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\begin{document}

\begin{center}
\textbf{La d\'{e}composition en poids des alg\`{e}bres de Hopf}

\textit{Fr\'{e}d\'{e}ric Patras}

Ann. Inst. Fourier, Grenoble \textbf{43}, 4 (1993), pp. 1067--1087

\url{http://aif.cedram.org/cgi-bin/fitem?id=AIF_1993__43_4_1067_0} (published version)

\textbf{Errata and remarks} (by \textit{Darij Grinberg})

\bigskip
\end{center}

The following are remarks I have made while reading the above-cited paper by
Fr\'{e}d\'{e}ric Patras. I think it is an interesting and rather readable text
(despite some minor typos and tersely written proofs).

Some of the below remarks are just quick corrections of minor mistakes (at
least as far as I can tell; I can neither guarantee that these ``mistakes''
really are mistakes, nor that my ``corrections'' are correct!). Some others
are detailed expositions of certain proofs which have been only vaguely
sketched in Patras's paper. Finally, some others give alternative proofs for
results in Patras's paper (sometimes inserting additional results into
Patras's paper, to be used as lemmata later on).

Different remarks are separated by horizontal lines, like this:

\parbreak


\textbf{Page 1068:} I think ``aux endomorphismes $\psi^{k}$'' should be ``aux
endomorphismes $\Psi^{k}$'' here.

\parbreak


\textbf{Page 1069, Definition 1.1:} There is nothing wrong here, but I think
it would be helpful to notice that what Patras calls ``alg\`{e}bre de Hopf''
is \textit{not} the same as what modern-day algebraists call a Hopf algebra.
What Patras calls ``alg\`{e}bre de Hopf'' is a kind of super-version of a
graded bialgebra (not necessarily having an antipode!); in constrast, what
modern-day algebraists call a Hopf algebra is just a bialgebra with antipode.
(Nevertheless, I am going to use the words ``Hopf algebra'' for what Patras
calls ``alg\`{e}bre de Hopf'' in the following.)

\parbreak


\textbf{Page 1070, fifth line of this page:} Here, Patras write:

``Une big\`{e}bre gradu\'{e}e ou une alg\`{e}bre de Hopf est \textit{connexe}
si $H_{0}\cong K$.''

This definition is good when $K$ is a field, but in the general case when $K$
is a commutative ring, it is not a reasonable definition of ``connected''.
Since Patras, in his paper, always works over a field $K$, this is not a
problem for him, but I still prefer the following (in my opinion, better)
definition of ``connected'': A graded bialgebra or Hopf algebra $H$ is
\textit{connected} if and only if the map $\epsilon\mid_{H_{0}}:H_{0}%
\rightarrow K$ is an isomorphism.

Note that, when $K$ is a field, this definition is equivalent to Patras's
definition, because we have the following equivalence of assertions:%
\begin{align*}
&  \ \left(  H_{0}\cong K\text{ as }K\text{-vector spaces}\right) \\
&  \Longleftrightarrow\ \left(  \dim\left(  H_{0}\right)  =1\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{where }\dim V\text{ denotes the dimension
of any }K\text{-vector space }V\right) \\
&  \Longleftrightarrow\ \left(  \dim\left(  \operatorname*{Ker}\left(
\epsilon\mid_{H_{0}}\right)  \right)  =0\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since we know that the map }\epsilon\mid_{H_{0}}:H_{0}\rightarrow
K\text{ is surjective (because}\\
\left(  \epsilon\mid_{H_{0}}\right)  \left(  1\right)  =\epsilon\left(
1\right)  =1\text{ (by the axioms of a bialgebra, since }H\text{ is a
bialgebra)),}\\
\text{and thus (by the isomorphism theorem) }K\cong H_{0}\diagup
\operatorname*{Ker}\left(  \epsilon\mid_{H_{0}}\right)  \text{, so that}\\
\dim K=\dim\left(  H_{0}\diagup\operatorname*{Ker}\left(  \epsilon\mid_{H_{0}%
}\right)  \right)  =\dim H_{0}-\dim\left(  \operatorname*{Ker}\left(
\epsilon\mid_{H_{0}}\right)  \right)  \text{,}\\
\text{so that }\dim H_{0}=\underbrace{\dim K}_{=1}+\dim\left(
\operatorname*{Ker}\left(  \epsilon\mid_{H_{0}}\right)  \right)
=1+\dim\left(  \operatorname*{Ker}\left(  \epsilon\mid_{H_{0}}\right)
\right)  \text{,}\\
\text{and therefore the equation }\dim\left(  H_{0}\right)  =1\text{ is
equivalent to }\dim\left(  \operatorname*{Ker}\left(  \epsilon\mid_{H_{0}%
}\right)  \right)  =0
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \operatorname*{Ker}\left(  \epsilon\mid
_{H_{0}}\right)  =0\right)  \ \Longleftrightarrow\ \left(  \epsilon\mid
_{H_{0}}\text{ is injective}\right)  \ \Longleftrightarrow\ \left(
\epsilon\mid_{H_{0}}\text{ is bijective}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since we know that the map }\epsilon\mid_{H_{0}}:H_{0}\rightarrow
K\text{ is surjective (because}\\
\left(  \epsilon\mid_{H_{0}}\right)  \left(  1\right)  =\epsilon\left(
1\right)  =1\text{ (by the axioms of a bialgebra, since }H\text{ is a
bialgebra)),}\\
\text{and thus this map }\epsilon\mid_{H_{0}}\text{ is injective if and only
if it is bijective}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \epsilon\mid_{H_{0}}\text{ is an
isomorphism}\right)  .
\end{align*}


\parbreak


\textbf{Page 1070, two lines above Definition 1.2:} Here, Patras writes:

``[...] l'ensemble $\mathcal{L}\left(  H\right)  $ des endomorphismes
lin\'{e}aires de $H$ [...]''.

I don't think that $\mathcal{L}\left(  H\right)  $ denotes the set of all
linear endomorphisms of $H$ throughout the text. It seems to me that
$\mathcal{L}\left(  H\right)  $ indeed denotes the set of all linear
endomorphisms of $H$ when $H$ is just a bialgebra (not graded); however, when
$H$ is a graded bialgebra or an ``alg\`{e}bre de Hopf'' (I would translate
this by ``Hopf algebra'', but as I said, this does not mean what people
nowadays mean by a ``Hopf algebra''), $\mathcal{L}\left(  H\right)  $ denotes
the set of all \textit{graded}\footnote{A linear map $f:V\rightarrow W$
between graded vector spaces $V$ and $W$ is said to be \textit{graded} (or
\textit{compatible with the grading}) if, for every $n\in\mathbb{N}$, it
satisfies $f\left(  V_{n}\right)  \subseteq W_{n}$.} linear endomorphisms of
$H$.

Note that I might be wrong about this, and $\mathcal{L}\left(  H\right)  $
might indeed mean the set of \textit{all} linear endomorphisms of $H$
throughout the text. In this case, however, the homomorphism $\rho_{n}$
defined on page 1074 (``Notons $\rho_{n}$ l'homomorphisme de restriction de
$\mathcal{L}\left(  H\right)  $ dans $\mathcal{L}\left(  H\right)  _{n}$.'')
is not a simple restriction homomorphism (i. e., it is not just given by
$\rho_{n}\left(  f\right)  =f\mid_{\bigoplus\limits_{i=0}^{n}H_{i}}$ for every
$f\in\mathcal{L}\left(  H\right)  $), but instead requires a more subtle
definition: It must then be defined by%
\[
\left(  \rho_{n}\left(  f\right)  =\sum_{i=0}^{n}p_{i}\circ f\circ
p_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }f\in\mathcal{L}\left(  H\right)
\right)  ,
\]
where $p_{i}:H\rightarrow H$ denotes the map which sends every element of $H$
to its $i$-th graded component (seen again as an element of $H$).

\parbreak


\textbf{Page 1071, proof of Proposition 1.4:} Here, Patras writes: ``La
deuxi\`{e}me partie de la proposition se ram\`{e}ne \`{a} \'{e}tablir
l'\'{e}galit\'{e} :%
\[
\Delta^{\left[  k\right]  }\circ\Pi^{\left[  l\right]  }=\Pi_{\left(
k\right)  }^{\left[  l\right]  }\circ\left(  \Delta^{\left[  k\right]
}\right)  ^{\otimes l},
\]
qui est une cons\'{e}quence \`{a} peu pr\`{e}s imm\'{e}diate des axiomes de
structure des big\`{e}bres commutatives.'' This is not totally precise. The
identity $\Delta^{\left[  k\right]  }\circ\Pi^{\left[  l\right]  }%
=\Pi_{\left(  k\right)  }^{\left[  l\right]  }\circ\left(  \Delta^{\left[
k\right]  }\right)  ^{\otimes l}$ is true in any bialgebra, not only in
commutative ones (it follows from the axioms of a bialgebra by a double
induction over $k$ and $l$). However, deriving the ``deuxi\`{e}me partie de la
proposition'' from this identity requires the bialgebra to be commutative.
Here are the details of this derivation: We have $\left(  \Delta^{\left[
k\right]  }\right)  ^{\otimes l}\circ\Delta^{\left[  l\right]  }%
=\Delta^{\left[  lk\right]  }$ (this holds for any coalgebra, and can be
proven by induction using the coassociativity and counity axioms of a
coalgebra\footnote{This proof can be found in [P3] (Lemma II.8 of [P3], to be
precise).}) and $\Pi^{\left[  k\right]  }\circ\Pi_{\left(  k\right)
}^{\left[  l\right]  }=\Pi^{\left[  lk\right]  }$ (this holds for any
commutative algebra, and is easy to see - but doesn't generally hold for
noncommutative algebras!), so that%
\begin{align*}
&  \underbrace{\Psi^{k}}_{=I^{\ast k}=\Pi^{\left[  k\right]  }\circ I^{\otimes
k}\circ\Delta^{\left[  k\right]  }=\Pi^{\left[  k\right]  }\circ
\Delta^{\left[  k\right]  }}\circ\underbrace{\Psi^{l}}_{=I^{\ast l}%
=\Pi^{\left[  l\right]  }\circ I^{\otimes l}\circ\Delta^{\left[  l\right]
}=\Pi^{\left[  l\right]  }\circ\Delta^{\left[  l\right]  }}\\
&  =\Pi^{\left[  k\right]  }\circ\underbrace{\Delta^{\left[  k\right]  }%
\circ\Pi^{\left[  l\right]  }}_{=\Pi_{\left(  k\right)  }^{\left[  l\right]
}\circ\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes l}}\circ
\Delta^{\left[  l\right]  }=\underbrace{\Pi^{\left[  k\right]  }\circ
\Pi_{\left(  k\right)  }^{\left[  l\right]  }}_{=\Pi^{\left[  lk\right]  }%
}\circ\underbrace{\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes
l}\circ\Delta^{\left[  l\right]  }}_{=\Delta^{\left[  lk\right]  }}%
=\Pi^{\left[  lk\right]  }\circ\Delta^{\left[  lk\right]  }\\
&  =\Pi^{\left[  lk\right]  }\circ I^{\otimes lk}\circ\Delta^{\left[
lk\right]  }=I^{\ast lk}=\Psi^{lk},
\end{align*}
and this proves the second part of Proposition 1.4.

\parbreak


\textbf{Page 1072:} A typo: ``Notons $\Phi^{k}$ the $n$-i\`{e}me
endomorphisme'' should be ``Notons $\Phi^{k}$ the $k$-i\`{e}me endomorphisme''.

\parbreak


\textbf{Page 1073, proof of Proposition 2.3:} There is nothing wrong to be
corrected here, but I don't find the proof of this proposition as obvious as
Patras does, so let me write down this proof here:

\begin{proof}
[Proof of Proposition 2.3.]We can prove that any commuting $x\in M$ and $y\in
M$ satisfy $\log_{k}x+\log_{k}y=\log_{k}\left(  xy\right)  $ (this follows
from the well-known fact that $\log\left(  1+X\right)  +\log\left(
1+Y\right)  =\log\left(  \left(  1+X\right)  \left(  1+Y\right)  \right)  $ in
the ring $K\left[  \left[  X,Y\right]  \right]  $ of formal power series,
using the fact that the $K$-representation $A$ is unipotent of rank $k$).
Using this fact, we can prove (by induction over $n$) that every $x\in M$ and
$n\in\mathbb{N}$ satisfy $n\log_{k}x=\log_{k}\left(  x^{n}\right)  $.

However, for every $x\in M$, we have%
\[
\left(  \rho\circ\Phi^{n}\right)  \left(  x\right)  =\rho\left(
\underbrace{\Phi^{n}\left(  x\right)  }_{=x^{n}}\right)  =\rho\left(
x^{n}\right)
\]
and%
\begin{align*}
\left(  \sum\limits_{i=0}^{k-1}n^{i}\cdot\varepsilon^{i}\right)  \left(
x\right)   &  =\sum\limits_{i=0}^{k-1}n^{i}\cdot\underbrace{\varepsilon
^{i}\left(  x\right)  }_{=\dfrac{\left(  \log_{k}x\right)  ^{i}}{i!}}%
=\sum\limits_{i=0}^{k-1}n^{i}\cdot\dfrac{\left(  \log_{k}x\right)  ^{i}}{i!}\\
&  =\sum\limits_{i=0}^{k-1}\dfrac{\left(  n\log_{k}x\right)  ^{i}}{i!}%
=\exp_{k}\left(  \underbrace{n\log_{k}x}_{=\log_{k}\left(  x^{n}\right)
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\exp_{k}\left(
n\log_{k}x\right)  \right) \\
&  =\exp_{k}\left(  \log_{k}\left(  x^{n}\right)  \right)
=\underbrace{\left(  \exp_{k}\circ\log_{k}\right)  }_{\substack{=\rho
\\\text{(by Lemma 2.1)}}}\left(  x^{n}\right)  =\rho\left(  x^{n}\right)  .
\end{align*}
Hence, for every $x\in M$, we have $\left(  \rho\circ\Phi^{n}\right)  \left(
x\right)  =\rho\left(  x\right)  =\left(  \sum\limits_{i=0}^{k-1}n^{i}%
\cdot\varepsilon^{i}\right)  \left(  x\right)  $. Thus, $\rho\circ\Phi
^{n}=\sum\limits_{i=0}^{k-1}n^{i}\cdot\varepsilon^{i}$. This proves
Proposition 2.3.
\end{proof}

\parbreak


\textbf{Page 1074, proof of Lemma 3.1:} Let me add that the same argument
which Patras used to prove Lemma 3.1 can be used to prove a more general statement:

\begin{quote}
\textbf{Lemma 3.11.} Let $\rho_{n}^{\prime}:\operatorname*{Hom}\nolimits_{K}%
\left(  H,H\right)  \rightarrow\operatorname*{Hom}\nolimits_{K}\left(
\bigoplus\limits_{i=0}^{n}H_{i},H\right)  $ be the map which takes every
linear map $g\in\operatorname*{Hom}\nolimits_{K}\left(  H,H\right)  $ to the
restriction of $g$ to $\bigoplus\limits_{i=0}^{n}H_{i}$.

For every map $f\in\operatorname*{Hom}\nolimits_{K}\left(  H,H\right)  $
satisfying $f\left(  1\right)  =1$, we have
\[
\left(  \rho_{n}^{\prime}\left(  f-1\right)  \right)  ^{\ast\left(
n+1\right)  }=0
\]
(where $1$ denotes the unity of the $K$-algebra $\mathcal{L}\left(  H\right)
$, i. e., the map $\eta\circ\epsilon$).
\end{quote}

\begin{proof}
[Proof of Lemma 3.11.]Copy the proof of Lemma 3.1, replacing every occurence
of $\Psi^{k}$ by $f$, and replacing every occurence of $\rho_{n}$ by $\rho
_{n}^{\prime}$. This gives a proof of Lemma 3.11.
\end{proof}

Note that we replaced $\rho_{n}$ by $\rho_{n}^{\prime}$ in the statement of
Lemma 3.11 because we didn't want to require $f$ to be graded. If
$f\in\operatorname*{Hom}\nolimits_{K}\left(  H,H\right)  $ is a graded map,
then $\rho_{n}\left(  f\right)  $ is ``more or less the same'' as $\rho
_{n}^{\prime}\left(  f\right)  $ (the only difference between the maps
$\rho_{n}\left(  f\right)  $ and $\rho_{n}^{\prime}\left(  f\right)  $ is that
the codomain of $\rho_{n}\left(  f\right)  $ is $\bigoplus\limits_{i=0}%
^{n}H_{i}$, whereas the codomain of $\rho_{n}^{\prime}\left(  f\right)  $ is
the whole $H$). However, if $f$ is not a graded map, $\rho_{n}\left(
f\right)  $ is either not defined or not identic with $\rho_{n}^{\prime
}\left(  f\right)  $ (depending on how $\rho_{n}$ is defined: see my remark
about ``Page 1070, two lines above Definition 1.2'' above).

Here is a very useful consequence of Lemma 3.11:

\begin{quote}
\textbf{Lemma 3.12.} Let $f\in\operatorname*{Hom}\nolimits_{K}\left(
H,H\right)  $ be a map satisfying $f\left(  1\right)  =1$. Then, for every
$x\in H$, the infinite sum $\sum\limits_{n=1}^{\infty}\left(  -1\right)
^{n+1}\dfrac{\left(  f-1\right)  ^{\ast n}}{n}\left(  x\right)  $ has only
finitely many nonzero terms.
\end{quote}

\begin{proof}
[Proof of Lemma 3.12.]Let $x\in H$. Since $x\in H=\bigoplus\limits_{i\in
\mathbb{N}}H_{i}=\bigcup\limits_{j\in\mathbb{N}}\left(  \bigoplus
\limits_{i=0}^{j}H_{i}\right)  $, there exists some $j\in\mathbb{N}$ such that
$x\in\bigoplus\limits_{i=0}^{j}H_{i}$. Consider this $j$.

Recall that $\rho_{j}^{\prime}:\operatorname*{Hom}\nolimits_{K}\left(
H,H\right)  \rightarrow\operatorname*{Hom}\nolimits_{K}\left(  \bigoplus
\limits_{i=0}^{j}H_{i},H\right)  $ is the map which takes every linear map
$g\in\operatorname*{Hom}\nolimits_{K}\left(  H,H\right)  $ to the restriction
of $g$ to $\bigoplus\limits_{i=0}^{j}H_{i}$. Hence, for every $n\in\mathbb{N}%
$, the map $\rho_{j}^{\prime}\left(  \left(  f-1\right)  ^{\ast n}\right)  $
is the restriction of the map $\left(  f-1\right)  ^{\ast n}$ to
$\bigoplus\limits_{i=0}^{j}H_{i}$. Since $x\in\bigoplus\limits_{i=0}^{j}H_{i}%
$, this yields that $\left(  \rho_{j}^{\prime}\left(  \left(  f-1\right)
^{\ast n}\right)  \right)  \left(  x\right)  =\left(  f-1\right)  ^{\ast
n}\left(  x\right)  $ for every $n\in\mathbb{N}$. But we also have $\rho
_{j}^{\prime}\left(  \left(  f-1\right)  ^{\ast n}\right)  =\left(  \rho
_{j}^{\prime}\left(  f-1\right)  \right)  ^{\ast n}$ for every $n\in
\mathbb{N}$ (since $\rho_{j}^{\prime}$ is a $K$-algebra homomorphism).

But Lemma 3.11 (applied to $j$ instead of $n$) yields $\left(  \rho
_{j}^{\prime}\left(  f-1\right)  \right)  ^{\ast\left(  j+1\right)  }=0$.
Hence, every integer $n\geq j+1$ satisfies%
\begin{align*}
\left(  \rho_{j}^{\prime}\left(  f-1\right)  \right)  ^{\ast n}  &  =\left(
\rho_{j}^{\prime}\left(  f-1\right)  \right)  ^{\ast\left(  \left(
j+1\right)  +\left(  n-\left(  j+1\right)  \right)  \right)  }\\
&  =\underbrace{\left(  \rho_{j}^{\prime}\left(  f-1\right)  \right)
^{\ast\left(  j+1\right)  }}_{=0}\ast\left(  \rho_{j}^{\prime}\left(
f-1\right)  \right)  ^{\ast\left(  n-\left(  j+1\right)  \right)  }=0.
\end{align*}
Thus, every integer $n\geq j+1$ satisfies
\begin{align*}
&  \left(  -1\right)  ^{n+1}\dfrac{\left(  f-1\right)  ^{\ast n}}{n}\left(
x\right) \\
&  =\left(  -1\right)  ^{n+1}\dfrac{\left(  f-1\right)  ^{\ast n}\left(
x\right)  }{n}=\left(  -1\right)  ^{n+1}\dfrac{\left(  \rho_{j}^{\prime
}\left(  \left(  f-1\right)  ^{\ast n}\right)  \right)  \left(  x\right)  }%
{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \rho_{j}^{\prime}\left(
\left(  f-1\right)  ^{\ast n}\right)  \right)  \left(  x\right)  =\left(
f-1\right)  ^{\ast n}\left(  x\right)  \right) \\
&  =\left(  -1\right)  ^{n+1}\dfrac{\left(  \rho_{j}^{\prime}\left(
f-1\right)  \right)  ^{\ast n}\left(  x\right)  }{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho_{j}^{\prime}\left(  \left(
f-1\right)  ^{\ast n}\right)  =\left(  \rho_{j}^{\prime}\left(  f-1\right)
\right)  ^{\ast n}\right) \\
&  =\left(  -1\right)  ^{n+1}\dfrac{0}{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  \rho_{j}^{\prime}\left(  f-1\right)  \right)  ^{\ast
n}=0\text{ (due to }n\geq j+1\text{)}\right) \\
&  =0.
\end{align*}
This proves Lemma 3.12.
\end{proof}

\parbreak


\textbf{Page 1074:} Two lines above Proposition 3.2, Patras writes:
``$\varepsilon_{n}^{i}$ est donc un morphisme de $E$ dans $\mathcal{L}\left(
H\right)  _{n}$''. It would be helpful to emphasize that ``morphisme'' means a
morphism of sets here, not a morphism of monoids (unless I am missing something!).

\parbreak


\textbf{Page 1074:} Four lines above Proposition 3.2, Patras writes: ``Nous
noterons dans la suite $\varepsilon_{n}^{i}$, $1\leq i\leq n$, [...]''. I
think that considering the $\varepsilon_{n}^{i}$ only for $1\leq i\leq n$ (but
not for $i=0$) is a bad decision, since it leads to several minor mistakes
afterwards. For example, the first identity on page 1077,%
\[
\rho_{n}\left(  \Psi^{\zeta}\right)  =\rho_{n}\left(  \sum_{i=1}^{n}\zeta
^{i}\cdot e^{i}\right)  ,
\]
is not completely correct, since the sum on the right hand is missing an $i=0$
term, but as long as $e^{0}$ is not defined, this does not make much sense.
For another example, Definition 3.7 does not uniquely define $\Psi^{\zeta}$,
because $H$ is not the direct sum of all $H_{n}^{\left(  i\right)  }$ unless
we allow $i$ to be $0$.

I think the simplest way to clean up this mess is to define the maps
$\varepsilon_{n}^{i}$ for all $0\leq i\leq n$ in the same as way as they are
defined for all $1\leq i\leq n$ in the text. This yields that $\varepsilon
_{n}^{0}\left(  x\right)  =\dfrac{\left(  \log_{1}x\right)  ^{\ast0}}%
{0!}=\dfrac{1}{1}=1$ (where $1$ denotes the unity of $\mathcal{L}\left(
H\right)  $; this is the map $\eta\circ\epsilon$ (not the map $I$)). Thus, in
particular, $\varepsilon_{n}^{0}\left(  I\right)  =1$, so that
\begin{align*}
e_{n}^{0}  &  =\varepsilon_{n}^{0}\left(  I\right)  \mid_{H_{n}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }e_{n}^{0}\right) \\
&  =1\mid_{H_{n}}=%
\begin{cases}
1, & \text{if }n=0;\\
0, & \text{if }n\neq0
\end{cases}
\ \ .
\end{align*}
Hence, for $n\neq0$, we have $e_{n}^{0}=0$. This is why we don't have care
about $e_{n}^{0}$ when $n\neq0$. However, for $n=0$, we have $e_{0}^{0}=1$.

Now we have the following (in my opinion, slightly better) version of
Proposition 3.2:

\begin{quote}
\textbf{Proposition 3.2'.} For every $n\in\mathbb{N}$ (including the case
$n=0$), we have $\Psi_{n}^{k}=\sum\limits_{i=0}^{n}k^{i}\cdot e_{n}^{i}$.
\end{quote}

\begin{proof}
[Proof of Proposition 3.2'.]Let $n\in\mathbb{N}$. Then, by the definition of
$\Psi_{n}^{k}$, we have%
\begin{align*}
\Psi_{n}^{k}  &  =\Psi^{k}\mid_{H_{n}}=\underbrace{\left(  \Psi^{k}%
\mid_{_{\bigoplus\limits_{i=0}^{n}H_{i}}}\right)  }_{=\rho_{n}\left(  \Psi
^{k}\right)  }\mid_{H_{n}}=\rho_{n}\left(  \underbrace{\Psi^{k}}_{=\Phi
^{k}\left(  I\right)  }\right)  \mid_{H_{n}}\\
&  =\rho_{n}\left(  \Phi^{k}\left(  I\right)  \right)  \mid_{H_{n}}=\left(
\rho_{n}\circ\Phi^{k}\right)  \left(  I\right)  \mid_{H_{n}}=\left(
\sum_{i=0}^{n}k^{i}\cdot\varepsilon_{n}^{i}\right)  \left(  I\right)
\mid_{H_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\rho_{n}\circ\Phi^{k}=\sum\limits_{i=0}^{n}k^{i}\cdot
\varepsilon_{n}^{i}\text{ by Proposition 2.3 (applied to}\\
E\text{, }\mathcal{L}\left(  H\right)  _{n}\text{, }\rho_{n}\text{,
}n+1\text{, }k\text{ instead of }M\text{, }A\text{, }\rho\text{, }k\text{ and
}n\text{)}%
\end{array}
\right) \\
&  =\sum_{i=0}^{n}k^{i}\cdot\underbrace{\varepsilon_{n}^{i}\left(  I\right)
\mid_{H_{n}}}_{\substack{=e_{n}^{i}\\\text{(because this is}\\\text{how }%
e_{n}^{i}\text{ was defined)}}}=\sum_{i=0}^{n}k^{i}\cdot e_{n}^{i}.
\end{align*}
This proves Proposition 3.2'.
\end{proof}

Proposition 3.2 is merely an obvious consequence of Proposition 3.2':

\begin{proof}
[Proof of Proposition 3.2.]For every $n>0$, we have%
\begin{align*}
\Psi_{n}^{k}  &  =\sum_{i=0}^{n}k^{i}\cdot e_{n}^{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition 3.2'}\right) \\
&  =k^{0}\cdot\underbrace{e_{n}^{0}}_{=0\text{ (since }n\neq0\text{)}}%
+\sum_{i=1}^{n}k^{i}\cdot e_{n}^{i}=\sum_{i=1}^{n}k^{i}\cdot e_{n}^{i}.
\end{align*}
This proves Proposition 3.2.
\end{proof}

\parbreak


\textbf{Page 1075, Definition 3.3:} It would not harm to add here that
$e_{n}^{i}$ is understood to be $0$ if $i>n$. (Otherwise, $e_{n}^{i}$ would
not be defined at all for $i>n$.)

\parbreak


\textbf{Page 1075, proof of Proposition 3.4:} Patras' proof of Proposition 3.4
confines itself to one sentence: ``Les deux identit\'{e}s r\'{e}sultent
respectivement de 1.4 et 3.2, et de la d\'{e}finition 2.2 des projecteurs de
poids $i$.''

I don't think this enough, however indirectly ``r\'{e}sultent'' is meant. It
is indeed easy to conclude $e^{i}\ast e^{j}=\dbinom{i+j}{i}\cdot e^{i+j}$ from
1.4; however, concluding $e^{i}\circ e^{j}=\delta_{j}^{i}\cdot e^{i}$ from 3.2
is rather difficult. Here is how I would prove Proposition 3.4:

First, a rather standard combinatorial identity which we won't prove:

\begin{quote}
\textbf{Theorem 0.1.} Let $N\in\mathbb{N}$. Then, the equalities%
\begin{equation}
\sum_{k=0}^{N}\left(  -1\right)  ^{k}\dbinom{N}{k}k^{\ell}%
=0\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\left\{  0,1,...,N-1\right\}
\label{0.1-1}%
\end{equation}
and%
\begin{equation}
\sum_{k=0}^{N}\left(  -1\right)  ^{k}\dbinom{N}{k}k^{N}=\left(  -1\right)
^{N}N! \label{0.1-2}%
\end{equation}
are satisfied in $\mathbb{Z}$.
\end{quote}

(This Theorem 0.1 is, for example, the result of applying Theorem 1 of [DG1]
to $R=\mathbb{Z}$.)

This has, as a consequence, a kind of ``polynomials that are zero at all
nonnegative integers must be identically zero'' result for torsionfree abelian groups:

\begin{quote}
\textbf{Theorem 0.2.} Let $R$ be a torsionfree abelian group. Let
$n\in\mathbb{N}$. Let $\left(  \alpha_{0},\alpha_{1},...,\alpha_{n}\right)  $
and $\left(  \beta_{0},\beta_{1},...,\beta_{n}\right)  $ be two $\left(
n+1\right)  $-tuples of elements of $R$ such that every $k\in\mathbb{N}$
satisfies $\sum\limits_{m=0}^{n}k^{m}\alpha_{m}=\sum\limits_{m=0}^{n}%
k^{m}\beta_{m}$. Then, $\alpha_{m}=\beta_{m}$ for every $m\in\left\{
0,1,...,n\right\}  $.
\end{quote}

\begin{proof}
[Proof of Theorem 0.2.]We are going to prove that for every $\ell\in\left\{
0,1,...,n\right\}  $, we have%
\begin{equation}
\alpha_{n-\ell}=\beta_{n-\ell}. \label{0.2.pf.1}%
\end{equation}


\textit{Proof of (\ref{0.2.pf.1}).} We will prove (\ref{0.2.pf.1}) by strong
induction over $\ell$. A strong induction does not need an induction base, so
let us start with the induction step:

\textit{Induction step:} Let $L\in\left\{  0,1,...,n\right\}  $ be arbitrary.
Assume that (\ref{0.2.pf.1}) is already proven for all $\ell\in\left\{
0,1,...,n\right\}  $ satisfying $\ell<L$. Now we must prove (\ref{0.2.pf.1})
for $\ell=L$.

We have
\begin{equation}
\alpha_{n-\ell}=\beta_{n-\ell}\text{ for every }\ell\in\left\{
0,1,...,n\right\}  \text{ satisfying }\ell<L \label{0.2.pf.2}%
\end{equation}
(since (\ref{0.2.pf.1}) is already proven for all $\ell\in\left\{
0,1,...,n\right\}  $ satisfying $\ell<L$).

Let $k\in\mathbb{N}$ be arbitrary. Then,%
\begin{align}
\sum\limits_{m=0}^{n}k^{m}\alpha_{m}  &  =\sum\limits_{m=0}^{n-L}k^{m}%
\alpha_{m}+\sum\limits_{m=n-L+1}^{n}k^{m}\alpha_{m}=\sum\limits_{m=0}%
^{n-L}k^{m}\alpha_{m}+\sum\limits_{\ell=0}^{L-1}k^{n-\ell}\alpha_{n-\ell
}\label{0.2.pf.3}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }n-\ell\text{ for
}m\text{ in the second sum}\right) \nonumber
\end{align}
and similarly%
\begin{equation}
\sum\limits_{m=0}^{n}k^{m}\beta_{m}=\sum\limits_{m=0}^{n-L}k^{m}\beta_{m}%
+\sum\limits_{\ell=0}^{L-1}k^{n-\ell}\beta_{n-\ell}. \label{0.2.pf.4}%
\end{equation}
Subtracting (\ref{0.2.pf.4}) from (\ref{0.2.pf.3}), we get%
\begin{align*}
\sum\limits_{m=0}^{n}k^{m}\alpha_{m}-\sum\limits_{m=0}^{n}k^{m}\beta_{m}  &
=\left(  \sum\limits_{m=0}^{n-L}k^{m}\alpha_{m}+\sum\limits_{\ell=0}%
^{L-1}k^{n-\ell}\underbrace{\alpha_{n-\ell}}_{\substack{=\beta_{n-\ell
}\\\text{(by (\ref{0.2.pf.2}),}\\\text{since }\ell<L\text{)}}}\right)
-\left(  \sum\limits_{m=0}^{n-L}k^{m}\beta_{m}+\sum\limits_{\ell=0}%
^{L-1}k^{n-\ell}\beta_{n-\ell}\right) \\
&  =\left(  \sum\limits_{m=0}^{n-L}k^{m}\alpha_{m}+\sum\limits_{\ell=0}%
^{L-1}k^{n-\ell}\beta_{n-\ell}\right)  -\left(  \sum\limits_{m=0}^{n-L}%
k^{m}\beta_{m}+\sum\limits_{\ell=0}^{L-1}k^{n-\ell}\beta_{n-\ell}\right) \\
&  =\sum\limits_{m=0}^{n-L}k^{m}\alpha_{m}-\sum\limits_{m=0}^{n-L}k^{m}%
\beta_{m}.
\end{align*}
Hence,%
\[
\sum\limits_{m=0}^{n-L}k^{m}\alpha_{m}-\sum\limits_{m=0}^{n-L}k^{m}\beta
_{m}=\underbrace{\sum\limits_{m=0}^{n}k^{m}\alpha_{m}}_{=\sum\limits_{m=0}%
^{n}k^{m}\beta_{m}}-\sum\limits_{m=0}^{n}k^{m}\beta_{m}=\sum\limits_{m=0}%
^{n}k^{m}\beta_{m}-\sum\limits_{m=0}^{n}k^{m}\beta_{m}=0.
\]
In other words,
\begin{equation}
\sum\limits_{m=0}^{n-L}k^{m}\alpha_{m}=\sum\limits_{m=0}^{n-L}k^{m}\beta_{m}.
\label{0.2.pf.5}%
\end{equation}


Now, forget that we fixed $k$. We thus have shown (\ref{0.2.pf.5}) for every
$k\in\mathbb{N}$.

