mathfrak{j}$ and $f=\mathfrak{q}$) shows that
$\left(  \operatorname*{symlift}\mathfrak{j}\right)  \circ\mathfrak{q}%
\in\mathfrak{g}\left(  H,\operatorname*{Sym}\left(  \mathfrak{e}\left(
H\right)  \right)  \right)  $ and $\left(  \operatorname*{symlift}%
\mathfrak{j}\right)  \circ e^{\ast\mathfrak{q}}=e^{\ast\left(  \left(
\operatorname*{symlift}\mathfrak{j}\right)  \circ\mathfrak{q}\right)  }$.

Now,%
\begin{align*}
\left(  \operatorname*{symlift}\mathfrak{j}\right)  \circ e^{\ast
\mathfrak{q}}  &  =e^{\ast\left(  \left(  \operatorname*{symlift}%
\mathfrak{j}\right)  \circ\mathfrak{q}\right)  }=e^{\ast\mathfrak{e}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \operatorname*{symlift}%
\mathfrak{j}\right)  \circ\mathfrak{q}=\mathfrak{e}\right) \\
&  =\operatorname*{id}.
\end{align*}
Since $e^{\ast\mathfrak{q}}$ is surjective, we can thus easily obtain
$e^{\ast\mathfrak{q}}\circ\left(  \operatorname*{symlift}\mathfrak{j}\right)
=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} Let $x\in
\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $. Thus,
$x\in\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)
=e^{\ast\mathfrak{q}}\left(  H\right)  $ (since the map $e^{\ast\mathfrak{q}}$
is surjective). In other words, there exists some $y\in H$ such that
$x=e^{\ast\mathfrak{q}}\left(  y\right)  $. Consider this $y$. Now, $\left(
\left(  \operatorname*{symlift}\mathfrak{j}\right)  \circ e^{\ast\mathfrak{q}%
}\right)  \left(  y\right)  =\left(  \operatorname*{symlift}\mathfrak{j}%
\right)  \left(  \underbrace{e^{\ast\mathfrak{q}}\left(  y\right)  }%
_{=x}\right)  =\left(  \operatorname*{symlift}\mathfrak{j}\right)  \left(
x\right)  $, so that $\left(  \operatorname*{symlift}\mathfrak{j}\right)
\left(  x\right)  =\underbrace{\left(  \left(  \operatorname*{symlift}%
\mathfrak{j}\right)  \circ e^{\ast\mathfrak{q}}\right)  }_{=\operatorname*{id}%
}\left(  y\right)  =\operatorname*{id}\left(  y\right)  =y$. Now,%
\[
\left(  e^{\ast\mathfrak{q}}\circ\left(  \operatorname*{symlift}%
\mathfrak{j}\right)  \right)  \left(  x\right)  =e^{\ast\mathfrak{q}}\left(
\underbrace{\left(  \operatorname*{symlift}\mathfrak{j}\right)  \left(
x\right)  }_{=y}\right)  =e^{\ast\mathfrak{q}}\left(  y\right)
=x=\operatorname*{id}\left(  x\right)  .
\]
\par
Now, forget that we fixed $x$. We thus have shown that $\left(  e^{\ast
\mathfrak{q}}\circ\left(  \operatorname*{symlift}\mathfrak{j}\right)  \right)
\left(  x\right)  =\operatorname*{id}\left(  x\right)  $ for each
$x\in\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $. In
other words, $e^{\ast\mathfrak{q}}\circ\left(  \operatorname*{symlift}%
\mathfrak{j}\right)  =\operatorname*{id}$.}.

We now have shown the two equalities $\left(  \operatorname*{symlift}%
\mathfrak{j}\right)  \circ e^{\ast\mathfrak{q}}=\operatorname*{id}$ and
$e^{\ast\mathfrak{q}}\circ\left(  \operatorname*{symlift}\mathfrak{j}\right)
=\operatorname*{id}$. Combining these equalities, we conclude that the maps
$\operatorname*{symlift}\mathfrak{j}:\operatorname*{Sym}\left(  \mathfrak{e}%
\left(  H\right)  \right)  \rightarrow H$ and $e^{\ast\mathfrak{q}%
}:H\rightarrow\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)
\right)  $ are mutually inverse. Furthermore, we have already shown that these
two maps are $k$-algebra homomorphisms. This completes the proof of
Theorem~\ref{Theorem1702} \textbf{(d)}.

\textbf{(e)} Theorem~\ref{Theorem1702} \textbf{(d)} shows that the two
$k$-algebra homomorphisms $\operatorname*{symlift}\mathfrak{j}%
:\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  \rightarrow
H$ and $e^{\ast\mathfrak{q}}:H\rightarrow\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  $ are mutually inverse. Hence, these
two $k$-algebra homomorphisms are invertible, and thus are $k$-algebra
isomorphisms. In particular, $\operatorname*{symlift}\mathfrak{j}%
:\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  \rightarrow
H$ is a $k$-algebra isomorphism. Thus, $\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  \cong H$ as $k$-algebras. This proves
Theorem~\ref{Theorem1702} \textbf{(e).}
\end{proof}

\begin{remark}
\label{Remark1751} It is possible to slightly improve
Theorem~\ref{Theorem1702}: Namely, the symmetric algebra $\operatorname*{Sym}%
\left(  \mathfrak{e}\left(  H\right)  \right)  $ canonically becomes a
filtered $k$-algebra\footnote{Indeed, whenever $V$ is a filtered $k$-vector
space, its symmetric algebra $\operatorname*{Sym}V$ becomes a filtered
$k$-algebra. In our situation, we can apply this to $V=\mathfrak{e}\left(
H\right)  $ (indeed, the vector space $\mathfrak{e}\left(  H\right)  $ becomes
filtered by virtue of being a subspace of $H$), and thus conclude that
$\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $
canonically becomes a filtered $k$-algebra.}. With respect to this filtration,
the isomorphism $\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)
\right)  \cong H$ claimed in Theorem~\ref{Theorem1702} \textbf{(e)} is
actually an isomorphism of \textbf{filtered} algebras, since both
homomorphisms $\operatorname*{symlift}\mathfrak{j}:\operatorname*{Sym}\left(
\mathfrak{e}\left(  H\right)  \right)  \rightarrow H$ and $e^{\ast
\mathfrak{q}}:H\rightarrow\operatorname*{Sym}\left(  \mathfrak{e}\left(
H\right)  \right)  $ respect the filtration. Proving this is not too difficult
(it is mostly an issue of bookkeeping).
\end{remark}

\section{\label{S62}More on symmetric algebras}

Next, we shall show some more properties of the symmetric algebra
$\operatorname*{Sym}V$ of a $k$-vector space $V$. We begin with the following fact:

\begin{lemma}
\label{Lemma1801} Let $k$ be a field of characteristic $0$. Let $V$ and $W$ be
two $k$-vector spaces. Let $f:\operatorname*{Sym}V\rightarrow W$ and
$g:\operatorname*{Sym}V\rightarrow W$ be two $k$-linear maps. Assume that%
\begin{equation}
f\left(  a^{n}\right)  =g\left(  a^{n}\right)  \ \ \ \ \ \ \ \ \ \ \text{for
each }a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \text{ and
}n\in\mathbb{N}. \label{1801.ass}%
\end{equation}
\footnote{See Definition~\ref{Definition1701} for the definition of the map
$\operatorname*{syminc}\nolimits_{V}:V\rightarrow\operatorname*{Sym}V$.} Then,
$f=g$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1801}.]Let $A$ be the $k$-algebra
$\operatorname*{Sym}V$. Recall that $\operatorname*{syminc}\nolimits_{V}$ is a
$k$-linear map $V\rightarrow\operatorname*{Sym}V$. In other words,
$\operatorname*{syminc}\nolimits_{V}$ is a $k$-linear map $V\rightarrow A$
(since $\operatorname*{Sym}V=A$).

Let $U=\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $. Then, $U$ is a
$k$-vector subspace of $A$ (since $\operatorname*{syminc}\nolimits_{V}$ is a
$k$-linear map $V\rightarrow A$). Also, $xy-yx\in U$ for any $x\in U$ and
$y\in U$\ \ \ \ \footnote{\textit{Proof.} Let $x\in U$ and $y\in U$. Then,
$x\in U\subseteq A$ and $y\in U\subseteq A$. But the $k$-algebra $A$ is
commutative (since $A=\operatorname*{Sym}V$ is the symmetric algebra of a
$k$-vector space). Hence, $xy=yx$ (since $x\in A$ and $y\in A$). Thus,
$xy-yx=0\in U$ (since $U$ is a $k$-vector subspace of $A$). Qed.}.

Consider the tensor algebra $\otimes V$. We identify $V$ with a $k$-vector
subspace of $\otimes V$ by viewing $V$ as the addend $V^{\otimes1}$ in the
direct sum $\bigoplus_{n\in\mathbb{N}}V^{\otimes n}=\otimes V$. Thus, the
canonical injection $V\rightarrow\otimes V$ becomes an inclusion map.

Recall that the symmetric algebra $\operatorname*{Sym}V$ is defined as a
quotient algebra of $\otimes V$. Let $\pi$ be the canonical projection
$\otimes V\rightarrow\operatorname*{Sym}V$. Then, $\pi$ is a surjective
$k$-algebra homomorphism.

The canonical map $\operatorname*{syminc}\nolimits_{V}:V\rightarrow
\operatorname*{Sym}V$ factors through the projection $\pi:\otimes
V\rightarrow\operatorname*{Sym}V$. More precisely, $\operatorname*{syminc}%
\nolimits_{V}\left(  x\right)  =\pi\left(  x\right)  $ for every $x\in V$
(this follows from the definition of $\operatorname*{syminc}\nolimits_{V}$).
Now,%
\begin{align}
U  &  =\operatorname*{syminc}\nolimits_{V}\left(  V\right)  =\left\{
\underbrace{\operatorname*{syminc}\nolimits_{V}\left(  x\right)  }%
_{=\pi\left(  x\right)  }\ \mid\ x\in V\right\}  =\left\{  \pi\left(
x\right)  \ \mid\ x\in V\right\} \nonumber\\
&  =\pi\left(  V\right)  . \label{1801.pf.2}%
\end{align}


By the definition of the tensor algebra, we have $\otimes V=\bigoplus
_{n\in\mathbb{N}}V^{\otimes n}$. Thus, $\otimes V=\bigoplus_{n\in\mathbb{N}%
}V^{\otimes n}=\sum_{n\in\mathbb{N}}V^{\otimes n}$ (since direct sums are sums).

For every $n\in\mathbb{N}$, we have $V^{\otimes n}=V^{n}$ as $k$-vector
subspaces of $\otimes V$ (where $V^{n}$ means $\underbrace{V\cdot V\cdot
\cdots\cdot V}_{n\text{ times}}$, as usual)\ \ \ \ \footnote{\textit{Proof.}
Let $n\in\mathbb{N}$. The $n$-th tensor power $V^{\otimes n}$ is spanned by
all pure tensors. In other words,%
\begin{align*}
V^{\otimes n}  &  =\left\langle \underbrace{v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}}_{\substack{=v_{1}v_{2}\cdots v_{n}\\\text{(because the
multiplication on }\otimes V\text{ is the}\\\text{tensor product, so we
have}\\v_{1}v_{2}\cdots v_{n}=v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\text{)}}}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\right\rangle \\
&  =\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\rangle .
\end{align*}
Compared with%
\[
V^{n}=\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\rangle ,
\]
this yields $V^{\otimes n}=V^{n}$, qed.}. Hence, $\sum_{n\in\mathbb{N}%
}V^{\otimes n}=\sum_{n\in\mathbb{N}}V^{n}$. Thus, $\otimes V=\sum
_{n\in\mathbb{N}}V^{\otimes n}=\sum_{n\in\mathbb{N}}V^{n}$.

Now, the map $\pi$ is surjective (since it is a projection). Hence,%
\begin{align*}
A  &  =\pi\left(  \underbrace{\otimes V}_{=\sum_{n\in\mathbb{N}}V^{n}}\right)
=\pi\left(  \sum_{n\in\mathbb{N}}V^{n}\right)  =\sum_{n\in\mathbb{N}}\left(
\underbrace{\pi\left(  V\right)  }_{\substack{=U\\\text{(by (\ref{1801.pf.2}%
))}}}\right)  ^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  =\sum_{n\in\mathbb{N}}U^{n}.
\end{align*}


Recall that $f:\operatorname*{Sym}V\rightarrow W$ and $g:\operatorname*{Sym}%
V\rightarrow W$ are two $k$-linear maps. In other words, $f:A\rightarrow W$
and $g:A\rightarrow W$ are two $k$-linear maps (since $A=\operatorname*{Sym}V$).

We have $f\left(  v^{n}\right)  =g\left(  v^{n}\right)  $ for every $v\in U$
and $n\in\mathbb{N}$\ \ \ \ \footnote{\textit{Proof.} Let $v\in U$ and
$n\in\mathbb{N}$. Then, $v\in U=\operatorname*{syminc}\nolimits_{V}\left(
V\right)  $. Hence, (\ref{1801.ass}) (applied to $a=v$) yields $f\left(
v^{n}\right)  =g\left(  v^{n}\right)  $. Qed.}. Hence,
Corollary~\ref{Corollary1147} (applied to $U$ instead of $V$) yields that
$f=g$. This proves Lemma~\ref{Lemma1801}.
\end{proof}

An easy corollary of Lemma~\ref{Lemma1801} (somewhat more suited to certain
applications) is the following fact:

\begin{lemma}
\label{Lemma1802} Let $k$ be a field of characteristic $0$. Let $V$ and $W$ be
two $k$-vector spaces. Let $f:\operatorname*{Sym}V\rightarrow W$ and
$g:\operatorname*{Sym}V\rightarrow W$ be two $k$-linear maps. Assume that%
\begin{equation}
f\left(  a^{n}\right)  =g\left(  a^{n}\right)  \label{1802.ass}%
\end{equation}
for each $a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)
$\ \ \ \ \footnote{See Definition~\ref{Definition1701} for the definition of
the map $\operatorname*{syminc}\nolimits_{V}:V\rightarrow\operatorname*{Sym}%
V$.} and each positive integer $n\in\mathbb{N}$. Assume further that $f\left(
1\right)  =g\left(  1\right)  $. Then, $f=g$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1802}.]We have $f\left(  a^{n}\right)  =g\left(
a^{n}\right)  $ for each $a\in\operatorname*{syminc}\nolimits_{V}\left(
V\right)  $ and $n\in\mathbb{N}$\ \ \ \ \footnote{\textit{Proof.} Let
$a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ and $n\in
\mathbb{N}$. We must show that $f\left(  a^{n}\right)  =g\left(  a^{n}\right)
$.
\par
Clearly, $n$ is an integer. If $n$ is positive, then $f\left(  a^{n}\right)
=g\left(  a^{n}\right)  $ follows immediately from (\ref{1802.ass}). Thus, for
the rest of this proof, we can WLOG assume that $n$ is not positive. Assume
this.
\par
We have $n\leq0$ (since $n$ is not positive). Hence, $n=0$ (since
$n\in\mathbb{N}$). Thus, $a^{n}=a^{0}=1$. Hence, $f\left(  \underbrace{a^{n}%
}_{=1}\right)  =f\left(  1\right)  =g\left(  \underbrace{1}_{=a^{n}}\right)
=g\left(  a^{n}\right)  $. Qed.}. Thus, Lemma~\ref{Lemma1801} shows that
$f=g$. This proves Lemma~\ref{Lemma1802}.
\end{proof}

Next, we recall the notion of a \textit{symmetric} map:

\begin{definition}
\label{Definition1803} Let $V$ and $W$ be two sets. Let $n\in\mathbb{N}$. A
map $f:V^{\times n}\rightarrow W$ is said to be \textit{symmetric} if each
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ and each
$\gamma\in S_{n}$ satisfy $f\left(  v_{\gamma\left(  1\right)  }%
,v_{\gamma\left(  2\right)  },\ldots,v_{\gamma\left(  n\right)  }\right)
=f\left(  v_{1},v_{2},\ldots,v_{n}\right)  $. (Recall that $S_{n}$ denotes the
$n$-th symmetric group.)
\end{definition}

Now, we can state the well-known universal property of the $n$-th symmetric power:

\begin{proposition}
\label{Proposition1804} Let $k$ be a field. Let $V$ be a $k$-vector space. Let
$n\in\mathbb{N}$. Let $\pi_{n}$ be the canonical projection $V^{\otimes
n}\rightarrow\operatorname*{Sym}\nolimits^{n}V$. Let $W$ be any $k$-vector
space. Let $f:V^{\times n}\rightarrow W$ be any symmetric $k$-multilinear map.
Then, there exists a unique $k$-linear map $f_{\operatorname*{Sym}%
}:\operatorname*{Sym}\nolimits^{n}V\rightarrow W$ such that every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfies
$f_{\operatorname*{Sym}}\left(  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1804}.]Proposition~\ref{Proposition1804}
is precisely \cite[Corollary 95]{tensorext} (in the particular case where $k$
is a field)\footnote{Notice that the projection that we denote by $\pi_{n}$
has been called $\operatorname*{sym}\nolimits_{V,n}$ in \cite[Corollary
95]{tensorext}.}.
\end{proof}

We are now ready for a general construction:

\begin{proposition}
\label{Proposition1805} Let $k$ be a field of characteristic $0$. Let $A$ be a
$k$-algebra. Let $V$ be a $k$-vector subspace of $A$. Then, there exists a
unique $k$-linear map $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow A$
such that every $n\in\mathbb{N}$ and every $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}$ satisfy%
\begin{align}
&  \operatorname*{spr}\left(  \operatorname*{syminc}\nolimits_{V}\left(
v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)
\cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  \right)
\nonumber\\
&  =\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }%
v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }.
\label{1805.eq}%
\end{align}

\end{proposition}

Before we prove Proposition~\ref{Proposition1805}, let us recall the universal
property of direct sums of vector spaces:

\begin{proposition}
\label{Proposition1806} Let $k$ be a field. Let $T$ be a set. Let $\left(
U_{t}\right)  _{t\in T}$ be a family of $k$-vector spaces. For each $t\in T$,
let $\iota_{t}:U_{t}\rightarrow\bigoplus_{s\in T}U_{s}$ be the canonical
inclusion of the addend $U_{t}$ into the direct sum $\bigoplus_{s\in T}U_{s}$.

Let $W$ be a $k$-vector space. For each $t\in T$, let $f_{t}:U_{t}\rightarrow
W$ be a $k$-linear map. Then, there exists a unique $k$-linear map
$f:\bigoplus_{s\in T}U_{s}\rightarrow W$ such that each $t\in T$ satisfies
$f_{t}=f\circ\iota_{t}$.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1805}.]Let $n\in\mathbb{N}$. Let
$\pi_{n}$ be the canonical projection $V^{\otimes n}\rightarrow
\operatorname*{Sym}\nolimits^{n}V$. This projection $\pi_{n}$ is $k$-linear
and surjective (since it is a projection).

Let us define a map $\varphi_{n}:V^{\times n}\rightarrow A$ by%
\[
\left(  \varphi_{n}\left(  v_{1},v_{2},\ldots,v_{n}\right)  =\dfrac{1}{n!}%
\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right)  .
\]
Then, this map $\varphi_{n}$ is clearly $n$-multilinear. Moreover, this map
$\varphi_{n}$ is symmetric\footnote{\textit{Proof.} Let $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ and $\gamma\in S_{n}$.
\par
Recall that $S_{n}$ is a group. Hence, the map $S_{n}\rightarrow
S_{n},\ \sigma\mapsto\gamma\circ\sigma$ is a bijection (since $\gamma\in
S_{n}$). But the definition of $\varphi_{n}$ yields%
\begin{align*}
\varphi_{n}\left(  v_{\gamma\left(  1\right)  },v_{\gamma\left(  2\right)
},\ldots,v_{\gamma\left(  n\right)  }\right)   &  =\dfrac{1}{n!}\sum
_{\sigma\in S_{n}}\underbrace{v_{\gamma\left(  \sigma\left(  1\right)
\right)  }v_{\gamma\left(  \sigma\left(  2\right)  \right)  }\cdots
v_{\gamma\left(  \sigma\left(  n\right)  \right)  }}_{\substack{=v_{\left(
\gamma\circ\sigma\right)  \left(  1\right)  }v_{\left(  \gamma\circ
\sigma\right)  \left(  2\right)  }\cdots v_{\left(  \gamma\circ\sigma\right)
\left(  n\right)  }\\\text{(since }v_{\gamma\left(  \sigma\left(  i\right)
\right)  }=v_{\left(  \gamma\circ\sigma\right)  \left(  i\right)  }\text{ for
each }i\in\left\{  1,2,\ldots,n\right\}  \text{)}}}\\
&  =\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\left(  \gamma\circ\sigma\right)
\left(  1\right)  }v_{\left(  \gamma\circ\sigma\right)  \left(  2\right)
}\cdots v_{\left(  \gamma\circ\sigma\right)  \left(  n\right)  }=\dfrac{1}%
{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(
2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\end{align*}
(here, we have substituted $\sigma$ for $\gamma\circ\sigma$ in the sum, since
the map $S_{n}\rightarrow S_{n},\ \sigma\mapsto\gamma\circ\sigma$ is a
bijection). Comparing this with%
\[
\varphi_{n}\left(  v_{1},v_{2},\ldots,v_{n}\right)  =\dfrac{1}{n!}\sum
_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\varphi_{n}\right)  ,
\]
we obtain $\varphi_{n}\left(  v_{\gamma\left(  1\right)  },v_{\gamma\left(
2\right)  },\ldots,v_{\gamma\left(  n\right)  }\right)  =\varphi_{n}\left(
v_{1},v_{2},\ldots,v_{n}\right)  $.
\par
Now, forget that we fixed $\left(  v_{1},v_{2},\ldots,v_{n}\right)  $ and
$\gamma$. We thus have shown that each $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}$ and each $\gamma\in S_{n}$ satisfy
$\varphi_{n}\left(  v_{\gamma\left(  1\right)  },v_{\gamma\left(  2\right)
},\ldots,v_{\gamma\left(  n\right)  }\right)  =\varphi_{n}\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  $. In other words, the map $\varphi_{n}$ is
symmetric (since the map $\varphi_{n}$ is symmetric if and only if each
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ and each
$\gamma\in S_{n}$ satisfy $\varphi_{n}\left(  v_{\gamma\left(  1\right)
},v_{\gamma\left(  2\right)  },\ldots,v_{\gamma\left(  n\right)  }\right)
=\varphi_{n}\left(  v_{1},v_{2},\ldots,v_{n}\right)  $ (by the definition of
\textquotedblleft symmetric\textquotedblright)).}. Hence,
Proposition~\ref{Proposition1804} (applied to $W=A$ and $f=\varphi_{n}$) shows
that there exists a unique $k$-linear map $f_{\operatorname*{Sym}%
}:\operatorname*{Sym}\nolimits^{n}V\rightarrow A$ such that every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfies
$f_{\operatorname*{Sym}}\left(  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \right)  =\varphi_{n}\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  $. Consider this map $f_{\operatorname*{Sym}}$, and
denote it by $\psi_{n}$.

Thus, $\psi_{n}$ is a $k$-linear map $f_{\operatorname*{Sym}}%
:\operatorname*{Sym}\nolimits^{n}V\rightarrow A$ such that every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfies
$f_{\operatorname*{Sym}}\left(  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \right)  =\varphi_{n}\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  $. In other words, $\psi_{n}$ is a $k$-linear map
$\operatorname*{Sym}\nolimits^{n}V\rightarrow A$ and has the property that
every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfies%
\begin{equation}
\psi_{n}\left(  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  \right)  =\varphi_{n}\left(  v_{1},v_{2},\ldots,v_{n}\right)  .
\label{1805.pf.psin-1}%
\end{equation}


Now, each $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$
satisfies%
\begin{equation}
\operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  =\pi_{n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
\label{1805.pf.syminc-prod}%
\end{equation}
\footnote{\textit{Proof of (\ref{1805.pf.syminc-prod}):} Let $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$.
\par
Consider the tensor algebra $\otimes V$. We identify $V$ with a $k$-vector
subspace of $\otimes V$ by viewing $V$ as the addend $V^{\otimes1}$ in the
direct sum $\bigoplus_{n\in\mathbb{N}}V^{\otimes n}=\otimes V$. Thus, the
canonical injection $V\rightarrow\otimes V$ becomes an inclusion map.
\par
Recall that the symmetric algebra $\operatorname*{Sym}V$ is defined as a
quotient algebra of $\otimes V$. Let $\pi$ be the canonical projection
$\otimes V\rightarrow\operatorname*{Sym}V$. Then, $\pi$ is a surjective
$k$-algebra homomorphism.
\par
The projection $\pi_{n}:V^{\otimes n}\rightarrow\operatorname*{Sym}%
\nolimits^{n}V$ is a restriction of the projection $\pi:\otimes V\rightarrow
\operatorname*{Sym}V$ (provided that we regard $V^{\otimes n}$ and
$\operatorname*{Sym}\nolimits^{n}V$ as $k$-vector subspaces of $\otimes V$ and
$\operatorname*{Sym}V$, respectively). Thus,%
\begin{equation}
\pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
=\pi\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  .
\label{1805.pf.psin-2.pf.0}%
\end{equation}
\par
The canonical map $\operatorname*{syminc}\nolimits_{V}:V\rightarrow
\operatorname*{Sym}V$ factors through the projection $\pi:\otimes
V\rightarrow\operatorname*{Sym}V$. More precisely,
\begin{equation}
\operatorname*{syminc}\nolimits_{V}\left(  x\right)  =\pi\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }x\in V \label{1805.pf.psin-2.pf.1}%
\end{equation}
(this follows from the definition of $\operatorname*{syminc}\nolimits_{V}$).
Now, each $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies the equality
$\operatorname*{syminc}\nolimits_{V}\left(  v_{i}\right)  =\pi\left(
v_{i}\right)  $ (by (\ref{1805.pf.psin-2.pf.1}), applied to $x=v_{i}$).
Multiplying these equalities for all $i\in\left\{  1,2,\ldots,n\right\}  $, we
obtain%
\begin{align*}
&  \operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right) \\
&  =\pi\left(  v_{1}\right)  \pi\left(  v_{2}\right)  \cdots\pi\left(
v_{n}\right) \\
&  =\pi\left(  \underbrace{v_{1}v_{2}\cdots v_{n}}_{\substack{=v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\\\text{(since the multiplication in
the}\\\text{algebra }\otimes V\text{ is the tensor product)}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  =\pi\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  =\pi
_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
\end{align*}
(by (\ref{1805.pf.psin-2.pf.0})). This proves (\ref{1805.pf.syminc-prod}).}
and%
\begin{equation}
\psi_{n}\left(  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  \right)  =\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }
\label{1805.pf.psin-3}%
\end{equation}
\footnote{\textit{Proof of (\ref{1805.pf.psin-3}):} Let $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{\times n}$. Then, (\ref{1805.pf.psin-1})
yields%
\[
\psi_{n}\left(  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  \right)  =\varphi_{n}\left(  v_{1},v_{2},\ldots,v_{n}\right)
=\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }%
v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\]
(by the definition of $\varphi_{n}$). This proves (\ref{1805.pf.psin-3}).}.

Now, forget that we fixed $n$. Thus, for each $n\in\mathbb{N}$, we have
constructed a $k$-linear map $\psi_{n}:\operatorname*{Sym}\nolimits^{n}%
V\rightarrow A$, and we have shown that each $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}$ satisfies (\ref{1805.pf.syminc-prod}) and
(\ref{1805.pf.psin-3}).

Recall that $\operatorname*{Sym}V=\bigoplus_{n\in\mathbb{N}}%
\operatorname*{Sym}\nolimits^{n}V=\bigoplus_{s\in\mathbb{N}}%
\operatorname*{Sym}\nolimits^{s}V$ (here, we have renamed the index $n$ as
$s$). For each $t\in\mathbb{N}$, let $\iota_{t}:\operatorname*{Sym}%
\nolimits^{t}V\rightarrow\bigoplus_{s\in\mathbb{N}}\operatorname*{Sym}%
\nolimits^{s}V$ be the canonical inclusion of the addend $\operatorname*{Sym}%
\nolimits^{t}V$ into the direct sum $\bigoplus_{s\in\mathbb{N}}%
\operatorname*{Sym}\nolimits^{s}V$. We shall use this inclusion to identify
$\operatorname*{Sym}\nolimits^{t}V$ with a subspace of $\bigoplus
_{s\in\mathbb{N}}\operatorname*{Sym}\nolimits^{s}V=\operatorname*{Sym}V$.
Thus,%
\begin{equation}
\iota_{t}\left(  q\right)  =q\ \ \ \ \ \ \ \ \ \ \text{for each }%
t\in\mathbb{N}\text{ and }q\in\operatorname*{Sym}\nolimits^{t}V.
\label{1805.pf.identify}%
\end{equation}


Proposition~\ref{Proposition1806} (applied to $T=\mathbb{N}$, $\left(
U_{t}\right)  _{t\in T}=\left(  \operatorname*{Sym}\nolimits^{t}V\right)
_{t\in\mathbb{N}}$, $W=A$ and $f_{t}=\psi_{t}$) shows that there exists a
unique $k$-linear map $f:\bigoplus_{s\in\mathbb{N}}\operatorname*{Sym}%
\nolimits^{s}V\rightarrow A$ such that each $t\in\mathbb{N}$ satisfies
$\psi_{t}=f\circ\iota_{t}$. Consider this $f$.

We know that $f$ is a $k$-linear map $\bigoplus_{s\in\mathbb{N}}%
\operatorname*{Sym}\nolimits^{s}V\rightarrow A$. In other words, $f$ is a
$k$-linear map $\operatorname*{Sym}V\rightarrow A$ (since $\operatorname*{Sym}%
V=\bigoplus_{s\in\mathbb{N}}\operatorname*{Sym}\nolimits^{s}V$).

Every $n\in\mathbb{N}$ and every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}$ satisfy%
\begin{align}
&  f\left(  \operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  \right) \nonumber\\
&  =\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }%
v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }
\label{1805.pf.it-is}%
\end{align}
\footnote{\textit{Proof of (\ref{1805.pf.it-is}):} Let $n\in\mathbb{N}$ and
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$. We must prove
(\ref{1805.pf.it-is}).
\par
Recall that each $t\in\mathbb{N}$ satisfies $\psi_{t}=f\circ\iota_{t}$.
Applying this to $t=n$, we obtain $\psi_{n}=f\circ\iota_{n}$.
\par
But (\ref{1805.pf.syminc-prod}) yields $\operatorname*{syminc}\nolimits_{V}%
\left(  v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)
\cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  =\pi_{n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \in\operatorname*{Sym}%
\nolimits^{n}V$. Thus, (\ref{1805.pf.identify}) (applied to $t=n$ and
$q=\operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  $) yields%
\[
\iota_{n}\left(  \operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  \right)
=\operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  .
\]
\par
Now,%
\begin{align*}
&  \underbrace{\psi_{n}}_{=f\circ\iota_{n}}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(
v_{2}\right)  \cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)
\right) \\
&  =\left(  f\circ\iota_{n}\right)  \left(  \operatorname*{syminc}%
\nolimits_{V}\left(  v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(
v_{2}\right)  \cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)
\right) \\
&  =f\left(  \underbrace{\iota_{n}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(
v_{2}\right)  \cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)
\right)  }_{=\operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  }\right) \\
&  =f\left(  \operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  \right)  .
\end{align*}
Hence,%
\begin{align*}
&  f\left(  \operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  \right) \\
&  =\psi_{n}\left(  \underbrace{\operatorname*{syminc}\nolimits_{V}\left(
v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)
\cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  }_{=\pi
_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  }\right) \\
&  =\psi_{n}\left(  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  \right)  =\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }.
\end{align*}
This proves (\ref{1805.pf.it-is}).}. Hence, there exists \textbf{at least one}
$k$-linear map $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow A$ such
that every $n\in\mathbb{N}$ and every $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}$ satisfy (\ref{1805.eq}) (namely,
$\operatorname*{spr}=f$).

Next, I shall show the following claim:

\textit{Claim 1:} Let $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow A$
be a $k$-linear map such that every $n\in\mathbb{N}$ and every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfy (\ref{1805.eq}).
Then, $\operatorname*{spr}=f$.

[\textit{Proof of Claim 1:} The maps $\operatorname*{spr}$ and $f$ are
$k$-linear. Thus, their difference $\operatorname*{spr}-f$ is $k$-linear as
well. Hence, $\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)  $ is a
$k$-vector subspace of $\operatorname*{Sym}V$.

Fix $n\in\mathbb{N}$. The $k$-vector space $V^{\otimes n}$ is spanned by its
pure tensors. In other words,%
\[
V^{\otimes n}=\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\rangle
.
\]
Thus,%
\begin{align*}
V^{\otimes n}  &  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\right\rangle \\
&  =\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle .
\end{align*}
Applying the map $\pi_{n}$ to both sides of this equality, we obtain%
\begin{align*}
\pi_{n}\left(  V^{\otimes n}\right)   &  =\pi_{n}\left(  \left\langle \left\{
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\}  \right\rangle \right) \\
&  =\left\langle \underbrace{\pi_{n}\left(  \left\{  v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\}  \right)  }_{=\left\{  \pi_{n}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\}  }\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{316.pf.3}), applied to }M=V^{\otimes n}\text{, }%
R=\operatorname*{Sym}\nolimits^{n}V\text{, }\phi=\pi_{n}\text{ and}\\
S=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\end{array}
\right) \\
&  =\left\langle \left\{  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{\times n}\right\}  \right\rangle .
\end{align*}
Comparing this with $\pi_{n}\left(  V^{\otimes n}\right)  =\operatorname*{Sym}%
\nolimits^{n}V$ (since the map $\pi_{n}$ is surjective), we obtain%
\begin{equation}
\operatorname*{Sym}\nolimits^{n}V=\left\langle \left\{  \pi_{n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}  \right\rangle .
\label{1805.pf.uniq-4}%
\end{equation}


Each $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfies%
\begin{align*}
&  \left(  \operatorname*{spr}-f\right)  \left(  \underbrace{\pi_{n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  }%
_{\substack{=\operatorname*{syminc}\nolimits_{V}\left(  v_{1}\right)
\operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  \\\text{(by
(\ref{1805.pf.syminc-prod}))}}}\right) \\
&  =\left(  \operatorname*{spr}-f\right)  \left(  \operatorname*{syminc}%
\nolimits_{V}\left(  v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(
v_{2}\right)  \cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)
\right) \\
&  =\underbrace{\operatorname*{spr}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(
v_{2}\right)  \cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)
\right)  }_{\substack{=\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\\\text{(by (\ref{1805.eq}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{f\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(
v_{2}\right)  \cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)
\right)  }_{\substack{=\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\\\text{(by (\ref{1805.pf.it-is}))}}}\\
&  =\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }%
v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }-\dfrac{1}%
{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(
2\right)  }\cdots v_{\sigma\left(  n\right)  }=0
\end{align*}
and therefore
\[
\pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
\in\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)  .
\]
In other words, we have%
\[
\left\{  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\subseteq\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)  .
\]
Hence, (\ref{284.pf.b.2}) (applied to $\operatorname*{Sym}V$, $\left\{
\pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}  $
and $\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)  $ instead of
$M$, $S$ and $Q$) yields%
\[
\left\langle \left\{  \pi_{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  \ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\right\}  \right\rangle \subseteq\operatorname*{Ker}\left(
\operatorname*{spr}-f\right)  .
\]
Now, (\ref{1805.pf.uniq-4}) becomes%
\begin{align*}
\operatorname*{Sym}\nolimits^{n}V  &  =\left\langle \left\{  \pi_{n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}  \right\rangle \\
&  \subseteq\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)  .
\end{align*}


Now, forget that we fixed $n$. We thus have shown that $\operatorname*{Sym}%
\nolimits^{n}V\subseteq\operatorname*{Ker}\left(  \operatorname*{spr}%
-f\right)  $ for each $n\in\mathbb{N}$. Hence,%
\[
\sum_{n\in\mathbb{N}}\underbrace{\operatorname*{Sym}\nolimits^{n}V}%
_{\subseteq\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)  }%
\subseteq\sum_{n\in\mathbb{N}}\operatorname*{Ker}\left(  \operatorname*{spr}%
-f\right)  \subseteq\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)
\]
(since $\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)  $ is a
$k$-vector space).

