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\ihead{On the square of the antipode}
\ohead{page \thepage}
\cfoot{}
\begin{document}
\title{On the square of the antipode in a connected filtered Hopf algebra [detailed version]}
\author{Darij Grinberg}
\date{
%TCIMACRO{\TeXButton{today}{\today}}%
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\maketitle
\begin{abstract}
\textbf{Abstract.} It is well-known that the antipode $S$ of a commutative or
cocommutative Hopf algebra satisfies $S^{2}=\operatorname*{id}$ (where
$S^{2}=S\circ S$). Recently, similar results have been obtained by Aguiar and
Lauve for connected graded Hopf algebras: Namely, if $H$ is a connected graded
Hopf algebra with grading $H=\bigoplus_{n\geq0}H_{n}$, then each positive
integer $n$ satisfies $\left( \operatorname*{id}-S^{2}\right) ^{n}\left(
H_{n}\right) =0$ and (even stronger) $\left( \left( \operatorname*{id}%
+S\right) \circ\left( \operatorname*{id}-S^{2}\right) ^{n-1}\right)
\left( H_{n}\right) =0$. For some specific $H$'s such as the
Malvenuto--Reutenauer Hopf algebra $\operatorname*{FQSym}$, the exponents can
be lowered.
In this note, we generalize these results in several directions: We replace
the base field by a commutative ring, replace the Hopf algebra by a coalgebra
(actually, a slightly more general object, with no coassociativity required),
and replace both $\operatorname*{id}$ and $S^{2}$ by \textquotedblleft
coalgebra homomorphisms\textquotedblright\ (of sorts). Specializing back to
connected graded Hopf algebras, we show that the exponent $n$ in Aguiar's
identity $\left( \operatorname*{id}-S^{2}\right) ^{n}\left( H_{n}\right)
=0$ can be lowered to $n-1$ (for $n>1$) if and only if $\left(
\operatorname*{id}-S^{2}\right) \left( H_{2}\right) =0$. (A sufficient
condition for this is that every pair of elements of $H_{1}$ commutes; this is
satisfied, e.g., for $\operatorname*{FQSym}$.)
\textbf{Keywords:} Hopf algebra, antipode, connected graded Hopf algebra,
combinatorial Hopf algebra.
\textbf{MSC subject classification:} 16T05, 16T30.
\end{abstract}
Consider, for simplicity, a connected graded Hopf algebra $H$ over a field (we
will soon switch to more general settings). Let $S$ be the antipode of $H$. A
classical result (e.g., \cite[Proposition 4.0.1 6)]{Sweedler} or
\cite[Corollary 3.3.11]{HaGuKi10} or \cite[Theorem 2.1.4 (vi)]{Abe80} or
\cite[Corollary 7.1.11]{Radfor12}) says that $S^{2}=\operatorname*{id}$ when
$H$ is commutative or cocommutative. (Here and in the following, powers are
composition powers; thus, $S^{2}$ means $S\circ S$.) In general,
$S^{2}=\operatorname*{id}$ need not hold. However, in \cite[Proposition
7]{AguLau14}, Aguiar and Lauve showed that $S^{2}$ is still locally unipotent,
and more precisely, we have
\[
\left( \operatorname*{id}-S^{2}\right) ^{n}\left( H_{n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for each }n>0,
\]
where $H_{n}$ denotes the $n$-th graded component of $H$. Later, Aguiar
\cite[Lemma 12.50]{Aguiar17} strengthened this equality to
\[
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{n-1}\right) \left( H_{n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for each }n>0.
\]
For specific combinatorially interesting Hopf algebras, even stronger results
hold; in particular,
\[
\left( \operatorname*{id}-S^{2}\right) ^{n-1}\left( H_{n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{holds for each }n>1
\]
when $H$ is the Malvenuto--Reutenauer Hopf algebra (see \cite[Example
8]{AguLau14}).
In this note, we will unify these results and transport them to a much more
general setting. First of all, the ground field will be replaced by an
arbitrary commutative ring; this generalization is not unexpected, but renders
the proof strategy of \cite[Proposition 7]{AguLau14} insufficient\footnote{In
fact, the proof in \cite[Proposition 7]{AguLau14} relies on the coradical
filtration of $H$ and its associated graded structure $\operatorname*{gr}H$.
If the base ring is a field, then $\operatorname*{gr}H$ is a well-defined
commutative Hopf algebra (see, e.g., \cite[Lemma 1]{AguLau14}), and thus the
antipode of $H$ can be viewed as a \textquotedblleft
deformation\textquotedblright\ of the antipode of $\operatorname*{gr}H$. But
the latter antipode does square to $\operatorname*{id}$ because
$\operatorname*{gr}H$ is commutative. Unfortunately, this proof does not
survive our generalization; in fact, even defining a Hopf algebra structure on
$\operatorname*{gr}H$ would likely require at least some flatness
assumptions.}. Second, we will replace the Hopf algebra by a coalgebra, or
rather by a more general structure that does not even require coassociativity.
The squared antipode $S^{2}$ will be replaced by an arbitrary
\textquotedblleft coalgebra\textquotedblright\ endomorphism\ $f$ (we are using
scare quotes because our structure is not really a coalgebra), and the
identity map by another such endomorphism\ $e$. Finally, the graded components
will be replaced by an arbitrary sequence of submodules satisfying certain
compatibility relations. We state the general result in Section
\ref{sec.general} and prove it in Section \ref{sec.pf.thm.id-f.gen}. In
Sections \ref{sec.cfc}--\ref{sec.hopf}, we progressively specialize this
result: first to connected filtered coalgebras with coalgebra endomorphisms
(in Section \ref{sec.cfc}), then to connected filtered Hopf algebras with
$S^{2}$ (in Section \ref{sec.filhopf}), and finally to connected graded Hopf
algebras with $S^{2}$ (in Section \ref{sec.hopf}). The latter specialization
covers the results of Aguiar and Lauve. (The Malvenuto--Reutenauer Hopf
algebra turns out to be a red herring; any connected graded Hopf algebra $H$
with the property that $ab=ba$ for all $a,b\in H_{1}$ will do.)
\subsubsection*{Acknowledgments}
I thank Marcelo Aguiar and Amy Pang for conversations I have learnt much from.
\subsubsection*{Remark on alternative versions}
This is the detailed version of the present note. The regular version (with
proofs in less detail) is available at
\[
\text{\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/antipode-squared.pdf}}%
\]
\section{Notations}
We will use the notions of coalgebras, bialgebras and Hopf algebras over a
commutative ring, as defined (e.g.) in \cite[Chapter 2]{Abe80}, \cite[Chapter
1]{GriRei}, \cite[Chapters 2, 3]{HaGuKi10}, \cite[Chapters 2, 5, 7]{Radfor12}
or \cite[Chapters I--IV]{Sweedler}. (In particular, our Hopf algebras are
\textbf{not} twisted by a $\mathbb{Z}/2$-grading as the topologists' ones
are.) We use the same notations for Hopf algebras as in \cite[Chapter
1]{GriRei}. In particular:
\begin{itemize}
\item We let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $.
\item \textquotedblleft Rings\textquotedblright\ and \textquotedblleft
algebras\textquotedblright\ are always required to be associative and have a unity.
\item We fix a commutative ring $\mathbf{k}$. The symbols \textquotedblleft%
$\otimes$\textquotedblright\ and \textquotedblleft$\operatorname*{End}%
$\textquotedblright\ shall always mean \textquotedblleft$\otimes_{\mathbf{k}}%
$\textquotedblright\ and \textquotedblleft$\operatorname*{End}%
\nolimits_{\mathbf{k}}$\textquotedblright, respectively. The unity of the ring
$\mathbf{k}$ will be called $1_{\mathbf{k}}$ or just $1$ if confusion is unlikely.
\item The comultiplication and the counit of a $\mathbf{k}$-coalgebra are
denoted by $\Delta$ and $\epsilon$.
\item \textquotedblleft Graded\textquotedblright\ $\mathbf{k}$-modules mean
$\mathbb{N}$-graded $\mathbf{k}$-modules. The base ring $\mathbf{k}$ itself is
not supposed to have any nontrivial grading.
\item The $n$-th graded component of a graded $\mathbf{k}$-module $V$ will be
called $V_{n}$. If $n<0$, then this is the zero submodule $0$.
\item A graded $\mathbf{k}$-Hopf algebra means a $\mathbf{k}$-Hopf algebra
that has a grading as a $\mathbf{k}$-module, and whose structure maps
(multiplication, unit, comultiplication and counit) are graded maps. (The
antipode is automatically graded in this case, by \cite[Exercise 1.4.29
(e)]{GriRei}.)
\item If $f$ is a map from a set to itself, and if $k\in\mathbb{N}$ is
arbitrary, then $f^{k}$ shall denote the map $\underbrace{f\circ f\circ
\cdots\circ f}_{k\text{ times}}$. (Thus, $f^{1}=f$ and $f^{0}%
=\operatorname*{id}$.)
\end{itemize}
\section{Theorems}
\subsection{\label{sec.general}The main theorem}
We can now state the main result of this note:
\begin{theorem}
\label{thm.id-f.gen}Let $D$ be a $\mathbf{k}$-module, and let $\left(
D_{1},D_{2},D_{3},\ldots\right) $ be a sequence of $\mathbf{k}$-submodules of
$D$. Let $\delta:D\rightarrow D\otimes D$ be any $\mathbf{k}$-linear map.
Let $e:D\rightarrow D$ and $f:D\rightarrow D$ be two $\mathbf{k}$-linear maps
such that%
\begin{align}
\operatorname*{Ker}\delta & \subseteq\operatorname*{Ker}\left( e-f\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.thm.id-f.gen.ass-Ker}\\
\left( f\otimes f\right) \circ\delta & =\delta\circ
f\ \ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.thm.id-f.gen.ass-mor}\\
\left( e\otimes e\right) \circ\delta & =\delta\circ
e\ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.thm.id-f.gen.ass-mor'}\\
f\circ e & =e\circ f. \label{pf.thm.id-f.gen.ass-comm}%
\end{align}
Let $p$ be a positive integer such that%
\begin{equation}
\left( e-f\right) \left( D_{1}+D_{2}+\cdots+D_{p}\right) =0.
\label{pf.thm.id-f.gen.ass-ann}%
\end{equation}
Assume furthermore that%
\begin{equation}
\delta\left( D_{n}\right) \subseteq\sum_{i=1}^{n-1}D_{i}\otimes
D_{n-i}\ \ \ \ \ \ \ \ \ \ \text{for each }n>p. \label{pf.thm.id-f.gen.ass-gr}%
\end{equation}
(Here, the \textquotedblleft$D_{i}\otimes D_{n-i}$\textquotedblright\ on the
right hand side means the image of $D_{i}\otimes D_{n-i}$ under the canonical
map $D_{i}\otimes D_{n-i}\rightarrow D\otimes D$ that is obtained by tensoring
the two inclusion maps $D_{i}\rightarrow D$ and $D_{n-i}\rightarrow D$
together. When $\mathbf{k}$ is not a field, this canonical map may fail to be injective.)
Then, for any integer $u>p$, we have%
\begin{equation}
\left( e-f\right) ^{u-p}\left( D_{u}\right) \subseteq\operatorname*{Ker}%
\delta\label{eq.thm.id-f.gen.clm1}%
\end{equation}
and%
\begin{equation}
\left( e-f\right) ^{u-p+1}\left( D_{u}\right) =0.
\label{eq.thm.id-f.gen.clm2}%
\end{equation}
\end{theorem}
As the statement of this theorem is not very intuitive, some explanations are
in order. The reader may think of the $D$ in Theorem \ref{thm.id-f.gen} as a
\textquotedblleft pre-coalgebra\textquotedblright, with $\delta$ being its
\textquotedblleft reduced coproduct\textquotedblright. Indeed, the easiest way
to obtain a nontrivial example is to fix a connected graded Hopf algebra $H$,
then define $D$ to be either $H$ or the \textquotedblleft positive
part\textquotedblright\ of $H$ (that is, the submodule $\bigoplus_{n>0}H_{n}$
of $H$), and define $\delta$ to be the map $x\mapsto\Delta\left( x\right)
-x\otimes1-1\otimes x+\epsilon\left( x\right) 1\otimes1$ (the so-called
\emph{reduced coproduct} of $H$). From this point of view,
$\operatorname*{Ker}\delta$ can be regarded as the set of \textquotedblleft
primitive\textquotedblright\ elements of $D$. The maps $f$ and $e$ can be
viewed as two commuting \textquotedblleft coalgebra
endomorphisms\textquotedblright\ of $D$ (indeed, the assumptions
(\ref{pf.thm.id-f.gen.ass-mor}) and (\ref{pf.thm.id-f.gen.ass-mor'}) are
essentially saying that $f$ and $e$ preserve the \textquotedblleft reduced
coproduct\textquotedblright\ $\delta$). The submodules $D_{1},D_{2}%
,D_{3},\ldots$ are an analogue of the (positive-degree) graded components of
$D$, while the assumption (\ref{pf.thm.id-f.gen.ass-gr}) says that the
\textquotedblleft reduced coproduct\textquotedblright\ $\delta$
\textquotedblleft respects the grading\textquotedblright\ (as is indeed the
case for connected graded Hopf algebras).
We stress that $p$ is allowed to be $1$ in Theorem \ref{thm.id-f.gen}; in this
case, the assumption (\ref{pf.thm.id-f.gen.ass-ann}) simplifies to $\left(
e-f\right) \left( 0\right) =0$, which is automatically true by the
linearity of $e-f$.
We shall prove Theorem \ref{thm.id-f.gen} in Section \ref{sec.pf.thm.id-f.gen}%
. First, however, let us explore its consequences for coalgebras and Hopf
algebras, recovering in particular the results of Aguiar and Lauve promised in
the introduction.
\subsection{\label{sec.cfc}Connected filtered coalgebras}
We begin by specializing Theorem \ref{thm.id-f.gen} to the setting of a
connected filtered coalgebra. There are several ways to define what a filtered
coalgebra is; ours is probably the most liberal:
\begin{definition}
\label{def.fil-coal}A \emph{filtered }$\mathbf{k}$\emph{-coalgebra} means a
$\mathbf{k}$-coalgebra $C$ equipped with an infinite sequence $\left(
C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $ of $\mathbf{k}$-submodules of
$C$ satisfying the following three conditions:
\begin{itemize}
\item We have%
\begin{equation}
C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}\subseteq\cdots.
\label{eq.def.fil-coal.chain}%
\end{equation}
\item We have%
\begin{equation}
\bigcup_{n\in\mathbb{N}}C_{\leq n}=C. \label{eq.def.fil-coal.union}%
\end{equation}
\item We have%
\begin{equation}
\Delta\left( C_{\leq n}\right) \subseteq\sum_{i=0}^{n}C_{\leq i}\otimes
C_{\leq n-i}\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}.
\label{eq.def.fil-coal.Del}%
\end{equation}
(Here, the \textquotedblleft$C_{\leq i}\otimes C_{\leq n-i}$\textquotedblright%
\ on the right hand side means the image of $C_{\leq i}\otimes C_{\leq n-i}$
under the canonical map $C_{\leq i}\otimes C_{\leq n-i}\rightarrow C\otimes C$
that is obtained by tensoring the two inclusion maps $C_{\leq i}\rightarrow C$
and $C_{\leq n-i}\rightarrow C$ together. When $\mathbf{k}$ is not a field,
this canonical map may fail to be injective.)
\end{itemize}
The sequence $\left( C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $ is called
the \emph{filtration} of the filtered $\mathbf{k}$-coalgebra $C$.
\end{definition}
A more categorically-minded person might replace the condition $\Delta\left(
C_{\leq n}\right) \subseteq\sum_{i=0}^{n}C_{\leq i}\otimes C_{\leq n-i}$ in
this definition by a stronger requirement (e.g., asking $\Delta$ to factor
through a linear map $C_{\leq n}\rightarrow\bigoplus_{i=0}^{n}C_{\leq
i}\otimes C_{\leq n-i}$, where the \textquotedblleft$\otimes$%
\textquotedblright\ signs now signify the actual tensor products rather than
their images in $C\otimes C$). However, we have no need for such stronger
requirements. Mercifully, all reasonable definitions of filtered $\mathbf{k}%
$-coalgebras agree when $\mathbf{k}$ is a field.
The condition (\ref{eq.def.fil-coal.union}) in Definition \ref{def.fil-coal}
shall never be used in the following; we merely state it to avoid muddling the
meaning of \textquotedblleft filtered $\mathbf{k}$-coalgebra\textquotedblright.
A graded $\mathbf{k}$-coalgebra $C$ automatically becomes a filtered
$\mathbf{k}$-coalgebra; indeed, we can define its filtration $\left(
C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $ by setting%
\[
C_{\leq n}=\bigoplus_{i=0}^{n}C_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }%
n\in\mathbb{N}.
\]
\begin{definition}
\label{def.con-fil-coal}Let $C$ be a filtered $\mathbf{k}$-coalgebra with
filtration $\left( C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $. Let
$1_{\mathbf{k}}$ denote the unity of the ring $\mathbf{k}$.
\textbf{(a)} The filtered $\mathbf{k}$-coalgebra $C$ is said to be
\emph{connected} if the restriction $\epsilon\mid_{C_{\leq0}}$ is a
$\mathbf{k}$-module isomorphism from $C_{\leq0}$ to $\mathbf{k}$.
\textbf{(b)} In this case, the element $\left( \epsilon\mid_{C_{\leq0}%
}\right) ^{-1}\left( 1_{\mathbf{k}}\right) \in C_{\leq0}$ is called the
\emph{unity} of $C$ and is denoted by $1_{C}$.
Now, assume that $C$ is a connected filtered $\mathbf{k}$-coalgebra.
\textbf{(c)} An element $x$ of $C$ is said to be \emph{primitive} if
$\Delta\left( x\right) =x\otimes1_{C}+1_{C}\otimes x$.
\textbf{(d)} The set of all primitive elements of $C$ is denoted by
$\operatorname*{Prim}C$.
\end{definition}
These notions of \textquotedblleft connected\textquotedblright,
\textquotedblleft unity\textquotedblright\ and \textquotedblleft
primitive\textquotedblright\ specialize to the commonly established concepts
of these names when $C$ is a graded $\mathbf{k}$-bialgebra. Indeed, Definition
\ref{def.con-fil-coal} \textbf{(b)} defines the unity $1_{C}$ of $C$ to be the
unique element of $C_{\leq0}$ that gets sent to $1_{\mathbf{k}}$ by the map
$\epsilon$; but this property is satisfied for the unity of a graded
$\mathbf{k}$-bialgebra as well. (We will repeat this argument in more detail
later on, in the proof of Proposition \ref{prop.fil-bial.1=1}.)
The following property of connected filtered $\mathbf{k}$-coalgebras will be
crucial for us:
\begin{proposition}
\label{prop.cfc.delta2}Let $C$ be a connected filtered $\mathbf{k}$-coalgebra
with filtration $\left( C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $.
Define a $\mathbf{k}$-linear map $\delta:C\rightarrow C\otimes C$ by setting
\[
\delta\left( c\right) :=\Delta\left( c\right) -c\otimes1_{C}-1_{C}\otimes
c+\epsilon\left( c\right) 1_{C}\otimes1_{C}\ \ \ \ \ \ \ \ \ \ \text{for
each }c\in C.
\]
Then:
\textbf{(a)} We have
\[
\delta\left( C_{\leq n}\right) \subseteq\sum_{i=1}^{n-1}C_{\leq i}\otimes
C_{\leq n-i}\ \ \ \ \ \ \ \ \ \ \text{for each }n>0.
\]
\textbf{(b)} If $f:C\rightarrow C$ is a $\mathbf{k}$-coalgebra homomorphism
satisfying $f\left( 1_{C}\right) =1_{C}$, then we have $\left( f\otimes
f\right) \circ\delta=\delta\circ f$.
\textbf{(c)} We have $\operatorname*{Prim}C=\left( \operatorname*{Ker}%
\delta\right) \cap\left( \operatorname*{Ker}\epsilon\right) $.
\textbf{(d)} The set $\operatorname*{Prim}C$ is a $\mathbf{k}$-submodule of
$C$.
\textbf{(e)} We have $\operatorname*{Ker}\delta=\mathbf{k}\cdot1_{C}%
+\operatorname*{Prim}C$.
\end{proposition}
We shall prove Proposition \ref{prop.cfc.delta2} in Section
\ref{sec.pf.prop.cfc.delta2}. The map $\delta$ in Proposition
\ref{prop.cfc.delta2} is called the \emph{reduced coproduct} of $C$.
Proposition \ref{prop.cfc.delta2} helps us apply Theorem \ref{thm.id-f.gen} to
filtered $\mathbf{k}$-coalgebras, resulting in the following:
\begin{corollary}
\label{cor.id-f.cfc}Let $C$ be a connected filtered $\mathbf{k}$-coalgebra
with filtration $\left( C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $.
Let $e:C\rightarrow C$ and $f:C\rightarrow C$ be two $\mathbf{k}$-coalgebra
homomorphisms such that%
\begin{align}
e\left( 1_{C}\right) & =1_{C}\ \ \ \ \ \ \ \ \ \ \text{and}\nonumber\\
f\left( 1_{C}\right) & =1_{C}\ \ \ \ \ \ \ \ \ \ \text{and}\nonumber\\
\operatorname*{Prim}C & \subseteq\operatorname*{Ker}\left( e-f\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\label{eq.cor.id-f.cfc.ass-Ker}\\
f\circ e & =e\circ f. \label{eq.cor.id-f.cfc.ass-comm}%
\end{align}
Let $p$ be a positive integer such that%
\begin{equation}
\left( e-f\right) \left( C_{\leq p}\right) =0.
\label{eq.cor.id-f.cfc.ass-ann}%
\end{equation}
Then:
\textbf{(a)} For any integer $u>p$, we have%
\begin{equation}
\left( e-f\right) ^{u-p}\left( C_{\leq u}\right) \subseteq
\operatorname*{Prim}C. \label{eq.cor.id-f.cfc.claim-1}%
\end{equation}
\textbf{(b)} For any integer $u\geq p$, we have%
\begin{equation}
\left( e-f\right) ^{u-p+1}\left( C_{\leq u}\right) =0.
\label{eq.cor.id-f.cfc.claim-2}%
\end{equation}
\end{corollary}
Corollary \ref{cor.id-f.cfc} results from an easy (although not completely
immediate) application of Theorem \ref{thm.id-f.gen} and Proposition
\ref{prop.cfc.delta2}. The detailed proof can be found in Section
\ref{sec.pf.cor.id-f.cfc}.
Specializing Corollary \ref{cor.id-f.cfc} further to the case of $p=1$, we can
obtain a nicer result:
\begin{corollary}
\label{cor.id-f.cfc1}Let $C$ be a connected filtered $\mathbf{k}$-coalgebra
with filtration $\left( C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $.
Let $e:C\rightarrow C$ and $f:C\rightarrow C$ be two $\mathbf{k}$-coalgebra
homomorphisms such that%
\begin{align*}
e\left( 1_{C}\right) & =1_{C}\ \ \ \ \ \ \ \ \ \ \text{and}\\
f\left( 1_{C}\right) & =1_{C}\ \ \ \ \ \ \ \ \ \ \text{and}\\
\operatorname*{Prim}C & \subseteq\operatorname*{Ker}\left( e-f\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\\
f\circ e & =e\circ f.
\end{align*}
Then:
\textbf{(a)} For any integer $u>1$, we have%
\[
\left( e-f\right) ^{u-1}\left( C_{\leq u}\right) \subseteq
\operatorname*{Prim}C.
\]
\textbf{(b)} For any positive integer $u$, we have%
\[
\left( e-f\right) ^{u}\left( C_{\leq u}\right) =0.
\]
\end{corollary}
See Section \ref{sec.pf.cor.id-f.cfc} for a proof of this corollary.