Now, $L\in\left\{  0,1,...,n\right\}  $ yields $n\geq L$, so that $n-L\geq0$.
Denote the nonnegative integer $n-L$ by $N$. Then, every $k\in\mathbb{N}$
satisfies%
\[
\sum\limits_{m=0}^{N}k^{m}\alpha_{m}=\sum\limits_{m=0}^{N}k^{m}\beta_{m}%
\]
(this is just the identity (\ref{0.2.pf.5}), rewritten using $N=n-L$). Thus,%
\begin{equation}
\sum_{k=0}^{N}\left(  -1\right)  ^{k}\dbinom{N}{k}\sum\limits_{m=0}^{N}%
k^{m}\alpha_{m}=\sum_{k=0}^{N}\left(  -1\right)  ^{k}\dbinom{N}{k}%
\sum\limits_{m=0}^{N}k^{m}\beta_{m}. \label{0.2.pf.6}%
\end{equation}
However, we have%
\begin{align*}
&  \sum_{k=0}^{N}\left(  -1\right)  ^{k}\dbinom{N}{k}\sum\limits_{m=0}%
^{N}k^{m}\alpha_{m}\\
&  =\sum\limits_{m=0}^{N}\alpha_{m}\sum_{k=0}^{N}\left(  -1\right)
^{k}\dbinom{N}{k}k^{m}=\sum\limits_{\ell=0}^{N}\alpha_{\ell}\sum_{k=0}%
^{N}\left(  -1\right)  ^{k}\dbinom{N}{k}k^{\ell}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }m\text{ as
}\ell\text{ in the first sum}\right) \\
&  =\sum\limits_{\ell=0}^{N-1}\alpha_{\ell}\underbrace{\sum_{k=0}^{N}\left(
-1\right)  ^{k}\dbinom{N}{k}k^{\ell}}_{\substack{=0\\\text{(by (\ref{0.1-1}%
))}}}+\alpha_{N}\underbrace{\sum_{k=0}^{N}\left(  -1\right)  ^{k}\dbinom{N}%
{k}k^{N}}_{\substack{=\left(  -1\right)  ^{N}N!\\\text{(by (\ref{0.1-2}))}}}\\
&  =\underbrace{\sum\limits_{\ell=0}^{N-1}\alpha_{\ell}0}_{=0}+\alpha
_{N}\left(  -1\right)  ^{N}N!=\alpha_{N}\left(  -1\right)  ^{N}N!=\left(
-1\right)  ^{N}N!\alpha_{N}%
\end{align*}
and similarly%
\[
\sum_{k=0}^{N}\left(  -1\right)  ^{k}\dbinom{N}{k}\sum\limits_{m=0}^{N}%
k^{m}\beta_{m}=\left(  -1\right)  ^{N}N!\beta_{N}.
\]
Using these two equalities, we find%
\begin{align*}
\left(  -1\right)  ^{N}N!\alpha_{N}  &  =\sum_{k=0}^{N}\left(  -1\right)
^{k}\dbinom{N}{k}\sum\limits_{m=0}^{N}k^{m}\alpha_{m}=\sum_{k=0}^{N}\left(
-1\right)  ^{k}\dbinom{N}{k}\sum\limits_{m=0}^{N}k^{m}\beta_{m}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{0.2.pf.6})}\right) \\
&  =\left(  -1\right)  ^{N}N!\beta_{N},
\end{align*}
so that%
\[
0=\left(  -1\right)  ^{N}N!\alpha_{N}-\left(  -1\right)  ^{N}N!\beta
_{N}=\left(  -1\right)  ^{N}N!\left(  \alpha_{N}-\beta_{N}\right)  .
\]
Since $\left(  -1\right)  ^{N}N!$ is a nonzero integer, this yields
$0=\alpha_{N}-\beta_{N}$ (since $R$ is torsionfree), so that $\alpha_{N}%
=\beta_{N}$. Since $N=n-L$, this rewrites as $\alpha_{n-L}=\beta_{n-L}$. In
other words, (\ref{0.2.pf.1}) is proven for $\ell=L$. This completes the
induction step. Thus, the induction proof of (\ref{0.2.pf.1}) is complete.

Now, for every $m\in\left\{  0,1,...,n\right\}  $, we have%
\begin{align*}
\alpha_{m}  &  =\alpha_{n-\left(  n-m\right)  }=\beta_{n-\left(  n-m\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{0.2.pf.1}), applied to }%
\ell=n-m\right) \\
&  =\beta_{m}.
\end{align*}
This proves Theorem 0.2.
\end{proof}

\begin{proof}
[Proof of Proposition 3.4.]We prove Proposition 3.4 in several steps.

\textbf{a)} For every $n\in\mathbb{N}$, every $k\in\mathbb{N}$ and every
$l\in\mathbb{N}$ we have%
\begin{equation}
\left(  \sum\limits_{i=0}^{n}k^{i}\cdot e_{n}^{i}\right)  \circ\left(
\sum\limits_{j=0}^{n}l^{j}\cdot e_{n}^{j}\right)  =\sum\limits_{i=0}%
^{n}\left(  kl\right)  ^{i}\cdot e_{n}^{i}. \label{3.4.pf.1}%
\end{equation}


\textit{Proof of (\ref{3.4.pf.1}).} Let $n\in\mathbb{N}$, $k\in\mathbb{N}$ and
$l\in\mathbb{N}$. Then, $\Psi^{k}\circ\Psi^{l}=\Psi^{kl}$ by Proposition 1.4.
This yields $\Psi_{n}^{k}\circ\Psi_{n}^{l}=\Psi_{n}^{kl}$ (since $\Psi_{n}%
^{k}$, $\Psi_{n}^{l}$ and $\Psi_{n}^{kl}$ are just the restrictions of
$\Psi^{k}$, $\Psi^{l}$ and $\Psi^{kl}$ to $H_{n}$).

Proposition 3.2' yields $\Psi_{n}^{k}=\sum\limits_{i=0}^{n}k^{i}\cdot
e_{n}^{i}$. Proposition 3.2' (applied to $l$ instead of $k$) yields $\Psi
_{n}^{l}=\sum\limits_{i=0}^{n}l^{i}\cdot e_{n}^{i}=\sum\limits_{j=0}^{n}%
l^{j}\cdot e_{n}^{j}$ (here we renamed the index $i$ as $j$ in the sum).
Proposition 3.2' (applied to $kl$ instead of $k$) yields $\Psi_{n}^{kl}%
=\sum\limits_{i=0}^{n}\left(  kl\right)  ^{i}\cdot e_{n}^{i}$. Thus,%
\[
\underbrace{\left(  \sum\limits_{i=0}^{n}k^{i}\cdot e_{n}^{i}\right)  }%
_{=\Psi_{n}^{k}}\circ\underbrace{\left(  \sum\limits_{j=0}^{n}l^{j}\cdot
e_{n}^{j}\right)  }_{=\Psi_{n}^{l}}=\Psi_{n}^{k}\circ\Psi_{n}^{l}=\Psi
_{n}^{kl}=\sum\limits_{i=0}^{n}\left(  kl\right)  ^{i}\cdot e_{n}^{i},
\]
so that (\ref{3.4.pf.1}) is proven.

\textbf{b)} For every $n\in\mathbb{N}$, every $l\in\mathbb{N}$ and every
$i\in\left\{  0,1,...,n\right\}  $, we have%
\begin{equation}
e_{n}^{i}\circ\left(  \sum\limits_{j=0}^{n}l^{j}\cdot e_{n}^{j}\right)
=l^{i}\cdot e_{n}^{i}. \label{3.4.pf.2}%
\end{equation}


\textit{Proof of (\ref{3.4.pf.2}).} Let $n\in\mathbb{N}$ and $l\in\mathbb{N}$.
For every $k\in\mathbb{N}$, we have%
\begin{align*}
\sum\limits_{m=0}^{n}k^{m}\left(  e_{n}^{m}\circ\left(  \sum\limits_{j=0}%
^{n}l^{j}\cdot e_{n}^{j}\right)  \right)   &  =\left(  \sum\limits_{m=0}%
^{n}k^{m}\cdot e_{n}^{m}\right)  \circ\left(  \sum\limits_{j=0}^{n}l^{j}\cdot
e_{n}^{j}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }K\text{-linear maps
is }K\text{-bilinear}\right) \\
&  =\left(  \sum\limits_{i=0}^{n}k^{i}\cdot e_{n}^{i}\right)  \circ\left(
\sum\limits_{j=0}^{n}l^{j}\cdot e_{n}^{j}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }m\text{ as
}i\text{ in the first sum}\right) \\
&  =\sum\limits_{i=0}^{n}\left(  kl\right)  ^{i}\cdot e_{n}^{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{3.4.pf.1})}\right) \\
&  =\sum\limits_{m=0}^{n}\underbrace{\left(  kl\right)  ^{m}}_{=k^{m}l^{m}%
}\cdot e_{n}^{m}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }i\text{ as
}m\text{ in the sum}\right) \\
&  =\sum\limits_{m=0}^{n}k^{m}l^{m}\cdot e_{n}^{m}.
\end{align*}
Thus, Theorem 0.2 (applied to $R=\operatorname*{End}\nolimits_{K}\left(
H_{n}\right)  $, $\alpha_{m}=e_{n}^{m}\circ\left(  \sum\limits_{j=0}^{n}%
l^{j}\cdot e_{n}^{j}\right)  $ and $\beta_{m}=l^{m}\cdot e_{n}^{m}$) yields
that $e_{n}^{m}\circ\left(  \sum\limits_{j=0}^{n}l^{j}\cdot e_{n}^{j}\right)
=l^{m}\cdot e_{n}^{m}$ for every $m\in\left\{  0,1,...,n\right\}  $. Renaming
the index $m$ as $i$ in this result, we obtain: $e_{n}^{i}\circ\left(
\sum\limits_{j=0}^{n}l^{j}\cdot e_{n}^{j}\right)  =l^{i}\cdot e_{n}^{i}$ for
every $i\in\left\{  0,1,...,n\right\}  $. Thus, (\ref{3.4.pf.2}) is proven.

\textbf{c)} For every $n\in\mathbb{N}$, every $i\in\left\{  0,1,...,n\right\}
$ and every $j\in\left\{  0,1,...,n\right\}  $, we have%
\begin{equation}
e_{n}^{i}\circ e_{n}^{j}=\delta_{j}^{i}\cdot e_{n}^{i}. \label{3.4.pf.3}%
\end{equation}


\textit{Proof of (\ref{3.4.pf.3}).} Let $n\in\mathbb{N}$ and $i\in\left\{
0,1,...,n\right\}  $. For every $k\in\mathbb{N}$, we have%
\begin{align*}
\sum\limits_{m=0}^{n}k^{m}\delta_{m}^{i}  &  =\sum\limits_{m\in\left\{
0,1,...,n\right\}  }k^{m}\delta_{m}^{i}=\underbrace{\sum
\limits_{\substack{m\in\left\{  0,1,...,n\right\}  ;\\m=i}}k^{m}\delta_{m}%
^{i}}_{=k^{i}\delta_{i}^{i}}+\sum\limits_{\substack{m\in\left\{
0,1,...,n\right\}  ;\\m\neq i}}k^{m}\underbrace{\delta_{m}^{i}}%
_{\substack{=0\\\text{(since }m\neq i\text{)}}}\\
&  =k^{i}\underbrace{\delta_{i}^{i}}_{=1}+\underbrace{\sum
\limits_{\substack{m\in\left\{  0,1,...,n\right\}  ;\\m\neq i}}k^{m}0}%
_{=0}=k^{i}.
\end{align*}
Now, for every $k\in\mathbb{N}$, we have%
\begin{align*}
\sum\limits_{m=0}^{n}k^{m}e_{n}^{i}\circ e_{n}^{m}  &  =e_{n}^{i}\circ\left(
\sum\limits_{m=0}^{n}k^{m}\cdot e_{n}^{m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }K\text{-linear maps
is }K\text{-bilinear}\right) \\
&  =e_{n}^{i}\circ\left(  \sum\limits_{j=0}^{n}k^{j}\cdot e_{n}^{j}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }m\text{ as
}j\text{ in the sum}\right) \\
&  =\underbrace{k^{i}}_{=\sum\limits_{m=0}^{n}k^{m}\delta_{m}^{i}}\cdot
e_{n}^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{3.4.pf.2}), applied to
}k\text{ instead of }l\right) \\
&  =\left(  \sum\limits_{m=0}^{n}k^{m}\delta_{m}^{i}\right)  \cdot e_{n}%
^{i}=\sum\limits_{m=0}^{n}k^{m}\delta_{m}^{i}\cdot e_{n}^{i}.
\end{align*}
Thus, Theorem 0.2 (applied to $R=\operatorname*{End}\nolimits_{K}\left(
H_{n}\right)  $, $\alpha_{m}=e_{n}^{i}\circ e_{n}^{m}$ and $\beta_{m}%
=\delta_{m}^{i}\cdot e_{n}^{i}$) yields that $e_{n}^{i}\circ e_{n}^{m}%
=\delta_{m}^{i}\cdot e_{n}^{i}$ for every $m\in\left\{  0,1,...,n\right\}  $.
Renaming the index $m$ as $j$ in this result, we obtain: $e_{n}^{i}\circ
e_{n}^{j}=\delta_{j}^{i}\cdot e_{n}^{i}$ for every $j\in\left\{
0,1,...,n\right\}  $. Thus, (\ref{3.4.pf.3}) is proven.

\textbf{d)} We have%
\begin{equation}
e_{n}^{i}\circ e_{n}^{j}=\delta_{j}^{i}\cdot e_{n}^{i}%
\ \ \ \ \ \ \ \ \ \ \text{ for every }i\in\mathbb{N}\text{, }j\in
\mathbb{N}\text{ and }n\in\mathbb{N}. \label{3.4.pf.4}%
\end{equation}


\textit{Proof of (\ref{3.4.pf.4}).} Let $i\in\mathbb{N}$, $j\in\mathbb{N}$ and
$n\in\mathbb{N}$. We must have one of the following three cases:

\textit{Case 1:} We have $i\in\left\{  0,1,...,n\right\}  $ and $j\in\left\{
0,1,...,n\right\}  $.

\textit{Case 2:} We have $i\notin\left\{  0,1,...,n\right\}  $.

\textit{Case 3:} We have $i\in\left\{  0,1,...,n\right\}  $ and $j\notin%
\left\{  0,1,...,n\right\}  $.

In Case 1, we notice that (\ref{3.4.pf.4}) follows directly from
(\ref{3.4.pf.3}).

In Case 2, we find that $e_{n}^{i}$ is $0$, and thus (\ref{3.4.pf.4}) is
trivially true.

In Case 3, both $e_{n}^{j}$ and $\delta_{j}^{i}$ are $0$ (in fact, $\delta
_{j}^{i}=0$ because $i\neq j$), and thus (\ref{3.4.pf.4}) is trivially true.

Hence, we have seen that (\ref{3.4.pf.4}) holds in each of the three Cases 1,
2 and 3. This proves (\ref{3.4.pf.4}).

\textbf{e)} For every $i\in\mathbb{N}$ and $j\in\mathbb{N}$, we have
$e^{i}\circ e^{j}=\delta_{j}^{i}\cdot e^{i}$.

\textit{Proof.} Let $i\in\mathbb{N}$ and $j\in\mathbb{N}$. The maps $e^{i}$
and $e^{j}$ are defined as graded endomorphisms of $\mathcal{L}\left(
H\right)  $, whose $n$-th graded components are $e_{n}^{i}$ and $e_{n}^{j}$
(respectively) for each $n\in\mathbb{N}$. Hence, in order to prove that
$e^{i}\circ e^{j}=\delta_{j}^{i}\cdot e^{i}$, it is enough to show that
$e_{n}^{i}\circ e_{n}^{j}=\delta_{j}^{i}\cdot e_{n}^{i}$ for every
$n\in\mathbb{N}$. But this has already been shown in (\ref{3.4.pf.4}). Thus,
we are done proving that $e^{i}\circ e^{j}=\delta_{j}^{i}\cdot e^{i}$.

\textbf{f)} For every $i\in\mathbb{N}$, we have%
\begin{equation}
e^{i}=\dfrac{\left(  \log I\right)  ^{\ast i}}{i!}, \label{3.4.pf.6}%
\end{equation}
where $\log I$ is to be understood as the result of applying the formal power
series of the logarithm to $I$. (This result is well-defined, since for every
$x\in H$, only finitely many terms of the formal power series $\left(  \log
I\right)  \left(  x\right)  $ are nonzero.)

\textit{Proof.} Let $i\in\mathbb{N}$. Also, let $n\in\mathbb{N}$ be arbitrary.

By the definition of $\rho_{n}$, every $f\in E$ such that $f\left(  1\right)
=1$ satisfies%
\begin{align*}
&  \rho_{n}\left(  \log f\right) \\
&  =\rho_{n}\left(  \sum_{m\geq1}\left(  -1\right)  ^{m+1}\dfrac{\left(
f-1\right)  ^{m}}{m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\log f=\sum_{m\geq1}\left(
-1\right)  ^{m+1}\dfrac{\left(  f-1\right)  ^{m}}{m}\text{ by the definition
of the logarithm}\right) \\
&  =\sum_{m\geq1}\left(  -1\right)  ^{m+1}\dfrac{\left(  \rho_{n}\left(
f\right)  -1\right)  ^{m}}{m}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\rho_{n}\text{ is a }K\text{-algebra homomorphism, and can}\\
\text{easily be seen to commute with reasonable infinite series}%
\end{array}
\right) \\
&  =\underbrace{\sum_{\substack{m\geq1;\\m<n+1}}}_{=\sum\limits_{m=1}^{n}%
}\left(  -1\right)  ^{m+1}\dfrac{\left(  \rho_{n}\left(  f\right)  -1\right)
^{m}}{m}+\sum_{\substack{m\geq1;\\m\geq n+1}}\left(  -1\right)  ^{m+1}%
\underbrace{\dfrac{\left(  \rho_{n}\left(  f\right)  -1\right)  ^{m}}{m}%
}_{\substack{=0\\\text{(since Lemma 3.1 yields}\\\left(  \rho_{n}\left(
f\right)  -1\right)  ^{n+1}=0\text{, so that}\\\left(  \rho_{n}\left(
f\right)  -1\right)  ^{m}=0\text{ (since }m\geq n+1\text{))}}}\\
&  =\sum\limits_{m=1}^{n}\left(  -1\right)  ^{m+1}\dfrac{\left(  \rho
_{n}\left(  f\right)  -1\right)  ^{m}}{m}+\underbrace{\sum_{\substack{m\geq
1;\\m\geq n+1}}\left(  -1\right)  ^{m+1}0}_{=0}=\sum\limits_{m=1}^{n}\left(
-1\right)  ^{m+1}\dfrac{\left(  \rho_{n}\left(  f\right)  -1\right)  ^{m}}%
{m}\\
&  =\log_{n+1}f\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\log_{n+1}f=\sum\limits_{m=1}%
^{n}\left(  -1\right)  ^{m+1}\dfrac{\left(  \rho_{n}\left(  f\right)
-1\right)  ^{m}}{m}\text{ by the definition of }\log_{n+1}f\right)  .
\end{align*}
Applied to $f=I$, this yields $\rho_{n}\left(  \log I\right)  =\log_{n+1}I$.

However, every graded map $g:H\rightarrow H$ satisfies $g\mid_{H_{n}}=\left(
g\mid_{_{\bigoplus\limits_{i=0}^{n}H_{i}}}\right)  \mid_{H_{n}}$%
\ \ \ \ \footnote{In fact, if we restrict the map $g$ to $H_{n}$, we get the
same result as if we restrict the map $g$ to $\bigoplus\limits_{i=0}^{n}H_{i}$
first and then restrict this restriction to $H_{n}$.} and $\rho_{n}\left(
g\right)  =g\mid_{_{\bigoplus\limits_{i=0}^{n}H_{i}}}$ (because this is how
$\rho_{n}$ was defined). Hence, every graded map $g:H\rightarrow H$ satisfies
$g\mid_{H_{n}}=\underbrace{\left(  g\mid_{_{\bigoplus\limits_{i=0}^{n}H_{i}}%
}\right)  }_{=\rho_{n}\left(  g\right)  }\mid_{H_{n}}=\rho_{n}\left(
g\right)  \mid_{H_{n}}$. Applied to $g=\dfrac{\left(  \log I\right)  ^{\ast
i}}{i!}$, this yields
\begin{align*}
&  \dfrac{\left(  \log I\right)  ^{\ast i}}{i!}\mid_{H_{n}}\\
&  =\rho_{n}\left(  \dfrac{\left(  \log I\right)  ^{\ast i}}{i!}\right)
\mid_{H_{n}}=\dfrac{\left(  \rho_{n}\left(  \log I\right)  \right)  ^{\ast i}%
}{i!}\mid_{H_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho_{n}\text{ is a }%
K\text{-algebra homomorphism, so that }\rho_{n}\left(  \dfrac{\left(  \log
I\right)  ^{\ast i}}{i!}\right)  =\dfrac{\left(  \rho_{n}\left(  \log
I\right)  \right)  ^{\ast i}}{i!}\right) \\
&  =\dfrac{\left(  \log_{n+1}I\right)  ^{\ast i}}{i!}\mid_{H_{n}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho_{n}\left(  \log I\right)
=\log_{n+1}I\right)  .
\end{align*}
Combined with%
\begin{align*}
e^{i}\mid_{H_{n}}  &  =e_{n}^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }e^{i}\right) \\
&  =\underbrace{\varepsilon_{n}^{i}\left(  I\right)  }_{\substack{=\dfrac
{\left(  \log_{n+1}I\right)  ^{\ast i}}{i!}\\\text{(by the definition of
}\varepsilon_{n}^{i}\text{)}}}\mid_{H_{n}}=\dfrac{\left(  \log_{n+1}I\right)
^{\ast i}}{i!}\mid_{H_{n}},
\end{align*}
this yields $e^{i}\mid_{H_{n}}=\dfrac{\left(  \log_{n+1}I\right)  ^{\ast i}%
}{i!}\mid_{H_{n}}=\dfrac{\left(  \log I\right)  ^{\ast i}}{i!}\mid_{H_{n}}$.

Now forget that we fixed $n\in\mathbb{N}$. We thus have proven that every
$n\in\mathbb{N}$ satisfies $e^{i}\mid_{H_{n}}=\dfrac{\left(  \log I\right)
^{\ast i}}{i!}\mid_{H_{n}}$. Therefore, $e^{i}=\dfrac{\left(  \log I\right)
^{\ast i}}{i!}$. We have thus proven (\ref{3.4.pf.6}).

\textbf{g)} For every $i\in\mathbb{N}$ and $j\in\mathbb{N}$, we have
$e^{i}\ast e^{j}=\dbinom{i+j}{i}\cdot e^{i+j}$.

\textit{Proof.} Let $i\in\mathbb{N}$ and $j\in\mathbb{N}$. By (\ref{3.4.pf.6}%
), we have $e^{i}=\dfrac{\left(  \log I\right)  ^{\ast i}}{i!}$. By
(\ref{3.4.pf.6}) (applied to $j$ instead of $i$), we have $e^{j}%
=\dfrac{\left(  \log I\right)  ^{\ast j}}{j!}$. By (\ref{3.4.pf.6}) (applied
to $i+j$ instead of $i$), we have $e^{i+j}=\dfrac{\left(  \log I\right)
^{\ast\left(  i+j\right)  }}{\left(  i+j\right)  !}$. Now,%
\begin{align*}
\underbrace{e^{i}}_{=\dfrac{\left(  \log I\right)  ^{\ast i}}{i!}}%
\ast\underbrace{e^{j}}_{=\dfrac{\left(  \log I\right)  ^{\ast j}}{j!}}  &
=\dfrac{\left(  \log I\right)  ^{\ast i}}{i!}\ast\dfrac{\left(  \log I\right)
^{\ast j}}{j!}\\
&  =\underbrace{\dfrac{\left(  i+j\right)  !}{i!j!}}_{=\dbinom{i+j}{i}}%
\cdot\underbrace{\dfrac{\left(  \log I\right)  ^{\ast\left(  i+j\right)  }%
}{\left(  i+j\right)  !}}_{=e^{i+j}}=\dbinom{i+j}{i}\cdot e^{i+j},
\end{align*}
qed.

From the results of steps \textbf{e)} and \textbf{g)}, we conclude that
Proposition 3.4 holds.
\end{proof}

\parbreak


\textbf{Page 1076, Corollary 3.6:} Here is an alternative proof of Corollary
3.6 (without using eigenspaces).

Before we even begin proving this corollary, let us record some lemmata that
could just as well have been stated (and proven) in Section 1:

\textbf{a)} First, an elementary lemma:

\begin{quote}
\textbf{Lemma 1.5.} For any two $K$-coalgebras $C$ and $D$, any two
$K$-algebras $A$ and $B$, and any four $K$-linear maps $p:C\rightarrow A$,
$q:C\rightarrow A$, $r:D\rightarrow B$ and $s:D\rightarrow B$, we have%
\[
\left(  p\otimes r\right)  \ast\left(  q\otimes s\right)  =\left(  p\ast
q\right)  \otimes\left(  r\ast s\right)  .
\]
(Here, $\left(  p\otimes r\right)  \ast\left(  q\otimes s\right)  $ means the
convolution of the two $K$-linear maps $p\otimes r:C\otimes D\rightarrow
A\otimes B$ and $q\otimes s:C\otimes D\rightarrow A\otimes B$.)
\end{quote}

This lemma is easy to prove (particularly if you are using Sweedler's
notation, but even without it).

The following two lemmata are easy consequences of Lemma 1.5:

\begin{quote}
\textbf{Lemma 1.6.} For any two $K$-coalgebras $C$ and $D$, any two
$K$-algebras $A$ and $B$, and any two $K$-linear maps $f:C\rightarrow A$ and
$g:D\rightarrow B$, we have%
\[
\left(  f\otimes1_{\operatorname*{Hom}\nolimits_{K}\left(  D,B\right)
}\right)  \ast\left(  1_{\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
}\otimes g\right)  =f\otimes g=\left(  1_{\operatorname*{Hom}\nolimits_{K}%
\left(  C,A\right)  }\otimes g\right)  \ast\left(  f\otimes
1_{\operatorname*{Hom}\nolimits_{K}\left(  D,B\right)  }\right)  .
\]
\footnote{Here and in the following, for any $K$-algebra $U$, we denote by
$1_{U}$ the unity of $U$.} Here, both $\operatorname*{Hom}\nolimits_{K}\left(
D,B\right)  $ and $\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ are
made into $K$-algebras by the convolution.

\textbf{Lemma 1.7.} For any two $K$-coalgebras $C$ and $D$, any two
$K$-algebras $A$ and $B$, and any two $K$-linear maps $f:C\rightarrow A$ and
$g:C\rightarrow A$, we have%
\[
\left(  f\otimes1_{\operatorname*{Hom}\nolimits_{K}\left(  D,B\right)
}\right)  \ast\left(  g\otimes1_{\operatorname*{Hom}\nolimits_{K}\left(
D,B\right)  }\right)  =\left(  f\ast g\right)  \otimes1_{\operatorname*{Hom}%
\nolimits_{K}\left(  D,B\right)  }.
\]
Here, $\operatorname*{Hom}\nolimits_{K}\left(  D,B\right)  $ is made into a
$K$-algebra by the convolution.
\end{quote}

By repeated application of Lemma 1.7, we get:

\begin{quote}
\textbf{Lemma 1.8.} \textbf{(a)} For any two $K$-coalgebras $C$ and $D$, any
two $K$-algebras $A$ and $B$, any $n\in\mathbb{N}$ and any $K$-linear map
$f:C\rightarrow A$, we have%
\[
\left(  f\otimes1_{\operatorname*{Hom}\nolimits_{K}\left(  D,B\right)
}\right)  ^{\ast n}=f^{\ast n}\otimes1_{\operatorname*{Hom}\nolimits_{K}%
\left(  D,B\right)  }.
\]
Here, $\operatorname*{Hom}\nolimits_{K}\left(  D,B\right)  $ is made into a
$K$-algebra by the convolution.

\textbf{(b)} For any two $K$-coalgebras $C$ and $D$, any two $K$-algebras $A$
and $B$, any $n\in\mathbb{N}$ and any $K$-linear map $g:D\rightarrow B$, we
have%
\[
\left(  1_{\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  }\otimes
g\right)  ^{\ast n}=1_{\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
}\otimes g^{\ast n}.
\]
Here, $\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ is made into a
$K$-algebra by the convolution.
\end{quote}

\begin{proof}
[Proof of Lemma 1.8.]Lemma 1.8 \textbf{(a)} is proven by induction over $n$
(and use of Lemma 1.7 in the induction step). Lemma 1.8 \textbf{(b)} is
completely analogous to Lemma 1.8 \textbf{(a)} (the only difference is the
order of the tensorands), so the proof is analogous as well. The details of
these proofs are left to the reader.
\end{proof}

\textbf{b)} On the other hand, we recall that for every connected graded
bialgebra $A$ and any $K$-linear map $f:A\rightarrow A$ satisfying $f\left(
1\right)  =1$, the $K$-linear map $\log f:A\rightarrow A$ is well-defined.
Namely, this map $\log f$ is defined by applying the formal power series of
the logarithm to $f$; in other words, $\log f$ is defined as the infinite sum
$\sum\limits_{n=1}^{\infty}\left(  -1\right)  ^{n+1}\dfrac{\left(  f-1\right)
^{\ast n}}{n}$ (where $1$ denotes the unity of the $K$-algebra $\mathcal{L}%
\left(  A\right)  $, i. e., the map $\eta\circ\epsilon$%
).\ \ \ \ \footnote{This sum is infinite, but it still gives us a well-defined
map $A\rightarrow A$, because \textit{for every }$x\in A$, the infinite sum
$\sum\limits_{n=1}^{\infty}\left(  -1\right)  ^{n+1}\dfrac{\left(  f-1\right)
^{\ast n}}{n}\left(  x\right)  $ has only finitely many nonzero terms (by
Lemma 3.12, applied to $A$ instead of $H$) and thus has a well-defined value
in $A$.}

The following lemma is easy to check:

\begin{quote}
\textbf{Lemma 1.9.} For every connected graded bialgebra $A$ and any two
$K$-linear maps $f:A\rightarrow A$ and $g:A\rightarrow A$ satisfying $f\left(
1\right)  =g\left(  1\right)  =1$ and $f\ast g=g\ast f$, we have $\log\left(
f\ast g\right)  =\log f+\log g$.
\end{quote}

In fact, Lemma 1.9 follows from the identity $\log\left(  \left(  1+X\right)
\left(  1+Y\right)  \right)  =\log\left(  1+X\right)  +\log\left(  1+Y\right)
$ in the ring $K\left[  \left[  X,Y\right]  \right]  $ of formal power series.