But
\begin{align*}
\operatorname*{Sym}V  &  =\bigoplus_{n\in\mathbb{N}}\operatorname*{Sym}%
\nolimits^{n}V=\sum_{n\in\mathbb{N}}\operatorname*{Sym}\nolimits^{n}%
V\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are sums}\right) \\
&  \subseteq\operatorname*{Ker}\left(  \operatorname*{spr}-f\right)  .
\end{align*}
Hence, $\operatorname*{spr}-f=0$. Thus, $\operatorname*{spr}=f$. This proves
Claim 1.]

Now, we contrast the following two facts:

\begin{itemize}
\item There exists \textbf{at least one} $k$-linear map $\operatorname*{spr}%
:\operatorname*{Sym}V\rightarrow A$ such that every $n\in\mathbb{N}$ and every
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfy
(\ref{1805.eq}).\footnote{Indeed, this has been proven above.}

\item There exists \textbf{at most one} $k$-linear map $\operatorname*{spr}%
:\operatorname*{Sym}V\rightarrow A$ such that every $n\in\mathbb{N}$ and every
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfy
(\ref{1805.eq}).\footnote{This is because each such map $\operatorname*{spr}$
must be equal to $f$ (by Claim 1).}
\end{itemize}

Combining these two facts, we conclude that there exists \textbf{a unique}
$k$-linear map $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow A$ such
that every $n\in\mathbb{N}$ and every $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}$ satisfy (\ref{1805.eq}). This proves
Proposition~\ref{Proposition1805}.
\end{proof}

Next, let us introduce a Hopf algebra structure on the symmetric algebra
$\operatorname*{Sym}V$ of any $k$-vector space $V$:

\begin{theorem}
\label{Theorem1807} Let $k$ be a field. Let $V$ be a $k$-vector space.

Then, there exists a unique $k$-bialgebra structure on the $k$-algebra
$\operatorname*{Sym}V$ whose comultiplication $\Delta_{\operatorname*{Sym}V}$
and counity $\varepsilon_{\operatorname*{Sym}V}$ satisfy%
\[
\left(
\begin{array}
[c]{r}%
\Delta_{\operatorname*{Sym}V}\left(  v\right)  =v\otimes1_{\operatorname*{Sym}%
V}+1_{\operatorname*{Sym}V}\otimes v\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \varepsilon_{\operatorname*{Sym}V}\left(  v\right)  =0\\
\text{for every }v\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)
\end{array}
\right)  .
\]

\end{theorem}

\begin{definition}
\label{Definition1808} Let $k$ be a field. Let $V$ be a $k$-vector space.

Then, when we speak of the \textquotedblleft$k$-bialgebra $\operatorname*{Sym}%
V$\textquotedblright, we mean the symmetric algebra $\operatorname*{Sym}V$
endowed with the unique $k$-bialgebra structure whose comultiplication
$\Delta_{\operatorname*{Sym}V}$ and counity $\varepsilon_{\operatorname*{Sym}%
V}$ satisfy%
\[
\left(
\begin{array}
[c]{r}%
\Delta_{\operatorname*{Sym}V}\left(  v\right)  =v\otimes1_{\operatorname*{Sym}%
V}+1_{\operatorname*{Sym}V}\otimes v\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \varepsilon_{\operatorname*{Sym}V}\left(  v\right)  =0\\
\text{for every }v\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)
\end{array}
\right)  .
\]
(This is well-defined due to Theorem~\ref{Theorem1807}.)
\end{definition}

Notice that Theorem~\ref{Theorem1807} can be viewed as a particular case of
Theorem~\ref{Theorem1311}, because the symmetric algebra $\operatorname*{Sym}%
V$ is the universal enveloping algebra $U\left(  \mathfrak{g}\right)  $ for
$\mathfrak{g}$ being the abelian Lie algebra on $V$\ \ \ \ \footnote{The
\textit{abelian Lie algebra} on a $k$-vector space $V$ is defined to be the
Lie algebra whose underlying vector space is $V$, and whose Lie bracket is
identically $0$.}. Thus, the $k$-bialgebra $\operatorname*{Sym}V$ (where $V$
is a $k$-vector space) is a particular case of the $k$-bialgebra $U\left(
\mathfrak{g}\right)  $ (where $\mathfrak{g}$ is a Lie algebra).

We observe a simple property of primitive elements in bialgebras:

\begin{proposition}
\label{Proposition1809a} Let $k$ be a field. Let $H$ be a bialgebra over $k$.
Let $a\in\operatorname*{Prim}H$ and $n\in\mathbb{N}$. Then,%
\[
\Delta_{H}\left(  a^{n}\right)  =\sum_{j=0}^{n}\dbinom{n}{j}a^{j}\otimes
a^{n-j}.
\]

\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1809a}.]We have $a\in
\operatorname*{Prim}H=\left(  \text{the set of all primitive elements of
}H\right)  $. In other words, the element $a$ of $H$ is primitive. In other
words, $\Delta\left(  a\right)  =a\otimes1_{H}+1_{H}\otimes a$ (by the
definition of \textquotedblleft primitive\textquotedblright).

By the axioms of a bialgebra, we know that $\Delta_{H}$ and $\varepsilon_{H}$
are $k$-algebra homomorphisms (since $H$ is a $k$-bialgebra).

Define two elements $f$ and $g$ of $H\otimes H$ by $f=a\otimes1_{H}$ and
$g=1_{H}\otimes a$. Then,%
\[
\underbrace{f}_{=a\otimes1_{H}}\underbrace{g}_{=1_{H}\otimes a}=\left(
a\otimes1_{H}\right)  \left(  1_{H}\otimes a\right)  =\underbrace{a1_{H}}%
_{=a}\otimes\underbrace{1_{H}a}_{=a}=a\otimes a.
\]
Comparing this with
\[
\underbrace{g}_{=1_{H}\otimes a}\underbrace{f}_{=a\otimes1_{H}}=\left(
1_{H}\otimes a\right)  \left(  a\otimes1_{H}\right)  =\underbrace{1_{H}a}%
_{=a}\otimes\underbrace{a1_{H}}_{=a}=a\otimes a,
\]
we obtain $fg=gf$. In other words, the two elements $f$ and $g$ of $H\otimes
H$ commute. Now, let $\mathfrak{H}$ be the $k$-subalgebra of $H\otimes H$
generated by $f$ and $g$. Then, the $k$-algebra $\mathfrak{H}$ is commutative
(by Corollary~\ref{Corollary47b}, applied to $A=H\otimes H$). Hence, we can
calculate inside $\mathfrak{H}$ as in any commutative algebra. In particular,
we can thus apply the binomial formula to $f$ and $g$. We thus obtain $\left(
f+g\right)  ^{i}=\sum_{j=0}^{i}\dbinom{i}{j}f^{j}g^{i-j}$ for every
$i\in\mathbb{N}$. Applying this to $i=n$, we find%
\begin{align*}
\left(  f+g\right)  ^{n}  &  =\sum_{j=0}^{n}\dbinom{n}{j}\underbrace{f^{j}%
}_{\substack{=\left(  a\otimes1_{H}\right)  ^{j}\\\text{(since }%
f=a\otimes1_{H}\text{)}}}\underbrace{g^{n-j}}_{\substack{=\left(  1_{H}\otimes
a\right)  ^{n-j}\\\text{(since }g=1_{H}\otimes a\text{)}}}=\sum_{j=0}%
^{n}\dbinom{n}{j}\underbrace{\left(  a\otimes1_{H}\right)  ^{j}}%
_{=a^{j}\otimes1_{H}}\underbrace{\left(  1_{H}\otimes a\right)  ^{n-j}%
}_{=1_{H}\otimes a^{n-j}}\\
&  =\sum_{j=0}^{n}\dbinom{n}{j}\underbrace{\left(  a^{j}\otimes1_{H}\right)
\left(  1_{H}\otimes a^{n-j}\right)  }_{=\left(  a^{j}1_{H}\right)
\otimes\left(  1_{H}a^{n-j}\right)  }\\
&  =\sum_{j=0}^{n}\dbinom{n}{j}\underbrace{\left(  a^{j}1_{H}\right)
}_{=a^{j}}\otimes\underbrace{\left(  1_{H}a^{n-j}\right)  }_{=a^{n-j}}%
=\sum_{j=0}^{n}\dbinom{n}{j}a^{j}\otimes a^{n-j}.
\end{align*}


But recall that $\Delta_{H}$ is a $k$-algebra homomorphism. Thus,%
\begin{align*}
\Delta_{H}\left(  a^{n}\right)   &  =\left(  \underbrace{\Delta_{H}\left(
a\right)  }_{=a\otimes1_{H}+1_{H}\otimes a}\right)  ^{n}=\left(
\underbrace{a\otimes1_{H}}_{=f}+\underbrace{1_{H}\otimes a}_{=g}\right)
^{n}\\
&  =\left(  f+g\right)  ^{n}=\sum_{j=0}^{n}\dbinom{n}{j}a^{j}\otimes a^{n-j}.
\end{align*}
This proves Proposition~\ref{Proposition1809a}.
\end{proof}

As a consequence of Proposition~\ref{Proposition1809a}, we obtain the
following formula holding in symmetric algebras:

\begin{corollary}
\label{Corollary1809b} Let $k$ be a field. Let $V$ be a $k$-vector space. Let
$a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ and $n\in
\mathbb{N}$. Then, the comultiplication $\Delta_{\operatorname*{Sym}V}$ of the
bialgebra $\operatorname*{Sym}V$ satisfies%
\[
\Delta_{\operatorname*{Sym}V}\left(  a^{n}\right)  =\sum_{j=0}^{n}\dbinom
{n}{j}a^{j}\otimes a^{n-j}.
\]

\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1809b}.]Let us first recall that
$\Delta_{\operatorname*{Sym}V}\left(  v\right)  =v\otimes
1_{\operatorname*{Sym}V}+1_{\operatorname*{Sym}V}\otimes v$ for every
$v\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ (by the definition
of $\Delta_{\operatorname*{Sym}V}$). Applying this to $v=a$, we obtain
$\Delta_{\operatorname*{Sym}V}\left(  a\right)  =a\otimes
1_{\operatorname*{Sym}V}+1_{\operatorname*{Sym}V}\otimes a$. In other words,
the element $a$ of $\operatorname*{Sym}V$ is primitive (by the definition of
\textquotedblleft primitive\textquotedblright). Thus,%
\[
a\in\left(  \text{the set of all primitive elements of }\operatorname*{Sym}%
V\right)  =\operatorname*{Prim}\left(  \operatorname*{Sym}V\right)  .
\]
Hence, Proposition~\ref{Proposition1809a} (applied to $H=\operatorname*{Sym}%
V$) yields $\Delta_{\operatorname*{Sym}V}\left(  a^{n}\right)  =\sum_{j=0}%
^{n}\dbinom{n}{j}a^{j}\otimes a^{n-j}$. This proves
Corollary~\ref{Corollary1809b}.
\end{proof}

Next, we apply Proposition~\ref{Proposition1805} to the situation of a
cocommutative filtered bialgebra. It turns out that in this situation, we can
say more:

\begin{proposition}
\label{Proposition1810} Let $k$ be a field of characteristic $0$. Let $H$ be a
bialgebra over $k$. Let $V$ denote the $k$-vector subspace
$\operatorname*{Prim}H$ of $H$. Proposition~\ref{Proposition1805} (applied to
$A=H$) yields that there exists a unique $k$-linear map $\operatorname*{spr}%
:\operatorname*{Sym}V\rightarrow H$ such that every $n\in\mathbb{N}$ and every
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfy%
\begin{align}
&  \operatorname*{spr}\left(  \operatorname*{syminc}\nolimits_{V}\left(
v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)
\cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  \right)
\nonumber\\
&  =\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }%
v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }.
\label{1811.eq}%
\end{align}
Consider this $\operatorname*{spr}$.

The map $\operatorname*{spr}$ is a $k$-coalgebra homomorphism.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1810}.]Let $a\in\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $. We are going to show that each
$n\in\mathbb{N}$ satisfies%
\begin{equation}
\left(  \left(  \operatorname*{spr}\otimes\operatorname*{spr}\right)
\circ\Delta_{\operatorname*{Sym}V}\right)  \left(  a^{n}\right)  =\left(
\Delta_{H}\circ\operatorname*{spr}\right)  \left(  a^{n}\right)
\label{1810.goal1}%
\end{equation}
and%
\begin{equation}
\varepsilon_{\operatorname*{Sym}V}\left(  a^{n}\right)  =\left(
\varepsilon_{H}\circ\operatorname*{spr}\right)  \left(  a^{n}\right)  .
\label{1810.goal2}%
\end{equation}
First, let us however make some preliminary work.

Recall that $\Delta_{\operatorname*{Sym}V}\left(  v\right)  =v\otimes
1_{\operatorname*{Sym}V}+1_{\operatorname*{Sym}V}\otimes v$ for every
$v\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ (by the definition
of $\Delta_{\operatorname*{Sym}V}$). Applying this to $v=a$, we obtain
$\Delta_{\operatorname*{Sym}V}\left(  a\right)  =a\otimes
1_{\operatorname*{Sym}V}+1_{\operatorname*{Sym}V}\otimes a$.

Recall that $\varepsilon_{\operatorname*{Sym}V}\left(  v\right)  =0$ for every
$v\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ (by the definition
of $\varepsilon_{\operatorname*{Sym}V}$). Applying this to $v=a$, we obtain
$\varepsilon_{\operatorname*{Sym}V}\left(  a\right)  =0$.

By the axioms of a bialgebra, we know that $\Delta_{\operatorname*{Sym}V}$ and
$\varepsilon_{\operatorname*{Sym}V}$ are $k$-algebra homomorphisms (since
$\operatorname*{Sym}V$ is a $k$-bialgebra).

By the axioms of a bialgebra, we know that $\Delta_{H}$ and $\varepsilon_{H}$
are $k$-algebra homomorphisms (since $H$ is a $k$-bialgebra).

We know that $a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $. In
other words, there exists some $w\in V$ such that $a=\operatorname*{syminc}%
\nolimits_{V}\left(  w\right)  $. Consider this $w$.

The equality (\ref{1811.eq}) (applied to $1$ and $\left(  w\right)  $ instead
of $n$ and $\left(  v_{1},v_{2},\ldots,v_{n}\right)  $) yields%
\begin{align*}
\operatorname*{spr}\left(  \operatorname*{syminc}\nolimits_{V}\left(
w\right)  \right)   &  =\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}=1}%
\sum_{\sigma\in S_{1}}w=\sum_{\sigma\in S_{1}}w=\sum_{\sigma\in\left\{
\operatorname*{id}\right\}  }w\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
S_{1}=\left\{  \operatorname*{id}\right\}  \right) \\
&  =w.
\end{align*}
Thus, $w=\operatorname*{spr}\left(  \underbrace{\operatorname*{syminc}%
\nolimits_{V}\left(  w\right)  }_{=a}\right)  =\operatorname*{spr}a$.

Let $n\in\mathbb{N}$. The equality (\ref{1811.eq}) (applied to $\left(
w,w,\ldots,w\right)  $ instead of $\left(  v_{1},v_{2},\ldots,v_{n}\right)  $)
yields%
\begin{align*}
&  \operatorname*{spr}\left(  \underbrace{\operatorname*{syminc}%
\nolimits_{V}\left(  w\right)  \operatorname*{syminc}\nolimits_{V}\left(
w\right)  \cdots\operatorname*{syminc}\nolimits_{V}\left(  w\right)
}_{n\text{ times }\operatorname*{syminc}\nolimits_{V}\left(  w\right)
}\right) \\
&  =\dfrac{1}{n!}\underbrace{\sum_{\sigma\in S_{n}}\underbrace{ww\cdots
w}_{n\text{ times }w}}_{=\left\vert S_{n}\right\vert \cdot\underbrace{ww\cdots
w}_{n\text{ times }w}}=\dfrac{1}{n!}\underbrace{\left\vert S_{n}\right\vert
}_{=n!}\cdot\underbrace{\underbrace{ww\cdots w}_{n\text{ times }w}}_{=w^{n}%
}=\dfrac{1}{n!}n!\cdot w^{n}=w^{n}.
\end{align*}
In view of $\underbrace{\operatorname*{syminc}\nolimits_{V}\left(  w\right)
\operatorname*{syminc}\nolimits_{V}\left(  w\right)  \cdots
\operatorname*{syminc}\nolimits_{V}\left(  w\right)  }_{n\text{ times
}\operatorname*{syminc}\nolimits_{V}\left(  w\right)  }=\left(
\underbrace{\operatorname*{syminc}\nolimits_{V}\left(  w\right)  }%
_{=a}\right)  ^{n}=a^{n}$, this rewrites as%
\begin{equation}
\operatorname*{spr}\left(  a^{n}\right)  =w^{n}. \label{1810.pf.spran}%
\end{equation}


Let us forget that we fixed $n$. We thus have proven (\ref{1810.pf.spran}) for
every $n\in\mathbb{N}$.

Let $n\in\mathbb{N}$. We have $w\in V=\operatorname*{Prim}H$. Hence,
Proposition~\ref{Proposition1809a} (applied to $w$ instead of $a$) yields%
\begin{equation}
\Delta_{H}\left(  w^{n}\right)  =\sum_{j=0}^{n}\dbinom{n}{j}w^{j}\otimes
w^{n-j}. \label{1810.pf.Delta-1}%
\end{equation}
But Corollary~\ref{Corollary1809b} yields%
\[
\Delta_{\operatorname*{Sym}V}\left(  a^{n}\right)  =\sum_{j=0}^{n}\dbinom
{n}{j}a^{j}\otimes a^{n-j}.
\]
Hence,%
\begin{align*}
&  \left(  \left(  \operatorname*{spr}\otimes\operatorname*{spr}\right)
\circ\Delta_{\operatorname*{Sym}V}\right)  \left(  a^{n}\right) \\
&  =\left(  \operatorname*{spr}\otimes\operatorname*{spr}\right)  \left(
\underbrace{\Delta_{\operatorname*{Sym}V}\left(  a^{n}\right)  }_{=\sum
_{j=0}^{n}\dbinom{n}{j}a^{j}\otimes a^{n-j}}\right) \\
&  =\left(  \operatorname*{spr}\otimes\operatorname*{spr}\right)  \left(
\sum_{j=0}^{n}\dbinom{n}{j}a^{j}\otimes a^{n-j}\right)  =\sum_{j=0}^{n}%
\dbinom{n}{j}\underbrace{\left(  \operatorname*{spr}\otimes\operatorname*{spr}%
\right)  \left(  a^{j}\otimes a^{n-j}\right)  }_{=\operatorname*{spr}\left(
a^{j}\right)  \otimes\operatorname*{spr}\left(  a^{n-j}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\operatorname*{spr}%
\otimes\operatorname*{spr}\text{ is }k\text{-linear}\right) \\
&  =\sum_{j=0}^{n}\dbinom{n}{j}\underbrace{\operatorname*{spr}\left(
a^{j}\right)  }_{\substack{=w^{j}\\\text{(by (\ref{1810.pf.spran}%
)}\\\text{(applied to }j\text{ instead of }n\text{))}}}\otimes
\underbrace{\operatorname*{spr}\left(  a^{n-j}\right)  }_{\substack{=w^{n-j}%
\\\text{(by (\ref{1810.pf.spran})}\\\text{(applied to }n-j\text{ instead of
}n\text{))}}}\\
&  =\sum_{j=0}^{n}\dbinom{n}{j}w^{j}\otimes w^{n-j}=\Delta_{H}\left(
\underbrace{w^{n}}_{\substack{=\operatorname*{spr}\left(  a^{n}\right)
\\\text{(by (\ref{1810.pf.spran}))}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{1810.pf.Delta-1})}\right) \\
&  =\Delta_{H}\left(  \operatorname*{spr}\left(  a^{n}\right)  \right)
=\left(  \Delta_{H}\circ\operatorname*{spr}\right)  \left(  a^{n}\right)  .
\end{align*}
In other words, (\ref{1810.goal1}) holds.

We have $w\in V=\operatorname*{Prim}H$. Thus, Remark~\ref{Remark21} (applied
to $x=w$) shows that $\varepsilon\left(  w\right)  =0$. Thus,
$\underbrace{\varepsilon_{H}}_{=\varepsilon}\left(  w\right)  =\varepsilon
\left(  w\right)  =0$.

Recall that $\varepsilon_{\operatorname*{Sym}V}$ is a $k$-algebra
homomorphism. Thus, $\varepsilon_{\operatorname*{Sym}V}\left(  a^{n}\right)
=\left(  \underbrace{\varepsilon_{\operatorname*{Sym}V}\left(  a\right)
}_{=0}\right)  ^{n}=0^{n}$.

On the other hand,
\begin{align*}
\left(  \varepsilon_{H}\circ\operatorname*{spr}\right)  \left(  a^{n}\right)
&  =\varepsilon_{H}\left(  \underbrace{\operatorname*{spr}\left(
a^{n}\right)  }_{\substack{=w^{n}\\\text{(by (\ref{1810.pf.spran}))}}}\right)
=\varepsilon_{H}\left(  w^{n}\right) \\
&  =\left(  \underbrace{\varepsilon_{H}\left(  w\right)  }_{=0}\right)
^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon_{H}\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  =0^{n}.
\end{align*}
Comparing this with $\varepsilon_{\operatorname*{Sym}V}\left(  a^{n}\right)
=0$, we find $\varepsilon_{\operatorname*{Sym}V}\left(  a^{n}\right)  =\left(
\varepsilon_{H}\circ\operatorname*{spr}\right)  \left(  a^{n}\right)  $. In
other words, (\ref{1810.goal2}) holds.

We have now proven that (\ref{1810.goal1}) and (\ref{1810.goal2}) hold.

Now, forget that we fixed $n$. We thus have proven that (\ref{1810.goal1}) and
(\ref{1810.goal2}) hold for each $n\in\mathbb{N}$.

Now, forget that we fixed $a$. We thus have proven that (\ref{1810.goal1}) and
(\ref{1810.goal2}) hold for each $a\in\operatorname*{syminc}\nolimits_{V}%
\left(  V\right)  $ and $n\in\mathbb{N}$.

In particular, (\ref{1810.goal1}) holds for each $a\in\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $ and $n\in\mathbb{N}$. In other words,
$\left(  \left(  \operatorname*{spr}\otimes\operatorname*{spr}\right)
\circ\Delta_{\operatorname*{Sym}V}\right)  \left(  a^{n}\right)  =\left(
\Delta_{H}\circ\operatorname*{spr}\right)  \left(  a^{n}\right)  $ for each
$a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ and $n\in
\mathbb{N}$. Hence, Lemma~\ref{Lemma1801} (applied to $W=H\otimes H$,
$f=\left(  \operatorname*{spr}\otimes\operatorname*{spr}\right)  \circ
\Delta_{\operatorname*{Sym}V}$ and $g=\Delta_{H}\circ\operatorname*{spr}$)
shows that $\left(  \operatorname*{spr}\otimes\operatorname*{spr}\right)
\circ\Delta_{\operatorname*{Sym}V}=\Delta_{H}\circ\operatorname*{spr}$. In
other words, $\Delta_{H}\circ\operatorname*{spr}=\left(  \operatorname*{spr}%
\otimes\operatorname*{spr}\right)  \circ\Delta_{\operatorname*{Sym}V}$.

Also, (\ref{1810.goal2}) holds for each $a\in\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $ and $n\in\mathbb{N}$. In other words,
$\varepsilon_{\operatorname*{Sym}V}\left(  a^{n}\right)  =\left(
\varepsilon_{H}\circ\operatorname*{spr}\right)  \left(  a^{n}\right)  $ for
each $a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ and
$n\in\mathbb{N}$. Hence, Lemma~\ref{Lemma1801} (applied to $W=k$,
$f=\varepsilon_{\operatorname*{Sym}V}$ and $g=\varepsilon_{H}\circ
\operatorname*{spr}$) shows that $\varepsilon_{\operatorname*{Sym}%
V}=\varepsilon_{H}\circ\operatorname*{spr}$. In other words, $\varepsilon
_{H}\circ\operatorname*{spr}=\varepsilon_{\operatorname*{Sym}V}$.

The two equalities $\Delta_{H}\circ\operatorname*{spr}=\left(
\operatorname*{spr}\otimes\operatorname*{spr}\right)  \circ\Delta
_{\operatorname*{Sym}V}$ and $\varepsilon_{H}\circ\operatorname*{spr}%
=\varepsilon_{\operatorname*{Sym}V}$ (combined) show that $\operatorname*{spr}%
$ is a $k$-coalgebra homomorphism (by the definition of a $k$-coalgebra
homomorphism). This proves Proposition~\ref{Proposition1810}.
\end{proof}

\begin{proposition}
\label{Proposition1811} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered cocommutative bialgebra over $k$. Let $V$ denote the
$k$-vector subspace $\operatorname*{Prim}H$ of $H$. Define the $k$-linear map
$\operatorname*{spr}:\operatorname*{Sym}V\rightarrow H$ as in
Proposition~\ref{Proposition1810}.

The map $\operatorname*{spr}$ is surjective.
\end{proposition}

\begin{proof}
[Proof of Proposition~\ref{Proposition1811}.]Theorem~\ref{Theorem280} yields%
\[
H=\sum_{i\in\mathbb{N}}\operatorname{symp}_{i}\left(
\underbrace{\operatorname*{Prim}H}_{=V}\right)  =\sum_{i\in\mathbb{N}%
}\operatorname{symp}_{i}V=\sum_{n\in\mathbb{N}}\operatorname{symp}_{n}V
\]
(here, we have renamed the summation index $i$ as $n$).

The set $\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  $ is a
$k$-vector subspace of $H$ (since $\operatorname*{spr}$ is a $k$-linear map).

But every $n\in\mathbb{N}$ satisfies $\operatorname{symp}_{n}V\subseteq
\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $n\in\mathbb{N}$. Then, every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}$ satisfies%
\begin{align*}
\dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma
\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }  &
=\operatorname*{spr}\left(  \underbrace{\operatorname*{syminc}\nolimits_{V}%
\left(  v_{1}\right)  \operatorname*{syminc}\nolimits_{V}\left(  v_{2}\right)
\cdots\operatorname*{syminc}\nolimits_{V}\left(  v_{n}\right)  }%
_{\in\operatorname*{Sym}V}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1811.eq})}\right) \\
&  \in\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  .
\end{align*}
In other words,%
\[
\left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(  1\right)
}v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\subseteq\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  .
\]
Hence, (\ref{284.pf.b.2}) (applied to $H$, $\left\{  \dfrac{1}{n!}\sum
_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)
}\cdots v_{\sigma\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{\times n}\right\}  $ and $\operatorname*{spr}\left(
\operatorname*{Sym}V\right)  $ instead of $M$, $S$ and $Q$) yields%
\[
\left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle \subseteq\operatorname*{spr}\left(  \operatorname*{Sym}V\right)
.
\]
Now, the definition of $\operatorname{symp}_{n}V$ yields%
\begin{align*}
\operatorname{symp}_{n}V  &  =\left\langle \dfrac{1}{n!}\sum_{\sigma\in S_{n}%
}v_{\sigma\left(  1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma
\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{\times n}\right\rangle \\
&  =\left\langle \left\{  \dfrac{1}{n!}\sum_{\sigma\in S_{n}}v_{\sigma\left(
1\right)  }v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times n}\right\}
\right\rangle \\
&  \subseteq\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  .
\end{align*}
Qed.}. Thus, $\sum_{n\in\mathbb{N}}\underbrace{\operatorname{symp}_{n}%
V}_{\subseteq\operatorname*{spr}\left(  \operatorname*{Sym}V\right)
}\subseteq\sum_{n\in\mathbb{N}}\operatorname*{spr}\left(  \operatorname*{Sym}%
V\right)  \subseteq\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  $
(since $\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  $ is a
$k$-vector space). Thus, $H=\sum_{n\in\mathbb{N}}\operatorname{symp}%
_{n}V\subseteq\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  $. In
other words, the map $\operatorname*{spr}$ is surjective. This proves
Proposition~\ref{Proposition1811}.
\end{proof}

Let us now state some further properties of the $k$-bialgebra
$\operatorname*{Sym}V$:

\begin{theorem}
\label{Theorem1812} Let $k$ be a field. Let $V$ be a $k$-vector space.

\textbf{(a)} We have $\operatorname*{Sym}V=\sum_{n\in\mathbb{N}}\left(
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \right)  ^{n}$.

\textbf{(b)} The family $\left(  \sum_{n=0}^{i}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  \right)  ^{n}\right)  _{i\in\mathbb{N}}$ is a
filtration of the $k$-vector space $\operatorname*{Sym}V$.

\textbf{(c)} The $k$-bialgebra $\operatorname*{Sym}V$ endowed with the
filtration $\left(  \sum_{n=0}^{i}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  \right)  ^{n}\right)  _{i\in\mathbb{N}}$ is a
connected filtered $k$-bialgebra.

\textbf{(d)} We have $\operatorname*{syminc}\nolimits_{V}\left(  V\right)
\subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}V\right)  $.

\textbf{(e)} The $k$-bialgebra $\operatorname*{Sym}V$ is cocommutative.

\textbf{(f)} The $k$-bialgebra $\operatorname*{Sym}V$ is a Hopf algebra.
\end{theorem}

Notice that Theorem~\ref{Theorem1812} can also be viewed as a particular case
of Theorem~\ref{Theorem1313} (since the $k$-bialgebra $\operatorname*{Sym}V$
(where $V$ is a $k$-vector space) is a particular case of the $k$-bialgebra
$U\left(  \mathfrak{g}\right)  $ (where $\mathfrak{g}$ is a Lie algebra)).
However, we shall give a self-contained proof of the parts of
Theorem~\ref{Theorem1812} that we will need:

\begin{proof}
[Partial proof of Theorem~\ref{Theorem1812}.]Recall that%
\begin{equation}
\Delta_{\operatorname*{Sym}V}\left(  v\right)  =v\otimes1_{\operatorname*{Sym}%
V}+1_{\operatorname*{Sym}V}\otimes v\ \ \ \ \ \ \ \ \ \ \text{for every }%
v\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \label{1812.pf.1}%
\end{equation}
(by the definition of the map $\Delta_{\operatorname*{Sym}V}$).

We have $\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \subseteq
\operatorname*{Prim}\left(  \operatorname*{Sym}V\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $v\in\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $. Then, $\Delta_{\operatorname*{Sym}V}\left(
v\right)  =v\otimes1_{\operatorname*{Sym}V}+1_{\operatorname*{Sym}V}\otimes v$
(according to (\ref{1812.pf.1})). In other words, the element $v$ of
$\operatorname*{Sym}V$ is primitive (by the definition of \textquotedblleft
primitive\textquotedblright). In other words, $v\in\operatorname*{Prim}\left(
\operatorname*{Sym}V\right)  $ (since $\operatorname*{Prim}\left(
\operatorname*{Sym}V\right)  $ is the set of all primitive elements of
$\operatorname*{Sym}V$).
\par
Now, forget that we fixed $v$. We thus have shown that every $v\in
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ satisfies
$v\in\operatorname*{Prim}\left(  \operatorname*{Sym}V\right)  $. In other
words, $\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \subseteq
\operatorname*{Prim}\left(  \operatorname*{Sym}V\right)  $, qed.}. This proves
Theorem~\ref{Theorem1812} \textbf{(d)}.

Let $\pi$ be the canonical projection $\otimes V\rightarrow\operatorname*{Sym}%
V$. Then, $\pi$ is a surjective $k$-algebra homomorphism. [Actually, $\pi$ is
a $k$-bialgebra homomorphism, but we don't need to know this.]

The canonical map $\operatorname*{syminc}\nolimits_{V}$ from $V$ to
$\operatorname*{Sym}V$ factors through the projection $\pi:\otimes
V\rightarrow\operatorname*{Sym}V$. More precisely, $\operatorname*{syminc}%
\nolimits_{V}\left(  x\right)  =\pi\left(  x\right)  $ for every $x\in V$
(this follows from the definition of $\operatorname*{syminc}\nolimits_{V}$).
Thus,%
\begin{equation}
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  =\left\{
\underbrace{\operatorname*{syminc}\nolimits_{V}\left(  x\right)  }%
_{=\pi\left(  x\right)  }\ \mid\ x\in V\right\}  =\left\{  \pi\left(
x\right)  \ \mid\ x\in V\right\}  =\pi\left(  V\right)  . \label{1812.pf.4}%
\end{equation}


By the definition of the tensor algebra, we have $\otimes V=\bigoplus
_{n\in\mathbb{N}}V^{\otimes n}$. Thus, $\otimes V=\bigoplus_{n\in\mathbb{N}%
}V^{\otimes n}=\sum_{n\in\mathbb{N}}V^{\otimes n}$ (since direct sums are sums).

For every $n\in\mathbb{N}$, we have $V^{\otimes n}=V^{n}$ as $k$-vector
subspaces of $\otimes V$ (where $V^{n}$ means $\underbrace{V\cdot V\cdot
\cdots\cdot V}_{n\text{ times}}$, as usual)\ \ \ \ \footnote{\textit{Proof.}
Let $n\in\mathbb{N}$. The $n$-th tensor power $V^{\otimes n}$ is spanned by
all pure tensors. In other words,%
\begin{align*}
V^{\otimes n}  &  =\left\langle \underbrace{v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}}_{\substack{=v_{1}v_{2}\cdots v_{n}\\\text{(because the
multiplication on }\otimes V\text{ is the}\\\text{tensor product, so we
have}\\v_{1}v_{2}\cdots v_{n}=v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\text{)}}}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{\times
n}\right\rangle \\
&  =\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\rangle .
\end{align*}
Compared with%
\[
V^{n}=\left\langle v_{1}v_{2}\cdots v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{\times n}\right\rangle ,
\]
this yields $V^{\otimes n}=V^{n}$, qed.}. Hence, $\sum_{n\in\mathbb{N}%
}V^{\otimes n}=\sum_{n\in\mathbb{N}}V^{n}$.

Now, $\otimes V=\sum_{n\in\mathbb{N}}V^{\otimes n}=\sum_{n\in\mathbb{N}}V^{n}%
$. Applying the map $\pi$ to this equality, we obtain%
\begin{align*}
\pi\left(  \otimes V\right)   &  =\pi\left(  \sum_{n\in\mathbb{N}}%
V^{n}\right)  =\sum_{n\in\mathbb{N}}\left(  \underbrace{\pi\left(  V\right)
}_{\substack{=\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \\\text{(by
(\ref{1812.pf.4}))}}}\right)  ^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\pi\text{ is a }k\text{-algebra homomorphism}\right) \\
&  =\sum_{n\in\mathbb{N}}\left(  \operatorname*{syminc}\nolimits_{V}\left(
V\right)  \right)  ^{n}.
\end{align*}
Since $\pi\left(  \otimes V\right)  =\operatorname*{Sym}V$ (because $\pi$ is
surjective), this rewrites as $\operatorname*{Sym}V=\sum_{n\in\mathbb{N}%
}\left(  \operatorname*{syminc}\nolimits_{V}\left(  V\right)  \right)  ^{n}$.
Thus, Theorem~\ref{Theorem1812} \textbf{(a)} is proven.