The particular case of Corollary \ref{cor.id-f.cfc1} for $e=\operatorname*{id}%
$ is particularly simple:
\begin{corollary}
\label{cor.id-f.cfc1id}Let $C$ be a connected filtered $\mathbf{k}$-coalgebra
with filtration $\left( C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $.
Let $f:C\rightarrow C$ be a $\mathbf{k}$-coalgebra homomorphism such that
\[
f\left( 1_{C}\right) =1_{C}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \operatorname*{Prim}C\subseteq\operatorname*{Ker}\left(
\operatorname*{id}-f\right) .
\]
Then:
\textbf{(a)} For any integer $u>1$, we have%
\[
\left( \operatorname*{id}-f\right) ^{u-1}\left( C_{\leq u}\right)
\subseteq\operatorname*{Prim}C.
\]
\textbf{(b)} For any positive integer $u$, we have%
\[
\left( \operatorname*{id}-f\right) ^{u}\left( C_{\leq u}\right) =0.
\]
\end{corollary}
Again, the proof of this corollary can be found in Section
\ref{sec.pf.cor.id-f.cfc}.
Note that Corollary \ref{cor.id-f.cfc1id} \textbf{(b)} is precisely
\cite[Theorem 37.1 \textbf{(a)}]{logid}.
\subsection{\label{sec.filhopf}Connected filtered bialgebras and Hopf
algebras}
We shall now apply our above results to connected filtered bialgebras and Hopf
algebras. We first define what we mean by these notions:
\begin{definition}
\label{def.fil-bial}\textbf{(a)} A \emph{filtered }$\mathbf{k}$%
\emph{-bialgebra} means a $\mathbf{k}$-bialgebra $H$ equipped with an infinite
sequence $\left( H_{\leq0},H_{\leq1},H_{\leq2},\ldots\right) $ of
$\mathbf{k}$-submodules of $H$ satisfying the following five conditions:
\begin{itemize}
\item We have%
\[
H_{\leq0}\subseteq H_{\leq1}\subseteq H_{\leq2}\subseteq\cdots.
\]
\item We have%
\[
\bigcup_{n\in\mathbb{N}}H_{\leq n}=H.
\]
\item We have
\[
\Delta\left( H_{\leq n}\right) \subseteq\sum_{i=0}^{n}H_{\leq i}\otimes
H_{\leq n-i}\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}.
\]
(Here, the \textquotedblleft$H_{\leq i}\otimes H_{\leq n-i}$\textquotedblright%
\ on the right hand side means the image of $H_{\leq i}\otimes H_{\leq n-i}$
under the canonical map $H_{\leq i}\otimes H_{\leq n-i}\rightarrow H\otimes H$
that is obtained by tensoring the two inclusion maps $H_{\leq i}\rightarrow H$
and $H_{\leq n-i}\rightarrow H$ together.)
\item We have $H_{\leq i}H_{\leq j}\subseteq H_{\leq i+j}$ for any
$i,j\in\mathbb{N}$. (Here, $H_{\leq i}H_{\leq j}$ denotes the $\mathbf{k}%
$-linear span of the set of all products $ab$ with $a\in H_{\leq i}$ and $b\in
H_{\leq j}$.)
\item The unity of the $\mathbf{k}$-algebra $H$ belongs to $H_{\leq0}$.{}
\end{itemize}
The sequence $\left( H_{\leq0},H_{\leq1},H_{\leq2},\ldots\right) $ is called
the \emph{filtration} of the filtered $\mathbf{k}$-bialgebra $H$.
\textbf{(b)} A \emph{filtered }$\mathbf{k}$\emph{-Hopf algebra} means a
filtered $\mathbf{k}$-bialgebra $H$ such that the $\mathbf{k}$-bialgebra $H$
is a Hopf algebra (i.e., has an antipode) and such that the antipode $S$ of
$H$ respects the filtration (i.e., satisfies $S\left( H_{\leq n}\right)
\subseteq H_{\leq n}$ for each $n\in\mathbb{N}$).
\end{definition}
The $H_{\leq i}H_{\leq j}\subseteq H_{\leq i+j}$ condition in Definition
\ref{def.fil-bial} \textbf{(a)} will not actually be used in what follows.
Thus, we could have omitted it; but this would have resulted in a less common
(and less well-behaved in other ways) concept of \textquotedblleft filtered
bialgebra\textquotedblright. Likewise, we have included the $S\left( H_{\leq
n}\right) \subseteq H_{\leq n}$ condition in Definition \ref{def.fil-bial}
\textbf{(b)}, even though we will never use it.
Every $\mathbf{k}$-bialgebra is automatically a $\mathbf{k}$-coalgebra. Thus,
every filtered $\mathbf{k}$-bialgebra is automatically a filtered $\mathbf{k}%
$-coalgebra. This allows the following definition:
\begin{definition}
A filtered $\mathbf{k}$-bialgebra $H$ is said to be \emph{connected} if the
filtered $\mathbf{k}$-coalgebra $H$ is connected.
\end{definition}
Thus, if $H$ is a connected filtered $\mathbf{k}$-bialgebra, then Definition
\ref{def.con-fil-coal} \textbf{(b)} defines a \textquotedblleft
unity\textquotedblright\ $1_{H}$ of $H$. This appears to cause an awkward
notational quandary, since $H$ already has a unity by virtue of being a
$\mathbf{k}$-algebra (and this latter unity is also commonly denoted by
$1_{H}$). Fortunately, this cannot cause any confusion, since these two
unities are identical, as the following proposition shows:
\begin{proposition}
\label{prop.fil-bial.1=1}Let $H$ be a connected filtered $\mathbf{k}%
$-bialgebra. Then, the unity $1_{H}$ defined according to Definition
\ref{def.con-fil-coal} \textbf{(b)} equals the unity of the $\mathbf{k}%
$-algebra $H$.
\end{proposition}
\begin{proof}
[Proof of Proposition \ref{prop.fil-bial.1=1}.]Let $1_{H}$ denote the unity
$1_{H}$ defined according to Definition \ref{def.con-fil-coal} \textbf{(b)}.
Let $\mathbf{1}$ denote the unity of the $\mathbf{k}$-algebra $H$. Thus, we
must show that $1_{H}=\mathbf{1}$.
We know that $H$ is a filtered $\mathbf{k}$-bialgebra. Hence, the unity of the
$\mathbf{k}$-algebra $H$ belongs to $H_{\leq0}$ (by Definition
\ref{def.fil-bial} \textbf{(a)}). In other words, $\mathbf{1}$ belongs to
$H_{\leq0}$ (since $\mathbf{1}$ is the unity of the $\mathbf{k}$-algebra $H$).
Thus, $\left( \epsilon\mid_{H_{\leq0}}\right) \left( \mathbf{1}\right) $
is well-defined.
The axioms of a $\mathbf{k}$-bialgebra yield $\epsilon\left( \mathbf{1}%
\right) =1_{\mathbf{k}}$ (since $H$ is a $\mathbf{k}$-bialgebra with unity
$\mathbf{1}$).
However, $1_{H}$ is defined to be $\left( \epsilon\mid_{H_{\leq0}}\right)
^{-1}\left( 1_{\mathbf{k}}\right) $ (by Definition \ref{def.fil-coal}
\textbf{(b)}, applied to $C=H$). Hence, $1_{H}=\left( \epsilon\mid_{H_{\leq
0}}\right) ^{-1}\left( 1_{\mathbf{k}}\right) $. On the other hand, we have
$\mathbf{1}=\left( \epsilon\mid_{H_{\leq0}}\right) ^{-1}\left(
1_{\mathbf{k}}\right) $ (since $\left( \epsilon\mid_{H_{\leq0}}\right)
\left( \mathbf{1}\right) =\epsilon\left( \mathbf{1}\right) =1_{\mathbf{k}%
}$). Comparing these two equalities, we obtain $1_{H}=\mathbf{1}$. As
explained above, this completes the proof of Proposition
\ref{prop.fil-bial.1=1}.
\end{proof}
In Definition \ref{def.con-fil-coal}, we have defined the notion of a
\textquotedblleft primitive element\textquotedblright\ of a connected filtered
$\mathbf{k}$-coalgebra $C$. In the same way, we can define a \textquotedblleft
primitive element\textquotedblright\ of a $\mathbf{k}$-bialgebra $H$ (using
the unity of the $\mathbf{k}$-algebra $H$ instead of $1_{C}$):
\begin{definition}
\label{def.prim-in-bialg}Let $H$ be a $\mathbf{k}$-bialgebra with unity
$1_{H}$.
\textbf{(a)} An element $x$ of $H$ is said to be \emph{primitive} if
$\Delta\left( x\right) =x\otimes1_{H}+1_{H}\otimes x$.
\textbf{(b)} The set of all primitive elements of $H$ is denoted by
$\operatorname*{Prim}H$.
\end{definition}
When $H$ is a connected filtered $\mathbf{k}$-bialgebra, Definition
\ref{def.prim-in-bialg} \textbf{(a)}\ agrees with Definition
\ref{def.con-fil-coal} \textbf{(c)}, since Proposition \ref{prop.fil-bial.1=1}
shows that the two meanings of $1_{H}$ are actually identical. Thus, when $H$
is a connected filtered $\mathbf{k}$-bialgebra, Definition
\ref{def.prim-in-bialg} \textbf{(b)}\ agrees with Definition
\ref{def.con-fil-coal} \textbf{(d)}. Hence, the notation $\operatorname*{Prim}%
H$ is unambiguous.
Next we state some basic properties of the antipode in a Hopf algebra that
will be used later on:
\begin{lemma}
\label{lem.bialg.antip-props}Let $H$ be a $\mathbf{k}$-Hopf algebra with unity
$1_{H}\in H$ and antipode $S\in\operatorname*{End}H$. Then:
\textbf{(a)} The map $S^{2}:H\rightarrow H$ is a $\mathbf{k}$-coalgebra homomorphism.
\textbf{(b)} We have $S\left( 1_{H}\right) =1_{H}$.
\textbf{(c)} We have $S\left( x\right) =-x$ for every primitive element $x$
of $H$.
\textbf{(d)} We have $S^{2}\left( x\right) =x$ for every primitive element
$x$ of $H$.
\end{lemma}
All parts of this lemma are proved in \cite[proof of Lemma 37.8]{logid} (at
least in the case when $\mathbf{k}$ is a field; but the proof applies equally
well in the general case). For the sake of completeness, we shall also give
the proof in Section \ref{sec.pf.filhopf}.
We can now state our main consequence for connected filtered Hopf algebras:
\begin{corollary}
\label{cor.id-S2.p}Let $H$ be a connected filtered $\mathbf{k}$-Hopf algebra
with filtration $\left( H_{\leq0},H_{\leq1},H_{\leq2},\ldots\right) $ and
antipode $S$.
Let $p$ be a positive integer such that%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) \left( H_{\leq p}\right) =0.
\label{eq.cor.id-S2.p.ass-ann}%
\end{equation}
Then:
\textbf{(a)} For any integer $u>p$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u-p}\left( H_{\leq u}\right)
\subseteq\operatorname*{Prim}H \label{eq.cor.id-S2.p.claim1}%
\end{equation}
and%
\begin{equation}
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-p}\right) \left( H_{\leq u}\right) =0.
\label{eq.cor.id-S2.p.claim3}%
\end{equation}
\textbf{(b)} For any integer $u\geq p$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u-p+1}\left( H_{\leq u}\right)
=0. \label{eq.cor.id-S2.p.claim2}%
\end{equation}
\end{corollary}
We shall derive this from Corollary \ref{cor.id-f.cfc} in Section
\ref{sec.filhopf}.
Specializing Corollary \ref{cor.id-S2.p} to $p=1$, we can easily obtain the following:
\begin{corollary}
\label{cor.id-S2.1}Let $H$ be a connected filtered $\mathbf{k}$-Hopf algebra
with filtration $\left( H_{\leq0},H_{\leq1},H_{\leq2},\ldots\right) $ and
antipode $S$. Then:
\textbf{(a)} For any integer $u>1$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u-1}\left( H_{\leq u}\right)
\subseteq\operatorname*{Prim}H \label{eq.cor.id-S2.1.claim1}%
\end{equation}
and%
\begin{equation}
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-1}\right) \left( H_{\leq u}\right) =0.
\label{eq.cor.id-S2.1.claim3}%
\end{equation}
\textbf{(b)} For any positive integer $u$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u}\left( H_{\leq u}\right) =0.
\label{eq.cor.id-S2.1.claim2}%
\end{equation}
\end{corollary}
Corollary \ref{cor.id-S2.1} \textbf{(b)} has already appeared in \cite[Theorem
37.7 \textbf{(a)}]{logid}. It can be derived from either Corollary
\ref{cor.id-S2.p} or Corollary \ref{cor.id-f.cfc1id}; we shall show the latter
derivation in Section \ref{sec.filhopf}.
\subsection{\label{sec.hopf}Connected graded Hopf algebras}
Let us now specialize our results even further to connected \textbf{graded}
Hopf algebras. We have already seen that any graded $\mathbf{k}$-coalgebra
automatically becomes a filtered $\mathbf{k}$-coalgebra. In the same way, any
graded $\mathbf{k}$-Hopf algebra automatically becomes a filtered $\mathbf{k}%
$-Hopf algebra. Moreover, a graded $\mathbf{k}$-Hopf algebra $H$ is connected
(in the sense that $H_{0}\cong\mathbf{k}$ as $\mathbf{k}$-modules) if and only
if the filtered $\mathbf{k}$-coalgebra $H$ is connected. (This follows easily
from \cite[Exercise 1.3.20 (e)]{GriRei}.) Thus, our above results for
connected filtered $\mathbf{k}$-Hopf algebras can be applied to connected
graded $\mathbf{k}$-Hopf algebras. From Corollary \ref{cor.id-S2.1}, we easily
obtain the following:
\begin{corollary}
\label{cor.id-S2.gr1}Let $H$ be a connected graded $\mathbf{k}$-Hopf algebra
with antipode $S$. Then, for any positive integer $u$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u-1}\left( H_{u}\right)
\subseteq\operatorname*{Prim}H \label{eq.cor.id-S2.gr1.claim1}%
\end{equation}
and%
\begin{equation}
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-1}\right) \left( H_{u}\right) =0
\label{eq.cor.id-S2.gr1.claim3}%
\end{equation}
and%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u}\left( H_{u}\right) =0.
\label{eq.cor.id-S2.gr1.claim2}%
\end{equation}
\end{corollary}
This is not an immediate consequence of Corollary \ref{cor.id-S2.1}, since the
condition \textquotedblleft$u$ is positive\textquotedblright\ is weaker than
the condition \textquotedblleft$u>1$\textquotedblright\ in Corollary
\ref{cor.id-S2.1} \textbf{(a)}; thus, deriving Corollary \ref{cor.id-S2.gr1}
from Corollary \ref{cor.id-S2.1} requires some extra work to account for the
case of $u=1$. We will give a detailed proof of Corollary \ref{cor.id-S2.gr1}
in Section \ref{sec.pf.hopf}.
The equality (\ref{eq.cor.id-S2.gr1.claim3}) in Corollary \ref{cor.id-S2.gr1}
yields \cite[Lemma 12.50]{Aguiar17}, whereas the equality
(\ref{eq.cor.id-S2.gr1.claim2}) yields \cite[Proposition 7]{AguLau14}. Next,
we apply Corollary \ref{cor.id-S2.p} to the graded setting:
\begin{corollary}
\label{cor.id-S2.grp}Let $H$ be a connected graded $\mathbf{k}$-Hopf algebra
with antipode $S$.
Let $p$ be a positive integer such that all $i\in\left\{ 2,3,\ldots
,p\right\} $ satisfy%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) \left( H_{i}\right) =0.
\label{eq.cor.id-S2.grp.ass-ann}%
\end{equation}
Then:
\textbf{(a)} For any integer $u>p$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u-p}\left( H_{\leq u}\right)
\subseteq\operatorname*{Prim}H \label{eq.cor.id-S2.grp.claim1}%
\end{equation}
and%
\begin{equation}
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-p}\right) \left( H_{\leq u}\right) =0.
\label{eq.cor.id-S2.grp.claim3}%
\end{equation}
\textbf{(b)} For any integer $u\geq p$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u-p+1}\left( H_{\leq u}\right)
=0. \label{eq.cor.id-S2.grp.claim2}%
\end{equation}
\end{corollary}
Again, the proof of Corollary \ref{cor.id-S2.grp} can be found in Section
\ref{sec.pf.hopf}.
The particular case of Corollary \ref{cor.id-S2.grp} for $p=2$ is the most
useful, as the condition (\ref{eq.cor.id-S2.grp.ass-ann}) boils down to the
equality $\left( \operatorname*{id}-S^{2}\right) \left( H_{2}\right) =0$
in this case, and the latter equality is satisfied rather frequently. Here is
one sufficient criterion (which we will prove in Section \ref{sec.pf.hopf}):
\begin{corollary}
\label{cor.id-S2.gr2}Let $H$ be a connected graded $\mathbf{k}$-Hopf algebra
with antipode $S$. Assume that%
\begin{equation}
ab=ba\ \ \ \ \ \ \ \ \ \ \text{for every }a,b\in H_{1}.
\label{eq.cor.id-S2.gr2.commH1}%
\end{equation}
Then:
\textbf{(a)} We have
\[
\left( \operatorname*{id}-S^{2}\right) \left( H_{2}\right) =0.
\]
\textbf{(b)} For any integer $u>2$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u-2}\left( H_{\leq u}\right)
\subseteq\operatorname*{Prim}H \label{eq.cor.id-S2.gr2.b.claim1}%
\end{equation}
and%
\begin{equation}
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-2}\right) \left( H_{\leq u}\right) =0.
\label{eq.cor.id-S2.gr2.b.claim3}%
\end{equation}
\textbf{(c)} For any integer $u>1$, we have%
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) ^{u-1}\left( H_{\leq u}\right) =0.
\label{eq.cor.id-S2.gr2.b.claim2}%
\end{equation}
\end{corollary}
The equality (\ref{eq.cor.id-S2.gr2.b.claim2}) in Corollary
\ref{cor.id-S2.gr2} generalizes \cite[Example 8]{AguLau14}. Indeed, if $H$ is
the Malvenuto--Reutenauer Hopf algebra\footnote{See \cite[\S 12.1]{Meliot17},
\cite[\S 7.1]{HaGuKi10} or \cite[\S 8.1]{GriRei} for the definition of this
Hopf algebra. (It is denoted $\operatorname*{FQSym}$ in \cite{Meliot17} and
\cite{GriRei}, and denoted $MPR$ in \cite{HaGuKi10}.)}, then the condition
(\ref{eq.cor.id-S2.gr2.commH1}) is satisfied (since $H_{1}$ is a free
$\mathbf{k}$-module of rank $1$ in this case); therefore, Corollary
\ref{cor.id-S2.gr2} \textbf{(c)} can be applied in this case, and we recover
\cite[Example 8]{AguLau14}. Likewise, we can obtain the same result if $H$ is
the Hopf algebra $\operatorname*{WQSym}$ of word quasisymmetric
functions\footnote{See (e.g.) \cite[\S 4.3.2]{MeNoTh13} for a definition of
this Hopf algebra.}.
It is worth noticing that the condition (\ref{eq.cor.id-S2.gr2.commH1}) is
only sufficient, but not necessary for (\ref{eq.cor.id-S2.gr2.b.claim2}). For
example, if $H$ is the tensor algebra of a free $\mathbf{k}$-module $V$ of
rank $\geq2$, then (\ref{eq.cor.id-S2.gr2.b.claim2}) holds (since $H$ is
cocommutative, so that $S^{2}=\operatorname*{id}$), but
(\ref{eq.cor.id-S2.gr2.commH1}) does not (since $u\otimes v\neq v\otimes u$ if
$u$ and $v$ are two distinct basis vectors of $V$).
An example of a connected graded Hopf algebra $H$ that does \textbf{not}
satisfy (\ref{eq.cor.id-S2.gr2.b.claim2}) (and thus does not satisfy
(\ref{eq.cor.id-S2.gr2.commH1}) either) is not hard to construct:
\begin{example}
Assume that the ring $\mathbf{k}$ is not trivial. Let $H$ be the free
$\mathbf{k}$-algebra with three generators $a,b,c$. We equip this $\mathbf{k}%
$-algebra $H$ with a grading, by requiring that its generators $a,b,c$ are
homogeneous of degrees $1,1,2$, respectively. Next, we define a
comultiplication $\Delta$ on $H$ by setting
\begin{align*}
\Delta\left( a\right) & =a\otimes1+1\otimes a;\\
\Delta\left( b\right) & =b\otimes1+1\otimes b;\\
\Delta\left( c\right) & =c\otimes1+a\otimes b+1\otimes c
\end{align*}
(where $1$ is the unity of $H$). Furthermore, we define a counit $\epsilon$ on
$H$ by setting $\epsilon\left( a\right) =\epsilon\left( b\right)
=\epsilon\left( c\right) =0$. It is straightforward to see that $H$ thus
becomes a connected graded $\mathbf{k}$-bialgebra, hence (by \cite[Proposition
1.4.16]{GriRei}) a connected graded $\mathbf{k}$-Hopf algebra. Its antipode
$S$ is easily seen to satisfy $S\left( c\right) =ab-c$ and $S^{2}\left(
c\right) =ba-ab+c\neq c$; thus, $\left( \operatorname*{id}-S^{2}\right)
\left( H_{2}\right) \neq0$. Hence, (\ref{eq.cor.id-S2.gr2.b.claim2}) does
not hold for $u=2$.
\end{example}
The Hopf algebra $H$ in this example is in fact an instance of a general
construction of connected graded $\mathbf{k}$-Hopf algebras that are
\textquotedblleft generic\textquotedblright\ (in the sense that their
structure maps satisfy no relations other than ones that hold in every
connected graded $\mathbf{k}$-Hopf algebra). This latter construction will be
elaborated upon in future work.
\begin{remark}
A brave reader might wonder whether the connectedness condition in Corollary
\ref{cor.id-S2.gr1} could be replaced by something weaker -- e.g., instead of
requiring $H$ to be connected, we might require that the subalgebra $H_{0}$ be
commutative. However, such a requirement would be insufficient. In fact, let
$\mathbf{k}=\mathbb{C}$. Then, for any integer $n>1$ and any primitive $n$-th
root of unity $q\in\mathbf{k}$, the Taft algebra $H_{n,q}$ defined in
\cite[\S 7.3]{Radfor12} can be viewed as a graded Hopf algebra (with $a\in
H_{0}$ and $x\in H_{1}$) whose subalgebra $H_{0}=\mathbf{k}\left[ a\right]
/\left( a^{n}-1\right) $ is commutative, but whose antipode $S$ does not
satisfy $\left( \operatorname*{id}-S^{2}\right) ^{k}\left( H_{1}\right)
=0$ for any $k\in\mathbb{N}$ (since $S^{2}\left( x\right) =q^{-1}x$ and
therefore $\left( \operatorname*{id}-S^{2}\right) ^{k}\left( x\right)
=\left( 1-q^{-1}\right) ^{k}x\neq0$ because $q^{-1}\neq1$).
\end{remark}
\section{Proofs}
We shall now prove all statements left unproved above.
We begin by stating three general facts from linear algebra, which will be
used several times in what follows:
\begin{lemma}
\label{lem.albeUV}Let $D$ be a $\mathbf{k}$-module. Let $U$ and $V$ be two
$\mathbf{k}$-submodules of $D$. Let $\alpha$ and $\beta$ be two elements of
$\operatorname*{End}D$. Then,%
\begin{equation}
\left( \alpha\otimes\beta\right) \left( U\otimes V\right) =\alpha\left(
U\right) \otimes\beta\left( V\right) . \label{pf.thm.id-f.gen.ass-Ker.abUV}%
\end{equation}
(Here, both $U\otimes V$ and $\alpha\left( U\right) \otimes\beta\left(
V\right) $ have to be understood as $\mathbf{k}$-submodules of $D\otimes D$,
specifically as the images of the \textquotedblleft actual\textquotedblright%
\ tensor products $U\otimes V$ and $\alpha\left( U\right) \otimes
\beta\left( V\right) $ under the canonical maps into $D\otimes D$.)