\textbf{c)} Next, we have:

\begin{quote}
\textbf{Lemma 1.10.} For any two connected graded $K$-bialgebras $A$ and $B$
and any $K$-linear map $f:A\rightarrow A$ satisfying $f\left(  1\right)  =1$,
we have $\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)  \left(
1\right)  =1$ and
\[
\log\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)  =\left(  \log
f\right)  \otimes1_{\mathcal{L}\left(  B\right)  }.
\]



\end{quote}

\begin{proof}
[Proof of Lemma 1.10.]Checking $\left(  f\otimes1_{\mathcal{L}\left(
B\right)  }\right)  \left(  1\right)  =1$ is very easy and left to the reader.
Now let us prove that $\log\left(  f\otimes1_{\mathcal{L}\left(  B\right)
}\right)  =\left(  \log f\right)  \otimes1_{\mathcal{L}\left(  B\right)  }$:

Let $n\in\mathbb{N}$. Then, $\left(  \left(  f-1_{\mathcal{L}\left(  A\right)
}\right)  \otimes1_{\operatorname*{Hom}\nolimits_{K}\left(  B,B\right)
}\right)  ^{\ast n}=\left(  f-1_{\mathcal{L}\left(  A\right)  }\right)  ^{\ast
n}\otimes1_{\operatorname*{Hom}\nolimits_{K}\left(  B,B\right)  }$ (by Lemma
1.8 \textbf{(a)}, applied to $A$, $B$ and $f-1_{\mathcal{L}\left(  A\right)
}$ instead of $C$, $D$ and $f$). Since $1_{\operatorname*{Hom}\nolimits_{K}%
\left(  B,B\right)  }=1_{\mathcal{L}\left(  B\right)  }$, this rewrites as
$\left(  \left(  f-1_{\mathcal{L}\left(  A\right)  }\right)  \otimes
1_{\mathcal{L}\left(  B\right)  }\right)  ^{\ast n}=\left(  f-1_{\mathcal{L}%
\left(  A\right)  }\right)  ^{\ast n}\otimes1_{\mathcal{L}\left(  B\right)  }%
$. However,%
\begin{align*}
f\otimes1_{\mathcal{L}\left(  B\right)  }-\underbrace{1_{\mathcal{L}\left(
A\otimes B\right)  }}_{=1_{\mathcal{L}\left(  A\right)  }\otimes
1_{\mathcal{L}\left(  B\right)  }}  &  =f\otimes1_{\mathcal{L}\left(
B\right)  }-1_{\mathcal{L}\left(  A\right)  }\otimes1_{\mathcal{L}\left(
B\right)  }\\
&  =\left(  f-1_{\mathcal{L}\left(  A\right)  }\right)  \otimes1_{\mathcal{L}%
\left(  B\right)  }%
\end{align*}
and thus%
\begin{align*}
\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }-1_{\mathcal{L}\left(
A\otimes B\right)  }\right)  ^{\ast n}  &  =\left(  \left(  f-1_{\mathcal{L}%
\left(  A\right)  }\right)  \otimes1_{\mathcal{L}\left(  B\right)  }\right)
^{\ast n}\\
&  =\left(  f-1_{\mathcal{L}\left(  A\right)  }\right)  ^{\ast n}%
\otimes1_{\mathcal{L}\left(  B\right)  }.
\end{align*}
Hence,%
\begin{align}
\dfrac{\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }-1_{\mathcal{L}\left(
A\otimes B\right)  }\right)  ^{\ast n}}{n}  &  =\dfrac{\left(
f-1_{\mathcal{L}\left(  A\right)  }\right)  ^{\ast n}\otimes1_{\mathcal{L}%
\left(  B\right)  }}{n}\nonumber\\
&  =\dfrac{\left(  f-1_{\mathcal{L}\left(  A\right)  }\right)  ^{\ast n}}%
{n}\otimes1_{\mathcal{L}\left(  B\right)  }. \label{1.10.pf.1}%
\end{align}


Forget that we fixed $n$. We thus have proved (\ref{1.10.pf.1}) for each
$n\in\mathbb{N}$.

By the definition of $\log\left(  f\otimes1_{\mathcal{L}\left(  B\right)
}\right)  $, we have%
\begin{align}
\log\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)   &
=\sum\limits_{n=1}^{\infty}\left(  -1\right)  ^{n+1}\underbrace{\dfrac{\left(
f\otimes1_{\mathcal{L}\left(  B\right)  }-1_{\mathcal{L}\left(  A\otimes
B\right)  }\right)  ^{\ast n}}{n}}_{\substack{=\dfrac{\left(  f-1_{\mathcal{L}%
\left(  A\right)  }\right)  ^{\ast n}}{n}\otimes1_{\mathcal{L}\left(
B\right)  }\\\text{(by (\ref{1.10.pf.1}))}}}\nonumber\\
&  =\sum\limits_{n=1}^{\infty}\left(  -1\right)  ^{n+1}\dfrac{\left(
f-1_{\mathcal{L}\left(  A\right)  }\right)  ^{\ast n}}{n}\otimes
1_{\mathcal{L}\left(  B\right)  }. \label{1.10.pf.2}%
\end{align}
On the other hand, by the definition of $\log f$, we have $\log f=\sum
\limits_{n=1}^{\infty}\left(  -1\right)  ^{n+1}\dfrac{\left(  f-1_{\mathcal{L}%
\left(  A\right)  }\right)  ^{\ast n}}{n}$, so that%
\[
\left(  \log f\right)  \otimes1_{\mathcal{L}\left(  B\right)  }=\left(
\sum\limits_{n=1}^{\infty}\left(  -1\right)  ^{n+1}\dfrac{\left(
f-1_{\mathcal{L}\left(  A\right)  }\right)  ^{\ast n}}{n}\right)
\otimes1_{\mathcal{L}\left(  B\right)  }=\sum\limits_{n=1}^{\infty}\left(
-1\right)  ^{n+1}\dfrac{\left(  f-1_{\mathcal{L}\left(  A\right)  }\right)
^{\ast n}}{n}\otimes1_{\mathcal{L}\left(  B\right)  }%
\]
(here, we are using the fact that the tensor product commutes with convergent
infinite sums; this can be easily checked). Comparing this with
(\ref{1.10.pf.2}), we find%
\[
\log\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)  =\left(  \log
f\right)  \otimes1_{\mathcal{L}\left(  B\right)  }.
\]
This proves Lemma 1.10.
\end{proof}

We now have:

\begin{quote}
\textbf{Lemma 1.11.} For any two connected graded $K$-bialgebras $A$ and $B$,
any $K$-linear map $f:A\rightarrow A$ satisfying $f\left(  1\right)  =1$, and
any $K$-linear map $g:B\rightarrow B$ satisfying $g\left(  1\right)  =1$, we
have $\left(  f\otimes g\right)  \left(  1\right)  =1$ and
\[
\log\left(  f\otimes g\right)  =\left(  \log f\right)  \otimes1_{\mathcal{L}%
\left(  B\right)  }+1_{\mathcal{L}\left(  A\right)  }\otimes\left(  \log
g\right)  .
\]



\end{quote}

\begin{proof}
[Proof of Lemma 1.11.]Again, checking that $\left(  f\otimes g\right)  \left(
1\right)  =1$ is very easy. Let us now prove that $\log\left(  f\otimes
g\right)  =\left(  \log f\right)  \otimes1_{\mathcal{L}\left(  B\right)
}+1_{\mathcal{L}\left(  A\right)  }\otimes\left(  \log g\right)  $.

By Lemma 1.6 (applied to $C=A$ and $D=B$), we get%
\[
\left(  f\otimes1_{\operatorname*{Hom}\nolimits_{K}\left(  B,B\right)
}\right)  \ast\left(  1_{\operatorname*{Hom}\nolimits_{K}\left(  A,A\right)
}\otimes g\right)  =f\otimes g=\left(  1_{\operatorname*{Hom}\nolimits_{K}%
\left(  A,A\right)  }\otimes g\right)  \ast\left(  f\otimes
1_{\operatorname*{Hom}\nolimits_{K}\left(  B,B\right)  }\right)  .
\]
Since $1_{\operatorname*{Hom}\nolimits_{K}\left(  A,A\right)  }=1_{\mathcal{L}%
\left(  A\right)  }$ and $1_{\operatorname*{Hom}\nolimits_{K}\left(
B,B\right)  }=1_{\mathcal{L}\left(  B\right)  }$, this rewrites as%
\[
\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)  \ast\left(
1_{\mathcal{L}\left(  A\right)  }\otimes g\right)  =f\otimes g=\left(
1_{\mathcal{L}\left(  A\right)  }\otimes g\right)  \ast\left(  f\otimes
1_{\mathcal{L}\left(  B\right)  }\right)  .
\]


By Lemma 1.10, we have $\left(  f\otimes1_{\mathcal{L}\left(  B\right)
}\right)  \left(  1\right)  =1$ and $\log\left(  f\otimes1_{\mathcal{L}\left(
B\right)  }\right)  =\left(  \log f\right)  \otimes1_{\mathcal{L}\left(
B\right)  }$. Similarly, $\left(  1_{\mathcal{L}\left(  A\right)  }\otimes
g\right)  \left(  1\right)  =1$ and $\log\left(  1_{\mathcal{L}\left(
A\right)  }\otimes g\right)  =1_{\mathcal{L}\left(  A\right)  }\otimes\left(
\log g\right)  $.

Since $\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)  \ast\left(
1_{\mathcal{L}\left(  A\right)  }\otimes g\right)  =\left(  1_{\mathcal{L}%
\left(  A\right)  }\otimes g\right)  \ast\left(  f\otimes1_{\mathcal{L}\left(
B\right)  }\right)  $, $\left(  f\otimes1_{\mathcal{L}\left(  B\right)
}\right)  \left(  1\right)  =1$ and $\left(  1_{\mathcal{L}\left(  A\right)
}\otimes g\right)  \left(  1\right)  =1$, we can apply Lemma 1.9 to
$f\otimes1_{\mathcal{L}\left(  B\right)  }$, $1_{\mathcal{L}\left(  A\right)
}\otimes g$ and $A\otimes B$ instead of $f$, $g$ and $A$. We obtain
$\log\left(  \left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)
\ast\left(  1_{\mathcal{L}\left(  A\right)  }\otimes g\right)  \right)
=\log\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)  +\log\left(
1_{\mathcal{L}\left(  A\right)  }\otimes g\right)  $. Hence,%
\begin{align*}
\log\left(  \underbrace{f\otimes g}_{=\left(  f\otimes1_{\mathcal{L}\left(
B\right)  }\right)  \ast\left(  1_{\mathcal{L}\left(  A\right)  }\otimes
g\right)  }\right)   &  =\log\left(  \left(  f\otimes1_{\mathcal{L}\left(
B\right)  }\right)  \ast\left(  1_{\mathcal{L}\left(  A\right)  }\otimes
g\right)  \right) \\
&  =\underbrace{\log\left(  f\otimes1_{\mathcal{L}\left(  B\right)  }\right)
}_{=\left(  \log f\right)  \otimes1_{\mathcal{L}\left(  B\right)  }%
}+\underbrace{\log\left(  1_{\mathcal{L}\left(  A\right)  }\otimes g\right)
}_{=1_{\mathcal{L}\left(  A\right)  }\otimes\left(  \log g\right)  }\\
&  =\left(  \log f\right)  \otimes1_{\mathcal{L}\left(  B\right)
}+1_{\mathcal{L}\left(  A\right)  }\otimes\left(  \log g\right)  .
\end{align*}
This proves Lemma 1.11.
\end{proof}

\textbf{d)} Now, finally, let us come to the alternative proof of Corollary 3.6:

\begin{proof}
[Alternative proof of Corollary 3.6.]Assume that $H$ is a graded bialgebra.
(The case of $H$ being a Hopf algebra is similar, in that the same argument
works but we have to interpret $H\otimes H$ as a tensor product of graded
superalgebras rather than as a plain tensor product of graded algebras.)

For every $i\in\mathbb{N}$, let $e_{H\otimes H}^{i}$ denote the $i$-th
``projecteur de poids $i$'' of the graded bialgebra (or Hopf algebra)
$H\otimes H$. Note that $e^{i}$ still denotes the $i$-th ``projecteur de poids
$i$'' of the graded bialgebra (or Hopf algebra) $H$.

By (\ref{3.4.pf.6}) (applied to $i=1$), we have $e^{1}=\dfrac{\left(  \log
I\right)  ^{\ast1}}{1!}=\dfrac{\log I}{1}=\log I$.

Note also that every $u\in\mathbb{N}$ satisfies%
\begin{align*}
e^{u}  &  =\dfrac{\left(  \log I\right)  ^{\ast u}}{u!}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{3.4.pf.6}), applied to }i=u\right)
\\
&  =\dfrac{\left(  e^{1}\right)  ^{\ast u}}{u!}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\log I=e^{1}\right)  ,
\end{align*}
so that%
\begin{equation}
\left(  e^{1}\right)  ^{\ast u}=u!\cdot e^{u}. \label{3.6.pf.2}%
\end{equation}


Denote by $I_{H\otimes H}$ the identity map $H\otimes H\rightarrow H\otimes
H$. Then,%
\begin{align*}
\log\left(  \underbrace{I_{H\otimes H}}_{=I\otimes I}\right)   &  =\log\left(
I\otimes I\right)  =\underbrace{\left(  \log I\right)  }_{=e^{1}}%
\otimes1_{\mathcal{L}\left(  H\right)  }+1_{\mathcal{L}\left(  H\right)
}\otimes\underbrace{\left(  \log I\right)  }_{=e^{1}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma 1.11, applied to }A=H\text{,
}B=H\text{, }f=I\text{ and }g=I\right) \\
&  =e^{1}\otimes1_{\mathcal{L}\left(  H\right)  }+1_{\mathcal{L}\left(
H\right)  }\otimes e^{1}.
\end{align*}


Now, for every $\ell\in\mathbb{N}$, we have%
\begin{align}
e_{H\otimes H}^{\ell}  &  =\dfrac{\left(  \log\left(  I_{H\otimes H}\right)
\right)  ^{\ast\ell}}{\ell!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{3.4.pf.6}), applied to }\ell\text{ and }H\otimes H\text{ instead of
}i\text{ and }H\right) \nonumber\\
&  =\dfrac{1}{\ell!}\left(  \underbrace{\log\left(  I_{H\otimes H}\right)
}_{=e^{1}\otimes1_{\mathcal{L}\left(  H\right)  }+1_{\mathcal{L}\left(
H\right)  }\otimes e^{1}}\right)  ^{\ast\ell}=\dfrac{1}{\ell!}%
\underbrace{\left(  e^{1}\otimes1_{\mathcal{L}\left(  H\right)  }%
+1_{\mathcal{L}\left(  H\right)  }\otimes e^{1}\right)  ^{\ast\ell}%
}_{\substack{=\sum\limits_{k=0}^{\ell}\dbinom{\ell}{k}\left(  e^{1}%
\otimes1_{\mathcal{L}\left(  H\right)  }\right)  ^{\ast k}\ast\left(
1_{\mathcal{L}\left(  H\right)  }\otimes e^{1}\right)  ^{\ast\left(
\ell-k\right)  }\\\text{(by the binomial formula)}}}\nonumber\\
&  =\dfrac{1}{\ell!}\sum\limits_{k=0}^{\ell}\dbinom{\ell}{k}%
\underbrace{\left(  e^{1}\otimes1_{\mathcal{L}\left(  H\right)  }\right)
^{\ast k}}_{\substack{=\left(  e^{1}\right)  ^{\ast k}\otimes1_{\mathcal{L}%
\left(  H\right)  }\\\text{(by Lemma 1.8 \textbf{(a)},}\\\text{applied to
}C=H\text{, }D=H\text{,}\\A=H\text{, }B=H\text{, }f=e^{1}\text{ and
}n=k\text{)}}}\ast\underbrace{\left(  1_{\mathcal{L}\left(  H\right)  }\otimes
e^{1}\right)  ^{\ast\left(  \ell-k\right)  }}_{\substack{=1_{\mathcal{L}%
\left(  H\right)  }\otimes\left(  e^{1}\right)  ^{\ast\left(  \ell-k\right)
}\\\text{(by Lemma 1.8 \textbf{(a)},}\\\text{applied to }C=H\text{,
}D=H\text{,}\\A=H\text{, }B=H\text{, }g=e^{1}\text{ and }n=\ell-k\text{)}%
}}\nonumber\\
&  =\dfrac{1}{\ell!}\sum\limits_{k=0}^{\ell}\dbinom{\ell}{k}\left(  \left(
e^{1}\right)  ^{\ast k}\otimes1_{\mathcal{L}\left(  H\right)  }\right)
\ast\left(  1_{\mathcal{L}\left(  H\right)  }\otimes\left(  e^{1}\right)
^{\ast\left(  \ell-k\right)  }\right) \nonumber\\
&  =\dfrac{1}{\ell!}\sum\limits_{k=0}^{\ell}\dbinom{\ell}{k}%
\underbrace{\left(  \left(  e^{1}\right)  ^{\ast k}\otimes
1_{\operatorname*{Hom}\nolimits_{K}\left(  H,H\right)  }\right)  \ast\left(
1_{\operatorname*{Hom}\nolimits_{K}\left(  H,H\right)  }\otimes\left(
e^{1}\right)  ^{\ast\left(  \ell-k\right)  }\right)  }_{\substack{=\left(
e^{1}\right)  ^{\ast k}\otimes\left(  e^{1}\right)  ^{\ast\left(
\ell-k\right)  }\\\text{(because Lemma 1.6 (applied to }C=H\text{, }D=H\text{,
}A=H\text{, }B=H\text{, }f=\left(  e^{1}\right)  ^{\ast k}\text{ and
}g=\left(  e^{1}\right)  ^{\ast\left(  \ell-k\right)  }\text{)}\\\text{yields
}\left(  \left(  e^{1}\right)  ^{\ast k}\otimes1_{\operatorname*{Hom}%
\nolimits_{K}\left(  H,H\right)  }\right)  \ast\left(  1_{\operatorname*{Hom}%
\nolimits_{K}\left(  H,H\right)  }\otimes\left(  e^{1}\right)  ^{\ast\left(
\ell-k\right)  }\right)  \\=\left(  e^{1}\right)  ^{\ast k}\otimes\left(
e^{1}\right)  ^{\ast\left(  \ell-k\right)  }=\left(  1_{\operatorname*{Hom}%
\nolimits_{K}\left(  H,H\right)  }\otimes\left(  e^{1}\right)  ^{\ast\left(
\ell-k\right)  }\right)  \ast\left(  \left(  e^{1}\right)  ^{\ast k}%
\otimes1_{\operatorname*{Hom}\nolimits_{K}\left(  H,H\right)  }\right)
\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }1_{\mathcal{L}\left(  H\right)
}=1_{\operatorname*{Hom}\nolimits_{K}\left(  H,H\right)  }\right) \nonumber\\
&  =\dfrac{1}{\ell!}\sum\limits_{k=0}^{\ell}\underbrace{\dbinom{\ell}{k}%
}_{=\dfrac{\ell!}{k!\left(  \ell-k\right)  !}}\underbrace{\left(
e^{1}\right)  ^{\ast k}}_{\substack{=k!\cdot e^{k}\\\text{(by (\ref{3.6.pf.2}%
), applied to }u=k\text{)}}}\otimes\underbrace{\left(  e^{1}\right)
^{\ast\left(  \ell-k\right)  }}_{\substack{=\left(  \ell-k\right)  !\cdot
e^{\ell-k}\\\text{(by (\ref{3.6.pf.2}), applied to }u=\ell-k\text{)}%
}}\nonumber\\
&  =\dfrac{1}{\ell!}\sum\limits_{k=0}^{\ell}\dfrac{\ell!}{k!\left(
\ell-k\right)  !}\cdot\left(  k!\cdot e^{k}\right)  \otimes\left(  \left(
\ell-k\right)  !\cdot e^{\ell-k}\right)  =\sum\limits_{k=0}^{\ell}e^{k}\otimes
e^{\ell-k}. \label{3.6.pf.3}%
\end{align}


Now, let us consider two cases:

\textit{Case 1:} The bialgebra $H$ is commutative.

\textit{Case 2:} The bialgebra $H$ is cocommutative.

Note that this is not really a case distinction, because in each of the two
cases we have to prove a \textit{different} claim: In Case 1, we have to prove
that the decomposition of Theorem 3.5 endows $H$ with a structure of a
bigraded algebra, whereas in Case 2, we have to prove that the decomposition
of Theorem 3.5 endows $H$ with a structure of a bigraded coalgebra.

First let us consider Case 1. In this case, $H$ is commutative. Thus, the map
$\Pi:H\otimes H\rightarrow H$ is an algebra homomorphism. Since $\Pi$ also is
a coalgebra homomorphism (by the axioms of a bialgebra) and a graded map
(since $H$ is a graded bialgebra), this yields that $\Pi$ is a graded
bialgebra homomorphism. Thus, since the definition of $e^{i}$ was functorial
with respect to $H$, the diagram%
\[%
%TCIMACRO{\TeXButton{consequence of functoriality}{\xymatrixcolsep{5pc}
%\xymatrix{
%H \otimes H \ar[d]_{\Pi} \ar[r]^{e^u_{H\otimes H}} & H \otimes H \ar[d]^{\Pi}
%\\
%H \ar[r]_{e^u} & H
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
H \otimes H \ar[d]_{\Pi} \ar[r]^{e^u_{H\otimes H}} & H \otimes H \ar[d]^{\Pi}
\\
H \ar[r]_{e^u} & H
}%
%EndExpansion
\]
is commutative for every $u\in\mathbb{N}$. In other words,
\begin{equation}
\Pi\circ e_{H\otimes H}^{u}=e^{u}\circ\Pi\ \ \ \ \ \ \ \ \ \ \text{for every
}u\in\mathbb{N}. \label{3.6.pf.C1}%
\end{equation}


Now, let $i\in\mathbb{N}$ and $j\in\mathbb{N}$ be arbitrary. By
(\ref{3.6.pf.3}) (applied to $\ell=i+j$), we have $e_{H\otimes H}^{i+j}%
=\sum\limits_{k=0}^{i+j}e^{k}\otimes e^{i+j-k}$, so that%
\begin{align}
e_{H\otimes H}^{i+j}\circ\left(  e^{i}\otimes e^{j}\right)   &  =\left(
\sum\limits_{k=0}^{i+j}e^{k}\otimes e^{i+j-k}\right)  \circ\left(
e^{i}\otimes e^{j}\right)  =\sum\limits_{k=0}^{i+j}\underbrace{\left(
e^{k}\otimes e^{i+j-k}\right)  \circ\left(  e^{i}\otimes e^{j}\right)
}_{=\left(  e^{k}\circ e^{i}\right)  \otimes\left(  e^{i+j-k}\circ
e^{j}\right)  }\nonumber\\
&  =\underbrace{\sum\limits_{k=0}^{i+j}}_{=\sum\limits_{k\in\left\{
0,1,...,i+j\right\}  }}\underbrace{\left(  e^{k}\circ e^{i}\right)
}_{\substack{=\delta_{i}^{k}e^{k}\\\text{(by Proposition 3.4, applied
to}\\k\text{ and }i\text{ instead of }i\text{ and }j\text{)}}}\otimes
\underbrace{\left(  e^{i+j-k}\circ e^{j}\right)  }_{\substack{=\delta
_{j}^{i+j-k}e^{i+j-k}\\\text{(by Proposition 3.4, applied to}\\i+j-k\text{ and
}j\text{ instead of }i\text{ and }j\text{)}}}\nonumber\\
&  =\sum\limits_{k\in\left\{  0,1,...,i+j\right\}  }\delta_{i}^{k}e^{k}%
\otimes\delta_{j}^{i+j-k}e^{i+j-k}\nonumber\\
&  =\underbrace{\sum\limits_{\substack{k\in\left\{  0,1,...,i+j\right\}
;\\k=i}}\delta_{i}^{k}e^{k}\otimes\delta_{j}^{i+j-k}e^{i+j-k}}%
_{\substack{=\delta_{i}^{i}e^{i}\otimes\delta_{j}^{i+j-i}e^{i+j-i}%
\\\text{(since }i\in\left\{  0,1,...,i+j\right\}  \text{)}}}+\sum
\limits_{\substack{k\in\left\{  0,1,...,i+j\right\}  ;\\k\neq i}%
}\underbrace{\delta_{i}^{k}e^{k}\otimes\delta_{j}^{i+j-k}e^{i+j-k}%
}_{\substack{=0\\\text{(since }k\neq i\text{, so that }\delta_{i}%
^{k}=0\text{)}}}\nonumber\\
&  =\underbrace{\delta_{i}^{i}}_{=1}e^{i}\otimes\underbrace{\delta_{j}%
^{i+j-i}}_{=\delta_{j}^{j}=1}\underbrace{e^{i+j-i}}_{=e^{j}}+\underbrace{\sum
\limits_{\substack{k\in\left\{  0,1,...,i+j\right\}  ;\\k\neq i}}0}_{=0}%
=e^{i}\otimes e^{j}+0\nonumber\\
&  =e^{i}\otimes e^{j}. \label{3.6.pf.8}%
\end{align}


Now, fix $n\in\mathbb{N}$ and $m\in\mathbb{N}$. By the definitions of
$H_{n}^{\left(  i\right)  }$ and $H_{m}^{\left(  j\right)  }$, we have
$H_{n}^{\left(  i\right)  }=e^{i}\left(  H_{n}\right)  $ and $H_{m}^{\left(
j\right)  }=e^{j}\left(  H_{m}\right)  $. Thus,%
\begin{align*}
\underbrace{H_{n}^{\left(  i\right)  }}_{=e^{i}\left(  H_{n}\right)  }%
\otimes\underbrace{H_{m}^{\left(  j\right)  }}_{=e^{j}\left(  H_{m}\right)  }
&  =e^{i}\left(  H_{n}\right)  \otimes e^{j}\left(  H_{m}\right)
=\underbrace{\left(  e^{i}\otimes e^{j}\right)  }_{\substack{=e_{H\otimes
H}^{i+j}\circ\left(  e^{i}\otimes e^{j}\right)  \\\text{(by (\ref{3.6.pf.8}%
))}}}\left(  H_{n}\otimes H_{m}\right) \\
&  =\left(  e_{H\otimes H}^{i+j}\circ\left(  e^{i}\otimes e^{j}\right)
\right)  \left(  H_{n}\otimes H_{m}\right)  =e_{H\otimes H}^{i+j}\left(
\underbrace{\left(  e^{i}\otimes e^{j}\right)  \left(  H_{n}\otimes
H_{m}\right)  }_{=e^{i}\left(  H_{n}\right)  \otimes e^{j}\left(
H_{m}\right)  }\right) \\
&  =e_{H\otimes H}^{i+j}\left(  \underbrace{e^{i}\left(  H_{n}\right)
}_{\substack{\subseteq H_{n}\\\text{(since }e^{i}\text{ is a graded map)}%
}}\otimes\underbrace{e^{j}\left(  H_{m}\right)  }_{\substack{\subseteq
H_{m}\\\text{(since }e^{j}\text{ is a graded map)}}}\right) \\
&  \subseteq e_{H\otimes H}^{i+j}\left(  H_{n}\otimes H_{m}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \Pi\left(  H_{n}^{\left(  i\right)  }\otimes H_{m}^{\left(  j\right)
}\right) \\
&  \subseteq\Pi\left(  e_{H\otimes H}^{i+j}\left(  H_{n}\otimes H_{m}\right)
\right)  =\underbrace{\left(  \Pi\circ e_{H\otimes H}^{i+j}\right)
}_{\substack{=e^{i+j}\circ\Pi\\\text{(by (\ref{3.6.pf.C1}), applied to
}u=i+j\text{)}}}\left(  H_{n}\otimes H_{m}\right) \\
&  =\left(  e^{i+j}\circ\Pi\right)  \left(  H_{n}\otimes H_{m}\right)
=e^{i+j}\underbrace{\left(  \Pi\left(  H_{n}\otimes H_{m}\right)  \right)
}_{\substack{=H_{n}\cdot H_{m}\\\text{(since }\Pi\text{ is the multiplication
map)}}}\\
&  =e^{i+j}\underbrace{\left(  H_{n}\cdot H_{m}\right)  }_{\substack{\subseteq
H_{n+m}\\\text{(since }H\text{ is a graded algebra)}}}\\
&  \subseteq e^{i+j}\left(  H_{n+m}\right)  =H_{n+m}^{\left(  i+j\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }H_{n+m}^{\left(  i+j\right)  }\text{
was defined as }e^{i+j}\left(  H_{n+m}\right)  \right)  .
\end{align*}


Let us now forget that we fixed $i$, $j$, $n$ and $m$. We have thus proved
that $\Pi\left(  H_{n}^{\left(  i\right)  }\otimes H_{m}^{\left(  j\right)
}\right)  \subseteq H_{n+m}^{\left(  i+j\right)  }$ for every $i\in\mathbb{N}%
$, $j\in\mathbb{N}$, $n\in\mathbb{N}$ and $m\in\mathbb{N}$. In other words, we
have proved that the multiplication map $\Pi$ is bigraded. Due to this (and to
the trivial fact that $1_{H}\in H_{0}^{\left(  0\right)  }$), the
decomposition of Theorem 3.5 endows $H$ with a structure of a bigraded
algebra. Thus, we have proven our claim in Case 1.

Next, let us consider Case 2. In this case, $H$ is cocommutative. Thus, the
map $\Delta:H\rightarrow H\otimes H$ is a coalgebra homomorphism. Since
$\Delta$ also is an algebra homomorphism (by the axioms of a bialgebra) and a
graded map (since $H$ is a graded bialgebra), this yields that $\Delta$ is a
graded bialgebra homomorphism. Thus, since the definition of $e^{i}$ was
functorial with respect to $H$, the diagram%
\[%
%TCIMACRO{\TeXButton{consequence of functoriality}{\xymatrixcolsep{5pc}
%\xymatrix{
%H \ar[r]^{e^u} \ar[d]_{\Delta} & H \ar[d]^{\Delta} \\
%H \otimes H \ar[r]_{e^u_{H\otimes H}} & H \otimes H
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
H \ar[r]^{e^u} \ar[d]_{\Delta} & H \ar[d]^{\Delta} \\
H \otimes H \ar[r]_{e^u_{H\otimes H}} & H \otimes H
}%
%EndExpansion
\]
is commutative for every $u\in\mathbb{N}$. In other words,
\begin{equation}
\Delta\circ e^{u}=e_{H\otimes H}^{u}\circ\Delta\ \ \ \ \ \ \ \ \ \ \text{for
every }u\in\mathbb{N}. \label{3.6.pf.C2}%
\end{equation}


Now, fix $n\in\mathbb{N}$ and $i\in\mathbb{N}$. Since $H$ is a graded
coalgebra, we have $\Delta\left(  H_{n}\right)  \subseteq\bigoplus
\limits_{m\leq n}\left(  H_{m}\otimes H_{n-m}\right)  =\sum\limits_{m\leq
n}H_{m}\otimes H_{n-m}$ (since direct sums are sums).

Let us notice that the direct sum $\bigoplus\limits_{m\leq n,\ j\leq i}\left(
H_{m}^{\left(  j\right)  }\otimes H_{n-m}^{\left(  i-j\right)  }\right)  $ is
well-defined.\footnote{\textit{Proof.} We know that $H=\bigoplus
\limits_{\left(  m,j\right)  \in\mathbb{N}^{2}}H_{m}^{\left(  j\right)  }$, so
that $H\otimes H=\left(  \bigoplus\limits_{\left(  m,j\right)  \in
\mathbb{N}^{2}}H_{m}^{\left(  j\right)  }\right)  \otimes H=\bigoplus
\limits_{\left(  m,j\right)  \in\mathbb{N}^{2}}\left(  H_{m}^{\left(
j\right)  }\otimes H\right)  $ (because tensor products commute with direct
sums). Thus, the direct sum $\bigoplus\limits_{m\leq n,\ j\leq i}\left(
H_{m}^{\left(  j\right)  }\otimes H\right)  $ is well-defined (because it is a
partial sum of the well-defined direct sum $\bigoplus\limits_{\left(
m,j\right)  \in\mathbb{N}^{2}}\left(  H_{m}^{\left(  j\right)  }\otimes
H\right)  $). Hence,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{whenever }\left(  x_{m,j}\right)  _{m\leq n,\ j\leq i}\text{ is a family
of elements of }H\otimes H\text{ such}\\
\text{that }\sum\limits_{m\leq n,\ j\leq i}x_{m,j}=0\text{ and such that every
}m\leq n\text{ and }j\leq i\text{ satisfy}\\
x_{m,j}\in H_{m}^{\left(  j\right)  }\otimes H\text{, then }\left(
x_{m,j}\right)  _{m\leq n,\ j\leq i}=\left(  0\right)  _{m\leq n,\ j\leq i}%
\end{array}
\right)  . \label{3.6.pf.triv}%
\end{equation}
As a consequence of this,%
\[
\left(
\begin{array}
[c]{c}%
\text{whenever }\left(  x_{m,j}\right)  _{m\leq n,\ j\leq i}\text{ is a family
of elements of }H\otimes H\text{ such}\\
\text{that }\sum\limits_{m\leq n,\ j\leq i}x_{m,j}=0\text{ and such that every
}m\leq n\text{ and }j\leq i\text{ satisfy}\\
x_{m,j}\in H_{m}^{\left(  j\right)  }\otimes H_{n-m}^{\left(  i-j\right)
}\text{, then }\left(  x_{m,j}\right)  _{m\leq n,\ j\leq i}=\left(  0\right)
_{m\leq n,\ j\leq i}%
\end{array}
\right)
\]
(because if $\left(  x_{m,j}\right)  _{m\leq n,\ j\leq i}$ is a family of
elements of $H\otimes H$ such that $\sum\limits_{m\leq n,\ j\leq i}x_{m,j}=0$
and such that every $m\leq n$ and $j\leq i$ satisfy $x_{m,j}\in H_{m}^{\left(
j\right)  }\otimes H_{n-m}^{\left(  i-j\right)  }$, then every $m\leq n$ and
$j\leq i$ satisfy $x_{m,j}\in H_{m}^{\left(  j\right)  }\otimes
\underbrace{H_{n-m}^{\left(  i-j\right)  }}_{\subseteq H}\subseteq
H_{m}^{\left(  j\right)  }\otimes H$, and thus we can apply (\ref{3.6.pf.triv}%
) to obtain $\left(  x_{m,j}\right)  _{m\leq n,\ j\leq i}=\left(  0\right)
_{m\leq n,\ j\leq i}$). In other words, the direct sum $\bigoplus
\limits_{m\leq n,\ j\leq i}\left(  H_{m}^{\left(  j\right)  }\otimes
H_{n-m}^{\left(  i-j\right)  }\right)  $ is well-defined, qed.} Moreover, it
satisfies%
\begin{align*}
\bigoplus\limits_{m\leq n,\ j\leq i}\left(  H_{m}^{\left(  j\right)  }\otimes
H_{n-m}^{\left(  i-j\right)  }\right)   &  =\sum\limits_{m\leq n,\ j\leq
i}\left(  \underbrace{H_{m}^{\left(  j\right)  }}_{\substack{=e^{j}\left(
H_{m}\right)  \\\text{(by the definition of }H_{m}^{\left(  j\right)
}\text{)}}}\otimes\underbrace{H_{n-m}^{\left(  i-j\right)  }}%
_{\substack{=e^{i-j}\left(  H_{n-m}\right)  \\\text{(by the definition of
}H_{n-m}^{\left(  i-j\right)  }\text{)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are sums}\right) \\
&  =\sum\limits_{m\leq n,\ j\leq i}e^{j}\left(  H_{m}\right)  \otimes
e^{i-j}\left(  H_{n-m}\right)  .
\end{align*}
However, by the definition of $H_{n}^{\left(  i\right)  }$, we have
$H_{n}^{\left(  i\right)  }=e^{i}\left(  H_{n}\right)  $, so that%
\begin{align*}
\Delta\left(  H_{n}^{\left(  i\right)  }\right)   &  =\Delta\left(
e^{i}\left(  H_{n}\right)  \right)  =\underbrace{\left(  \Delta\circ
e^{i}\right)  }_{\substack{=e_{H\otimes H}^{i}\circ\Delta\\\text{(by
(\ref{3.6.pf.C2}), applied to }u=i\text{)}}}\left(  H_{n}\right)  =\left(
e_{H\otimes H}^{i}\circ\Delta\right)  \left(  H_{n}\right) \\
&  =e_{H\otimes H}^{i}\underbrace{\left(  \Delta\left(  H_{n}\right)  \right)
}_{\subseteq\sum\limits_{m\leq n}H_{m}\otimes H_{n-m}}\subseteq e_{H\otimes
H}^{i}\left(  \sum\limits_{m\leq n}H_{m}\otimes H_{n-m}\right) \\
&  =\left(  \sum\limits_{k=0}^{i}e^{k}\otimes e^{i-k}\right)  \left(
\sum\limits_{m\leq n}H_{m}\otimes H_{n-m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since (\ref{3.6.pf.3}) (applied to }%
\ell=i\text{) yields }e_{H\otimes H}^{i}=\sum\limits_{k=0}^{i}e^{k}\otimes
e^{i-k}\right) \\
&  \subseteq\underbrace{\sum\limits_{k=0}^{i}\sum\limits_{m\leq n}}%
_{=\sum\limits_{m\leq n,\ k\leq i}}\underbrace{\left(  e^{k}\otimes
e^{i-k}\right)  \left(  H_{m}\otimes H_{n-m}\right)  }_{=e^{k}\left(
H_{m}\right)  \otimes e^{i-k}\left(  H_{n-m}\right)  }=\sum\limits_{m\leq
n,\ k\leq i}e^{k}\left(  H_{m}\right)  \otimes e^{i-k}\left(  H_{n-m}\right)
\\
&  =\sum\limits_{m\leq n,\ j\leq i}e^{j}\left(  H_{m}\right)  \otimes
e^{i-j}\left(  H_{n-m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }k\text{ as
}j\text{ in the sum}\right)  .
\end{align*}
Hence,%
\[
\Delta\left(  H_{n}^{\left(  i\right)  }\right)  \subseteq\sum\limits_{m\leq
n,\ j\leq i}e^{j}\left(  H_{m}\right)  \otimes e^{i-j}\left(  H_{n-m}\right)
=\bigoplus\limits_{m\leq n,\ j\leq i}\left(  H_{m}^{\left(  j\right)  }\otimes
H_{n-m}^{\left(  i-j\right)  }\right)  .
\]


Now forget that we fixed $n$ and $i$. We have shown that every $n\in
\mathbb{N}$ and $i\in\mathbb{N}$ satisfy $\Delta\left(  H_{n}^{\left(
i\right)  }\right)  \subseteq\bigoplus\limits_{m\leq n,\ j\leq i}\left(
H_{m}^{\left(  j\right)  }\otimes H_{n-m}^{\left(  i-j\right)  }\right)  $. In
other words, the comultiplication map $\Delta$ is bigraded. Due to this (and
to the easily checked fact that the map $\epsilon$ is $0$ on $H_{n}^{\left(
i\right)  }$ for all $\left(  n,i\right)  \neq\left(  0,0\right)  $), the
decomposition of Theorem 3.5 endows $H$ with a structure of a bigraded
coalgebra. Thus, we have proven our claim in Case 2.