We have
\[
\Delta_{\operatorname*{Sym}V}\left(  \operatorname*{syminc}\nolimits_{V}%
\left(  V\right)  \right)  \subseteq\operatorname*{syminc}\nolimits_{V}\left(
V\right)  \otimes\left(  k\cdot1_{\operatorname*{Sym}V}\right)  +\left(
k\cdot1_{\operatorname*{Sym}V}\right)  \otimes\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  .
\]
\footnote{\textit{Proof.} Every $v\in\operatorname*{syminc}\nolimits_{V}%
\left(  V\right)  $ satisfies%
\begin{align*}
\Delta_{\operatorname*{Sym}V}\left(  v\right)   &  =\underbrace{v}%
_{\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  }\otimes
\underbrace{1_{\operatorname*{Sym}V}}_{\in k\cdot1_{\operatorname*{Sym}V}%
}+\underbrace{1_{\operatorname*{Sym}V}}_{\in k\cdot1_{\operatorname*{Sym}V}%
}\otimes\underbrace{v}_{\in\operatorname*{syminc}\nolimits_{V}\left(
V\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{according to (\ref{1812.pf.1}%
)}\right) \\
&  \in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \otimes\left(
k\cdot1_{\operatorname*{Sym}V}\right)  +\left(  k\cdot1_{\operatorname*{Sym}%
V}\right)  \otimes\operatorname*{syminc}\nolimits_{V}\left(  V\right)  .
\end{align*}
In other words, $\Delta_{\operatorname*{Sym}V}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  \right)  \subseteq\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  \otimes\left(  k\cdot1_{\operatorname*{Sym}%
V}\right)  +\left(  k\cdot1_{\operatorname*{Sym}V}\right)  \otimes
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $, qed.} Combining this
with the equality $\operatorname*{Sym}V=\sum_{n\in\mathbb{N}}\left(
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \right)  ^{n}$, we see
that we can apply Proposition~\ref{Proposition1314} to $H=\operatorname*{Sym}%
V$ and $V=\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $.

Proposition~\ref{Proposition1314} \textbf{(a)} (applied to
$H=\operatorname*{Sym}V$ and $V=\operatorname*{syminc}\nolimits_{V}\left(
V\right)  $) yields that the family $\left(  \sum_{n=0}^{i}\left(
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \right)  ^{n}\right)
_{i\in\mathbb{N}}$ is a filtration of the $k$-vector space
$\operatorname*{Sym}V$. This proves Theorem~\ref{Theorem1812} \textbf{(b)}.

Proposition~\ref{Proposition1314} \textbf{(b)} (applied to
$H=\operatorname*{Sym}V$ and $V=\operatorname*{syminc}\nolimits_{V}\left(
V\right)  $) yields that the $k$-bialgebra $\operatorname*{Sym}V$ endowed with
the filtration $\left(  \sum_{n=0}^{i}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  \right)  ^{n}\right)  _{i\in\mathbb{N}}$ is a
connected filtered $k$-bialgebra. This proves Theorem~\ref{Theorem1812}
\textbf{(c)}.

Finally, Proposition~\ref{Proposition1314} \textbf{(c)} (applied to
$H=\operatorname*{Sym}V$ and $V=\operatorname*{syminc}\nolimits_{V}\left(
V\right)  $) yields that the $k$-bialgebra $\operatorname*{Sym}V$ is
cocommutative (since $\operatorname*{syminc}\nolimits_{V}\left(  V\right)
\subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}V\right)  $). This
proves Theorem~\ref{Theorem1812} \textbf{(e)}.

We are not going to prove Theorem~\ref{Theorem1812} \textbf{(f)} (as we are
not going to need it).
\end{proof}

Next, let us show a slight extension of Theorem~\ref{Theorem1201}:

\begin{corollary}
\label{Corollary1821} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra with
the same underlying $k$-coalgebra structure. Let $v\in\operatorname*{Prim}H$
and $n\in\mathbb{N}$. Then,%
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(  v^{n}\right)
=\delta_{n,1}v.
\]
Here, we are using the \textit{Kronecker delta notation} (i.e., whenever $u$
and $v$ are two objects, we set $\delta_{u,v}=%
\begin{cases}
1, & \text{if }u=v;\\
0, & \text{if }u\neq v
\end{cases}
$).
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1821}.]Recall that $\operatorname*{id}\in
G\left(  H,H\right)  $. Thus, $\operatorname*{Log}\operatorname*{id}%
\in\mathfrak{g}\left(  H,H\right)  $ (because $\operatorname*{Log}%
F\in\mathfrak{g}\left(  H,H\right)  $ for every $F\in G\left(  H,H\right)  $).
Hence,%
\[
\operatorname*{Log}\operatorname*{id}\in\mathfrak{g}\left(  H,H\right)
=\left\{  f\in\mathcal{L}\left(  H,H\right)  \ \mid\ f\left(  1_{H}\right)
=0\right\}
\]
(by the definition of $\mathfrak{g}\left(  H,H\right)  $). In other words,
$\operatorname*{Log}\operatorname*{id}$ is an element $f\in\mathcal{L}\left(
H,H\right)  $ satisfying $f\left(  1_{H}\right)  =0$. In other words,
$\operatorname*{Log}\operatorname*{id}$ is an element of $\mathcal{L}\left(
H,H\right)  $ and satisfies $\left(  \operatorname*{Log}\operatorname*{id}%
\right)  \left(  1_{H}\right)  =0$.

We are in one of the following three cases:

\textit{Case 1:} We have $n=0$.

\textit{Case 2:} We have $n=1$.

\textit{Case 3:} We have neither $n=0$ nor $n=1$.

Let us first consider Case 1. In this case, we have $n=0$. Hence, $v^{n}%
=v^{0}=1_{H}$. Applying the map $\operatorname*{Log}\operatorname*{id}$ to
both sides of this equality, we obtain $\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  v^{n}\right)  =\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  1_{H}\right)  =0$. Comparing this with
$\underbrace{\delta_{n,1}}_{\substack{=0\\\text{(since }n=0\neq1\text{)}}%
}v=0$, we obtain $\left(  \operatorname*{Log}\operatorname*{id}\right)
\left(  v^{n}\right)  =\delta_{n,1}v$. Thus, Corollary~\ref{Corollary1821} is
proven in Case 1.

Let us now consider Case 2. In this case, we have $n=1$. Hence, $v^{n}%
=v^{1}=v$. But Proposition~\ref{Proposition19} \textbf{(c)} (applied
to\textbf{ }$A=H$ and $F=\operatorname*{id}$) shows that $\left(
\operatorname*{Log}\operatorname*{id}\right)  \mid_{\operatorname*{Prim}%
H}=\operatorname*{id}\mid_{\operatorname*{Prim}H}$. But $v\in
\operatorname*{Prim}H$. Thus,
\[
\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(  v\right)
=\underbrace{\left(  \left(  \operatorname*{Log}\operatorname*{id}\right)
\mid_{\operatorname*{Prim}H}\right)  }_{=\operatorname*{id}\mid
_{\operatorname*{Prim}H}}\left(  v\right)  =\left(  \operatorname*{id}%
\mid_{\operatorname*{Prim}H}\right)  \left(  v\right)  =\operatorname*{id}%
\left(  v\right)  =v.
\]
Now, $\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(
\underbrace{v^{n}}_{=v}\right)  =\left(  \operatorname*{Log}\operatorname*{id}%
\right)  \left(  v\right)  =v$. Comparing this with $\underbrace{\delta_{n,1}%
}_{\substack{=1\\\text{(since }n=1\text{)}}}v=v$, we obtain $\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  v^{n}\right)
=\delta_{n,1}v$. Thus, Corollary~\ref{Corollary1821} is proven in Case 2.

Now, let us consider Case 3. In this case, we have neither $n=0$ nor $n=1$.
Thus, $n\notin\left\{  0,1\right\}  $. Combining $n\in\mathbb{N}$ with
$n\in\left\{  0,1\right\}  $, we find $n\in\mathbb{N}\setminus\left\{
0,1\right\}  =\left\{  2,3,4,\ldots\right\}  $, so that $n\geq2>1$. Thus,
Theorem~\ref{Theorem1201} shows that $\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  v^{n}\right)  =0$. But $n>1$ and thus
$n\neq1$. Hence, $\delta_{n,1}=0$. Comparing $\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  v^{n}\right)  =0$ with $\underbrace{\delta
_{n,1}}_{=0}v=0$, we obtain $\left(  \operatorname*{Log}\operatorname*{id}%
\right)  \left(  v^{n}\right)  =\delta_{n,1}v$. Thus,
Corollary~\ref{Corollary1821} is proven in Case 3.

We now have proven Corollary~\ref{Corollary1821} in each of the three Cases 1,
2 and 3. Thus, Corollary~\ref{Corollary1821} is proven (since these three
Cases cover all possibilities).
\end{proof}

Next, we shall show a result that resembles Theorem~\ref{Theorem1702} but
concerns cocommutative bialgebras rather than commutative bialgebras:

\begin{lemma}
\label{Lemma1831} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered cocommutative bialgebra over $k$. Consider the convolution
algebra $\mathcal{L}\left(  H,H\right)  $. Let $\mathfrak{e}$ denote the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $.

Let $\mathfrak{j}$ denote the inclusion map $\mathfrak{e}\left(  H\right)
\rightarrow H$. Define a map $\mathfrak{e}^{\prime}:H\rightarrow
\mathfrak{e}\left(  H\right)  $ by%
\[
\left(  \mathfrak{e}^{\prime}\left(  h\right)  =\mathfrak{e}\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H\right)  .
\]
\footnote{This is well-defined, since $\mathfrak{e}\left(  h\right)
\in\mathfrak{e}\left(  H\right)  $ for every $h\in H$.} This map
$\mathfrak{e}^{\prime}$ is clearly $k$-linear\footnote{In fact, this map
$\mathfrak{e}^{\prime}$ is obtained from the $k$-linear map $\mathfrak{e}$ by
changing the target to $\mathfrak{e}\left(  H\right)  $.}. Define a $k$-linear
map $\mathfrak{q}:H\rightarrow\operatorname*{Sym}\left(  \mathfrak{e}\left(
H\right)  \right)  $ by $\mathfrak{q}=\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }\circ\mathfrak{e}^{\prime}$.

Let $V$ denote the $k$-vector subspace $\operatorname*{Prim}H$ of $H$. Define
the $k$-linear map $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow H$ as
in Proposition~\ref{Proposition1810}.

Recall that $\operatorname*{Sym}V$ is a connected filtered $k$-bialgebra
(according to Theorem~\ref{Theorem1812} \textbf{(c)}). Hence, a $k$-linear map
$\operatorname*{Log}\operatorname{id}_{\operatorname*{Sym}V}%
:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$ is defined for this
$k$-bialgebra $\operatorname*{Sym}V$ (in the same way as the $k$-linear map
$\operatorname*{Log}\operatorname*{id}:H\rightarrow H$ is defined for the
$k$-bialgebra $H$).

\textbf{(a)} We have $\mathfrak{e}\left(  H\right)  =V$. Thus, $\mathfrak{q}$
is a $k$-linear map $H\rightarrow\operatorname*{Sym}V$, and the map
$\mathfrak{q}\circ\operatorname*{spr}:\operatorname*{Sym}V\rightarrow
\operatorname*{Sym}V$ is well-defined.

\textbf{(b)} We have $\mathfrak{q}\circ\operatorname*{spr}=\operatorname*{Log}%
\operatorname{id}_{\operatorname*{Sym}V}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{Lemma1831}.]Theorem~\ref{Theorem1} shows that the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $ is a
projection from $H$ to the subspace $\operatorname*{Prim}H$ of all primitive
elements of $H$. Thus, $\left(  \operatorname*{Log}\operatorname*{id}\right)
\left(  H\right)  =\operatorname*{Prim}H$. Now, $\underbrace{\mathfrak{e}%
}_{=\operatorname*{Log}\operatorname*{id}}\left(  H\right)  =\left(
\operatorname*{Log}\operatorname*{id}\right)  \left(  H\right)
=\operatorname*{Prim}H=V$.

Recall that $\mathfrak{q}$ is a $k$-linear map $H\rightarrow
\operatorname*{Sym}\left(  \mathfrak{e}\left(  H\right)  \right)  $. In other
words, $\mathfrak{q}$ is a $k$-linear map $H\rightarrow\operatorname*{Sym}V$
(since $\mathfrak{e}\left(  H\right)  =V$). Hence, the map $\mathfrak{q}%
\circ\operatorname*{spr}:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$
is well-defined (since $\operatorname*{spr}$ is a $k$-linear map
$\operatorname*{Sym}V\rightarrow H$). Altogether, we thus have proven
Lemma~\ref{Lemma1831} \textbf{(a)}.

\textbf{(b)} In the following, the symbol \textquotedblleft$\operatorname*{id}%
$\textquotedblright\ (without subscript) shall always mean the identity map
$\operatorname{id}_{H}:H\rightarrow H$ (not the identity map
$\operatorname{id}_{\operatorname*{Sym}V}:\operatorname*{Sym}V\rightarrow
\operatorname*{Sym}V$).

We know that the map $\mathfrak{q}\circ\operatorname*{spr}:\operatorname*{Sym}%
V\rightarrow\operatorname*{Sym}V$ is well-defined. Denote this map by $f$.
Thus, $f=\mathfrak{q}\circ\operatorname*{spr}$ is a map from
$\operatorname*{Sym}V$ to $\operatorname*{Sym}V$.

We have $f=\mathfrak{q}\circ\operatorname*{spr}$. Hence, $f$ is the
composition of two $k$-linear maps (since the two maps $\mathfrak{q}$ and
$\operatorname*{spr}$ are $k$-linear). Thus, $f$ itself is $k$-linear.

Define a $k$-linear map $g:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}%
V$ by $g=\operatorname*{Log}\operatorname{id}_{\operatorname*{Sym}V}$.

In the following, we shall use the \textit{Kronecker delta notation} (i.e.,
whenever $u$ and $v$ are two objects, we set $\delta_{u,v}=%
\begin{cases}
1, & \text{if }u=v;\\
0, & \text{if }u\neq v
\end{cases}
$).

Let $a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ and
$n\in\mathbb{N}$. We shall show that $f\left(  a^{n}\right)  =g\left(
a^{n}\right)  $.

We have $a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)
\subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}V\right)  $ (by
Theorem~\ref{Theorem1812} \textbf{(d)}). Hence, Corollary~\ref{Corollary1821}
(applied to $\operatorname*{Sym}V$ and $a$ instead of $H$ and $v$) shows that
$\left(  \operatorname*{Log}\operatorname{id}_{\operatorname*{Sym}V}\right)
\left(  a^{n}\right)  =\delta_{n,1}a$. Thus, $\underbrace{g}%
_{=\operatorname*{Log}\operatorname{id}_{\operatorname*{Sym}V}}\left(
a^{n}\right)  =\left(  \operatorname*{Log}\operatorname{id}%
_{\operatorname*{Sym}V}\right)  \left(  a^{n}\right)  =\delta_{n,1}a$.

We know that $a\in\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $. In
other words, there exists some $w\in V$ such that $a=\operatorname*{syminc}%
\nolimits_{V}\left(  w\right)  $. Consider this $w$.

We have $\operatorname*{spr}\left(  a^{n}\right)  =w^{n}$. (Indeed, this is
the equality (\ref{1810.pf.spran}) from the proof of
Proposition~\ref{Proposition1810}, and has already been shown during the proof
of Proposition~\ref{Proposition1810}.)

We have $w\in V=\operatorname*{Prim}H$. Hence, Corollary~\ref{Corollary1821}
(applied to $v=w$) shows that $\left(  \operatorname*{Log}\operatorname*{id}%
\right)  \left(  w^{n}\right)  =\delta_{n,1}w$.

The definition of $\mathfrak{e}^{\prime}$ yields $\mathfrak{e}^{\prime}\left(
w^{n}\right)  =\underbrace{\mathfrak{e}}_{=\operatorname*{Log}%
\operatorname*{id}}\left(  w^{n}\right)  =\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  w^{n}\right)  =\delta_{n,1}w$.

Now,%
\begin{align*}
\underbrace{f}_{=\mathfrak{q}\circ\operatorname*{spr}}\left(  a^{n}\right)
&  =\left(  \mathfrak{q}\circ\operatorname*{spr}\right)  \left(  a^{n}\right)
=\underbrace{\mathfrak{q}}_{=\operatorname*{syminc}\nolimits_{\mathfrak{e}%
\left(  H\right)  }\circ\mathfrak{e}^{\prime}}\left(
\underbrace{\operatorname*{spr}\left(  a^{n}\right)  }_{=w^{n}}\right)
=\left(  \operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }%
\circ\mathfrak{e}^{\prime}\right)  \left(  w^{n}\right) \\
&  =\underbrace{\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)
}}_{\substack{=\operatorname*{syminc}\nolimits_{V}\\\text{(since }%
\mathfrak{e}\left(  H\right)  =V\text{)}}}\left(  \underbrace{\mathfrak{e}%
^{\prime}\left(  w^{n}\right)  }_{=\delta_{n,1}w}\right)
=\operatorname*{syminc}\nolimits_{V}\left(  \delta_{n,1}w\right) \\
&  =\delta_{n,1}\underbrace{\operatorname*{syminc}\nolimits_{V}\left(
w\right)  }_{=a}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
\operatorname*{syminc}\nolimits_{V}\text{ is }k\text{-linear}\right) \\
&  =\delta_{n,1}a.
\end{align*}
Comparing this with $g\left(  a^{n}\right)  =\delta_{n,1}a$, we obtain
$f\left(  a^{n}\right)  =g\left(  a^{n}\right)  $.

Now, forget that we fixed $a$. We thus have shown that $f\left(  a^{n}\right)
=g\left(  a^{n}\right)  $ for each $a\in\operatorname*{syminc}\nolimits_{V}%
\left(  V\right)  $ and $n\in\mathbb{N}$. Thus, Lemma~\ref{Lemma1801} (applied
to $W=\operatorname*{Sym}V$) shows that $f=g$. Comparing this with
$f=\mathfrak{q}\circ\operatorname*{spr}$, we obtain $\mathfrak{q}%
\circ\operatorname*{spr}=f=g=\operatorname*{Log}\operatorname{id}%
_{\operatorname*{Sym}V}$. This proves Lemma~\ref{Lemma1831} \textbf{(b)}.
\end{proof}

\begin{theorem}
\label{Theorem1832} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered cocommutative bialgebra over $k$. Consider the convolution
algebra $\mathcal{L}\left(  H,H\right)  $. Let $\mathfrak{e}$ denote the map
$\operatorname*{Log}\operatorname*{id}\in\mathcal{L}\left(  H,H\right)  $.

Let $\mathfrak{j}$ denote the inclusion map $\mathfrak{e}\left(  H\right)
\rightarrow H$. Define a map $\mathfrak{e}^{\prime}:H\rightarrow
\mathfrak{e}\left(  H\right)  $ by%
\[
\left(  \mathfrak{e}^{\prime}\left(  h\right)  =\mathfrak{e}\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H\right)  .
\]
\footnote{This is well-defined, since $\mathfrak{e}\left(  h\right)
\in\mathfrak{e}\left(  H\right)  $ for every $h\in H$.} This map
$\mathfrak{e}^{\prime}$ is clearly $k$-linear\footnote{In fact, this map
$\mathfrak{e}^{\prime}$ is obtained from the $k$-linear map $\mathfrak{e}$ by
changing the target to $\mathfrak{e}\left(  H\right)  $.}. Define a $k$-linear
map $\mathfrak{q}:H\rightarrow\operatorname*{Sym}\left(  \mathfrak{e}\left(
H\right)  \right)  $ by $\mathfrak{q}=\operatorname*{syminc}%
\nolimits_{\mathfrak{e}\left(  H\right)  }\circ\mathfrak{e}^{\prime}$.

Let $V$ denote the $k$-vector subspace $\operatorname*{Prim}H$ of $H$. Define
the $k$-linear map $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow H$ as
in Proposition~\ref{Proposition1810}.

From Lemma~\ref{Lemma1831} \textbf{(a)}, we know that $\mathfrak{e}\left(
H\right)  =V$. Thus, $\mathfrak{q}$ is a $k$-linear map $H\rightarrow
\operatorname*{Sym}V$.

Recall that $\operatorname*{Sym}V$ is a connected filtered $k$-bialgebra
(according to Theorem~\ref{Theorem1812} \textbf{(c)}).

The maps $e^{\ast\mathfrak{q}}:H\rightarrow\operatorname*{Sym}V$ and
$\operatorname*{spr}:\operatorname*{Sym}V\rightarrow H$ are mutually inverse
$k$-coalgebra isomorphisms.
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{Theorem1832}.]Recall that $\mathfrak{q}$ is a
$k$-linear map $H\rightarrow\operatorname*{Sym}\left(  \mathfrak{e}\left(
H\right)  \right)  $. In other words, $\mathfrak{q}$ is a $k$-linear map
$H\rightarrow\operatorname*{Sym}V$ (since $\mathfrak{e}\left(  H\right)  =V$).

The map $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow H$ is a
$k$-coalgebra homomorphism (by Proposition~\ref{Proposition1810}) and is
surjective (by Proposition~\ref{Proposition1811}).

We have $\mathfrak{q}\in\mathfrak{g}\left(  H,\operatorname*{Sym}V\right)
$\ \ \ \ \footnote{\textit{Proof.} We have $\left(  \operatorname*{Log}%
\operatorname*{id}\right)  \left(  1_{H}\right)  =0$. (This has already been
proven in the proof of Corollary~\ref{Corollary1821}). Now, the definition of
$\mathfrak{e}^{\prime}$ yields $\mathfrak{e}^{\prime}\left(  1_{H}\right)
=\underbrace{\mathfrak{e}}_{=\operatorname*{Log}\operatorname*{id}}\left(
1_{H}\right)  =\left(  \operatorname*{Log}\operatorname*{id}\right)  \left(
1_{H}\right)  =0$. Furthermore, $\underbrace{\mathfrak{q}}%
_{=\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }%
\circ\mathfrak{e}^{\prime}}\left(  1_{H}\right)  =\left(
\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }%
\circ\mathfrak{e}^{\prime}\right)  \left(  1_{H}\right)
=\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }\left(
\underbrace{\mathfrak{e}^{\prime}\left(  1_{H}\right)  }_{=0}\right)
=\operatorname*{syminc}\nolimits_{\mathfrak{e}\left(  H\right)  }\left(
0\right)  =0$ (since the map $\operatorname*{syminc}\nolimits_{\mathfrak{e}%
\left(  H\right)  }$ is $k$-linear).
\par
But $\mathfrak{q}$ is a $k$-linear map $H\rightarrow\operatorname*{Sym}V$. In
other words, $\mathfrak{q}\in\mathcal{L}\left(  H,\operatorname*{Sym}V\right)
$.
\par
Thus, $\mathfrak{q}$ is an $f\in\mathcal{L}\left(  H,\operatorname*{Sym}%
V\right)  $ satisfying $f\left(  1_{H}\right)  =0$ (since $\mathfrak{q}\left(
1_{H}\right)  =0$). In other words, $\mathfrak{q}\in\left\{  f\in
\mathcal{L}\left(  H,\operatorname*{Sym}V\right)  \ \mid\ f\left(
1_{H}\right)  =0\right\}  $.
\par
But the definition of $\mathfrak{g}\left(  H,\operatorname*{Sym}V\right)  $
yields $\mathfrak{g}\left(  H,\operatorname*{Sym}V\right)  =\left\{
f\in\mathcal{L}\left(  H,\operatorname*{Sym}V\right)  \ \mid\ f\left(
1_{H}\right)  =0\right\}  $. Thus, $\mathfrak{q}\in\left\{  f\in
\mathcal{L}\left(  H,\operatorname*{Sym}V\right)  \ \mid\ f\left(
1_{H}\right)  =0\right\}  =\mathfrak{g}\left(  H,\operatorname*{Sym}V\right)
$.}. Hence, the map $e^{\ast\mathfrak{q}}:H\rightarrow\operatorname*{Sym}V$ is well-defined.

Recall that $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow H$ is a
$k$-coalgebra homomorphism. Also $\operatorname*{spr}\left(
1_{\operatorname*{Sym}V}\right)  =1_{H}$\ \ \ \ \footnote{\textit{Proof.} The
equality (\ref{1811.eq}) (applied to $n=0$ and $\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  =\left(  {}\right)  $) yields%
\begin{align*}
&  \operatorname*{spr}\left(  \left(  \text{empty product in }%
\operatorname*{Sym}V\right)  \right) \\
&  =\underbrace{\dfrac{1}{0!}}_{=\dfrac{1}{1}=1}\sum_{\sigma\in S_{0}%
}\underbrace{\left(  \text{empty product in }H\right)  }_{=1_{H}}=\sum
_{\sigma\in S_{0}}1_{H}=\underbrace{\left\vert S_{0}\right\vert }_{=0!=1}%
\cdot1_{H}=1_{H}.
\end{align*}
Comparing this with $\operatorname*{spr}\left(  \underbrace{\left(
\text{empty product in }\operatorname*{Sym}V\right)  }%
_{=1_{\operatorname*{Sym}V}}\right)  =\operatorname*{spr}\left(
1_{\operatorname*{Sym}V}\right)  $, we obtain $\operatorname*{spr}\left(
1_{\operatorname*{Sym}V}\right)  =1_{H}$.}.

Hence, Proposition~\ref{Proposition1101} \textbf{(e)} (applied to $H$,
$\operatorname*{Sym}V$, $\operatorname*{Sym}V$ and $\operatorname*{spr}$
instead of $C$, $D$, $A$ and $\varphi$) shows that every $f\in\mathfrak{g}%
\left(  H,\operatorname*{Sym}V\right)  $ satisfies $f\circ\operatorname*{spr}%
\in\mathfrak{g}\left(  \operatorname*{Sym}V,\operatorname*{Sym}V\right)  $ and
$e^{\ast\left(  f\circ\operatorname*{spr}\right)  }=e^{\ast f}\circ
\operatorname*{spr}$. Applying this to $f=\mathfrak{q}$, we obtain
$\mathfrak{q}\circ\operatorname*{spr}\in\mathfrak{g}\left(
\operatorname*{Sym}V,\operatorname*{Sym}V\right)  $ and $e^{\ast\left(
\mathfrak{q}\circ\operatorname*{spr}\right)  }=e^{\ast\mathfrak{q}}%
\circ\operatorname*{spr}$.

We have $\operatorname{id}_{\operatorname*{Sym}V}\in G\left(
\operatorname*{Sym}V,\operatorname*{Sym}V\right)  $ (since $\operatorname{id}%
_{B}\in G\left(  B,B\right)  $ for every $k$-bialgebra $B$). Thus,
Proposition~\ref{Proposition13} \textbf{(b)} (applied to $\operatorname*{Sym}%
V$, $\operatorname*{Sym}V$ and $\operatorname{id}_{\operatorname*{Sym}V}$
instead of $A$, $H$ and $F$) yields $e^{\ast\left(  \operatorname*{Log}%
\operatorname{id}_{\operatorname*{Sym}V}\right)  }=\operatorname{id}%
_{\operatorname*{Sym}V}$.

Lemma~\ref{Lemma1831} \textbf{(b)} yields $\mathfrak{q}\circ
\operatorname*{spr}=\operatorname*{Log}\operatorname{id}_{\operatorname*{Sym}%
V}$. Thus, $e^{\ast\left(  \mathfrak{q}\circ\operatorname*{spr}\right)
}=e^{\ast\left(  \operatorname*{Log}\operatorname{id}_{\operatorname*{Sym}%
V}\right)  }=\operatorname{id}_{\operatorname*{Sym}V}$. Comparing this with
$e^{\ast\left(  \mathfrak{q}\circ\operatorname*{spr}\right)  }=e^{\ast
\mathfrak{q}}\circ\operatorname*{spr}$, we obtain $e^{\ast\mathfrak{q}}%
\circ\operatorname*{spr}=\operatorname{id}_{\operatorname*{Sym}V}$.

Using the surjectivity of $\operatorname*{spr}$, it is now easy to conclude
that $\operatorname*{spr}\circ e^{\ast\mathfrak{q}}=\operatorname{id}_{H}%
$\ \ \ \ \footnote{\textit{Proof.} Let $h\in H$. Then, $h\in
H=\operatorname*{spr}\left(  \operatorname*{Sym}V\right)  $ (since the map
$\operatorname*{spr}$ is surjective). In other words, there exists some
$a\in\operatorname*{Sym}V$ such that $h=\operatorname*{spr}a$. Consider this
$a$.
\par
We have $\underbrace{\left(  e^{\ast\mathfrak{q}}\circ\operatorname*{spr}%
\right)  }_{=\operatorname{id}_{\operatorname*{Sym}V}}\left(  a\right)
=\operatorname{id}_{\operatorname*{Sym}V}\left(  a\right)  =a$, so that
$a=\left(  e^{\ast\mathfrak{q}}\circ\operatorname*{spr}\right)  \left(
a\right)  =e^{\ast\mathfrak{q}}\left(  \underbrace{\operatorname*{spr}a}%
_{=h}\right)  =e^{\ast\mathfrak{q}}\left(  h\right)  $. Now,%
\[
\left(  \operatorname*{spr}\circ e^{\ast\mathfrak{q}}\right)  \left(
h\right)  =\operatorname*{spr}\left(  \underbrace{e^{\ast\mathfrak{q}}\left(
h\right)  }_{=a}\right)  =\operatorname*{spr}a=h=\operatorname{id}_{H}\left(
h\right)  .
\]
\par
Now, forget that we fixed $h$. We thus have shown that $\left(
\operatorname*{spr}\circ e^{\ast\mathfrak{q}}\right)  \left(  h\right)
=\operatorname{id}_{H}\left(  h\right)  $ for each $h\in H$. In other words,
$\operatorname*{spr}\circ e^{\ast\mathfrak{q}}=\operatorname{id}_{H}$.}.

Combining the equalities $e^{\ast\mathfrak{q}}\circ\operatorname*{spr}%
=\operatorname{id}_{\operatorname*{Sym}V}$ and $\operatorname*{spr}\circ
e^{\ast\mathfrak{q}}=\operatorname{id}_{H}$, we conclude that the maps
$e^{\ast\mathfrak{q}}:H\rightarrow\operatorname*{Sym}V$ and
$\operatorname*{spr}:\operatorname*{Sym}V\rightarrow H$ are mutually inverse.
Thus, these maps are invertible. Furthermore, the inverse of
$\operatorname*{spr}$ is $\operatorname*{spr}\nolimits^{-1}=e^{\ast
\mathfrak{q}}$ (since the maps $e^{\ast\mathfrak{q}}:H\rightarrow
\operatorname*{Sym}V$ and $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow
H$ are mutually inverse).

The map $\operatorname*{spr}$ is a $k$-coalgebra isomorphism (since it is
invertible and it is a $k$-coalgebra homomorphism). Hence, its inverse
$\operatorname*{spr}\nolimits^{-1}$ is a $k$-coalgebra isomorphism as well.
Since $\operatorname*{spr}\nolimits^{-1}=e^{\ast\mathfrak{q}}$, this rewrites
as follows: The map $e^{\ast\mathfrak{q}}$ is a $k$-coalgebra isomorphism.
Altogether, we thus have shown that the maps $e^{\ast\mathfrak{q}%
}:H\rightarrow\operatorname*{Sym}V$ and $\operatorname*{spr}%
:\operatorname*{Sym}V\rightarrow H$ are mutually inverse $k$-coalgebra
isomorphisms. This proves Theorem~\ref{Theorem1832}.
\end{proof}

\begin{corollary}
\label{Corollary1833} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected filtered cocommutative bialgebra over $k$. Let $V$ denote the
$k$-vector subspace $\operatorname*{Prim}H$ of $H$. Define the $k$-linear map
$\operatorname*{spr}:\operatorname*{Sym}V\rightarrow H$ as in
Proposition~\ref{Proposition1810}. Then, $\operatorname*{spr}$ is a
$k$-coalgebra isomorphism.
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{Corollary1833}.]We shall use the notations introduced
in Theorem~\ref{Theorem1832}.

Theorem~\ref{Theorem1832} shows that the maps $e^{\ast\mathfrak{q}%
}:H\rightarrow\operatorname*{Sym}V$ and $\operatorname*{spr}%
:\operatorname*{Sym}V\rightarrow H$ are mutually inverse $k$-coalgebra
isomorphisms. Thus, in particular, $\operatorname*{spr}$ is a $k$-coalgebra
isomorphism. This proves Corollary~\ref{Corollary1833}.
\end{proof}

We notice that Corollary~\ref{Corollary1833} is closely related to
\cite[Corollary 4.4]{2}. Indeed, the \textquotedblleft canonical
map\textquotedblright\ in the corrected version of \cite[Corollary 4.4]%
{2}\footnote{The definition of the \textquotedblleft canonical
map\textquotedblright\ in \cite[Corollary 4.4]{2} is slightly wrong: The
\textquotedblleft$\bigoplus_{n\in\mathbf{N}}\iota^{\otimes n}\circ\Delta_{n}%
$\textquotedblright\ should be replaced by \textquotedblleft$\bigoplus
_{n\in\mathbf{N}}\dfrac{1}{n!}\iota^{\otimes n}\circ\Delta_{n}$%
\textquotedblright.} (applied to $A=H$) is more or less our map $e^{\ast
\mathfrak{q}}:H\rightarrow\operatorname*{Sym}V$, at least if we identify the
invariant space $\bigoplus_{n\in\mathbb{N}}\left(  \left(
\operatorname*{Prim}H\right)  ^{\otimes n}\right)  ^{S_{n}}\subseteq
\otimes\operatorname*{Prim}H$ with the symmetric algebra $\operatorname*{Sym}%
\left(  \operatorname*{Prim}H\right)  $.

\begin{remark}
\label{Remark1834} Again, it is possible to slightly improve
Theorem~\ref{Theorem1832} and Corollary~\ref{Corollary1833}: Namely, the
symmetric algebra $\operatorname*{Sym}V$ canonically becomes a filtered
$k$-algebra\footnote{Indeed, $V$ is a filtered $k$-vector space (by virtue of
being a subspace of the filtered $k$-vector space $H$), and thus its symmetric
algebra $\operatorname*{Sym}V$ becomes a filtered $k$-algebra.}. With respect
to this filtration, the maps $e^{\ast\mathfrak{q}}:H\rightarrow
\operatorname*{Sym}V$ and $\operatorname*{spr}:\operatorname*{Sym}V\rightarrow
H$ in Theorem~\ref{Theorem1832} are actually isomorphisms of \textbf{filtered}
$k$-coalgebras. Proving this is not too difficult (it is mostly an issue of bookkeeping).
\end{remark}

\section{\label{S63}Graded versions of Leray's theorem}

The notion of \textquotedblleft Leray's theorem\textquotedblright\ does (to my
knowledge) not refer to any particular fact; rather, it stands for a group of
results, each of which claims that a commutative bialgebra satisfying certain
conditions (e.g., connected filtered or connected graded) must be isomorphic
\textbf{as an algebra} to a symmetric algebra of a certain vector space.
Theorem~\ref{Theorem1702} is one of these results; but there are others. To
derive one other such result, let us first prepare by analyzing properties of
graded vector spaces.

\begin{lemma}
\label{lem.leray2.graded-v1} Let $k$ be a field. Let $V$ be a graded
$k$-vector space. For each $m\in\mathbb{N}$, we define a $k$-vector subspace
$V_{>m}$ of $V$ by $V_{>m}=\sum_{\substack{g\in\mathbb{N};\\g>m}}V_{g}$.

Let $f:V\rightarrow V$ be a $k$-linear map.\footnote{We do not require $f$ to
be graded.} Assume that%
\begin{equation}
f\left(  V_{n}\right)  \subseteq V_{>n}\ \ \ \ \ \ \ \ \ \ \text{for each
}n\in\mathbb{N}. \label{eq.lem.leray2.graded-v1.ass}%
\end{equation}
Then, the map $\operatorname*{id}\nolimits_{V}-f$ is injective.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.graded-v1}.]We shall use the notations
introduced in Remark~\ref{Remark220} and in Definition~\ref{Definition221}.