\end{lemma}
The proof of Lemma \ref{lem.albeUV} is straightforward and therefore omitted.
\begin{lemma}
\label{lem.albegadel}Let $D$ be a $\mathbf{k}$-module. Let $\alpha
,\beta,\gamma,\delta$ be four elements of $\operatorname*{End}D$. Then, in
$\operatorname*{End}\left( D\otimes D\right) $, we have%
\[
\left( \alpha\otimes\beta\right) \circ\left( \gamma\otimes\delta\right)
=\left( \alpha\circ\gamma\right) \otimes\left( \beta\circ\delta\right) .
\]
\end{lemma}
This fact can be verified easily by comparing how the left and the right hand
sides transform any given pure tensor $u\otimes v\in D\otimes D$. Again, we
leave the details to the reader.
\begin{lemma}
\label{lem.alber}Let $D$ be a $\mathbf{k}$-module. Let $\alpha$ and $\beta$ be
two elements of $\operatorname*{End}D$. Let $n\in\mathbb{N}$. Then, in
$\operatorname*{End}\left( D\otimes D\right) $, we have%
\[
\left( \alpha\otimes\beta\right) ^{n}=\alpha^{n}\otimes\beta^{n}.
\]
\end{lemma}
Lemma \ref{lem.alber} follows by induction on $k$ using Lemma
\ref{lem.albegadel}.
\subsection{\label{sec.pf.thm.id-f.gen}Proof of Theorem \ref{thm.id-f.gen}}
Our first goal is to prove Theorem \ref{thm.id-f.gen}.
\begin{proof}
[Proof of Theorem \ref{thm.id-f.gen}.]We shall prove
(\ref{eq.thm.id-f.gen.clm1}) and (\ref{eq.thm.id-f.gen.clm2}) by strong
induction on $u$:
\textit{Induction step:} Fix an integer $n>p$. Assume (as the induction
hypothesis) that (\ref{eq.thm.id-f.gen.clm1}) and (\ref{eq.thm.id-f.gen.clm2})
hold for all $up$ satisfying $up$ satisfying $uu-p$ satisfy%
\begin{equation}
g^{v}\left( D_{u}\right) =0 \label{pf.thm.id-f.gen.ass-Ker.IH4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.id-f.gen.ass-Ker.IH4}):} Let $uu-p$ be a positive integer. We must prove that
$g^{v}\left( D_{u}\right) =0$.
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $u\leq p$.
\par
\textit{Case 2:} We have $u>p$.
\par
Let us first consider Case 1. In this case, we have $u\leq p$. Hence,
$u\in\left\{ 1,2,\ldots,p\right\} $ (since $u$ is a positive integer) and
therefore $g\left( D_{u}\right) =0$ (by (\ref{pf.thm.id-f.gen.gDu=0})).
However, we have $v\geq1$ (since $v$ is a positive integer). Hence,
$g^{v}=g^{v-1}\circ g$, so that
\begin{align*}
g^{v}\left( D_{u}\right) & =\left( g^{v-1}\circ g\right) \left(
D_{u}\right) =g^{v-1}\left( \underbrace{g\left( D_{u}\right) }_{=0}\right)
\\
& =g^{v-1}\left( 0\right) =0\ \ \ \ \ \ \ \ \ \ \left( \text{since the map
}g^{v-1}\text{ is }\mathbf{k}\text{-linear}\right) .
\end{align*}
Hence, (\ref{pf.thm.id-f.gen.ass-Ker.IH4}) is proved in Case 1.
\par
Let us now consider Case 2. In this case, we have $u>p$. Therefore,
(\ref{pf.thm.id-f.gen.ass-Ker.IH3}) yields $\left( e-f\right) ^{u-p+1}%
\left( D_{u}\right) =0$. This rewrites as $g^{u-p+1}\left( D_{u}\right)
=0$ (since $g=e-f$). Set $u^{\prime}=u-p+1$. Thus, we have $g^{u^{\prime}%
}\left( D_{u}\right) =g^{u-p+1}\left( D_{u}\right) =0$.
\par
However, from $v>u-p$, we obtain $v\geq u-p+1$ (since $v$ and $u-p$ are
integers). In other words, $v\geq u^{\prime}$ (since $u^{\prime}=u-p+1$).
Thus, $g^{v}=g^{v-u^{\prime}}\circ g^{u^{\prime}}$, so that%
\begin{align*}
g^{v}\left( D_{u}\right) & =\left( g^{v-u^{\prime}}\circ g^{u^{\prime}%
}\right) \left( D_{u}\right) =g^{v-u^{\prime}}\left(
\underbrace{g^{u^{\prime}}\left( D_{u}\right) }_{=0}\right) \\
& =g^{v-u^{\prime}}\left( 0\right) =0\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }g^{v-u^{\prime}}\text{ is }\mathbf{k}\text{-linear}%
\right) .
\end{align*}
Hence, (\ref{pf.thm.id-f.gen.ass-Ker.IH4}) is proved in Case 2.
\par
We have now proved (\ref{pf.thm.id-f.gen.ass-Ker.IH4}) in both Cases 1 and 2.
Since these two Cases cover all possibilities, we thus conclude that
(\ref{pf.thm.id-f.gen.ass-Ker.IH4}) always holds.}.
Now, let $k=n-p$. Then, $k=n-p>0$ (since $n>p$), so that $k\in\mathbb{N}$
(since $k$ is an integer). Furthermore, (\ref{pf.thm.id-f.gen.ass-Ker.delgk=})
yields $\delta\circ g^{k}=h^{k}\circ\delta$. Thus,
\begin{align}
\left( \delta\circ g^{k}\right) \left( D_{n}\right) & =\left(
h^{k}\circ\delta\right) \left( D_{n}\right) =h^{k}\underbrace{\left(
\delta\left( D_{n}\right) \right) }_{\substack{\subseteq\sum_{i=1}%
^{n-1}D_{i}\otimes D_{n-i}\\\text{(by (\ref{pf.thm.id-f.gen.ass-gr}))}%
}}\nonumber\\
& \subseteq h^{k}\left( \sum_{i=1}^{n-1}D_{i}\otimes D_{n-i}\right)
\nonumber\\
& =\sum_{i=1}^{n-1}h^{k}\left( D_{i}\otimes D_{n-i}\right)
\label{pf.thm.id-f.gen.in-bigsum}%
\end{align}
(since the map $h^{k}$ is $\mathbf{k}$-linear).
We shall now prove that each $i\in\left\{ 1,2,\ldots,n-1\right\} $ and each
$r\in\left\{ 0,1,\ldots,k\right\} $ satisfy
\begin{equation}
\left( g^{r}\otimes g^{k-r}\right) \left( D_{i}\otimes D_{n-i}\right) =0.
\label{pf.thm.id-f.gen.ass-Ker.tens=0}%
\end{equation}
[\textit{Proof of (\ref{pf.thm.id-f.gen.ass-Ker.tens=0}):} Fix $i\in\left\{
1,2,\ldots,n-1\right\} $ and $r\in\left\{ 0,1,\ldots,k\right\} $. We must
prove (\ref{pf.thm.id-f.gen.ass-Ker.tens=0}).
We note that (\ref{pf.thm.id-f.gen.ass-Ker.abUV}) (applied to $\alpha=g^{r}$,
$\beta=g^{k-r}$, $U=D_{i}$ and $V=D_{n-i}$) yields%
\begin{equation}
\left( g^{r}\otimes g^{k-r}\right) \left( D_{i}\otimes D_{n-i}\right)
=g^{r}\left( D_{i}\right) \otimes g^{k-r}\left( D_{n-i}\right) .
\label{pf.thm.id-f.gen.ass-Ker.tens=0.pf.1}%
\end{equation}
We have $i\in\left\{ 1,2,\ldots,n-1\right\} $ and thus $1\leq i\leq n-1$, so
that $i\leq n-1i$, so that $n-i>0$. Moreover, from $i\geq1>0$,
we obtain $n-i0$ and $n-i>0$).
We are in one of the following three cases:
\textit{Case 1:} We have $r\geq k$.
\textit{Case 2:} We have $r\geq i$.
\textit{Case 3:} We have neither $r\geq k$ nor $r\geq i$.
Let us first consider Case 1. In this case, we have $r\geq k$. Thus, $r\geq
k=\underbrace{n}_{>i}-p>i-p$. Moreover, the integer $r$ is positive (since
$r\geq k>0$). Hence, (\ref{pf.thm.id-f.gen.ass-Ker.IH4}) (applied to $u=i$ and
$v=r$) yields $g^{r}\left( D_{i}\right) =0$ (since $r>i-p$). Now,
(\ref{pf.thm.id-f.gen.ass-Ker.tens=0.pf.1}) becomes%
\[
\left( g^{r}\otimes g^{k-r}\right) \left( D_{i}\otimes D_{n-i}\right)
=\underbrace{g^{r}\left( D_{i}\right) }_{=0}\otimes g^{k-r}\left(
D_{n-i}\right) =0\otimes g^{k-r}\left( D_{n-i}\right) =0.
\]
Thus, (\ref{pf.thm.id-f.gen.ass-Ker.tens=0}) is proved in Case 1.
Let us next consider Case 2. In this case, we have $r\geq i$. Thus, $r\geq
i>i-p$ (since $p>0$). Moreover, the integer $r$ is positive (since $r\geq
i>0$). Hence, (\ref{pf.thm.id-f.gen.ass-Ker.IH4}) (applied to $u=i$ and $v=r$)
yields $g^{r}\left( D_{i}\right) =0$ (since $r>i-p$). Now,
(\ref{pf.thm.id-f.gen.ass-Ker.tens=0.pf.1}) becomes%
\[
\left( g^{r}\otimes g^{k-r}\right) \left( D_{i}\otimes D_{n-i}\right)
=\underbrace{g^{r}\left( D_{i}\right) }_{=0}\otimes g^{k-r}\left(
D_{n-i}\right) =0\otimes g^{k-r}\left( D_{n-i}\right) =0.
\]
Thus, (\ref{pf.thm.id-f.gen.ass-Ker.tens=0}) is proved in Case 2.
Finally, let us consider Case 3. In this case, we have neither $r\geq k$ nor
$r\geq i$. In other words, we have $rn-p-i=n-i-p.
\]
Hence, (\ref{pf.thm.id-f.gen.ass-Ker.IH4}) (applied to $u=n-i$ and $v=k-r$)
yields $g^{k-r}\left( D_{n-i}\right) =0$. Now,
(\ref{pf.thm.id-f.gen.ass-Ker.tens=0.pf.1}) becomes%
\[
\left( g^{r}\otimes g^{k-r}\right) \left( D_{i}\otimes D_{n-i}\right)
=g^{r}\left( D_{i}\right) \otimes\underbrace{g^{k-r}\left( D_{n-i}\right)
}_{=0}=g^{r}\left( D_{i}\right) \otimes0=0.
\]
Thus, (\ref{pf.thm.id-f.gen.ass-Ker.tens=0}) is proved in Case 3.
We have now proved (\ref{pf.thm.id-f.gen.ass-Ker.tens=0}) in all three Cases
1, 2 and 3. Thus, the proof of (\ref{pf.thm.id-f.gen.ass-Ker.tens=0}) is complete.]
Now, each $i\in\left\{ 1,2,\ldots,n-1\right\} $ satisfies%
\begin{align}
\underbrace{h^{k}}_{\substack{=\sum_{r=0}^{k}h_{k,r}\\\text{(by
(\ref{pf.thm.id-f.gen.hk=}))}}}\left( D_{i}\otimes D_{n-i}\right) &
=\left( \sum_{r=0}^{k}h_{k,r}\right) \left( D_{i}\otimes D_{n-i}\right)
\nonumber\\
& \subseteq\sum_{r=0}^{k}\underbrace{h_{k,r}}_{\substack{=\dbinom{k}%
{r}\left( e^{k-r}\otimes f^{r}\right) \circ\left( g^{r}\otimes
g^{k-r}\right) \\\text{(by (\ref{pf.thm.id-f.gen.hkr=}))}}}\left(
D_{i}\otimes D_{n-i}\right) \nonumber\\
& =\sum_{r=0}^{k}\underbrace{\left( \dbinom{k}{r}\left( e^{k-r}\otimes
f^{r}\right) \circ\left( g^{r}\otimes g^{k-r}\right) \right) \left(
D_{i}\otimes D_{n-i}\right) }_{=\dbinom{k}{r}\left( e^{k-r}\otimes
f^{r}\right) \left( \left( g^{r}\otimes g^{k-r}\right) \left(
D_{i}\otimes D_{n-i}\right) \right) }\nonumber\\
& =\sum_{r=0}^{k}\dbinom{k}{r}\left( e^{k-r}\otimes f^{r}\right)
\underbrace{\left( \left( g^{r}\otimes g^{k-r}\right) \left( D_{i}\otimes
D_{n-i}\right) \right) }_{\substack{=0\\\text{(by
(\ref{pf.thm.id-f.gen.ass-Ker.tens=0}))}}}\nonumber\\
& =\sum_{r=0}^{k}\dbinom{k}{r}\underbrace{\left( e^{k-r}\otimes
f^{r}\right) \left( 0\right) }_{\substack{=0\\\text{(since the map }%
e^{k-r}\otimes f^{r}\text{ is }\mathbf{k}\text{-linear)}}}=\sum_{r=0}%
^{k}\underbrace{\dbinom{k}{r}0}_{=0}\nonumber\\
& =0. \label{pf.thm.id-f.gen.hkrof=0}%
\end{align}
Hence, (\ref{pf.thm.id-f.gen.in-bigsum}) becomes%
\[
\left( \delta\circ g^{k}\right) \left( D_{n}\right) \subseteq\sum
_{i=1}^{n-1}\underbrace{h^{k}\left( D_{i}\otimes D_{n-i}\right)
}_{\substack{\subseteq0\\\text{(by (\ref{pf.thm.id-f.gen.hkrof=0}))}%
}}\subseteq\sum_{i=1}^{n-1}0=0.
\]
From this, we can easily obtain
\[
g^{k}\left( D_{n}\right) \subseteq\operatorname*{Ker}\delta
\]
\footnote{\textit{Proof.} Let $x\in g^{k}\left( D_{n}\right) $. Thus, there
exists some $y\in D_{n}$ such that $x=g^{k}\left( y\right) $. Consider this
$y$. Applying the map $\delta$ to both sides of the equality $x=g^{k}\left(
y\right) $, we find%
\[
\delta\left( x\right) =\delta\left( g^{k}\left( y\right) \right)
=\left( \delta\circ g^{k}\right) \left( \underbrace{y}_{\in D_{n}}\right)
\in\left( \delta\circ g^{k}\right) \left( D_{n}\right) \subseteq0,
\]
so that $\delta\left( x\right) =0$. Hence, $x\in\operatorname*{Ker}\delta$.
\par
Now, forget that we fixed $x$. We thus have shown that $x\in
\operatorname*{Ker}\delta$ for each $x\in g^{k}\left( D_{n}\right) $. In
other words, $g^{k}\left( D_{n}\right) \subseteq\operatorname*{Ker}\delta$%
.}. Since $g=e-f$ and $k=n-p$, we can rewrite this as follows:%
\begin{equation}
\left( e-f\right) ^{n-p}\left( D_{n}\right) \subseteq\operatorname*{Ker}%
\delta. \label{pf.thm.id-f.gen.done1}%
\end{equation}
However, we have $\operatorname*{Ker}\delta\subseteq\operatorname*{Ker}\left(
e-f\right) $ (by (\ref{pf.thm.id-f.gen.ass-Ker})) and therefore $\left(
e-f\right) \left( \operatorname*{Ker}\delta\right) =0$%
\ \ \ \ \footnote{\textit{Proof.} Let $y\in\left( e-f\right) \left(
\operatorname*{Ker}\delta\right) $. Thus, there exists some $x\in
\operatorname*{Ker}\delta$ such that $y=\left( e-f\right) \left( x\right)
$. Consider this $x$.
\par
We have $x\in\operatorname*{Ker}\delta\subseteq\operatorname*{Ker}\left(
e-f\right) $, so that $\left( e-f\right) \left( x\right) =0$. Hence,
$y=\left( e-f\right) \left( x\right) =0$.
\par
Forget that we fixed $y$. We thus have shown that $y=0$ for each $y\in\left(
e-f\right) \left( \operatorname*{Ker}\delta\right) $. In other words,
$\left( e-f\right) \left( \operatorname*{Ker}\delta\right) \subseteq0$.
Hence, $\left( e-f\right) \left( \operatorname*{Ker}\delta\right) =0$
(since $\left( e-f\right) \left( \operatorname*{Ker}\delta\right) $ is a
$\mathbf{k}$-module).}. Thus,%
\begin{align*}
\underbrace{\left( e-f\right) ^{n-p+1}}_{=\left( e-f\right) \circ\left(
e-f\right) ^{n-p}}\left( D_{n}\right) & =\left( \left( e-f\right)
\circ\left( e-f\right) ^{n-p}\right) \left( D_{n}\right) \\
& =\left( e-f\right) \left( \underbrace{\left( e-f\right) ^{n-p}\left(
D_{n}\right) }_{\substack{\subseteq\operatorname*{Ker}\delta\\\text{(by
(\ref{pf.thm.id-f.gen.done1}))}}}\right) \subseteq\left( e-f\right) \left(
\operatorname*{Ker}\delta\right) =0.
\end{align*}
In other words,%
\begin{equation}
\left( e-f\right) ^{n-p+1}\left( D_{n}\right) =0.
\label{pf.thm.id-f.gen.done2}%
\end{equation}
We have now proved the relations (\ref{pf.thm.id-f.gen.done1}) and
(\ref{pf.thm.id-f.gen.done2}). In other words, (\ref{eq.thm.id-f.gen.clm1})
and (\ref{eq.thm.id-f.gen.clm2}) hold for $u=n$. This completes the induction
step. Thus, we have proved by strong induction that
(\ref{eq.thm.id-f.gen.clm1}) and (\ref{eq.thm.id-f.gen.clm2}) hold for all
integers $u>p$. This proves Theorem \ref{thm.id-f.gen}.
\end{proof}
\subsection{\label{sec.pf.prop.cfc.delta2}Proof of Proposition
\ref{prop.cfc.delta2}}
Our next goal is to prove Proposition \ref{prop.cfc.delta2}. We shall work
towards this goal by proving some lemmas. First, we note a simple consequence
of the axioms of a $\mathbf{k}$-coalgebra: \Needspace{15pc}
\begin{remark}
\label{rmk.cfc.can}Let $C$ be any $\mathbf{k}$-coalgebra. Let
$\operatorname*{can}\nolimits_{1}:C\otimes\mathbf{k}\rightarrow C$ be the
$\mathbf{k}$-module isomorphism sending $c\otimes1$ to $c$ for each $c\in C$.
Let $\operatorname*{can}\nolimits_{2}:\mathbf{k}\otimes C\rightarrow C$ be the
$\mathbf{k}$-module isomorphism sending $1\otimes c$ to $c$ for each $c\in C$.
Recall that $C$ is a $\mathbf{k}$-coalgebra, and therefore satisfies the
axioms of a $\mathbf{k}$-coalgebra. Hence, in particular, the diagram%
\[
\xymatrixcolsep{5pc}\xymatrix{
C \otimes \kk \ar[r]^-{\operatorname{can}_1} & C & \kk \otimes C \ar[l]_-{\operatorname{can}_2} \\
C \otimes C \ar[u]^{\id \otimes \epsilon} & C \ar[l]^-{\Delta} \ar[u]^\id \ar[r]_-{\Delta} & C \otimes C \ar[u]^{\epsilon \otimes \id}
}
\]
is commutative (since this is one of the axioms of a $\mathbf{k}$-coalgebra).
In other words, we have%
\begin{align}
\operatorname*{can}\nolimits_{1}\circ\left( \operatorname*{id}\otimes
\epsilon\right) \circ\Delta & =\operatorname*{id}%
\ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.lem.cfc.idbar.a.coalg1}\\
\operatorname*{can}\nolimits_{2}\circ\left( \epsilon\otimes\operatorname*{id}%
\right) \circ\Delta & =\operatorname*{id}. \label{pf.lem.cfc.idbar.a.coalg2}%
\end{align}
\end{remark}
Next, we state a simple property of a slight generalization of primitive
elements in a coalgebra:
\begin{lemma}
\label{lem.coalg.primitive-e0}Let $C$ be any $\mathbf{k}$-coalgebra. Let
$a,b,d\in C$ be three elements satisfying $\epsilon\left( a\right) =1$ and
$\epsilon\left( b\right) =1$ and $\Delta\left( d\right) =d\otimes
a+b\otimes d$. Then, $\epsilon\left( d\right) =0$.
\end{lemma}
We shall later apply Lemma \ref{lem.coalg.primitive-e0} to the case when
$a=b=1_{C}$ (and $C$ is either a connected filtered $\mathbf{k}$-coalgebra or
a $\mathbf{k}$-bialgebra, so that $1_{C}$ does make sense); however, it is not
any harder to prove it in full generality:
\begin{proof}
[Proof of Lemma \ref{lem.coalg.primitive-e0}.]Let us first observe that the
map%
\begin{align}
C\times C & \rightarrow C\otimes C,\nonumber\\
\left( u,v\right) & \mapsto u\otimes v
\label{pf.lem.coalg.primitive-e0.tensoring}%
\end{align}
is $\mathbf{k}$-bilinear. (This follows straight from the definition of the
tensor product.)
Applying the map $\operatorname*{id}\otimes\epsilon:C\otimes C\rightarrow
C\otimes\mathbf{k}$ to both sides of the equality $\Delta\left( d\right)
=d\otimes a+b\otimes d$, we obtain%
\begin{align}
\left( \operatorname*{id}\otimes\epsilon\right) \left( \Delta\left(
d\right) \right) & =\left( \operatorname*{id}\otimes\epsilon\right)
\left( d\otimes a+b\otimes d\right) \nonumber\\
& =\underbrace{\left( \operatorname*{id}\otimes\epsilon\right) \left(
d\otimes a\right) }_{=\operatorname*{id}\left( d\right) \otimes
\epsilon\left( a\right) }+\underbrace{\left( \operatorname*{id}%
\otimes\epsilon\right) \left( b\otimes d\right) }_{=\operatorname*{id}%
\left( b\right) \otimes\epsilon\left( d\right) }\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since the map }\operatorname*{id}%
\otimes\epsilon\text{ is }\mathbf{k}\text{-linear}\right) \nonumber\\
& =\underbrace{\operatorname*{id}\left( d\right) }_{=d}\otimes
\underbrace{\epsilon\left( a\right) }_{=1}+\underbrace{\operatorname*{id}%
\left( b\right) }_{=b}\otimes\underbrace{\epsilon\left( d\right)
}_{=\epsilon\left( d\right) 1}\nonumber\\
& =d\otimes1+\underbrace{b\otimes\epsilon\left( d\right) 1}%
_{\substack{=b\epsilon\left( d\right) \otimes1\\\text{(since }%
\epsilon\left( d\right) \text{ is a scalar in }\mathbf{k}\text{ and
thus}\\\text{can be moved past the }\otimes\text{ sign)}}}=d\otimes
1+b\epsilon\left( d\right) \otimes1\nonumber\\
& =\left( d+b\epsilon\left( d\right) \right) \otimes1
\label{pf.lem.coalg.primitive-e0.1}%
\end{align}
(since the map (\ref{pf.lem.coalg.primitive-e0.tensoring}) is $\mathbf{k}$-bilinear).