The proof of Corollary 3.6 is now complete (since the claim of Corollary 3.6
has been proven in each of the Cases 1 and 2).
\end{proof}

\parbreak


\textbf{Page 1076, Definition 3.7:} As explained above, this definition is
only correct if $i$ is allowed to be $0$.

\parbreak


\textbf{Page 1077, Proof of Proposition 3.8:} The formulas in this proof are
slightly wrong: Replace the $\sum\limits_{i=1}^{n}$ sign by $\sum
\limits_{i=0}^{n}$. Also, replace each of the $\sum\limits_{i,j=1}^{n}$ signs
by $\sum\limits_{i,j=0}^{n}$.

\parbreak


\textbf{Page 1077, Proposition 3.9:} Let me give some hints for the proof of
this proposition.

First, in order to prove that ``Les op\'{e}rations caract\'{e}ristiques, les
op\'{e}rations caract\'{e}ristiques g\'{e}n\'{e}ralis\'{e}es et les
projecteurs de poids $i$ sur $H$ et $H^{\star\operatorname*{gr}}$ sont alors
deux \`{a} deux adjoints'', the main step to make is to show that $f^{\star
}\ast g^{\star}=\left(  f\ast g\right)  ^{\star}$ for any two graded
$K$-linear maps $f:H\rightarrow H$ and $g:H\rightarrow H$. This is easy to
show (using the definition of the convolution: $f\ast g=\Pi\circ\left(
f\otimes g\right)  \circ\Delta$). Once this is shown, it yields that $\left(
f^{\ast i}\right)  ^{\star}=\left(  f^{\star}\right)  ^{\ast i}$ for every
graded $K$-linear map $f:H\rightarrow H$ and every $i\in\mathbb{N}$, and that
$\left(  \log f\right)  ^{\star}=\log\left(  f^{\star}\right)  $ whenever both
sides of this equation are well-defined, etc. - and ultimately the adjointness
part of Proposition 3.9.

Now to the proof that $\left(  H^{\left(  i\right)  }\right)  ^{\perp
}=\bigoplus\limits_{j\neq i}H^{\star\operatorname*{gr}\left(  j\right)  }$:

For every $n\in\mathbb{N}$, let $e^{n\star}$ denote the ``projecteur de poids
$i$ associ\'{e} \`{a} $H^{\star\operatorname*{gr}}$''. We know that
$e^{n\star}$ is adjoint to $e^{n}$ for every $n\in\mathbb{N}$.

For every $f\in H^{\star\operatorname*{gr}}$, we have the following
equivalence of assertions:%
\begin{align*}
\left(  f\in\bigoplus\limits_{j\neq i}H^{\star\operatorname*{gr}\left(
j\right)  }\right)  \  &  \Longleftrightarrow\ \left(  e^{i\star}\left(
f\right)  =0\right)  \ \Longleftrightarrow\ \left(  f\circ e^{i}=0\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }e^{i\star}\text{ is adjoint to
}e^{i}\text{, so that }e^{i\star}\left(  f\right)  =f\circ e^{i}\right) \\
&  \Longleftrightarrow\ \left(  f\left(  e^{i}\left(  H\right)  \right)
=0\right)  \ \Longleftrightarrow\ \left(  f\left(  H^{\left(  i\right)
}\right)  =0\right)  \ \Longleftrightarrow\ \left(  f\in\left(  H^{\left(
i\right)  }\right)  ^{\perp}\right)  .
\end{align*}
Hence, $\bigoplus\limits_{j\neq i}H^{\star\operatorname*{gr}\left(  j\right)
}=\left(  H^{\left(  i\right)  }\right)  ^{\perp}$, thus completing our proof
of Proposition 3.9.

\parbreak


\textbf{Section 3:} Let me add a lemma into this section which I am going to
use further below:

\begin{quote}
\textbf{Lemma 3.10.} Let $n\in\mathbb{N}$ and $i\in\left\{  0,1,...,n\right\}
$. There exists some $N\in\mathbb{N}$ and some elements $\alpha_{0}$,
$\alpha_{1}$, $...$, $\alpha_{N}$ of $K$ such that every graded bialgebra or
Hopf algebra $H$ satisfies%
\begin{equation}
e_{n}^{i}=\sum\limits_{k=0}^{N}\alpha_{k}\Psi_{n}^{k}. \label{3.10}%
\end{equation}
(Note that we are using the notations of Section 3 here, i. e., the $e_{n}%
^{i}$ and $\Psi_{n}^{k}$ in Lemma 3.10 are the endomorphisms of $H_{n}$
defined on page 1074.)
\end{quote}

\begin{proof}
[Proof of Lemma 3.10.]By the definition of $e_{n}^{i}$, we have%
\begin{align*}
e_{n}^{i}  &  =\underbrace{\varepsilon_{n}^{i}\left(  I\right)  }%
_{\substack{=\dfrac{\left(  \log_{n+1}\left(  I\right)  \right)  ^{\ast i}%
}{i!}\\\text{(by the definition of }\varepsilon_{n}^{i}\text{)}}}\mid_{H_{n}%
}=\dfrac{\left(  \log_{n+1}\left(  I\right)  \right)  ^{\ast i}}{i!}%
\mid_{H_{n}}\\
&  =\dfrac{1}{i!}\left(  \underbrace{\log_{n+1}\left(  I\right)
}_{\substack{=\sum\limits_{m=1}^{n}\left(  -1\right)  ^{m+1}\dfrac{\left(
\rho_{n}\left(  I\right)  -1\right)  ^{\ast m}}{m}\\\text{(by the definition
of }\log_{n+1}\text{)}}}\right)  ^{\ast i}\mid_{H_{n}}=\dfrac{1}{i!}\left(
\sum\limits_{m=1}^{n}\left(  -1\right)  ^{m+1}\dfrac{\left(  \rho_{n}\left(
I\right)  -1\right)  ^{\ast m}}{m}\right)  ^{\ast i}\mid_{H_{n}}\\
&  =\rho_{n}\left(  \dfrac{1}{i!}\left(  \sum\limits_{m=1}^{n}\left(
-1\right)  ^{m+1}\dfrac{\left(  I-1\right)  ^{\ast m}}{m}\right)  ^{\ast
i}\right)  \mid_{H_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because }\rho_{n}\text{ is a
}K\text{-algebra homomorphism}\right) \\
&  =\dfrac{1}{i!}\left(  \sum\limits_{m=1}^{n}\left(  -1\right)  ^{m+1}%
\dfrac{\left(  I-1\right)  ^{\ast m}}{m}\right)  ^{\ast i}\mid_{H_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because any }f\in\mathcal{L}\left(  H\right)  \text{ satisfies }\rho
_{n}\left(  f\right)  \mid_{H_{n}}=f\mid_{H_{n}}\\
\text{(since }\rho_{n}\left(  f\right)  \text{ is the restriction of }f\text{
to }\bigoplus\limits_{i=0}^{n}H_{i}\text{, and restricting this restriction}\\
\text{further to }H_{n}\text{ gives the same result as just restricting
}f\text{ itself to }H_{n}\text{)}%
\end{array}
\right)
\end{align*}
in every graded bialgebra or Hopf algebra $H$. But $\dfrac{1}{i!}\left(
\sum\limits_{m=1}^{n}\left(  -1\right)  ^{m+1}\dfrac{\left(  I-1\right)
^{\ast m}}{m}\right)  ^{\ast i}$ can be seen as a polynomial (whose
coefficients lie in $K$ and don't depend on $H$) applied to $I$. In other
words, there exists some $N\in\mathbb{N}$ and some appropriate elements
$\alpha_{0}$, $\alpha_{1}$, $...$, $\alpha_{N}$ of $K$ (which don't depend on
$H$) such that $\dfrac{1}{i!}\left(  \sum\limits_{m=1}^{n}\left(  -1\right)
^{m+1}\dfrac{\left(  I-1\right)  ^{\ast m}}{m}\right)  ^{\ast i}%
=\sum\limits_{k=0}^{N}\alpha_{k}I^{\ast k}$. Consider this $N$ and these
elements $\alpha_{0}$, $\alpha_{1}$, $...$, $\alpha_{N}$ of $K$. Then, in
every graded bialgebra or Hopf algebra $H$, we have%
\begin{align*}
e_{n}^{i}  &  =\underbrace{\dfrac{1}{i!}\left(  \sum\limits_{m=1}^{n}\left(
-1\right)  ^{m+1}\dfrac{\left(  I-1\right)  ^{\ast m}}{m}\right)  ^{\ast i}%
}_{=\sum\limits_{k=0}^{N}\alpha_{k}I^{\ast k}}\mid_{H_{n}}=\left(
\sum\limits_{k=0}^{N}\alpha_{k}I^{\ast k}\right)  \mid_{H_{n}}\\
&  =\sum\limits_{k=0}^{N}\alpha_{k}\left(  \underbrace{I^{\ast k}}_{=\Psi^{k}%
}\mid_{H_{n}}\right)  =\sum\limits_{k=0}^{N}\alpha_{k}\underbrace{\left(
\Psi^{k}\mid_{H_{n}}\right)  }_{=\Psi_{n}^{k}}=\sum\limits_{k=0}^{N}\alpha
_{k}\Psi_{n}^{k},
\end{align*}
thus proving Lemma 3.10.
\end{proof}

\parbreak


\textbf{Page 1078:} One line above Lemma 4.1, Patras writes:
``\'{e}l\'{e}ments de $H$''. This should be ``\'{e}l\'{e}ments $x$ de $H$''.

\parbreak


\textbf{Page 1078, proof of Lemma 4.1:} I wouldn't agree that ``L'inclusion
$\operatorname*{Prim}H\subset H^{\left(  1\right)  }$ est imm\'{e}diate.''

Here is how I would prove that $\operatorname*{Prim}H\subset H^{\left(
1\right)  }$:

\textbf{a)} First, a lemma:%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{For any }K\text{-algebra }A\text{ and any two }K\text{-linear maps
}f:H\rightarrow A\text{ and }g:H\rightarrow A\\
\text{satisfying }f\left(  1\right)  =g\left(  1\right)  =0\text{, we have
}\left(  f\ast g\right)  \left(  1\right)  =0\text{ and }\left(  f\ast
g\right)  \left(  \operatorname*{Prim}H\right)  =0
\end{array}
\right)  . \label{4.1.pf.1}%
\end{equation}


\begin{proof}
[Proof of (\ref{4.1.pf.1}).]Let $A$ be a $K$-algebra, and let $f:H\rightarrow
A$ and $g:H\rightarrow A$ be two $K$-linear maps satisfying $f\left(
1\right)  =g\left(  1\right)  =0$. Let $x\in\operatorname*{Prim}H$. Then, $x$
is primitive (by the definition of $\operatorname*{Prim}H$), so that
$\Delta\left(  x\right)  =x\otimes1+1\otimes x$. Now, by the definition of
convolution, $f\ast g=\Pi\circ\left(  f\otimes g\right)  \circ\Delta$, so that%
\begin{align*}
\left(  f\ast g\right)  \left(  x\right)   &  =\left(  \Pi\circ\left(
f\otimes g\right)  \circ\Delta\right)  \left(  x\right)  =\Pi\left(  \left(
f\otimes g\right)  \underbrace{\left(  \Delta\left(  x\right)  \right)
}_{=x\otimes1+1\otimes x}\right) \\
&  =\Pi\left(  \underbrace{\left(  f\otimes g\right)  \left(  x\otimes
1+1\otimes x\right)  }_{=f\left(  x\right)  \otimes g\left(  1\right)
+f\left(  1\right)  \otimes g\left(  x\right)  }\right) \\
&  =\Pi\left(  f\left(  x\right)  \otimes\underbrace{g\left(  1\right)  }%
_{=0}+\underbrace{f\left(  1\right)  }_{=0}\otimes g\left(  x\right)  \right)
=\Pi\left(  \underbrace{f\left(  x\right)  \otimes0+0\otimes g\left(
x\right)  }_{=0}\right)  =0.
\end{align*}


Now forget that we fixed $x$. We thus have shown that $\left(  f\ast g\right)
\left(  x\right)  =0$ for every $x\in\operatorname*{Prim}H$. Thus, $\left(
f\ast g\right)  \left(  \operatorname*{Prim}H\right)  =0$. Besides,%
\begin{align*}
\underbrace{\left(  f\ast g\right)  }_{=\Pi\circ\left(  f\otimes g\right)
\circ\Delta}\left(  1\right)   &  =\left(  \Pi\circ\left(  f\otimes g\right)
\circ\Delta\right)  \left(  1\right)  =\Pi\left(  \left(  f\otimes g\right)
\underbrace{\left(  \Delta\left(  1\right)  \right)  }_{=1\otimes1}\right)
=\Pi\left(  \underbrace{\left(  f\otimes g\right)  \left(  1\otimes1\right)
}_{=f\left(  1\right)  \otimes g\left(  1\right)  }\right) \\
&  =\Pi\left(  \underbrace{f\left(  1\right)  }_{=0}\otimes
\underbrace{g\left(  1\right)  }_{=0}\right)  =\Pi\left(  0\otimes0\right)
=0.
\end{align*}
Thus, (\ref{4.1.pf.1}) is proven.
\end{proof}

\textbf{b)} As a consequence of the previous lemma, we can prove the next
lemma:%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{For any }K\text{-algebra }A\text{, any integer }i\geq2\text{ and any
}K\text{-linear map }f:H\rightarrow A\\
\text{satisfying }f\left(  1\right)  =0\text{, we have }f^{\ast i}\left(
1\right)  =0\text{ and }f^{\ast i}\left(  \operatorname*{Prim}H\right)  =0
\end{array}
\right)  . \label{4.1.pf.2}%
\end{equation}


\begin{proof}
[Proof of (\ref{4.1.pf.2}).]We will prove (\ref{4.1.pf.2}) by induction over
$i$ (where the induction base is the case $i=2$):

\textit{Induction base:} For any $K$-algebra $A$ and any $K$-linear map
$f:H\rightarrow A$ satisfying $f\left(  1\right)  =0$, we have%
\[
\underbrace{f^{\ast2}}_{=f\ast f}\left(  1\right)  =\left(  f\ast f\right)
\left(  1\right)  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{4.1.pf.1}),
applied to }g=f\right)
\]
and%
\[
\underbrace{f^{\ast2}}_{=f\ast f}\left(  \operatorname*{Prim}H\right)
=\left(  f\ast f\right)  \left(  \operatorname*{Prim}H\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{4.1.pf.1}), applied to
}g=f\right)  .
\]
In other words, (\ref{4.1.pf.2}) holds for $i=2$. This completes the induction base.

\textit{Induction step:} Let $n$ be an integer $\geq2$. Assume that
(\ref{4.1.pf.2}) holds for $i=n$. We must prove that (\ref{4.1.pf.2}) also
holds for $i=n+1$.

Let $A$ be a $K$-algebra, and $f:H\rightarrow A$ be a $K$-linear map
satisfying $f\left(  1\right)  =0$. We assumed that (\ref{4.1.pf.2}) holds for
$i=n$. Hence, by (\ref{4.1.pf.2}) (applied to $i=n$), we get $f^{\ast
n}\left(  1\right)  =0$ and $f^{\ast n}\left(  \operatorname*{Prim}H\right)
=0$. Since $f\left(  1\right)  =f^{\ast n}\left(  1\right)  =0$, we can apply
(\ref{4.1.pf.1}) to $g=f^{\ast n}$, and conclude that $\left(  f\ast f^{\ast
n}\right)  \left(  1\right)  =0$ and $\left(  f\ast f^{\ast n}\right)  \left(
\operatorname*{Prim}H\right)  =0$. Since $f\ast f^{\ast n}=f^{\ast\left(
n+1\right)  }$, this rewrites as follows: We have $f^{\ast\left(  n+1\right)
}\left(  1\right)  =0$ and $f^{\ast\left(  n+1\right)  }\left(
\operatorname*{Prim}H\right)  =0$.

Forget that we fixed $A$ and $f$. We thus have showed that for any $K$-algebra
$A$ and any $K$-linear map $f:H\rightarrow A$ satisfying $f\left(  1\right)
=0$, we have $f^{\ast\left(  n+1\right)  }\left(  1\right)  =0$ and
$f^{\ast\left(  n+1\right)  }\left(  \operatorname*{Prim}H\right)  =0$. In
other words, we have proven that (\ref{4.1.pf.2}) holds for $i=n+1$. This
completes the induction step. Thus, the induction proof of (\ref{4.1.pf.2}) is complete.
\end{proof}

\textbf{c)} Now, we will prove:%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{For any }K\text{-algebra }A\text{ and any }K\text{-linear map
}f:H\rightarrow A\text{ satisfying}\\
f\left(  1\right)  =1\text{, we have }\left(  \log f\right)  \mid
_{\operatorname*{Prim}H}=f\mid_{\operatorname*{Prim}H}.
\end{array}
\right)  \label{4.1.pf.3}%
\end{equation}
Note that, in this assertion, $\log f$ is defined by applying the formal power
series of the logarithm to $f$; in other words, $\log f$ is defined as the
infinite sum $\sum\limits_{n=1}^{\infty}\left(  -1\right)  ^{n+1}%
\dfrac{\left(  f-1\right)  ^{\ast n}}{n}$ (where $1$ denotes the unity of the
$K$-algebra $\operatorname*{Hom}\nolimits_{K}\left(  H,A\right)
$\ \ \ \ \footnote{This $K$-algebra $\operatorname*{Hom}\nolimits_{K}\left(
H,A\right)  $ is the $K$-vector space of all $K$-linear maps $H\rightarrow A$,
with convolution as multiplication.}, i. e., the map $\eta\circ\epsilon$).
This sum is infinite, but it still gives us a well-defined map $H\rightarrow
A$, because \textit{for every }$x\in H$, the infinite sum $\sum\limits_{n=1}%
^{\infty}\left(  -1\right)  ^{n+1}\dfrac{\left(  f-1\right)  ^{\ast n}}%
{n}\left(  x\right)  $ has only finitely many nonzero terms\footnote{This
follows from Lemma 3.12 if $A=H$, and is proven exactly in the same way in the
general case.} and thus has a well-defined value in $A$.

\begin{proof}
[Proof of (\ref{4.1.pf.3}).]Let $A$ be a $K$-algebra, and let $f:H\rightarrow
A$ be a $K$-linear map satisfying $f\left(  1\right)  =1$. Then, $\left(
f-1\right)  \left(  1\right)  =\underbrace{f\left(  1\right)  }_{=1}%
-\underbrace{1\left(  1\right)  }_{=1}=1-1=0$. Hence, for any integer $i\geq
2$, we have
\begin{equation}
\left(  f-1\right)  ^{\ast i}\left(  1\right)  =0\text{ and }\left(
f-1\right)  ^{\ast i}\left(  \operatorname*{Prim}H\right)  =0 \label{4.1.pf.4}%
\end{equation}
(by (\ref{4.1.pf.2}), applied to $f-1$ instead of $f$).

Now let $x\in\operatorname*{Prim}H$. Then,%
\begin{equation}
\text{for any integer }i\geq2\text{, we have }\left(  f-1\right)  ^{\ast
i}\left(  x\right)  =0 \label{4.1.pf.5}%
\end{equation}
(since for any integer $i\geq2$, we have $\left(  f-1\right)  ^{\ast i}\left(
\operatorname*{Prim}H\right)  =0$ (by (\ref{4.1.pf.4})), but $\left(
f-1\right)  ^{\ast i}\left(  x\right)  \in\left(  f-1\right)  ^{\ast i}\left(
\operatorname*{Prim}H\right)  $ (since $x\in\operatorname*{Prim}H$), so that
$\left(  f-1\right)  ^{\ast i}\left(  x\right)  \in\left(  f-1\right)  ^{\ast
i}\left(  \operatorname*{Prim}H\right)  =0$ and thus $\left(  f-1\right)
^{\ast i}\left(  x\right)  =0$).

On the other hand, $x\in\operatorname*{Prim}H$, so that $x$ is primitive, and
thus $\epsilon\left(  x\right)  =0$ (indeed, there is a well-known fact that
every primitive element $\xi$ of any bialgebra satisfies $\epsilon\left(
\xi\right)  =0$).

Recall that (by the definition of $\log f$) we have%
\begin{align*}
\log f  &  =\sum\limits_{n=1}^{\infty}\left(  -1\right)  ^{n+1}\dfrac{\left(
f-1\right)  ^{\ast n}}{n}=\sum\limits_{i=1}^{\infty}\left(  -1\right)
^{i+1}\dfrac{\left(  f-1\right)  ^{\ast i}}{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }n\text{ as
}i\text{ in the sum}\right)  ,
\end{align*}
so that%
\begin{align*}
\left(  \log f\right)  \left(  x\right)   &  =\left(  \sum\limits_{i=1}%
^{\infty}\left(  -1\right)  ^{i+1}\dfrac{\left(  f-1\right)  ^{\ast i}}%
{i}\right)  \left(  x\right)  =\sum\limits_{i=1}^{\infty}\left(  -1\right)
^{i+1}\dfrac{\left(  f-1\right)  ^{\ast i}\left(  x\right)  }{i}\\
&  =\underbrace{\left(  -1\right)  ^{1+1}}_{=1}\underbrace{\dfrac{\left(
f-1\right)  ^{\ast1}\left(  x\right)  }{1}}_{=\left(  f-1\right)  ^{\ast
1}\left(  x\right)  =\left(  f-1\right)  \left(  x\right)  }+\sum
\limits_{i=2}^{\infty}\left(  -1\right)  ^{i+1}\underbrace{\dfrac{\left(
f-1\right)  ^{\ast i}\left(  x\right)  }{i}}_{\substack{=0\\\text{(since
}\left(  f-1\right)  ^{\ast i}\left(  x\right)  =0\\\text{(by (\ref{4.1.pf.5}%
), since }i\geq2\text{))}}}\\
&  =\left(  f-1\right)  \left(  x\right)  +\underbrace{\sum\limits_{i=2}%
^{\infty}\left(  -1\right)  ^{i+1}0}_{=0}=\left(  f-1\right)  \left(
x\right)  =f\left(  x\right)  -\underbrace{1}_{=\eta\circ\epsilon}\left(
x\right) \\
&  =f\left(  x\right)  -\underbrace{\left(  \eta\circ\epsilon\right)  \left(
x\right)  }_{=\eta\left(  \epsilon\left(  x\right)  \right)  }=f\left(
x\right)  -\eta\left(  \underbrace{\epsilon\left(  x\right)  }_{=0}\right)
=f\left(  x\right)  -\underbrace{\eta\left(  0\right)  }_{=0}=f\left(
x\right)  .
\end{align*}


Now forget that we fixed $x$. We thus have shown that every $x\in
\operatorname*{Prim}H$ satisfies $\left(  \log f\right)  \left(  x\right)
=f\left(  x\right)  $. In other words, $\left(  \log f\right)  \mid
_{\operatorname*{Prim}H}=f\mid_{\operatorname*{Prim}H}$. This proves
(\ref{4.1.pf.3}).
\end{proof}

\textbf{d)} Now to the proof of $\operatorname*{Prim}H\subset H^{\left(
1\right)  }$:

\begin{proof}
[Proof of $\operatorname*{Prim}H\subset H^{\left(  1\right)  }$.]First, we
notice that every $i\in\mathbb{N}$ satisfies $e^{i}\left(  H\right)
=H^{\left(  i\right)  }$. This is because, for every $i\in\mathbb{N}$, we have%
\begin{align*}
e^{i}\left(  H\right)   &  =\bigoplus\limits_{n\in\mathbb{N}}e_{n}^{i}\left(
H_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because the }n\text{-th graded component
of }e^{i}\text{ is }e_{n}^{i}\text{ (by Definition 3.3)}\right) \\
&  =\bigoplus\limits_{n\geq i}\underbrace{e_{n}^{i}\left(  H_{n}\right)
}_{=H_{n}^{\left(  i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we removed the addends with }n<i\text{ from our direct sum;}\\
\text{this did not change the sum (since }e_{n}^{i}=0\text{ when }n<i\text{)}%
\end{array}
\right) \\
&  =\bigoplus\limits_{n\geq i}H_{n}^{\left(  i\right)  }=H^{\left(  i\right)
}.
\end{align*}
Applied to $i=1$, this yields $e^{1}\left(  H\right)  =H^{\left(  1\right)  }$.

On the other hand, by (\ref{3.4.pf.6}) (applied to $i=1$), we have
$e^{1}=\dfrac{\left(  \log I\right)  ^{\ast1}}{1!}=\dfrac{\log I}{1}=\log I$.
Thus, $e^{1}\mid_{\operatorname*{Prim}H}=\left(  \log I\right)  \mid
_{\operatorname*{Prim}H}=I\mid_{\operatorname*{Prim}H}$ (by (\ref{4.1.pf.3}),
applied to $f=I$ and $A=H$). Hence, for every $x\in\operatorname*{Prim}H$, we
have $e^{1}\left(  x\right)  =I\left(  x\right)  $. Thus, for every
$x\in\operatorname*{Prim}H$, we have $x=I\left(  x\right)  =e^{1}\left(
x\right)  \in e^{1}\left(  H\right)  =H^{\left(  1\right)  }$. In other words,
$\operatorname*{Prim}H\subset H^{\left(  1\right)  }$, qed.
\end{proof}

\parbreak


\textbf{Page 1079, line 7:} Typo in this line: ``sructure'' should be ``structure''.

\parbreak


\textbf{Page 1079, the equation after line 7:} This equation is%
\[
\forall x\in A_{n},\ \forall y\in A_{m},\ \left[  x,y\right]  =x\otimes
y-\left(  -1\right)  ^{n\cdot m}y\otimes x.
\]
This should instead be%
\[
\forall x\in A_{n},\ \forall y\in A_{m},\ \left[  x,y\right]  =xy-\left(
-1\right)  ^{n\cdot m}yx.
\]


\parbreak


\textbf{Page 1079, Proposition 4.3:} A proof of Proposition 4.3 is sketched in
[C] (more precisely, in the proof of Theorem 3.8.3 in [C]).

\parbreak


\textbf{Page 1080, Section 5:} Here, Patras claims that%
\begin{equation}
\Delta\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  =\sum
\limits_{p+q=k}\sum\limits_{\beta}\operatorname*{sgn}\left(  \beta\right)
\left(  x_{i_{\beta\left(  1\right)  }}\otimes...\otimes x_{i_{\beta\left(
p\right)  }}\right)  \otimes\left(  x_{i_{\beta\left(  p+1\right)  }}%
\otimes...\otimes x_{i_{\beta\left(  p+q\right)  }}\right)  , \label{shuf.1}%
\end{equation}
where the $\sum\limits_{\beta}$ sum is over all ``op\'{e}rateurs
d'interclassement d'indices $p$ et $q$''\ \ \ \ \footnote{I think that the
proper English translation of the notion ``op\'{e}rateur d'interclassement
d'indices $p$ et $q$'' is ``$\left(  p,q\right)  $-shuffle''; but I am not
sure about this. Anyway there seem to exist at least three non-equivalent
definitions of ``$\left(  p,q\right)  $-shuffle'' in literature, so one should
be careful when using this notion.}. He defines an ``op\'{e}rateur
d'interclassement d'indices $p$ et $q$'' as an element $\sigma$ of $S_{p+q}$
satisfying%
\begin{equation}
\sigma^{-1}\left(  1\right)  <...<\sigma^{-1}\left(  p\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \sigma^{-1}\left(
p+1\right)  <...<\sigma^{-1}\left(  p+q\right)  . \label{shuf.2}%
\end{equation}


This is wrong. There are two ways to correct this mistake: Either replace the
formula (\ref{shuf.1}) by%
\begin{equation}
\Delta\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  =\sum
\limits_{p+q=k}\sum\limits_{\beta}\operatorname*{sgn}\left(  \beta\right)
\left(  x_{i_{\beta^{-1}\left(  1\right)  }}\otimes...\otimes x_{i_{\beta
^{-1}\left(  p\right)  }}\right)  \otimes\left(  x_{i_{\beta^{-1}\left(
p+1\right)  }}\otimes...\otimes x_{i_{\beta^{-1}\left(  p+q\right)  }}\right)
, \label{shuf.3}%
\end{equation}
where the $\sum\limits_{\beta}$ sum is still over all ``op\'{e}rateurs
d'interclassement d'indices $p$ et $q$''. Or replace the formula
(\ref{shuf.2}) (in the definition of an ``op\'{e}rateur d'interclassement
d'indices $p$ et $q$'') by
\[
\sigma\left(  1\right)  <...<\sigma\left(  p\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \sigma\left(  p+1\right)
<...<\sigma\left(  p+q\right)  .
\]


In the following, I am going to assume that the mistake has been corrected in
the first way (i. e., that the formula (\ref{shuf.1}) has been replaced by
(\ref{shuf.3})).

\parbreak


\textbf{Page 1080, Section 5:} Just a remark on $T\left(  X\right)  $. Patras
gave the formula (\ref{shuf.1}) for the comultiplication of $Th\left(
X\right)  $, and we corrected it to (\ref{shuf.3}). There is an analogous
formula for the comultiplication of $T\left(  X\right)  $, namely:%
\begin{equation}
\Delta\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  =\sum
\limits_{p+q=k}\sum\limits_{\beta}\left(  x_{i_{\beta^{-1}\left(  1\right)  }%
}\otimes...\otimes x_{i_{\beta^{-1}\left(  p\right)  }}\right)  \otimes\left(
x_{i_{\beta^{-1}\left(  p+1\right)  }}\otimes...\otimes x_{i_{\beta
^{-1}\left(  p+q\right)  }}\right)  . \label{shuf.TV}%
\end{equation}
Here, the $\sum\limits_{\beta}$ sum is still over all ``op\'{e}rateurs
d'interclassement d'indices $p$ et $q$''.