If $n\in\mathbb{N}$ and $m\in\mathbb{N}$ satisfy $m\leq n$, then%
\begin{equation}
p_{m,V}\left(  V_{>n}\right)  =0 \label{pf.lem.leray2.graded-v1.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-v1.1}):} Let
$n\in\mathbb{N}$ and $m\in\mathbb{N}$ be such that $m\leq n$. The definition
of $V_{>n}$ yields $V_{>n}=\sum_{\substack{g\in\mathbb{N};\\g>n}}V_{g}$.
\par
Let $g\in\mathbb{N}$ be such that $g>n$. Then, $g>n\geq m$ (since $m\leq n$).
Hence, $g\neq m$. Thus, (\ref{221.0}) (applied to $m$ and $g$ instead of $n$
and $m$) yields $p_{m,V}\mid_{V_{g}}=0$. Thus, $\underbrace{\left(
p_{m,V}\mid_{V_{g}}\right)  }_{=0}\left(  V_{g}\right)  =0\left(
V_{g}\right)  =0$. Comparing this with $\left(  p_{m,V}\mid_{V_{g}}\right)
\left(  V_{g}\right)  =p_{m,V}\left(  V_{g}\right)  $, we obtain
$p_{m,V}\left(  V_{g}\right)  =0$.
\par
Now, forget that we fixed $g$. We thus have shown that $p_{m,V}\left(
V_{g}\right)  =0$ for each $g\in\mathbb{N}$ satisfying $g>n$. Hence,
$\sum_{\substack{g\in\mathbb{N};\\g>n}}\underbrace{p_{m,V}\left(
V_{g}\right)  }_{=0}=\sum_{\substack{g\in\mathbb{N};\\g>n}}0=0$.
\par
But recall that $V_{>n}=\sum_{\substack{g\in\mathbb{N};\\g>n}}V_{g}$. Applying
the map $p_{m,V}$ to both sides of this equality, we find%
\begin{align*}
p_{m,V}\left(  V_{>n}\right)   &  =p_{m,V}\left(  \sum_{\substack{g\in
\mathbb{N};\\g>n}}V_{g}\right)  =\sum_{\substack{g\in\mathbb{N};\\g>n}%
}p_{m,V}\left(  V_{g}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}p_{m,V}\text{ is }k\text{-linear}\right) \\
&  =0.
\end{align*}
This proves (\ref{pf.lem.leray2.graded-v1.1}).}.

Each $m\in\mathbb{N}$ satisfies%
\begin{equation}
p_{m,V}\left(  \sum_{\substack{g\in\mathbb{N};\\g\geq m}}f\left(
V_{g}\right)  \right)  =0 \label{pf.lem.leray2.graded-v1.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-v1.2}):} Let
$m\in\mathbb{N}$. Let $g\in\mathbb{N}$ be such that $g\geq m$. Hence, $m\leq
g$. But (\ref{eq.lem.leray2.graded-v1.ass}) (applied to $n=g$) yields
$f\left(  V_{g}\right)  \subseteq V_{>g}$. Hence, $p_{m,V}\left(
\underbrace{f\left(  V_{g}\right)  }_{\subseteq V_{>g}}\right)  \subseteq
p_{m,V}\left(  V_{>g}\right)  =0$ (by (\ref{pf.lem.leray2.graded-v1.1})
(applied to $n=g$)). Hence, $p_{m,V}\left(  f\left(  V_{g}\right)  \right)
=0$.
\par
Now, forget that we fixed $g$. We thus have shown that%
\begin{equation}
p_{m,V}\left(  f\left(  V_{g}\right)  \right)  =0\ \ \ \ \ \ \ \ \ \ \text{for
each }g\in\mathbb{N}\text{ satisfying }g\geq m.
\label{pf.lem.leray2.graded-v1.2.pf.1}%
\end{equation}
\par
Now, the map $p_{m,V}$ is $k$-linear. Hence,%
\[
p_{m,V}\left(  \sum_{\substack{g\in\mathbb{N};\\g\geq m}}f\left(
V_{g}\right)  \right)  =\sum_{\substack{g\in\mathbb{N};\\g\geq m}%
}\underbrace{p_{m,V}\left(  f\left(  V_{g}\right)  \right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.leray2.graded-v1.2.pf.1}))}}%
}=\sum_{\substack{g\in\mathbb{N};\\g\geq m}}0=0.
\]
This proves (\ref{pf.lem.leray2.graded-v1.2}).}.

Now, let $x\in\operatorname*{Ker}\left(  \operatorname*{id}\nolimits_{V}%
-f\right)  $ be arbitrary. We shall prove that $x=0$.

We have $x\in\operatorname*{Ker}\left(  \operatorname*{id}\nolimits_{V}%
-f\right)  $. In other words, $x$ is an element of $V$ satisfying $\left(
\operatorname*{id}\nolimits_{V}-f\right)  \left(  x\right)  =0$. Comparing
$\left(  \operatorname*{id}\nolimits_{V}-f\right)  \left(  x\right)  =0$ with
$\left(  \operatorname*{id}\nolimits_{V}-f\right)  \left(  x\right)
=\underbrace{\operatorname*{id}\nolimits_{V}\left(  x\right)  }_{=x}-f\left(
x\right)  =x-f\left(  x\right)  $, we obtain $x-f\left(  x\right)  =0$. In
other words, $x=f\left(  x\right)  $.

But (\ref{221.vsum}) (applied to $v=x$) yields%
\begin{equation}
x=\sum_{\ell\in\mathbb{N}}p_{\ell,V}\left(  x\right)  =\sum_{g\in\mathbb{N}%
}p_{g,V}\left(  x\right)  \label{pf.lem.leray2.graded-v1.3}%
\end{equation}
(here, we have renamed the summation index $\ell$ as $g$).

Next, we claim that every $g\in\mathbb{N}$ satisfies%
\begin{equation}
p_{g,V}\left(  x\right)  =0. \label{pf.lem.leray2.graded-v1.pkvx=0}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.leray2.graded-v1.pkvx=0}):} We will prove
(\ref{pf.lem.leray2.graded-v1.pkvx=0}) by strong induction over $g$:

\textit{Induction step:} Let $m\in\mathbb{N}$. Assume that
(\ref{pf.lem.leray2.graded-v1.pkvx=0}) holds for every $g<m$. We must then
prove that (\ref{pf.lem.leray2.graded-v1.pkvx=0}) holds for $g=m$.

We have assumed that (\ref{pf.lem.leray2.graded-v1.pkvx=0}) holds for every
$g<m$. In other words, for every $g\in\mathbb{N}$ satisfying $g<m$, we have%
\begin{equation}
p_{g,V}\left(  x\right)  =0. \label{pf.lem.leray2.graded-v1.pkvx=0.pf.1}%
\end{equation}


Now, (\ref{pf.lem.leray2.graded-v1.3}) becomes%
\begin{align*}
x  &  =\sum_{g\in\mathbb{N}}p_{g,V}\left(  x\right)  =\sum_{\substack{g\in
\mathbb{N};\\g<m}}\underbrace{p_{g,V}\left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.leray2.graded-v1.pkvx=0.pf.1}))}}%
}+\sum_{\substack{g\in\mathbb{N};\\g\geq m}}p_{g,V}\left(  x\right)
=\underbrace{\sum_{\substack{g\in\mathbb{N};\\g<m}}0}_{=0}+\sum
_{\substack{g\in\mathbb{N};\\g\geq m}}p_{g,V}\left(  x\right) \\
&  =\sum_{\substack{g\in\mathbb{N};\\g\geq m}}p_{g,V}\left(  \underbrace{x}%
_{\in V}\right) \\
&  \in\sum_{\substack{g\in\mathbb{N};\\g\geq m}}\underbrace{p_{g,V}\left(
V\right)  }_{\substack{=V_{g}\\\text{(by (\ref{221.im}) (applied to
}n=g\text{))}}}=\sum_{\substack{g\in\mathbb{N};\\g\geq m}}V_{g}.
\end{align*}
But recall that%
\[
x=f\left(  \underbrace{x}_{\in\sum_{\substack{g\in\mathbb{N};\\g\geq m}}V_{g}%
}\right)  \in f\left(  \sum_{\substack{g\in\mathbb{N};\\g\geq m}}V_{g}\right)
=\sum_{\substack{g\in\mathbb{N};\\g\geq m}}f\left(  V_{g}\right)
\]
(since the map $f$ is $k$-linear). Applying the map $p_{m,V}$ to both sides of
this relation, we obtain%
\[
p_{m,V}\left(  x\right)  \in p_{m,V}\left(  \sum_{\substack{g\in
\mathbb{N};\\g\geq m}}f\left(  V_{g}\right)  \right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.leray2.graded-v1.2}%
)}\right)  .
\]
In other words, $p_{m,V}\left(  x\right)  =0$. In other words,
(\ref{pf.lem.leray2.graded-v1.pkvx=0}) holds for $g=m$. This completes the
induction step. Thus, the proof of (\ref{pf.lem.leray2.graded-v1.pkvx=0}) by
induction is complete.]

Now, (\ref{pf.lem.leray2.graded-v1.3}) becomes%
\[
x=\sum_{g\in\mathbb{N}}\underbrace{p_{g,V}\left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.leray2.graded-v1.pkvx=0}))}}%
}=\sum_{g\in\mathbb{N}}0=0.
\]


Now, forget that we fixed $x$. We thus have proven that $x=0$ for each
$x\in\operatorname*{Ker}\left(  \operatorname*{id}\nolimits_{V}-f\right)  $.
In other words, $\operatorname*{Ker}\left(  \operatorname*{id}\nolimits_{V}%
-f\right)  \subseteq0$. Thus, $\operatorname*{Ker}\left(  \operatorname*{id}%
\nolimits_{V}-f\right)  =0$. Hence, the map $\operatorname*{id}\nolimits_{V}%
-f$ is injective (since this map is $k$-linear (since the maps
$\operatorname*{id}\nolimits_{V}$ and $f$ are $k$-linear)). This proves
Lemma~\ref{lem.leray2.graded-v1}.
\end{proof}

Our next few lemmas will rely on the notations introduced in
Convention~\ref{Convention230}.

\begin{lemma}
\label{lem.graded-vec.WinVn}Let $k$ be a field. Let $V$ be a graded $k$-vector
space. Let $U$ be a subset of $V$. Let $n\in\mathbb{N}$.

Consider the map $p_{n,V}:V\rightarrow V$ defined in
Definition~\ref{Definition221}.

\textbf{(a)} If $U\subseteq V_{n}$, then $p_{n,V}\left(  U\right)  =U$.

\textbf{(b)} Let $m\in\mathbb{N}$ be such that $m\neq n$. If $U\subseteq
V_{m}$, then $p_{n,V}\left(  U\right)  =0$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.graded-vec.WinVn}.]\textbf{(a)} Assume that
$U\subseteq V_{n}$. The equality (\ref{221.id}) yields $p_{n,V}\mid_{V_{n}%
}=\operatorname{id}_{V}\mid_{V_{n}}$.

But each $x\in U$ satisfies%
\begin{align}
p_{n,V}\left(  x\right)   &  =\underbrace{\left(  p_{n,V}\mid_{V_{n}}\right)
}_{=\operatorname{id}_{V}\mid_{V_{n}}}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }x\in U\subseteq V_{n}\right)
\nonumber\\
&  =\left(  \operatorname{id}_{V}\mid_{V_{n}}\right)  \left(  x\right)
=\operatorname{id}_{V}\left(  x\right)  =x.
\label{pf.lem.graded-vec.WinVn.a.1}%
\end{align}
Now,%
\[
p_{n,V}\left(  U\right)  =\left\{  \underbrace{p_{n,V}\left(  x\right)
}_{\substack{=x\\\text{(by (\ref{pf.lem.graded-vec.WinVn.a.1}))}}}\ \mid\ x\in
U\right\}  =\left\{  x\ \mid\ x\in U\right\}  =U.
\]
This proves Lemma~\ref{lem.graded-vec.WinVn} \textbf{(a)}.

\textbf{(b)} Assume that $U\subseteq V_{m}$. But recall that $m\neq n$. Thus,
$n\neq m$. Hence, (\ref{221.0}) yields $p_{n,V}\mid_{V_{m}}=0$. But%
\begin{align*}
p_{n,V}\left(  \underbrace{U}_{\subseteq V_{m}}\right)   &  \subseteq
p_{n,V}\left(  V_{m}\right)  =\underbrace{\left(  p_{n,V}\mid_{V_{m}}\right)
}_{=0}\left(  V_{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
p_{n,V}\mid_{V_{m}}\right)  \left(  V_{m}\right)  =p_{n,V}\left(
V_{m}\right)  \right) \\
&  =0\left(  V_{m}\right)  =0.
\end{align*}
In other words, $p_{n,V}\left(  U\right)  =0$. This proves
Lemma~\ref{lem.graded-vec.WinVn} \textbf{(b)}.
\end{proof}

\begin{lemma}
\label{lem.graded-alg.genby1}Let $k$ be a field. Let $A$ be a graded $k$-algebra.

\textbf{(a)} Each $n\in\mathbb{N}$ satisfies $\left(  A_{1}\right)
^{n}\subseteq A_{n}$.

\textbf{(b)} Assume that $A=\operatorname*{AlgGen}\nolimits_{k}\left(
A_{1}\right)  $. Then, each $n\in\mathbb{N}$ satisfies $A_{n}=\left(
A_{1}\right)  ^{n}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.graded-alg.genby1}.]\textbf{(a) }Since $A$ is a
graded $k$-algebra, we have
\begin{equation}
A_{\ell}A_{m}\subseteq A_{\ell+m}\ \ \ \ \ \ \ \ \ \ \text{for any }\ell
\in\mathbb{N}\text{ and }m\in\mathbb{N}. \label{pf.lem.graded-alg.genby1.AlAm}%
\end{equation}


Each $n\in\mathbb{N}$ satisfies%
\begin{equation}
\left(  A_{1}\right)  ^{n}\subseteq A_{n}
\label{pf.lem.graded-alg.genby1.A1nAn}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.graded-alg.genby1.A1nAn}):} We shall
prove (\ref{pf.lem.graded-alg.genby1.A1nAn}) by induction over $n$:
\par
\textit{Induction base:} We have $1_{A}\in A_{0}$ (since $A$ is a graded
$k$-algebra). Now, $\left(  A_{1}\right)  ^{0}=k\cdot\underbrace{1_{A}}_{\in
A_{0}}\subseteq k\cdot A_{0}\subseteq A_{0}$ (since $A_{0}$ is a $k$-vector
subspace of $A$). In other words, (\ref{pf.lem.graded-alg.genby1.A1nAn}) holds
for $n=0$. This completes the induction base.
\par
\textit{Induction step:} Let $N$ be a positive integer. Assume that
(\ref{pf.lem.graded-alg.genby1.A1nAn}) holds for $n=N-1$. We must prove that
(\ref{pf.lem.graded-alg.genby1.A1nAn}) holds for $n=N$.
\par
We know that $N-1\in\mathbb{N}$ (since $N$ is a positive integer). Hence,
(\ref{pf.lem.graded-alg.genby1.AlAm}) (applied to $\ell=N-1$ and $m=1$) yields
$A_{N-1}A_{1}\subseteq A_{\left(  N-1\right)  +1}=A_{N}$ (since $\left(
N-1\right)  +1=N$).
\par
But we have assumed that (\ref{pf.lem.graded-alg.genby1.A1nAn}) holds for
$n=N-1$. In other words, we have $\left(  A_{1}\right)  ^{N-1}\subseteq
A_{N-1}$. Now,%
\[
\left(  A_{1}\right)  ^{N}=\underbrace{\left(  A_{1}\right)  ^{N-1}%
}_{\subseteq A_{N-1}}A_{1}\subseteq A_{N-1}A_{1}\subseteq A_{N}.
\]
In other words, (\ref{pf.lem.graded-alg.genby1.A1nAn}) holds for $n=N$. This
completes the induction step.
\par
Thus, the induction proof of (\ref{pf.lem.graded-alg.genby1.A1nAn}) is
complete.}. This proves Lemma~\ref{lem.graded-alg.genby1} \textbf{(a)}.

\textbf{(b)} Recall the following basic fact (recorded in
Convention~\ref{Convention230}): If $S$ is a subset of the $k$-algebra $A$,
then the $k$-subalgebra of $A$ generated by $S$ is the $k$-vector subspace
$\sum_{\ell\in\mathbb{N}}\left\langle S\right\rangle ^{\ell}$ of $A$. In other
words, if $S$ is a subset of the $k$-algebra $A$, then
\[
\left(  \text{the }k\text{-subalgebra of }A\text{ generated by }S\right)
=\sum_{\ell\in\mathbb{N}}\left\langle S\right\rangle ^{\ell}.
\]
Applying this to $S=A_{1}$, we obtain%
\[
\left(  \text{the }k\text{-subalgebra of }A\text{ generated by }A_{1}\right)
=\sum_{\ell\in\mathbb{N}}\left\langle A_{1}\right\rangle ^{\ell}.
\]


On the other hand, if $W$ is a $k$-vector space, and if $S$ is a $k$-vector
subspace of $W$, then $\left\langle S\right\rangle =S$.\ \ \ \ \footnote{This
is an elementary fact from linear algebra.} Applying this to $W=A$ and
$S=A_{1}$, we obtain $\left\langle A_{1}\right\rangle =A_{1}$ (since $A_{1}$
is a $k$-vector subspace of $A$).

But $\operatorname*{AlgGen}\nolimits_{k}\left(  A_{1}\right)  $ is defined as
the $k$-subalgebra of $A$ generated by $A_{1}$. Hence,%
\begin{align*}
\operatorname*{AlgGen}\nolimits_{k}\left(  A_{1}\right)   &  =\left(
\text{the }k\text{-subalgebra of }A\text{ generated by }A_{1}\right)
=\sum_{\ell\in\mathbb{N}}\underbrace{\left\langle A_{1}\right\rangle ^{\ell}%
}_{\substack{=\left(  A_{1}\right)  ^{\ell}\\\text{(since }\left\langle
A_{1}\right\rangle =A_{1}\text{)}}}\\
&  =\sum_{\ell\in\mathbb{N}}\left(  A_{1}\right)  ^{\ell}.
\end{align*}
The assumptions of Lemma~\ref{lem.graded-alg.genby1} now yield
\begin{equation}
A=\operatorname*{AlgGen}\nolimits_{k}\left(  A_{1}\right)  =\sum_{\ell
\in\mathbb{N}}\left(  A_{1}\right)  ^{\ell}.
\label{pf.lem.graded-alg.genby1.A=sum}%
\end{equation}


We shall use the notations introduced in Remark~\ref{Remark220} and in
Definition~\ref{Definition221}.

Now, fix $n\in\mathbb{N}$. For each $\ell\in\mathbb{N}$ satisfying $\ell\neq
n$, we have%
\begin{equation}
p_{n,A}\left(  \left(  A_{1}\right)  ^{\ell}\right)  =0
\label{pf.lem.graded-alg.genby1.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.graded-alg.genby1.3}):} Let $\ell
\in\mathbb{N}$ satisfy $\ell\neq n$. But (\ref{pf.lem.graded-alg.genby1.A1nAn}%
) (applied to $\ell$ instead of $n$) shows that $\left(  A_{1}\right)  ^{\ell
}\subseteq A_{\ell}$. Hence, Lemma~\ref{lem.graded-vec.WinVn} \textbf{(b)}
(applied to $A$, $\left(  A_{1}\right)  ^{\ell}$ and $\ell$ instead of $V$,
$U$ and $m$) yields $p_{n,A}\left(  \left(  A_{1}\right)  ^{\ell}\right)  =0$.
This proves (\ref{pf.lem.graded-alg.genby1.3}).}. On the other hand,%
\begin{equation}
p_{n,A}\left(  \left(  A_{1}\right)  ^{n}\right)  =\left(  A_{1}\right)  ^{n}
\label{pf.lem.graded-alg.genby1.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.graded-alg.genby1.4}):} The relation
(\ref{pf.lem.graded-alg.genby1.A1nAn}) yields $\left(  A_{1}\right)
^{n}\subseteq A_{n}$. Hence, Lemma~\ref{lem.graded-vec.WinVn} \textbf{(a)}
(applied to $A$ and $\left(  A_{1}\right)  ^{n}$ instead of $V$ and $U$)
yields $p_{n,A}\left(  \left(  A_{1}\right)  ^{n}\right)  =\left(
A_{1}\right)  ^{n}$. This proves (\ref{pf.lem.graded-alg.genby1.4}).}.

Applying the map $p_{n,A}$ to both sides of the equality
(\ref{pf.lem.graded-alg.genby1.A=sum}), we obtain%
\begin{align*}
p_{n,A}\left(  A\right)   &  =p_{n,A}\left(  \sum_{\ell\in\mathbb{N}}\left(
A_{1}\right)  ^{\ell}\right)  =\sum_{\ell\in\mathbb{N}}p_{n,A}\left(  \left(
A_{1}\right)  ^{\ell}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}p_{n,A}\text{ is }k\text{-linear}\right) \\
&  =\underbrace{p_{n,A}\left(  \left(  A_{1}\right)  ^{n}\right)
}_{\substack{=\left(  A_{1}\right)  ^{n}\\\text{(by
(\ref{pf.lem.graded-alg.genby1.4}))}}}+\sum_{\substack{\ell\in\mathbb{N}%
;\\\ell\neq n}}\underbrace{p_{n,A}\left(  \left(  A_{1}\right)  ^{\ell
}\right)  }_{\substack{=0\\\text{(by (\ref{pf.lem.graded-alg.genby1.3}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}\ell=n\text{ from the sum}\right) \\
&  =\left(  A_{1}\right)  ^{n}+\underbrace{\sum_{\substack{\ell\in
\mathbb{N};\\\ell\neq n}}0}_{=0}=\left(  A_{1}\right)  ^{n}.
\end{align*}
Thus, $\left(  A_{1}\right)  ^{n}=p_{n,A}\left(  A\right)  =A_{n}$ (by
(\ref{221.im}) (applied to $V=A$)). In other words, $A_{n}=\left(
A_{1}\right)  ^{n}$. This proves Lemma~\ref{lem.graded-alg.genby1}
\textbf{(b)}.
\end{proof}

\begin{lemma}
\label{lem.leray.graded-a2}Let $k$ be a field. Let $A$ and $B$ be
$k$-algebras. Let $f:A\rightarrow B$ and $g:A\rightarrow B$ be two $k$-algebra
homomorphisms. Let $U$ and $V$ be two $k$-vector subspaces of $A$. Then,%
\[
\left(  g-f\right)  \left(  UV\right)  \subseteq\left(  \left(  g-f\right)
\left(  U\right)  \right)  \cdot g\left(  V\right)  +f\left(  U\right)
\cdot\left(  \left(  g-f\right)  \left(  V\right)  \right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray.graded-a2}.]The map $g-f$ is $k$-linear (since
the maps $g$ and $f$ are $k$-linear). Hence, $\left(  g-f\right)  \left(
U\right)  $ and $\left(  g-f\right)  \left(  V\right)  $ are $k$-vector
subspaces of $B$. Also, $g\left(  V\right)  $ is a $k$-vector subspace of $B$
(since the map $g$ is $k$-linear), and $f\left(  U\right)  $ is a $k$-vector
subspace of $B$ (since the map $f$ is $k$-linear). Hence,%
\[
\left(  \left(  g-f\right)  \left(  U\right)  \right)  \cdot g\left(
V\right)  +f\left(  U\right)  \cdot\left(  \left(  g-f\right)  \left(
V\right)  \right)
\]
is a $k$-vector subspace of $B$.

Each $\left(  u,v\right)  \in U\times V$ satisfies%
\begin{equation}
\left(  g-f\right)  \left(  uv\right)  \in\left(  \left(  g-f\right)  \left(
U\right)  \right)  \cdot g\left(  V\right)  +f\left(  U\right)  \cdot\left(
\left(  g-f\right)  \left(  V\right)  \right)
\label{pf.lem.leray.graded-a2.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray.graded-a2.1}):} Let $\left(
u,v\right)  \in U\times V$. Thus, $u\in U$ and $v\in V$.
\par
Now,%
\begin{align*}
\left(  g-f\right)  \left(  uv\right)   &  =\underbrace{g\left(  uv\right)
}_{\substack{=g\left(  u\right)  g\left(  v\right)  \\\text{(since }g\text{ is
a }k\text{-algebra}\\\text{homomorphism)}}}-\underbrace{f\left(  uv\right)
}_{\substack{=f\left(  u\right)  f\left(  v\right)  \\\text{(since }f\text{ is
a }k\text{-algebra}\\\text{homomorphism)}}}=g\left(  u\right)  g\left(
v\right)  -f\left(  u\right)  f\left(  v\right) \\
&  =\underbrace{\left(  g\left(  u\right)  g\left(  v\right)  -f\left(
u\right)  g\left(  v\right)  \right)  }_{=\left(  g\left(  u\right)  -f\left(
u\right)  \right)  \cdot g\left(  v\right)  }+\underbrace{\left(  f\left(
u\right)  g\left(  v\right)  -f\left(  u\right)  f\left(  v\right)  \right)
}_{=f\left(  u\right)  \cdot\left(  g\left(  v\right)  -f\left(  v\right)
\right)  }\\
&  =\underbrace{\left(  g\left(  u\right)  -f\left(  u\right)  \right)
}_{=\left(  g-f\right)  \left(  u\right)  }\cdot g\left(  v\right)  +f\left(
u\right)  \cdot\underbrace{\left(  g\left(  v\right)  -f\left(  v\right)
\right)  }_{=\left(  g-f\right)  \left(  v\right)  }\\
&  =\left(  g-f\right)  \left(  \underbrace{u}_{\in U}\right)  \cdot g\left(
\underbrace{v}_{\in V}\right)  +f\left(  \underbrace{u}_{\in U}\right)
\cdot\left(  g-f\right)  \left(  \underbrace{v}_{\in V}\right) \\
&  \in\left(  \left(  g-f\right)  \left(  U\right)  \right)  \cdot g\left(
V\right)  +f\left(  U\right)  \cdot\left(  \left(  g-f\right)  \left(
V\right)  \right)  .
\end{align*}
This proves (\ref{pf.lem.leray.graded-a2.1}).}. In other words,%
\begin{align*}
&  \left\{  \left(  g-f\right)  \left(  uv\right)  \ \mid\ \left(  u,v\right)
\in U\times V\right\} \\
&  \subseteq\left(  \left(  g-f\right)  \left(  U\right)  \right)  \cdot
g\left(  V\right)  +f\left(  U\right)  \cdot\left(  \left(  g-f\right)
\left(  V\right)  \right)  .
\end{align*}
Thus, (\ref{284.pf.b.2}) (applied to $B$, $\left\{  \left(  g-f\right)
\left(  uv\right)  \ \mid\ \left(  u,v\right)  \in U\times V\right\}  $ and
$\left(  \left(  g-f\right)  \left(  U\right)  \right)  \cdot g\left(
V\right)  +f\left(  U\right)  \cdot\left(  \left(  g-f\right)  \left(
V\right)  \right)  $ instead of $M$, $S$ and $Q$) shows that%
\begin{align}
&  \left\langle \left\{  \left(  g-f\right)  \left(  uv\right)  \ \mid
\ \left(  u,v\right)  \in U\times V\right\}  \right\rangle \nonumber\\
&  \subseteq\left(  \left(  g-f\right)  \left(  U\right)  \right)  \cdot
g\left(  V\right)  +f\left(  U\right)  \cdot\left(  \left(  g-f\right)
\left(  V\right)  \right)  . \label{pf.lem.leray.graded-a2.2}%
\end{align}


The definition of $UV$ yields
\begin{equation}
UV=\left\langle uv\ \mid\ \left(  u,v\right)  \in U\times V\right\rangle
=\left\langle \left\{  uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}
\right\rangle . \label{pf.lem.leray.graded-a2.3}%
\end{equation}
Thus, (\ref{316.pf.3}) (applied to $A$, $B$, $g-f$ and $\left\{
uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}  $ instead of $M$, $R$,
$\phi$ and $S$) yields%
\begin{align}
\left(  g-f\right)  \left(  \left\langle \left\{  uv\ \mid\ \left(
u,v\right)  \in U\times V\right\}  \right\rangle \right)   &  =\left\langle
\left\{  \left(  g-f\right)  \left(  uv\right)  \ \mid\ \left(  u,v\right)
\in U\times V\right\}  \right\rangle \nonumber\\
&  \subseteq\left(  \left(  g-f\right)  \left(  U\right)  \right)  \cdot
g\left(  V\right)  +f\left(  U\right)  \cdot\left(  \left(  g-f\right)
\left(  V\right)  \right)  \label{pf.lem.leray.graded-a2.5}%
\end{align}
(by (\ref{pf.lem.leray.graded-a2.2})).

Now, applying the map $g-f$ to both sides of the equality
(\ref{pf.lem.leray.graded-a2.3}), we obtain%
\begin{align*}
\left(  g-f\right)  \left(  UV\right)   &  =\left(  g-f\right)  \left(
\left\langle \left\{  uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}
\right\rangle \right) \\
&  \subseteq\left(  \left(  g-f\right)  \left(  U\right)  \right)  \cdot
g\left(  V\right)  +f\left(  U\right)  \cdot\left(  \left(  g-f\right)
\left(  V\right)  \right)
\end{align*}
(by (\ref{pf.lem.leray.graded-a2.5})). This proves
Lemma~\ref{lem.leray.graded-a2}.
\end{proof}

As an application of the above lemmata, we can show a criterion for
endomorphisms of a graded algebra to be injective:

\begin{lemma}
\label{lem.leray2.graded-a1}Let $k$ be a field. Let $A$ be a graded
$k$-algebra. For each $m\in\mathbb{N}$, we define a $k$-vector subspace
$A_{>m}$ of $A$ by $A_{>m}=\sum_{\substack{g\in\mathbb{N};\\g>m}}A_{g}$.

Let $f:A\rightarrow A$ be a $k$-algebra homomorphism.\footnote{We do not
require $f$ to be graded.} Assume that each $x\in A_{1}$ satisfies%
\begin{equation}
f\left(  x\right)  -x\in A_{>1}. \label{eq.lem.leray2.graded-a1.ass}%
\end{equation}


\textbf{(a)} We have $\left(  \operatorname*{id}\nolimits_{A}-f\right)
\left(  \left(  A_{1}\right)  ^{n}\right)  \subseteq A_{>n}$ for each
$n\in\mathbb{N}$.

\textbf{(b)} Assume that $A=\operatorname*{AlgGen}\nolimits_{k}\left(
A_{1}\right)  $. Then, the map $f$ is injective.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.graded-a1}.]The map $\operatorname*{id}%
\nolimits_{A}-f$ is $k$-linear (since both maps $\operatorname*{id}%
\nolimits_{A}$ and $f$ are $k$-linear). Notice that%
\begin{equation}
\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(  A_{1}\right)
\subseteq A_{>1} \label{pf.lem.leray2.graded-a1.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-a1.1}):} Each $x\in
A_{1}$ satisfies%
\begin{align*}
\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(  x\right)   &
=\underbrace{\operatorname*{id}\nolimits_{A}\left(  x\right)  }_{=x}-f\left(
x\right)  =x-f\left(  x\right)  =-\underbrace{\left(  f\left(  x\right)
-x\right)  }_{\substack{\in A_{>1}\\\text{(by
(\ref{eq.lem.leray2.graded-a1.ass}))}}}\in-A_{>1}\subseteq A_{>1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }A_{>1}\text{ is a }k\text{-vector
subspace of }A\right)  .
\end{align*}
In other words, $\left\{  \left(  \operatorname*{id}\nolimits_{A}-f\right)
\left(  x\right)  \ \mid\ x\in A_{1}\right\}  \subseteq A_{>1}$. But $\left(
\operatorname*{id}\nolimits_{A}-f\right)  \left(  A_{1}\right)  =\left\{
\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(  x\right)
\ \mid\ x\in A_{1}\right\}  \subseteq A_{>1}$. This proves
(\ref{pf.lem.leray2.graded-a1.1}).}.

Since $A$ is a graded $k$-algebra, we have
\begin{equation}
A_{\ell}A_{m}\subseteq A_{\ell+m}\ \ \ \ \ \ \ \ \ \ \text{for any }\ell
\in\mathbb{N}\text{ and }m\in\mathbb{N}. \label{pf.lem.leray2.graded-a1.AlAm}%
\end{equation}
Thus, each positive integer $N$ satisfies%
\begin{equation}
A_{>1}\cdot A_{N-1}\subseteq A_{>N} \label{pf.lem.leray2.graded-a1.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-a1.2}):} Let $N$ be a
positive integer. Thus, $N-1\in\mathbb{N}$. The definition of $A_{>N}$ yields
$A_{>N}=\sum_{\substack{g\in\mathbb{N};\\g>N}}A_{g}$.
\par
But the definition of $A_{>1}$ yields $A_{>1}=\sum_{\substack{g\in
\mathbb{N};\\g>1}}A_{g}$. Hence,%
\begin{align*}
\underbrace{A_{>1}}_{=\sum_{\substack{g\in\mathbb{N};\\g>1}}A_{g}}\cdot
A_{N-1}  &  =\left(  \sum_{\substack{g\in\mathbb{N};\\g>1}}A_{g}\right)  \cdot
A_{N-1}=\sum_{\substack{g\in\mathbb{N};\\g>1}}\underbrace{A_{g}A_{N-1}%
}_{\substack{\subseteq A_{g+\left(  N-1\right)  }\\\text{(by
(\ref{pf.lem.leray2.graded-a1.AlAm})}\\\text{(applied to }\ell=g\text{ and
}m=N-1\text{))}}}\\
&  \subseteq\sum_{\substack{g\in\mathbb{N};\\g>1}}A_{g+\left(  N-1\right)  }\\
&  =\sum_{\substack{g\in\mathbb{N};\\g>1+\left(  N-1\right)  }}A_{g}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }g+\left(
N-1\right)  \text{ for }g\text{ in the sum}\right) \\
&  =\sum_{\substack{g\in\mathbb{N};\\g>N}}A_{g}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }1+\left(  N-1\right)  =N\right) \\
&  =A_{>N}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A_{>N}=\sum_{\substack{g\in
\mathbb{N};\\g>N}}A_{g}\right)  .
\end{align*}
This proves (\ref{pf.lem.leray2.graded-a1.2}).}.