Define the maps $\operatorname*{can}\nolimits_{1}$ and $\operatorname*{can}%
\nolimits_{2}$ as in Remark \ref{rmk.cfc.can}. Applying the map
$\operatorname*{can}\nolimits_{1}$ to both sides of the equality
(\ref{pf.lem.coalg.primitive-e0.1}), we obtain%
\[
\operatorname*{can}\nolimits_{1}\left( \left( \operatorname*{id}%
\otimes\epsilon\right) \left( \Delta\left( d\right) \right) \right)
=\operatorname*{can}\nolimits_{1}\left( \left( d+b\epsilon\left( d\right)
\right) \otimes1\right) =d+b\epsilon\left( d\right)
\]
(by the definition of $\operatorname*{can}\nolimits_{1}$). Hence,%
\[
d+b\epsilon\left( d\right) =\operatorname*{can}\nolimits_{1}\left( \left(
\operatorname*{id}\otimes\epsilon\right) \left( \Delta\left( d\right)
\right) \right) =\underbrace{\left( \operatorname*{can}\nolimits_{1}%
\circ\left( \operatorname*{id}\otimes\epsilon\right) \circ\Delta\right)
}_{\substack{=\operatorname*{id}\\\text{(by (\ref{pf.lem.cfc.idbar.a.coalg1}%
))}}}\left( d\right) =\operatorname*{id}\left( d\right) =d.
\]
Subtracting $d$ from both sides of this equality, we obtain $b\epsilon\left(
d\right) =0$. Applying the map $\epsilon$ to both sides of this equality, we
find $\epsilon\left( b\epsilon\left( d\right) \right) =0$. In view of%
\begin{align*}
\epsilon\left( \underbrace{b\epsilon\left( d\right) }_{=\epsilon\left(
d\right) b}\right) & =\epsilon\left( \epsilon\left( d\right) b\right)
=\epsilon\left( d\right) \underbrace{\epsilon\left( b\right) }%
_{=1}\ \ \ \ \ \ \ \ \ \ \left( \text{since the map }\epsilon\text{ is
}\mathbf{k}\text{-linear}\right) \\
& =\epsilon\left( d\right) ,
\end{align*}
this rewrites as $\epsilon\left( d\right) =0$. This proves Lemma
\ref{lem.coalg.primitive-e0}.
\end{proof}
Next, let us define a \textquotedblleft reduced identity map\textquotedblright%
\ $\idbar$ for any connected filtered $\mathbf{k}$-coalgebra $C$, and explore
some of its properties:
\begin{lemma}
\label{lem.cfc.idbar}Let $C$ be a connected filtered $\mathbf{k}$-coalgebra
with filtration $\left( C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $.
Define a $\mathbf{k}$-linear map $\idbar:C\rightarrow C$ by setting%
\[
\idbar\left( c\right) :=c-\epsilon\left( c\right) 1_{C}%
\ \ \ \ \ \ \ \ \ \ \text{for each }c\in C.
\]
Define a $\mathbf{k}$-linear map $\delta:C\rightarrow C\otimes C$ by setting
\[
\delta\left( c\right) :=\Delta\left( c\right) -c\otimes1_{C}-1_{C}\otimes
c+\epsilon\left( c\right) 1_{C}\otimes1_{C}\ \ \ \ \ \ \ \ \ \ \text{for
each }c\in C.
\]
Then:
\textbf{(a)} We have $\delta=\left( \idbar\otimes\idbar\right) \circ\Delta$.
(Here, of course, $\Delta$ denotes the comultiplication of $C$.)
\textbf{(b)} We have $\idbar\left( C_{\leq n}\right) \subseteq C_{\leq n}$
for each $n\in\mathbb{N}$.
\textbf{(c)} We have $\idbar\left( C_{\leq0}\right) =0$.
\end{lemma}
\begin{proof}
[Proof of Lemma \ref{lem.cfc.idbar}.]Let us first observe that the map%
\begin{align}
C\times C & \rightarrow C\otimes C,\nonumber\\
\left( u,v\right) & \mapsto u\otimes v
\label{pf.lem.cfc.idbar.tensoring-map}%
\end{align}
is $\mathbf{k}$-bilinear. (This follows straight from the definition of the
tensor product.)
\textbf{(a)} Let $c\in C$. The element $\Delta\left( c\right) $ is a tensor
in $C\otimes C$, and thus can be written in the form%
\begin{equation}
\Delta\left( c\right) =\sum_{i=1}^{m}\lambda_{i}c_{i}\otimes d_{i}
\label{pf.lem.cfc.idbar.a.1}%
\end{equation}
for some $m\in\mathbb{N}$, some $\lambda_{1},\lambda_{2},\ldots,\lambda_{m}%
\in\mathbf{k}$, some $c_{1},c_{2},\ldots,c_{m}\in C$ and some $d_{1}%
,d_{2},\ldots,d_{m}\in C$. Consider this $m$, these $\lambda_{1},\lambda
_{2},\ldots,\lambda_{m}$, these $c_{1},c_{2},\ldots,c_{m}$ and these
$d_{1},d_{2},\ldots,d_{m}$.
Now, define the maps $\operatorname*{can}\nolimits_{1}$ and
$\operatorname*{can}\nolimits_{2}$ as in Remark \ref{rmk.cfc.can}. Applying
the map $\operatorname*{id}\otimes\epsilon$ to both sides of the equality
(\ref{pf.lem.cfc.idbar.a.1}), we obtain%
\begin{align*}
\left( \operatorname*{id}\otimes\epsilon\right) \left( \Delta\left(
c\right) \right) & =\left( \operatorname*{id}\otimes\epsilon\right)
\left( \sum_{i=1}^{m}\lambda_{i}c_{i}\otimes d_{i}\right) \\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\left( \operatorname*{id}%
\otimes\epsilon\right) \left( c_{i}\otimes d_{i}\right) }%
_{=\operatorname*{id}\left( c_{i}\right) \otimes\epsilon\left(
d_{i}\right) }\ \ \ \ \ \ \ \ \ \ \left( \text{since the map }%
\operatorname*{id}\otimes\epsilon\text{ is }\mathbf{k}\text{-linear}\right) \\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\operatorname*{id}\left(
c_{i}\right) }_{=c_{i}}\otimes\underbrace{\epsilon\left( d_{i}\right)
}_{=\epsilon\left( d_{i}\right) 1}=\sum_{i=1}^{m}\lambda_{i}%
\underbrace{c_{i}\otimes\epsilon\left( d_{i}\right) 1}_{\substack{=c_{i}%
\epsilon\left( d_{i}\right) \otimes1\\\text{(since }\epsilon\left(
d_{i}\right) \text{ is a scalar in }\mathbf{k}\text{ and thus}\\\text{can be
moved past the }\otimes\text{ sign)}}}\\
& =\sum_{i=1}^{m}\lambda_{i}c_{i}\epsilon\left( d_{i}\right) \otimes1.
\end{align*}
Applying the map $\operatorname*{can}\nolimits_{1}$ to both sides of this
equality, we obtain%
\begin{align*}
\operatorname*{can}\nolimits_{1}\left( \left( \operatorname*{id}%
\otimes\epsilon\right) \left( \Delta\left( c\right) \right) \right) &
=\operatorname*{can}\nolimits_{1}\left( \sum_{i=1}^{m}\lambda_{i}%
c_{i}\epsilon\left( d_{i}\right) \otimes1\right) \\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\operatorname*{can}\nolimits_{1}%
\left( c_{i}\epsilon\left( d_{i}\right) \otimes1\right) }%
_{\substack{=c_{i}\epsilon\left( d_{i}\right) \\\text{(by the definition of
}\operatorname*{can}\nolimits_{1}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\operatorname*{can}\nolimits_{1}\text{ is }%
\mathbf{k}\text{-linear}\right) \\
& =\sum_{i=1}^{m}\lambda_{i}c_{i}\epsilon\left( d_{i}\right) .
\end{align*}
Comparing this with
\[
\operatorname*{can}\nolimits_{1}\left( \left( \operatorname*{id}%
\otimes\epsilon\right) \left( \Delta\left( c\right) \right) \right)
=\underbrace{\left( \operatorname*{can}\nolimits_{1}\circ\left(
\operatorname*{id}\otimes\epsilon\right) \circ\Delta\right) }%
_{\substack{=\operatorname*{id}\\\text{(by (\ref{pf.lem.cfc.idbar.a.coalg1}%
))}}}\left( c\right) =\operatorname*{id}\left( c\right) =c,
\]
we obtain%
\begin{equation}
\sum_{i=1}^{m}\lambda_{i}c_{i}\epsilon\left( d_{i}\right) =c.
\label{pf.lem.cfc.idbar.a.coalg1c}%
\end{equation}
Furthermore, applying the map $\epsilon\otimes\operatorname*{id}$ to both
sides of the equality (\ref{pf.lem.cfc.idbar.a.1}), we obtain%
\begin{align*}
\left( \epsilon\otimes\operatorname*{id}\right) \left( \Delta\left(
c\right) \right) & =\left( \epsilon\otimes\operatorname*{id}\right)
\left( \sum_{i=1}^{m}\lambda_{i}c_{i}\otimes d_{i}\right) \\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\left( \epsilon\otimes
\operatorname*{id}\right) \left( c_{i}\otimes d_{i}\right) }_{=\epsilon
\left( c_{i}\right) \otimes\operatorname*{id}\left( d_{i}\right)
}\ \ \ \ \ \ \ \ \ \ \left( \text{since the map }\epsilon\otimes
\operatorname*{id}\text{ is }\mathbf{k}\text{-linear}\right) \\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\epsilon\left( c_{i}\right)
}_{=1\epsilon\left( c_{i}\right) }\otimes\underbrace{\operatorname*{id}%
\left( d_{i}\right) }_{=d_{i}}=\sum_{i=1}^{m}\lambda_{i}%
\underbrace{1\epsilon\left( c_{i}\right) \otimes d_{i}}_{\substack{=1\otimes
\epsilon\left( c_{i}\right) d_{i}\\\text{(since }\epsilon\left(
c_{i}\right) \text{ is a scalar in }\mathbf{k}\text{ and thus}\\\text{can be
moved past the }\otimes\text{ sign)}}}\\
& =\sum_{i=1}^{m}\lambda_{i}1\otimes\epsilon\left( c_{i}\right) d_{i}.
\end{align*}
Applying the map $\operatorname*{can}\nolimits_{2}$ to both sides of this
equality, we obtain%
\begin{align*}
\operatorname*{can}\nolimits_{2}\left( \left( \epsilon\otimes
\operatorname*{id}\right) \left( \Delta\left( c\right) \right) \right)
& =\operatorname*{can}\nolimits_{2}\left( \sum_{i=1}^{m}\lambda_{i}%
1\otimes\epsilon\left( c_{i}\right) d_{i}\right) \\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\operatorname*{can}\nolimits_{2}%
\left( 1\otimes\epsilon\left( c_{i}\right) d_{i}\right) }%
_{\substack{=\epsilon\left( c_{i}\right) d_{i}\\\text{(by the definition of
}\operatorname*{can}\nolimits_{2}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\operatorname*{can}\nolimits_{2}\text{ is }%
\mathbf{k}\text{-linear}\right) \\
& =\sum_{i=1}^{m}\lambda_{i}\epsilon\left( c_{i}\right) d_{i}.
\end{align*}
Comparing this with
\[
\operatorname*{can}\nolimits_{2}\left( \left( \epsilon\otimes
\operatorname*{id}\right) \left( \Delta\left( c\right) \right) \right)
=\underbrace{\left( \operatorname*{can}\nolimits_{2}\circ\left(
\epsilon\otimes\operatorname*{id}\right) \circ\Delta\right) }%
_{\substack{=\operatorname*{id}\\\text{(by (\ref{pf.lem.cfc.idbar.a.coalg2}%
))}}}\left( c\right) =\operatorname*{id}\left( c\right) =c,
\]
we obtain%
\begin{equation}
\sum_{i=1}^{m}\lambda_{i}\epsilon\left( c_{i}\right) d_{i}=c.
\label{pf.lem.cfc.idbar.a.coalg2c}%
\end{equation}
Applying the map $\idbar$ to both sides of this equality, we obtain%
\[
\idbar\left( \sum_{i=1}^{m}\lambda_{i}\epsilon\left( c_{i}\right)
d_{i}\right) =\idbar\left( c\right) =c-\epsilon\left( c\right) 1_{C}%
\]
(by the definition of $\idbar$). Hence,%
\begin{equation}
c-\epsilon\left( c\right) 1_{C}=\idbar\left( \sum_{i=1}^{m}\lambda
_{i}\epsilon\left( c_{i}\right) d_{i}\right) =\sum_{i=1}^{m}\lambda
_{i}\epsilon\left( c_{i}\right) \idbar\left( d_{i}\right)
\label{pf.lem.cfc.idbar.a.coalg2cid}%
\end{equation}
(since the map $\idbar$ is $\mathbf{k}$-linear).
Now, applying the map $\idbar\otimes\idbar$ to both sides of the equality
(\ref{pf.lem.cfc.idbar.a.1}), we obtain%
\begin{align}
& \left( \idbar\otimes\idbar\right) \left( \Delta\left( c\right) \right)
\nonumber\\
& =\left( \idbar\otimes\idbar\right) \left( \sum_{i=1}^{m}\lambda_{i}%
c_{i}\otimes d_{i}\right) \nonumber\\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\left( \idbar\otimes\idbar\right)
\left( c_{i}\otimes d_{i}\right) }_{=\idbar\left( c_{i}\right)
\otimes\idbar\left( d_{i}\right) }\ \ \ \ \ \ \ \ \ \ \left( \text{since
the map }\idbar\otimes\idbar\text{ is }\mathbf{k}\text{-linear}\right)
\nonumber\\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\idbar\left( c_{i}\right)
}_{\substack{=c_{i}-\epsilon\left( c_{i}\right) 1_{C}\\\text{(by the
definition of }\idbar\text{)}}}\otimes\idbar\left( d_{i}\right) \nonumber\\
& =\sum_{i=1}^{m}\lambda_{i}\underbrace{\left( c_{i}-\epsilon\left(
c_{i}\right) 1_{C}\right) \otimes\idbar\left( d_{i}\right) }%
_{\substack{=c_{i}\otimes\idbar\left( d_{i}\right) -\epsilon\left(
c_{i}\right) 1_{C}\otimes\idbar\left( d_{i}\right) \\\text{(since the map
(\ref{pf.lem.cfc.idbar.tensoring-map}) is }\mathbf{k}\text{-bilinear)}%
}}\nonumber\\
& =\sum_{i=1}^{m}\lambda_{i}\left( c_{i}\otimes\idbar\left( d_{i}\right)
-\epsilon\left( c_{i}\right) 1_{C}\otimes\idbar\left( d_{i}\right) \right)
\nonumber\\
& =\sum_{i=1}^{m}\lambda_{i}c_{i}\otimes\idbar\left( d_{i}\right)
-\sum_{i=1}^{m}\lambda_{i}\epsilon\left( c_{i}\right) 1_{C}\otimes
\idbar\left( d_{i}\right) . \label{pf.lem.cfc.idbar.a.3}%
\end{align}
We shall now separately simplify the two sums on the right hand side of this equality.
Indeed, we have
\begin{align*}
& \sum_{i=1}^{m}\lambda_{i}c_{i}\otimes\underbrace{\idbar\left(
d_{i}\right) }_{\substack{=d_{i}-\epsilon\left( d_{i}\right) 1_{C}%
\\\text{(by the definition of }\idbar\text{)}}}\\
& =\sum_{i=1}^{m}\underbrace{\lambda_{i}c_{i}\otimes\left( d_{i}%
-\epsilon\left( d_{i}\right) 1_{C}\right) }_{\substack{=\lambda_{i}%
c_{i}\otimes d_{i}-\lambda_{i}c_{i}\otimes\epsilon\left( d_{i}\right)
1_{C}\\\text{(since the map (\ref{pf.lem.cfc.idbar.tensoring-map}) is
}\mathbf{k}\text{-bilinear)}}}\\
& =\sum_{i=1}^{m}\left( \lambda_{i}c_{i}\otimes d_{i}-\lambda_{i}%
c_{i}\otimes\epsilon\left( d_{i}\right) 1_{C}\right) \\
& =\underbrace{\sum_{i=1}^{m}\lambda_{i}c_{i}\otimes d_{i}}%
_{\substack{=\Delta\left( c\right) \\\text{(by (\ref{pf.lem.cfc.idbar.a.1}%
))}}}-\sum_{i=1}^{m}\underbrace{\lambda_{i}c_{i}\otimes\epsilon\left(
d_{i}\right) 1_{C}}_{\substack{=\lambda_{i}c_{i}\epsilon\left( d_{i}\right)
\otimes1_{C}\\\text{(since }\epsilon\left( d_{i}\right) \text{ is a scalar
in }\mathbf{k}\text{ and thus}\\\text{can be moved past the }\otimes\text{
sign)}}}\\
& =\Delta\left( c\right) -\underbrace{\sum_{i=1}^{m}\lambda_{i}%
c_{i}\epsilon\left( d_{i}\right) \otimes1_{C}}_{\substack{=\left(
\sum_{i=1}^{m}\lambda_{i}c_{i}\epsilon\left( d_{i}\right) \right)
\otimes1_{C}\\\text{(since the map (\ref{pf.lem.cfc.idbar.tensoring-map}) is
}\mathbf{k}\text{-bilinear)}}}=\Delta\left( c\right) -\underbrace{\left(
\sum_{i=1}^{m}\lambda_{i}c_{i}\epsilon\left( d_{i}\right) \right)
}_{\substack{=c\\\text{(by (\ref{pf.lem.cfc.idbar.a.coalg1c}))}}}\otimes
1_{C}\\
& =\Delta\left( c\right) -c\otimes1_{C}%
\end{align*}
and%
\begin{align*}
& \sum_{i=1}^{m}\underbrace{\lambda_{i}\epsilon\left( c_{i}\right) 1_{C}%
}_{=1_{C}\lambda_{i}\epsilon\left( c_{i}\right) }\otimes\idbar\left(
d_{i}\right) \\
& =\sum_{i=1}^{m}\underbrace{1_{C}\lambda_{i}\epsilon\left( c_{i}\right)
\otimes\idbar\left( d_{i}\right) }_{\substack{=1_{C}\otimes\lambda
_{i}\epsilon\left( c_{i}\right) \idbar\left( d_{i}\right) \\\text{(since
}\lambda_{i}\epsilon\left( c_{i}\right) \text{ is a scalar in }%
\mathbf{k}\text{ and thus}\\\text{can be moved past the }\otimes\text{ sign)}%
}}\\
& =\sum_{i=1}^{m}1_{C}\otimes\lambda_{i}\epsilon\left( c_{i}\right)
\idbar\left( d_{i}\right) =1_{C}\otimes\underbrace{\left( \sum_{i=1}%
^{m}\lambda_{i}\epsilon\left( c_{i}\right) \idbar\left( d_{i}\right)
\right) }_{\substack{=c-\epsilon\left( c\right) 1_{C}\\\text{(by
(\ref{pf.lem.cfc.idbar.a.coalg2cid}))}}}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since the map
(\ref{pf.lem.cfc.idbar.tensoring-map}) is }\mathbf{k}\text{-bilinear}\right)
\\
& =1_{C}\otimes\left( c-\epsilon\left( c\right) 1_{C}\right)
=1_{C}\otimes c-\underbrace{1_{C}\otimes\epsilon\left( c\right) 1_{C}%
}_{\substack{=1_{C}\epsilon\left( c\right) \otimes1_{C}\\\text{(since
}\epsilon\left( c\right) \text{ is a scalar in }\mathbf{k}\text{ and
thus}\\\text{can be moved past the }\otimes\text{ sign)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since the map
(\ref{pf.lem.cfc.idbar.tensoring-map}) is }\mathbf{k}\text{-bilinear}\right)
\\
& =1_{C}\otimes c-\underbrace{1_{C}\epsilon\left( c\right) }_{=\epsilon
\left( c\right) 1_{C}}\otimes1_{C}=1_{C}\otimes c-\epsilon\left( c\right)
1_{C}\otimes1_{C}.
\end{align*}
In light of these two equalities, we can rewrite (\ref{pf.lem.cfc.idbar.a.3})
as%
\[
\left( \idbar\otimes\idbar\right) \left( \Delta\left( c\right) \right)
=\left( \Delta\left( c\right) -c\otimes1_{C}\right) -\left( 1_{C}\otimes
c-\epsilon\left( c\right) 1_{C}\otimes1_{C}\right) .
\]
Comparing this with%
\begin{align*}
\delta\left( c\right) & =\Delta\left( c\right) -c\otimes1_{C}%
-1_{C}\otimes c+\epsilon\left( c\right) 1_{C}\otimes1_{C}%
\ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\delta\right) \\
& =\left( \Delta\left( c\right) -c\otimes1_{C}\right) -\left(
1_{C}\otimes c-\epsilon\left( c\right) 1_{C}\otimes1_{C}\right) ,
\end{align*}
we obtain $\delta\left( c\right) =\left( \idbar\otimes\idbar\right)
\left( \Delta\left( c\right) \right) =\left( \left( \idbar\otimes
\idbar\right) \circ\Delta\right) \left( c\right) $.
Forget that we fixed $c$. We thus have shown that $\delta\left( c\right)
=\left( \left( \idbar\otimes\idbar\right) \circ\Delta\right) \left(
c\right) $ for each $c\in C$. In other words, $\delta=\left( \idbar\otimes
\idbar\right) \circ\Delta$. This proves Lemma \ref{lem.cfc.idbar}
\textbf{(a)}.
\textbf{(b)} Let $n\in\mathbb{N}$. Let $c\in C_{\leq n}$. We have $1_{C}\in
C_{\leq0}$ (by Definition \ref{def.con-fil-coal} \textbf{(b)}). However,
(\ref{eq.def.fil-coal.chain}) yields $C_{\leq0}\subseteq C_{\leq1}\subseteq
C_{\leq2}\subseteq\cdots$ (since $C$ is a filtered $\mathbf{k}$-coalgebra).
Therefore, $C_{\leq0}\subseteq C_{\leq n}$. Thus, $1_{C}\in C_{\leq0}\subseteq
C_{\leq n}$. Now, the definition of $\idbar$ yields%
\[
\idbar\left( c\right) =\underbrace{c}_{\in C_{\leq n}}-\epsilon\left(
c\right) \underbrace{1_{C}}_{\in C_{\leq n}}\in C_{\leq n}-\epsilon\left(
c\right) C_{\leq n}\subseteq C_{\leq n}%
\]
(since $C_{\leq n}$ is a $\mathbf{k}$-module).
Forget that we fixed $c$. We thus have shown that $\idbar\left( c\right) \in
C_{\leq n}$ for each $c\in C_{\leq n}$. In other words, we have $\idbar\left(
C_{\leq n}\right) \subseteq C_{\leq n}$. This proves Lemma
\ref{lem.cfc.idbar} \textbf{(b)}.
\textbf{(c)} The filtered $\mathbf{k}$-coalgebra $C$ is connected. In other
words, the restriction $\epsilon\mid_{C_{\leq0}}$ is a $\mathbf{k}$-module
isomorphism from $C_{\leq0}$ to $\mathbf{k}$ (by Definition
\ref{def.con-fil-coal} \textbf{(a)}). Hence, this restriction $\epsilon
\mid_{C_{\leq0}}$ is bijective, and thus injective. Also, we have $1_{C}\in
C_{\leq0}$ (by Definition \ref{def.con-fil-coal} \textbf{(b)}).
Now, let $c\in C_{\leq0}$. Hence, $\left( \epsilon\mid_{C_{\leq0}}\right)
\left( c\right) $ is well-defined. Definition \ref{def.con-fil-coal}
\textbf{(b)} yields $1_{C}=\left( \epsilon\mid_{C_{\leq0}}\right)
^{-1}\left( 1_{\mathbf{k}}\right) $. Thus, $\left( \epsilon\mid_{C_{\leq0}%
}\right) \left( 1_{C}\right) =1_{\mathbf{k}}$. In other words,
$\epsilon\left( 1_{C}\right) =1_{\mathbf{k}}$ (since $\left( \epsilon
\mid_{C_{\leq0}}\right) \left( 1_{C}\right) =\epsilon\left( 1_{C}\right)
$).