\parbreak


\textbf{Page 1081, Proof of Proposition 5.1:} Here, Patras writes:

``En explicitant les formules pour le produit et le coproduit dans $T\left(
X\right)  $ (resp. $Th\left(  X\right)  $) en termes d'op\'{e}rateurs
d'interclassement, on v\'{e}rifie facilement que :%
\[
\Psi_{n}^{k}\in K\left[  S_{n}\right]  \hookrightarrow\operatorname*{End}%
\left(  T_{n}\left(  X\right)  \right)
\]
(resp. :%
\[
\Psi_{n}^{k}\in K\left[  S_{n}\right]  \hookrightarrow\operatorname*{End}%
\left(  Th_{n}\left(  X\right)  \right)  \text{ ).''}%
\]


Let me detail this argument:

First, let us work in $T\left(  X\right)  $. By repeated application of
(\ref{shuf.TV}), we see that any $\ell\in\mathbb{N}$, any $k\in\mathbb{N}$ and
any $\left(  i_{1},i_{2},...,i_{\ell}\right)  \in\left\{  1,2,...,n\right\}
^{\ell}$ satisfy%
\begin{align}
&  \Delta^{\left[  k\right]  }\left(  x_{i_{1}}\otimes...\otimes x_{i_{\ell}%
}\right) \nonumber\\
&  =\sum\limits_{p_{1}+p_{2}+...+p_{k}=\ell}\sum\limits_{\substack{\sigma\in
S_{\ell};\\\sigma\left(  1\right)  <\sigma\left(  2\right)  <...<\sigma\left(
p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)  <\sigma\left(  p_{1}+2\right)
<...<\sigma\left(  p_{1}+p_{2}\right)  ;\\\sigma\left(  p_{1}+p_{2}+1\right)
<\sigma\left(  p_{1}+p_{2}+2\right)  <...<\sigma\left(  p_{1}+p_{2}%
+p_{3}\right)  ;\\...;\\\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+1\right)
<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+2\right)  <...<\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{i_{\sigma\left(  1\right)  }}\otimes
...\otimes x_{i_{\sigma\left(  p_{1}\right)  }}\right)  \otimes\left(
x_{i_{\sigma\left(  p_{1}+1\right)  }}\otimes...\otimes x_{i_{\sigma\left(
p_{1}+p_{2}\right)  }}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \otimes\left(  x_{i_{\sigma\left(  p_{1}+p_{2}%
+1\right)  }}\otimes...\otimes x_{i_{\sigma\left(  p_{1}+p_{2}+p_{3}\right)
}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \otimes...\otimes\left(  x_{i_{\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+1\right)  }}\otimes...\otimes x_{i_{\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\right)  . \label{symm.1}%
\end{align}
Hence, any $k\in\mathbb{N}$ and any $\left(  i_{1},i_{2},...,i_{n}\right)
\in\left\{  1,2,...,n\right\}  ^{n}$ satisfy%
\begin{align*}
&  \Psi_{n}^{k}\left(  x_{i_{1}}\otimes...\otimes x_{i_{n}}\right) \\
&  =\underbrace{\Psi^{k}}_{=I^{\ast k}=\Pi^{\left[  k\right]  }\circ
\Delta^{\left[  k\right]  }}\left(  x_{i_{1}}\otimes...\otimes x_{i_{n}%
}\right)  =\left(  \Pi^{\left[  k\right]  }\circ\Delta^{\left[  k\right]
}\right)  \left(  x_{i_{1}}\otimes...\otimes x_{i_{n}}\right) \\
&  =\Pi^{\left[  k\right]  }\left(  \Delta^{\left[  k\right]  }\left(
x_{i_{1}}\otimes...\otimes x_{i_{n}}\right)  \right) \\
&  =\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum\limits_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  <\sigma\left(  2\right)  <...<\sigma\left(
p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)  <\sigma\left(  p_{1}+2\right)
<...<\sigma\left(  p_{1}+p_{2}\right)  ;\\\sigma\left(  p_{1}+p_{2}+1\right)
<\sigma\left(  p_{1}+p_{2}+2\right)  <...<\sigma\left(  p_{1}+p_{2}%
+p_{3}\right)  ;\\...;\\\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+1\right)
<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+2\right)  <...<\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{i_{\sigma\left(  1\right)  }}\otimes
...\otimes x_{i_{\sigma\left(  p_{1}\right)  }}\right)  \cdot\left(
x_{i_{\sigma\left(  p_{1}+1\right)  }}\otimes...\otimes x_{i_{\sigma\left(
p_{1}+p_{2}\right)  }}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  x_{i_{\sigma\left(  p_{1}+p_{2}+1\right)
}}\otimes...\otimes x_{i_{\sigma\left(  p_{1}+p_{2}+p_{3}\right)  }}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot...\cdot\left(  x_{i_{\sigma\left(  p_{1}%
+p_{2}+...+p_{k-1}+1\right)  }}\otimes...\otimes x_{i_{\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{symm.1}) and since }\Pi\text{ is
the multiplication map}\right) \\
&  =\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum\limits_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  <\sigma\left(  2\right)  <...<\sigma\left(
p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)  <\sigma\left(  p_{1}+2\right)
<...<\sigma\left(  p_{1}+p_{2}\right)  ;\\\sigma\left(  p_{1}+p_{2}+1\right)
<\sigma\left(  p_{1}+p_{2}+2\right)  <...<\sigma\left(  p_{1}+p_{2}%
+p_{3}\right)  ;\\...;\\\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+1\right)
<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+2\right)  <...<\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \underbrace{x_{i_{\sigma\left(  1\right)  }}\otimes
x_{i_{\sigma\left(  2\right)  }}\otimes...\otimes x_{i_{\sigma\left(
n\right)  }}}_{=\sigma^{-1}\left(  x_{i_{1}}\otimes...\otimes x_{i_{n}%
}\right)  }\\
&  =\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum\limits_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  <\sigma\left(  2\right)  <...<\sigma\left(
p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)  <\sigma\left(  p_{1}+2\right)
<...<\sigma\left(  p_{1}+p_{2}\right)  ;\\\sigma\left(  p_{1}+p_{2}+1\right)
<\sigma\left(  p_{1}+p_{2}+2\right)  <...<\sigma\left(  p_{1}+p_{2}%
+p_{3}\right)  ;\\...;\\\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+1\right)
<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+2\right)  <...<\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \sigma^{-1}\left(  x_{i_{1}}\otimes...\otimes x_{i_{n}%
}\right)  .
\end{align*}
Thus, for every $k\in\mathbb{N}$, the map $\Psi_{n}^{k}\in\operatorname*{End}%
\left(  T_{n}\left(  X\right)  \right)  $ is the image of the element%
\[
\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum\limits_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  <\sigma\left(  2\right)  <...<\sigma\left(
p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)  <\sigma\left(  p_{1}+2\right)
<...<\sigma\left(  p_{1}+p_{2}\right)  ;\\\sigma\left(  p_{1}+p_{2}+1\right)
<\sigma\left(  p_{1}+p_{2}+2\right)  <...<\sigma\left(  p_{1}+p_{2}%
+p_{3}\right)  ;\\...;\\\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+1\right)
<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+2\right)  <...<\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\sigma^{-1}\in K\left[  S_{n}\right]
\]
under the map $K\left[  S_{n}\right]  \hookrightarrow\operatorname*{End}%
\left(  T_{n}\left(  X\right)  \right)  $. Similarly, we can show the
analogous result for $Th_{n}\left(  X\right)  $ instead of $T_{n}\left(
X\right)  $: Namely, for every $k\in\mathbb{N}$, the map $\Psi_{n}^{k}%
\in\operatorname*{End}\left(  Th_{n}\left(  X\right)  \right)  $ is the image
of the element%
\[
\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum\limits_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  <\sigma\left(  2\right)  <...<\sigma\left(
p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)  <\sigma\left(  p_{1}+2\right)
<...<\sigma\left(  p_{1}+p_{2}\right)  ;\\\sigma\left(  p_{1}+p_{2}+1\right)
<\sigma\left(  p_{1}+p_{2}+2\right)  <...<\sigma\left(  p_{1}+p_{2}%
+p_{3}\right)  ;\\...;\\\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+1\right)
<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+2\right)  <...<\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\operatorname*{sgn}\left(
\sigma\right)  \cdot\sigma^{-1}\in K\left[  S_{n}\right]
\]
under the map $K\left[  S_{n}\right]  \hookrightarrow\operatorname*{End}%
\left(  Th_{n}\left(  X\right)  \right)  $.

\parbreak


\textbf{Page 1082, Proof of Proposition 5.1:} Here, Patras writes:

``Le reste de la proposition ne pr\'{e}sente pas de difficult\'{e}s.''

Let me try to make this part of the proof a bit more precise. Namely, let me
show that the idempotents $e_{n}^{i}$ and $f_{n}^{i}$ of the algebra $K\left[
S_{n}\right]  $ are obtained from each other by the involution%
\begin{align*}
K\left[  S_{n}\right]   &  \rightarrow K\left[  S_{n}\right] \\
\sigma &  \mapsto\operatorname*{sgn}\left(  \sigma\right)  \cdot\sigma
\end{align*}
of the $K$-algebra $K\left[  S_{n}\right]  $.

In fact, let $\operatorname*{inv}$ denote the $K$-vector space homomorphism%
\begin{align*}
K\left[  S_{n}\right]   &  \rightarrow K\left[  S_{n}\right] \\
\sigma &  \mapsto\operatorname*{sgn}\left(  \sigma\right)  \cdot
\sigma\ \ \ \ \ \ \ \ \ \ \left(  \text{for every }\sigma\in S_{n}\right)  .
\end{align*}
It is easy to see that $\operatorname*{inv}$ is a $K$-algebra homomorphism
(this is more or less because $\operatorname*{sgn}:S_{n}\rightarrow\left\{
-1,1\right\}  $ is a group homomorphism) and an involution (since $\left(
\operatorname*{sgn}\left(  \sigma\right)  \right)  ^{2}=1$ for every
$\sigma\in S_{n}$). We must now prove that $\operatorname*{inv}\left(
f_{n}^{i}\right)  =e_{n}^{i}$. We will do this in two steps:

\textbf{a)} For every $k\in\mathbb{N}$, let $\Psi_{T,n}^{k}$ denote the
element of $K\left[  S_{n}\right]  $ whose image under the map $K\left[
S_{n}\right]  \hookrightarrow\operatorname*{End}\left(  T_{n}\left(  X\right)
\right)  $ is the $\Psi_{n}^{k}$ of $T\left(  X\right)  $ (where by
``$\Psi_{n}^{k}$ of $T\left(  X\right)  $'', we mean the map $\Psi_{n}^{k}$
defined with respect to the graded bialgebra $T\left(  X\right)  $). Then, for
every $k\in\mathbb{N}$, we have%
\begin{equation}
\Psi_{T,n}^{k}=\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum
\limits_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)  <\sigma\left(
2\right)  <...<\sigma\left(  p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)
<\sigma\left(  p_{1}+2\right)  <...<\sigma\left(  p_{1}+p_{2}\right)
;\\\sigma\left(  p_{1}+p_{2}+1\right)  <\sigma\left(  p_{1}+p_{2}+2\right)
<...<\sigma\left(  p_{1}+p_{2}+p_{3}\right)  ;\\...;\\\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+1\right)  <\sigma\left(  p_{1}+p_{2}+...+p_{k-1}%
+2\right)  <...<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }%
}\sigma^{-1}. \label{5.1.pf.1}%
\end{equation}
\footnote{This is because we showed above that for every $k\in\mathbb{N}$, the
map $\Psi_{n}^{k}\in\operatorname*{End}\left(  T_{n}\left(  X\right)  \right)
$ is the image of the element%
\[
\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum\limits_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  <\sigma\left(  2\right)  <...<\sigma\left(
p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)  <\sigma\left(  p_{1}+2\right)
<...<\sigma\left(  p_{1}+p_{2}\right)  ;\\\sigma\left(  p_{1}+p_{2}+1\right)
<\sigma\left(  p_{1}+p_{2}+2\right)  <...<\sigma\left(  p_{1}+p_{2}%
+p_{3}\right)  ;\\...;\\\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+1\right)
<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+2\right)  <...<\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\sigma^{-1}\in K\left[  S_{n}\right]
\]
under the map $K\left[  S_{n}\right]  \hookrightarrow\operatorname*{End}%
\left(  T_{n}\left(  X\right)  \right)  $.}

Similarly, for every $k\in\mathbb{N}$, let $\Psi_{Th,n}^{k}$ denote the
element of $K\left[  S_{n}\right]  $ whose image under the map $K\left[
S_{n}\right]  \hookrightarrow\operatorname*{End}\left(  Th_{n}\left(
X\right)  \right)  $ is the $\Psi_{n}^{k}$ of $Th\left(  X\right)  $ (where by
``$\Psi_{n}^{k}$ of $Th\left(  X\right)  $'', we mean the map $\Psi_{n}^{k}$
defined with respect to the Hopf algebra $Th\left(  X\right)  $). Then, for
every $k\in\mathbb{N}$, we have%
\begin{equation}
\Psi_{Th,n}^{k}=\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum
\limits_{\substack{\sigma\in S_{n};\\\sigma\left(  1\right)  <\sigma\left(
2\right)  <...<\sigma\left(  p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)
<\sigma\left(  p_{1}+2\right)  <...<\sigma\left(  p_{1}+p_{2}\right)
;\\\sigma\left(  p_{1}+p_{2}+1\right)  <\sigma\left(  p_{1}+p_{2}+2\right)
<...<\sigma\left(  p_{1}+p_{2}+p_{3}\right)  ;\\...;\\\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+1\right)  <\sigma\left(  p_{1}+p_{2}+...+p_{k-1}%
+2\right)  <...<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)
}}\operatorname*{sgn}\left(  \sigma\right)  \cdot\sigma^{-1}. \label{5.1.pf.2}%
\end{equation}
\footnote{This is because we showed above that for every $k\in\mathbb{N}$, the
map $\Psi_{n}^{k}\in\operatorname*{End}\left(  Th_{n}\left(  X\right)
\right)  $ is the image of the element%
\[
\sum\limits_{p_{1}+p_{2}+...+p_{k}=n}\sum\limits_{\substack{\sigma\in
S_{n};\\\sigma\left(  1\right)  <\sigma\left(  2\right)  <...<\sigma\left(
p_{1}\right)  ;\\\sigma\left(  p_{1}+1\right)  <\sigma\left(  p_{1}+2\right)
<...<\sigma\left(  p_{1}+p_{2}\right)  ;\\\sigma\left(  p_{1}+p_{2}+1\right)
<\sigma\left(  p_{1}+p_{2}+2\right)  <...<\sigma\left(  p_{1}+p_{2}%
+p_{3}\right)  ;\\...;\\\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+1\right)
<\sigma\left(  p_{1}+p_{2}+...+p_{k-1}+2\right)  <...<\sigma\left(
p_{1}+p_{2}+...+p_{k-1}+p_{k}\right)  }}\operatorname*{sgn}\left(
\sigma\right)  \cdot\sigma^{-1}\in K\left[  S_{n}\right]
\]
under the map $K\left[  S_{n}\right]  \hookrightarrow\operatorname*{End}%
\left(  Th_{n}\left(  X\right)  \right)  $.}

Comparing (\ref{5.1.pf.1}) with (\ref{5.1.pf.2}), we immediately see that
$\operatorname*{inv}\left(  \Psi_{T,n}^{k}\right)  =\Psi_{Th,n}^{k}$.

\textbf{b)} Fix some $n\in\mathbb{N}$ and some $i\in\left\{
0,1,...,n\right\}  $. Consider the $N$ and the elements $\alpha_{0}$,
$\alpha_{1}$, $...$, $\alpha_{N}$ whose existence is guaranteed by Lemma 3.10.

(Note that the equality (\ref{3.10}) is a concretization of Patras' claim that
``Ce morphisme peut, d'apr\`{e}s 1.3 et 3.1, se r\'{e}\'{e}crire comme une
combinaison lin\'{e}aire finie d'endomorphismes caract\'{e}ristiques'' on page
1081. Patras' assertion doesn't make it clear that the coefficients of this
``combinaison lin\'{e}aire'' don't depend on $H$, but our Lemma 3.10 does, and
we are going to use this now.)

Applying (\ref{3.10}) to $H=T\left(  X\right)  $, we get $f_{n}^{i}%
=\sum\limits_{k=0}^{N}\alpha_{k}\Psi_{T,n}^{k}$. Applying (\ref{3.10}) to
$H=Th\left(  X\right)  $, we get $e_{n}^{i}=\sum\limits_{k=0}^{N}\alpha
_{k}\Psi_{Th,n}^{k}$. Thus,%
\begin{align*}
\operatorname*{inv}\left(  f_{n}^{i}\right)   &  =\operatorname*{inv}\left(
\sum\limits_{k=0}^{N}\alpha_{k}\Psi_{T,n}^{k}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f_{n}^{i}=\sum\limits_{k=0}^{N}%
\alpha_{k}\Psi_{T,n}^{k}\right) \\
&  =\sum\limits_{k=0}^{N}\alpha_{k}\underbrace{\operatorname*{inv}\left(
\Psi_{T,n}^{k}\right)  }_{=\Psi_{Th,n}^{k}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\operatorname*{inv}\text{ is }K\text{-linear}\right) \\
&  =\sum\limits_{k=0}^{N}\alpha_{k}\Psi_{Th,n}^{k}=e_{n}^{i},
\end{align*}
qed.

\parbreak


\textbf{Page 1082, three lines above Lemma 5.2:} Patras writes: ``Par
d\'{e}finition, l'alg\`{e}bre de Lie libre (resp. gradu\'{e}e) est la plus
petite sous-alg\`{e}bre de Lie (resp. gradu\'{e}e) de $T\left(  X\right)  $
(resp. de $Th\left(  X\right)  $) contenant $X$.'' I don't think this really
follows from the definition (but it follows from the
Poincar\'{e}-Birkhoff-Witt theorem)\footnote{The paper [BF] gives a different
proof of the fact that the free Lie algebra over $X$ is the smallest Lie
subalgebra of $T\left(  X\right)  $ containing $X$. This proof doesn't use the
Poincar\'{e}-Birkhoff-Witt theorem but still is far from being trivial.}.

\parbreak


\textbf{Page 1082:} On this page, it should be said somewhere that $Th^{\star
}\left(  X\right)  $ is just an abbreviation for $Th^{\star\operatorname*{gr}%
}\left(  X\right)  $ (and not the dual of $Th\left(  X\right)  $ as an
ungraded $K$-vector space).

\parbreak


\textbf{Page 1082:} On the penultimate line of page 1082, Patras writes:
``l'alg\`{e}bre gradu\'{e}e $\mathcal{T}$''. This is inaccurate, since
$\mathcal{T}$ is not a graded algebra but the \textit{completion} of a graded
algebra (with respect to the canonical topology induced by the grading).
Fortunately this does not prevent the conclusion (that the logarithm of $S$ is
well-defined) from being valid (it actually would not be valid if
$\mathcal{T}$ was just a graded algebra!).

\parbreak


\textbf{Page 1083, proof of Proposition 5.3:} On the second line of this page,
there is a typo: $Tg\left(  X\right)  $ should be $Th\left(  X\right)  $.

\parbreak


\textbf{Page 1083, proof of Proposition 5.3:} Here, I don't understand why
Patras claims that%
\[
Q_{i_{1},...,i_{k}}^{\left(  m\right)  }=\Pi^{\left[  m\right]  }\circ\left(
I-\eta\circ\epsilon\right)  ^{\otimes m}\circ\Delta^{\left[  m\right]
}\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  .
\]


But I think there is an alternative proof of Proposition 5.3 anyway. It is
rather simple, but it uses a lot of notation and commonplace facts from linear
algebra. So let us begin with some definitions that could just as well stand
in a linear algebra text:

\begin{quote}
\textbf{Definition 0.3.} Whenever $V$ and $W$ are two $K$-vector spaces, then
we denote by $\rho_{V,W}$ the $K$-linear map%
\begin{align*}
V^{\star}\otimes W  &  \rightarrow\operatorname*{Hom}\nolimits_{K}\left(
V,W\right)  ,\\
f\otimes\xi &  \mapsto\left(  \text{the map }V\rightarrow W\text{ which sends
every }x\in V\text{ to }f\left(  x\right)  \xi\right)
\end{align*}
(where $V^{\star}$ denotes the dual of $V$). This map $\rho_{V,W}$ is
injective (this can be proven by standard linear algebra, i. e., working with
bases). In general, it is not surjective (but it is surjective if $\dim
V<\infty$).
\end{quote}

Next, a similarly elementary definition related to graded vector spaces:

\begin{quote}
\textbf{Definition 0.4.} \textbf{(a)} Whenever $V$ and $W$ are two graded
$K$-vector spaces, we let $\operatorname*{dirsum}\nolimits_{V,W}$ denote the
canonical injection $\prod\limits_{i\in\mathbb{N}}\operatorname*{Hom}%
\nolimits_{K}\left(  V_{i},W_{i}\right)  \rightarrow\operatorname*{Hom}%
\nolimits_{K}\left(  V,W\right)  $ which takes every family $\left(
f_{i}\right)  _{i\in\mathbb{N}}\in\prod\limits_{i\in\mathbb{N}}%
\operatorname*{Hom}\nolimits_{K}\left(  V_{i},W_{i}\right)  $ of maps to the
direct sum $\bigoplus\limits_{i\in\mathbb{N}}f_{i}:\bigoplus\limits_{i\in
\mathbb{N}}V_{i}\rightarrow\bigoplus\limits_{i\in\mathbb{N}}W_{i}$ (this
direct sum is, of course, a map $V\rightarrow W$, since $\bigoplus
\limits_{i\in\mathbb{N}}V_{i}=V$ and $\bigoplus\limits_{i\in\mathbb{N}}%
W_{i}=W$).

\textbf{(b)} Whenever $V$ is a graded $K$-vector space and $W$ is a $K$-vector
space, let us define a topology on the $K$-vector space $\operatorname*{Hom}%
\nolimits_{K}\left(  V,W\right)  $ as follows\footnote{Keep in mind that
$\operatorname*{Hom}\nolimits_{K}\left(  V,W\right)  $ is the space of all
$K$-linear maps (not only the graded ones) from $V$ to $W$.}: Let
$\mathbb{N}_{-}$ denote the set $\left\{  -1,0,1,2,...\right\}  $. For every
$i\in\mathbb{N}_{-}$, let $V^{\left(  i\right)  }$ denote the subspace
$V_{0}+V_{1}+...+V_{i}$ of $V$. For any $i\in\mathbb{N}_{-}$ and any
$g\in\operatorname*{Hom}\nolimits_{K}\left(  V^{\left(  i\right)  },W\right)
$, let $\operatorname*{Hom}\nolimits_{K,i,g}\left(  V,W\right)  $ denote the
subset%
\[
\left\{  f\in\operatorname*{Hom}\nolimits_{K}\left(  V,W\right)  \ \mid
\ f\mid_{V^{\left(  i\right)  }}=g\right\}  \ \ \ \ \ \ \ \ \ \ \text{of}%
\ \ \ \ \ \ \ \ \ \ \operatorname*{Hom}\nolimits_{K}\left(  V,W\right)  .
\]
Then, we define the topology on the $K$-vector space $\operatorname*{Hom}%
\nolimits_{K}\left(  V,W\right)  $ to be the topology generated by the sets
$\operatorname*{Hom}\nolimits_{K,i,g}\left(  V,W\right)  $ with $i\in
\mathbb{N}_{-}$ and $g\in\operatorname*{Hom}\nolimits_{K}\left(  V^{\left(
i\right)  },W\right)  $ (as open sets). (Note that $\operatorname*{Hom}%
\nolimits_{K,-1,0}\left(  V,W\right)  =\operatorname*{Hom}\nolimits_{K}\left(
V,W\right)  $.) This topology will be called the \textit{right degree
topology} on $\operatorname*{Hom}\nolimits_{K}\left(  V,W\right)  $. (I am
pretty sure that this topology has a different, more standard name, but I
don't know it.)

Let us summarize a few (easy-to-prove) properties of this right degree topology:

Let $V$ be a graded $K$-vector space and $W$ a $K$-vector space. First of all,
the right degree topology makes $\operatorname*{Hom}\nolimits_{K}\left(
V,W\right)  $ into a Hausdorff topological space, so it makes sense to speak
of ``the limit'' of a convergent sequence. Second, the right degree topology
makes $\operatorname*{Hom}\nolimits_{K}\left(  V,W\right)  $ into a complete
space, as can be easily seen. Furthermore, it is easy to see that a sequence
$\left(  f_{i}\right)  _{i\in\mathbb{N}}$ of $K$-linear maps $f_{i}%
:V\rightarrow W$ converges to a $K$-linear map $f:V\rightarrow W$ (with
respect to the right degree topology on $\operatorname*{Hom}\nolimits_{K}%
\left(  V,W\right)  $) if and only if for every $i\in\mathbb{N}$, there exists
some $N\in\mathbb{N}$ such that every $n\geq N$ satisfies $f_{n}%
\mid_{V^{\left(  i\right)  }}=f\mid_{V^{\left(  i\right)  }}$. Thus, an
infinite sum $\sum\limits_{i\in\mathbb{N}}g_{i}$ of $K$-linear maps
$g_{i}:V\rightarrow W$ converges to a $K$-linear map $g:V\rightarrow W$ if and
only if for every $i\in\mathbb{N}$, there exists some $N\in\mathbb{N}$ such
that
\begin{align*}
&  g_{N}\left(  V^{\left(  i\right)  }\right)  =g_{N+1}\left(  V^{\left(
i\right)  }\right)  =g_{N+2}\left(  V^{\left(  i\right)  }\right)  =...=0\\
&  \text{and}\ \ \ \ \ \ \ \ \ \ \left(  \sum\limits_{\substack{i\in
\mathbb{N};\\i\leq N-1}}g_{i}\right)  \mid_{V^{\left(  i\right)  }}%
=g\mid_{V^{\left(  i\right)  }}.
\end{align*}
Consequently, it is easy to see that an infinite sum $\sum\limits_{i\in
\mathbb{N}}g_{i}$ of $K$-linear maps $g_{i}:V\rightarrow W$ converges if and
only if for every $i\in\mathbb{N}$, there exists some $N\in\mathbb{N}$ such
that
\[
g_{N}\left(  V^{\left(  i\right)  }\right)  =g_{N+1}\left(  V^{\left(
i\right)  }\right)  =g_{N+2}\left(  V^{\left(  i\right)  }\right)  =...=0.
\]
(In this case, the value of this infinite sum is the function $g$ that sends
every $x\in V$ to $\sum\limits_{i\in\mathbb{N}}g_{i}\left(  x\right)  $; here,
the infinite sum $\sum\limits_{i\in\mathbb{N}}g_{i}\left(  x\right)  $ has a
well-defined value since all but finitely many of its terms are zero.)

It is easy to see that if $V$ and $W$ are two graded $K$-vector spaces, then
the map $\operatorname*{dirsum}\nolimits_{V,W}$ is continuous, where the
topology on the space $\prod\limits_{i\in\mathbb{N}}\operatorname*{Hom}%
\nolimits_{K}\left(  V_{i},W_{i}\right)  $ is the canonical one obtained by
seeing $\prod\limits_{i\in\mathbb{N}}\operatorname*{Hom}\nolimits_{K}\left(
V_{i},W_{i}\right)  $ as the completion of the graded vector space
$\bigoplus\limits_{i\in\mathbb{N}}\operatorname*{Hom}\nolimits_{K}\left(
V_{i},W_{i}\right)  $, whereas the topology on $\operatorname*{Hom}%
\nolimits_{K}\left(  V,W\right)  $ is the right degree topology.
\end{quote}

Next, let us introduce a certain subspace of the tensor product of two graded spaces:

\begin{quote}
\textbf{Definition 0.5.} Let $V$ and $W$ be two graded $K$-vector spaces. Let
$V\underset{=}{\otimes}W$ denote the $K$-vector subspace $\sum\limits_{i\in
\mathbb{N}}V_{i}\otimes W_{i}$ of $V\otimes W$. (Here, for every
$i\in\mathbb{N}$, we consider $V_{i}\otimes W_{i}$ as a subspace of $V\otimes
W$, because $V_{i}$ is a subspace of $V$ and $W_{i}$ is a subspace of $W$.)

Here is an alternative definition of $V\underset{=}{\otimes}W$: Since
$V=\bigoplus\limits_{i\in\mathbb{N}}V_{i}$ and $W=\bigoplus\limits_{j\in
\mathbb{N}}W_{j}$, we have%
\[
V\otimes W=\left(  \bigoplus\limits_{i\in\mathbb{N}}V_{i}\right)
\otimes\left(  \bigoplus\limits_{j\in\mathbb{N}}W_{j}\right)  =\bigoplus
\limits_{\left(  i,j\right)  \in\mathbb{N}^{2}}\left(  V_{i}\otimes
W_{j}\right)
\]
(since tensor products commute with direct sums). Now, let
$V\underset{=}{\otimes}W$ denote the subspace $\bigoplus
\limits_{\substack{\left(  i,j\right)  \in\mathbb{N}^{2};\\i=j}}\left(
V_{i}\otimes W_{j}\right)  $ of this direct sum $\bigoplus\limits_{\left(
i,j\right)  \in\mathbb{N}^{2}}\left(  V_{i}\otimes W_{j}\right)  $.

From this alternative description of $V\underset{=}{\otimes}W$ as the space
$\bigoplus\limits_{\substack{\left(  i,j\right)  \in\mathbb{N}^{2}%
;\\i=j}}\left(  V_{i}\otimes W_{j}\right)  $, it is clear that
$V\underset{=}{\otimes}W\cong\bigoplus\limits_{i\in\mathbb{N}}\left(
V_{i}\otimes W_{i}\right)  $. More precisely, the map%
\begin{align*}
\bigoplus\limits_{i\in\mathbb{N}}\left(  V_{i}\otimes W_{i}\right)   &
\rightarrow V\underset{=}{\otimes}W,\\
\left(  a_{i}\right)  _{i\in\mathbb{N}}  &  \mapsto\sum\limits_{i\in
\mathbb{N}}a_{i}%
\end{align*}
is a well-defined canonical isomorphism of graded $K$-vector spaces. We denote
this isomorphism by $\operatorname*{combine}\nolimits_{V,W}$.

The completion of the graded $K$-vector space $\bigoplus\limits_{i\in
\mathbb{N}}\left(  V_{i}\otimes W_{i}\right)  $ is $\prod\limits_{i\in
\mathbb{N}}\left(  V_{i}\otimes W_{i}\right)  $. We denote the completion of
the graded $K$-vector space $V\underset{=}{\otimes}W$ by
$V\widehat{\underset{=}{\otimes}}W$. Let also
$\widehat{\operatorname*{combine}\nolimits_{V,W}}$ denote the completion of
the isomorphism $\operatorname*{combine}\nolimits_{V,W}:\bigoplus
\limits_{i\in\mathbb{N}}\left(  V_{i}\otimes W_{i}\right)  \rightarrow
V\underset{=}{\otimes}W$. Since $\operatorname*{combine}\nolimits_{V,W}$ is an
isomorphism of graded $K$-vector spaces, its completion
$\widehat{\operatorname*{combine}\nolimits_{V,W}}$ is an isomorphism
$\prod\limits_{i\in\mathbb{N}}\left(  V_{i}\otimes W_{i}\right)  \rightarrow
V\widehat{\underset{=}{\otimes}}W$ of topological $K$-vector spaces. This
isomorphism $\widehat{\operatorname*{combine}\nolimits_{V,W}}$ is the map%
\begin{align*}
\prod\limits_{i\in\mathbb{N}}\left(  V_{i}\otimes W_{i}\right)   &
\rightarrow V\widehat{\underset{=}{\otimes}}W,\\
\left(  a_{i}\right)  _{i\in\mathbb{N}}  &  \mapsto\sum\limits_{i\in
\mathbb{N}}a_{i}%
\end{align*}
(where the sum $\sum\limits_{i\in\mathbb{N}}a_{i}$ is automatically convergent
by the completeness of $V\widehat{\underset{=}{\otimes}}W$).

The completion of the canonical inclusion $V\underset{=}{\otimes}W\rightarrow
V\otimes W$ is an injective continuous map $V\widehat{\underset{=}{\otimes}%
}W\rightarrow V\widehat{\otimes}W$ of $K$-vector spaces. This injective map
allows us to consider $V\widehat{\underset{=}{\otimes}}W$ as a $K$-vector
subspace of $V\widehat{\otimes}W$. We are going to do so.
\end{quote}

These three definitions were purely linear-algebraical (and topological to the
extent they involved completions). It is time to tie in some algebras and
coalgebras into this. First, the purely algebraic part:

\begin{quote}
\textbf{Proposition 1.12.} Let $A$ and $B$ be two graded $K$-algebras. Then,
$A\underset{=}{\otimes}B$ is a graded $K$-subalgebra of $A\otimes B$, and
$A\widehat{\underset{=}{\otimes}}B$ is a $K$-subalgebra of $A\widehat{\otimes
}B$.
\end{quote}

The proof of Proposition 1.12 is a completely straightforward check (it boils
down to showing that $1_{A\otimes B}\in A_{0}\otimes B_{0}$ and that $\left(
A_{i}\otimes B_{i}\right)  \cdot\left(  A_{j}\otimes B_{j}\right)  \subseteq
A_{i+j}\otimes B_{i+j}$ for all $i\in\mathbb{N}$ and $j\in\mathbb{N}$, where
we consider $A_{u}\otimes B_{v}$ as a $K$-vector subspace of $A\otimes B$ for
all $u\in\mathbb{N}$ and $v\in\mathbb{N}$).