Also,%
\begin{equation}
f\left(  A_{1}\right)  \subseteq A_{>0} \label{pf.lem.leray2.graded-a1.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-a1.3}):} Let $y\in
f\left(  A_{1}\right)  $. Thus, there exists some $x\in A_{1}$ such that
$y=f\left(  x\right)  $. Consider this $y$.
\par
From (\ref{eq.lem.leray2.graded-a1.ass}), we obtain $f\left(  x\right)  -x\in
A_{>1}$, so that $f\left(  x\right)  \in x+A_{>1}$.
\par
But the definition of $A_{>1}$ yields $A_{>1}=\sum_{\substack{g\in
\mathbb{N};\\g>1}}A_{g}$. Meanwhile, the definition of $A_{>0}$ yields%
\[
A_{>0}=\sum_{\substack{g\in\mathbb{N};\\g>0}}A_{g}=A_{1}+A_{2}+A_{3}%
+\cdots=A_{1}+\underbrace{\left(  A_{2}+A_{3}+A_{4}+\cdots\right)  }%
_{=\sum_{\substack{g\in\mathbb{N};\\g>1}}A_{g}=A_{>1}}=A_{1}+A_{>1}.
\]
\par
But $y=f\left(  x\right)  \in\underbrace{x}_{\in A_{1}}+A_{>1}\in A_{1}%
+A_{>1}=A_{>0}$.
\par
Now, forget that we fixed $y$. We thus have proven that $y\in A_{>0}$ for each
$y\in f\left(  A_{1}\right)  $. In other words, $f\left(  A_{1}\right)
\subseteq A_{>0}$. This proves (\ref{pf.lem.leray2.graded-a1.3}).}.
Furthermore, every positive integer $h$ satisfies%
\begin{equation}
A_{h}\cdot A_{>N-1}\subseteq A_{>N} \label{pf.lem.leray2.graded-a1.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-a1.4}):} Let $h$ be a
positive integer. The definition of $A_{>N-1}$ yields $A_{>N-1}=\sum
_{\substack{g\in\mathbb{N};\\g>N-1}}A_{g}$. But the definition of $A_{>N}$
yields $A_{>N}=\sum_{\substack{g\in\mathbb{N};\\g>N}}A_{g}$.
\par
But $h\geq1$ (since $h$ is a positive integer) and thus $\left(  N-1\right)
+\underbrace{h}_{\geq1}\geq\left(  N-1\right)  +1=N$. Hence, every
$g\in\mathbb{N}$ satisfying $g>\left(  N-1\right)  +h$ also satisfies $g>N$
(since $g>\left(  N-1\right)  +h\geq N$). Thus, the sum $\sum_{\substack{g\in
\mathbb{N};\\g>\left(  N-1\right)  +h}}A_{g}$ is a subsum of the sum
$\sum_{\substack{g\in\mathbb{N};\\g>N}}A_{g}$. Hence, $\sum_{\substack{g\in
\mathbb{N};\\g>\left(  N-1\right)  +h}}A_{g}\subseteq\sum_{\substack{g\in
\mathbb{N};\\g>N}}A_{g}=A_{>N}$.
\par
Now,%
\begin{align*}
A_{h}\cdot\underbrace{A_{>N-1}}_{=\sum_{\substack{g\in\mathbb{N}%
;\\g>N-1}}A_{g}}  &  =A_{h}\cdot\left(  \sum_{\substack{g\in\mathbb{N}%
;\\g>N-1}}A_{g}\right)  =\sum_{\substack{g\in\mathbb{N};\\g>N-1}%
}\underbrace{A_{h}A_{g}}_{\substack{\subseteq A_{h+g}\\\text{(by
(\ref{pf.lem.leray2.graded-a1.AlAm})}\\\text{(applied to }\ell=h\text{ and
}m=g\text{))}}}\\
&  \subseteq\sum_{\substack{g\in\mathbb{N};\\g>N-1}}\underbrace{A_{h+g}%
}_{=A_{g+h}}=\sum_{\substack{g\in\mathbb{N};\\g>N-1}}A_{g+h}\\
&  =\sum_{\substack{g\in\mathbb{N};\\g>\left(  N-1\right)  +h}}A_{g}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }g+h\text{ for
}g\text{ in the sum}\right) \\
&  \subseteq A_{>N}.
\end{align*}
This proves (\ref{pf.lem.leray2.graded-a1.4}).}. Thus,%
\begin{equation}
A_{>0}\cdot A_{>N-1}\subseteq A_{>N} \label{pf.lem.leray2.graded-a1.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-a1.5}):} The definition
of $A_{>0}$ yields%
\[
A_{>0}=\sum_{\substack{g\in\mathbb{N};\\g>0}}A_{g}=\sum_{\substack{h\in
\mathbb{N};\\h>0}}A_{h}%
\]
(here, we have renamed the summation index $g$ as $h$). Thus,%
\[
\underbrace{A_{>0}}_{=\sum_{\substack{h\in\mathbb{N};\\h>0}}A_{h}}\cdot
A_{>N-1}=\left(  \sum_{\substack{h\in\mathbb{N};\\h>0}}A_{h}\right)  \cdot
A_{>N-1}=\sum_{\substack{h\in\mathbb{N};\\h>0}}\underbrace{A_{h}\cdot
A_{>N-1}}_{\substack{\subseteq A_{>N}\\\text{(by
(\ref{pf.lem.leray2.graded-a1.4}))}}}\subseteq\sum_{\substack{h\in
\mathbb{N};\\h>0}}A_{>N}\subseteq A_{>N}%
\]
(since $A_{>N}$ is a $k$-vector subspace of $A$). This proves
(\ref{pf.lem.leray2.graded-a1.5}).}.

\textbf{(a)} We shall prove Lemma~\ref{lem.leray2.graded-a1} \textbf{(a)} by
induction over $n$:

\textit{Induction base:} We have $\left(  \operatorname*{id}\nolimits_{A}%
-f\right)  \left(  \left(  A_{1}\right)  ^{0}\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} Recall that $f$ is a $k$-algebra
homomorphism. Hence, $f\left(  1_{A}\right)  =1_{A}$. Now,%
\[
\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(  1_{A}\right)
=\underbrace{\operatorname*{id}\nolimits_{A}\left(  1_{A}\right)  }_{=1_{A}%
}-\underbrace{f\left(  1_{A}\right)  }_{=1_{A}}=1_{A}-1_{A}=0.
\]
\par
Now,%
\begin{align*}
\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(  \underbrace{\left(
A_{1}\right)  ^{0}}_{=k\cdot1_{A}}\right)   &  =\left(  \operatorname*{id}%
\nolimits_{A}-f\right)  \left(  k\cdot1_{A}\right)  =k\cdot\underbrace{\left(
\operatorname*{id}\nolimits_{A}-f\right)  \left(  1_{A}\right)  }%
_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\operatorname*{id}%
\nolimits_{A}-f\text{ is }k\text{-linear}\right) \\
&  =k\cdot0=0,
\end{align*}
qed.}. Hence, $\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(
\left(  A_{1}\right)  ^{0}\right)  =0\subseteq A_{>0}$ (since $A_{>0}$ is a
$k$-vector space). In other words, Lemma~\ref{lem.leray2.graded-a1}
\textbf{(a)} holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that
Lemma~\ref{lem.leray2.graded-a1} \textbf{(a)} holds for $n=N-1$. We must show
that Lemma~\ref{lem.leray2.graded-a1} \textbf{(a)} holds for $n=N$.

We have $N-1\in\mathbb{N}$ (since $N$ is a positive integer). Thus,
Lemma~\ref{lem.graded-alg.genby1} \textbf{(a)} (applied to $n=N-1$) yields
$\left(  A_{1}\right)  ^{N-1}\subseteq A_{N-1}$.

Also, recall that Lemma~\ref{lem.leray2.graded-a1} \textbf{(a)} holds for
$n=N-1$. In other words, we have $\left(  \operatorname*{id}\nolimits_{A}%
-f\right)  \left(  \left(  A_{1}\right)  ^{N-1}\right)  \subseteq A_{>N-1}$.

Now, both $\operatorname*{id}\nolimits_{A}:A\rightarrow A$ and $f:A\rightarrow
A$ are $k$-algebra homomorphisms. Thus, Lemma~\ref{lem.leray.graded-a2}
(applied to $B=A$, $U=A_{1}$ and $V=\left(  A_{1}\right)  ^{N-1}$) shows that%
\begin{align*}
\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(  A_{1}\left(
A_{1}\right)  ^{N-1}\right)   &  \subseteq\underbrace{\left(  \left(
\operatorname*{id}\nolimits_{A}-f\right)  \left(  A_{1}\right)  \right)
}_{\substack{\subseteq A_{>1}\\\text{(by (\ref{pf.lem.leray2.graded-a1.1}))}%
}}\cdot\underbrace{\operatorname*{id}\nolimits_{A}\left(  \left(
A_{1}\right)  ^{N-1}\right)  }_{=\left(  A_{1}\right)  ^{N-1}\subseteq
A_{N-1}}+\underbrace{f\left(  A_{1}\right)  }_{\substack{\subseteq
A_{>0}\\\text{(by (\ref{pf.lem.leray2.graded-a1.3}))}}}\cdot
\underbrace{\left(  \left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(
\left(  A_{1}\right)  ^{N-1}\right)  \right)  }_{\subseteq A_{>N-1}}\\
&  \subseteq\underbrace{A_{>1}\cdot A_{N-1}}_{\substack{\subseteq
A_{>N}\\\text{(by (\ref{pf.lem.leray2.graded-a1.2}))}}}+\underbrace{A_{>0}%
\cdot A_{>N-1}}_{\substack{\subseteq A_{>N}\\\text{(by
(\ref{pf.lem.leray2.graded-a1.5}))}}}\\
&  \subseteq A_{>N}+A_{>N}\subseteq A_{>N}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }A_{>N}\text{ is a }k\text{-vector space}\right)  .
\end{align*}
Since $A_{1}\left(  A_{1}\right)  ^{N-1}=\left(  A_{1}\right)  ^{N}$, this
rewrites as $\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(  \left(
A_{1}\right)  ^{N}\right)  \subseteq A_{>N}$. In other words,
Lemma~\ref{lem.leray2.graded-a1} \textbf{(a)} holds for $n=N$. This completes
the induction step. Thus, Lemma~\ref{lem.leray2.graded-a1} \textbf{(a)} is
proven by induction.

\textbf{(b)} For each $n\in\mathbb{N}$, we have%
\[
\left(  \operatorname*{id}\nolimits_{A}-f\right)  \left(  \underbrace{A_{n}%
}_{\substack{=\left(  A_{1}\right)  ^{n}\\\text{(by
Lemma~\ref{lem.graded-alg.genby1} \textbf{(b)})}}}\right)  =\left(
\operatorname*{id}\nolimits_{A}-f\right)  \left(  \left(  A_{1}\right)
^{n}\right)  \subseteq A_{>n}%
\]
(by Lemma~\ref{lem.leray2.graded-a1} \textbf{(a)}). Thus,
Lemma~\ref{lem.leray2.graded-v1} (applied to $A$ and $\operatorname*{id}%
\nolimits_{A}-f$ instead of $V$ and $f$) shows that the map
$\operatorname*{id}\nolimits_{A}-\left(  \operatorname*{id}\nolimits_{A}%
-f\right)  $ is injective. Since $\operatorname*{id}\nolimits_{A}-\left(
\operatorname*{id}\nolimits_{A}-f\right)  =f$, this rewrites as follows: The
map $f$ is injective. This proves Lemma~\ref{lem.leray2.graded-a1}
\textbf{(b)}.
\end{proof}

We shall now prove a simple lemma about homogeneous subspaces:

\begin{lemma}
\label{lem.leray2.graded-v2-homg}Let $k$ be a field. Let $V$ be a graded
$k$-vector space. Let $W$ be a homogeneous subspace of $V$. Then,
$p_{n,V}\left(  W\right)  =W\cap V_{n}$ for any $n\in\mathbb{N}$. (Here, the
map $p_{n,V}:V\rightarrow V$ is defined as in Definition~\ref{Definition221}.)
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.graded-v2-homg}.]Recall that $W$ is a
homogeneous subspace of $V$ if and only if $W=\bigoplus_{n\in\mathbb{N}%
}\left(  W\cap V_{n}\right)  $ (by the definition of a \textquotedblleft
homogeneous subspace\textquotedblright). Hence, $W=\bigoplus_{n\in\mathbb{N}%
}\left(  W\cap V_{n}\right)  $ (since $W$ is a homogeneous subspace of $V$).
Thus,%
\begin{align}
W  &  =\bigoplus_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  =\sum
_{n\in\mathbb{N}}\left(  W\cap V_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since direct sums are sums}\right) \nonumber\\
&  =\sum_{m\in\mathbb{N}}\left(  W\cap V_{m}\right)
\label{pf.lem.leray2.graded-v2-homg.4}%
\end{align}
(here, we renamed the summation index $n$ as $m$).

Let $n\in\mathbb{N}$. For any $m\in\mathbb{N}$ satisfying $m\neq n$, we have%
\begin{equation}
p_{n,V}\left(  W\cap V_{m}\right)  =0 \label{pf.lem.leray2.graded-v2-homg.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-v2-homg.1}):} Let
$m\in\mathbb{N}$ be such that $m\neq n$. Thus,
Lemma~\ref{lem.graded-vec.WinVn} \textbf{(b)} (applied to $W\cap V_{m}$
instead of $U$) yields $p_{n,V}\left(  W\cap V_{m}\right)  =0$ (since $W\cap
V_{m}\subseteq V_{m}$). This proves (\ref{pf.lem.leray2.graded-v2-homg.1}).}.
But%
\begin{equation}
p_{n,V}\left(  W\cap V_{n}\right)  =W\cap V_{n}
\label{pf.lem.leray2.graded-v2-homg.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-v2-homg.2}):}
Lemma~\ref{lem.graded-vec.WinVn} \textbf{(a)} (applied to $U=W\cap V_{n}$)
yields $p_{n,V}\left(  W\cap V_{n}\right)  =W\cap V_{n}$ (since $W\cap
V_{n}\subseteq V_{n}$). This proves (\ref{pf.lem.leray2.graded-v2-homg.2}).}.

Applying the map $p_{n,V}$ to both sides of the equality
(\ref{pf.lem.leray2.graded-v2-homg.4}), we obtain%
\begin{align*}
p_{n,V}\left(  W\right)   &  =p_{n,V}\left(  \sum_{m\in\mathbb{N}}\left(
W\cap V_{m}\right)  \right)  =\sum_{m\in\mathbb{N}}p_{n,V}\left(  W\cap
V_{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }p_{n,V}\text{
is }k\text{-linear}\right) \\
&  =\underbrace{p_{n,V}\left(  W\cap V_{n}\right)  }_{\substack{=W\cap
V_{n}\\\text{(by (\ref{pf.lem.leray2.graded-v2-homg.2}))}}}+\sum
_{\substack{m\in\mathbb{N};\\m\neq n}}\underbrace{p_{n,V}\left(  W\cap
V_{m}\right)  }_{\substack{=0\\\text{(by (\ref{pf.lem.leray2.graded-v2-homg.1}%
))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}m=n\text{ from the sum}\right) \\
&  =\left(  W\cap V_{n}\right)  +\underbrace{\sum_{\substack{m\in
\mathbb{N};\\m\neq n}}0}_{=0}=W\cap V_{n}.
\end{align*}
This proves Lemma~\ref{lem.leray2.graded-v2-homg}.
\end{proof}

Next, we shall show a criterion for generating sets of graded algebras:

\begin{lemma}
\label{lem.leray2.graded-a2-gen}Let $k$ be a field. Let $A$ be a graded
$k$-algebra such that $A_{0}=k\cdot1_{A}$. For each $m\in\mathbb{N}$, we
define a $k$-vector subspace $A_{>m}$ of $A$ by $A_{>m}=\sum_{\substack{g\in
\mathbb{N};\\g>m}}A_{g}$.

Let $V$ be a homogeneous subspace of $A$ such that $A_{>0}=V+\left(
A_{>0}\right)  ^{2}$. Then:

\textbf{(a)} Every positive integer $n$ satisfies $A_{n}\subseteq\left(  V\cap
A_{n}\right)  +\sum_{g=1}^{n-1}A_{g}A_{n-g}$.

\textbf{(b)} We have $A=\operatorname*{AlgGen}\nolimits_{k}V$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.graded-a2-gen}.]\textbf{(a)} We shall use the
notations introduced in Remark~\ref{Remark220} and in
Definition~\ref{Definition221}.

Since $A$ is a graded $k$-algebra, we have
\begin{equation}
A_{\ell}A_{m}\subseteq A_{\ell+m}\ \ \ \ \ \ \ \ \ \ \text{for any }\ell
\in\mathbb{N}\text{ and }m\in\mathbb{N}.
\label{pf.lem.leray2.graded-a2-gen.AlAm}%
\end{equation}


Let $n\mathbb{\ }$be a positive integer. Define a $k$-vector subspace $Q$ of
$A$ by $Q=\sum_{g=1}^{n-1}A_{g}A_{n-g}$. Then, every two positive integers
$\ell$ and $m$ satisfy%
\begin{equation}
p_{n,A}\left(  A_{\ell}A_{m}\right)  \subseteq Q
\label{pf.lem.leray2.graded-a2-gen.a.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-a2-gen.a.1}):} Let
$\ell$ and $m$ be two positive integers. We are in one of the following two
cases:
\par
\textit{Case 1:} We have $\ell+m=n$.
\par
\textit{Case 2:} We have $\ell+m\neq n$.
\par
Let us first consider Case 1. In this case, we have $\ell+m=n$. Thus,
$m=n-\ell$. Also, $\ell>0$ (since $\ell$ is a positive integer) and $m>0$
(since $m$ is a positive integer). Also, $n=\ell+\underbrace{m}_{>0}>\ell$, so
that $\ell<n$. Combining this with $\ell>0$, we obtain $\ell\in\left\{
1,2,\ldots,n-1\right\}  $ (since $\ell$ is a positive integer). Hence,
$A_{\ell}A_{n-\ell}$ is an addend of the sum $\sum_{g=1}^{n-1}A_{g}A_{n-g}$.
Thus, $A_{\ell}A_{n-\ell}\subseteq\sum_{g=1}^{n-1}A_{g}A_{n-g}=Q$.
\par
Now, (\ref{pf.lem.leray2.graded-a2-gen.AlAm}) yields $A_{\ell}A_{m}\subseteq
A_{\ell+m}=A_{n}$ (since $\ell+m=n$). Hence, Lemma~\ref{lem.graded-vec.WinVn}
\textbf{(a)} (applied to $A$ and $A_{\ell}A_{m}$ instead of $V$ and $U$) shows
that $p_{n,A}\left(  A_{\ell}A_{m}\right)  =A_{\ell}A_{m}=A_{\ell}A_{n-\ell}$
(since $m=n-\ell$). Hence, $p_{n,A}\left(  A_{\ell}A_{m}\right)  =A_{\ell
}A_{n-\ell}\subseteq Q$. Thus, (\ref{pf.lem.leray2.graded-a2-gen.a.1}) is
proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $\ell+m\neq n$. Now,
(\ref{pf.lem.leray2.graded-a2-gen.AlAm}) yields $A_{\ell}A_{m}\subseteq
A_{\ell+m}$. Hence, Lemma~\ref{lem.graded-vec.WinVn} \textbf{(b)} (applied to
$A$, $A_{\ell}A_{m}$ and $\ell+m$ instead of $V$, $U$ and $m$) shows that
$p_{n,A}\left(  A_{\ell}A_{m}\right)  =0$ (since $\ell+m\neq n$). Thus,
$p_{n,A}\left(  A_{\ell}A_{m}\right)  =0\subseteq Q$ (since $Q$ is a
$k$-vector space). Thus, (\ref{pf.lem.leray2.graded-a2-gen.a.1}) is proven in
Case 2.
\par
We have now proven (\ref{pf.lem.leray2.graded-a2-gen.a.1}) in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, we thus conclude
that (\ref{pf.lem.leray2.graded-a2-gen.a.1}) always holds.}. Hence,%
\begin{equation}
p_{n,A}\left(  \left(  A_{>0}\right)  ^{2}\right)  \subseteq Q
\label{pf.lem.leray2.graded-a2-gen.a.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-a2-gen.a.2}):} The
definition of $A_{>0}$ yields $A_{>0}=\sum_{\substack{g\in\mathbb{N}%
;\\g>0}}A_{g}=\sum_{\substack{m\in\mathbb{N};\\m>0}}A_{m}$ (here, we have
renamed the summation index $g$ as $m$). Now,%
\begin{align*}
\left(  A_{>0}\right)  ^{2}  &  =\underbrace{A_{>0}}_{=\sum_{\substack{g\in
\mathbb{N};\\g>0}}A_{g}}\underbrace{A_{>0}}_{=\sum_{\substack{m\in
\mathbb{N};\\m>0}}A_{m}}=\left(  \sum_{\substack{g\in\mathbb{N};\\g>0}%
}A_{g}\right)  \left(  \sum_{\substack{m\in\mathbb{N};\\m>0}}A_{m}\right) \\
&  =\sum_{\substack{g\in\mathbb{N};\\g>0}}\sum_{\substack{m\in\mathbb{N}%
;\\m>0}}A_{g}A_{m}.
\end{align*}
Applying the map $p_{n,A}$ to both sides of this relation, we obtain%
\begin{align*}
p_{n,A}\left(  \left(  A_{>0}\right)  ^{2}\right)   &  =p_{n,A}\left(
\sum_{\substack{g\in\mathbb{N};\\g>0}}\sum_{\substack{m\in\mathbb{N}%
;\\m>0}}A_{g}A_{m}\right)  =\sum_{\substack{g\in\mathbb{N};\\g>0}%
}\underbrace{p_{n,A}\left(  \sum_{\substack{m\in\mathbb{N};\\m>0}}A_{g}%
A_{m}\right)  }_{\substack{=\sum_{\substack{m\in\mathbb{N};\\m>0}%
}p_{n,A}\left(  A_{g}A_{m}\right)  \\\text{(since the map }p_{n,A}\text{ is
}k\text{-linear)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
p_{n,A}\text{ is }k\text{-linear}\right) \\
&  =\sum_{\substack{g\in\mathbb{N};\\g>0}}\sum_{\substack{m\in\mathbb{N}%
;\\m>0}}\underbrace{p_{n,A}\left(  A_{g}A_{m}\right)  }_{\substack{\subseteq
Q\\\text{(by (\ref{pf.lem.leray2.graded-a2-gen.a.1}) (applied to }%
\ell=g\text{))}}}\subseteq\sum_{\substack{g\in\mathbb{N};\\g>0}%
}\underbrace{\sum_{\substack{m\in\mathbb{N};\\m>0}}Q}_{\substack{\subseteq
Q\\\text{(since }Q\text{ is a }k\text{-vector space)}}}\\
&  \subseteq\sum_{\substack{g\in\mathbb{N};\\g>0}}Q\subseteq
Q\ \ \ \ \ \ \ \ \ \ \left(  \text{since }Q\text{ is a }k\text{-vector
space}\right)  .
\end{align*}
}.

Lemma~\ref{lem.leray2.graded-v2-homg} (applied to $A$ and $V$ instead of $V$
and $W$) yields $p_{n,A}\left(  V\right)  =V\cap A_{n}$.

But $A_{n}\subseteq p_{n,A}\left(  A_{>0}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} The definition of $A_{>0}$ yields
$A_{>0}=\sum_{\substack{g\in\mathbb{N};\\g>0}}A_{g}$. But $n$ is a positive
integer; thus, $n\in\mathbb{N}$ and $n>0$. Hence, $A_{n}$ is an addend of the
sum $\sum_{\substack{g\in\mathbb{N};\\g>0}}A_{g}$ (namely, the addend for
$g=n$). Thus, $A_{n}\subseteq\sum_{\substack{g\in\mathbb{N};\\g>0}%
}A_{g}=A_{>0}$. Hence, $p_{n,A}\left(  \underbrace{A_{n}}_{\subseteq A_{>0}%
}\right)  \subseteq p_{n,A}\left(  A_{>0}\right)  $.
\par
But Lemma~\ref{lem.graded-vec.WinVn} \textbf{(a)} (applied to $A$ and $A_{n}$
instead of $V$ and $U$) shows that $p_{n,A}\left(  A_{n}\right)  =A_{n}$
(since $A_{n}\subseteq A_{n}$). Thus, $A_{n}=p_{n,A}\left(  A_{n}\right)
\subseteq p_{n,A}\left(  A_{>0}\right)  $.}. Hence,%
\begin{align*}
A_{n}  &  \subseteq p_{n,A}\left(  \underbrace{A_{>0}}_{=V+\left(
A_{>0}\right)  ^{2}}\right)  =p_{n,A}\left(  V+\left(  A_{>0}\right)
^{2}\right)  =\underbrace{p_{n,A}\left(  V\right)  }_{=V\cap A_{n}%
}+\underbrace{p_{n,A}\left(  \left(  A_{>0}\right)  ^{2}\right)
}_{\substack{\subseteq Q\\\text{(by (\ref{pf.lem.leray2.graded-a2-gen.a.2}))}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }p_{n,A}\text{ is
}k\text{-linear}\right) \\
&  \subseteq\left(  V\cap A_{n}\right)  +\underbrace{Q}_{=\sum_{g=1}%
^{n-1}A_{g}A_{n-g}}=\left(  V\cap A_{n}\right)  +\sum_{g=1}^{n-1}A_{g}A_{n-g}.
\end{align*}
This proves Lemma~\ref{lem.leray2.graded-a2-gen} \textbf{(a)}.

\textbf{(b)} Let $R=\operatorname*{AlgGen}\nolimits_{k}V$. Then, $R$ is a
$k$-subalgebra of $A$ (since $\operatorname*{AlgGen}\nolimits_{k}V$ is a
$k$-subalgebra of $A$), and thus a $k$-vector subspace of $A$.

Lemma~\ref{lem.AlgGen.Sinit} (applied to $S=V$) yields $V\subseteq
\operatorname*{AlgGen}\nolimits_{k}V=R$.

Now, we are going to prove that%
\begin{equation}
A_{n}\subseteq R\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}.
\label{pf.lem.leray2.graded-a2-gen.b.goal}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.leray2.graded-a2-gen.b.goal}):} We shall prove
(\ref{pf.lem.leray2.graded-a2-gen.b.goal}) by strong induction over $n$:

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
(\ref{pf.lem.leray2.graded-a2-gen.b.goal}) holds for every $n<N$. We must now
show that (\ref{pf.lem.leray2.graded-a2-gen.b.goal}) holds for $n=N$.

If $N=0$, then this is obvious\footnote{\textit{Proof.} Assume that $N=0$.
Thus, $A_{N}=A_{0}=k\cdot1_{A}$. But $R$ is a $k$-subalgebra of $A$; thus,
$1_{A}\in R$. Also, $R$ is a $k$-vector subspace of $A$ (since $R$ is a
$k$-subalgebra of $A$); thus, $k\cdot R\subseteq R$. Now, $A_{N}%
=k\cdot\underbrace{1_{A}}_{\in R}\subseteq k\cdot R\subseteq R$. In other
words, (\ref{pf.lem.leray2.graded-a2-gen.b.goal}) holds for $n=N$. Qed.}.
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 $N=0$). Combining this with
$N\in\mathbb{N}$, we obtain $N\in\mathbb{N}\setminus\left\{  0\right\}
=\left\{  1,2,3,\ldots\right\}  $. In other words, $N$ is a positive integer.
Hence, Lemma~\ref{lem.leray2.graded-a2-gen} \textbf{(a)} (applied to $n=N$)
yields
\begin{equation}
A_{N}\subseteq\left(  V\cap A_{N}\right)  +\sum_{g=1}^{N-1}A_{g}A_{N-g}.
\label{pf.lem.leray2.graded-a2-gen.b.goal.pf.1}%
\end{equation}


But we have assumed that (\ref{pf.lem.leray2.graded-a2-gen.b.goal}) holds for
every $n<N$. In other words, for each $n\in\mathbb{N}$ satisfying $n<N$, we
have%
\begin{equation}
A_{n}\subseteq R. \label{pf.lem.leray2.graded-a2-gen.b.goal.pf.indhyp}%
\end{equation}


Now, each $g\in\left\{  1,2,\ldots,N-1\right\}  $ satisfies%
\begin{equation}
A_{g}A_{N-g}\subseteq R \label{pf.lem.leray2.graded-a2-gen.b.goal.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.graded-a2-gen.b.goal.pf.2}):}
Let $g\in\left\{  1,2,\ldots,N-1\right\}  $. Thus, $g\leq N-1<N$. Also,
$g\in\left\{  1,2,\ldots,N-1\right\}  \subseteq\mathbb{N}$. Hence,
(\ref{pf.lem.leray2.graded-a2-gen.b.goal.pf.indhyp}) (applied to $n=g$) yields
$A_{g}\subseteq R$. But we also have $g>0$ (since $g\in\left\{  1,2,\ldots
,N-1\right\}  $) and thus $N-\underbrace{g}_{>0}<N$. On the other hand, from
$g<N$, we obtain $N-g>0$, so that $N-g\in\mathbb{N}$. Thus,
(\ref{pf.lem.leray2.graded-a2-gen.b.goal.pf.indhyp}) (applied to $n=N-g$)
yields $A_{N-g}\subseteq R$. Now,%
\[
\underbrace{A_{g}}_{\subseteq R}\underbrace{A_{N-g}}_{\subseteq R}\subseteq
RR\subseteq R\ \ \ \ \ \ \ \ \ \ \left(  \text{since }R\text{ is a
}k\text{-subalgebra of }A\right)  .
\]
This proves (\ref{pf.lem.leray2.graded-a2-gen.b.goal.pf.2}).}. Finally,
(\ref{pf.lem.leray2.graded-a2-gen.b.goal.pf.1}) becomes%
\begin{align*}
A_{N}  &  \subseteq\underbrace{\left(  V\cap A_{N}\right)  }_{\subseteq
V\subseteq R}+\sum_{g=1}^{N-1}\underbrace{A_{g}A_{N-g}}_{\substack{\subseteq
R\\\text{(by (\ref{pf.lem.leray2.graded-a2-gen.b.goal.pf.2}))}}}\subseteq
R+\underbrace{\sum_{g=1}^{N-1}R}_{\substack{\subseteq R\\\text{(since }R\text{
is a }k\text{-vector subspace of }A\text{)}}}\\
&  \subseteq R+R\subseteq R\ \ \ \ \ \ \ \ \ \ \left(  \text{since }R\text{ is
a }k\text{-vector subspace of }A\right)  .
\end{align*}
In other words, (\ref{pf.lem.leray2.graded-a2-gen.b.goal}) holds for $n=N$.
Thus, the induction step is complete. This completes the induction proof of
(\ref{pf.lem.leray2.graded-a2-gen.b.goal}).]

Now recall that $A$ is a graded $k$-algebra. Hence, $A=\bigoplus
_{n\in\mathbb{N}}A_{n}=\sum_{n\in\mathbb{N}}A_{n}$ (since direct sums are
sums). Hence,%
\[
A=\sum_{n\in\mathbb{N}}\underbrace{A_{n}}_{\substack{\subseteq R\\\text{(by
(\ref{pf.lem.leray2.graded-a2-gen.b.goal}))}}}\subseteq\sum_{n\in\mathbb{N}%
}R\subseteq R
\]
(since $R$ is a $k$-vector subspace of $A$). Combined with $R\subseteq A$,
this yields $A=R$. Thus, $A=R=\operatorname*{AlgGen}\nolimits_{k}V$. This
proves Lemma~\ref{lem.leray2.graded-a2-gen} \textbf{(b)}.
\end{proof}

\begin{lemma}
\label{lem.leray2.Sym-gen}Let $k$ be a field. Let $V$ be a $k$-vector space.
The $k$-algebra $\operatorname*{Sym}V$ is generated by its subset
$\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $\ \ \ \ \footnote{We
are using the notations introduced in Definition~\ref{Definition1701}.}.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.Sym-gen}.]Let $R$ be the $k$-subalgebra of
$\operatorname*{Sym}V$ generated by $\operatorname*{syminc}\nolimits_{V}%
\left(  V\right)  $. Then, clearly, $R$ is a $k$-subalgebra of
$\operatorname*{Sym}V$ and contains $\operatorname*{syminc}\nolimits_{V}%
\left(  V\right)  $ as a subset. But Theorem~\ref{Theorem1812} \textbf{(a)}
yields%
\[
\operatorname*{Sym}V=\sum_{n\in\mathbb{N}}\left(
\underbrace{\operatorname*{syminc}\nolimits_{V}\left(  V\right)
}_{\substack{\subseteq R\\\text{(since }R\text{ contains }%
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \\\text{as a subset)}%
}}\right)  ^{n}\subseteq\sum_{n\in\mathbb{N}}\underbrace{R^{n}}%
_{\substack{\subseteq R\\\text{(since }R\text{ is a }k\text{-algebra)}%
}}\subseteq\sum_{n\in\mathbb{N}}R\subseteq R
\]
(since $R$ is a $k$-vector subspace of $\operatorname*{Sym}V$ (since $R$ is a
$k$-subalgebra of $\operatorname*{Sym}V$)). Combining this with the obvious
relation $R\subseteq\operatorname*{Sym}V$, we obtain $\operatorname*{Sym}V=R$.

Now, recall that $R$ is the $k$-subalgebra of $\operatorname*{Sym}V$ generated
by $\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $. In other words,
$\operatorname*{Sym}V$ is the $k$-subalgebra of $\operatorname*{Sym}V$
generated by $\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $ (since
$\operatorname*{Sym}V=R$). In other words, the $k$-algebra
$\operatorname*{Sym}V$ is generated by its subset $\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $. This proves Lemma~\ref{lem.leray2.Sym-gen}.
\end{proof}

After these basic facts, let us again return to coalgebras and exponentials:

\begin{lemma}
\label{lem.leray2.abstract}Let $k$ be a field of characteristic $0$. Let $H$
be a connected filtered $k$-coalgebra. Let $A$ be a graded $k$-algebra. For
each $m\in\mathbb{N}$, we define a $k$-vector subspace $A_{>m}$ of $A$ by
$A_{>m}=\sum_{\substack{g\in\mathbb{N};\\g>m}}A_{g}$.

Let $\mathfrak{a}\in\mathfrak{g}\left(  H,A\right)  $ be such that
$\mathfrak{a}\left(  H\right)  \subseteq A_{1}$. Notice that the $k$-linear
map $e^{\ast\mathfrak{a}}:H\rightarrow A$ is well-defined (since
$\mathfrak{a}\in\mathfrak{g}\left(  H,A\right)  $).

We have $e^{\ast\mathfrak{a}}\left(  x\right)  -\mathfrak{a}\left(  x\right)
\in A_{>1}$ for each $x\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.abstract}.]Remark~\ref{Remark7k} (applied to
$C=H$) yields that $H$ is a unital coalgebra with unity $1_{H}=\left(
\varepsilon_{H}\mid_{H_{\leq0}}\right)  ^{-1}\left(  1\right)  $. In other
words, $\left(  H,1_{H}\right)  $ is a unital coalgebra, where $1_{H}=\left(
\varepsilon_{H}\mid_{H_{\leq0}}\right)  ^{-1}\left(  1\right)  $. In other
words, $H$ is a $k$-coalgebra and $1_{H}$ is an element of $H$ satisfying
$\Delta_{H}\left(  1_{H}\right)  =1_{H}\otimes1_{H}$ and $\varepsilon
_{H}\left(  1_{H}\right)  =1$ (by the definition of a \textquotedblleft unital
coalgebra\textquotedblright).