Set $d=\epsilon\left( c\right) 1_{C}$. Then, $d=\epsilon\left( c\right)
\underbrace{1_{C}}_{\in C_{\leq0}}\in\epsilon\left( c\right) C_{\leq
0}\subseteq C_{\leq0}$ (since $C_{\leq0}$ is a $\mathbf{k}$-module). Thus,
$\left( \epsilon\mid_{C_{\leq0}}\right) \left( d\right) $ is well-defined.
Comparing
\[
\left( \epsilon\mid_{C_{\leq0}}\right) \left( c\right) =\epsilon\left(
c\right)
\]
with%
\begin{align*}
\left( \epsilon\mid_{C_{\leq0}}\right) \left( d\right) & =\epsilon
\left( \underbrace{d}_{=\epsilon\left( c\right) 1_{C}}\right)
=\epsilon\left( \epsilon\left( c\right) 1_{C}\right) \\
& =\epsilon\left( c\right) \underbrace{\epsilon\left( 1_{C}\right)
}_{=1_{\mathbf{k}}}\ \ \ \ \ \ \ \ \ \ \left( \text{since the map }%
\epsilon\text{ is }\mathbf{k}\text{-linear}\right) \\
& =\epsilon\left( c\right) ,
\end{align*}
we obtain $\left( \epsilon\mid_{C_{\leq0}}\right) \left( c\right) =\left(
\epsilon\mid_{C_{\leq0}}\right) \left( d\right) $.
However, the map $\epsilon\mid_{C_{\leq0}}$ is injective. In other words, if
$u$ and $v$ are two elements of $C_{\leq0}$ satisfying $\left( \epsilon
\mid_{C_{\leq0}}\right) \left( u\right) =\left( \epsilon\mid_{C_{\leq0}%
}\right) \left( v\right) $, then $u=v$. Applying this to $u=c$ and $v=d$,
we obtain $c=d$ (since $\left( \epsilon\mid_{C_{\leq0}}\right) \left(
c\right) =\left( \epsilon\mid_{C_{\leq0}}\right) \left( d\right) $). Now,
the definition of $\idbar$ yields%
\[
\idbar\left( c\right) =c-\underbrace{\epsilon\left( c\right) 1_{C}%
}_{\substack{=d\\\text{(by the definition of }d\text{)}}%
}=c-d=0\ \ \ \ \ \ \ \ \ \ \left( \text{since }c=d\right) .
\]
Forget that we have fixed $c$. We thus have shown that $\idbar\left(
c\right) =0$ for each $c\in C_{\leq0}$. In other words, $\idbar\left(
C_{\leq0}\right) =0$. This proves Lemma \ref{lem.cfc.idbar} \textbf{(c)}.
\end{proof}
\begin{proof}
[Proof of Proposition \ref{prop.cfc.delta2}.]\textbf{(a)} Define a
$\mathbf{k}$-linear map $\idbar:C\rightarrow C$ as in Lemma
\ref{lem.cfc.idbar}. Then, Lemma \ref{lem.cfc.idbar} \textbf{(a)} yields
$\delta=\left( \idbar\otimes\idbar\right) \circ\Delta$.
Now, let $n>0$ be an integer. Then, (\ref{eq.def.fil-coal.Del}) yields%
\begin{equation}
\Delta\left( C_{\leq n}\right) \subseteq\sum_{i=0}^{n}C_{\leq i}\otimes
C_{\leq n-i} \label{pf.prop.cfc.delta2.a.1}%
\end{equation}
(since $C$ is a filtered $\mathbf{k}$-coalgebra). Now,
\begin{align*}
& \underbrace{\delta}_{=\left( \idbar\otimes\idbar\right) \circ\Delta
}\left( C_{\leq n}\right) \\
& =\left( \left( \idbar\otimes\idbar\right) \circ\Delta\right) \left(
C_{\leq n}\right) =\left( \idbar\otimes\idbar\right) \underbrace{\left(
\Delta\left( C_{\leq n}\right) \right) }_{\substack{\subseteq\sum_{i=0}%
^{n}C_{\leq i}\otimes C_{\leq n-i}\\\text{(by (\ref{pf.prop.cfc.delta2.a.1}%
))}}}\\
& \subseteq\left( \idbar\otimes\idbar\right) \left( \sum_{i=0}^{n}C_{\leq
i}\otimes C_{\leq n-i}\right) \\
& =\sum_{i=0}^{n}\underbrace{\left( \idbar\otimes\idbar\right) \left(
C_{\leq i}\otimes C_{\leq n-i}\right) }_{\substack{=\idbar\left( C_{\leq
i}\right) \otimes\idbar\left( C_{\leq n-i}\right) \\\text{(by
(\ref{pf.thm.id-f.gen.ass-Ker.abUV}), applied to }D=C\text{, }\alpha
=\idbar\text{, }\beta=\idbar\text{, }U=C_{\leq i}\text{ and }V=C_{\leq
n-i}\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since the map }\idbar\otimes\idbar\text{
is }\mathbf{k}\text{-linear}\right) \\
& =\sum_{i=0}^{n}\idbar\left( C_{\leq i}\right) \otimes\idbar\left(
C_{\leq n-i}\right) \\
& =\idbar\left( C_{\leq0}\right) \otimes\idbar\left( C_{\leq n-0}\right)
+\sum_{i=1}^{n-1}\idbar\left( C_{\leq i}\right) \otimes\idbar\left( C_{\leq
n-i}\right) +\idbar\left( C_{\leq n}\right) \otimes
\underbrace{\idbar\left( C_{\leq n-n}\right) }_{\substack{=\idbar\left(
C_{\leq0}\right) \\\text{(since }n-n=0\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addends for }i=0\text{ and for }i=n\text{
from the}\\
\text{sum (and these are indeed two distinct addends, since }n>0\text{)}%
\end{array}
\right) \\
& =\underbrace{\idbar\left( C_{\leq0}\right) }_{\substack{=0\\\text{(by
Lemma \ref{lem.cfc.idbar} \textbf{(c)})}}}\otimes\idbar\left( C_{\leq
n-0}\right) +\sum_{i=1}^{n-1}\idbar\left( C_{\leq i}\right) \otimes
\idbar\left( C_{\leq n-i}\right) +\idbar\left( C_{\leq n}\right)
\otimes\underbrace{\idbar\left( C_{\leq0}\right) }_{\substack{=0\\\text{(by
Lemma \ref{lem.cfc.idbar} \textbf{(c)})}}}\\
& =\underbrace{0\otimes\idbar\left( C_{\leq n-0}\right) }_{=0}+\sum
_{i=1}^{n-1}\idbar\left( C_{\leq i}\right) \otimes\idbar\left( C_{\leq
n-i}\right) +\underbrace{\idbar\left( C_{\leq n}\right) \otimes0}_{=0}\\
& =\sum_{i=1}^{n-1}\underbrace{\idbar\left( C_{\leq i}\right)
}_{\substack{\subseteq C_{\leq i}\\\text{(by Lemma \ref{lem.cfc.idbar}
\textbf{(b)},}\\\text{applied to }i\text{ instead of }n\text{)}}%
}\otimes\underbrace{\idbar\left( C_{\leq n-i}\right) }_{\substack{\subseteq
C_{\leq n-i}\\\text{(by Lemma \ref{lem.cfc.idbar} \textbf{(b)},}%
\\\text{applied to }n-i\text{ instead of }n\text{)}}}\\
& \subseteq\sum_{i=1}^{n-1}C_{\leq i}\otimes C_{\leq n-i}.
\end{align*}
This proves Proposition \ref{prop.cfc.delta2} \textbf{(a)}.
\textbf{(b)} Let $f:C\rightarrow C$ be a $\mathbf{k}$-coalgebra homomorphism
satisfying $f\left( 1_{C}\right) =1_{C}$. Thus, $f$ is a $\mathbf{k}%
$-coalgebra homomorphism; in other words, $f$ is a $\mathbf{k}$-linear map
satisfying $\left( f\otimes f\right) \circ\Delta=\Delta\circ f$ and
$\epsilon=\epsilon\circ f$ (by the definition of a \textquotedblleft%
$\mathbf{k}$-coalgebra homomorphism\textquotedblright).
Let $c\in C$. Then,%
\begin{align*}
& \left( \left( f\otimes f\right) \circ\delta\right) \left( c\right) \\
& =\left( f\otimes f\right) \left( \underbrace{\delta\left( c\right)
}_{\substack{=\Delta\left( c\right) -c\otimes1_{C}-1_{C}\otimes
c+\epsilon\left( c\right) 1_{C}\otimes1_{C}\\\text{(by the definition of
}\delta\text{)}}}\right) \\
& =\left( f\otimes f\right) \left( \Delta\left( c\right) -c\otimes
1_{C}-1_{C}\otimes c+\epsilon\left( c\right) 1_{C}\otimes1_{C}\right) \\
& =\underbrace{\left( f\otimes f\right) \left( \Delta\left( c\right)
\right) }_{=\left( \left( f\otimes f\right) \circ\Delta\right) \left(
c\right) }-\underbrace{\left( f\otimes f\right) \left( c\otimes
1_{C}\right) }_{=f\left( c\right) \otimes f\left( 1_{C}\right)
}-\underbrace{\left( f\otimes f\right) \left( 1_{C}\otimes c\right)
}_{=f\left( 1_{C}\right) \otimes f\left( c\right) }+\epsilon\left(
c\right) \underbrace{\left( f\otimes f\right) \left( 1_{C}\otimes
1_{C}\right) }_{=f\left( 1_{C}\right) \otimes f\left( 1_{C}\right) }\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since the map }f\otimes f\text{ is
}\mathbf{k}\text{-linear}\right) \\
& =\underbrace{\left( \left( f\otimes f\right) \circ\Delta\right)
}_{=\Delta\circ f}\left( c\right) -f\left( c\right) \otimes
\underbrace{f\left( 1_{C}\right) }_{=1_{C}}-\underbrace{f\left(
1_{C}\right) }_{=1_{C}}\otimes f\left( c\right) +\underbrace{\epsilon
}_{=\epsilon\circ f}\left( c\right) \underbrace{f\left( 1_{C}\right)
}_{=1_{C}}\otimes\underbrace{f\left( 1_{C}\right) }_{=1_{C}}\\
& =\underbrace{\left( \Delta\circ f\right) \left( c\right) }%
_{=\Delta\left( f\left( c\right) \right) }-f\left( c\right) \otimes
1_{C}-1_{C}\otimes f\left( c\right) +\underbrace{\left( \epsilon\circ
f\right) \left( c\right) }_{=\epsilon\left( f\left( c\right) \right)
}1_{C}\otimes1_{C}\\
& =\Delta\left( f\left( c\right) \right) -f\left( c\right) \otimes
1_{C}-1_{C}\otimes f\left( c\right) +\epsilon\left( f\left( c\right)
\right) 1_{C}\otimes1_{C}.
\end{align*}
Comparing this with%
\begin{align*}
\left( \delta\circ f\right) \left( c\right) & =\delta\left( f\left(
c\right) \right) =\Delta\left( f\left( c\right) \right) -f\left(
c\right) \otimes1_{C}-1_{C}\otimes f\left( c\right) +\epsilon\left(
f\left( c\right) \right) 1_{C}\otimes1_{C}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\delta\right) ,
\end{align*}
we obtain $\left( \left( f\otimes f\right) \circ\delta\right) \left(
c\right) =\left( \delta\circ f\right) \left( c\right) $.
Forget that we fixed $c$. We thus have shown that $\left( \left( f\otimes
f\right) \circ\delta\right) \left( c\right) =\left( \delta\circ f\right)
\left( c\right) $ for each $c\in C$. In other words, $\left( f\otimes
f\right) \circ\delta=\delta\circ f$. This proves Proposition
\ref{prop.cfc.delta2} \textbf{(b)}.
\textbf{(c)} Definition \ref{def.con-fil-coal} \textbf{(b)} yields
$1_{C}=\left( \epsilon\mid_{C_{\leq0}}\right) ^{-1}\left( 1_{\mathbf{k}%
}\right) $. Thus, $\left( \epsilon\mid_{C_{\leq0}}\right) \left(
1_{C}\right) =1_{\mathbf{k}}$. In other words, $\epsilon\left( 1_{C}\right)
=1$ (since $\left( \epsilon\mid_{C_{\leq0}}\right) \left( 1_{C}\right)
=\epsilon\left( 1_{C}\right) $ and $1_{\mathbf{k}}=1$).
Let $c\in\left( \operatorname*{Ker}\delta\right) \cap\left(
\operatorname*{Ker}\epsilon\right) $. Thus, $c\in\left( \operatorname*{Ker}%
\delta\right) \cap\left( \operatorname*{Ker}\epsilon\right) \subseteq
\operatorname*{Ker}\delta$, so that $\delta\left( c\right) =0$. Moreover,
$c\in\left( \operatorname*{Ker}\delta\right) \cap\left( \operatorname*{Ker}%
\epsilon\right) \subseteq\operatorname*{Ker}\epsilon$, so that $\epsilon
\left( c\right) =0$. However, from $\delta\left( c\right) =0$, we obtain%
\begin{align*}
0 & =\delta\left( c\right) =\Delta\left( c\right) -c\otimes1_{C}%
-1_{C}\otimes c+\underbrace{\epsilon\left( c\right) }_{=0}1_{C}\otimes
1_{C}\ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\delta\right) \\
& =\Delta\left( c\right) -c\otimes1_{C}-1_{C}\otimes c+\underbrace{0\cdot
1_{C}\otimes1_{C}}_{=0}\\
& =\Delta\left( c\right) -c\otimes1_{C}-1_{C}\otimes c.
\end{align*}
In other words, $\Delta\left( c\right) =c\otimes1_{C}+1_{C}\otimes c$. In
other words, the element $c$ of $C$ is primitive (by the definition of
\textquotedblleft primitive\textquotedblright). In other words, $c\in
\operatorname*{Prim}C$ (since $\operatorname*{Prim}C$ is defined as the set of
all primitive elements of $C$).
Forget that we fixed $c$. We thus have shown that $c\in\operatorname*{Prim}C$
for each $c\in\left( \operatorname*{Ker}\delta\right) \cap\left(
\operatorname*{Ker}\epsilon\right) $. In other words, $\left(
\operatorname*{Ker}\delta\right) \cap\left( \operatorname*{Ker}%
\epsilon\right) \subseteq\operatorname*{Prim}C$.
Now, let $d\in\operatorname*{Prim}C$. Thus, the element $d$ of $C$ is
primitive (since $\operatorname*{Prim}C$ is defined as the set of all
primitive elements of $C$). In other words, $\Delta\left( d\right)
=d\otimes1_{C}+1_{C}\otimes d$ (by the definition of \textquotedblleft
primitive\textquotedblright). Hence, Lemma \ref{lem.coalg.primitive-e0}
(applied to $1_{C}$ and $1_{C}$ instead of $a$ and $b$) yields $\epsilon
\left( d\right) =0$ (since $\epsilon\left( 1_{C}\right) =1$). Hence,
$d\in\operatorname*{Ker}\epsilon$.
Furthermore, the definition of $\delta$ yields%
\[
\delta\left( d\right) =\underbrace{\Delta\left( d\right) -d\otimes
1_{C}-1_{C}\otimes d}_{\substack{=0\\\text{(since }\Delta\left( d\right)
=d\otimes1_{C}+1_{C}\otimes d\text{)}}}+\epsilon\left( d\right) 1_{C}%
\otimes1_{C}=\underbrace{\epsilon\left( d\right) }_{=0}1_{C}\otimes1_{C}=0.
\]
Hence, $d\in\operatorname*{Ker}\delta$. Combining this with $d\in
\operatorname*{Ker}\epsilon$, we obtain $d\in\left( \operatorname*{Ker}%
\delta\right) \cap\left( \operatorname*{Ker}\epsilon\right) $.
Forget that we fixed $d$. We thus have shown that $d\in\left(
\operatorname*{Ker}\delta\right) \cap\left( \operatorname*{Ker}%
\epsilon\right) $ for each $d\in\operatorname*{Prim}C$. In other words,
$\operatorname*{Prim}C\subseteq\left( \operatorname*{Ker}\delta\right)
\cap\left( \operatorname*{Ker}\epsilon\right) $. Combining this with
$\left( \operatorname*{Ker}\delta\right) \cap\left( \operatorname*{Ker}%
\epsilon\right) \subseteq\operatorname*{Prim}C$, we obtain
$\operatorname*{Prim}C=\left( \operatorname*{Ker}\delta\right) \cap\left(
\operatorname*{Ker}\epsilon\right) $. This proves Proposition
\ref{prop.cfc.delta2} \textbf{(c)}.
\textbf{(d)} Proposition \ref{prop.cfc.delta2} \textbf{(c)} yields
$\operatorname*{Prim}C=\left( \operatorname*{Ker}\delta\right) \cap\left(
\operatorname*{Ker}\epsilon\right) $.
The maps $\delta$ and $\epsilon$ are $\mathbf{k}$-linear. Hence, their kernels
$\operatorname*{Ker}\delta$ and $\operatorname*{Ker}\epsilon$ are $\mathbf{k}%
$-submodules of $C$. The intersection $\left( \operatorname*{Ker}%
\delta\right) \cap\left( \operatorname*{Ker}\epsilon\right) $ of these two
kernels must therefore be a $\mathbf{k}$-submodule of $C$ as well. In other
words, $\operatorname*{Prim}C$ is a $\mathbf{k}$-submodule of $C$ (since
$\operatorname*{Prim}C=\left( \operatorname*{Ker}\delta\right) \cap\left(
\operatorname*{Ker}\epsilon\right) $). This proves Proposition
\ref{prop.cfc.delta2} \textbf{(d)}.
\textbf{(e)} We first observe that $1_{C}\in C_{\leq0}$ (by Definition
\ref{def.con-fil-coal} \textbf{(b)}). However, (\ref{eq.def.fil-coal.chain})
yields $C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}\subseteq\cdots$ (since
$C$ is a filtered $\mathbf{k}$-coalgebra). Therefore, $C_{\leq0}\subseteq
C_{\leq1}$. Thus, $1_{C}\in C_{\leq0}\subseteq C_{\leq1}$. Hence,%
\begin{align*}
\delta\left( 1_{C}\right) & \in\delta\left( C_{\leq1}\right)
\subseteq\sum_{i=1}^{1-1}C_{\leq i}\otimes C_{\leq1-i}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by Proposition \ref{prop.cfc.delta2}
\textbf{(a)}, applied to }n=1\right) \\
& =\left( \text{empty sum}\right) =0.
\end{align*}
In other words, $\delta\left( 1_{C}\right) =0$.
Definition \ref{def.con-fil-coal} \textbf{(b)} yields $1_{C}=\left(
\epsilon\mid_{C_{\leq0}}\right) ^{-1}\left( 1_{\mathbf{k}}\right) $. Thus,
$\left( \epsilon\mid_{C_{\leq0}}\right) \left( 1_{C}\right) =1_{\mathbf{k}%
}$. In other words, $\epsilon\left( 1_{C}\right) =1$ (since $\left(
\epsilon\mid_{C_{\leq0}}\right) \left( 1_{C}\right) =\epsilon\left(
1_{C}\right) $ and $1_{\mathbf{k}}=1$).
Let $u\in\operatorname*{Ker}\delta$. Thus, $\delta\left( u\right) =0$. Set
$v=u-\epsilon\left( u\right) 1_{C}$. Then,%
\begin{align*}
\delta\left( \underbrace{v}_{=u-\epsilon\left( u\right) 1_{C}}\right) &
=\delta\left( u-\epsilon\left( u\right) 1_{C}\right) =\underbrace{\delta
\left( u\right) }_{=0}-\epsilon\left( u\right) \underbrace{\delta\left(
1_{C}\right) }_{=0}\ \ \ \ \ \ \ \ \ \ \left( \text{since the map }%
\delta\text{ is }\mathbf{k}\text{-linear}\right) \\
& =0-\epsilon\left( u\right) 0=0,
\end{align*}
so that $v\in\operatorname*{Ker}\delta$. Furthermore,%
\begin{align*}
\epsilon\left( \underbrace{v}_{=u-\epsilon\left( u\right) 1_{C}}\right)
& =\epsilon\left( u-\epsilon\left( u\right) 1_{C}\right) =\epsilon\left(
u\right) -\epsilon\left( u\right) \underbrace{\epsilon\left( 1_{C}\right)
}_{=1}\ \ \ \ \ \ \ \ \ \ \left( \text{since the map }\epsilon\text{ is
}\mathbf{k}\text{-linear}\right) \\
& =\epsilon\left( u\right) -\epsilon\left( u\right) =0,
\end{align*}
so that $v\in\operatorname*{Ker}\epsilon$. Combining this with $v\in
\operatorname*{Ker}\delta$, we obtain $v\in\left( \operatorname*{Ker}%
\delta\right) \cap\left( \operatorname*{Ker}\epsilon\right)
=\operatorname*{Prim}C$ (by Proposition \ref{prop.cfc.delta2} \textbf{(c)}).
Now, from $v=u-\epsilon\left( u\right) 1_{C}$, we obtain%
\[
u=\underbrace{\epsilon\left( u\right) }_{\in\mathbf{k}}1_{C}+\underbrace{v}%
_{\in\operatorname*{Prim}C}\in\mathbf{k}\cdot1_{C}+\operatorname*{Prim}C.
\]
Forget that we fixed $u$. We thus have shown that $u\in\mathbf{k}\cdot
1_{C}+\operatorname*{Prim}C$ for each $u\in\operatorname*{Ker}\delta$. In
other words,%
\begin{equation}
\operatorname*{Ker}\delta\subseteq\mathbf{k}\cdot1_{C}+\operatorname*{Prim}C.
\label{pf.prop.cfc.delta2.e.oneside}%
\end{equation}
On the other hand, let $w\in\mathbf{k}\cdot1_{C}+\operatorname*{Prim}C$. Thus,
we can write $w$ in the form $w=x+y$ for some $x\in\mathbf{k}\cdot1_{C}$ and
some $y\in\operatorname*{Prim}C$. Consider these $x$ and $y$. We have%
\begin{align*}
y & \in\operatorname*{Prim}C=\left( \operatorname*{Ker}\delta\right)
\cap\left( \operatorname*{Ker}\epsilon\right) \ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition \ref{prop.cfc.delta2} \textbf{(c)}}\right) \\
& \subseteq\operatorname*{Ker}\delta,
\end{align*}
so that $\delta\left( y\right) =0$.
We have $x\in\mathbf{k}\cdot1_{C}$; in other words, $x=\lambda\cdot1_{C}$ for
some $\lambda\in\mathbf{k}$. Consider this $\lambda$. Now, $w=\underbrace{x}%
_{=\lambda\cdot1_{C}}+y=\lambda\cdot1_{C}+y$. Applying the map $\delta$ to
both sides of this equality, we obtain%
\begin{align*}
\delta\left( w\right) & =\delta\left( \lambda\cdot1_{C}+y\right)
=\lambda\cdot\underbrace{\delta\left( 1_{C}\right) }_{=0}+\underbrace{\delta
\left( y\right) }_{=0}\ \ \ \ \ \ \ \ \ \ \left( \text{since the map
}\delta\text{ is }\mathbf{k}\text{-linear}\right) \\
& =\lambda\cdot0+0=0.
\end{align*}
In other words, $w\in\operatorname*{Ker}\delta$.
Forget that we fixed $w$. We thus have shown that $w\in\operatorname*{Ker}%
\delta$ for each $w\in\mathbf{k}\cdot1_{C}+\operatorname*{Prim}C$. In other
words,%
\[
\mathbf{k}\cdot1_{C}+\operatorname*{Prim}C\subseteq\operatorname*{Ker}\delta.
\]
Combining this with (\ref{pf.prop.cfc.delta2.e.oneside}), we obtain
$\operatorname*{Ker}\delta=\mathbf{k}\cdot1_{C}+\operatorname*{Prim}C$. This
proves Proposition \ref{prop.cfc.delta2} \textbf{(e)}.