Now here is something more interesting:

\begin{quote}
\textbf{Proposition 1.13.} Let $C$ be a locally-finite\footnote{Here, a graded
$K$-vector space $V$ is said to be \textit{locally-finite} if for every
$n\in\mathbb{N}$, the $n$-th graded component of $V$ is a finite-dimensional
$K$-vector space. So our notion of ``locally finite'' is exactly what Patras
calls ``de type fini''.} graded $K$-coalgebra, and let $A$ be a graded
$K$-algebra. Since $C$ is a locally-finite graded $K$-coalgebra, the graded
dual $C^{\star\operatorname*{gr}}$ becomes a graded $K$-algebra.

Consider the continuous $K$-vector space isomorphism
$\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}:\prod\limits_{i\in\mathbb{N}}\left(  C_{i}^{\star\operatorname*{gr}}\otimes
A_{i}\right)  \rightarrow C^{\star\operatorname*{gr}}%
\widehat{\underset{=}{\otimes}}A$.

For every $i\in\mathbb{N}$, we have an injective map $\rho_{C_{i},A_{i}%
}:\left(  C_{i}\right)  ^{\star}\otimes A_{i}\rightarrow\operatorname*{Hom}%
\nolimits_{K}\left(  C_{i},A_{i}\right)  $. Since $\left(  C_{i}\right)
^{\star}=C_{i}^{\star\operatorname*{gr}}$ for every $i\in\mathbb{N}$, this
rewrites as follows: For every $i\in\mathbb{N}$, we have an injective map
$\rho_{C_{i},A_{i}}:C_{i}^{\star\operatorname*{gr}}\otimes A_{i}%
\rightarrow\operatorname*{Hom}\nolimits_{K}\left(  C_{i},A_{i}\right)  $.
Thus, the product $\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}%
:\prod\limits_{i\in\mathbb{N}}\left(  C_{i}^{\star\operatorname*{gr}}\otimes
A_{i}\right)  \rightarrow\prod\limits_{i\in\mathbb{N}}\operatorname*{Hom}%
\nolimits_{K}\left(  C_{i},A_{i}\right)  $ of these maps is also injective.

Finally, consider the $K$-vector space injection $\operatorname*{dirsum}%
\nolimits_{C,A}:\prod\limits_{i\in\mathbb{N}}\operatorname*{Hom}%
\nolimits_{K}\left(  C_{i},A_{i}\right)  \rightarrow\operatorname*{Hom}%
\nolimits_{K}\left(  C,A\right)  $.

Denote by $\phi_{C,A}$ the map%
\[
\operatorname*{dirsum}\nolimits_{C,A}\circ\left(  \prod\limits_{i\in
\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \circ\widehat{\operatorname*{combine}%
\nolimits_{C^{\star\operatorname*{gr}},A}}^{-1}:C^{\star\operatorname*{gr}%
}\widehat{\underset{=}{\otimes}}A\rightarrow\operatorname*{Hom}\nolimits_{K}%
\left(  C,A\right)  .
\]
Clearly, this map $\phi_{C,A}$ makes the diagram%
\[%
%TCIMACRO{\TeXButton{square diagram}{\xymatrixcolsep{7pc}
%\xymatrix{
%\prod\limits_{i\in\mathbb N} \left(C_i^{\star\gr} \otimes A_i\right
%) \arinj[r]^{\prod\limits_{i\in\mathbb N} \rho_{C_i,A_i} } \ar[d]^{\cong
%}_{\widehat{\combine_{C^{\star\gr},A}}} & \prod\limits_{i\in\mathbb N}
%\Hom_K\left(C_i, A_i\right) \arinj[d]^{\dirsum_{C,A}} \\
%C^{\star\gr} \widehat{\underset{=}{\otimes}} A \ar[r]_{\phi_{C,A}}
%& \Hom_K\left(C,A\right)
%}}}%
%BeginExpansion
\xymatrixcolsep{7pc}
\xymatrix{
\prod\limits_{i\in\mathbb N} \left(C_i^{\star\gr} \otimes A_i\right
) \arinj[r]^{\prod\limits_{i\in\mathbb N} \rho_{C_i,A_i} } \ar[d]^{\cong
}_{\widehat{\combine_{C^{\star\gr},A}}} & \prod\limits_{i\in\mathbb N}
\Hom_K\left(C_i, A_i\right) \arinj[d]^{\dirsum_{C,A}} \\
C^{\star\gr} \widehat{\underset{=}{\otimes}} A \ar[r]_{\phi_{C,A}}
& \Hom_K\left(C,A\right)
}%
%EndExpansion
\]
commute.

\textbf{(a)} This map $\phi_{C,A}$ is an injective and continuous $K$-algebra
homomorphism. (Here, the $K$-algebra structure on $C^{\star\operatorname*{gr}%
}\widehat{\underset{=}{\otimes}}A$ is given by the fact that $C^{\star
\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$ is a $K$-subalgebra of
$C^{\star\operatorname*{gr}}\widehat{\otimes}A$, whereas the $K$-algebra
structure on $\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ is defined
to be the convolution.)

\textbf{(b)} For every $n\in\mathbb{N}$, every $\gamma\in C_{n}^{\star
\operatorname*{gr}}$ and every $a\in A_{n}$, we have%
\begin{equation}
\phi_{C,A}\left(  \gamma\otimes a\right)  =\left(  \text{the map }C\rightarrow
A\text{ which sends every }x\in C\text{ to }\gamma\left(  x\right)  a\right)
. \label{1.13.b.1}%
\end{equation}


In other words, for every $n\in\mathbb{N}$, every $\gamma\in C_{n}%
^{\star\operatorname*{gr}}$ and every $a\in A_{n}$, we have%
\begin{equation}
\left(  \phi_{C,A}\left(  \gamma\otimes a\right)  \right)  \left(  x\right)
=\gamma\left(  x\right)  a\ \ \ \ \ \ \ \ \ \ \text{for every }x\in C\text{.}
\label{1.13.b.2}%
\end{equation}


\textbf{(c)} Let $f:C\rightarrow A$ be a graded $K$-linear map. For every
$k\in\mathbb{N}$, let $\left(  c_{k,\ell}\right)  _{\ell\in I_{k}}$ be a basis
of the $K$-vector space $C_{k}$, and let $\left(  c_{k,\ell}^{\star}\right)
_{\ell\in I_{k}}$ be the basis of $C_{k}^{\star\operatorname*{gr}}$ dual to
this basis $\left(  c_{k,\ell}\right)  _{\ell\in I_{k}}$.\ \ \ \ \footnote{At
this place, we are using the condition that $C$ is locally-finite. (In fact,
since $C$ is locally-finite, the space $C_{k}$ is finite-dimensional, so that
every basis of $C_{k}$ has a dual basis of $\left(  C_{k}\right)  ^{\star
}=C_{k}^{\star\operatorname*{gr}}$.)} Then,%
\[
f=\phi_{C,A}\left(  \sum\limits_{k\in\mathbb{N}}\sum\limits_{\ell\in I_{k}%
}c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell}\right)  \right)  .
\]



\end{quote}

\begin{proof}
[Proof of Proposition 1.13.]\textbf{(b)} Let $n\in\mathbb{N}$, $\gamma\in
C_{n}^{\star\operatorname*{gr}}$ and $a\in A_{n}$ be arbitrary.

Define a family $\left(  t_{i}\right)  _{i\in\mathbb{N}}\in\prod
\limits_{i\in\mathbb{N}}\left(  C_{i}^{\star\operatorname*{gr}}\otimes
A_{i}\right)  $ by
\[
\left(  t_{i}=%
\begin{cases}
\gamma\otimes a, & \text{if }i=n;\\
0, & \text{if }i\neq n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}\right)  .
\]
Then, $t_{n}=\gamma\otimes a$, whereas every $i\neq n$ satisfies $t_{i}=0$.

Since we know that $\widehat{\operatorname*{combine}\nolimits_{C^{\star
\operatorname*{gr}},A}}$ is the map%
\begin{align*}
\prod\limits_{i\in\mathbb{N}}\left(  C_{i}^{\star\operatorname*{gr}}\otimes
A_{i}\right)   &  \rightarrow C^{\star\operatorname*{gr}}%
\widehat{\underset{=}{\otimes}}A,\\
\left(  a_{i}\right)  _{i\in\mathbb{N}}  &  \mapsto\sum\limits_{i\in
\mathbb{N}}a_{i},
\end{align*}
we have%
\begin{align}
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}\left(  \left(  a_{i}\right)  _{i\in\mathbb{N}}\right)   &  =\sum
\limits_{i\in\mathbb{N}}a_{i}\label{1.13.pf.0}\\
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a_{i}\right)  _{i\in\mathbb{N}}
&  \in\prod\limits_{i\in\mathbb{N}}\left(  C_{i}^{\star\operatorname*{gr}%
}\otimes A_{i}\right)  .\nonumber
\end{align}
Applied to $\left(  a_{i}\right)  _{i\in\mathbb{N}}=\left(  t_{i}\right)
_{i\in\mathbb{N}}$, this yields%
\[
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}\left(  \left(  t_{i}\right)  _{i\in\mathbb{N}}\right)  =\sum\limits_{i\in
\mathbb{N}}t_{i}=\underbrace{\sum\limits_{\substack{i\in\mathbb{N}%
;\\i=n}}t_{i}}_{=t_{n}=\gamma\otimes a}+\underbrace{\sum
\limits_{\substack{i\in\mathbb{N};\\i\neq n}}t_{i}}%
_{\substack{=0\\\text{(since every }i\neq n\text{ satisfies }t_{i}=0\text{)}%
}}=\gamma\otimes a.
\]
Thus,%
\[
\left(  t_{i}\right)  _{i\in\mathbb{N}}=\widehat{\operatorname*{combine}%
\nolimits_{C^{\star\operatorname*{gr}},A}}^{-1}\left(  \gamma\otimes a\right)
.
\]


On the other hand, by the definition of $\prod\limits_{i\in\mathbb{N}}%
\rho_{C_{i},A_{i}}$, we have $\left(  \prod\limits_{i\in\mathbb{N}}\rho
_{C_{i},A_{i}}\right)  \left(  \left(  t_{i}\right)  _{i\in\mathbb{N}}\right)
=\left(  \rho_{C_{i},A_{i}}\left(  t_{i}\right)  \right)  _{i\in\mathbb{N}}$,
so that%
\begin{align*}
\operatorname*{dirsum}\nolimits_{C,A}\left(  \left(  \prod\limits_{i\in
\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \left(  \left(  t_{i}\right)
_{i\in\mathbb{N}}\right)  \right)   &  =\operatorname*{dirsum}\nolimits_{C,A}%
\left(  \left(  \rho_{C_{i},A_{i}}\left(  t_{i}\right)  \right)
_{i\in\mathbb{N}}\right)  =\bigoplus\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}%
}\left(  t_{i}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\operatorname*{dirsum}\nolimits_{C,A}\right)  .
\end{align*}
In other words,%
\begin{align*}
\bigoplus\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\left(  t_{i}\right)   &
=\operatorname*{dirsum}\nolimits_{C,A}\left(  \left(  \prod\limits_{i\in
\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \underbrace{\left(  \left(
t_{i}\right)  _{i\in\mathbb{N}}\right)  }_{=\widehat{\operatorname*{combine}%
\nolimits_{C^{\star\operatorname*{gr}},A}}^{-1}\left(  \gamma\otimes a\right)
}\right) \\
&  =\operatorname*{dirsum}\nolimits_{C,A}\left(  \left(  \prod\limits_{i\in
\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \left(  \widehat{\operatorname*{combine}%
\nolimits_{C^{\star\operatorname*{gr}},A}}^{-1}\left(  \gamma\otimes a\right)
\right)  \right) \\
&  =\underbrace{\left(  \operatorname*{dirsum}\nolimits_{C,A}\circ\left(
\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \circ
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}\right)  }_{=\phi_{C,A}}\left(  \gamma\otimes a\right)  =\phi
_{C,A}\left(  \gamma\otimes a\right)  .
\end{align*}


Recall that the map $\rho_{C_{n},A_{n}}$ was defined as the $K$-linear map%
\begin{align*}
\left(  C_{n}\right)  ^{\star}\otimes A_{n}  &  \rightarrow\operatorname*{Hom}%
\nolimits_{K}\left(  C_{n},A_{n}\right)  ,\\
f\otimes\xi &  \mapsto\left(  \text{the map }C_{n}\rightarrow A_{n}\text{
which sends every }x\in C_{n}\text{ to }f\left(  x\right)  \xi\right)  .
\end{align*}
Hence, $\rho_{C_{n},A_{n}}\left(  \gamma\otimes a\right)  =\left(  \text{the
map }C_{n}\rightarrow A_{n}\text{ which sends every }x\in C_{n}\text{ to
}\gamma\left(  x\right)  a\right)  $. Hence,%
\begin{equation}
\left(  \rho_{C_{n},A_{n}}\left(  \gamma\otimes a\right)  \right)  \left(
y\right)  =\gamma\left(  y\right)  a\ \ \ \ \ \ \ \ \ \ \text{for every }y\in
C_{n}. \label{1.13.pf.1}%
\end{equation}


Now let $x\in C$ be arbitrary. For every $i\in\mathbb{N}$, let $x_{i}$ be the
$i$-th graded component of $x$. Since every nonnegative integer $i\neq n$
satisfies $t_{i}=0$, it is clear that every nonnegative integer $i\neq n$
satisfies $\left(  \rho_{C_{i},A_{i}}\left(  t_{i}\right)  \right)  \left(
x_{i}\right)  =\underbrace{\left(  \rho_{C_{i},A_{i}}\left(  0\right)
\right)  }_{=0}\left(  x_{i}\right)  =0\left(  x_{i}\right)  =0$. Thus,
\begin{equation}
\sum\limits_{\substack{i\in\mathbb{N};\\i\neq n}}\left(  \rho_{C_{i},A_{i}%
}\left(  t_{i}\right)  \right)  \left(  x_{i}\right)  =\sum
\limits_{\substack{i\in\mathbb{N};\\i\neq n}}0=0. \label{1.13.pf.1b}%
\end{equation}


On the other hand, $\gamma\in C_{n}^{\star\operatorname*{gr}}$, so that every
nonnegative integer $i\neq n$ satisfies $\gamma\left(  C_{i}\right)  =0$ (by
the definition of the graded dual $C^{\star\operatorname*{gr}}$ and its
grading) and therefore%
\begin{equation}
\gamma\left(  x_{i}\right)  =0 \label{1.13.pf.1c}%
\end{equation}
(since $x_{i}\in C_{i}$). However, we have $x=\sum_{i\in\mathbb{N}}x_{i}$
(since each $x_{i}$ is the $i$-th graded component of $x$), so that
\begin{align}
\gamma\left(  x\right)   &  =\gamma\left(  \sum_{i\in\mathbb{N}}x_{i}\right)
=\sum_{i\in\mathbb{N}}\gamma\left(  x_{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\gamma\text{ is }K\text{-linear}\right) \nonumber\\
&  =\underbrace{\sum_{\substack{i\in\mathbb{N};\\i=n}}\gamma\left(
x_{i}\right)  }_{=\gamma\left(  x_{n}\right)  }+\sum_{\substack{i\in
\mathbb{N};\\i\neq n}}\underbrace{\gamma\left(  x_{i}\right)  }%
_{\substack{=0\\\text{(by (\ref{1.13.pf.1c}))}}}=\gamma\left(  x_{n}\right)
+\underbrace{\sum_{\substack{i\in\mathbb{N};\\i\neq n}}0}_{=0}\nonumber\\
&  =\gamma\left(  x_{n}\right)  . \label{1.13.pf.1d}%
\end{align}


Now,
\begin{align*}
\underbrace{\left(  \phi_{C,A}\left(  \gamma\otimes a\right)  \right)
}_{=\bigoplus\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\left(  t_{i}\right)
}\left(  x\right)   &  =\left(  \bigoplus\limits_{i\in\mathbb{N}}\rho
_{C_{i},A_{i}}\left(  t_{i}\right)  \right)  \left(  x\right)  =\sum
\limits_{i\in\mathbb{N}}\left(  \rho_{C_{i},A_{i}}\left(  t_{i}\right)
\right)  \left(  x_{i}\right) \\
&  =\underbrace{\sum\limits_{\substack{i\in\mathbb{N};\\i=n}}\left(
\rho_{C_{i},A_{i}}\left(  t_{i}\right)  \right)  \left(  x_{i}\right)
}_{=\left(  \rho_{C_{n},A_{n}}\left(  t_{n}\right)  \right)  \left(
x_{n}\right)  }+\underbrace{\sum\limits_{\substack{i\in\mathbb{N};\\i\neq
n}}\left(  \rho_{C_{i},A_{i}}\left(  t_{i}\right)  \right)  \left(
x_{i}\right)  }_{\substack{=0\\\text{(by (\ref{1.13.pf.1b}))}}}\\
&  =\left(  \rho_{C_{n},A_{n}}\underbrace{\left(  t_{n}\right)  }%
_{=\gamma\otimes a}\right)  \left(  x_{n}\right)  =\left(  \rho_{C_{n},A_{n}%
}\left(  \gamma\otimes a\right)  \right)  \left(  x_{n}\right) \\
&  =\underbrace{\gamma\left(  x_{n}\right)  }_{\substack{=\gamma\left(
x\right)  \\\text{(by (\ref{1.13.pf.1d}))}}}a\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{1.13.pf.1}), applied to }x_{n}\text{ instead of }y\right) \\
&  =\gamma\left(  x\right)  a.
\end{align*}


Now forget that we fixed $x$. We have just proven that every $x\in C$
satisfies $\left(  \phi_{C,A}\left(  \gamma\otimes a\right)  \right)  \left(
x\right)  =\gamma\left(  x\right)  a$. Hence,%
\[
\phi_{C,A}\left(  \gamma\otimes a\right)  =\left(  \text{the map }C\rightarrow
A\text{ which sends every }x\in C\text{ to }\gamma\left(  x\right)  a\right)
.
\]
This proves Proposition 1.13 \textbf{(b)}.

\textbf{(a)} We know that the maps $\operatorname*{dirsum}\nolimits_{C,A}$,
$\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}$ and
$\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}$ are injective\footnote{\textit{Proof.} The map $\operatorname*{dirsum}%
\nolimits_{C,A}$ is injective by Definition 0.4 \textbf{(a)}. The map
$\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}$ is injective because the
product of injective maps is injective, and because the map $\rho_{C_{i}%
,A_{i}}$ is injective for each $i\in\mathbb{N}$ (by Definition 0.3). The map
$\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}$ is injective because it is the inverse of an isomorphism.}. Hence,
their composition $\operatorname*{dirsum}\nolimits_{C,A}\circ\left(
\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \circ
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}$ must also be injective. Since this composition $\operatorname*{dirsum}%
\nolimits_{C,A}\circ\left(  \prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}%
}\right)  \circ\widehat{\operatorname*{combine}\nolimits_{C^{\star
\operatorname*{gr}},A}}^{-1}$ is the map $\phi_{C,A}$, this yields that
$\phi_{C,A}$ is injective.

We know that the maps $\operatorname*{dirsum}\nolimits_{C,A}$, $\prod
\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}$ and
$\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}$ are continuous\footnote{\textit{Proof.} The map
$\operatorname*{dirsum}\nolimits_{C,A}$ is continuous by Definition 0.4
\textbf{(b)}. The map $\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}$ is
continuous because it is the completion of a graded map (namely, of the graded
map $\bigoplus\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}$). The map
$\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}$ is continuous because it is the inverse of an isomorphism of
topological $K$-vector spaces.}. Hence, their composition
$\operatorname*{dirsum}\nolimits_{C,A}\circ\left(  \prod\limits_{i\in
\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \circ\widehat{\operatorname*{combine}%
\nolimits_{C^{\star\operatorname*{gr}},A}}^{-1}$ must also be continuous.
Since this composition $\operatorname*{dirsum}\nolimits_{C,A}\circ\left(
\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \circ
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}$ is the map $\phi_{C,A}$, this yields that $\phi_{C,A}$ is continuous.

So we now know that $\phi_{C,A}$ is an injective and continuous $K$-vector
space homomorphism. We still need to check that $\phi_{C,A}$ is a $K$-algebra homomorphism.

Let us first show that%
\begin{equation}
\phi_{C,A}\left(  \gamma\otimes a\right)  \ast\phi_{C,A}\left(  \delta\otimes
b\right)  =\phi_{C,A}\left(  \left(  \gamma\ast\delta\right)  \otimes
ab\right)  \label{1.13.pf.2}%
\end{equation}
for all $n\in\mathbb{N}$, $m\in\mathbb{N}$, $a\in A_{n}$, $b\in A_{m}$,
$\gamma\in C_{n}^{\star\operatorname*{gr}}$ and $\delta\in C_{m}%
^{\star\operatorname*{gr}}$.

\textit{Proof of (\ref{1.13.pf.2}).} Let $n\in\mathbb{N}$, $m\in\mathbb{N}$,
$a\in A_{n}$, $b\in A_{m}$, $\gamma\in C_{n}^{\star\operatorname*{gr}}$ and
$\delta\in C_{m}^{\star\operatorname*{gr}}$ be arbitrary.

Let $y\in C$. Since $\Delta\left(  y\right)  \in C\otimes C$, we can write the
tensor $\Delta\left(  y\right)  $ in the form $\Delta\left(  y\right)
=\sum\limits_{\ell=1}^{L}\lambda_{\ell}c_{\ell}\otimes d_{\ell}$ for some
$L\in\mathbb{N}$, some elements $\lambda_{1}$, $\lambda_{2}$, $...$,
$\lambda_{L}$ of $K$, some elements $c_{1}$, $c_{2}$, $...$, $c_{L}$ of $C$,
and some elements $d_{1}$, $d_{2}$, $...$, $d_{L}$ of $C$. Consider this $L$,
these $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{L}$, these $c_{1}$,
$c_{2}$, $...$, $c_{L}$, and these $d_{1}$, $d_{2}$, $...$, $d_{L}$.

By the definition of the convolution,%
\[
\phi_{C,A}\left(  \gamma\otimes a\right)  \ast\phi_{C,A}\left(  \delta\otimes
b\right)  =\mu_{A}\circ\left(  \phi_{C,A}\left(  \gamma\otimes a\right)
\otimes\phi_{C,A}\left(  \delta\otimes b\right)  \right)  \circ\Delta_{C},
\]
so that%
\begin{align*}
&  \left(  \phi_{C,A}\left(  \gamma\otimes a\right)  \ast\phi_{C,A}\left(
\delta\otimes b\right)  \right)  \left(  y\right) \\
&  =\left(  \mu_{A}\circ\left(  \phi_{C,A}\left(  \gamma\otimes a\right)
\otimes\phi_{C,A}\left(  \delta\otimes b\right)  \right)  \circ\Delta
_{C}\right)  \left(  y\right) \\
&  =\left(  \mu_{A}\circ\left(  \phi_{C,A}\left(  \gamma\otimes a\right)
\otimes\phi_{C,A}\left(  \delta\otimes b\right)  \right)  \right)
\underbrace{\left(  \Delta_{C}\left(  y\right)  \right)  }_{=\sum
\limits_{\ell=1}^{L}\lambda_{\ell}c_{\ell}\otimes d_{\ell}}\\
&  =\left(  \mu_{A}\circ\left(  \phi_{C,A}\left(  \gamma\otimes a\right)
\otimes\phi_{C,A}\left(  \delta\otimes b\right)  \right)  \right)  \left(
\sum\limits_{\ell=1}^{L}\lambda_{\ell}c_{\ell}\otimes d_{\ell}\right) \\
&  =\mu_{A}\underbrace{\left(  \left(  \phi_{C,A}\left(  \gamma\otimes
a\right)  \otimes\phi_{C,A}\left(  \delta\otimes b\right)  \right)  \left(
\sum\limits_{\ell=1}^{L}\lambda_{\ell}c_{\ell}\otimes d_{\ell}\right)
\right)  }_{\substack{=\sum\limits_{\ell=1}^{L}\lambda_{\ell}\left(
\phi_{C,A}\left(  \gamma\otimes a\right)  \right)  \left(  c_{\ell}\right)
\otimes\left(  \phi_{C,A}\left(  \delta\otimes b\right)  \right)  \left(
d_{\ell}\right)  \\\text{(by the definition of }\phi_{C,A}\left(
\gamma\otimes a\right)  \otimes\phi_{C,A}\left(  \delta\otimes b\right)
\text{)}}}\\
&  =\mu_{A}\left(  \sum\limits_{\ell=1}^{L}\lambda_{\ell}\left(  \phi
_{C,A}\left(  \gamma\otimes a\right)  \right)  \left(  c_{\ell}\right)
\otimes\left(  \phi_{C,A}\left(  \delta\otimes b\right)  \right)  \left(
d_{\ell}\right)  \right) \\
&  =\sum\limits_{\ell=1}^{L}\lambda_{\ell}\underbrace{\left(  \phi
_{C,A}\left(  \gamma\otimes a\right)  \right)  \left(  c_{\ell}\right)
}_{\substack{=\gamma\left(  c_{\ell}\right)  a\\\text{(by (\ref{1.13.b.2}),
applied to}\\c_{\ell}\text{ instead of }x\text{)}}}\cdot\underbrace{\left(
\phi_{C,A}\left(  \delta\otimes b\right)  \right)  \left(  d_{\ell}\right)
}_{\substack{=\delta\left(  d_{\ell}\right)  b\\\text{(by (\ref{1.13.b.2}),
applied to}\\d_{\ell}\text{, }\delta\text{ and }b\text{ instead of }x\text{,
}\gamma\text{ and }a\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{A}\text{ is the
multiplication map}\right) \\
&  =\sum\limits_{\ell=1}^{L}\lambda_{\ell}\gamma\left(  c_{\ell}\right)
a\delta\left(  d_{\ell}\right)  b=\sum\limits_{\ell=1}^{L}\lambda_{\ell}%
\gamma\left(  c_{\ell}\right)  \delta\left(  d_{\ell}\right)  \cdot ab.
\end{align*}
Since
\begin{align*}
&  \underbrace{\left(  \gamma\ast\delta\right)  }_{\substack{=\mu_{K}%
\circ\left(  \gamma\otimes\delta\right)  \circ\Delta_{C}\\\text{(by the
definition of convolution)}}}\left(  y\right) \\
&  =\left(  \mu_{K}\circ\left(  \gamma\otimes\delta\right)  \circ\Delta
_{C}\right)  \left(  y\right)  =\mu_{K}\left(  \left(  \gamma\otimes
\delta\right)  \underbrace{\left(  \Delta_{C}\left(  y\right)  \right)
}_{=\sum\limits_{\ell=1}^{L}\lambda_{\ell}c_{\ell}\otimes d_{\ell}}\right)
=\mu_{K}\underbrace{\left(  \left(  \gamma\otimes\delta\right)  \left(
\sum\limits_{\ell=1}^{L}\lambda_{\ell}c_{\ell}\otimes d_{\ell}\right)
\right)  }_{\substack{=\sum\limits_{\ell=1}^{L}\lambda_{\ell}\gamma\left(
c_{\ell}\right)  \otimes\delta\left(  d_{\ell}\right)  \\\text{(by the
definition of }\gamma\otimes\delta\text{)}}}\\
&  =\mu_{K}\left(  \sum\limits_{\ell=1}^{L}\lambda_{\ell}\gamma\left(
c_{\ell}\right)  \otimes\delta\left(  d_{\ell}\right)  \right)  =\sum
\limits_{\ell=1}^{L}\lambda_{\ell}\gamma\left(  c_{\ell}\right)  \delta\left(
d_{\ell}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{K}\text{ is the
multiplication map}\right)  ,
\end{align*}
this becomes%
\[
\left(  \phi_{C,A}\left(  \gamma\otimes a\right)  \ast\phi_{C,A}\left(
\delta\otimes b\right)  \right)  \left(  y\right)  =\underbrace{\sum
\limits_{\ell=1}^{L}\lambda_{\ell}\gamma\left(  c_{\ell}\right)  \delta\left(
d_{\ell}\right)  }_{=\left(  \gamma\ast\delta\right)  \left(  y\right)  }\cdot
ab=\left(  \gamma\ast\delta\right)  \left(  y\right)  \cdot ab.
\]
Compared with
\begin{align*}
\left(  \phi_{C,A}\left(  \left(  \gamma\ast\delta\right)  \otimes ab\right)
\right)  \left(  y\right)   &  =\left(  \gamma\ast\delta\right)  \left(
y\right)  \cdot ab\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1.13.b.2}), applied to }y\text{,
}\gamma\ast\delta\text{ and }ab\text{ instead of }x\text{, }\gamma\text{ and
}a\right)  ,
\end{align*}
this yields%
\[
\left(  \phi_{C,A}\left(  \gamma\otimes a\right)  \ast\phi_{C,A}\left(
\delta\otimes b\right)  \right)  \left(  y\right)  =\left(  \phi_{C,A}\left(
\left(  \gamma\ast\delta\right)  \otimes ab\right)  \right)  \left(  y\right)
.
\]


Now forget that we fixed $y$. We thus have shown that every $y\in C$ satisfies
$\left(  \phi_{C,A}\left(  \gamma\otimes a\right)  \ast\phi_{C,A}\left(
\delta\otimes b\right)  \right)  \left(  y\right)  =\left(  \phi_{C,A}\left(
\left(  \gamma\ast\delta\right)  \otimes ab\right)  \right)  \left(  y\right)
$. In other words, $\phi_{C,A}\left(  \gamma\otimes a\right)  \ast\phi
_{C,A}\left(  \delta\otimes b\right)  =\phi_{C,A}\left(  \left(  \gamma
\ast\delta\right)  \otimes ab\right)  $. This proves (\ref{1.13.pf.2}).

Next let us show a slightly more general fact: Let us show that%
\begin{equation}
\phi_{C,A}\left(  t\right)  \ast\phi_{C,A}\left(  s\right)  =\phi_{C,A}\left(
ts\right)  \label{1.13.pf.3}%
\end{equation}
for all $n\in\mathbb{N}$, $m\in\mathbb{N}$, $t\in C_{n}^{\star
\operatorname*{gr}}\otimes A_{n}$ and $s\in C_{m}^{\star\operatorname*{gr}%
}\otimes A_{m}$. Here, the product $ts$ is to be understood as a product in
the $K$-algebra $C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$
(which contains $t$ and $s$ because both $C_{n}^{\star\operatorname*{gr}%
}\otimes A_{n}$ and $C_{m}^{\star\operatorname*{gr}}\otimes A_{m}$ canonically
inject into $C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$).

\textit{Proof of (\ref{1.13.pf.3}).} Let $n\in\mathbb{N}$, $m\in\mathbb{N}$,
$t\in C_{n}^{\star\operatorname*{gr}}\otimes A_{n}$ and $s\in C_{m}%
^{\star\operatorname*{gr}}\otimes A_{m}$ be arbitrary.

Since $t$ is a tensor in $C_{n}^{\star\operatorname*{gr}}\otimes A_{n}$, we
can write $t$ in the form $t=\sum\limits_{i=1}^{I}\lambda_{i}\gamma_{i}\otimes
a_{i}$ for some $I\in\mathbb{N}$, some elements $\lambda_{1}$, $\lambda_{2}$,
$...$, $\lambda_{I}$ of $K$, some elements $\gamma_{1}$, $\gamma_{2}$, $...$,
$\gamma_{I}$ of $C_{n}^{\star\operatorname*{gr}}$, and some elements $a_{1}$,
$a_{2}$, $...$, $a_{I}$ of $A_{n}$. Consider this $I$, these $\lambda_{1}$,
$\lambda_{2}$, $...$, $\lambda_{I}$, these $\gamma_{1}$, $\gamma_{2}$, $...$,
$\gamma_{I}$, and these $a_{1}$, $a_{2}$, $...$, $a_{I}$.

Since $s$ is a tensor in $C_{m}^{\star\operatorname*{gr}}\otimes A_{m}$, we
can write $s$ in the form $s=\sum\limits_{j=1}^{J}\nu_{j}\delta_{j}\otimes
b_{j}$ for some $J\in\mathbb{N}$, some elements $\nu_{1}$, $\nu_{2}$, $...$,
$\nu_{J}$ of $K$, some elements $\delta_{1}$, $\delta_{2}$, $...$, $\delta
_{J}$ of $C_{m}^{\star\operatorname*{gr}}$, and some elements $b_{1}$, $b_{2}%
$, $...$, $b_{J}$ of $A_{m}$. Consider this $J$, these $\nu_{1}$, $\nu_{2}$,
$...$, $\nu_{J}$, these $\delta_{1}$, $\delta_{2}$, $...$, $\delta_{J}$, and
these $b_{1}$, $b_{2}$, $...$, $b_{J}$.