Since $A$ is a graded $k$-algebra, we have
\begin{equation}
A_{\ell}A_{m}\subseteq A_{\ell+m}\ \ \ \ \ \ \ \ \ \ \text{for any }\ell
\in\mathbb{N}\text{ and }m\in\mathbb{N}. \label{pf.lem.leray2.abstract.a.AlAm}%
\end{equation}


Each $i\in\mathbb{N}$ satisfies%
\begin{equation}
\mathfrak{a}^{\ast i}\left(  H\right)  \subseteq A_{i}
\label{pf.lem.leray2.abstract.a.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.abstract.a.1}):} We shall prove
(\ref{pf.lem.leray2.abstract.a.1}) by induction over $i$:
\par
\textit{Induction base:} Recall the map $e_{H,A}$ defined in
Definition~\ref{def.eHA}. Its definition yields $e_{H,A}=\eta_{A}%
\circ\varepsilon_{H}$. Thus, each $h\in H$ satisfies
\begin{align*}
\underbrace{e_{H,A}}_{=\eta_{A}\circ\varepsilon_{H}}\left(  h\right)   &
=\left(  \eta_{A}\circ\varepsilon_{H}\right)  \left(  h\right)  =\eta
_{A}\left(  \varepsilon_{H}\left(  h\right)  \right)  =\varepsilon_{H}\left(
h\right)  \cdot\underbrace{1_{A}}_{\substack{\in A_{0}\\\text{(since
}A\\\text{is a graded }k\text{-algebra)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of the map }\eta_{A}\right) \\
&  \in\varepsilon_{H}\left(  h\right)  \cdot A_{0}\subseteq A_{0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A_{0}\text{ is a }k\text{-vector
subspace of }A\right)  .
\end{align*}
In other words, $e_{H,A}\left(  H\right)  \subseteq A_{0}$. Now,
$\underbrace{\mathfrak{a}^{\ast0}}_{=e_{H,A}}\left(  H\right)  =e_{H,A}\left(
H\right)  \subseteq A_{0}$. In other words, (\ref{pf.lem.leray2.abstract.a.1})
holds for $i=0$. This completes the induction base.
\par
\textit{Induction step:} Let $j\in\mathbb{N}$. Assume that
(\ref{pf.lem.leray2.abstract.a.1}) holds for $i=j$. We must prove that
(\ref{pf.lem.leray2.abstract.a.1}) holds for $i=j+1$.
\par
We have assumed that (\ref{pf.lem.leray2.abstract.a.1}) holds for $i=j$. In
other words, we have $\mathfrak{a}^{\ast j}\left(  H\right)  \subseteq A_{j}$.
But%
\[
\mathfrak{a}^{\ast\left(  j+1\right)  }=\mathfrak{a}\ast\mathfrak{a}^{\ast
j}=\mu_{A}\circ\left(  \mathfrak{a}\otimes\mathfrak{a}^{\otimes j}\right)
\circ\Delta_{H}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
convolution}\right)  .
\]
Thus,%
\begin{align*}
&  \underbrace{\mathfrak{a}^{\ast\left(  j+1\right)  }}_{=\mu_{A}\circ\left(
\mathfrak{a}\otimes\mathfrak{a}^{\otimes j}\right)  \circ\Delta_{H}}\left(
H\right) \\
&  =\left(  \mu_{A}\circ\left(  \mathfrak{a}\otimes\mathfrak{a}^{\otimes
j}\right)  \circ\Delta_{H}\right)  \left(  H\right)  =\mu_{A}\left(  \left(
\mathfrak{a}\otimes\mathfrak{a}^{\otimes j}\right)  \left(  \underbrace{\Delta
_{H}\left(  H\right)  }_{\subseteq H\otimes H}\right)  \right) \\
&  \subseteq\mu_{A}\left(  \underbrace{\left(  \mathfrak{a}\otimes
\mathfrak{a}^{\otimes j}\right)  \left(  H\otimes H\right)  }_{\subseteq
\mathfrak{a}\left(  H\right)  \otimes\mathfrak{a}^{\otimes j}\left(  H\right)
}\right)  \subseteq\mu_{A}\left(  \mathfrak{a}\left(  H\right)  \otimes
\mathfrak{a}^{\otimes j}\left(  H\right)  \right) \\
&  =\underbrace{\mathfrak{a}\left(  H\right)  }_{\subseteq A_{1}}%
\cdot\underbrace{\mathfrak{a}^{\otimes j}\left(  H\right)  }_{\subseteq A_{j}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma~\ref{lem.alg.muUV} (applied to
}U=\mathfrak{a}\left(  H\right)  \text{ and }V=\mathfrak{a}^{\otimes j}\left(
H\right)  \text{)}\right) \\
&  \subseteq A_{1}A_{j}\subseteq A_{1+j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.leray2.abstract.a.AlAm}) (applied to }\ell=1\text{ and
}m=j\text{)}\right) \\
&  =A_{j+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }1+j=j+1\right)  .
\end{align*}
In other words, (\ref{pf.lem.leray2.abstract.a.1}) holds for $i=j+1$. This
completes the induction step. Thus, the induction proof of
(\ref{pf.lem.leray2.abstract.a.1}) is complete.}. Thus, every integer $i>1$
satisfies%
\begin{equation}
\mathfrak{a}^{\ast i}\left(  H\right)  \subseteq A_{>1}
\label{pf.lem.leray2.abstract.a.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.abstract.a.2}):} Let $i>1$ be
an integer. Thus, $i\in\mathbb{N}$. Hence, (\ref{pf.lem.leray2.abstract.a.1})
yields $\mathfrak{a}^{\ast i}\left(  H\right)  \subseteq A_{i}$.
\par
But the definition of $A_{>1}$ yields $A_{>1}=\sum_{\substack{g\in
\mathbb{N};\\g>1}}A_{g}$.
\par
Now, $i\in\mathbb{N}$ satisfies $i>1$. Hence, $A_{i}$ is an addend of the sum
$\sum_{\substack{g\in\mathbb{N};\\g>1}}A_{g}$ (namely, the addend for $g=i$).
Hence, $A_{i}\subseteq\sum_{\substack{g\in\mathbb{N};\\g>1}}A_{g}=A_{>1}$.
Now, $\mathfrak{a}^{\ast i}\left(  H\right)  \subseteq A_{i}\subseteq A_{>1}$.
This proves (\ref{pf.lem.leray2.abstract.a.2}).}.

Now, let $x\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Thus,
\[
x\in\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \subseteq
H=\bigcup_{n\in\mathbb{N}}H_{\leq n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}H\text{ is filtered}\right)  .
\]
In other words, there exists some $n\in\mathbb{N}$ satisfying $x\in H_{\leq
n}$. Consider this $x$. Each integer $i>n+1$ satisfies%
\begin{equation}
\mathfrak{a}^{\ast i}\left(  x\right)  =0 \label{pf.lem.leray2.abstract.a.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.abstract.a.4}):} Let $i$ be an
integer such that $i>n+1$. Then, $i>n+1>n\geq0$, so that $i\in\mathbb{N}$.
Thus, Remark~\ref{Remark9} (applied to $f=\mathfrak{a}$) yields $\mathfrak{a}%
^{\ast i}\left(  H_{\leq n}\right)  =0$. Now, $x\in H_{\leq n}$, so that
$\mathfrak{a}^{\ast i}\left(  \underbrace{x}_{\in H_{\leq n}}\right)
\in\mathfrak{a}^{\ast i}\left(  H_{\leq n}\right)  =0$. In other words,
$\mathfrak{a}^{\ast i}\left(  x\right)  =0$. This proves
(\ref{pf.lem.leray2.abstract.a.4}).}. Also,%
\begin{equation}
\mathfrak{a}^{\ast0}\left(  x\right)  =0 \label{pf.lem.leray2.abstract.a.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.abstract.a.5}):} Recall the map
$e_{H,A}$ defined in Definition~\ref{def.eHA}. Its definition yields
$e_{H,A}=\eta_{A}\circ\varepsilon_{H}$. But
\[
\underbrace{\mathfrak{a}^{\ast0}}_{=e_{H,A}=\eta_{A}\circ\varepsilon_{H}%
}\left(  x\right)  =\left(  \eta_{A}\circ\varepsilon_{H}\right)  \left(
x\right)  =\eta_{A}\left(  \underbrace{\varepsilon_{H}\left(  x\right)
}_{\substack{=0\\\text{(since }x\in\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \text{)}}}\right)  =\eta_{A}\left(  0\right)  =0
\]
(since $\eta_{A}$ is a $k$-linear map).}. Note that $\underbrace{n}_{\geq
0}+1\geq1$.

The definition of $e^{\ast\mathfrak{a}}$ yields%
\begin{align*}
e^{\ast\mathfrak{a}}\left(  x\right)   &  =\sum_{i\geq0}\underbrace{\dfrac
{\mathfrak{a}^{\ast i}\left(  x\right)  }{i!}}_{=\dfrac{1}{i!}\mathfrak{a}%
^{\ast i}\left(  x\right)  }=\sum_{i\geq0}\dfrac{1}{i!}\mathfrak{a}^{\ast
i}\left(  x\right)  =\underbrace{\sum_{\substack{i\geq0;\\i\leq n+1}}}%
_{=\sum_{i=0}^{n+1}}\dfrac{1}{i!}\mathfrak{a}^{\ast i}\left(  x\right)
+\sum_{\substack{i\geq0;\\i>n+1}}\dfrac{1}{i!}\underbrace{\mathfrak{a}^{\ast
i}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{pf.lem.leray2.abstract.a.4}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }i\geq0\text{ satisfies either
}i\leq n+1\text{ or }i>n+1\text{ (but not both)}\right) \\
&  =\sum_{i=0}^{n+1}\dfrac{1}{i!}\mathfrak{a}^{\ast i}\left(  x\right)
+\underbrace{\sum_{\substack{i\geq0;\\i>n+1}}\dfrac{1}{i!}0}_{=0}=\sum
_{i=0}^{n+1}\dfrac{1}{i!}\mathfrak{a}^{\ast i}\left(  x\right)  =\dfrac{1}%
{0!}\underbrace{\mathfrak{a}^{\ast0}\left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.leray2.abstract.a.5}))}}}+\sum
_{i=1}^{n+1}\dfrac{1}{i!}\mathfrak{a}^{\ast i}\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=0\text{ from the sum}\right) \\
&  =\underbrace{\dfrac{1}{0!}0}_{=0}+\sum_{i=1}^{n+1}\dfrac{1}{i!}%
\mathfrak{a}^{\ast i}\left(  x\right)  =\sum_{i=1}^{n+1}\dfrac{1}%
{i!}\mathfrak{a}^{\ast i}\left(  x\right)  =\underbrace{\dfrac{1}{1!}%
}_{=\dfrac{1}{1}=1}\underbrace{\mathfrak{a}^{\ast1}}_{=\mathfrak{a}}\left(
x\right)  +\sum_{i=2}^{n+1}\dfrac{1}{i!}\mathfrak{a}^{\ast i}\left(  x\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=0\text{ from the sum}\\
\text{(this is allowed since }n+1\geq1\text{)}%
\end{array}
\right) \\
&  =\mathfrak{a}\left(  x\right)  +\sum_{i=2}^{n+1}\dfrac{1}{i!}%
\mathfrak{a}^{\ast i}\left(  x\right)  .
\end{align*}
Subtracting $\mathfrak{a}\left(  x\right)  $ from both sides of this equality,
we find%
\[
e^{\ast\mathfrak{a}}\left(  x\right)  -\mathfrak{a}\left(  x\right)
=\sum_{i=2}^{n+1}\dfrac{1}{i!}\mathfrak{a}^{\ast i}\left(  \underbrace{x}_{\in
H}\right)  \in\sum_{i=2}^{n+1}\dfrac{1}{i!}\underbrace{\mathfrak{a}^{\ast
i}\left(  H\right)  }_{\substack{\subseteq A_{>1}\\\text{(by
(\ref{pf.lem.leray2.abstract.a.2}))}}}\subseteq\sum_{i=2}^{n+1}\dfrac{1}%
{i!}A_{>1}\subseteq A_{>1}%
\]
(since $A_{>1}$ is a $k$-vector subspace of $A$). This proves
Lemma~\ref{lem.leray2.abstract}.
\end{proof}

\begin{lemma}
\label{lem.leray2.abstract2}Let $k$ be a field of characteristic $0$. Let $H$
be a connected filtered $k$-coalgebra and, at the same time, a $k$-bialgebra
with the same underlying $k$-coalgebra structure. Let $A$ be a commutative
graded $k$-algebra satisfying $A=\operatorname*{AlgGen}\nolimits_{k}\left(
A_{1}\right)  $.

Let $\mathfrak{b}:A\rightarrow H$ be a $k$-algebra homomorphism satisfying
$\mathfrak{b}\left(  A_{1}\right)  \subseteq\operatorname*{Ker}\left(
\varepsilon_{H}\right)  $. Let $\mathfrak{a}\in\mathfrak{g}\left(  H,A\right)
$ be an $\left(  \varepsilon_{H},\varepsilon_{H}\right)  $-derivation
satisfying $\mathfrak{a}\left(  H\right)  \subseteq A_{1}$. Notice that the
$k$-linear map $e^{\ast\mathfrak{a}}:H\rightarrow A$ is well-defined (since
$\mathfrak{a}\in\mathfrak{g}\left(  H,A\right)  $).

Assume further that%
\begin{equation}
\left(  \mathfrak{a}\circ\mathfrak{b}\right)  \left(  x\right)
=x\ \ \ \ \ \ \ \ \ \ \text{for each }x\in A_{1}.
\label{eq.lem.leray2.abstract2.ass}%
\end{equation}
Then, both maps $e^{\ast\mathfrak{a}}\circ\mathfrak{b}:A\rightarrow A$ and
$\mathfrak{b}:A\rightarrow H$ are injective.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.abstract2}.]Lemma~\ref{Lemma121} (applied to
$f=\mathfrak{a}$) yields that $e^{\ast\mathfrak{a}}$ is a $k$-algebra homomorphism.

The two maps $e^{\ast\mathfrak{a}}$ and $\mathfrak{b}$ are $k$-algebra
homomorphisms. Hence, their composition $e^{\ast\mathfrak{a}}\circ
\mathfrak{b}:A\rightarrow A$ is a $k$-algebra homomorphism as well (since the
composition of two $k$-algebra homomorphisms always is a $k$-algebra homomorphism).

For each $m\in\mathbb{N}$, we define a $k$-vector subspace $A_{>m}$ of $A$ by
$A_{>m}=\sum_{\substack{g\in\mathbb{N};\\g>m}}A_{g}$.

Each $x\in A_{1}$ satisfies $\left(  e^{\ast\mathfrak{a}}\circ\mathfrak{b}%
\right)  \left(  x\right)  -x\in A_{>1}$\ \ \ \ \footnote{\textit{Proof.} Let
$x\in A_{1}$. Then, (\ref{eq.lem.leray2.abstract2.ass}) yields $\left(
\mathfrak{a}\circ\mathfrak{b}\right)  \left(  x\right)  =x$. Hence, $x=\left(
\mathfrak{a}\circ\mathfrak{b}\right)  \left(  x\right)  =\mathfrak{a}\left(
\mathfrak{b}\left(  x\right)  \right)  $.
\par
But $\mathfrak{b}\left(  \underbrace{x}_{\in A_{1}}\right)  \in\mathfrak{b}%
\left(  A_{1}\right)  \subseteq\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  $. Thus, Lemma~\ref{lem.leray2.abstract} (applied to
$\mathfrak{b}\left(  x\right)  $ instead of $x$) yields $e^{\ast\mathfrak{a}%
}\left(  \mathfrak{b}\left(  x\right)  \right)  -\mathfrak{a}\left(
\mathfrak{b}\left(  x\right)  \right)  \in A_{>1}$. Thus,%
\[
\underbrace{\left(  e^{\ast\mathfrak{a}}\circ\mathfrak{b}\right)  \left(
x\right)  }_{=e^{\ast\mathfrak{a}}\left(  \mathfrak{b}\left(  x\right)
\right)  }-\underbrace{x}_{=\mathfrak{a}\left(  \mathfrak{b}\left(  x\right)
\right)  }=e^{\ast\mathfrak{a}}\left(  \mathfrak{b}\left(  x\right)  \right)
-\mathfrak{a}\left(  \mathfrak{b}\left(  x\right)  \right)  \in A_{>1}.
\]
Qed.}. Thus, Lemma~\ref{lem.leray2.graded-a1} \textbf{(b)} (applied to
$f=e^{\ast\mathfrak{a}}\circ\mathfrak{b}$) shows that the map $e^{\ast
\mathfrak{a}}\circ\mathfrak{b}$ is injective.

Next, let us recall the following basic fact about maps: If $X$, $Y$ and $Z$
are three sets, and if $\alpha:X\rightarrow Y$ and $\beta:Y\rightarrow Z$ are
two maps such that $\beta\circ\alpha$ is injective, then the map $\alpha$ is
injective.\footnote{\textit{Proof.} Let $X$, $Y$ and $Z$ be three sets. Let
$\alpha:X\rightarrow Y$ and $\beta:Y\rightarrow Z$ be two maps such that
$\beta\circ\alpha$ is injective. We must show that the map $\alpha$ is
injective.
\par
Let $u$ and $v$ be two elements of $X$ such that $\alpha\left(  u\right)
=\alpha\left(  v\right)  $. Then, $\left(  \beta\circ\alpha\right)  \left(
u\right)  =\beta\left(  \underbrace{\alpha\left(  u\right)  }_{=\alpha\left(
v\right)  }\right)  =\beta\left(  \alpha\left(  v\right)  \right)  =\left(
\beta\circ\alpha\right)  \left(  v\right)  $. Since the map $\beta\circ\alpha$
is injective, we obtain $u=v$ from this.
\par
Now, forget that we fixed $u$ and $v$. We thus have shown that if $u$ and $v$
are two elements of $X$ such that $\alpha\left(  u\right)  =\alpha\left(
v\right)  $, then $u=v$. In other words, the map $\alpha$ is injective. Qed.}
Applying this to $X=A$, $Y=H$, $Z=A$, $\alpha=\mathfrak{b}$ and $\beta
=e^{\ast\mathfrak{a}}$, we conclude that the map $\mathfrak{b}$ is injective
(since the map $e^{\ast\mathfrak{a}}\circ\mathfrak{b}$ is injective).

We thus have shown that both maps $e^{\ast\mathfrak{a}}\circ\mathfrak{b}%
:A\rightarrow A$ and $\mathfrak{b}:A\rightarrow H$ are injective. This proves
Lemma~\ref{lem.leray2.abstract2}.
\end{proof}

\begin{lemma}
\label{lem.leray2.H0ker}Let $k$ be a field. Let $C$ be a graded $k$-coalgebra.
For each $m\in\mathbb{N}$, we define a $k$-vector subspace $C_{>m}$ of $C$ by
$C_{>m}=\sum_{\substack{g\in\mathbb{N};\\g>m}}C_{g}$. Then:

\textbf{(a)} We have $C_{>0}\subseteq\operatorname*{Ker}\left(  \varepsilon
_{C}\right)  $.

\textbf{(b)} Assume that the graded $k$-coalgebra $C$ is connected. Then,
$C_{>0}=\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.H0ker}.]The map $\varepsilon_{C}$ is
$k$-linear. Hence, its kernel $\operatorname*{Ker}\left(  \varepsilon
_{C}\right)  $ is a $k$-vector subspace of $C$.

Each positive integer $g$ satisfies%
\begin{equation}
C_{g}\subseteq\operatorname*{Ker}\left(  \varepsilon_{C}\right)
\label{pf.lem.leray2.H0ker.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.H0ker.1}):} Let $g$ be a
positive integer. Hence, $g\in\mathbb{N}$.
\par
Let us give $k$ the usual grading (the one where $k_{0}=k$ and $k_{n}=0$ for
all positive $n\in\mathbb{N}$).
\par
Recall that $C$ is a graded $k$-coalgebra. Thus, its counity map
$\varepsilon_{C}:C\rightarrow k$ is graded. In other words, $\varepsilon
_{C}\left(  C_{n}\right)  \subseteq k_{n}$ for every $n\in\mathbb{N}$.
Applying this to $n=g$, we obtain $\varepsilon_{C}\left(  C_{g}\right)
\subseteq k_{g}$.
\par
But recall that $k_{n}=0$ for all positive $n\in\mathbb{N}$ (by the definition
of the grading on $k$). Applying this to $n=g$, we conclude that $k_{g}=0$.
Thus, $\varepsilon_{C}\left(  C_{g}\right)  \subseteq k_{g}=0$, so that
$\varepsilon_{C}\left(  C_{g}\right)  =0$ and thus $C_{g}\subseteq
\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $. This proves
(\ref{pf.lem.leray2.H0ker.1}).}.

Now, the definition of $C_{>0}$ yields%
\begin{align}
C_{>0}  &  =\sum_{\substack{g\in\mathbb{N};\\g>0}}\underbrace{C_{g}%
}_{\substack{\subseteq\operatorname*{Ker}\left(  \varepsilon_{C}\right)
\\\text{(by (\ref{pf.lem.leray2.H0ker.1}))}}}\label{pf.lem.leray2.H0ker.2}\\
&  \subseteq\sum_{\substack{g\in\mathbb{N};\\g>0}}\operatorname*{Ker}\left(
\varepsilon_{C}\right)  \subseteq\operatorname*{Ker}\left(  \varepsilon
_{C}\right) \nonumber
\end{align}
(since $\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $ is a $k$-vector
subspace of $C$). This proves Lemma~\ref{lem.leray2.H0ker} \textbf{(a)}.

\textbf{(b)} We know that $C$ is a graded $k$-coalgebra. Thus,
Proposition~\ref{Proposition206} shows that $\left(  C,\left(  C_{\leq
n}\right)  _{n\geq0}\right)  $ is a filtered $k$-coalgebra. As usual, we shall
refer to this filtered $k$-coalgebra $\left(  C,\left(  C_{\leq n}\right)
_{n\geq0}\right)  $ simply as \textquotedblleft the filtered $k$-coalgebra
$C$\textquotedblright.

Remark~\ref{Remark213} shows that the graded $k$-coalgebra $C$ is connected if
and only if the filtered $k$-coalgebra $C$ is connected. Thus, the filtered
$k$-coalgebra $C$ is connected (since the graded $k$-coalgebra $C$ is connected).

Remark~\ref{Remark7k} yields that $C$ is a unital coalgebra with unity
$1_{C}=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)
$. In other words, $\left(  C,1_{C}\right)  $ is a unital coalgebra, where
$1_{C}=\left(  \varepsilon_{C}\mid_{C_{\leq0}}\right)  ^{-1}\left(  1\right)
$. In other words, $C$ is a $k$-coalgebra and $1_{C}$ is an element of $C$
satisfying $\Delta_{C}\left(  1_{C}\right)  =1_{C}\otimes1_{C}$ and
$\varepsilon_{C}\left(  1_{C}\right)  =1$ (by the definition of a
\textquotedblleft unital coalgebra\textquotedblright).

Remark~\ref{Remark7m} yields that $C$ is connected if and only if $C_{\leq
0}=k\cdot1_{C}$. Thus, $C_{\leq0}=k\cdot1_{C}$ (since $C$ is connected). By
the definition of $C_{\leq0}$, we have $C_{\leq0}=\bigoplus_{\ell=0}%
^{0}C_{\ell}=C_{0}$. Thus, $C_{0}=C_{\leq0}=k\cdot1_{C}$.

Now, $C$ is graded; thus,%
\begin{align*}
C  &  =\bigoplus_{g\in\mathbb{N}}C_{g}=C_{0}\oplus\underbrace{\left(
\bigoplus_{\substack{g\in\mathbb{N};\\g>0}}C_{g}\right)  }_{\substack{=\sum
_{\substack{g\in\mathbb{N};\\g>0}}C_{g}\\\text{(since direct sums are sums)}%
}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend}\\
\text{for }g=0\text{ from the direct sum}%
\end{array}
\right) \\
&  =C_{0}\oplus\underbrace{\left(  \sum_{\substack{g\in\mathbb{N};\\g>0}%
}C_{g}\right)  }_{\substack{=C_{>0}\\\text{(by (\ref{pf.lem.leray2.H0ker.2}%
))}}}=C_{0}\oplus C_{>0}.
\end{align*}


Now, let $x\in\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $. Then,
$x\in\operatorname*{Ker}\left(  \varepsilon_{C}\right)  \subseteq
C=C_{0}\oplus C_{>0}=C_{0}+C_{>0}$ (since direct sums are sums). In other
words, there exist $y\in C_{0}$ and $z\in C_{>0}$ such that $x=y+z$. Consider
these $y$ and $z$.

We have $y\in C_{0}=k\cdot1_{C}$. Thus, there exists some $\lambda\in k$ such
that $y=\lambda\cdot1_{C}$. Consider this $\lambda$.

We have $z\in C_{>0}\subseteq\operatorname*{Ker}\left(  \varepsilon
_{C}\right)  $ (by Lemma~\ref{lem.leray2.H0ker} \textbf{(a)}) and thus
$\varepsilon_{C}\left(  z\right)  =0$. Also, $\varepsilon_{C}\left(  x\right)
=0$ (since $x\in\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $). Thus,%
\begin{align*}
0  &  =\varepsilon_{C}\left(  \underbrace{x}_{=y+z}\right)  =\varepsilon
_{C}\left(  y+z\right)  =\varepsilon_{C}\left(  y\right)
+\underbrace{\varepsilon_{C}\left(  z\right)  }_{=0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\varepsilon_{C}\text{ is
}k\text{-linear}\right) \\
&  =\varepsilon_{C}\left(  \underbrace{y}_{=\lambda\cdot1_{C}}\right)
=\varepsilon_{C}\left(  \lambda\cdot1_{C}\right)  =\lambda\cdot
\underbrace{\varepsilon_{C}\left(  1_{C}\right)  }_{=1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\varepsilon_{C}\text{ is
}k\text{-linear}\right) \\
&  =\lambda.
\end{align*}
Thus, $\lambda=0$, so that $y=\underbrace{\lambda}_{=0}\cdot1_{C}=0$. Hence,
$x=\underbrace{y}_{=0}+z=z\in C_{>0}$.

Now, forget that we fixed $x$. We thus have proven that $x\in C_{>0}$ for each
$x\in\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $. In other words,
$\operatorname*{Ker}\left(  \varepsilon_{C}\right)  \subseteq C_{>0}$. But
Lemma~\ref{lem.leray2.H0ker} \textbf{(a)} yields $C_{>0}\subseteq
\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $. Combining this with
$\operatorname*{Ker}\left(  \varepsilon_{C}\right)  \subseteq C_{>0}$, we
obtain $C_{>0}=\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $. This
proves Lemma~\ref{lem.leray2.H0ker} \textbf{(b)}.
\end{proof}

\begin{lemma}
\label{lem.leray2.exist-alpha}Let $k$ be a field. Let $C$ be a unital
coalgebra. Let $V$ and $W$ be two $k$-vector subspaces of $C$ such that
$\operatorname*{Ker}\left(  \varepsilon_{C}\right)  =V\oplus W$. Then, there
exists a $k$-linear map $\alpha:C\rightarrow V$ that satisfies $\alpha\left(
W+k\cdot1_{C}\right)  =0$ and $\alpha\mid_{V}=\operatorname*{id}\nolimits_{V}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray2.exist-alpha}.]Recall that $\operatorname*{Ker}%
\left(  \varepsilon_{C}\right)  =V\oplus W$. In particular, $V\oplus W$ is a
well-defined internal direct sum. Let $\mathfrak{p}:V\oplus W\rightarrow V$ be
the canonical projection from this direct sum to its first addend $V$. Thus,
$\mathfrak{p}\mid_{V}=\operatorname*{id}\nolimits_{V}$ and $\mathfrak{p}%
\mid_{W}=0$.

The map $\mathfrak{p}$ is a $k$-linear map $V\oplus W\rightarrow V$ (since it
is a canonical projection from a direct sum to one of its addends). In other
words, the map $\mathfrak{p}$ is a $k$-linear map $\operatorname*{Ker}\left(
\varepsilon_{C}\right)  \rightarrow V$ (since $C_{>0}=V\oplus W$).

We know that $\left(  C,1_{C}\right)  $ is a unital coalgebra. In other words,
$C$ is a $k$-coalgebra and $1_{C}$ is an element of $C$ satisfying $\Delta
_{C}\left(  1_{C}\right)  =1_{C}\otimes1_{C}$ and $\varepsilon_{C}\left(
1_{C}\right)  =1$ (by the definition of a \textquotedblleft unital
coalgebra\textquotedblright).

Consider the $k$-linear map $e_{C,C}:C\rightarrow C$. (Here, we are using the
notation introduced in Definition~\ref{def.eHA2}.)

Define a $k$-linear map $\xi:C\rightarrow C$ by $\xi=\operatorname*{id}%
\nolimits_{C}-e_{C,C}$.\ \ \ \ \footnote{This is indeed a $k$-linear map,
since both maps $\operatorname*{id}\nolimits_{C}$ and $e_{C,C}$ are
$k$-linear.} Then, each $c\in C$ satisfies $\xi\left(  c\right)
\in\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $c\in C$. Then, the definition of
$e_{C,C}$ yields $e_{C,C}=\eta_{C}\circ\varepsilon_{C}$, so that%
\[
\underbrace{e_{C,C}}_{=\eta_{C}\circ\varepsilon_{C}}\left(  c\right)  =\left(
\eta_{C}\circ\varepsilon_{C}\right)  \left(  c\right)  =\eta_{C}\left(
\varepsilon_{C}\left(  c\right)  \right)  =\varepsilon_{C}\left(  c\right)
\cdot1_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\eta
_{C}\right)  .
\]
Now,%
\begin{align*}
\varepsilon_{C}\left(  \underbrace{\xi}_{=\operatorname*{id}\nolimits_{C}%
-e_{C,C}}\left(  c\right)  \right)   &  =\varepsilon_{C}\left(
\underbrace{\left(  \operatorname*{id}\nolimits_{C}-e_{C,C}\right)  \left(
c\right)  }_{=\operatorname*{id}\nolimits_{C}\left(  c\right)  -e_{C,C}\left(
c\right)  }\right)  =\varepsilon_{C}\left(  \underbrace{\operatorname*{id}%
\nolimits_{C}\left(  c\right)  }_{=c}-\underbrace{e_{C,C}\left(  c\right)
}_{=\varepsilon_{C}\left(  c\right)  \cdot1_{C}}\right) \\
&  =\varepsilon_{C}\left(  c\right)  -\underbrace{\varepsilon_{C}\left(
\varepsilon_{C}\left(  c\right)  \cdot1_{C}\right)  }_{\substack{=\varepsilon
_{C}\left(  c\right)  \cdot\varepsilon_{C}\left(  1_{C}\right)  \\\text{(since
the map }\varepsilon_{C}\text{ is }k\text{-linear)}}}=\varepsilon_{C}\left(
c\right)  -\varepsilon_{C}\left(  c\right)  \cdot\underbrace{\varepsilon
_{C}\left(  1_{C}\right)  }_{=1}\\
&  =\varepsilon_{C}\left(  c\right)  -\varepsilon_{C}\left(  c\right)  =0.
\end{align*}
In other words, $\xi\left(  c\right)  \in\operatorname*{Ker}\left(
\varepsilon_{C}\right)  $. Qed.}. Hence, we can define a map $\xi^{\prime
}:C\rightarrow\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $ by%
\[
\left(  \xi^{\prime}\left(  c\right)  =\xi\left(  c\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }c\in C\right)  .
\]
Consider this map $\xi^{\prime}$. This map $\xi^{\prime}$ is obtained from
$\xi$ by restricting the codomain to $\operatorname*{Ker}\left(
\varepsilon_{C}\right)  $ (since $\xi^{\prime}\left(  c\right)  =\xi\left(
c\right)  $ for every $c\in C$). Hence, this map $\xi^{\prime}$ is $k$-linear
(since the map $\xi$ is $k$-linear).

Thus, $\xi^{\prime}$ is a $k$-linear map $C\rightarrow\operatorname*{Ker}%
\left(  \varepsilon_{C}\right)  $, whereas $\mathfrak{p}$ is a $k$-linear map
$\operatorname*{Ker}\left(  \varepsilon_{C}\right)  \rightarrow V$. Hence, the
composition $\mathfrak{p}\circ\xi^{\prime}$ of these two maps is a $k$-linear
map $C\rightarrow V$.

Furthermore,
\begin{equation}
\xi^{\prime}\left(  x\right)  =x\ \ \ \ \ \ \ \ \ \ \text{for each }%
x\in\operatorname*{Ker}\left(  \varepsilon_{C}\right)
\label{pf.lem.leray2.exist-alpha.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray2.exist-alpha.3}):} Let
$x\in\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $. Thus, $x\in C$ and
$\varepsilon_{C}\left(  x\right)  =0$.
\par
But the definition of $e_{C,C}$ yields $e_{C,C}=\eta_{C}\circ\varepsilon_{C}$,
so that%
\[
\underbrace{e_{C,C}}_{=\eta_{C}\circ\varepsilon_{C}}\left(  x\right)  =\left(
\eta_{C}\circ\varepsilon_{C}\right)  \left(  x\right)  =\eta_{C}\left(
\underbrace{\varepsilon_{C}\left(  x\right)  }_{=0}\right)  =\eta_{C}\left(
0\right)  =0
\]
(since the map $\eta_{C}$ is $k$-linear). Now, the definition of $\xi^{\prime
}$ yields%
\[
\xi^{\prime}\left(  x\right)  =\underbrace{\xi}_{=\operatorname*{id}%
\nolimits_{C}-e_{C,C}}\left(  x\right)  =\left(  \operatorname*{id}%
\nolimits_{C}-e_{C,C}\right)  \left(  x\right)
=\underbrace{\operatorname*{id}\nolimits_{C}\left(  x\right)  }_{=x}%
-\underbrace{e_{C,C}\left(  x\right)  }_{=0}=x.
\]
Qed.}. Hence, $\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \left(  W\right)
=0$\ \ \ \ \footnote{\textit{Proof.} Let $x\in W$. Then, $x\in W\subseteq
V\oplus W=\operatorname*{Ker}\left(  \varepsilon_{C}\right)  $. Thus,
(\ref{pf.lem.leray2.exist-alpha.3}) shows that $\xi^{\prime}\left(  x\right)
=x$. But $x\in W$; therefore, $\left(  \mathfrak{p}\mid_{W}\right)  \left(
x\right)  $ is well-defined. We have $\underbrace{\left(  \mathfrak{p}\mid
_{W}\right)  }_{=0}\left(  x\right)  =0\left(  x\right)  =0$. Comparing this
with $\left(  \mathfrak{p}\mid_{W}\right)  \left(  x\right)  =\mathfrak{p}%
\left(  x\right)  $, we obtain $\mathfrak{p}\left(  x\right)  =0$.
\par
Now, $\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \left(  x\right)
=\mathfrak{p}\left(  \underbrace{\xi^{\prime}\left(  x\right)  }_{=x}\right)
=\mathfrak{p}\left(  x\right)  =0$. Hence, $x\in\operatorname*{Ker}\left(
\mathfrak{p}\circ\xi^{\prime}\right)  $.
\par
Let us now forget that we fixed $x$. We thus have proven that $x\in
\operatorname*{Ker}\left(  \mathfrak{p}\circ\xi^{\prime}\right)  $ for each
$x\in W$. In other words, $W\subseteq\operatorname*{Ker}\left(  \mathfrak{p}%
\circ\xi^{\prime}\right)  $. In other words, $\left(  \mathfrak{p}\circ
\xi^{\prime}\right)  \left(  W\right)  =0$.} and $\left(  \mathfrak{p}\circ
\xi^{\prime}\right)  \left(  k\cdot1_{C}\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} The definition of $e_{C,C}$ yields
$e_{C,C}=\eta_{C}\circ\varepsilon_{C}$, so that%
\begin{align*}
\underbrace{e_{C,C}}_{=\eta_{C}\circ\varepsilon_{C}}\left(  1_{C}\right)   &
=\left(  \eta_{C}\circ\varepsilon_{C}\right)  \left(  1_{C}\right)  =\eta
_{C}\left(  \underbrace{\varepsilon_{C}\left(  1_{C}\right)  }_{=1}\right)
=\eta_{C}\left(  1\right) \\
&  =1\cdot1_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map
}\eta_{C}\right) \\
&  =1_{C}.
\end{align*}
Now, the definition of $\xi^{\prime}$ yields%
\[
\xi^{\prime}\left(  1_{C}\right)  =\underbrace{\xi}_{=\operatorname*{id}%
\nolimits_{C}-e_{C,C}}\left(  1_{C}\right)  =\left(  \operatorname*{id}%
\nolimits_{C}-e_{C,C}\right)  \left(  1_{C}\right)
=\underbrace{\operatorname*{id}\nolimits_{C}\left(  1_{C}\right)  }_{=1_{C}%
}-\underbrace{e_{C,C}\left(  1_{C}\right)  }_{=1_{C}}=1_{C}-1_{C}=0.
\]
Since the map $\xi^{\prime}$ is $k$-linear, we have $\xi^{\prime}\left(
k\cdot1_{C}\right)  =k\cdot\underbrace{\xi^{\prime}\left(  1_{C}\right)
}_{=0}=k\cdot0=0$. Now, $\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \left(
k\cdot1_{C}\right)  =\mathfrak{p}\left(  \underbrace{\xi^{\prime}\left(
k\cdot1_{C}\right)  }_{=0}\right)  =\mathfrak{p}\left(  0\right)  =0$ (since
the map $\mathfrak{p}$ is $k$-linear).}. Hence, $\left(  \mathfrak{p}\circ
\xi^{\prime}\right)  \left(  W+k\cdot1_{C}\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} The map $\mathfrak{p}\circ\xi^{\prime}$ is
$k$-linear. Thus,%
\[
\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \left(  W+k\cdot1_{C}\right)
=\underbrace{\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \left(  W\right)
}_{=0}+\underbrace{\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \left(
k\cdot1_{C}\right)  }_{=0}=0+0=0.
\]
}. Furthermore, $\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \mid
_{V}=\operatorname*{id}\nolimits_{V}$\ \ \ \ \footnote{\textit{Proof.} Let
$x\in V$. Then, $x\in V\subseteq V\oplus W=\operatorname*{Ker}\left(
\varepsilon_{C}\right)  $. Thus, (\ref{pf.lem.leray2.exist-alpha.3}) shows
that $\xi^{\prime}\left(  x\right)  =x$. But $x\in V$; thus, the element
$\left(  \mathfrak{p}\mid_{V}\right)  \left(  x\right)  $ is well-defined.
Hence, $\underbrace{\left(  \mathfrak{p}\mid_{V}\right)  }%
_{=\operatorname*{id}\nolimits_{V}}\left(  x\right)  =\operatorname*{id}%
\nolimits_{V}\left(  x\right)  =x$, so that $x=\left(  \mathfrak{p}\mid
_{V}\right)  \left(  x\right)  =\mathfrak{p}\left(  x\right)  $. Hence,
$\mathfrak{p}\left(  x\right)  =x$.
\par
Now,%
\[
\left(  \left(  \mathfrak{p}\circ\xi^{\prime}\right)  \mid_{V}\right)  \left(
x\right)  =\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \left(  x\right)
=\mathfrak{p}\left(  \underbrace{\xi^{\prime}\left(  x\right)  }_{=x}\right)
=\mathfrak{p}\left(  x\right)  =x=\operatorname*{id}\nolimits_{V}\left(
x\right)  .
\]
\par
Now, forget that we fixed $x$. We thus have shown that $\left(  \left(
\mathfrak{p}\circ\xi^{\prime}\right)  \mid_{V}\right)  \left(  x\right)
=\operatorname*{id}\nolimits_{V}\left(  x\right)  $ for each $x\in V$. In
other words, $\left(  \mathfrak{p}\circ\xi^{\prime}\right)  \mid
_{V}=\operatorname*{id}\nolimits_{V}$.}. Hence, there exists a $k$-linear map
$\alpha:C\rightarrow V$ that satisfies $\alpha\left(  W+k\cdot1_{C}\right)
=0$ and $\alpha\mid_{V}=\operatorname*{id}\nolimits_{V}$ (namely,
$\alpha=\mathfrak{p}\circ\xi^{\prime}$). This proves
Lemma~\ref{lem.leray2.exist-alpha}.
\end{proof}

We are now ready to prove another form of Leray's theorem:

\begin{theorem}
\label{thm.leray2.main} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected graded commutative bialgebra over $k$. For each $m\in\mathbb{N}$, we
define a $k$-vector subspace $H_{>m}$ of $H$ by $H_{>m}=\sum_{\substack{g\in
\mathbb{N};\\g>m}}H_{g}$. Let $V$ be a homogeneous subspace of $H$ such that
$H_{>0}=V\oplus\left(  H_{>0}\right)  ^{2}$.