\end{proof}
\subsection{\label{sec.pf.cor.id-f.cfc}Proofs of the corollaries from Section
\ref{sec.cfc}}
\begin{proof}
[Proof of Corollary \ref{cor.id-f.cfc}.]We have $\left( e-f\right) \left(
1_{C}\right) =\underbrace{e\left( 1_{C}\right) }_{=1_{C}}%
-\underbrace{f\left( 1_{C}\right) }_{=1_{C}}=1_{C}-1_{C}=0$. Hence,
$\mathbf{k}\cdot1_{C}\subseteq\operatorname*{Ker}\left( e-f\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $x\in\mathbf{k}\cdot1_{C}$. Thus,
$x=\lambda\cdot1_{C}$ for some $\lambda\in\mathbf{k}$. Consider this $\lambda
$. Now, applying the map $e-f$ to both sides of the equality $x=\lambda
\cdot1_{C}$, we obtain%
\begin{align*}
\left( e-f\right) \left( x\right) & =\left( e-f\right) \left(
\lambda\cdot1_{C}\right) =\lambda\cdot\underbrace{\left( e-f\right) \left(
1_{C}\right) }_{=0}\ \ \ \ \ \ \ \ \ \ \left( \text{since the map }e-f\text{
is }\mathbf{k}\text{-linear}\right) \\
& =0.
\end{align*}
In other words, $x\in\operatorname*{Ker}\left( e-f\right) $.
\par
Forget that we fixed $x$. We thus have shown that $x\in\operatorname*{Ker}%
\left( e-f\right) $ for each $x\in\mathbf{k}\cdot1_{C}$. In other words,
$\mathbf{k}\cdot1_{C}\subseteq\operatorname*{Ker}\left( e-f\right) $.}.
Clearly, $\left( C_{\leq0},C_{\leq1},C_{\leq2},\ldots\right) $ is a sequence
of $\mathbf{k}$-submodules of $C$ (by the definition of a filtered
$\mathbf{k}$-coalgebra). Thus, $\left( C_{\leq1},C_{\leq2},C_{\leq3}%
,\ldots\right) $ is a sequence of $\mathbf{k}$-submodules of $C$ as well.
Define the $\mathbf{k}$-linear map $\delta:C\rightarrow C\otimes C$ as in
Proposition \ref{prop.cfc.delta2}. The map $f$ is a $\mathbf{k}$-coalgebra
homomorphism satisfying $f\left( 1_{C}\right) =1_{C}$. Thus, Proposition
\ref{prop.cfc.delta2} \textbf{(b)} yields that $\left( f\otimes f\right)
\circ\delta=\delta\circ f$. The same argument (applied to $e$ instead of $f$)
yields $\left( e\otimes e\right) \circ\delta=\delta\circ e$. Moreover,
Proposition \ref{prop.cfc.delta2} \textbf{(e)} yields
\begin{align*}
\operatorname*{Ker}\delta & =\underbrace{\mathbf{k}\cdot1_{C}}_{\subseteq
\operatorname*{Ker}\left( e-f\right) }+\underbrace{\operatorname*{Prim}%
C}_{\substack{\subseteq\operatorname*{Ker}\left( e-f\right) \\\text{(by
(\ref{eq.cor.id-f.cfc.ass-Ker}))}}}\subseteq\operatorname*{Ker}\left(
e-f\right) +\operatorname*{Ker}\left( e-f\right) \\
& \subseteq\operatorname*{Ker}\left( e-f\right) \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\operatorname*{Ker}\left( e-f\right) \text{ is a }%
\mathbf{k}\text{-module}\right) .
\end{align*}
However, Proposition \ref{prop.cfc.delta2} \textbf{(a)} shows that%
\[
\delta\left( C_{\leq n}\right) \subseteq\sum_{i=1}^{n-1}C_{\leq i}\otimes
C_{\leq n-i}\ \ \ \ \ \ \ \ \ \ \text{for each }n>0.
\]
Hence,%
\[
\delta\left( C_{\leq n}\right) \subseteq\sum_{i=1}^{n-1}C_{\leq i}\otimes
C_{\leq n-i}\ \ \ \ \ \ \ \ \ \ \text{for each }n>p.
\]
Moreover, the definition of a filtered $\mathbf{k}$-coalgebra yields
$C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}\subseteq\cdots$. Hence, each
$i\in\left\{ 1,2,\ldots,p\right\} $ satisfies
\begin{equation}
C_{\leq i}\subseteq C_{\leq p} \label{pf.cor.id-f.cfc}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.id-f.cfc}):} Let $i\in\left\{
1,2,\ldots,p\right\} $. Thus, $i\in\mathbb{N}$ and $i\leq p$. However, we
have $C_{\leq0}\subseteq C_{\leq1}\subseteq C_{\leq2}\subseteq\cdots$. In
other words, if $a$ and $b$ are two elements of $\mathbb{N}$ satisfying $a\leq
b$, then $C_{\leq a}\subseteq C_{\leq b}$. Applying this to $a=i$ and $b=p$,
we obtain $C_{\leq i}\subseteq C_{\leq p}$ (since $i\leq p$). This proves
(\ref{pf.cor.id-f.cfc}).}. Thus,
\[
C_{\leq1}+C_{\leq2}+\cdots+C_{\leq p}=\sum_{i=1}^{p}\underbrace{C_{\leq i}%
}_{\substack{\subseteq C_{\leq p}\\\text{(by (\ref{pf.cor.id-f.cfc}))}%
}}\subseteq\sum_{i=1}^{p}C_{\leq p}\subseteq C_{\leq p}%
\]
(since $C_{\leq p}$ is a $\mathbf{k}$-module). Therefore,%
\[
\left( e-f\right) \left( \underbrace{C_{\leq1}+C_{\leq2}+\cdots+C_{\leq p}%
}_{\subseteq C_{\leq p}}\right) \subseteq\left( e-f\right) \left( C_{\leq
p}\right) =0
\]
(by (\ref{eq.cor.id-f.cfc.ass-ann})), so that $\left( e-f\right) \left(
C_{\leq1}+C_{\leq2}+\cdots+C_{\leq p}\right) =0$.
Hence, Theorem \ref{thm.id-f.gen} (applied to $D=C$ and $D_{i}=C_{\leq i}$)
shows that for any integer $u>p$, we have%
\begin{equation}
\left( e-f\right) ^{u-p}\left( C_{\leq u}\right) \subseteq
\operatorname*{Ker}\delta\label{pf.cor.id-f.cfc.claim-1a}%
\end{equation}
and%
\begin{equation}
\left( e-f\right) ^{u-p+1}\left( C_{\leq u}\right) =0.
\label{pf.cor.id-f.cfc.claim-1b}%
\end{equation}
We are now close to proving both parts of Corollary \ref{cor.id-f.cfc}. Let us
begin with part \textbf{(a)}:
\textbf{(a)} Let $u>p$ be an integer. Then, $u-p>0$ (since $u>p$), so that
$u-p\geq1$ (since $u-p$ is an integer). Thus, $\left( e-f\right)
^{u-p}=\left( e-f\right) \circ\left( e-f\right) ^{u-p-1}$. However, $f$ is
a $\mathbf{k}$-coalgebra homomorphism, and thus satisfies $\epsilon\circ
f=\epsilon$ (by the definition of a $\mathbf{k}$-coalgebra homomorphism).
Similarly, $\epsilon\circ e=\epsilon$. Since composition of $\mathbf{k}%
$-linear maps is a $\mathbf{k}$-bilinear operation (and since the maps
$\epsilon$, $e$ and $f$ are $\mathbf{k}$-linear), we have
\[
\epsilon\circ\left( e-f\right) =\underbrace{\epsilon\circ e}_{=\epsilon
}-\underbrace{\epsilon\circ f}_{=\epsilon}=\epsilon-\epsilon=0.
\]
Thus,%
\[
\epsilon\circ\underbrace{\left( e-f\right) ^{u-p}}_{=\left( e-f\right)
\circ\left( e-f\right) ^{u-p-1}}=\underbrace{\epsilon\circ\left(
e-f\right) }_{=0}\circ\left( e-f\right) ^{u-p-1}=0\circ\left( e-f\right)
^{u-p-1}=0.
\]
Therefore, $\left( e-f\right) ^{u-p}\left( C_{\leq u}\right)
\subseteq\operatorname*{Ker}\epsilon$\ \ \ \ \footnote{\textit{Proof.} Let
$z\in\left( e-f\right) ^{u-p}\left( C_{\leq u}\right) $. Thus, $z$ can be
written in the form $z=\left( e-f\right) ^{u-p}\left( x\right) $ for some
$x\in C_{\leq u}$. Consider this $x$. From $z=\left( e-f\right)
^{u-p}\left( x\right) $, we obtain $\epsilon\left( z\right) =\epsilon
\left( \left( e-f\right) ^{u-p}\left( x\right) \right)
=\underbrace{\left( \epsilon\circ\left( e-f\right) ^{u-p}\right) }%
_{=0}\left( x\right) =0\left( x\right) =0$. In other words, $z\in
\operatorname*{Ker}\epsilon$.
\par
Forget that we fixed $z$. We thus have shown that $z\in\operatorname*{Ker}%
\epsilon$ for each $z\in\left( e-f\right) ^{u-p}\left( C_{\leq u}\right)
$. In other words, $\left( e-f\right) ^{u-p}\left( C_{\leq u}\right)
\subseteq\operatorname*{Ker}\epsilon$.}. Combining this with
(\ref{pf.cor.id-f.cfc.claim-1a}), we obtain $\left( e-f\right) ^{u-p}\left(
C_{\leq u}\right) \subseteq\left( \operatorname*{Ker}\delta\right)
\cap\left( \operatorname*{Ker}\epsilon\right) =\operatorname*{Prim}C$ (by
Proposition \ref{prop.cfc.delta2} \textbf{(c)}). This proves Corollary
\ref{cor.id-f.cfc} \textbf{(a)}.
\textbf{(b)} Let $u\geq p$ be an integer. We must prove that $\left(
e-f\right) ^{u-p+1}\left( C_{\leq u}\right) =0$. If $u>p$, then this
follows from (\ref{pf.cor.id-f.cfc.claim-1b}). Thus, for the rest of this
proof, we WLOG assume that we don't have $u>p$. Hence, $u\leq p$. Combining
this with $u\geq p$, we obtain $u=p$. Thus,
\[
\left( e-f\right) ^{u-p+1}\left( C_{\leq u}\right) =\underbrace{\left(
e-f\right) ^{p-p+1}}_{=\left( e-f\right) ^{1}=e-f}\left( C_{\leq
p}\right) =\left( e-f\right) \left( C_{\leq p}\right) =0
\]
(by (\ref{eq.cor.id-f.cfc.ass-ann})). This proves Corollary \ref{cor.id-f.cfc}
\textbf{(b)}.
\end{proof}
\begin{proof}
[Proof of Corollary \ref{cor.id-f.cfc1}.]Define the $\mathbf{k}$-linear map
$\delta:C\rightarrow C\otimes C$ as in Proposition \ref{prop.cfc.delta2}. Just
as we did in the proof of Corollary \ref{cor.id-f.cfc}, we can show that
$\operatorname*{Ker}\delta\subseteq\operatorname*{Ker}\left( e-f\right) $.
However, Proposition \ref{prop.cfc.delta2} \textbf{(a)} (applied to $n=1$)
yields%
\[
\delta\left( C_{\leq1}\right) \subseteq\sum_{i=1}^{1-1}C_{\leq i}\otimes
C_{\leq1-i}=\left( \text{empty sum}\right) =0.
\]
Hence, it is easy to see that $C_{\leq1}\subseteq\operatorname*{Ker}\delta
$\ \ \ \ \footnote{\textit{Proof.} Let $x\in C_{\leq1}$. Thus, $\delta\left(
x\right) \in\delta\left( C_{\leq1}\right) \subseteq0$, so that
$\delta\left( x\right) =0$. In other words, $x\in\operatorname*{Ker}\delta$.
\par
Forget that we fixed $x$. We thus have shown that $x\in\operatorname*{Ker}%
\delta$ for each $x\in C_{\leq1}$. In other words, $C_{\leq1}\subseteq
\operatorname*{Ker}\delta$.}. Consequently, $C_{\leq1}\subseteq
\operatorname*{Ker}\delta\subseteq\operatorname*{Ker}\left( e-f\right) $.
Thus,
\begin{equation}
\left( e-f\right) \left( C_{\leq1}\right) =0 \label{pf.cor.id-f.cfc1.ec=0}%
\end{equation}
\footnote{\textit{Proof.} For each $x\in C_{\leq1}$, we have $\left(
e-f\right) \left( x\right) =0$ (since $x\in C_{\leq1}\subseteq
\operatorname*{Ker}\left( e-f\right) $). In other words, we have $\left(
e-f\right) \left( C_{\leq1}\right) =0$.}. Hence, we can apply Corollary
\ref{cor.id-f.cfc} to $p=1$.
Therefore, applying Corollary \ref{cor.id-f.cfc} \textbf{(a)} to $p=1$, we
conclude the following: For any integer $u>1$, we have%
\begin{equation}
\left( e-f\right) ^{u-1}\left( C_{\leq u}\right) \subseteq
\operatorname*{Prim}C. \label{pf.cor.id-f.cfc1.goal-1a}%
\end{equation}
This proves Corollary \ref{cor.id-f.cfc1} \textbf{(a)}. It remains to prove
Corollary \ref{cor.id-f.cfc1} \textbf{(b)}:
\textbf{(b)} Let $u$ be a positive integer. Thus, $u\geq1$. Hence, Corollary
\ref{cor.id-f.cfc} \textbf{(b)} (applied to $p=1$) shows that $\left(
e-f\right) ^{u-1+1}\left( C_{\leq u}\right) =0$ (since we know that we can
apply Corollary \ref{cor.id-f.cfc} to $p=1$). In view of $u-1+1=u$, this
rewrites as $\left( e-f\right) ^{u}\left( C_{\leq u}\right) =0$. This
proves Corollary \ref{cor.id-f.cfc1} \textbf{(b)}.
\end{proof}
\begin{proof}
[Proof of Corollary \ref{cor.id-f.cfc1id}.]Clearly, $\operatorname*{id}%
:C\rightarrow C$ is a $\mathbf{k}$-coalgebra homomorphism such that
$\operatorname*{id}\left( 1_{C}\right) =1_{C}$. Furthermore, $f\circ
\operatorname*{id}=f=\operatorname*{id}\circ f$. Hence, we can apply Corollary
\ref{cor.id-f.cfc1} to $e=\operatorname*{id}$. In particular, Corollary
\ref{cor.id-f.cfc1} \textbf{(a)} (applied to $e=\operatorname*{id}$) yields
that for any integer $u>1$, we have%
\[
\left( \operatorname*{id}-f\right) ^{u-1}\left( C_{\leq u}\right)
\subseteq\operatorname*{Prim}C.
\]
This proves Corollary \ref{cor.id-f.cfc1id} \textbf{(a)}. Furthermore,
Corollary \ref{cor.id-f.cfc1} \textbf{(b)} (applied to $e=\operatorname*{id}$)
yields that for any positive integer $u$, we have%
\[
\left( \operatorname*{id}-f\right) ^{u}\left( C_{\leq u}\right) =0.
\]
This proves Corollary \ref{cor.id-f.cfc1id} \textbf{(b)}.
\end{proof}
\subsection{\label{sec.pf.filhopf}Proofs for Section \ref{sec.filhopf}}
Before we prove the claims left unproved in Section \ref{sec.filhopf}, let us
recall the defining property of the antipode of a Hopf algebra: \Needspace{15pc}
\begin{remark}
\label{rmk.hopf.antipode-pro}Let $H$ be a $\mathbf{k}$-Hopf algebra with
antipode $S$. Let $1_{H}$ denote the unity of the $\mathbf{k}$-algebra $H$.
Let $m:H\otimes H\rightarrow H$ be the $\mathbf{k}$-linear map that sends each
pure tensor $x\otimes y\in H\otimes H$ to the product $xy\in H$. Let
$u:\mathbf{k}\rightarrow H$ be the $\mathbf{k}$-linear map that sends
$1_{\mathbf{k}}$ to $1_{H}$. Then, the diagram%
\[%
%TCIMACRO{\TeXButton{antipode diagram}{\xymatrix{
%&H \otimes H \ar[rr]^{S \otimes\id_H}& &H \otimes H \ar[dr]^m& \\
%H \ar[ur]^\Delta\ar[rr]^{\epsilon} \ar[dr]_\Delta& &\kk\ar[rr]^{u} & & H\\
%&H \otimes H \ar[rr]_{\id_H \otimes S}& &H \otimes H \ar[ur]_m& \\
%}
%}}%
%BeginExpansion
\xymatrix{
&H \otimes H \ar[rr]^{S \otimes\id_H}& &H \otimes H \ar[dr]^m& \\
H \ar[ur]^\Delta\ar[rr]^{\epsilon} \ar[dr]_\Delta& &\kk\ar[rr]^{u} & & H\\
&H \otimes H \ar[rr]_{\id_H \otimes S}& &H \otimes H \ar[ur]_m& \\
}
%EndExpansion
\]
commutes.\footnotemark\ In other words, we have%
\begin{align}
m\circ\left( S\otimes\operatorname*{id}\nolimits_{H}\right) \circ\Delta &
=u\circ\epsilon\ \ \ \ \ \ \ \ \ \ \text{and}%
\label{eq.rmk.hopf.antipode-pro.1}\\
m\circ\left( \operatorname*{id}\nolimits_{H}\otimes S\right) \circ\Delta &
=u\circ\epsilon. \label{eq.rmk.hopf.antipode-pro.2}%
\end{align}
\end{remark}
\footnotetext{Indeed, this is the diagram (1.4.3) in \cite{GriRei}.}
\begin{proof}
[Proof of Lemma \ref{lem.bialg.antip-props}.]\textbf{(a)} Let $T:H\otimes
H\rightarrow H\otimes H$ be the $\mathbf{k}$-linear map that sends each pure
tensor $x\otimes y\in H\otimes H$ to $y\otimes x$. This map $T$ is known as
the \emph{twist map}. It is obviously an involution, i.e., satisfies
$T^{2}=\operatorname*{id}$. Furthermore, it is easy to see that any two
$\mathbf{k}$-linear maps $\alpha,\beta\in\operatorname*{End}H$ satisfy%
\begin{equation}
\left( \alpha\otimes\beta\right) \circ T=T\circ\left( \beta\otimes
\alpha\right) . \label{pf.lem.bialg.antip-props.a.T-comm}%
\end{equation}
Furthermore, it is well-known (see, e.g., \cite[Theorem 2.1.4 (iii) and
(iv)]{Abe80} or \cite[Exercise 1.4.28]{GriRei} or, in an equivalent form,
\cite[Proposition 7.1.9 (b)]{Radfor12}) that the antipode $S$ of $H$ is a
$\mathbf{k}$-coalgebra anti-endomorphism, i.e., that it satisfies%
\[
\Delta\circ S=T\circ\left( S\otimes S\right) \circ\Delta
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \epsilon\circ S=\epsilon.
\]
Now,%
\begin{align*}
\Delta\circ\underbrace{S^{2}}_{=S\circ S} & =\underbrace{\Delta\circ
S}_{=T\circ\left( S\otimes S\right) \circ\Delta}\circ S=T\circ\left(
S\otimes S\right) \circ\underbrace{\Delta\circ S}_{=T\circ\left( S\otimes
S\right) \circ\Delta}\\
& =T\circ\underbrace{\left( S\otimes S\right) \circ T}_{\substack{=T\circ
\left( S\otimes S\right) \\\text{(by
(\ref{pf.lem.bialg.antip-props.a.T-comm}), applied to }\alpha=S\text{ and
}\beta=S\text{)}}}\circ\left( S\otimes S\right) \circ\Delta\\
& =\underbrace{T\circ T}_{=T^{2}=\operatorname*{id}}\circ\underbrace{\left(
S\otimes S\right) \circ\left( S\otimes S\right) }_{\substack{=\left(
S\circ S\right) \otimes\left( S\circ S\right) \\\text{(by Lemma
\ref{lem.albegadel})}}}\circ\Delta=\left( \underbrace{\left( S\circ
S\right) }_{=S^{2}}\otimes\underbrace{\left( S\circ S\right) }_{=S^{2}%
}\right) \circ\Delta\\
& =\left( S^{2}\otimes S^{2}\right) \circ\Delta
\end{align*}
and%
\[
\epsilon\circ\underbrace{S^{2}}_{=S\circ S}=\underbrace{\epsilon\circ
S}_{=\epsilon}\circ S=\epsilon\circ S=\epsilon.
\]
These two equalities show that $S^{2}$ is a $\mathbf{k}$-coalgebra
homomorphism (since $S^{2}$ is a $\mathbf{k}$-linear map from $H$ to $H$).
This proves Lemma \ref{lem.bialg.antip-props} \textbf{(a)}.
\textbf{(b)} The axioms of a $\mathbf{k}$-bialgebra yield $\epsilon\left(
1_{H}\right) =1_{\mathbf{k}}$ (since $H$ is a $\mathbf{k}$-bialgebra) and
$\Delta\left( 1_{H}\right) =1_{H}\otimes1_{H}$ (likewise).
Define the maps $m$ and $u$ as in Remark \ref{rmk.hopf.antipode-pro}. Applying
both sides of the equality (\ref{eq.rmk.hopf.antipode-pro.1}) to $1_{H}$, we
obtain%
\[
\left( m\circ\left( S\otimes\operatorname*{id}\nolimits_{H}\right)
\circ\Delta\right) \left( 1_{H}\right) =\left( u\circ\epsilon\right)
\left( 1_{H}\right) =u\left( \underbrace{\epsilon\left( 1_{H}\right)
}_{=1_{\mathbf{k}}}\right) =u\left( 1_{\mathbf{k}}\right) =1_{H}%
\]
(by the definition of $u$). Hence,%
\begin{align*}
1_{H} & =\left( m\circ\left( S\otimes\operatorname*{id}\nolimits_{H}%
\right) \circ\Delta\right) \left( 1_{H}\right) =m\left( \left(
S\otimes\operatorname*{id}\nolimits_{H}\right) \left( \underbrace{\Delta
\left( 1_{H}\right) }_{=1_{H}\otimes1_{H}}\right) \right) \\
& =m\left( \underbrace{\left( S\otimes\operatorname*{id}\nolimits_{H}%
\right) \left( 1_{H}\otimes1_{H}\right) }_{=S\left( 1_{H}\right)
\otimes\operatorname*{id}\nolimits_{H}\left( 1_{H}\right) }\right)
=m\left( S\left( 1_{H}\right) \otimes\operatorname*{id}\nolimits_{H}\left(
1_{H}\right) \right) \\
& =S\left( 1_{H}\right) \cdot\underbrace{\operatorname*{id}\nolimits_{H}%
\left( 1_{H}\right) }_{=1_{H}}\ \ \ \ \ \ \ \ \ \ \left( \text{by the
definition of }m\right) \\
& =S\left( 1_{H}\right) .
\end{align*}
This proves Lemma \ref{lem.bialg.antip-props} \textbf{(b)}.
\textbf{(c)} This is well-known (see, e.g., \cite[Proposition 1.4.17]%
{GriRei}). For the sake of completeness, let us nevertheless give a proof:
Let $x$ be a primitive element of $H$. Thus, $\Delta\left( x\right)
=x\otimes1_{H}+1_{H}\otimes x$ (by the definition of \textquotedblleft
primitive\textquotedblright). Moreover, the axioms of a $\mathbf{k}$-bialgebra
yield $\epsilon\left( 1_{H}\right) =1$ (since $H$ is a $\mathbf{k}%
$-bialgebra). Hence, Lemma \ref{lem.coalg.primitive-e0} (applied to $C=H$,
$a=1_{H}$, $b=1_{H}$ and $d=x$) yields $\epsilon\left( x\right) =0$.