Now,%
\begin{align}
\phi_{C,A}\underbrace{\left(  t\right)  }_{=\sum\limits_{i=1}^{I}\lambda
_{i}\gamma_{i}\otimes a_{i}}\ast\phi_{C,A}\underbrace{\left(  s\right)
}_{=\sum\limits_{j=1}^{J}\nu_{j}\delta_{j}\otimes b_{j}}  &  =\underbrace{\phi
_{C,A}\left(  \sum\limits_{i=1}^{I}\lambda_{i}\gamma_{i}\otimes a_{i}\right)
}_{\substack{=\sum\limits_{i=1}^{I}\lambda_{i}\phi_{C,A}\left(  \gamma
_{i}\otimes a_{i}\right)  \\\text{(since }\phi_{C,A}\text{ is }%
K\text{-linear)}}}\ast\underbrace{\phi_{C,A}\left(  \sum\limits_{j=1}^{J}%
\nu_{j}\delta_{j}\otimes b_{j}\right)  }_{\substack{=\sum\limits_{j=1}^{J}%
\nu_{j}\phi_{C,A}\left(  \delta_{j}\otimes b_{j}\right)  \\\text{(since }%
\phi_{C,A}\text{ is }K\text{-linear)}}}\nonumber\\
&  =\left(  \sum\limits_{i=1}^{I}\lambda_{i}\phi_{C,A}\left(  \gamma
_{i}\otimes a_{i}\right)  \right)  \ast\left(  \sum\limits_{j=1}^{J}\nu
_{j}\phi_{C,A}\left(  \delta_{j}\otimes b_{j}\right)  \right) \nonumber\\
&  =\sum\limits_{i=1}^{I}\sum\limits_{j=1}^{J}\lambda_{i}\nu_{j}%
\underbrace{\phi_{C,A}\left(  \gamma_{i}\otimes a_{i}\right)  \ast\phi
_{C,A}\left(  \delta_{j}\otimes b_{j}\right)  }_{\substack{=\phi_{C,A}\left(
\left(  \gamma_{i}\ast\delta_{j}\right)  \otimes a_{i}b_{j}\right)
\\\text{(by (\ref{1.13.pf.2}), applied to }\gamma_{i}\text{, }a_{i}\text{,
}\delta_{j}\text{ and }b_{j}\\\text{instead of }\gamma\text{, }a\text{,
}\delta\text{ and }b\text{)}}}\nonumber\\
&  =\sum\limits_{i=1}^{I}\sum\limits_{j=1}^{J}\lambda_{i}\nu_{j}\phi
_{C,A}\left(  \left(  \gamma_{i}\ast\delta_{j}\right)  \otimes a_{i}%
b_{j}\right)  . \label{1.13.pf.4}%
\end{align}
Meanwhile, multiplying the equations $t=\sum\limits_{i=1}^{I}\lambda_{i}%
\gamma_{i}\otimes a_{i}$ and $s=\sum\limits_{j=1}^{J}\nu_{j}\delta_{j}\otimes
b_{j}$, we obtain%
\begin{align*}
ts  &  =\left(  \sum\limits_{i=1}^{I}\lambda_{i}\gamma_{i}\otimes
a_{i}\right)  \left(  \sum\limits_{j=1}^{J}\nu_{j}\delta_{j}\otimes
b_{j}\right)  =\sum\limits_{i=1}^{I}\sum\limits_{j=1}^{J}\lambda_{i}\nu
_{j}\underbrace{\left(  \gamma_{i}\otimes a_{i}\right)  \left(  \delta
_{j}\otimes b_{j}\right)  }_{\substack{=\left(  \gamma_{i}\ast\delta
_{j}\right)  \otimes a_{i}b_{j}\\\text{(by the definition of the product in
}C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A\text{)}}}\\
&  =\sum\limits_{i=1}^{I}\sum\limits_{j=1}^{J}\lambda_{i}\nu_{j}\left(
\gamma_{i}\ast\delta_{j}\right)  \otimes a_{i}b_{j},
\end{align*}
so that%
\[
\phi_{C,A}\left(  ts\right)  =\phi_{C,A}\left(  \sum\limits_{i=1}^{I}%
\sum\limits_{j=1}^{J}\lambda_{i}\nu_{j}\left(  \gamma_{i}\ast\delta
_{j}\right)  \otimes a_{i}b_{j}\right)  =\sum\limits_{i=1}^{I}\sum
\limits_{j=1}^{J}\lambda_{i}\nu_{j}\phi_{C,A}\left(  \left(  \gamma_{i}%
\ast\delta_{j}\right)  \otimes a_{i}b_{j}\right)
\]
(since $\phi_{C,A}$ is $K$-linear). Compared to (\ref{1.13.pf.4}), this yields
$\phi_{C,A}\left(  t\right)  \ast\phi_{C,A}\left(  s\right)  =\phi
_{C,A}\left(  ts\right)  $. This proves (\ref{1.13.pf.3}).

Next we are going to show that%
\begin{equation}
\phi_{C,A}\left(  t\right)  \ast\phi_{C,A}\left(  s\right)  =\phi_{C,A}\left(
ts\right)  \label{1.13.pf.5}%
\end{equation}
for any $t\in C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$ and
$s\in C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$.

\textit{Proof of (\ref{1.13.pf.5}).} Let $t\in C^{\star\operatorname*{gr}%
}\widehat{\underset{=}{\otimes}}A$ and $s\in C^{\star\operatorname*{gr}%
}\widehat{\underset{=}{\otimes}}A$ be arbitrary.

Write the family $\widehat{\operatorname*{combine}\nolimits_{C^{\star
\operatorname*{gr}},A}}^{-1}\left(  t\right)  \in\prod\limits_{i\in\mathbb{N}%
}\left(  C_{i}^{\star\operatorname*{gr}}\otimes A_{i}\right)  $ in the form
$\left(  t_{i}\right)  _{i\in\mathbb{N}}$. Then,
\begin{align*}
t  &  =\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}%
},A}}\left(  \left(  t_{i}\right)  _{i\in\mathbb{N}}\right)  =\sum
\limits_{i\in\mathbb{N}}t_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1.13.pf.0}), applied to }\left(  a_{i}\right)  _{i\in\mathbb{N}}=\left(
t_{i}\right)  _{i\in\mathbb{N}}\right) \\
&  =\sum\limits_{n\in\mathbb{N}}t_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{here,
we renamed the index }i\text{ as }n\right)  ,
\end{align*}
so that $\phi_{C,A}\left(  t\right)  =\phi_{C,A}\left(  \sum\limits_{n\in
\mathbb{N}}t_{n}\right)  =\sum\limits_{n\in\mathbb{N}}\phi_{C,A}\left(
t_{n}\right)  $ (since $\phi_{C,A}$ is a continuous $K$-vector space homomorphism).

Write the family $\widehat{\operatorname*{combine}\nolimits_{C^{\star
\operatorname*{gr}},A}}^{-1}\left(  s\right)  \in\prod\limits_{i\in\mathbb{N}%
}\left(  C_{i}^{\star\operatorname*{gr}}\otimes A_{i}\right)  $ in the form
$\left(  s_{i}\right)  _{i\in\mathbb{N}}$. Then,%
\begin{align*}
s  &  =\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}%
},A}}\left(  \left(  s_{i}\right)  _{i\in\mathbb{N}}\right)  =\sum
\limits_{i\in\mathbb{N}}s_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{1.13.pf.0}), applied to }\left(  a_{i}\right)  _{i\in\mathbb{N}}=\left(
s_{i}\right)  _{i\in\mathbb{N}}\right) \\
&  =\sum\limits_{m\in\mathbb{N}}s_{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{here,
we renamed the index }i\text{ as }m\right)  ,
\end{align*}
so that $\phi_{C,A}\left(  s\right)  =\phi_{C,A}\left(  \sum\limits_{m\in
\mathbb{N}}s_{m}\right)  =\sum\limits_{m\in\mathbb{N}}\phi_{C,A}\left(
s_{m}\right)  $ (since $\phi_{C,A}$ is a continuous $K$-vector space homomorphism).

Now it is time to notice that $\operatorname*{Hom}\nolimits_{K}\left(
C,A\right)  $ is a topological $K$-algebra, i. e., the convolution is a
continuous map $\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
\times\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  \rightarrow
\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $.\ \ \ \ \footnote{This
fact does not even require all conditions of Proposition 1.13. As long as $C$
is a graded $K$-coalgebra (not necessarily locally-finite) and $A$ is a
$K$-algebra (not necessarily graded), the convolution algebra
$\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ (with the right degree
topology) is a topological $K$-algebra. This follows from Lemma 1.14 further
below.}

Furthermore, $C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$ is
a topological $K$-algebra (in fact, for any two $K$-algebras $\mathfrak{A}$
and $\mathfrak{B}$, the $K$-algebra $\mathfrak{A}\widehat{\underset{=}{\otimes
}}\mathfrak{B}$ is a topological $K$-algebra).

Multiplying the equalities $\phi_{C,A}\left(  t\right)  =\sum\limits_{n\in
\mathbb{N}}\phi_{C,A}\left(  t_{n}\right)  $ and $\phi_{C,A}\left(  s\right)
=\sum\limits_{m\in\mathbb{N}}\phi_{C,A}\left(  s_{m}\right)  $, we obtain%
\begin{align}
\phi_{C,A}\left(  t\right)  \ast\phi_{C,A}\left(  s\right)   &  =\left(
\sum\limits_{n\in\mathbb{N}}\phi_{C,A}\left(  t_{n}\right)  \right)
\ast\left(  \sum\limits_{m\in\mathbb{N}}\phi_{C,A}\left(  s_{m}\right)
\right) \nonumber\\
&  =\sum\limits_{n\in\mathbb{N}}\sum\limits_{m\in\mathbb{N}}%
\underbrace{\left(  \phi_{C,A}\left(  t_{n}\right)  \right)  \ast\left(
\phi_{C,A}\left(  s_{m}\right)  \right)  }_{\substack{=\phi_{C,A}\left(
t_{n}s_{m}\right)  \\\text{(by (\ref{1.13.pf.3}), applied to }t_{n}\text{ and
}s_{m}\text{ instead of }t\text{ and }s\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  \text{ is a
topological }K\text{-algebra, so that}\\
\text{there is a ``distributive law'' for}\\
\text{convergent infinite sums in }\operatorname*{Hom}\nolimits_{K}\left(
C,A\right)
\end{array}
\right) \nonumber\\
&  =\sum\limits_{n\in\mathbb{N}}\sum\limits_{m\in\mathbb{N}}\phi_{C,A}\left(
t_{n}s_{m}\right)  . \label{1.13.pf.5.pf.3}%
\end{align}
On the other hand, multiplying the equalities $t=\sum\limits_{n\in\mathbb{N}%
}t_{n}$ and $s=\sum\limits_{m\in\mathbb{N}}s_{m}$, we obtain%
\begin{align*}
ts  &  =\left(  \sum\limits_{n\in\mathbb{N}}t_{n}\right)  \ast\left(
\sum\limits_{m\in\mathbb{N}}s_{m}\right)  =\sum\limits_{n\in\mathbb{N}}%
\sum\limits_{m\in\mathbb{N}}t_{n}\ast s_{m}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A\text{
is a topological }K\text{-algebra, so that}\\
\text{there is a ``distributive law'' for convergent infinite sums in
}C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A
\end{array}
\right)  ,
\end{align*}
so that%
\begin{align*}
\phi_{C,A}\left(  ts\right)   &  =\phi_{C,A}\left(  \sum\limits_{n\in
\mathbb{N}}\sum\limits_{m\in\mathbb{N}}t_{n}s_{m}\right)  =\sum\limits_{n\in
\mathbb{N}}\sum\limits_{m\in\mathbb{N}}\phi_{C,A}\left(  t_{n}s_{m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\phi_{C,A}\text{ is a continuous
}K\text{-linear map}\right) \\
&  =\phi_{C,A}\left(  t\right)  \ast\phi_{C,A}\left(  s\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1.13.pf.5.pf.3})}\right)  .
\end{align*}
Thus, (\ref{1.13.pf.5}) is proven.

Finally, let us recall that we denote the unity of a $K$-algebra $U$ by
$1_{U}$. Then, $1_{C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}%
}A}=1_{C^{\star\operatorname*{gr}}}\otimes1_{A}$, so that%
\begin{align*}
\phi_{C,A}\left(  1_{C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes
}}A}\right)   &  =\phi_{C,A}\left(  1_{C^{\star\operatorname*{gr}}}%
\otimes1_{A}\right) \\
&  =\left(  \text{the map }C\rightarrow A\text{ which sends every }x\in
C\text{ to }\underbrace{1_{C^{\star\operatorname*{gr}}}}_{=\epsilon}\left(
x\right)  1_{A}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1.13.b.1}), applied to
}n=0\text{, }\gamma=1_{C^{\star\operatorname*{gr}}}\text{ and }a=1_{A}\right)
\\
&  =\left(  \text{the map }C\rightarrow A\text{ which sends every }x\in
C\text{ to }\underbrace{\epsilon\left(  x\right)  1_{A}}_{=\eta\left(
\epsilon\left(  x\right)  \right)  =\left(  \eta\circ\epsilon\right)  \left(
x\right)  }\right) \\
&  =\left(  \text{the map }C\rightarrow A\text{ which sends every }x\in
C\text{ to }\left(  \eta\circ\epsilon\right)  \left(  x\right)  \right) \\
&  =\eta\circ\epsilon=1_{\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
}.
\end{align*}
Combined with the fact that (\ref{1.13.pf.5}) holds for any $t\in
C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$ and $s\in
C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$, this yields that
$\phi_{C,A}$ is a $K$-algebra homomorphism. This completes the proof of
Proposition 1.13 \textbf{(b)}.

\textbf{(c)} First of all, applying (\ref{1.13.pf.0}) to $\left(
a_{i}\right)  _{i\in\mathbb{N}}=\left(  \sum\limits_{\ell\in I_{i}}c_{i,\ell
}^{\star}\otimes f\left(  c_{i,\ell}\right)  \right)  _{i\in\mathbb{N}}$, we
obtain%
\[
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}\left(  \left(  \sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\otimes f\left(
c_{i,\ell}\right)  \right)  _{i\in\mathbb{N}}\right)  =\sum\limits_{i\in
\mathbb{N}}\sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\otimes f\left(
c_{i,\ell}\right)  =\sum\limits_{k\in\mathbb{N}}\sum\limits_{\ell\in I_{k}%
}c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell}\right)
\]
(here, we renamed the index $i$ as $k$ in the first sum). Thus,%
\[
\left(  \sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\otimes f\left(
c_{i,\ell}\right)  \right)  _{i\in\mathbb{N}}=\widehat{\operatorname*{combine}%
\nolimits_{C^{\star\operatorname*{gr}},A}}^{-1}\left(  \sum\limits_{k\in
\mathbb{N}}\sum\limits_{\ell\in I_{k}}c_{k,\ell}^{\star}\otimes f\left(
c_{k,\ell}\right)  \right)  .
\]


Now let $x\in C$ be arbitrary. For every $i\in\mathbb{N}$, let $x_{i}$ be the
$i$-th graded component of $x$. Then, $x=\sum\limits_{i\in\mathbb{N}}x_{i}$.

Since $\phi_{C,A}=\operatorname*{dirsum}\nolimits_{C,A}\circ\left(
\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \circ
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}$, we have%
\begin{align*}
&  \phi_{C,A}\left(  \sum\limits_{k\in\mathbb{N}}\sum\limits_{\ell\in I_{k}%
}c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell}\right)  \right) \\
&  =\left(  \operatorname*{dirsum}\nolimits_{C,A}\circ\left(  \prod
\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \circ
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}\right)  \left(  \sum\limits_{k\in\mathbb{N}}\sum\limits_{\ell\in I_{k}%
}c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell}\right)  \right) \\
&  =\operatorname*{dirsum}\nolimits_{C,A}\left(  \left(  \prod\limits_{i\in
\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \underbrace{\left(
\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}^{-1}\left(  \sum\limits_{k\in\mathbb{N}}\sum\limits_{\ell\in I_{k}}%
c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell}\right)  \right)  \right)
}_{=\left(  \sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\otimes f\left(
c_{i,\ell}\right)  \right)  _{i\in\mathbb{N}}}\right) \\
&  =\operatorname*{dirsum}\nolimits_{C,A}\underbrace{\left(  \left(
\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\right)  \left(  \sum
\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell}\right)
\right)  _{i\in\mathbb{N}}\right)  }_{\substack{=\left(  \rho_{C_{i},A_{i}%
}\left(  \sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\otimes f\left(
c_{i,\ell}\right)  \right)  \right)  _{i\in\mathbb{N}}\\\text{(by the
definition of }\prod\limits_{i\in\mathbb{N}}\rho_{C_{i},A_{i}}\text{)}}}\\
&  =\operatorname*{dirsum}\nolimits_{C,A}\left(  \left(  \rho_{C_{i},A_{i}%
}\left(  \sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\otimes f\left(
c_{i,\ell}\right)  \right)  \right)  _{i\in\mathbb{N}}\right)  =\bigoplus
\limits_{i\in\mathbb{N}}\underbrace{\left(  \rho_{C_{i},A_{i}}\left(
\sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell
}\right)  \right)  \right)  }_{\substack{=\sum\limits_{\ell\in I_{i}}%
\rho_{C_{i},A_{i}}\left(  c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell
}\right)  \right)  \\\text{(since }\rho_{C_{i},A_{i}}\text{ is }%
K\text{-linear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\operatorname*{dirsum}\nolimits_{C,A}\right) \\
&  =\bigoplus\limits_{i\in\mathbb{N}}\left(  \sum\limits_{\ell\in I_{i}}%
\rho_{C_{i},A_{i}}\left(  c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell
}\right)  \right)  \right)  ,
\end{align*}
so that%
\begin{align}
&  \left(  \phi_{C,A}\left(  \sum\limits_{k\in\mathbb{N}}\sum\limits_{\ell\in
I_{k}}c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell}\right)  \right)  \right)
\left(  x\right) \nonumber\\
&  =\left(  \bigoplus\limits_{i\in\mathbb{N}}\left(  \sum\limits_{\ell\in
I_{i}}\rho_{C_{i},A_{i}}\left(  c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell
}\right)  \right)  \right)  \right)  \left(  x\right) \nonumber\\
&  =\sum\limits_{i\in\mathbb{N}}\left(  \sum\limits_{\ell\in I_{i}}\rho
_{C_{i},A_{i}}\left(  c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell}\right)
\right)  \right)  \left(  x_{i}\right) \nonumber\\
&  =\sum\limits_{i\in\mathbb{N}}\sum\limits_{\ell\in I_{i}}\left(  \rho
_{C_{i},A_{i}}\left(  c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell}\right)
\right)  \right)  \left(  x_{i}\right)  . \label{1.13.pf.7}%
\end{align}


Recall that, for every $i\in\mathbb{N}$, the map $\rho_{C_{i},A_{i}}$ was
defined as the $K$-linear map%
\begin{align*}
\left(  C_{i}\right)  ^{\star}\otimes A_{i}  &  \rightarrow\operatorname*{Hom}%
\nolimits_{K}\left(  C_{i},A_{i}\right)  ,\\
f\otimes\xi &  \mapsto\left(  \text{the map }C_{i}\rightarrow A_{i}\text{
which sends every }x\in C_{i}\text{ to }f\left(  x\right)  \xi\right)  .
\end{align*}
Hence, for every $i\in\mathbb{N}$ and every $\ell\in I_{i}$, we have
\[
\rho_{C_{i},A_{i}}\left(  c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell
}\right)  \right)  =\left(  \text{the map }C_{i}\rightarrow A_{i}\text{ which
sends every }x\in C_{i}\text{ to }c_{i,\ell}^{\star}\left(  x\right)  f\left(
c_{i,\ell}\right)  \right)  .
\]
Hence, for every $i\in\mathbb{N}$ and every $\ell\in I_{i}$, we have%
\begin{equation}
\left(  \rho_{C_{i},A_{i}}\left(  c_{i,\ell}^{\star}\otimes f\left(
c_{i,\ell}\right)  \right)  \right)  \left(  x_{i}\right)  =c_{i,\ell}^{\star
}\left(  x_{i}\right)  f\left(  c_{i,\ell}\right)  \label{1.13.pf.8}%
\end{equation}
(because $x_{i}\in C_{i}$). On the other hand, for every $i\in\mathbb{N}$,
every $y\in C_{i}$ satisfies $\sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star
}\left(  y\right)  c_{i,\ell}=y$ (because $\left(  c_{i,\ell}^{\star}\right)
_{\ell\in I_{i}}$ is the basis of $C_{i}^{\star\operatorname*{gr}}$ dual to
the basis $\left(  c_{i,\ell}\right)  _{\ell\in I_{i}}$ of $C_{i}$). Applied
to $y=x_{i}$, this yields
\begin{equation}
\sum\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\left(  x_{i}\right)  c_{i,\ell
}=x_{i} \label{1.13.pf.9}%
\end{equation}
for every $i\in\mathbb{N}$.

The equality (\ref{1.13.pf.7}) now becomes%
\begin{align*}
&  \left(  \phi_{C,A}\left(  \sum\limits_{k\in\mathbb{N}}\sum\limits_{\ell\in
I_{k}}c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell}\right)  \right)  \right)
\left(  x\right) \\
&  =\sum\limits_{i\in\mathbb{N}}\sum\limits_{\ell\in I_{i}}\underbrace{\left(
\rho_{C_{i},A_{i}}\left(  c_{i,\ell}^{\star}\otimes f\left(  c_{i,\ell
}\right)  \right)  \right)  \left(  x_{i}\right)  }_{\substack{=c_{i,\ell
}^{\star}\left(  x_{i}\right)  f\left(  c_{i,\ell}\right)  \\\text{(by
(\ref{1.13.pf.8}))}}}=\sum\limits_{i\in\mathbb{N}}\underbrace{\sum
\limits_{\ell\in I_{i}}c_{i,\ell}^{\star}\left(  x_{i}\right)  f\left(
c_{i,\ell}\right)  }_{\substack{=f\left(  \sum\limits_{\ell\in I_{i}}%
c_{i,\ell}^{\star}\left(  x_{i}\right)  c_{i,\ell}\right)  \\\text{(since
}f\text{ is }K\text{-linear)}}}\\
&  =\sum\limits_{i\in\mathbb{N}}f\underbrace{\left(  \sum\limits_{\ell\in
I_{i}}c_{i,\ell}^{\star}\left(  x_{i}\right)  c_{i,\ell}\right)
}_{\substack{=x_{i}\\\text{(by (\ref{1.13.pf.9}))}}}=\sum\limits_{i\in
\mathbb{N}}f\left(  x_{i}\right)  =f\underbrace{\left(  \sum\limits_{i\in
\mathbb{N}}x_{i}\right)  }_{=x}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}f\text{ is }K\text{-linear}\right) \\
&  =f\left(  x\right)  .
\end{align*}


Now forget that we fixed $x$. We have thus proven that every $x\in C$
satisfies $\left(  \phi_{C,A}\left(  \sum\limits_{k\in\mathbb{N}}%
\sum\limits_{\ell\in I_{k}}c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell
}\right)  \right)  \right)  \left(  x\right)  =f\left(  x\right)  $. In other
words, \newline$\phi_{C,A}\left(  \sum\limits_{k\in\mathbb{N}}\sum
\limits_{\ell\in I_{k}}c_{k,\ell}^{\star}\otimes f\left(  c_{k,\ell}\right)
\right)  =f$. This proves Proposition 1.13 \textbf{(c)}.
\end{proof}

This completes the proof of Proposition 1.13, but we still need to show the
following fact that we used:

\begin{quote}
\textbf{Lemma 1.14.} Let $C$ be a graded $K$-coalgebra, and let $A$ be a
$K$-algebra. Consider the convolution $K$-algebra $\operatorname*{Hom}%
\nolimits_{K}\left(  C,A\right)  $, endowed with the right degree topology.
Then, $\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ is a topological
$K$-algebra, i. e., the convolution is a continuous map $\operatorname*{Hom}%
\nolimits_{K}\left(  C,A\right)  \times\operatorname*{Hom}\nolimits_{K}\left(
C,A\right)  \rightarrow\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $.
\end{quote}

Before we prove Lemma 1.14, two very trivial facts:

\begin{quote}
\textbf{Lemma 1.15.} Let $C$ be a graded $K$-coalgebra. Let $\mathbb{N}_{-}$
denote the set $\left\{  -1,0,1,2,...\right\}  $. For every $i\in
\mathbb{N}_{-}$, let $C^{\left(  i\right)  }$ denote the subspace $C_{0}%
+C_{1}+...+C_{i}$ of $C$. Then, for every $i\in\mathbb{N}_{-}$, the subset
$C^{\left(  i\right)  }$ is a subcoalgebra of $C$.

\textbf{Lemma 1.16.} Let $C$ be a $K$-coalgebra. Let $D$ be a subcoalgebra of
$C$. Let $A$ be a $K$-algebra. Let $f:C\rightarrow A$ and $g:C\rightarrow A$
be two $K$-linear maps. Then, $\left(  f\ast g\right)  \mid_{D}=\left(
f\mid_{D}\right)  \ast\left(  g\mid_{D}\right)  $.
\end{quote}

\begin{proof}
[Proof of Lemma 1.15.]Let $i\in\mathbb{N}_{-}$. We have%
\begin{equation}
C_{k}\subseteq C^{\left(  i\right)  }\text{ for every }k\in\left\{
0,1,...,i\right\}  . \label{1.15.pf.1}%
\end{equation}
\footnote{\textit{Proof.} Let $k\in\left\{  0,1,...,i\right\}  $. Then,
$C_{k}$ is an addend of the sum $C_{0}+C_{1}+...+C_{i}$. Hence, $C_{k}%
\subseteq C_{0}+C_{1}+...+C_{i}=C^{\left(  i\right)  }$, and thus
(\ref{1.15.pf.1}) is proven.}

Since $C$ is a graded coalgebra, we have $\Delta\left(  C_{n}\right)
\subseteq\sum\limits_{\ell=0}^{n}C_{\ell}\otimes C_{n-\ell}$ for every
$n\in\mathbb{N}$. Now, $C^{\left(  i\right)  }=C_{0}+C_{1}+...+C_{i}%
=\sum\limits_{n=0}^{i}C_{n}$, so that%
\begin{align*}
\Delta\left(  C^{\left(  i\right)  }\right)   &  =\Delta\left(  \sum
\limits_{n=0}^{i}C_{n}\right)  =\sum\limits_{n=0}^{i}\underbrace{\Delta\left(
C_{n}\right)  }_{\subseteq\sum\limits_{\ell=0}^{n}C_{\ell}\otimes C_{n-\ell}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is }K\text{-linear}%
\right) \\
&  \subseteq\sum\limits_{n=0}^{i}\sum\limits_{\ell=0}^{n}\underbrace{C_{\ell}%
}_{\substack{\subseteq C^{\left(  i\right)  }\\\text{(by (\ref{1.15.pf.1})
(applied to }k=\ell\text{),}\\\text{because }\ell\leq n\leq i\text{ and thus
}\ell\in\left\{  0,1,...,i\right\}  \text{)}}}\otimes\underbrace{C_{n-\ell}%
}_{\substack{\subseteq C^{\left(  i\right)  }\\\text{(by (\ref{1.15.pf.1})
(applied to }k=n-\ell\text{),}\\\text{because }n-\ell\leq n\leq i\text{ and
thus }n-\ell\in\left\{  0,1,...,i\right\}  \text{)}}}\\
&  \subseteq\sum\limits_{n=0}^{i}\sum\limits_{\ell=0}^{n}C^{\left(  i\right)
}\otimes C^{\left(  i\right)  }\subseteq C^{\left(  i\right)  }\otimes
C^{\left(  i\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }C^{\left(  i\right)  }\otimes
C^{\left(  i\right)  }\text{ is a }K\text{-vector space}\right)  .
\end{align*}
Hence, $C^{\left(  i\right)  }$ is a subcoalgebra of $C$. This proves Lemma 1.15.
\end{proof}

\begin{proof}
[Proof of Lemma 1.16.]Let $\iota$ denote the canonical inclusion map
$D\rightarrow C$. Then, $\iota$ is a $K$-coalgebra homomorphism, so that
$\Delta_{C}\circ\iota=\left(  \iota\otimes\iota\right)  \circ\Delta_{D}$.
However, by the definition of convolution, we have the two equalities $f\ast
g=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}$ and $\left(
f\mid_{D}\right)  \ast\left(  g\mid_{D}\right)  =\mu_{A}\circ\left(  \left(
f\mid_{D}\right)  \otimes\left(  g\mid_{D}\right)  \right)  \circ\Delta_{D}$.
Besides, $f\mid_{D}=f\circ\iota$ (because $\iota$ is the inclusion map
$D\rightarrow C$) and $g\mid_{D}=g\circ\iota$ (for the same reason). Now,%
\begin{align*}
\left(  f\ast g\right)  \mid_{D}  &  =\underbrace{\left(  f\ast g\right)
}_{=\mu_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}}\circ
\iota\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\iota\text{ is the inclusion map
}D\rightarrow C\right) \\
&  =\mu_{A}\circ\left(  f\otimes g\right)  \circ\underbrace{\Delta_{C}%
\circ\iota}_{=\left(  \iota\otimes\iota\right)  \circ\Delta_{D}}=\mu_{A}%
\circ\underbrace{\left(  f\otimes g\right)  \circ\left(  \iota\otimes
\iota\right)  }_{=\left(  f\circ\iota\right)  \otimes\left(  g\circ
\iota\right)  }\circ\Delta_{D}\\
&  =\mu_{A}\circ\left(  \underbrace{\left(  f\circ\iota\right)  }_{=f\mid_{D}%
}\otimes\underbrace{\left(  g\circ\iota\right)  }_{=g\mid_{D}}\right)
\circ\Delta_{D}=\mu_{A}\circ\left(  \left(  f\mid_{D}\right)  \otimes\left(
g\mid_{D}\right)  \right)  \circ\Delta_{D}\\
&  =\left(  f\mid_{D}\right)  \ast\left(  g\mid_{D}\right)  .
\end{align*}
This proves Lemma 1.16.
\end{proof}

\begin{proof}
[Proof of Lemma 1.14.]Let $\mathbb{N}_{-}$ denote the set $\left\{
-1,0,1,2,...\right\}  $. For every $i\in\mathbb{N}_{-}$, let $C^{\left(
i\right)  }$ denote the subspace $C_{0}+C_{1}+...+C_{i}$ of $C$. For any
$i\in\mathbb{N}_{-}$ and any $g\in\operatorname*{Hom}\nolimits_{K}\left(
C^{\left(  i\right)  },A\right)  $, let $\operatorname*{Hom}\nolimits_{K,i,g}%
\left(  C,A\right)  $ denote the subset%
\[
\left\{  f\in\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  \ \mid
\ f\mid_{C^{\left(  i\right)  }}=g\right\}  \ \ \ \ \ \ \ \ \ \ \text{of}%
\ \ \ \ \ \ \ \ \ \ \operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  .
\]


By the definition of the right degree topology (in Definition 0.4
\textbf{(b)}), the right degree topology on $\operatorname*{Hom}%
\nolimits_{K}\left(  C,A\right)  $ is the topology generated by the sets
$\operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  $ with $i\in
\mathbb{N}_{-}$ and $g\in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(
i\right)  },A\right)  $ (as open sets). In other words, the set%
\[
\left\{  \operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  \ \mid
\ i\in\mathbb{N}_{-};\ g\in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(
i\right)  },A\right)  \right\}
\]
is a basis of the right degree topology on $\operatorname*{Hom}\nolimits_{K}%
\left(  C,A\right)  $.

We recall a very easy fact from general topology:%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{If }\mathfrak{A}\text{ and }\mathfrak{B}\text{ are two topological
spaces, and }\mathcal{S}\text{ is a basis of the topology on }\mathfrak{B}%
\text{,}\\
\text{and if }T:\mathfrak{A}\rightarrow\mathfrak{B}\text{ is a map such that
}\left(  \text{the set }T^{-1}\left(  U\right)  \text{ is open for every }%
U\in\mathcal{S}\right)  \text{,}\\
\text{then }T\text{ is continuous}%
\end{array}
\right)  . \label{1.14.pf.1}%
\end{equation}
(In brief, this fact means that the continuity of a map needs not be checked
on every open set - it is enough to check it on the open sets of a basis.)

Let $\mathfrak{A}$ be the topological space $\operatorname*{Hom}%
\nolimits_{K}\left(  C,A\right)  \times\operatorname*{Hom}\nolimits_{K}\left(
C,A\right)  $. Let $\mathfrak{B}$ be the topological space
$\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $. Let
$T:\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  \times
\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  \rightarrow
\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ be the product map of
the $K$-algebra $\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $. (In
other words, $T$ is the map which takes any $\left(  f,g\right)
\in\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  \times
\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ to the convolution
$f\ast g$.) We are going to prove that $T$ is continuous.

Note that $\mathfrak{B}=\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $,
so that $\mathfrak{B}\times\mathfrak{B}=\operatorname*{Hom}\nolimits_{K}%
\left(  C,A\right)  \times\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
=\mathfrak{A}$.

Denote by $\mathcal{S}$ the set $\left\{  \operatorname*{Hom}\nolimits_{K,i,g}%
\left(  C,A\right)  \ \mid\ i\in\mathbb{N}_{-};\ g\in\operatorname*{Hom}%
\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)  \right\}  $.