Let $\mathfrak{i}$ denote the inclusion map $V\rightarrow H$. Let
$\mathfrak{b}$ denote the $k$-algebra homomorphism $\operatorname*{symlift}%
\mathfrak{i}:\operatorname*{Sym}V\rightarrow H$\ \ \ \ \footnote{We are using
the notations introduced in Definition~\ref{Definition1701}.}.

\textbf{(a)} This map $\mathfrak{b}$ is a $k$-algebra isomorphism.

\textbf{(b)} Let $\alpha:H\rightarrow V$ be a $k$-linear map that satisfies
$\alpha\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0$ and $\alpha\mid_{V}=\operatorname*{id}%
\nolimits_{V}$. Define a $k$-linear map $\mathfrak{a}:H\rightarrow
\operatorname*{Sym}V$ by $\mathfrak{a}=\operatorname*{syminc}\nolimits_{V}%
\circ\alpha$. Then, $\mathfrak{a}\in\mathfrak{g}\left(  H,\operatorname*{Sym}%
V\right)  $, and thus the $k$-linear map $e^{\ast\mathfrak{a}}:H\rightarrow
\operatorname*{Sym}V$ is well-defined. This map $e^{\ast\mathfrak{a}%
}:H\rightarrow\operatorname*{Sym}V$ is a $k$-algebra isomorphism.\footnote{The
maps $\mathfrak{b}$ and $e^{\ast\mathfrak{a}}$ are not mutually inverse in
general.}
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{thm.leray2.main}.]Recall that $H$ is a $k$-bialgebra.
Thus, $\varepsilon_{H}:H\rightarrow k$ is a $k$-algebra homomorphism (by the
axioms of a $k$-bialgebra).

We know that $H$ is a graded $k$-bialgebra. Thus,
Proposition~\ref{Proposition210} shows that $\left(  H,\left(  H_{\leq
n}\right)  _{n\geq0}\right)  $ is a filtered $k$-bialgebra. As usual, we shall
refer to this filtered $k$-bialgebra $\left(  H,\left(  H_{\leq n}\right)
_{n\geq0}\right)  $ simply as \textquotedblleft the filtered $k$-bialgebra
$H$\textquotedblright.

Remark~\ref{Remark213} (applied to $C=H$) shows that the graded $k$-coalgebra
$H$ is connected if and only if the filtered $k$-coalgebra $H$ is connected.
Thus, the filtered $k$-coalgebra $H$ is connected (since the graded
$k$-coalgebra $H$ is connected).

Remark~\ref{Remark7k} (applied to $C=H$) yields that $H$ is a unital coalgebra
with unity $1_{H}=\left(  \varepsilon_{H}\mid_{H_{\leq0}}\right)  ^{-1}\left(
1\right)  $. In other words, $\left(  H,1_{H}\right)  $ is a unital coalgebra,
where $1_{H}=\left(  \varepsilon_{H}\mid_{H_{\leq0}}\right)  ^{-1}\left(
1\right)  $.

\textbf{(b)} We have $\mathfrak{a}\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} We have%
\begin{align*}
\underbrace{\mathfrak{a}}_{=\operatorname*{syminc}\nolimits_{V}\circ\alpha
}\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)   &  =\left(  \operatorname*{syminc}\nolimits_{V}%
\circ\alpha\right)  \left(  \left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right) \\
&  =\operatorname*{syminc}\nolimits_{V}\left(  \underbrace{\alpha\left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  }_{=0}\right)  =\operatorname*{syminc}\nolimits_{V}%
\left(  0\right)  =0
\end{align*}
(since the map $\operatorname*{syminc}\nolimits_{V}$ is $k$-linear).}. But
Theorem~\ref{Theorem115} (applied to $A=\operatorname*{Sym}V$) shows that
$\mathfrak{a}$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation if and only if $\mathfrak{a}\left(  \left(  \operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$. Hence,
$\mathfrak{a}$ is an $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation (since $\mathfrak{a}\left(  \left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}+k\cdot1_{H}\right)  =0$). Hence,
Proposition~\ref{Proposition1735} (applied to $A=\operatorname*{Sym}V$ and
$h=\mathfrak{a}$) shows that $\mathfrak{a}\in\mathfrak{g}\left(
H,\operatorname*{Sym}V\right)  $. Thus, the $k$-linear map $e^{\ast
\mathfrak{a}}:H\rightarrow\operatorname*{Sym}V$ is well-defined.

We have $\mathfrak{a}\mid_{V}=\operatorname*{syminc}\nolimits_{V}%
$\ \ \ \ \footnote{\textit{Proof.} Let $v\in V$. Then, $\left(  \mathfrak{a}%
\mid_{V}\right)  \left(  v\right)  =\underbrace{\mathfrak{a}}%
_{=\operatorname*{syminc}\nolimits_{V}\circ\alpha}\left(  v\right)  =\left(
\operatorname*{syminc}\nolimits_{V}\circ\alpha\right)  \left(  v\right)
=\operatorname*{syminc}\nolimits_{V}\left(  \alpha\left(  v\right)  \right)
$. But from $v\in V$, we obtain $\left(  \alpha\mid_{V}\right)  \left(
v\right)  =\alpha\left(  v\right)  $, so that $\alpha\left(  v\right)
=\underbrace{\left(  \alpha\mid_{V}\right)  }_{=\operatorname*{id}%
\nolimits_{V}}\left(  v\right)  =\operatorname*{id}\nolimits_{V}\left(
v\right)  =v$. Thus, $\left(  \mathfrak{a}\mid_{V}\right)  \left(  v\right)
=\operatorname*{syminc}\nolimits_{V}\left(  \underbrace{\alpha\left(
v\right)  }_{=v}\right)  =\operatorname*{syminc}\nolimits_{V}\left(  v\right)
$.
\par
Now, forget that we fixed $v$. We thus have proven that $\left(
\mathfrak{a}\mid_{V}\right)  \left(  v\right)  =\operatorname*{syminc}%
\nolimits_{V}\left(  v\right)  $ for each $v\in V$. In other words,
$\mathfrak{a}\mid_{V}=\operatorname*{syminc}\nolimits_{V}$.}. Also,
$\mathfrak{a}\left(  H\right)  =\operatorname*{syminc}\nolimits_{V}\left(
V\right)  $\ \ \ \ \footnote{\textit{Proof.} Recall that $\alpha\mid
_{V}=\operatorname*{id}\nolimits_{V}$; that is, $\operatorname*{id}%
\nolimits_{V}=\alpha\mid_{V}$.
\par
Now, each $v\in V$ satisfies $v=\underbrace{\operatorname*{id}\nolimits_{V}%
}_{=\alpha\mid_{V}}\left(  v\right)  =\left(  \alpha\mid_{V}\right)  \left(
v\right)  =\alpha\left(  \underbrace{v}_{\in V\subseteq H}\right)  \in
\alpha\left(  H\right)  $. Thus, $V\subseteq\alpha\left(  H\right)  $.
Combining this with $\alpha\left(  H\right)  \subseteq V$ (which is obvious,
since $\alpha$ is a map $H\rightarrow V$), we obtain $V=\alpha\left(
H\right)  $. Thus, $\alpha\left(  H\right)  =V$.
\par
But $\underbrace{\mathfrak{a}}_{=\operatorname*{syminc}\nolimits_{V}%
\circ\alpha}\left(  H\right)  =\left(  \operatorname*{syminc}\nolimits_{V}%
\circ\alpha\right)  \left(  H\right)  =\operatorname*{syminc}\nolimits_{V}%
\left(  \underbrace{\alpha\left(  H\right)  }_{=V}\right)
=\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $.}.

Lemma~\ref{lem.leray2.Sym-gen} yields that the $k$-algebra
$\operatorname*{Sym}V$ is generated by its subset $\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $. In other words, the $k$-algebra
$\operatorname*{Sym}V$ is generated by its subset $\mathfrak{a}\left(
H\right)  $ (since $\mathfrak{a}\left(  H\right)  =\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $). In other words, the subset $\mathfrak{a}%
\left(  H\right)  $ of $\operatorname*{Sym}V$ generates the $k$-algebra
$\operatorname*{Sym}V$. Hence, Corollary~\ref{Corollary1734} (applied to
$A=\operatorname*{Sym}V$ and $f=\mathfrak{a}$) shows that the map
$e^{\ast\mathfrak{a}}:H\rightarrow\operatorname*{Sym}V$ is a surjective
$k$-algebra homomorphism.

It is well-known that $\operatorname*{Sym}V$ is a graded $k$-algebra, with
grading given by
\[
\left(  \operatorname*{Sym}V\right)  _{n}=\operatorname*{Sym}\nolimits^{n}%
V\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}.
\]
\footnote{Notice that this grading has \textbf{nothing} to do with the grading
on $H$.} In particular, $\left(  \operatorname*{Sym}V\right)  _{1}%
=\operatorname*{Sym}\nolimits^{1}V=\operatorname*{syminc}\nolimits_{V}\left(
V\right)  $.

We have $\mathfrak{b}\circ\operatorname*{syminc}\nolimits_{V}=\mathfrak{i}%
$\ \ \ \ \footnote{\textit{Proof.} Recall that $\operatorname*{symlift}%
\mathfrak{i}$ is the unique $k$-algebra homomorphism $\Phi:\operatorname*{Sym}%
V\rightarrow H$ satisfying $\Phi\circ\operatorname*{syminc}\nolimits_{V}%
=\mathfrak{i}$ (because this is how $\operatorname*{symlift}\mathfrak{i}$ was
defined). Thus, $\operatorname*{symlift}\mathfrak{i}$ is a $k$-algebra
homomorphism $\operatorname*{Sym}V\rightarrow H$ and satisfies $\left(
\operatorname*{symlift}\mathfrak{i}\right)  \circ\operatorname*{syminc}%
\nolimits_{V}=\mathfrak{i}$.
\par
Now, $\underbrace{\mathfrak{b}}_{=\operatorname*{symlift}\mathfrak{i}}%
\circ\operatorname*{syminc}\nolimits_{V}=\left(  \operatorname*{symlift}%
\mathfrak{i}\right)  \circ\operatorname*{syminc}\nolimits_{V}=\mathfrak{i}$.}.

Recall that the $k$-algebra $\operatorname*{Sym}V$ is generated by its subset
$\operatorname*{syminc}\nolimits_{V}\left(  V\right)  $. In other words, the
$k$-subalgebra of $\operatorname*{Sym}V$ generated by $\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $ is $\operatorname*{Sym}V$. In other words,
$\operatorname*{AlgGen}\nolimits_{k}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  \right)  $ is $\operatorname*{Sym}%
V$\ \ \ \ \footnote{since $\operatorname*{AlgGen}\nolimits_{k}\left(
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \right)  $ is the
$k$-subalgebra of $\operatorname*{Sym}V$ generated by $\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $ (by the definition of $\operatorname*{AlgGen}%
\nolimits_{k}\left(  \operatorname*{syminc}\nolimits_{V}\left(  V\right)
\right)  $)}. In other words, $\operatorname*{AlgGen}\nolimits_{k}\left(
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  \right)
=\operatorname*{Sym}V$. Thus,%
\[
\operatorname*{Sym}V=\operatorname*{AlgGen}\nolimits_{k}\left(
\underbrace{\operatorname*{syminc}\nolimits_{V}\left(  V\right)  }_{=\left(
\operatorname*{Sym}V\right)  _{1}}\right)  =\operatorname*{AlgGen}%
\nolimits_{k}\left(  \left(  \operatorname*{Sym}V\right)  _{1}\right)  .
\]
Moreover, it is well-known that the $k$-algebra $\operatorname*{Sym}V$ is commutative.

The $k$-algebra homomorphism $\mathfrak{b}:\operatorname*{Sym}V\rightarrow H$
satisfies $\mathfrak{b}\left(  \left(  \operatorname*{Sym}V\right)
_{1}\right)  \subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)
$\ \ \ \ \footnote{\textit{Proof.} We have%
\[
\mathfrak{b}\left(  \underbrace{\left(  \operatorname*{Sym}V\right)  _{1}%
}_{=\operatorname*{syminc}\nolimits_{V}\left(  V\right)  }\right)
=\mathfrak{b}\left(  \operatorname*{syminc}\nolimits_{V}\left(  V\right)
\right)  =\underbrace{\left(  \mathfrak{b}\circ\operatorname*{syminc}%
\nolimits_{V}\right)  }_{=\mathfrak{i}}\left(  V\right)  =\mathfrak{i}\left(
V\right)  =V
\]
(since $\mathfrak{i}$ is an inclusion map).
\par
Lemma~\ref{lem.leray2.H0ker} \textbf{(a)} (applied to $C=H$) shows that
$H_{>0}\subseteq\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $.
\par
But recall that $H_{>0}=V\oplus\left(  H_{>0}\right)  ^{2}\supseteq V$, so
that $V\subseteq H_{>0}\subseteq\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  $. Thus, $\mathfrak{b}\left(  \left(  \operatorname*{Sym}%
V\right)  _{1}\right)  =V\subseteq\operatorname*{Ker}\left(  \varepsilon
_{H}\right)  $.}. The $\left(  \varepsilon_{H},\varepsilon_{H}\right)
$-derivation $\mathfrak{a}\in\mathfrak{g}\left(  H,\operatorname*{Sym}%
V\right)  $ satisfies $\mathfrak{a}\left(  H\right)  \subseteq\left(
\operatorname*{Sym}V\right)  _{1}$\ \ \ \ \footnote{\textit{Proof.} We have
$\underbrace{\mathfrak{a}}_{=\operatorname*{syminc}\nolimits_{V}\circ\alpha
}\left(  H\right)  =\left(  \operatorname*{syminc}\nolimits_{V}\circ
\alpha\right)  \left(  H\right)  =\operatorname*{syminc}\nolimits_{V}\left(
\underbrace{\alpha\left(  H\right)  }_{\subseteq V}\right)  \subseteq
\operatorname*{syminc}\nolimits_{V}\left(  V\right)  =\left(
\operatorname*{Sym}V\right)  _{1}$.}. Furthermore, we have%
\[
\left(  \mathfrak{a}\circ\mathfrak{b}\right)  \left(  x\right)
=x\ \ \ \ \ \ \ \ \ \ \text{for each }x\in\left(  \operatorname*{Sym}V\right)
_{1}%
\]
\footnote{\textit{Proof.} Let $x\in\left(  \operatorname*{Sym}V\right)  _{1}$.
Then, $x\in\left(  \operatorname*{Sym}V\right)  _{1}=\operatorname*{syminc}%
\nolimits_{V}\left(  V\right)  $. In other words, there exists some $w\in V$
such that $x=\operatorname*{syminc}\nolimits_{V}\left(  w\right)  $. Consider
this $w$.
\par
From $w\in V$, we obtain $\alpha\left(  w\right)  =\underbrace{\left(
\alpha\mid_{V}\right)  }_{=\operatorname*{id}\nolimits_{V}}\left(  w\right)
=\operatorname*{id}\nolimits_{V}\left(  w\right)  =w$.
\par
Applying the map $\mathfrak{b}$ to both sides of the equality
$x=\operatorname*{syminc}\nolimits_{V}\left(  w\right)  $, we find%
\[
\mathfrak{b}\left(  x\right)  =\mathfrak{b}\left(  \operatorname*{syminc}%
\nolimits_{V}\left(  w\right)  \right)  =\underbrace{\left(  \mathfrak{b}%
\circ\operatorname*{syminc}\nolimits_{V}\right)  }_{=\mathfrak{i}}\left(
w\right)  =\mathfrak{i}\left(  w\right)  =w
\]
(since $\mathfrak{i}$ is merely an inclusion map). Now,%
\[
\left(  \mathfrak{a}\circ\mathfrak{b}\right)  \left(  x\right)
=\underbrace{\mathfrak{a}}_{=\operatorname*{syminc}\nolimits_{V}\circ\alpha
}\left(  \underbrace{\mathfrak{b}\left(  x\right)  }_{=w}\right)  =\left(
\operatorname*{syminc}\nolimits_{V}\circ\alpha\right)  \left(  w\right)
=\operatorname*{syminc}\nolimits_{V}\left(  \underbrace{\alpha\left(
w\right)  }_{=w}\right)  =\operatorname*{syminc}\nolimits_{V}\left(  w\right)
=x.
\]
Qed.}. Hence, Lemma~\ref{lem.leray2.abstract2} (applied to
$A=\operatorname*{Sym}V$) shows that both maps $e^{\ast\mathfrak{a}}%
\circ\mathfrak{b}:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$ and
$\mathfrak{b}:A\rightarrow\operatorname*{Sym}V$ are injective.

On the other hand, $H_{>0}=V\oplus\left(  H_{>0}\right)  ^{2}=V+\left(
H_{>0}\right)  ^{2}$ (since direct sums are sums). Furthermore, $H_{0}%
=k\cdot1_{H}$ (by Proposition~\ref{Proposition921}). Thus,
Lemma~\ref{lem.leray2.graded-a2-gen} \textbf{(b)} (applied to $A=H$) yields
$H=\operatorname*{AlgGen}\nolimits_{k}V$.

Now, $V=\mathfrak{i}\left(  V\right)  $\ \ \ \ \footnote{\textit{Proof.}
Recall that $\mathfrak{i}$ is merely an inclusion map. Hence, $\mathfrak{i}%
\left(  V\right)  =V$, so that $V=\mathfrak{i}\left(  V\right)  $.}. Hence,
$H=\operatorname*{AlgGen}\nolimits_{k}\underbrace{V}_{=\mathfrak{i}\left(
V\right)  }=\operatorname*{AlgGen}\nolimits_{k}\left(  \mathfrak{i}\left(
V\right)  \right)  $.

Now, recall that $\operatorname*{AlgGen}\nolimits_{k}\left(  \mathfrak{i}%
\left(  V\right)  \right)  $ is the $k$-subalgebra of $H$ generated by
$\mathfrak{i}\left(  V\right)  $ (by the definition of $\operatorname*{AlgGen}%
\nolimits_{k}\left(  \mathfrak{i}\left(  V\right)  \right)  $). In other
words, $H$ is the $k$-subalgebra of $H$ generated by $\mathfrak{i}\left(
V\right)  $ (since $H=\operatorname*{AlgGen}\nolimits_{k}\left(
\mathfrak{i}\left(  V\right)  \right)  $). In other words, the image
$\mathfrak{i}\left(  V\right)  $ generates the $k$-algebra $H$. Hence,
Lemma~\ref{Lemma1721} \textbf{(a)} (applied to $A=H$ and $\varphi
=\mathfrak{i}$) shows that the $k$-algebra homomorphism
$\operatorname*{symlift}\mathfrak{i}:\operatorname*{Sym}V\rightarrow H$ is
surjective. In other words, the $k$-algebra homomorphism $\mathfrak{b}%
:\operatorname*{Sym}V\rightarrow H$ is surjective (since $\mathfrak{b}%
=\operatorname*{symlift}\mathfrak{i}$).

We now know that the map $\mathfrak{b}$ is both injective and surjective.
Hence, $\mathfrak{b}$ is bijective. Thus, $\mathfrak{b}$ is invertible.
Therefore,%
\begin{equation}
\mathfrak{b}\text{ is a }k\text{-algebra isomorphism}
\label{pf.thm.leray2.main.b.b-is-iso}%
\end{equation}
(since $\mathfrak{b}$ is an invertible $k$-algebra homomorphism). Hence, the
inverse $\mathfrak{b}^{-1}$ of $\mathfrak{b}$ is well-defined, and also is a
$k$-algebra isomorphism. This map $\mathfrak{b}^{-1}$ is invertible (since it
is a $k$-algebra isomorphism), thus bijective, thus injective.

Recall that the map $e^{\ast\mathfrak{a}}\circ\mathfrak{b}$ is surjective.
Thus, both maps $e^{\ast\mathfrak{a}}\circ\mathfrak{b}$ and $\mathfrak{b}%
^{-1}$ are injective. Therefore, their composition $\left(  e^{\ast
\mathfrak{a}}\circ\mathfrak{b}\right)  \circ\mathfrak{b}^{-1}$ is also
injective (since the composition of any two injective maps is injective). In
other words, the map $e^{\ast\mathfrak{a}}$ is injective (since $\left(
e^{\ast\mathfrak{a}}\circ\mathfrak{b}\right)  \circ\mathfrak{b}^{-1}%
=e^{\ast\mathfrak{a}}\circ\underbrace{\mathfrak{b}\circ\mathfrak{b}^{-1}%
}_{=\operatorname*{id}}=e^{\ast\mathfrak{a}}$).

We now know that the map $e^{\ast\mathfrak{a}}$ is both injective and
surjective. Thus, $e^{\ast\mathfrak{a}}$ is bijective. Hence, $e^{\ast
\mathfrak{a}}$ is invertible. Thus, $e^{\ast\mathfrak{a}}$ is a $k$-algebra
isomorphism (since $e^{\ast\mathfrak{a}}$ is an invertible $k$-algebra
homomorphism). This proves Theorem~\ref{thm.leray2.main} \textbf{(b)}.

\textbf{(a)} We are going to apply the fact
(\ref{pf.thm.leray2.main.b.b-is-iso}), which we have proven during our above
proof of Theorem~\ref{thm.leray2.main} \textbf{(b)}. However, in order to do
this, we must ensure that we are in the situation of
Theorem~\ref{thm.leray2.main} \textbf{(b)}; that is, we must construct a
$k$-linear map $\alpha:H\rightarrow V$ with the properties that $\alpha\left(
\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}+k\cdot1_{H}\right)  =0$ and $\alpha\mid_{V}=\operatorname*{id}%
\nolimits_{V}$.

However, this is easy: Lemma~\ref{lem.leray2.H0ker} \textbf{(b)} (applied to
$C=H$) shows that $H_{>0}=\operatorname*{Ker}\left(  \varepsilon_{H}\right)
$. Hence, $\operatorname*{Ker}\left(  \varepsilon_{H}\right)  =H_{>0}%
=V\oplus\left(  H_{>0}\right)  ^{2}$. Thus, Lemma~\ref{lem.leray2.exist-alpha}
(applied to $C=H$ and $W=\left(  H_{>0}\right)  ^{2}$) shows that there exists
a $k$-linear map $\alpha:H\rightarrow V$ that satisfies $\alpha\left(  \left(
H_{>0}\right)  ^{2}+k\cdot1_{H}\right)  =0$ and $\alpha\mid_{V}%
=\operatorname*{id}\nolimits_{V}$. Consider this $\alpha$. We have
$\alpha\left(  \left(  H_{>0}\right)  ^{2}+k\cdot1_{H}\right)  =0$. Since
$H_{>0}=\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $, this rewrites as
$\alpha\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)  =0$. Thus, we now know that $\alpha$ is a
$k$-linear map $H\rightarrow V$ satisfying $\alpha\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}+k\cdot
1_{H}\right)  =0$ and $\alpha\mid_{V}=\operatorname*{id}\nolimits_{V}$.
Therefore, the hypotheses of Theorem~\ref{thm.leray2.main} \textbf{(b)} are
satisfied. Thus, the fact (\ref{pf.thm.leray2.main.b.b-is-iso}) shows that
$\mathfrak{b}$ is a $k$-algebra isomorphism. This proves
Theorem~\ref{thm.leray2.main} \textbf{(a)}.
\end{proof}

\begin{remark}
\label{rmk.leray2.graded}Consider the situation of
Theorem~\ref{thm.leray2.main}. We can say slightly more: Namely, the
$k$-algebra isomorphism $\mathfrak{b}:\operatorname*{Sym}V\rightarrow H$ is
graded, provided that the $k$-algebra $\operatorname*{Sym}V$ has been equipped
with the appropriate grading. (To equip $\operatorname*{Sym}V$ with the
appropriate grading, we have to regard $V$ as a graded $k$-vector
space\footnote{Indeed, $V$ is canonically graded, since it is a homogeneous
subspace of the graded $k$-vector space $H$.}, and extend this grading to
$\operatorname*{Sym}V$. Note that this is \textbf{not} the grading on
$\operatorname*{Sym}V$ that was used in the proof of
Theorem~\ref{thm.leray2.main}.)

The proof of this fact is left to the reader; it is fairly straightforward.
Similarly, in the situation of Theorem~\ref{thm.leray2.main} \textbf{(b)}, the
$k$-algebra isomorphism $e^{\ast\mathfrak{a}}:H\rightarrow\operatorname*{Sym}%
V$ is graded as well (where, again, $\operatorname*{Sym}V$ is equipped with
the appropriate grading).
\end{remark}

We shall next transform Theorem~\ref{thm.leray2.main} into a form more suited
for applications. But first, let us prove a simple property of graded maps,
similar in spirit to Remark~\ref{Remark301}:

\begin{remark}
\label{rmk.graded.im} Let $k$ be a field. Let $U$ and $V$ be two graded
$k$-vector spaces. Let $f:U\rightarrow V$ be a graded $k$-linear map. Then,
$f\left(  U\right)  $ is a homogeneous subspace of $V$.
\end{remark}

\begin{proof}
[Proof of Remark~\ref{rmk.graded.im}.]If $W$ is any $k$-vector subspace of
$V$, then the internal direct sum $\bigoplus_{n\in\mathbb{N}}\left(  W\cap
V_{n}\right)  $ is always well-defined\footnote{This was proven during
Definition~\ref{Definition300}.}. Applying this to $W=f\left(  U\right)  $, we
conclude that the internal direct sum $\bigoplus_{n\in\mathbb{N}}\left(
f\left(  U\right)  \cap V_{n}\right)  $ is well-defined.

But $U$ is graded. Thus, $U=\bigoplus_{n\in\mathbb{N}}U_{n}=\sum
_{n\in\mathbb{N}}U_{n}$ (since direct sums are sums).

Also, the map $f$ is graded. In other words, every $n\in\mathbb{N}$ satisfies
\begin{equation}
f\left(  U_{n}\right)  \subseteq V_{n} \label{pf.rmk.graded.im.1}%
\end{equation}
(by the definition of a graded map). Hence, every $n\in\mathbb{N}$ satisfies%
\begin{equation}
f\left(  U_{n}\right)  \subseteq f\left(  U\right)  \cap V_{n}
\label{pf.rmk.graded.im.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.rmk.graded.im.2}):} Let $n\in\mathbb{N}$.
Then, (\ref{pf.rmk.graded.im.1}) yields $f\left(  U\right)  \subseteq V_{n}$.
Combining $f\left(  \underbrace{U_{n}}_{\subseteq U}\right)  \subseteq
f\left(  U\right)  $ with $f\left(  U\right)  \subseteq V_{n}$, we obtain
$f\left(  U_{n}\right)  \subseteq f\left(  U\right)  \cap V_{n}$. This proves
(\ref{pf.rmk.graded.im.2}).}.

Applying the map $f$ to both sides of the equality $U=\sum_{n\in\mathbb{N}%
}U_{n}$, we obtain%
\begin{align*}
f\left(  U\right)   &  =f\left(  \sum_{n\in\mathbb{N}}U_{n}\right)
=\sum_{n\in\mathbb{N}}\underbrace{f\left(  U_{n}\right)  }%
_{\substack{\subseteq f\left(  U\right)  \cap V_{n}\\\text{(by
(\ref{pf.rmk.graded.im.2}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}f\text{ is }k\text{-linear}\right) \\
&  \subseteq\sum_{n\in\mathbb{N}}\left(  f\left(  U\right)  \cap V_{n}\right)
.
\end{align*}
Combining this with%
\[
\sum_{n\in\mathbb{N}}\underbrace{\left(  f\left(  U\right)  \cap V_{n}\right)
}_{\subseteq f\left(  U\right)  }\subseteq\sum_{n\in\mathbb{N}}f\left(
U\right)  \subseteq f\left(  U\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}f\left(  U\right)  \text{ is a }k\text{-vector space}\right)  ,
\]
we obtain%
\begin{equation}
f\left(  U\right)  =\sum_{n\in\mathbb{N}}\left(  f\left(  U\right)  \cap
V_{n}\right)  . \label{pf.rmk.graded.im.5}%
\end{equation}


But recall that the internal direct sum $\bigoplus_{n\in\mathbb{N}}\left(
f\left(  U\right)  \cap V_{n}\right)  $ is well-defined. It satisfies%
\begin{align*}
\bigoplus_{n\in\mathbb{N}}\left(  f\left(  U\right)  \cap V_{n}\right)   &
=\sum_{n\in\mathbb{N}}\left(  f\left(  U\right)  \cap V_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are sums}\right) \\
&  =f\left(  U\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.rmk.graded.im.5})}\right)  .
\end{align*}
Thus, $f\left(  U\right)  =\bigoplus_{n\in\mathbb{N}}\left(  f\left(
U\right)  \cap V_{n}\right)  $.

But $f\left(  U\right)  $ is a $k$-vector subspace of $V$ (since
$f:U\rightarrow V$ is a $k$-linear map). Hence, $f\left(  U\right)  $ is a
homogeneous subspace of $V$ if and only if $f\left(  U\right)  =\bigoplus
_{n\in\mathbb{N}}\left(  f\left(  U\right)  \cap V_{n}\right)  $ (by the
definition of a \textquotedblleft homogeneous subspace\textquotedblright).
Thus, $f\left(  U\right)  $ is a homogeneous subspace of $V$ (since we know
that $f\left(  U\right)  =\bigoplus_{n\in\mathbb{N}}\left(  f\left(  U\right)
\cap V_{n}\right)  $). This proves Remark~\ref{rmk.graded.im}.
\end{proof}

We can now re-package Theorem~\ref{thm.leray2.main} into the following corollary:

\begin{corollary}
\label{cor.leray2.ver2} Let $k$ be a field of characteristic $0$. Let $H$ be a
connected graded commutative bialgebra over $k$.

Let $\mathfrak{d}:H\rightarrow H$ be a graded $k$-linear map that is a
projection (i.e., it satisfies $\mathfrak{d}\circ\mathfrak{d}=\mathfrak{d}$)
and satisfies $\operatorname*{Ker}\mathfrak{d}=H_{0}+\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}$.

Let $V=\mathfrak{d}\left(  H\right)  $. Let $\mathfrak{i}$ denote the
inclusion map $V\rightarrow H$. Let $\mathfrak{b}$ denote the $k$-algebra
homomorphism $\operatorname*{symlift}\mathfrak{i}:\operatorname*{Sym}%
V\rightarrow H$\ \ \ \ \footnote{We are using the notations introduced in
Definition~\ref{Definition1701}.}.

\textbf{(a)} This map $\mathfrak{b}$ is a $k$-algebra isomorphism.

\textbf{(b)} Let $\alpha:H\rightarrow V$ be the map defined by $\left(
\alpha\left(  x\right)  =\mathfrak{d}\left(  x\right)  \text{ for each }x\in
H\right)  $. It is easy to see that this map $\alpha$ is well-defined and
$k$-linear. Define a $k$-linear map $\mathfrak{a}:H\rightarrow
\operatorname*{Sym}V$ by $\mathfrak{a}=\operatorname*{syminc}\nolimits_{V}%
\circ\alpha$. Then, $\mathfrak{a}\in\mathfrak{g}\left(  H,\operatorname*{Sym}%
V\right)  $, and thus the $k$-linear map $e^{\ast\mathfrak{a}}:H\rightarrow
\operatorname*{Sym}V$ is well-defined. This map $e^{\ast\mathfrak{a}%
}:H\rightarrow\operatorname*{Sym}V$ is a $k$-algebra isomorphism.\footnote{The
maps $\mathfrak{b}$ and $e^{\ast\mathfrak{a}}$ are not mutually inverse in
general.}
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{cor.leray2.ver2}.]For each $m\in\mathbb{N}$, we
define a $k$-vector subspace $H_{>m}$ of $H$ by $H_{>m}=\sum_{\substack{g\in
\mathbb{N};\\g>m}}H_{g}$. Notice that $H=H_{0}+H_{>0}$%
\ \ \ \ \footnote{\textit{Proof.} The $k$-vector space $H$ is graded. Hence,
$H=\bigoplus_{g\in\mathbb{N}}H_{g}=\sum_{g\in\mathbb{N}}H_{g}$ (since direct
sums are sums).
\par
The definition of $H_{>0}$ yields $H_{>0}=\sum_{\substack{g\in\mathbb{N}%
;\\g>0}}H_{g}$. But
\begin{align*}
H  &  =\sum_{g\in\mathbb{N}}H_{g}=H_{0}+\underbrace{\sum_{\substack{g\in
\mathbb{N};\\g>0}}H_{g}}_{=H_{>0}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend}\\
\text{for }g=0\text{ from the sum}%
\end{array}
\right) \\
&  =H_{0}+H_{>0}.
\end{align*}
}.

Proposition~\ref{Proposition921} yields $H_{0}=k\cdot1_{H}$.

Remark~\ref{rmk.graded.im} (applied to $H$, $H$ and $\mathfrak{d}$ instead of
$U$, $V$ and $f$) shows that $\mathfrak{d}\left(  H\right)  $ is a homogeneous
subspace of $H$. In other words, $V$ is a homogeneous subspace of $H$ (since
$V=\mathfrak{d}\left(  H\right)  $).