Define the maps $m$ and $u$ as in Remark \ref{rmk.hopf.antipode-pro}. Applying
both sides of the equality (\ref{eq.rmk.hopf.antipode-pro.1}) to $x$, we
obtain%
\[
\left( m\circ\left( S\otimes\operatorname*{id}\nolimits_{H}\right)
\circ\Delta\right) \left( x\right) =\left( u\circ\epsilon\right) \left(
x\right) =u\left( \underbrace{\epsilon\left( x\right) }_{=0}\right)
=u\left( 0\right) =0
\]
(since the map $u$ is $\mathbf{k}$-linear). Hence,%
\begin{align*}
0 & =\left( m\circ\left( S\otimes\operatorname*{id}\nolimits_{H}\right)
\circ\Delta\right) \left( x\right) \\
& =m\left( \left( S\otimes\operatorname*{id}\nolimits_{H}\right) \left(
\underbrace{\Delta\left( x\right) }_{=x\otimes1_{H}+1_{H}\otimes x}\right)
\right) =m\left( \underbrace{\left( S\otimes\operatorname*{id}%
\nolimits_{H}\right) \left( x\otimes1_{H}+1_{H}\otimes x\right)
}_{\substack{=\left( S\otimes\operatorname*{id}\nolimits_{H}\right) \left(
x\otimes1_{H}\right) +\left( S\otimes\operatorname*{id}\nolimits_{H}\right)
\left( 1_{H}\otimes x\right) \\\text{(since the map }S\otimes
\operatorname*{id}\nolimits_{H}\text{ is }\mathbf{k}\text{-linear)}}}\right)
\\
& =m\left( \underbrace{\left( S\otimes\operatorname*{id}\nolimits_{H}%
\right) \left( x\otimes1_{H}\right) }_{=S\left( x\right) \otimes
\operatorname*{id}\nolimits_{H}\left( 1_{H}\right) }+\underbrace{\left(
S\otimes\operatorname*{id}\nolimits_{H}\right) \left( 1_{H}\otimes x\right)
}_{=S\left( 1_{H}\right) \otimes\operatorname*{id}\nolimits_{H}\left(
x\right) }\right) \\
& =m\left( S\left( x\right) \otimes\underbrace{\operatorname*{id}%
\nolimits_{H}\left( 1_{H}\right) }_{=1_{H}}+\underbrace{S\left(
1_{H}\right) }_{\substack{=1_{H}\\\text{(by Lemma \ref{lem.bialg.antip-props}
\textbf{(b)})}}}\otimes\underbrace{\operatorname*{id}\nolimits_{H}\left(
x\right) }_{=x}\right) =m\left( S\left( x\right) \otimes1_{H}%
+1_{H}\otimes x\right) \\
& =\underbrace{m\left( S\left( x\right) \otimes1_{H}\right)
}_{\substack{=S\left( x\right) \cdot1_{H}\\\text{(by the definition of
}m\text{)}}}+\underbrace{m\left( 1_{H}\otimes x\right) }_{\substack{=1_{H}%
\cdot x\\\text{(by the definition of }m\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }m\text{ is }\mathbf{k}\text{-linear}\right) \\
& =S\left( x\right) \cdot1_{H}+1_{H}\cdot x=S\left( x\right) +x.
\end{align*}
Hence, $S\left( x\right) =-x$. This proves Lemma \ref{lem.bialg.antip-props}
\textbf{(c)}.
\textbf{(d)} Let $x$ be a primitive element of $H$. Then, Lemma
\ref{lem.bialg.antip-props} \textbf{(c)} yields $S\left( x\right) =-x$. Now,%
\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 }\mathbf{k}\text{-linear}\right) \\
& =-\left( -x\right) =x.
\end{align*}
This proves Lemma \ref{lem.bialg.antip-props} \textbf{(d)}.
\end{proof}
\begin{proof}
[Proof of Corollary \ref{cor.id-S2.p}.]We know that $H$ is a connected
filtered $\mathbf{k}$-Hopf algebra, thus a connected filtered $\mathbf{k}%
$-bialgebra and therefore a connected filtered $\mathbf{k}$-coalgebra.
Furthermore, Proposition \ref{prop.fil-bial.1=1} shows that the unity $1_{H}$
defined according to Definition \ref{def.con-fil-coal} \textbf{(b)} equals the
unity of the $\mathbf{k}$-algebra $H$. Thus, the notion of a \textquotedblleft
primitive element\textquotedblright\ of $H$ does not depend on whether we
regard $H$ as a $\mathbf{k}$-bialgebra or as a connected filtered $\mathbf{k}$-coalgebra.
Lemma \ref{lem.bialg.antip-props} \textbf{(b)} yields $S\left( 1_{H}\right)
=1_{H}$. Hence,
\[
\underbrace{S^{2}}_{=S\circ S}\left( 1_{H}\right) =\left( S\circ S\right)
\left( 1_{H}\right) =S\left( \underbrace{S\left( 1_{H}\right) }_{=1_{H}%
}\right) =S\left( 1_{H}\right) =1_{H}.
\]
Moreover, Lemma \ref{lem.bialg.antip-props} \textbf{(a)} yields that the map
$S^{2}:H\rightarrow H$ is a $\mathbf{k}$-coalgebra homomorphism. Of course,
the map $\operatorname*{id}:H\rightarrow H$ is a $\mathbf{k}$-coalgebra
homomorphism as well, and satisfies $\operatorname*{id}\left( 1_{H}\right)
=1_{H}$. Furthermore, Lemma \ref{lem.bialg.antip-props} \textbf{(d)} entails
that $\operatorname*{Prim}H\subseteq\operatorname*{Ker}\left(
\operatorname*{id}-S^{2}\right) $\ \ \ \ \footnote{\textit{Proof.} Let
$x\in\operatorname*{Prim}H$. Thus, $x$ is a primitive element of $H$ (since
$\operatorname*{Prim}H$ is defined as the set of all primitive elements of
$H$). Therefore, Lemma \ref{lem.bialg.antip-props} \textbf{(d)} yields
$S^{2}\left( x\right) =x$. Hence, $\left( \operatorname*{id}-S^{2}\right)
\left( x\right) =\underbrace{\operatorname*{id}\left( x\right) }%
_{=x}-\underbrace{S^{2}\left( x\right) }_{=x}=x-x=0$, so that $x\in
\operatorname*{Ker}\left( \operatorname*{id}-S^{2}\right) $.
\par
Forget that we fixed $x$. We thus have shown that $x\in\operatorname*{Ker}%
\left( \operatorname*{id}-S^{2}\right) $ for each $x\in\operatorname*{Prim}%
H$. In other words, we have $\operatorname*{Prim}H\subseteq\operatorname*{Ker}%
\left( \operatorname*{id}-S^{2}\right) $.}. Moreover, $S^{2}\circ
\operatorname*{id}=S^{2}=\operatorname*{id}\circ S^{2}$. Furthermore, $p$ is a
positive integer and satisfies $\left( \operatorname*{id}-S^{2}\right)
\left( H_{\leq p}\right) =0$ (by (\ref{eq.cor.id-S2.p.ass-ann})). Hence, we
can apply Corollary \ref{cor.id-f.cfc} to $C=H$ and $C_{\leq i}=H_{\leq i}$
and $e=\operatorname*{id}$ and $f=S^{2}$. Let us do this now.
Corollary \ref{cor.id-f.cfc} \textbf{(b)} (applied to $C=H$ and $C_{\leq
i}=H_{\leq i}$ and $e=\operatorname*{id}$ and $f=S^{2}$) shows that for any
integer $u\geq p$, we have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-p+1}\left( H_{\leq u}\right)
=0.
\]
This proves Corollary \ref{cor.id-S2.p} \textbf{(b)}. It remains to prove
Corollary \ref{cor.id-S2.p} \textbf{(a)}:
\textbf{(a)} Let $u>p$ be any integer. Then, Corollary \ref{cor.id-f.cfc}
\textbf{(a)} (applied to $C=H$ and $C_{\leq i}=H_{\leq i}$ and
$e=\operatorname*{id}$ and $f=S^{2}$) shows that%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-p}\left( H_{\leq u}\right)
\subseteq\operatorname*{Prim}H.
\]
This proves (\ref{eq.cor.id-S2.p.claim1}). It remains to prove
(\ref{eq.cor.id-S2.p.claim3}).
First, we shall show that $\left( \operatorname*{id}+S\right) \left(
\operatorname*{Prim}H\right) =0$.
Indeed, let $x\in\operatorname*{Prim}H$. Thus, $x$ is a primitive element of
$H$ (since $\operatorname*{Prim}H$ was defined as the set of all primitive
elements of $H$). Thus, Lemma \ref{lem.bialg.antip-props} \textbf{(c)} yields
$S\left( x\right) =-x$. Hence, $\left( \operatorname*{id}+S\right) \left(
x\right) =\underbrace{\operatorname*{id}\left( x\right) }_{=x}%
+\underbrace{S\left( x\right) }_{=-x}=x+\left( -x\right) =0$.
Forget that we fixed $x$. We thus have shown that $\left( \operatorname*{id}%
+S\right) \left( x\right) =0$ for each $x\in\operatorname*{Prim}H$. In
other words, $\left( \operatorname*{id}+S\right) \left(
\operatorname*{Prim}H\right) =0$.
Now,
\begin{align*}
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-p}\right) \left( H_{\leq u}\right) & =\left(
\operatorname*{id}+S\right) \left( \underbrace{\left( \operatorname*{id}%
-S^{2}\right) ^{u-p}\left( H_{\leq u}\right) }_{\substack{\subseteq
\operatorname*{Prim}H\\\text{(by (\ref{eq.cor.id-S2.p.claim1}))}}}\right) \\
& \subseteq\left( \operatorname*{id}+S\right) \left( \operatorname*{Prim}%
H\right) =0.
\end{align*}
Therefore, $\left( \left( \operatorname*{id}+S\right) \circ\left(
\operatorname*{id}-S^{2}\right) ^{u-p}\right) \left( H_{\leq u}\right) =0$
(since $\left( \left( \operatorname*{id}+S\right) \circ\left(
\operatorname*{id}-S^{2}\right) ^{u-p}\right) \left( H_{\leq u}\right) $
is a $\mathbf{k}$-module). This proves (\ref{eq.cor.id-S2.p.claim3}). Thus,
the proof of Corollary \ref{cor.id-S2.p} \textbf{(a)} is complete.
\end{proof}
\begin{proof}
[Proof of Corollary \ref{cor.id-S2.1}.]We know that $H$ is a connected
filtered $\mathbf{k}$-Hopf algebra, thus a connected filtered $\mathbf{k}%
$-bialgebra and therefore a connected filtered $\mathbf{k}$-coalgebra.
Furthermore, Proposition \ref{prop.fil-bial.1=1} shows that the unity $1_{H}$
defined according to Definition \ref{def.con-fil-coal} \textbf{(b)} equals the
unity of the $\mathbf{k}$-algebra $H$. Thus, the notion of a \textquotedblleft
primitive element\textquotedblright\ of $H$ does not depend on whether we
regard $H$ as a $\mathbf{k}$-bialgebra or as a connected filtered $\mathbf{k}$-coalgebra.
In our above proof of Corollary \ref{cor.id-S2.p}, we have already shown that
\begin{itemize}
\item we have $S^{2}\left( 1_{H}\right) =1_{H}$;
\item the map $S^{2}:H\rightarrow H$ is a $\mathbf{k}$-coalgebra homomorphism;
\item we have $\operatorname*{Prim}H\subseteq\operatorname*{Ker}\left(
\operatorname*{id}-S^{2}\right) $.
\end{itemize}
Hence, we can apply Corollary \ref{cor.id-f.cfc1id} to $C=H$ and $C_{\leq
i}=H_{\leq i}$ and $f=S^{2}$. Let us do this now.
Corollary \ref{cor.id-f.cfc1id} \textbf{(b)} (applied to $C=H$ and $C_{\leq
i}=H_{\leq i}$ and $f=S^{2}$) shows that for any positive integer $u$, we have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u}\left( H_{\leq u}\right) =0.
\]
This proves Corollary \ref{cor.id-S2.1} \textbf{(b)}. It remains to prove
Corollary \ref{cor.id-S2.1} \textbf{(a)}:
\textbf{(a)} Let $u>1$ be any integer. Then, Corollary \ref{cor.id-f.cfc1id}
\textbf{(a)} (applied to $C=H$ and $C_{\leq i}=H_{\leq i}$ and $f=S^{2}$)
shows that%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-1}\left( H_{\leq u}\right)
\subseteq\operatorname*{Prim}H.
\]
This proves (\ref{eq.cor.id-S2.1.claim1}). It remains to prove
(\ref{eq.cor.id-S2.1.claim3}).
We have $\left( \operatorname*{id}+S\right) \left( \operatorname*{Prim}%
H\right) =0$ (indeed, we have already shown this in our above proof of
Corollary \ref{cor.id-S2.p} \textbf{(a)}). Now,
\begin{align*}
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-1}\right) \left( H_{\leq u}\right) & =\left(
\operatorname*{id}+S\right) \left( \underbrace{\left( \operatorname*{id}%
-S^{2}\right) ^{u-1}\left( H_{\leq u}\right) }_{\substack{\subseteq
\operatorname*{Prim}H\\\text{(by (\ref{eq.cor.id-S2.1.claim1}))}}}\right) \\
& \subseteq\left( \operatorname*{id}+S\right) \left( \operatorname*{Prim}%
H\right) =0.
\end{align*}
Therefore, $\left( \left( \operatorname*{id}+S\right) \circ\left(
\operatorname*{id}-S^{2}\right) ^{u-1}\right) \left( H_{\leq u}\right) =0$
(since $\left( \left( \operatorname*{id}+S\right) \circ\left(
\operatorname*{id}-S^{2}\right) ^{u-1}\right) \left( H_{\leq u}\right) $
is a $\mathbf{k}$-module). This proves (\ref{eq.cor.id-S2.1.claim3}). Thus,
the proof of Corollary \ref{cor.id-S2.1} \textbf{(a)} is complete.
\end{proof}
\subsection{\label{sec.pf.hopf}Proofs for Section \ref{sec.hopf}}
We shall next focus on proving the claims left unproven in Section
\ref{sec.hopf}. Before we do so, let us first collect a few basic properties
of connected graded Hopf algebras into a lemma for convenience:
\begin{lemma}
\label{lem.cghopf.basics}Let $H$ be a connected graded $\mathbf{k}$-Hopf
algebra with unity $1_{H}$ and antipode $S$. Then:
\textbf{(a)} If $n$ is a positive integer, and if $x$ is an element of $H_{n}%
$, then we have%
\[
\Delta\left( x\right) =1_{H}\otimes x+x\otimes1_{H}%
+w\ \ \ \ \ \ \ \ \ \ \text{for some }w\in\sum_{k=1}^{n-1}H_{k}\otimes
H_{n-k}.
\]
\textbf{(b)} We have $H_{1}\subseteq\operatorname*{Prim}H$.
\textbf{(c)} We have $S\left( ab\right) =ba$ for any $a,b\in H_{1}$.
\end{lemma}
\begin{proof}
[Proof of Lemma \ref{lem.cghopf.basics}.]\textbf{(a)} This follows from
\cite[Exercise 1.3.20 (h)]{GriRei} (applied to $A=H$). (Note that what we are
calling $w$ is denoted by $\Delta_{+}\left( x\right) $ in \cite[Exercise
1.3.20 (h)]{GriRei}.) It also appears in \cite[Proposition II.1.1]{Mancho06}
and in \cite[Theorem 2.18]{Preiss16} (using the notation $\widetilde{\Delta
}\left( x\right) $ for $\Delta\left( x\right) -1_{H}\otimes x-x\otimes
1_{H}$).
\textbf{(b)} Let $x\in H_{1}$. Then, $x$ is an element of $H_{1}$. Hence,
Lemma \ref{lem.cghopf.basics} \textbf{(a)} (applied to $n=1$) yields that we
have%
\[
\Delta\left( x\right) =1_{H}\otimes x+x\otimes1_{H}%
+w\ \ \ \ \ \ \ \ \ \ \text{for some }w\in\sum_{k=1}^{1-1}H_{k}\otimes
H_{1-k}.
\]
Consider this $w$. We have%
\[
w\in\sum_{k=1}^{1-1}H_{k}\otimes H_{1-k}=\left( \text{empty sum}\right) =0,
\]
so that $w=0$. Hence, $\Delta\left( x\right) =1_{H}\otimes x+x\otimes
1_{H}+\underbrace{w}_{=0}=1_{H}\otimes x+x\otimes1_{H}=x\otimes1_{H}%
+1_{H}\otimes x$. In other words, the element $x$ of $H$ is primitive (by the
definition of a \textquotedblleft primitive\textquotedblright\ element). In
other words, $x\in\operatorname*{Prim}H$ (since $\operatorname*{Prim}H$ is
defined as the set of all primitive elements of $H$).
Forget that we fixed $x$. We thus have shown that $x\in\operatorname*{Prim}H$
for each $x\in H_{1}$. In other words, $H_{1}\subseteq\operatorname*{Prim}H$.
This proves Lemma \ref{lem.cghopf.basics} \textbf{(b)}.
\textbf{(c)} Let $a,b\in H_{1}$. Then, $a\in H_{1}\subseteq
\operatorname*{Prim}H$ (by Lemma \ref{lem.cghopf.basics} \textbf{(b)}). In
other words, the element $a$ of $H$ is primitive (since $\operatorname*{Prim}%
H$ is defined as the set of all primitive elements of $H$). Hence, $S\left(
a\right) =-a$ (by Lemma \ref{lem.bialg.antip-props} \textbf{(c)}, applied to
$x=a$). Similarly, $S\left( b\right) =-b$. However, it is well-known (see,
e.g., \cite[Proposition 1.4.10]{GriRei} or \cite[Proposition 7.1.9
(a)]{Radfor12}) that the antipode $S$ of $H$ is a $\mathbf{k}$-algebra
anti-endomorphism, i.e., that it satisfies $S\left( 1_{H}\right) =1_{H}$
and
\begin{equation}
S\left( uv\right) =S\left( v\right) S\left( u\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }u,v\in H. \label{pf.lem.cghopf.basics.c.uv}%
\end{equation}
Applying (\ref{pf.lem.cghopf.basics.c.uv}) to $u=a$ and $v=b$, we obtain
$S\left( ab\right) =\underbrace{S\left( b\right) }_{=-b}%
\underbrace{S\left( a\right) }_{=-a}=\left( -b\right) \left( -a\right)
=ba$. This proves Lemma \ref{lem.cghopf.basics} \textbf{(c)}.
\end{proof}
\begin{proof}
[Proof of Corollary \ref{cor.id-S2.gr1}.]As we know, the graded $\mathbf{k}%
$-Hopf algebra $H$ automatically becomes a filtered $\mathbf{k}$-Hopf algebra
with filtration $\left( H_{\leq0},H_{\leq1},H_{\leq2},\ldots\right) $
defined by setting%
\[
H_{\leq n}:=\bigoplus_{i=0}^{n}H_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }%
n\in\mathbb{N}.
\]
This filtered $\mathbf{k}$-Hopf algebra $H$ is connected, since $H_{\leq
0}=H_{0}$. Thus, Corollary \ref{cor.id-S2.1} can be applied.
Let $u$ be a positive integer. Then, $H_{u}\subseteq H_{\leq u}$%
\ \ \ \ \footnote{\textit{Proof.} The definition of $H_{\leq u}$ yields
$H_{\leq u}=\bigoplus_{i=0}^{u}H_{i}$. However, $H_{u}\subseteq\bigoplus
_{i=0}^{u}H_{i}$ (since $H_{u}$ is an addend of the direct sum $\bigoplus
_{i=0}^{u}H_{i}$). In view of $H_{\leq u}=\bigoplus_{i=0}^{u}H_{i}$, this
rewrites as $H_{u}\subseteq H_{\leq u}$.}.
Now, we must prove the three relations (\ref{eq.cor.id-S2.gr1.claim1}),
(\ref{eq.cor.id-S2.gr1.claim3}) and (\ref{eq.cor.id-S2.gr1.claim2}). The third
one is the easiest: We have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u}\left( \underbrace{H_{u}%
}_{\subseteq H_{\leq u}}\right) \subseteq\left( \operatorname*{id}%
-S^{2}\right) ^{u}\left( H_{\leq u}\right) =0
\]
(by Corollary \ref{cor.id-S2.1} \textbf{(b)}) and therefore $\left(
\operatorname*{id}-S^{2}\right) ^{u}\left( H_{u}\right) =0$ (since $\left(
\operatorname*{id}-S^{2}\right) ^{u}\left( H_{u}\right) $ is a $\mathbf{k}%
$-module). This proves (\ref{eq.cor.id-S2.gr1.claim2}).
We shall now focus on proving (\ref{eq.cor.id-S2.gr1.claim1}). Indeed, if
$u>1$, then (\ref{eq.cor.id-S2.gr1.claim1}) follows from%
\begin{align*}
\left( \operatorname*{id}-S^{2}\right) ^{u-1}\left( \underbrace{H_{u}%
}_{\subseteq H_{\leq u}}\right) & \subseteq\left( \operatorname*{id}%
-S^{2}\right) ^{u-1}\left( H_{\leq u}\right) \\
& \subseteq\operatorname*{Prim}H\ \ \ \ \ \ \ \ \ \ \left( \text{by
(\ref{eq.cor.id-S2.1.claim1}), since }u>1\right) .
\end{align*}
Thus, for the rest of this proof of (\ref{eq.cor.id-S2.gr1.claim1}), we WLOG
assume that we don't have $u>1$. Hence, we have $u=1$ (since $u$ is a positive
integer). Therefore, $u-1=0$, so that $\left( \operatorname*{id}%
-S^{2}\right) ^{u-1}=\left( \operatorname*{id}-S^{2}\right) ^{0}%
=\operatorname*{id}$. Thus,
\begin{align*}
\underbrace{\left( \operatorname*{id}-S^{2}\right) ^{u-1}}%
_{=\operatorname*{id}}\left( H_{u}\right) & =\operatorname*{id}\left(
H_{u}\right) =H_{u}=H_{1}\ \ \ \ \ \ \ \ \ \ \left( \text{since }u=1\right)
\\
& \subseteq\operatorname*{Prim}H\ \ \ \ \ \ \ \ \ \ \left( \text{by Lemma
\ref{lem.cghopf.basics} \textbf{(b)}}\right) .
\end{align*}
This completes our proof of (\ref{eq.cor.id-S2.gr1.claim1}).
Now, it remains to prove (\ref{eq.cor.id-S2.gr1.claim3}). We have $\left(
\operatorname*{id}+S\right) \left( \operatorname*{Prim}H\right) =0$
(indeed, we have already shown this in our above proof of Corollary
\ref{cor.id-S2.p} \textbf{(a)}). Now,
\begin{align*}
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-1}\right) \left( H_{u}\right) & =\left(
\operatorname*{id}+S\right) \left( \underbrace{\left( \operatorname*{id}%
-S^{2}\right) ^{u-1}\left( H_{u}\right) }_{\substack{\subseteq
\operatorname*{Prim}H\\\text{(by (\ref{eq.cor.id-S2.gr1.claim1}))}}}\right) \\
& \subseteq\left( \operatorname*{id}+S\right) \left( \operatorname*{Prim}%
H\right) =0.
\end{align*}
Therefore, $\left( \left( \operatorname*{id}+S\right) \circ\left(
\operatorname*{id}-S^{2}\right) ^{u-1}\right) \left( H_{u}\right) =0$
(since $\left( \left( \operatorname*{id}+S\right) \circ\left(
\operatorname*{id}-S^{2}\right) ^{u-1}\right) \left( H_{u}\right) $ is a
$\mathbf{k}$-module). This proves (\ref{eq.cor.id-S2.gr1.claim3}).
Thus, we have shown all three relations (\ref{eq.cor.id-S2.gr1.claim1}),
(\ref{eq.cor.id-S2.gr1.claim3}) and (\ref{eq.cor.id-S2.gr1.claim2}). This
completes the proof of Corollary \ref{cor.id-S2.gr1}.
\end{proof}
\begin{proof}
[Proof of Corollary \ref{cor.id-S2.grp}.]Let $1_{H}$ denote the unity of the
$\mathbf{k}$-algebra $H$.
As we know, the graded $\mathbf{k}$-Hopf algebra $H$ automatically becomes a
filtered $\mathbf{k}$-Hopf algebra with filtration $\left( H_{\leq0}%
,H_{\leq1},H_{\leq2},\ldots\right) $ defined by setting%
\[
H_{\leq n}:=\bigoplus_{i=0}^{n}H_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }%
n\in\mathbb{N}.
\]
This filtered $\mathbf{k}$-Hopf algebra $H$ is connected (because the graded
$\mathbf{k}$-Hopf algebra $H$ is connected, and because $H_{\leq0}=H_{0}$).
We know that $p$ is a positive integer; thus, both $0$ and $1$ are elements of
the set $\left\{ 0,1,\ldots,p\right\} $.
Now, we shall show that
\begin{equation}
\left( \operatorname*{id}-S^{2}\right) \left( H_{\leq p}\right) =0.
\label{pf.cor.id-S2.grp.1}%
\end{equation}
[\textit{Proof of (\ref{pf.cor.id-S2.grp.1}):} It is easy to see that $\left(
\operatorname*{id}-S^{2}\right) \left( H_{0}\right) =0$%
\ \ \ \ \footnote{\textit{Proof.} In our above proof of Corollary
\ref{cor.id-S2.p}, we have already shown that $S^{2}\left( 1_{H}\right)
=1_{H}$. Hence,%
\[
\left( \operatorname*{id}-S^{2}\right) \left( 1_{H}\right)
=\underbrace{\operatorname*{id}\left( 1_{H}\right) }_{=1_{H}}%
-\underbrace{S^{2}\left( 1_{H}\right) }_{=1_{H}}=1_{H}-1_{H}=0.
\]
\par
Define a $\mathbf{k}$-linear map $\idbar:H\rightarrow H$ by setting%
\[
\idbar\left( c\right) :=c-\epsilon\left( c\right) 1_{H}%
\ \ \ \ \ \ \ \ \ \ \text{for each }c\in H.
\]
Then, Lemma \ref{lem.cfc.idbar} \textbf{(c)} (applied to $H$ and $H_{\leq i}$
instead of $C$ and $C_{\leq i}$) yields $\idbar\left( H_{\leq0}\right) =0$.
\par
The definition of $H_{\leq0}$ yields $H_{\leq0}=\bigoplus_{i=0}^{0}H_{i}%
=H_{0}$. Thus, $H_{0}=H_{\leq0}$. Applying the map $\idbar$ to both sides of
this equality, we obtain $\idbar\left( H_{0}\right) =\idbar\left( H_{\leq
0}\right) =0$.
\par
Now, let $c\in H_{0}$. Then, the definition of $\idbar$ yields $\idbar\left(
c\right) =c-\epsilon\left( c\right) 1_{H}$. On the other hand, we have
$\idbar\left( c\right) =0$ (since $\idbar\left( \underbrace{c}_{\in H_{0}%
}\right) \in\idbar\left( H_{0}\right) =0$). Comparing these two equalities,
we obtain $c-\epsilon\left( c\right) 1_{H}=0$. In other words,
$c=\epsilon\left( c\right) 1_{H}$. Now, applying the map $\operatorname*{id}%
-S^{2}$ to both sides of this equality, we obtain%
\begin{align*}
\left( \operatorname*{id}-S^{2}\right) \left( c\right) & =\left(
\operatorname*{id}-S^{2}\right) \left( \epsilon\left( c\right)
1_{H}\right) \\
& =\epsilon\left( c\right) \cdot\underbrace{\left( \operatorname*{id}%
-S^{2}\right) \left( 1_{H}\right) }_{=0}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\operatorname*{id}-S^{2}\text{ is }\mathbf{k}%
\text{-linear}\right) \\
& =0.
\end{align*}
\par
Forget that we fixed $c$. We thus have shown that $\left( \operatorname*{id}%
-S^{2}\right) \left( c\right) =0$ for each $c\in H_{0}$. In other words,
$\left( \operatorname*{id}-S^{2}\right) \left( H_{0}\right) =0$.}.
Furthermore, $\left( \operatorname*{id}-S^{2}\right) \left( H_{1}\right)
=0$\ \ \ \ \footnote{\textit{Proof.} Let $x\in H_{1}$. Then, $x\in
H_{1}\subseteq\operatorname*{Prim}H$ (by Lemma \ref{lem.cghopf.basics}
\textbf{(b)}). In other words, the element $x$ of $H$ is primitive (since
$\operatorname*{Prim}H$ is defined as the set of all primitive elements of
$H$). Hence, Lemma \ref{lem.bialg.antip-props} \textbf{(d)} yields
$S^{2}\left( x\right) =x$. Now,%
\[
\left( \operatorname*{id}-S^{2}\right) \left( x\right)
=\underbrace{\operatorname*{id}\left( x\right) }_{=x}-\underbrace{S^{2}%
\left( x\right) }_{=x}=x-x=0.
\]
\par
Forget that we fixed $x$. We thus have shown that $\left( \operatorname*{id}%
-S^{2}\right) \left( x\right) =0$ for each $x\in H_{1}$. In other words, we
have $\left( \operatorname*{id}-S^{2}\right) \left( H_{1}\right) =0$.}.
Now, the definition of $H_{\leq p}$ yields $H_{\leq p}=\bigoplus_{i=0}%
^{p}H_{i}=\sum_{i=0}^{p}H_{i}$ (since direct sums are sums). Applying the map
$\operatorname*{id}-S^{2}$ to both sides of this equality, we obtain%
\begin{align*}
\left( \operatorname*{id}-S^{2}\right) \left( H_{\leq p}\right) &
=\left( \operatorname*{id}-S^{2}\right) \left( \sum_{i=0}^{p}H_{i}\right)
\\
& =\sum_{i=0}^{p}\left( \operatorname*{id}-S^{2}\right) \left(
H_{i}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since the map }%
\operatorname*{id}-S^{2}\text{ is }\mathbf{k}\text{-linear}\right) \\
& =\underbrace{\left( \operatorname*{id}-S^{2}\right) \left( H_{0}\right)
}_{=0}+\underbrace{\left( \operatorname*{id}-S^{2}\right) \left(
H_{1}\right) }_{=0}+\sum_{i=2}^{p}\underbrace{\left( \operatorname*{id}%
-S^{2}\right) \left( H_{i}\right) }_{\substack{=0\\\text{(by
(\ref{eq.cor.id-S2.grp.ass-ann}))}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addends for }i=0\\
\text{and for }i=1\text{ from the sum (since both }0\text{ and }1\\
\text{are elements of the set }\left\{ 0,1,\ldots,p\right\} \text{)}%
\end{array}
\right) \\
& =\underbrace{0+0}_{=0}+\underbrace{\sum_{i=2}^{p}0}_{=0}=0.
\end{align*}
This proves (\ref{pf.cor.id-S2.grp.1}).]
Hence, we can apply Corollary \ref{cor.id-S2.p}. In particular, Corollary
\ref{cor.id-S2.p} \textbf{(a)} shows that for any integer $u>p$, we have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-p}\left( H_{\leq u}\right)
\subseteq\operatorname*{Prim}H
\]
and%
\[
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-p}\right) \left( H_{\leq u}\right) =0.
\]
This proves Corollary \ref{cor.id-S2.grp} \textbf{(a)}.
Furthermore, Corollary \ref{cor.id-S2.p} \textbf{(b)} shows that for any
integer $u\geq p$, we have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-p+1}\left( H_{\leq u}\right)
=0.
\]
This proves Corollary \ref{cor.id-S2.grp} \textbf{(b)}.
\end{proof}
\begin{proof}
[Proof of Corollary \ref{cor.id-S2.gr2}.]\textbf{(a)} Let $1_{H}$ denote the
unity of the $\mathbf{k}$-algebra $H$. Define the maps $m$ and $u$ as in
Remark \ref{rmk.hopf.antipode-pro}.
Let $x\in H_{2}$. Then, $x$ is an element of $H_{2}$. Hence, Lemma
\ref{lem.cghopf.basics} \textbf{(a)} (applied to $n=2$) yields that we have%
\[
\Delta\left( x\right) =1_{H}\otimes x+x\otimes1_{H}%
+w\ \ \ \ \ \ \ \ \ \ \text{for some }w\in\sum_{k=1}^{2-1}H_{k}\otimes
H_{2-k}.
\]
Consider this $w$. We have%
\begin{align*}
w & \in\sum_{k=1}^{2-1}H_{k}\otimes H_{2-k}=\sum_{k=1}^{1}H_{k}\otimes
H_{2-k}\ \ \ \ \ \ \ \ \ \ \left( \text{since }2-1=1\right) \\
& =H_{1}\otimes H_{2-1}=H_{1}\otimes H_{1}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }2-1=1\right) .
\end{align*}
Therefore, $w$ is a tensor in $H_{1}\otimes H_{1}$. Hence, $w$ can be written
in the form
\begin{equation}
w=\sum_{i=1}^{k}\lambda_{i}a_{i}\otimes b_{i} \label{pf.cor.id-S2.gr2.a.3}%
\end{equation}
for some $k\in\mathbb{N}$, some $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}%
\in\mathbf{k}$, some $a_{1},a_{2},\ldots,a_{k}\in H_{1}$ and some $b_{1}%
,b_{2},\ldots,b_{k}\in H_{1}$. Consider this $k$, these $\lambda_{1}%
,\lambda_{2},\ldots,\lambda_{k}$, these $a_{1},a_{2},\ldots,a_{k}$ and these
$b_{1},b_{2},\ldots,b_{k}$.
We have $a_{1},a_{2},\ldots,a_{k}\in H_{1}$. Thus, for each $i\in\left\{
1,2,\ldots,k\right\} $, we have $a_{i}\in H_{1}\subseteq\operatorname*{Prim}%
H$ (by Lemma \ref{lem.cghopf.basics} \textbf{(b)}) and therefore%
\begin{equation}
S\left( a_{i}\right) =-a_{i} \label{pf.cor.id-S2.gr2.a.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.id-S2.gr2.a.4}):} Let $i\in\left\{
1,2,\ldots,k\right\} $. Then, $a_{i}\in H_{1}\subseteq\operatorname*{Prim}H$
(by Lemma \ref{lem.cghopf.basics} \textbf{(b)}). In other words, the element
$a_{i}$ of $H$ is primitive (since $\operatorname*{Prim}H$ is defined as the
set of all primitive elements of $H$). Therefore, Lemma
\ref{lem.bialg.antip-props} \textbf{(c)} (applied to $a_{i}$ instead of $x$)
yields $S\left( a_{i}\right) =-a_{i}$. This proves
(\ref{pf.cor.id-S2.gr2.a.4}).}. Moreover, for each $i\in\left\{
1,2,\ldots,k\right\} $, we have
\begin{equation}
S\left( a_{i}b_{i}\right) =a_{i}b_{i} \label{pf.cor.id-S2.gr2.a.Saibi}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.id-S2.gr2.a.Saibi}):} Let
$i\in\left\{ 1,2,\ldots,k\right\} $. Then, $a_{i}\in H_{1}$ (since
$a_{1},a_{2},\ldots,a_{k}\in H_{1}$) and $b_{i}\in H_{1}$ (since $b_{1}%
,b_{2},\ldots,b_{k}\in H_{1}$). Hence, (\ref{eq.cor.id-S2.gr2.commH1})
(applied to $a=a_{i}$ and $b=b_{i}$) yields $a_{i}b_{i}=b_{i}a_{i}$. On the
other hand, Lemma \ref{lem.cghopf.basics} \textbf{(c)} (applied to $a=a_{i}$
and $b=b_{i}$) yields $S\left( a_{i}b_{i}\right) =b_{i}a_{i}$. Comparing
these two equalities, we obtain $S\left( a_{i}b_{i}\right) =a_{i}b_{i}$.
This proves (\ref{pf.cor.id-S2.gr2.a.Saibi}).}.
Applying the map $S\otimes\operatorname*{id}\nolimits_{H}:H\otimes
H\rightarrow H\otimes H$ to both sides of the equality
(\ref{pf.cor.id-S2.gr2.a.3}), we obtain%
\begin{align}
\left( S\otimes\operatorname*{id}\nolimits_{H}\right) \left( w\right) &
=\left( S\otimes\operatorname*{id}\nolimits_{H}\right) \left( \sum
_{i=1}^{k}\lambda_{i}a_{i}\otimes b_{i}\right) =\sum_{i=1}^{k}\lambda
_{i}\underbrace{\left( S\otimes\operatorname*{id}\nolimits_{H}\right)
\left( a_{i}\otimes b_{i}\right) }_{=S\left( a_{i}\right) \otimes
\operatorname*{id}\nolimits_{H}\left( b_{i}\right) }\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since the map }S\otimes\operatorname*{id}%
\nolimits_{H}\text{ is }\mathbf{k}\text{-linear}\right) \nonumber\\
& =\sum_{i=1}^{k}\lambda_{i}\underbrace{S\left( a_{i}\right) }%
_{\substack{=-a_{i}\\\text{(by (\ref{pf.cor.id-S2.gr2.a.4}))}}}\otimes
\underbrace{\operatorname*{id}\nolimits_{H}\left( b_{i}\right) }_{=b_{i}%
}=\sum_{i=1}^{k}\lambda_{i}\underbrace{\left( -a_{i}\right) \otimes b_{i}%
}_{\substack{=-a_{i}\otimes b_{i}\\\text{(since the map }H\times H\rightarrow
H\otimes H,\ \left( u,v\right) \mapsto u\otimes v\\\text{is }\mathbf{k}%
\text{-bilinear (by the definition of the}\\\text{tensor product))}%
}}\nonumber\\
& =\sum_{i=1}^{k}\lambda_{i}\left( -a_{i}\otimes b_{i}\right)
=-\underbrace{\sum_{i=1}^{k}\lambda_{i}a_{i}\otimes b_{i}}%
_{\substack{=w\\\text{(by (\ref{pf.cor.id-S2.gr2.a.3}))}}}\nonumber\\
& =-w. \label{pf.cor.id-S2.gr2.a.5}%
\end{align}
The Hopf algebra $H$ is graded. Hence, its counit $\epsilon$ is a graded map
from $H$ to $\mathbf{k}$ (by the definition of a graded Hopf algebra). In
other words, $\epsilon\left( H_{i}\right) \subseteq\mathbf{k}_{i}$ for each
$i\in\mathbb{N}$. Thus, $\epsilon\left( H_{2}\right) \subseteq\mathbf{k}%
_{2}=0$ (since the graded $\mathbf{k}$-module $\mathbf{k}$ is concentrated in
degree $0$). Therefore, $\epsilon\left( \underbrace{x}_{\in H_{2}}\right)
\in\epsilon\left( H_{2}\right) \subseteq0$, so that $\epsilon\left(
x\right) =0$.
Lemma \ref{lem.bialg.antip-props} \textbf{(b)} yields $S\left( 1_{H}\right)
=1_{H}$.
Applying both sides of the equality (\ref{eq.rmk.hopf.antipode-pro.1}) to $x$,
we obtain%
\[
\left( m\circ\left( S\otimes\operatorname*{id}\nolimits_{H}\right)
\circ\Delta\right) \left( x\right) =\left( u\circ\epsilon\right) \left(
x\right) =u\left( \underbrace{\epsilon\left( x\right) }_{=0}\right)
=u\left( 0\right) =0
\]
(since the map $u$ is $\mathbf{k}$-linear). Therefore,%
\begin{align*}
0 & =\left( m\circ\left( S\otimes\operatorname*{id}\nolimits_{H}\right)
\circ\Delta\right) \left( x\right) \\
& =m\left( \left( S\otimes\operatorname*{id}\nolimits_{H}\right) \left(
\underbrace{\Delta\left( x\right) }_{=1_{H}\otimes x+x\otimes1_{H}%
+w}\right) \right) \\
& =m\left( \underbrace{\left( S\otimes\operatorname*{id}\nolimits_{H}%
\right) \left( 1_{H}\otimes x+x\otimes1_{H}+w\right) }_{\substack{=\left(
S\otimes\operatorname*{id}\nolimits_{H}\right) \left( 1_{H}\otimes x\right)
+\left( S\otimes\operatorname*{id}\nolimits_{H}\right) \left( x\otimes
1_{H}\right) +\left( S\otimes\operatorname*{id}\nolimits_{H}\right) \left(
w\right) \\\text{(since the map }S\otimes\operatorname*{id}\nolimits_{H}%
\text{ is }\mathbf{k}\text{-linear)}}}\right) \\
& =m\left( \underbrace{\left( S\otimes\operatorname*{id}\nolimits_{H}%
\right) \left( 1_{H}\otimes x\right) }_{=S\left( 1_{H}\right)
\otimes\operatorname*{id}\nolimits_{H}\left( x\right) }+\underbrace{\left(
S\otimes\operatorname*{id}\nolimits_{H}\right) \left( x\otimes1_{H}\right)
}_{=S\left( x\right) \otimes\operatorname*{id}\nolimits_{H}\left(
1_{H}\right) }+\underbrace{\left( S\otimes\operatorname*{id}\nolimits_{H}%
\right) \left( w\right) }_{\substack{=-w\\\text{(by
(\ref{pf.cor.id-S2.gr2.a.5}))}}}\right) \\
& =m\left( \underbrace{S\left( 1_{H}\right) }_{=1_{H}}\otimes
\underbrace{\operatorname*{id}\nolimits_{H}\left( x\right) }_{=x}+S\left(
x\right) \otimes\underbrace{\operatorname*{id}\nolimits_{H}\left(
1_{H}\right) }_{=1_{H}}+\left( -w\right) \right) \\
& =m\left( 1_{H}\otimes x+S\left( x\right) \otimes1_{H}+\left( -w\right)
\right) =\underbrace{m\left( 1_{H}\otimes x\right) }_{\substack{=1_{H}%
x\\\text{(by the definition}\\\text{of the map }m\text{)}}%
}+\underbrace{m\left( S\left( x\right) \otimes1_{H}\right) }%
_{\substack{=S\left( x\right) \cdot1_{H}\\\text{(by the definition}%
\\\text{of the map }m\text{)}}}+\underbrace{m\left( -w\right) }%
_{\substack{=-m\left( w\right) \\\text{(since the map }m\\\text{is
}\mathbf{k}\text{-linear)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since the map }m\text{ is }%
\mathbf{k}\text{-linear}\right) \\
& =\underbrace{1_{H}x}_{=x}+\underbrace{S\left( x\right) \cdot1_{H}%
}_{=S\left( x\right) }+\left( -m\left( w\right) \right) =x+S\left(
x\right) +\left( -m\left( w\right) \right) =x+S\left( x\right)
-m\left( w\right) .
\end{align*}
Solving this equality for $S\left( x\right) $, we obtain%
\begin{equation}
S\left( x\right) =m\left( w\right) -x. \label{pf.cor.id-S2.gr2.a.Sx=}%
\end{equation}
Applying the map $m:H\otimes H\rightarrow H$ to both sides of the equality
(\ref{pf.cor.id-S2.gr2.a.3}), we obtain%
\begin{align}
m\left( w\right) & =m\left( \sum_{i=1}^{k}\lambda_{i}a_{i}\otimes
b_{i}\right) \nonumber\\
& =\sum_{i=1}^{k}\lambda_{i}\underbrace{m\left( a_{i}\otimes b_{i}\right)
}_{\substack{=a_{i}b_{i}\\\text{(by the definition of the map }m\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left( \text{since the map }m\text{ is }\mathbf{k}%
\text{-linear}\right) \nonumber\\
& =\sum_{i=1}^{k}\lambda_{i}a_{i}b_{i}. \label{pf.cor.id-S2.gr2.a.mw=}%
\end{align}
Applying the map $S$ to both sides of this equality, we obtain%
\begin{align}
S\left( m\left( w\right) \right) & =S\left( \sum_{i=1}^{k}\lambda
_{i}a_{i}b_{i}\right) =\sum_{i=1}^{k}\lambda_{i}\underbrace{S\left(
a_{i}b_{i}\right) }_{\substack{=a_{i}b_{i}\\\text{(by
(\ref{pf.cor.id-S2.gr2.a.Saibi}))}}}\ \ \ \ \ \ \ \ \ \ \left( \text{since
the map }S\text{ is }\mathbf{k}\text{-linear}\right) \nonumber\\
& =\sum_{i=1}^{k}\lambda_{i}a_{i}b_{i}=m\left( w\right)
\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.cor.id-S2.gr2.a.mw=})}\right) .
\label{pf.cor.id-S2.gr2.a.Smw=}%
\end{align}
Now, applying the map $S$ to both sides of the equality
(\ref{pf.cor.id-S2.gr2.a.Sx=}), we obtain%
\begin{align*}
S\left( S\left( x\right) \right) & =S\left( m\left( w\right)
-x\right) =\underbrace{S\left( m\left( w\right) \right) }%
_{\substack{=m\left( w\right) \\\text{(by (\ref{pf.cor.id-S2.gr2.a.Smw=}))}%
}}-\underbrace{S\left( x\right) }_{\substack{=m\left( w\right)
-x\\\text{(by (\ref{pf.cor.id-S2.gr2.a.Sx=}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }S\text{ is }\mathbf{k}\text{-linear}\right) \\
& =m\left( w\right) -\left( m\left( w\right) -x\right) =x.
\end{align*}
Now,%
\[
\left( \operatorname*{id}-S^{2}\right) \left( x\right)
=\underbrace{\operatorname*{id}\left( x\right) }_{=x}-\underbrace{S^{2}%
}_{=S\circ S}\left( x\right) =x-\underbrace{\left( S\circ S\right) \left(
x\right) }_{=S\left( S\left( x\right) \right) =x}=x-x=0.
\]
Forget that we fixed $x$. We thus have shown that $\left( \operatorname*{id}%
-S^{2}\right) \left( x\right) =0$ for each $x\in H_{2}$. In other words,
$\left( \operatorname*{id}-S^{2}\right) \left( H_{2}\right) =0$. This
proves Corollary \ref{cor.id-S2.gr2} \textbf{(a)}.
Now we know that $\left( \operatorname*{id}-S^{2}\right) \left(
H_{2}\right) =0$ (by Corollary \ref{cor.id-S2.gr2} \textbf{(a)}). In other
words, $\left( \operatorname*{id}-S^{2}\right) \left( H_{i}\right) =0$
holds for $i=2$. In other words, all $i\in\left\{ 2,3,\ldots,2\right\} $
satisfy $\left( \operatorname*{id}-S^{2}\right) \left( H_{i}\right) =0$
(since the only $i\in\left\{ 2,3,\ldots,2\right\} $ is $2$). Hence, we can
apply Corollary \ref{cor.id-S2.grp} to $p=2$.
Thus, Corollary \ref{cor.id-S2.grp} \textbf{(a)} (applied to $p=2$) yields
that for any integer $u>2$, we have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-2}\left( H_{\leq u}\right)
\subseteq\operatorname*{Prim}H
\]
and%
\[
\left( \left( \operatorname*{id}+S\right) \circ\left( \operatorname*{id}%
-S^{2}\right) ^{u-2}\right) \left( H_{\leq u}\right) =0.
\]
This proves Corollary \ref{cor.id-S2.gr2} \textbf{(b)}.
Furthermore, Corollary \ref{cor.id-S2.grp} \textbf{(b)} (applied to $p=2$)
yields that for any integer $u\geq2$, we have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-2+1}\left( H_{\leq u}\right)
=0.
\]
In other words, for any integer $u\geq2$, we have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-1}\left( H_{\leq u}\right) =0
\]
(since $u-2+1=u-1$). In other words, for any integer $u>1$, we have%
\[
\left( \operatorname*{id}-S^{2}\right) ^{u-1}\left( H_{\leq u}\right) =0
\]
(since \textquotedblleft$u>1$\textquotedblright\ is equivalent to
\textquotedblleft$u\geq2$\textquotedblright\ when $u$ is an integer). This
proves Corollary \ref{cor.id-S2.gr2} \textbf{(c)}.
\end{proof}
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\end{document}