Lemma~\ref{lem.leray2.H0ker} \textbf{(b)} (applied to $C=H$) shows that
$H_{>0}=\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $. Also, $\left(
H_{>0}\right)  ^{2}\subseteq H_{>0}$\ \ \ \ \footnote{\textit{Proof.} Recall
that $H$ is a $k$-bialgebra. Thus, the map $\varepsilon_{H}$ is a $k$-algebra
homomorphism (by the axioms of a $k$-bialgebra). Hence, its kernel
$\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $ is an ideal of $H$. In
other words, $H_{>0}$ is an ideal of $H$ (since $H_{>0}=\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $). Hence, $HH_{>0}\subseteq H_{>0}$ and
$H_{>0}H\subseteq H_{>0}$. Now, $\left(  H_{>0}\right)  ^{2}=H_{>0}%
\underbrace{H_{>0}}_{\subseteq H}\subseteq H_{>0}H\subseteq H_{>0}$.}.

Recall that the map $\mathfrak{d}$ is graded. Hence,
\begin{equation}
\mathfrak{d}\left(  H_{>0}\right)  \subseteq H_{>0}
\label{pf.cor.leray2.ver2.2}%
\end{equation}
\ \ \ \ \footnote{\textit{Proof.} The map $\mathfrak{d}$ is graded. In other
words, every $n\in\mathbb{N}$ satisfies
\begin{equation}
\mathfrak{d}\left(  H_{n}\right)  \subseteq H_{n}
\label{pf.cor.leray2.ver2.2.pf.1}%
\end{equation}
(by the definition of a graded map). Now, the definition of $H_{>0}$ yields
$H_{>0}=\sum_{\substack{g\in\mathbb{N};\\g>0}}H_{g}$. Applying the map
$\mathfrak{d}$ to both sides of this equality, we obtain%
\begin{align*}
\mathfrak{d}\left(  H_{>0}\right)   &  =\mathfrak{d}\left(  \sum
_{\substack{g\in\mathbb{N};\\g>0}}H_{g}\right)  =\sum_{\substack{g\in
\mathbb{N};\\g>0}}\underbrace{\mathfrak{d}\left(  H_{g}\right)  }%
_{\substack{\subseteq H_{g}\\\text{(by (\ref{pf.cor.leray2.ver2.2.pf.1}%
)}\\\text{(applied to }n=g\text{))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
the map }\mathfrak{d}\text{ is }k\text{-linear}\right) \\
&  \subseteq\sum_{\substack{g\in\mathbb{N};\\g>0}}H_{g}=H_{>0}.
\end{align*}
}.

Recall that $\mathfrak{d}$ is a projection. In other words, $\mathfrak{d}%
\circ\mathfrak{d}=\mathfrak{d}$.

Now, it is easy to see that $H_{>0}\subseteq V+\left(  H_{>0}\right)  ^{2}%
$\ \ \ \ \footnote{\textit{Proof.} Let $x\in H_{>0}$. The map $\mathfrak{d}$
is $k$-linear; thus, we have%
\[
\mathfrak{d}\left(  x-\mathfrak{d}\left(  x\right)  \right)  =\mathfrak{d}%
\left(  x\right)  -\underbrace{\mathfrak{d}\left(  \mathfrak{d}\left(
x\right)  \right)  }_{=\left(  \mathfrak{d}\circ\mathfrak{d}\right)  \left(
x\right)  }=\mathfrak{d}\left(  x\right)  -\underbrace{\left(  \mathfrak{d}%
\circ\mathfrak{d}\right)  }_{=\mathfrak{d}}\left(  x\right)  =\mathfrak{d}%
\left(  x\right)  -\mathfrak{d}\left(  x\right)  =0.
\]
In other words, $x-\mathfrak{d}\left(  x\right)  \in\operatorname*{Ker}%
\mathfrak{d}$. In light of $\operatorname*{Ker}\mathfrak{d}=H_{0}+\left(
\underbrace{\operatorname*{Ker}\left(  \varepsilon_{H}\right)  }_{=H_{>0}%
}\right)  ^{2}=H_{0}+\left(  H_{>0}\right)  ^{2}$, this rewrites as
$x-\mathfrak{d}\left(  x\right)  \in H_{0}+\left(  H_{>0}\right)  ^{2}$. In
other words, there exist $y\in H_{0}$ and $z\in\left(  H_{>0}\right)  ^{2}$
such that $x-\mathfrak{d}\left(  x\right)  =y+z$. Consider these $y$ and $z$.
\par
We shall show that $y=0$. Indeed, we have $y\in H_{0}=k\cdot1_{H}$; thus,
there exists some $\lambda\in k$ such that $y=\lambda\cdot1_{H}$. Consider
this $\lambda$.
\par
We have $z\in\left(  H_{>0}\right)  ^{2}\subseteq H_{>0}=\operatorname*{Ker}%
\left(  \varepsilon_{H}\right)  $ and thus $\varepsilon_{H}\left(  z\right)
=0$.
\par
Recall that $H$ is a $k$-bialgebra. Hence, $\varepsilon_{H}\left(
1_{H}\right)  =1$ (by the axioms of a $k$-bialgebra). Now,%
\begin{align*}
\varepsilon_{H}\left(  \underbrace{y}_{=\lambda\cdot1_{H}}\right)   &
=\varepsilon_{H}\left(  \lambda\cdot1_{H}\right)  =\lambda\cdot
\underbrace{\varepsilon_{H}\left(  1_{H}\right)  }_{=1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\varepsilon_{H}\text{ is
}k\text{-linear}\right) \\
&  =\lambda.
\end{align*}
\par
We have $\mathfrak{d}\left(  \underbrace{x}_{\in H_{>0}}\right)
\subseteq\mathfrak{d}\left(  H_{>0}\right)  \subseteq H_{>0}$ (by
(\ref{pf.cor.leray2.ver2.2})). Thus, $\mathfrak{d}\left(  x\right)  \in
H_{>0}=\operatorname*{Ker}\left(  \varepsilon_{H}\right)  $, so that
$\varepsilon_{H}\left(  \mathfrak{d}\left(  x\right)  \right)  =0$. Now, let
us apply the map $\varepsilon_{H}$ to both sides of the equality
$x-\mathfrak{d}\left(  x\right)  =y+z$. We thus obtain%
\begin{align*}
\varepsilon_{H}\left(  x-\mathfrak{d}\left(  x\right)  \right)   &
=\varepsilon_{H}\left(  y+z\right)  =\underbrace{\varepsilon_{H}\left(
y\right)  }_{=\lambda}+\underbrace{\varepsilon_{H}\left(  z\right)  }%
_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\varepsilon\text{ is
}k\text{-linear}\right) \\
&  =\lambda.
\end{align*}
Hence,%
\[
\lambda=\varepsilon_{H}\left(  x-\mathfrak{d}\left(  x\right)  \right)
=\underbrace{\varepsilon_{H}\left(  x\right)  }_{\substack{=0\\\text{(since
}x\in H_{>0}=\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \text{)}%
}}-\underbrace{\varepsilon_{H}\left(  \mathfrak{d}\left(  x\right)  \right)
}_{=0}=0.
\]
Hence, $y=\underbrace{\lambda}_{=0}\cdot1_{H}=0$. Thus, $x-\mathfrak{d}\left(
x\right)  =\underbrace{y}_{=0}+z=z$, so that $x=\mathfrak{d}\left(
\underbrace{x}_{\in H}\right)  +\underbrace{z}_{\in\left(  H_{>0}\right)
^{2}}\in\underbrace{\mathfrak{d}\left(  H\right)  }_{=V}+\left(
H_{>0}\right)  ^{2}=V+\left(  H_{>0}\right)  ^{2}$.
\par
Now, forget that we fixed $x$. We thus have shown that each $x\in H_{>0}$
satisfies $x\in V+\left(  H_{>0}\right)  ^{2}$. In other words, $H_{>0}%
\subseteq V+\left(  H_{>0}\right)  ^{2}$.} and $V+\left(  H_{>0}\right)
^{2}\subseteq H_{>0}$\ \ \ \ \footnote{\textit{Proof.} We have $H_{0}\subseteq
H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}=\operatorname*{Ker}\mathfrak{d}$ and thus $\mathfrak{d}\left(
H_{0}\right)  =0$.
\par
But%
\begin{align*}
V  &  =\mathfrak{d}\left(  \underbrace{H}_{=H_{0}+H_{>0}}\right)
=\mathfrak{d}\left(  H_{0}+H_{>0}\right)  =\underbrace{\mathfrak{d}\left(
H_{0}\right)  }_{=0}+\mathfrak{d}\left(  H_{>0}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mathfrak{d}\text{ is
}k\text{-linear}\right) \\
&  =\mathfrak{d}\left(  H_{>0}\right)  \subseteq H_{>0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.cor.leray2.ver2.2})}\right)  .
\end{align*}
Now,%
\[
\underbrace{V}_{\subseteq H_{>0}}+\underbrace{\left(  H_{>0}\right)  ^{2}%
}_{\subseteq H_{>0}}\subseteq H_{>0}+H_{>0}\subseteq H_{>0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }H_{>0}\text{ is a }k\text{-vector
space}\right)  .
\]
}. Combining these two relations, we obtain $H_{>0}=V+\left(  H_{>0}\right)
^{2}$.

But $V\cap\left(  H_{>0}\right)  ^{2}=0$\ \ \ \ \footnote{\textit{Proof.} Let
$x\in V\cap\left(  H_{>0}\right)  ^{2}$. Then, $x\in V\cap\left(
H_{>0}\right)  ^{2}\subseteq V=\mathfrak{d}\left(  H\right)  $. Thus, there
exists some $y\in H$ such that $x=\mathfrak{d}\left(  y\right)  $. Consider
this $y$.
\par
Now, $\mathfrak{d}\left(  \underbrace{x}_{=\mathfrak{d}\left(  y\right)
}\right)  =\mathfrak{d}\left(  \mathfrak{d}\left(  y\right)  \right)
=\underbrace{\left(  \mathfrak{d}\circ\mathfrak{d}\right)  }_{=\mathfrak{d}%
}\left(  y\right)  =\mathfrak{d}\left(  y\right)  =x$.
\par
But $x\in V\cap\left(  H_{>0}\right)  ^{2}\subseteq\left(  \underbrace{H_{>0}%
}_{=\operatorname*{Ker}\left(  \varepsilon_{H}\right)  }\right)  ^{2}=\left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}\subseteq
H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}=\operatorname*{Ker}\mathfrak{d}$, so that $\mathfrak{d}\left(  x\right)
=0$. Comparing this with $\mathfrak{d}\left(  x\right)  =x$, we obtain $x=0$.
\par
Now, forget that we fixed $x$. We thus have shown that every $x\in
V\cap\left(  H_{>0}\right)  ^{2}$ satisfies $x=0$. In other words,
$V\cap\left(  H_{>0}\right)  ^{2}=0$.}. Hence, the internal direct sum
$V\oplus\left(  H_{>0}\right)  ^{2}$ is well-defined. This internal direct sum
satisfies $V\oplus\left(  H_{>0}\right)  ^{2}=V+\left(  H_{>0}\right)  ^{2}$.
Comparing this with $H_{>0}=V+\left(  H_{>0}\right)  ^{2}$, we obtain
$H_{>0}=V\oplus\left(  H_{>0}\right)  ^{2}$. Theorem~\ref{thm.leray2.main}
\textbf{(a)} thus yields that the map $\mathfrak{b}$ is a $k$-algebra
isomorphism. This proves Corollary~\ref{cor.leray2.ver2} \textbf{(a)}.

\textbf{(b)} Let us first prove that the map $\alpha$ is well-defined and $k$-linear.

Every $x\in H$ satisfies $\mathfrak{d}\left(  \underbrace{x}_{\in H}\right)
\in\mathfrak{d}\left(  H\right)  =V$. Thus, the map $\alpha:H\rightarrow V$ is
well-defined (since this map is defined by $\left(  \alpha\left(  x\right)
=\mathfrak{d}\left(  x\right)  \text{ for each }x\in H\right)  $).
Furthermore, the map $\alpha$ is $k$-linear\footnote{\textit{Proof.} This map
$\alpha$ is obtained from $\mathfrak{d}$ by restricting the codomain to $V$
(since $\alpha\left(  x\right)  =\mathfrak{d}\left(  x\right)  $ for every
$x\in H$). Hence, this map $\alpha$ is $k$-linear (since the map
$\mathfrak{d}$ is $k$-linear).}.

Now, it remains to show that $\mathfrak{a}\in\mathfrak{g}\left(
H,\operatorname*{Sym}V\right)  $, and that the map $e^{\ast\mathfrak{a}%
}:H\rightarrow\operatorname*{Sym}V$ is a $k$-algebra isomorphism.

For every subset $T$ of $H$, we have%
\[
\alpha\left(  T\right)  =\left\{  \underbrace{\alpha\left(  x\right)
}_{\substack{=\mathfrak{d}\left(  x\right)  \\\text{(by the definition of
}\alpha\text{)}}}\ \mid\ x\in T\right\}  =\left\{  \mathfrak{d}\left(
x\right)  \ \mid\ x\in T\right\}  =\mathfrak{d}\left(  T\right)  .
\]
Applying this to $T=\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}$, we obtain%
\begin{align*}
\alpha\left(  \left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+k\cdot1_{H}\right)   &  =\mathfrak{d}\left(  \left(
\operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)  ^{2}%
+\underbrace{k\cdot1_{H}}_{=H_{0}}\right)  =\mathfrak{d}\left(
\underbrace{\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)
\right)  ^{2}+H_{0}}_{=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon
_{H}\right)  \right)  ^{2}=\operatorname*{Ker}\mathfrak{d}}\right) \\
&  =\mathfrak{d}\left(  \operatorname*{Ker}\mathfrak{d}\right)  =0.
\end{align*}
Moreover, $\alpha\mid_{V}=\operatorname*{id}\nolimits_{V}$%
\ \ \ \footnote{\textit{Proof.} Let $x\in V$. The definition of $\alpha$
yields $\alpha\left(  x\right)  =\mathfrak{d}\left(  x\right)  $.
\par
But $x\in V=\mathfrak{d}\left(  H\right)  $. Thus, there exists some $y\in H$
such that $x=\mathfrak{d}\left(  y\right)  $. Consider this $y$.
\par
Now, $\mathfrak{d}\left(  \underbrace{x}_{=\mathfrak{d}\left(  y\right)
}\right)  =\mathfrak{d}\left(  \mathfrak{d}\left(  y\right)  \right)
=\underbrace{\left(  \mathfrak{d}\circ\mathfrak{d}\right)  }_{=\mathfrak{d}%
}\left(  y\right)  =\mathfrak{d}\left(  y\right)  =x$. Hence, $\alpha\left(
x\right)  =\mathfrak{d}\left(  x\right)  =x$.
\par
Finally, $x\in V$, so that $\left(  \alpha\mid_{V}\right)  \left(  x\right)
=\alpha\left(  x\right)  =x=\operatorname*{id}\nolimits_{V}\left(  x\right)
$.
\par
Now, forget that we fixed $x$. We thus have shown that $\left(  \alpha\mid
_{V}\right)  \left(  x\right)  =\operatorname*{id}\nolimits_{V}\left(
x\right)  $ for each $x\in V$. In other words, $\alpha\mid_{V}%
=\operatorname*{id}\nolimits_{V}$.}. Theorem~\ref{thm.leray2.main}
\textbf{(b)} thus shows that $\mathfrak{a}\in\mathfrak{g}\left(
H,\operatorname*{Sym}V\right)  $, and that the map $e^{\ast\mathfrak{a}%
}:H\rightarrow\operatorname*{Sym}V$ is a $k$-algebra isomorphism. This proves
Corollary~\ref{cor.leray2.ver2} \textbf{(b)}.
\end{proof}

Corollary~\ref{cor.leray2.ver2} can be applied to various choices of the map
$\mathfrak{d}$. One obvious choice is the Eulerian idempotent
$\operatorname*{Log}\operatorname*{id}$; however, this results in nothing new,
but just a weaker version of Theorem~\ref{Theorem1702} \textbf{(d)}%
\footnote{It is a weaker version for two reasons: Firstly, it requires $H$ to
be graded (not just filtered); secondly, it does not prove that the maps
$\operatorname*{symlift}\mathfrak{j}$ and $e^{\ast\mathfrak{q}}$ are mutually
inverse.}. However, we can also apply it to the Dynkin idempotents, and then
we obtain something new. Actually, let us first extend
Theorem~\ref{Theorem931}:

\begin{theorem}
\label{thm.leray2.931} Let $k$ be a field of characteristic $0$. Let $H$ be a
commutative graded $k$-bialgebra. Let $E_{H}$ be defined according to
Definition~\ref{Definition701}.

Let $P:H\rightarrow H$ and $Q:H\rightarrow H$ be two graded $k$-algebra
homomorphisms satisfying $\varepsilon_{H}\circ P=\varepsilon_{H}$,
$\varepsilon_{H}\circ Q=\varepsilon_{H}$ and $P\ast\operatorname{id}_{H}\ast
Q=e_{H,H}$. Here, the map $e_{H,H}$ is defined to be the map $\eta_{H}%
\circ\varepsilon_{H}:H\rightarrow H$ (this definition of the map $e_{H,H}$ is
identical with the definition of the map $e_{H,A}$ in Definition~\ref{def.eHA}).

Let $\mathfrak{d}$ be the map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast
E_{H}\ast Q\right)  :H\rightarrow H$.

\textbf{(a)} The map $\mathfrak{d}$ is a projection (i.e., it satisfies
$\mathfrak{d}\circ\mathfrak{d}=\mathfrak{d}$) and satisfies
$\operatorname*{Ker}\mathfrak{d}=H_{0}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$.

\textbf{(b)} The map $\mathfrak{d}$ is graded.

Let $V=\mathfrak{d}\left(  H\right)  $. Let $\mathfrak{i}$ denote the
inclusion map $V\rightarrow H$. Let $\mathfrak{b}$ denote the $k$-algebra
homomorphism $\operatorname*{symlift}\mathfrak{i}:\operatorname*{Sym}%
V\rightarrow H$\ \ \ \ \footnote{We are using the notations introduced in
Definition~\ref{Definition1701}.}.

\textbf{(c)} This map $\mathfrak{b}$ is a $k$-algebra isomorphism.

\textbf{(d)} Let $\alpha:H\rightarrow V$ be the map defined by $\left(
\alpha\left(  x\right)  =\mathfrak{d}\left(  x\right)  \text{ for each }x\in
H\right)  $. It is easy to see that this map $\alpha$ is well-defined and
$k$-linear. Define a $k$-linear map $\mathfrak{a}:H\rightarrow
\operatorname*{Sym}V$ by $\mathfrak{a}=\operatorname*{syminc}\nolimits_{V}%
\circ\alpha$. Then, $\mathfrak{a}\in\mathfrak{g}\left(  H,\operatorname*{Sym}%
V\right)  $, and thus the $k$-linear map $e^{\ast\mathfrak{a}}:H\rightarrow
\operatorname*{Sym}V$ is well-defined. This map $e^{\ast\mathfrak{a}%
}:H\rightarrow\operatorname*{Sym}V$ is a $k$-algebra isomorphism.\footnote{The
maps $\mathfrak{b}$ and $e^{\ast\mathfrak{a}}$ are not mutually inverse in
general.}
\end{theorem}

\begin{proof}
[Proof of Theorem~\ref{thm.leray2.931}.]Theorem~\ref{Theorem931} shows that
the map $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  $
is a projection such that $\operatorname*{Ker}\left(  E_{H}%
^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  \right)
=H_{0}+\left(  \operatorname*{Ker}\left(  \varepsilon_{H}\right)  \right)
^{2}$. Since $E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast
Q\right)  =\mathfrak{d}$ (because $\mathfrak{d}$ was defined to be
$E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)  $), this
rewrites as follows: The map $\mathfrak{d}$ is a projection such that
$\operatorname*{Ker}\mathfrak{d}=H_{0}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$. This proves
Theorem~\ref{thm.leray2.931} \textbf{(a)}.

In the proof of Theorem~\ref{Theorem912}, we have shown that the map $P\ast
E_{H}\ast Q$ is graded. Furthermore, Corollary~\ref{Corollary724} \textbf{(c)}
(applied to $H$ instead of $V$) shows that the map $E_{H}^{\operatorname*{inv}%
}$ is graded.

Now, recall that $\mathfrak{d}=E_{H}^{\operatorname*{inv}}\circ\left(  P\ast
E_{H}\ast Q\right)  $. Hence, the map $\mathfrak{d}$ is the composition of the
two graded maps $E_{H}^{\operatorname*{inv}}$ and $P\ast E_{H}\ast Q$ (since
we know that both maps $E_{H}^{\operatorname*{inv}}$ and $P\ast E_{H}\ast Q$
are graded), and therefore itself is a graded map (since any composition of
two graded maps must be a graded map). This proves
Theorem~\ref{thm.leray2.931} \textbf{(b)}.

We thus conclude that the hypotheses of Corollary~\ref{cor.leray2.ver2} are satisfied.

\textbf{(c)} Thus, Corollary~\ref{cor.leray2.ver2} \textbf{(a)} shows that the
map $\mathfrak{b}$ is a $k$-algebra isomorphism. This proves
Theorem~\ref{thm.leray2.931} \textbf{(c)}.

\textbf{(d)} Furthermore, Corollary~\ref{cor.leray2.ver2} \textbf{(b)} shows
that the map $\alpha$ is well-defined and $k$-linear, and that $\mathfrak{a}%
\in\mathfrak{g}\left(  H,\operatorname*{Sym}V\right)  $, and that the map
$e^{\ast\mathfrak{a}}:H\rightarrow\operatorname*{Sym}V$ is a $k$-algebra
isomorphism. This proves Theorem~\ref{thm.leray2.931} \textbf{(d)}.
\end{proof}

Let us now specialize Theorem~\ref{thm.leray2.931} to the actual Dynkin idempotents:

\begin{corollary}
\label{cor.leray2.dynkin} Let $k$ be a field of characteristic $0$. Let $H$ be
a commutative graded $k$-Hopf algebra. Let $S$ be the antipode of $H$. Define
a map $E_{H}:H\rightarrow H$ according to Definition~\ref{Definition701}.
Define a map $E_{H}^{\operatorname*{inv}}:H\rightarrow H$ according to
Definition~\ref{Definition721}.

Let $\mathfrak{d}:H\rightarrow H$ be one of the two maps $E_{H}%
^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  $ and $E_{H}%
^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  $.

\textbf{(a)} The map $\mathfrak{d}$ is a projection (i.e., it satisfies
$\mathfrak{d}\circ\mathfrak{d}=\mathfrak{d}$) and satisfies
$\operatorname*{Ker}\mathfrak{d}=H_{0}+\left(  \operatorname*{Ker}\left(
\varepsilon_{H}\right)  \right)  ^{2}$.

\textbf{(b)} The map $\mathfrak{d}$ is graded.

Let $V=\mathfrak{d}\left(  H\right)  $. Let $\mathfrak{i}$ denote the
inclusion map $V\rightarrow H$. Let $\mathfrak{b}$ denote the $k$-algebra
homomorphism $\operatorname*{symlift}\mathfrak{i}:\operatorname*{Sym}%
V\rightarrow H$\ \ \ \ \footnote{We are using the notations introduced in
Definition~\ref{Definition1701}.}.

\textbf{(c)} This map $\mathfrak{b}$ is a $k$-algebra isomorphism.

\textbf{(d)} Let $\alpha:H\rightarrow V$ be the map defined by $\left(
\alpha\left(  x\right)  =\mathfrak{d}\left(  x\right)  \text{ for each }x\in
H\right)  $. It is easy to see that this map $\alpha$ is well-defined and
$k$-linear. Define a $k$-linear map $\mathfrak{a}:H\rightarrow
\operatorname*{Sym}V$ by $\mathfrak{a}=\operatorname*{syminc}\nolimits_{V}%
\circ\alpha$. Then, $\mathfrak{a}\in\mathfrak{g}\left(  H,\operatorname*{Sym}%
V\right)  $, and thus the $k$-linear map $e^{\ast\mathfrak{a}}:H\rightarrow
\operatorname*{Sym}V$ is well-defined. This map $e^{\ast\mathfrak{a}%
}:H\rightarrow\operatorname*{Sym}V$ is a $k$-algebra isomorphism.\footnote{The
maps $\mathfrak{b}$ and $e^{\ast\mathfrak{a}}$ are not mutually inverse in
general.}
\end{corollary}

\begin{proof}
[Proof of Corollary~\ref{cor.leray2.dynkin}.]Just as in the proof of
Corollary~\ref{Corollary908}, we can prove the following facts:

\begin{itemize}
\item The map $S$ is a $k$-algebra homomorphism from $H$ to $H$.

\item The map $e_{H,H}$ is a $k$-algebra homomorphism (where $e_{H,H}$ is
defined as according to Definition~\ref{def.eHA}).

\item We have $S=\operatorname{id}_{H}^{\ast\left(  -1\right)  }$.

\item We have $S\ast\operatorname{id}_{H}\ast e_{H,H}=e_{H,H}$ and
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}$.
\end{itemize}

Just as in the proof of Corollary~\ref{Corollary913}, we can prove the
following facts:

\begin{itemize}
\item We have $\varepsilon_{H}\circ e_{H,H}=\varepsilon_{H}$ and
$\varepsilon_{H}\circ S=\varepsilon_{H}$.

\item The maps $e_{H,H}$ and $S$ are graded.
\end{itemize}

Now,
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{there exist two graded }k\text{-algebra homomorphisms }P:H\rightarrow
H\\
\text{and }Q:H\rightarrow H\text{ satisfying }\varepsilon_{H}\circ
P=\varepsilon_{H}\text{, }\varepsilon_{H}\circ Q=\varepsilon_{H}\text{,}\\
P\ast\operatorname{id}_{H}\ast Q=e_{H,H}\text{ and }\mathfrak{d}%
=E_{H}^{\operatorname*{inv}}\circ\left(  P\ast E_{H}\ast Q\right)
\end{array}
\right)  \label{pf.cor.leray2.dynkin.existsPQ}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.leray2.dynkin.existsPQ}):} Recall that
$e_{H,H}$ is the unity of the convolution algebra $\mathcal{L}\left(
H,H\right)  $. Thus, $e_{H,H}\ast E_{H}=E_{H}$ and $E_{H}\ast e_{H,H}=E_{H}$.
\par
We know that $\mathfrak{d}$ is one of the two maps $E_{H}^{\operatorname*{inv}%
}\circ\left(  E_{H}\ast S\right)  $ and $E_{H}^{\operatorname*{inv}}%
\circ\left(  S\ast E_{H}\right)  $. Thus, we are in one of the following two
cases:
\par
\textit{Case 1:} We have $\mathfrak{d}=E_{H}^{\operatorname*{inv}}\circ\left(
E_{H}\ast S\right)  $.
\par
\textit{Case 2:} We have $\mathfrak{d}=E_{H}^{\operatorname*{inv}}\circ\left(
S\ast E_{H}\right)  $.
\par
Let us first consider Case 1. In this case, we have $\mathfrak{d}%
=E_{H}^{\operatorname*{inv}}\circ\left(  E_{H}\ast S\right)  $. Thus,
$\mathfrak{d}=E_{H}^{\operatorname*{inv}}\circ\left(  \underbrace{E_{H}%
}_{=e_{H,H}\ast E_{H}}\ast S\right)  =E_{H}^{\operatorname*{inv}}\circ\left(
e_{H,H}\ast E_{H}\ast S\right)  $.
\par
Altogether, we thus know that $e_{H,H}:H\rightarrow H$ and $S:H\rightarrow H$
are two graded $k$-algebra homomorphisms satisfying $\varepsilon_{H}\circ
e_{H,H}=\varepsilon_{H}$, $\varepsilon_{H}\circ S=\varepsilon_{H}$,
$e_{H,H}\ast\operatorname{id}_{H}\ast S=e_{H,H}$ and $\mathfrak{d}%
=E_{H}^{\operatorname*{inv}}\circ\left(  e_{H,H}\ast E_{H}\ast S\right)  $.
Hence, there exist two graded $k$-algebra homomorphisms $P:H\rightarrow H$ and
$Q:H\rightarrow H$ satisfying $\varepsilon_{H}\circ P=\varepsilon_{H}$,
$\varepsilon_{H}\circ Q=\varepsilon_{H}$, $P\ast\operatorname{id}_{H}\ast
Q=e_{H,H}$ and $\mathfrak{d}=E_{H}^{\operatorname*{inv}}\circ\left(  P\ast
E_{H}\ast Q\right)  $ (namely, $P=e_{H,H}$ and $Q=S$). Thus,
(\ref{pf.cor.leray2.dynkin.existsPQ}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $\mathfrak{d}=E_{H}%
^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\right)  $. Thus, $\mathfrak{d}%
=E_{H}^{\operatorname*{inv}}\circ\left(  S\ast\underbrace{E_{H}}_{=E_{H}\ast
e_{H,H}}\right)  =E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\ast
e_{H,H}\right)  $.
\par
Altogether, we thus know that $S:H\rightarrow H$ and $e_{H,H}:H\rightarrow H$
are two graded $k$-algebra homomorphisms satisfying $\varepsilon_{H}\circ
S=\varepsilon_{H}$, $\varepsilon_{H}\circ e_{H,H}=\varepsilon_{H}$,
$S\ast\operatorname{id}_{H}\ast e_{H,H}=e_{H,H}$ and $\mathfrak{d}%
=E_{H}^{\operatorname*{inv}}\circ\left(  S\ast E_{H}\ast e_{H,H}\right)  $.
Hence, there exist two graded $k$-algebra homomorphisms $P:H\rightarrow H$ and
$Q:H\rightarrow H$ satisfying $\varepsilon_{H}\circ P=\varepsilon_{H}$,
$\varepsilon_{H}\circ Q=\varepsilon_{H}$, $P\ast\operatorname{id}_{H}\ast
Q=e_{H,H}$ and $\mathfrak{d}=E_{H}^{\operatorname*{inv}}\circ\left(  P\ast
E_{H}\ast Q\right)  $ (namely, $P=S$ and $Q=e_{H,H}$). Thus,
(\ref{pf.cor.leray2.dynkin.existsPQ}) is proven in Case 2.
\par
We have thus proven (\ref{pf.cor.leray2.dynkin.existsPQ}) in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, we thus conclude
that (\ref{pf.cor.leray2.dynkin.existsPQ}) always holds.}. Consider these $P$
and $Q$. Thus, the hypotheses of Theorem~\ref{thm.leray2.931} are satisfied.
Hence, the four parts of Corollary~\ref{cor.leray2.dynkin} follow immediately
from the corresponding four parts of Theorem~\ref{thm.leray2.931}.
\end{proof}

\section{\label{S-commring2}A final remark on commutative rings}

Let us make one final remark about generalizing the preceding results:

In \S \ref{S11}, we noticed that most of the results in \S \ref{S0a}%
-\S \ref{S10}, and most of the proofs, do not require $k$ to be a field; only
minor modifications are required to make them work when $k$ is just a
commutative ring with unity. Something similar can be said about the results
of \S \ref{S12}-\S \ref{S63}: All theorems stated in \S \ref{S12}-\S \ref{S63}
still hold if $k$ is just a commutative ring with unity, as long as the
following replacements are made:

\begin{itemize}
\item Any occurrence of ``$k$-vector space'' must be replaced by ``$k$-module''.

\item Any requirement that $k$ be a field of characteristic $0$ must be
replaced by a requirement that $k$ be a commutative $\mathbb{Q}$-algebra.

\item The definition of the notion of a ``filtered $k$-coalgebra'' has to be
replaced by the Definition~\ref{Definition80}.
\end{itemize}

Moreover, all of the proofs in \S \ref{S12}-\S \ref{S63} can be easily
adjusted to the general case when $k$ is just a commutative ring with unity.
The meaning of the word \textquotedblleft adjust\textquotedblright\ here (i.
e., the exact changes that have to be made to the proofs) is the same as the
one explained in \S \ref{S11}.

Note that, while the notion of a \textquotedblleft filtered $k$%
-coalgebra\textquotedblright\ took a little bit of thinking to be adjusted to
the case when $k$ is a ring, the notion of a \textquotedblleft graded
$k$-coalgebra\textquotedblright\ generalizes in the most obvious way, since
direct sums commute with tensor products over arbitrary rings (not only over fields).

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[Mancho06]{1}Dominique Manchon, \textit{Hopf algebras, from basics to
applications to renormalization}, Revised and updated version, may 2006,
arXiv:math/0408405v2.\newline\url{http://arxiv.org/abs/math/0408405v2}

\bibitem[PatReu98]{2}Fr\'{e}d\'{e}ric Patras and Christophe Reutenauer,
\textit{Higher Lie idempotents}, November 18, 1998.\newline\url{http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.25.1416}

\bibitem[QEDMO09]{3}\textit{6th QEDMO 2009, Problem 4 (the Cauchy identity)},
with Solution by Darij Grinberg.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/QEDMO6P4long.pdf}

\bibitem[PatReu00]{4}Fr\'{e}d\'{e}ric Patras and Christophe Reutenauer,
\textit{On Dynkin and Klyacho idempotents in graded bialgebras}.\newline\url{http://www-irma.u-strasbg.fr/annexes/publications/pdf/01029.pdf}

\bibitem[DMTCN13]{5}G\'{e}rard Henry Edmond Duchamp, Vincel Hoang Ngoc Minh,
Christophe Tollu, B\`{u}i Chi\^{e}n, Nguyen Hoang Nghia, \textit{Combinatorics
of deformed shuffle Hopf algebras}, arXiv:1302.5391v7.\newline\url{http://arxiv.org/abs/1302.5391v7}

\bibitem[EGHLSVY]{6}Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai
Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina,
\textit{Introduction to representation theory}, arXiv:0901.0827v5.\newline\url{http://arxiv.org/abs/0901.0827v5}

\bibitem[Grinbe08]{7}Darij Grinberg, \textit{St. Petersburg 2003: An
alternating sum of zero-sum subset numbers}, version 14 March 2008.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/StPeters2003.pdf}

\bibitem[DPR13]{8}Persi Diaconis, C. Y. Amy Pang, Arun Ram, \textit{Hopf
algebras and Markov chains: Two examples and a theory}, version of 30 December
2014.\newline\url{https://amypang.github.io/papers/hpmc.pdf}

\bibitem[Cartie06]{9}Pierre Cartier, \textit{A Primer of Hopf algebras},
September 2006, IHES/M/06/40.\newline\url{http://preprints.ihes.fr/2006/M/M-06-40.pdf}

\bibitem[AguLau14]{AguLau14}Marcelo Aguiar, Aaron Lauve, \textit{The
characteristic polynomial of the Adams operators on graded connected Hopf
algebras}, Algebra Number Theory 9 (2015), pp. 547--583. A preprint is
available at \href{https://arxiv.org/abs/1403.7584v2}{arXiv:1403.7584v2}.

\bibitem[Grinbe17]{detnotes}Darij Grinberg, \textit{Notes on the combinatorial
fundamentals of algebra}, Version of 25 May 2021.\newline%
\url{http://www.cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf} \newline The
numbering of theorems and formulas in this link might shift when the project
gets updated; for a \textquotedblleft frozen\textquotedblright\ version whose
numbering is guaranteed to match that in the citations above, see
\url{https://github.com/darijgr/detnotes/releases/tag/2021-05-25} or
\href{https://arxiv.org/abs/2008.09862v2}{arXiv:2008.09862v2}.

\bibitem[Grinbe15]{tensorext}Darij Grinberg, \textit{A few classical results
on tensor, symmetric and exterior powers}, version 0.3 (June 13,
2017).\newline\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/tensorext.pdf}

\bibitem[Schnei15]{Schneider-Hopf}Hans-J\"{u}rgen Schneider,
\textit{Hopfalgebren} (lecture notes taken by Darij Grinberg).\newline\url{https://sites.google.com/site/darijgrinberg/hopfalgebren}
\end{thebibliography}


\end{document}