We know that $\left\{  \operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)
\ \mid\ i\in\mathbb{N}_{-};\ g\in\operatorname*{Hom}\nolimits_{K}\left(
C^{\left(  i\right)  },A\right)  \right\}  $ is a basis of the topology on
$\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $. Since $\left\{
\operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  \ \mid\ i\in
\mathbb{N}_{-};\ g\in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(
i\right)  },A\right)  \right\}  =\mathcal{S}$ and $\operatorname*{Hom}%
\nolimits_{K}\left(  C,A\right)  =\mathfrak{B}$, this rewrites as follows: The
set $\mathcal{S}$ is a basis of the topology on $\mathfrak{B}$.

Now we will prove that%
\begin{equation}
\text{the set }T^{-1}\left(  U\right)  \text{ is open for every }%
U\in\mathcal{S}\text{.} \label{1.14.pf.2}%
\end{equation}


In order to prove this, we will show that every $i\in\mathbb{N}_{-}$ and every
$g\in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
$ satisfy%
\begin{align}
&  T^{-1}\left(  \operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)
\right) \nonumber\\
&  =\bigcup_{\substack{\left(  g_{1},g_{2}\right)  \in\operatorname*{Hom}%
\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)  \times
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)
. \label{1.14.pf.3}%
\end{align}


\textit{Proof of (\ref{1.14.pf.3}).} Let $i\in\mathbb{N}_{-}$ and
$g\in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
$.

\textbf{a)} Let $s$ be an element of $T^{-1}\left(  \operatorname*{Hom}%
\nolimits_{K,i,g}\left(  C,A\right)  \right)  $. Then, $s\in
\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  \times\operatorname*{Hom}%
\nolimits_{K}\left(  C,A\right)  $, so we can write $s$ in the form $s=\left(
s_{1},s_{2}\right)  $ for some $s_{1}\in\operatorname*{Hom}\nolimits_{K}%
\left(  C,A\right)  $ and $s_{2}\in\operatorname*{Hom}\nolimits_{K}\left(
C,A\right)  $. Consider these $s_{1}$ and $s_{2}$. Let $t_{1}=s_{1}%
\mid_{C^{\left(  i\right)  }}$ and $t_{2}=s_{2}\mid_{C^{\left(  i\right)  }}$.

By Lemma 1.15, we know that $C^{\left(  i\right)  }$ is a subcoalgebra of $C$.
Thus, Lemma 1.16 (applied to $C^{\left(  i\right)  }$, $s_{1}$ and $s_{2}$
instead of $D$, $f$ and $g$) yields%
\begin{equation}
\left(  s_{1}\ast s_{2}\right)  \mid_{C^{\left(  i\right)  }}%
=\underbrace{\left(  s_{1}\mid_{C^{\left(  i\right)  }}\right)  }_{=t_{1}}%
\ast\underbrace{\left(  s_{2}\mid_{C^{\left(  i\right)  }}\right)  }_{=t_{2}%
}=t_{1}\ast t_{2}. \label{1.14.pf.3.pf.1}%
\end{equation}


Since $s=\left(  s_{1},s_{2}\right)  $, we have $T\left(  s\right)  =T\left(
s_{1},s_{2}\right)  =s_{1}\ast s_{2}$ (because $T$ is the product map of the
$K$-algebra $\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $). Thus,%
\begin{align*}
s_{1}\ast s_{2}  &  =T\left(  s\right)  \in\operatorname*{Hom}%
\nolimits_{K,i,g}\left(  C,A\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}s\in T^{-1}\left(  \operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)
\right)  \right) \\
&  =\left\{  f\in\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
\ \mid\ f\mid_{C^{\left(  i\right)  }}=g\right\}  ,
\end{align*}
so that $\left(  s_{1}\ast s_{2}\right)  \mid_{C^{\left(  i\right)  }}=g$.
Comparing this with (\ref{1.14.pf.3.pf.1}), we find $t_{1}\ast t_{2}=g$. On
the other hand, the definition of $\operatorname*{Hom}\nolimits_{K,i,t_{1}%
}\left(  C,A\right)  $ says that%
\[
\operatorname*{Hom}\nolimits_{K,i,t_{1}}\left(  C,A\right)  =\left\{
f\in\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  \ \mid\ f\mid
_{C^{\left(  i\right)  }}=t_{1}\right\}  .
\]
Hence, from $s_{1}\mid_{C^{\left(  i\right)  }}=t_{1}$, we get $s_{1}%
\in\left\{  f\in\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
\ \mid\ f\mid_{C^{\left(  i\right)  }}=t_{1}\right\}  =\operatorname*{Hom}%
\nolimits_{K,i,t_{1}}\left(  C,A\right)  $. Similarly, $s_{2}\in
\operatorname*{Hom}\nolimits_{K,i,t_{2}}\left(  C,A\right)  $.

Since $s_{1}\in\operatorname*{Hom}\nolimits_{K,i,t_{1}}\left(  C,A\right)  $
and $s_{2}\in\operatorname*{Hom}\nolimits_{K,i,t_{2}}\left(  C,A\right)  $, we
have \newline$\left(  s_{1},s_{2}\right)  \in\operatorname*{Hom}%
\nolimits_{K,i,t_{1}}\left(  C,A\right)  \times\operatorname*{Hom}%
\nolimits_{K,i,t_{2}}\left(  C,A\right)  $. Hence,%
\begin{align*}
s  &  =\left(  s_{1},s_{2}\right)  \in\operatorname*{Hom}\nolimits_{K,i,t_{1}%
}\left(  C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,t_{2}}\left(
C,A\right) \\
&  \subseteq\bigcup_{\substack{\left(  g_{1},g_{2}\right)  \in
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
\times\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }t_{1}\ast t_{2}=g\right)  .
\end{align*}


Now forget that we fixed $s$. We have proven that every $s\in T^{-1}\left(
\operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  \right)  $ satisfies
$s\in\bigcup\limits_{\substack{\left(  g_{1},g_{2}\right)  \in
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
\times\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)
$. In other words,%
\begin{align}
&  T^{-1}\left(  \operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)
\right) \nonumber\\
&  \subseteq\bigcup_{\substack{\left(  g_{1},g_{2}\right)  \in
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
\times\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)
. \label{1.14.pf.3a}%
\end{align}


\textbf{b)} Now let $z$ be an element of $\bigcup\limits_{\substack{\left(
g_{1},g_{2}\right)  \in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(
i\right)  },A\right)  \times\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(
i\right)  },A\right)  ;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}%
\nolimits_{K,i,g_{1}}\left(  C,A\right)  \times\operatorname*{Hom}%
\nolimits_{K,i,g_{2}}\left(  C,A\right)  $. Then, there exists some $\left(
\gamma_{1},\gamma_{2}\right)  \in\operatorname*{Hom}\nolimits_{K}\left(
C^{\left(  i\right)  },A\right)  \times\operatorname*{Hom}\nolimits_{K}\left(
C^{\left(  i\right)  },A\right)  $ such that $\gamma_{1}\ast\gamma_{2}=g$ and
$z\in\operatorname*{Hom}\nolimits_{K,i,\gamma_{1}}\left(  C,A\right)
\times\operatorname*{Hom}\nolimits_{K,i,\gamma_{2}}\left(  C,A\right)  $.
Consider this $\left(  \gamma_{1},\gamma_{2}\right)  $.

Since $z\in\operatorname*{Hom}\nolimits_{K,i,\gamma_{1}}\left(  C,A\right)
\times\operatorname*{Hom}\nolimits_{K,i,\gamma_{2}}\left(  C,A\right)  $, we
can write $z$ in the form $z=\left(  z_{1},z_{2}\right)  $ for some $z_{1}%
\in\operatorname*{Hom}\nolimits_{K,i,\gamma_{1}}\left(  C,A\right)  $ and
$z_{2}\in\operatorname*{Hom}\nolimits_{K,i,\gamma_{2}}\left(  C,A\right)  $.
Consider these $z_{1}$ and $z_{2}$.

Since
\[
z_{1}\in\operatorname*{Hom}\nolimits_{K,i,\gamma_{1}}\left(  C,A\right)
=\left\{  f\in\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
\ \mid\ f\mid_{C^{\left(  i\right)  }}=\gamma_{1}\right\}
\]
(by the definition of $\operatorname*{Hom}\nolimits_{K,i,\gamma_{1}}\left(
C,A\right)  $), we have $z_{1}\mid_{C^{\left(  i\right)  }}=\gamma_{1}$.
Similarly, $z_{2}\mid_{C^{\left(  i\right)  }}=\gamma_{2}$.

Recall that $C^{\left(  i\right)  }$ is a subcoalgebra of $C$. Hence, Lemma
1.16 (applied to $C^{\left(  i\right)  }$, $z_{1}$ and $z_{2}$ instead of $D$,
$f$ and $g$) yields%
\begin{equation}
\left(  z_{1}\ast z_{2}\right)  \mid_{C^{\left(  i\right)  }}%
=\underbrace{\left(  z_{1}\mid_{C^{\left(  i\right)  }}\right)  }_{=\gamma
_{1}}\ast\underbrace{\left(  z_{2}\mid_{C^{\left(  i\right)  }}\right)
}_{=\gamma_{2}}=\gamma_{1}\ast\gamma_{2}=g. \label{1.14.pf.3b.pf.1}%
\end{equation}


Now, $z=\left(  z_{1},z_{2}\right)  $ yields $T\left(  z\right)  =T\left(
z_{1},z_{2}\right)  =z_{1}\ast z_{2}$ (since $T$ is the product map of the
$K$-algebra $\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $), so that%
\[
\left(  T\left(  z\right)  \right)  \mid_{C^{\left(  i\right)  }}=\left(
z_{1}\ast z_{2}\right)  \mid_{C^{\left(  i\right)  }}%
=g\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1.14.pf.3b.pf.1})}\right)  ,
\]
so that $T\left(  z\right)  \in\left\{  f\in\operatorname*{Hom}\nolimits_{K}%
\left(  C,A\right)  \ \mid\ f\mid_{C^{\left(  i\right)  }}=g\right\}
=\operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  $ and thus $z\in
T^{-1}\left(  \operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  \right)
$.

Now forget that we fixed $z$. We have thus shown that every \newline%
$z\in\bigcup\limits_{\substack{\left(  g_{1},g_{2}\right)  \in
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
\times\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)
$ satisfies $z\in T^{-1}\left(  \operatorname*{Hom}\nolimits_{K,i,g}\left(
C,A\right)  \right)  $. In other words,%
\begin{align*}
&  \bigcup\limits_{\substack{\left(  g_{1},g_{2}\right)  \in
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
\times\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)
\\
&  \subseteq T^{-1}\left(  \operatorname*{Hom}\nolimits_{K,i,g}\left(
C,A\right)  \right)  .
\end{align*}
Combining this relation with (\ref{1.14.pf.3a}), we obtain (\ref{1.14.pf.3}).
This proves (\ref{1.14.pf.3}).

\textit{Proof of (\ref{1.14.pf.2}).} We have $\mathcal{S}=\left\{
\operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  \ \mid\ i\in
\mathbb{N}_{-};\ g\in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(
i\right)  },A\right)  \right\}  $. Hence,%
\begin{equation}
\operatorname*{Hom}\nolimits_{K,i,h}\left(  C,A\right)  \in\mathcal{S}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}_{-}\text{ and }%
h\in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)  .
\label{1.14.pf.5}%
\end{equation}


Now let $U\in\mathcal{S}$. Then, $U\in\mathcal{S}=\left\{  \operatorname*{Hom}%
\nolimits_{K,i,g}\left(  C,A\right)  \ \mid\ i\in\mathbb{N}_{-};\ g\in
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
\right\}  $. Hence, there exists some $i\in\mathbb{N}_{-}$ and some
$g\in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
$ such that $U=\operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  $.
Consider these $i$ and $g$.

For every $\left(  g_{1},g_{2}\right)  \in\operatorname*{Hom}\nolimits_{K}%
\left(  C^{\left(  i\right)  },A\right)  \times\operatorname*{Hom}%
\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)  $, the subset
$\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(  C,A\right)  \times
\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)  $ of
$\mathfrak{A}$ is open.\footnote{\textit{Proof.} Let $\left(  g_{1}%
,g_{2}\right)  \in\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)
},A\right)  \times\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)
},A\right)  $. Then, $\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \in\mathcal{S}$ (by (\ref{1.14.pf.5}), applied to $h=g_{1}$) and
$\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)  \in\mathcal{S}$
(by (\ref{1.14.pf.5}), applied to $h=g_{2}$).
\par
Since $\mathcal{S}$ is a basis of the topology on $\mathfrak{B}$, every
element of $\mathcal{S}$ is an open subset of $\mathfrak{B}$. Thus, both
$\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(  C,A\right)  $ and
$\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)  $ are open
subsets of $\mathfrak{B}$ (since $\operatorname*{Hom}\nolimits_{K,i,g_{1}%
}\left(  C,A\right)  \in\mathcal{S}$ and $\operatorname*{Hom}%
\nolimits_{K,i,g_{2}}\left(  C,A\right)  \in\mathcal{S}$). Hence (by the
definition of the product topology), their cartesian product
$\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(  C,A\right)  \times
\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)  $ is an open
subset of $\mathfrak{B}\times\mathfrak{B}=\mathfrak{A}$, qed.} Thus, the set%
\[
\bigcup_{\substack{\left(  g_{1},g_{2}\right)  \in\operatorname*{Hom}%
\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)  \times
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)
\]
is a union of open sets, and therefore open itself. Since%
\begin{align*}
&  T^{-1}\left(  U\right)  =T^{-1}\left(  \operatorname*{Hom}\nolimits_{K,i,g}%
\left(  C,A\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{because
}U=\operatorname*{Hom}\nolimits_{K,i,g}\left(  C,A\right)  \right) \\
&  =\bigcup_{\substack{\left(  g_{1},g_{2}\right)  \in\operatorname*{Hom}%
\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)  \times
\operatorname*{Hom}\nolimits_{K}\left(  C^{\left(  i\right)  },A\right)
;\\g_{1}\ast g_{2}=g}}\operatorname*{Hom}\nolimits_{K,i,g_{1}}\left(
C,A\right)  \times\operatorname*{Hom}\nolimits_{K,i,g_{2}}\left(  C,A\right)
\end{align*}
(by (\ref{1.14.pf.3})), this rewrites as follows: The set $T^{-1}\left(
U\right)  $ is open. This proves (\ref{1.14.pf.2}).

Now that (\ref{1.14.pf.2}) is proven, we can apply (\ref{1.14.pf.1}) and
conclude that $T$ is continuous. Since $T:\operatorname*{Hom}\nolimits_{K}%
\left(  C,A\right)  \times\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)
\rightarrow\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ is the
product map of the $K$-algebra $\operatorname*{Hom}\nolimits_{K}\left(
C,A\right)  $, this shows that the product map of the $K$-algebra
$\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ is continuous. Thus,
$\operatorname*{Hom}\nolimits_{K}\left(  C,A\right)  $ is a topological
$K$-algebra (since we already know that $\operatorname*{Hom}\nolimits_{K}%
\left(  C,A\right)  $ is a topological $K$-vector space). This proves Lemma 1.14.
\end{proof}

Now, finally, the reason why were doing all of this:

\begin{proof}
[Alternative proof of Proposition 5.3.]First of all, let $C=Th\left(
X\right)  $ and $A=Th\left(  X\right)  $.

For every graded $K$-linear map $f:C\rightarrow A$, let $S_{f}$ denote the
element%
\[
\sum\limits_{k\in\mathbb{N}}\sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)
\in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes
x_{i_{k}}^{\star}\right)  \otimes f\left(  x_{i_{1}}\otimes...\otimes
x_{i_{k}}\right)  \in\mathcal{T}.
\]


Consider the elements $T\in\mathcal{T}$ and $S\in\mathcal{T}$ defined on page
1082 of Patras's paper. These elements were defined by%
\[
T=\sum\limits_{k\in\mathbb{N}^{\ast}}\sum\limits_{\left(  i_{1},i_{2}%
,...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}%
}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  \otimes\left(  x_{i_{1}%
}\otimes...\otimes x_{i_{k}}\right)
\]
and $S=1\otimes1+T$.

Recall that $I$ denotes the identity map $\operatorname*{id}%
\nolimits_{Th\left(  X\right)  }$ of $Th\left(  X\right)  $. Thus, $I$ is a
graded $K$-linear map $C\rightarrow A$ (since $C=Th\left(  X\right)  $ and
$A=Th\left(  X\right)  $), and we have%
\begin{align*}
S_{I}  &  =\sum\limits_{k\in\mathbb{N}}\sum\limits_{\left(  i_{1}%
,i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}%
}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  \otimes
\underbrace{I\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  }_{=\left(
x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }S_{I}\right) \\
&  =\sum\limits_{k\in\mathbb{N}}\sum\limits_{\left(  i_{1},i_{2}%
,...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}%
}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  \otimes\left(  x_{i_{1}%
}\otimes...\otimes x_{i_{k}}\right) \\
&  =\underbrace{\sum\limits_{\left(  i_{1},i_{2},...,i_{0}\right)  \in\left\{
1,2,...,n\right\}  ^{0}}\left(  x_{i_{1}}^{\star}\otimes...\otimes x_{i_{0}%
}^{\star}\right)  \otimes\left(  x_{i_{1}}\otimes...\otimes x_{i_{0}}\right)
}_{\substack{=\left(  \text{empty tensor product}\right)  \otimes\left(
\text{empty tensor product}\right)  \\\text{(since there exists only one
}\left(  i_{1},i_{2},...,i_{0}\right)  \in\left\{  1,2,...,n\right\}
^{0}\text{, and this}\\\left(  i_{1},i_{2},...,i_{0}\right)  \text{ is the
empty }0\text{-tuple)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{k\in\mathbb{N}^{\ast}}%
\sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)  \in\left\{
1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes x_{i_{k}%
}^{\star}\right)  \otimes\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)
}_{=T}\\
&  =\underbrace{\left(  \text{empty tensor product}\right)  }_{=1}%
\otimes\underbrace{\left(  \text{empty tensor product}\right)  }%
_{=1}+T=1\otimes1+T=S.
\end{align*}


Now let us show that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for every graded }K\text{-linear map }f:C\rightarrow A\text{, the
element}\\
S_{f}\text{ lies in }C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes
}}A\text{ and satisfies }\phi_{C,A}\left(  S_{f}\right)  =f
\end{array}
\right)  . \label{5.3.pf.1}%
\end{equation}


\textit{Proof of (\ref{5.3.pf.1}).} Let $f:C\rightarrow A$ be a graded
$K$-linear map. For every $k\in\mathbb{N}$, we have
\begin{align*}
&  \sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)  \in\left\{
1,2,...,n\right\}  ^{k}}\underbrace{\left(  x_{i_{1}}^{\star}\otimes...\otimes
x_{i_{k}}^{\star}\right)  }_{\in C_{k}^{\star\operatorname*{gr}}}%
\otimes\underbrace{f\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)
}_{\substack{\in A_{k}\\\text{(since }x_{i_{1}}\otimes...\otimes x_{i_{k}}\in
C_{k}\\\text{and since }f\text{ is graded)}}}\\
&  \in\sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)  \in\left\{
1,2,...,n\right\}  ^{k}}C_{k}^{\star\operatorname*{gr}}\otimes A_{k}\subseteq
C_{k}^{\star\operatorname*{gr}}\otimes A_{k}%
\end{align*}
(since $C_{k}^{\star\operatorname*{gr}}\otimes A_{k}$ is a $K$-vector space).
Hence,
\[
\left(  \sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)  \in\left\{
1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes x_{i_{k}%
}^{\star}\right)  \otimes f\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}%
}\right)  \right)  _{k\in\mathbb{N}}\in\prod\limits_{k\in\mathbb{N}}\left(
C_{k}^{\star\operatorname*{gr}}\otimes A_{k}\right)  .
\]


Now, recall (from Definition 0.5) that $\widehat{\operatorname*{combine}%
\nolimits_{C^{\star\operatorname*{gr}},A}}$ is the map%
\begin{align*}
\prod\limits_{i\in\mathbb{N}}\left(  C_{i}^{\star\operatorname*{gr}}\otimes
A_{i}\right)   &  \rightarrow C^{\star\operatorname*{gr}}%
\widehat{\underset{=}{\otimes}}A,\\
\left(  a_{i}\right)  _{i\in\mathbb{N}}  &  \mapsto\sum\limits_{i\in
\mathbb{N}}a_{i}.
\end{align*}
Renaming the indices $i$ as $k$ in this formula, we see that
$\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}}$
is the map%
\begin{align*}
\prod\limits_{k\in\mathbb{N}}\left(  C_{k}^{\star\operatorname*{gr}}\otimes
A_{k}\right)   &  \rightarrow C^{\star\operatorname*{gr}}%
\widehat{\underset{=}{\otimes}}A,\\
\left(  a_{k}\right)  _{k\in\mathbb{N}}  &  \mapsto\sum\limits_{k\in
\mathbb{N}}a_{k}.
\end{align*}
Thus,%
\begin{align*}
&  \widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}},A}%
}\left(  \left(  \sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)
\in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes
x_{i_{k}}^{\star}\right)  \otimes f\left(  x_{i_{1}}\otimes...\otimes
x_{i_{k}}\right)  \right)  _{k\in\mathbb{N}}\right) \\
&  =\sum\limits_{k\in\mathbb{N}}\sum\limits_{\left(  i_{1},i_{2}%
,...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}%
}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  \otimes f\left(
x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  =S_{f},
\end{align*}
so that%
\begin{align*}
S_{f}  &  =\widehat{\operatorname*{combine}\nolimits_{C^{\star
\operatorname*{gr}},A}}\left(  \left(  \sum\limits_{\left(  i_{1}%
,i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}%
}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  \otimes f\left(
x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  \right)  _{k\in\mathbb{N}}\right)
\\
&  \in\widehat{\operatorname*{combine}\nolimits_{C^{\star\operatorname*{gr}%
},A}}\left(  \prod\limits_{k\in\mathbb{N}}\left(  C_{k}^{\star
\operatorname*{gr}}\otimes A_{k}\right)  \right)  \subseteq C^{\star
\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A.
\end{align*}
Thus, we have proven the first part of the claim of (\ref{5.3.pf.1}).

Now to the second part: For every $k\in\mathbb{N}$, notice that $\left(
x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  _{\left(  i_{1},i_{2}%
,...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}$ is a basis of the
$K$-vector space $C_{k}$ (since $C=Th\left(  X\right)  $), and $\left(
x_{i_{1}}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  _{\left(
i_{1},i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}$ is the
basis of the $K$-vector space $C_{k}^{\star\operatorname*{gr}}$ dual to this
basis $\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  _{\left(
i_{1},i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}$. Hence,
Proposition 1.13 \textbf{(c)} (applied to $\left(  c_{k,\ell}\right)
_{\ell\in I_{k}}=\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)
_{\left(  i_{1},i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}$
and $\left(  c_{k,\ell}^{\star}\right)  _{\ell\in I_{k}}=\left(  x_{i_{1}%
}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  _{\left(  i_{1}%
,i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}$) yields that%
\[
f=\phi_{C,A}\underbrace{\left(  \sum\limits_{k\in\mathbb{N}}\sum
\limits_{\left(  i_{1},i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}
^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)
\otimes f\left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  \right)
}_{=S_{f}}=\phi_{C,A}\left(  S_{f}\right)  .
\]


Thus we have shown that $S_{f}\in C^{\star\operatorname*{gr}}%
\widehat{\underset{=}{\otimes}}A$ and that $\phi_{C,A}\left(  S_{f}\right)
=f$. Hence, (\ref{5.3.pf.1}) is proven.

Applying (\ref{5.3.pf.1}) to $f=I$, we conclude that $S_{I}$ lies in
$C^{\star\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$ and satisfies
$\phi_{C,A}\left(  S_{I}\right)  =I$.

Since $\phi_{C,A}$ commutes with taking the logarithm (because $\phi_{C,A}$ is
a continuous $K$-algebra homomorphism, and because every continuous
$K$-algebra homomorphism commutes with taking the logarithm), we have
$\log\left(  \phi_{C,A}\left(  S_{I}\right)  \right)  =\phi_{C,A}\left(
\log\left(  S_{I}\right)  \right)  $. Thus,%
\[
\log\underbrace{I}_{=\phi_{C,A}\left(  S_{I}\right)  }=\log\left(  \phi
_{C,A}\left(  S_{I}\right)  \right)  =\phi_{C,A}\left(  \log
\underbrace{\left(  S_{I}\right)  }_{=S}\right)  =\phi_{C,A}\left(  \log
S\right)  .
\]
However, applying (\ref{5.3.pf.1}) to $f=\log I$ (this is allowed since $\log
I$ is graded), we conclude that $S_{\log I}$ lies in $C^{\star
\operatorname*{gr}}\widehat{\underset{=}{\otimes}}A$ and satisfies $\phi
_{C,A}\left(  S_{\log I}\right)  =\log I$. Thus, $\phi_{C,A}\left(  S_{\log
I}\right)  =\log I=\phi_{C,A}\left(  \log S\right)  $. Since $\phi_{C,A}$ is
injective (by Proposition 1.13 \textbf{(a)}), this yields $S_{\log I}=\log S$.
Hence,%
\begin{align}
\log S  &  =S_{\log I}=\sum\limits_{k\in\mathbb{N}}\sum\limits_{\left(
i_{1},i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}\left(
x_{i_{1}}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  \otimes\left(
\log I\right)  \left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }S_{\log I}\right)
\nonumber\\
&  =\underbrace{\sum\limits_{\left(  i_{1},i_{2},...,i_{0}\right)  \in\left\{
1,2,...,n\right\}  ^{0}}\left(  x_{i_{1}}^{\star}\otimes...\otimes x_{i_{0}%
}^{\star}\right)  \otimes\left(  \log I\right)  \left(  x_{i_{1}}%
\otimes...\otimes x_{i_{0}}\right)  }_{\substack{=\left(  \text{empty tensor
product}\right)  \otimes\left(  \log I\right)  \left(  \text{empty tensor
product}\right)  \\\text{(since there exists only one }\left(  i_{1}%
,i_{2},...,i_{0}\right)  \in\left\{  1,2,...,n\right\}  ^{0}\text{, and
this}\\\left(  i_{1},i_{2},...,i_{0}\right)  \text{ is the empty
}0\text{-tuple)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{k\in\mathbb{N}^{\ast}}\sum
\limits_{\left(  i_{1},i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}
^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)
\otimes\left(  \log I\right)  \left(  x_{i_{1}}\otimes...\otimes x_{i_{k}%
}\right) \nonumber\\
&  =\underbrace{\left(  \text{empty tensor product}\right)  }_{=1}%
\otimes\left(  \log I\right)  \underbrace{\left(  \text{empty tensor
product}\right)  }_{=1}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{k\in\mathbb{N}^{\ast}}%
}_{=\sum\limits_{k\geq1}}\sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)
\in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes
x_{i_{k}}^{\star}\right)  \otimes\left(  \log I\right)  \left(  x_{i_{1}%
}\otimes...\otimes x_{i_{k}}\right) \nonumber\\
&  =\underbrace{1\otimes\left(  \log I\right)  \left(  1\right)  }_{=0\text{
(since }\left(  \log I\right)  \left(  1\right)  =0\text{)}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{k\geq1}\sum\limits_{\left(  i_{1}%
,i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}%
}^{\star}\otimes...\otimes x_{i_{k}}^{\star}\right)  \otimes\left(  \log
I\right)  \left(  x_{i_{1}}\otimes...\otimes x_{i_{k}}\right) \nonumber\\
&  =\sum\limits_{k\geq1}\sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)
\in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes
x_{i_{k}}^{\star}\right)  \otimes\left(  \log I\right)  \left(  x_{i_{1}%
}\otimes...\otimes x_{i_{k}}\right)  . \label{5.3.pf.2}%
\end{align}


Now let us remember that Patras (on page 1082-1083) defines the elements
$Q_{i_{1},...,i_{k}}$ (for all $k\in\mathbb{N}^{\ast}$ and all $\left(
i_{1},i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}$) by
decomposing $\log S$ into the form%
\[
\log S=\sum\limits_{k\geq1}\sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)
\in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes
x_{i_{k}}^{\star}\right)  \otimes Q_{i_{1},...,i_{k}}.
\]
Due to (\ref{5.3.pf.2}), this decomposition is clearly given by%
\[
\log S=\sum\limits_{k\geq1}\sum\limits_{\left(  i_{1},i_{2},...,i_{k}\right)
\in\left\{  1,2,...,n\right\}  ^{k}}\left(  x_{i_{1}}^{\star}\otimes...\otimes
x_{i_{k}}^{\star}\right)  \otimes\left(  \log I\right)  \left(  x_{i_{1}%
}\otimes...\otimes x_{i_{k}}\right)  .
\]
Hence, $Q_{i_{1},...,i_{k}}=\left(  \log I\right)  \left(  x_{i_{1}}%
\otimes...\otimes x_{i_{k}}\right)  $ for every $k\in\mathbb{N}^{\ast}$ and
every $\left(  i_{1},i_{2},...,i_{k}\right)  \in\left\{  1,2,...,n\right\}
^{k}$.

Now, denote the connected cocommutative graded Hopf algebra $Th\left(
X\right)  $ by $H$. By (\ref{3.4.pf.6}) (applied to $i=1$), we have
$e^{1}=\dfrac{\left(  \log I\right)  ^{\ast1}}{1!}=\dfrac{\log I}{1}=\log I$.
Now, every $k\in\mathbb{N}^{\ast}$ and every $\left(  i_{1},i_{2}%
,...,i_{k}\right)  \in\left\{  1,2,...,n\right\}  ^{k}$ satisfy%
\begin{align*}
Q_{i_{1},...,i_{k}}  &  =\underbrace{\left(  \log I\right)  }_{=e^{1}}\left(
x_{i_{1}}\otimes...\otimes x_{i_{k}}\right)  =e^{1}\left(  x_{i_{1}}%
\otimes...\otimes x_{i_{k}}\right)  =e_{k}^{1}\left(  x_{i_{1}}\otimes
...\otimes x_{i_{k}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }x_{i_{1}}\otimes...\otimes
x_{i_{k}}\in H_{k}\right) \\
&  \in e_{k}^{1}\left(  H\right)  =H_{k}^{\left(  1\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }H_{k}^{\left(  1\right)  }\text{ was
defined as }e_{k}^{1}\left(  H\right)  \right) \\
&  \subseteq H^{\left(  1\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }H^{\left(  1\right)  }\text{ was defined as }\bigoplus
\limits_{n\geq1}H_{n}^{\left(  1\right)  }\text{, and thus contains }%
H_{k}^{\left(  1\right)  }\\
\text{(since }H_{k}^{\left(  1\right)  }\text{ is an addend of the sum
}\bigoplus\limits_{n\geq1}H_{n}^{\left(  1\right)  }\text{)}%
\end{array}
\right) \\
&  =\operatorname*{Prim}H\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma 4.1, since
}H\text{ is connected cocommutative graded}\right) \\
&  =\operatorname*{Prim}\left(  Th\left(  X\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }H=Th\left(  X\right)  \right)  .
\end{align*}
This proves both parts of Proposition 5.3.
\end{proof}

\parbreak


\textbf{Page 1085, proof of Lemma 6.3:} This proof is an induction proof, but
the induction base (i. e., the case $n=1$) is missing. Fortunately, this is
not much of a problem since the induction base can be done by a slight
simplification of the argument from the induction step.

The proof itself is nice, but let me rewrite it for more clarity. We start
with a lemma:

\begin{quote}
\textbf{Lemma 6.5.} Let $K$ be a field of characteristic $p$, where $p$ is a
prime. Let $C$ be a $K$-vector space, and let $D$ be a $K$-vector subspace of
$C$. Let $\alpha:C\rightarrow C$ be a $K$-linear map such that%
\begin{equation}
\left(  \alpha\left(  x\right)  =x\ \ \ \ \ \ \ \ \ \ \text{for every }x\in
D\right)  \label{6.5.1}%
\end{equation}
and%
\begin{equation}
\left(  \alpha\left(  x\right)  \equiv x\operatorname{mod}%
D\ \ \ \ \ \ \ \ \ \ \text{for every }x\in C\right)  . \label{6.5.2}%
\end{equation}
Then, $\alpha^{p}=\operatorname*{id}\nolimits_{C}$.
\end{quote}

\begin{proof}
[Proof of Lemma 6.5.]We claim that for every $\ell\in\mathbb{N}$, we have%
\begin{equation}
\alpha^{\ell}\left(  x\right)  =x+\ell\left(  \alpha\left(  x\right)
-x\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }x\in C. \label{6.5.pf.1}%
\end{equation}


\textit{Proof of (\ref{6.5.pf.1}).} We are going to prove (\ref{6.5.pf.1}) by
induction over $\ell$:

\textit{Induction base:} We have $\underbrace{\alpha^{0}}_{=\operatorname*{id}%
}\left(  x\right)  =\operatorname*{id}\left(  x\right)  =x=x+0\left(
\alpha\left(  x\right)  -x\right)  $ for every $x\in C$. In other words,
(\ref{6.5.pf.1}) holds for $\ell=0$. This completes the induction base.

\textit{Induction step:} Let $L\in\mathbb{N}$. Assume that (\ref{6.5.pf.1})
holds for $\ell=L$. We now must prove (\ref{6.5.pf.1}) for $\ell=L+1$.

Since (\ref{6.5.pf.1}) holds for $\ell=L$, we have $\alpha^{L}\left(
x\right)  =x+L\left(  \alpha\left(  x\right)  -x\right)  $ for every $x\in C$.
On the other hand, every $x\in C$ satisfies $\alpha\left(  x\right)  \equiv
x\operatorname{mod}D$ (by (\ref{6.5.2})) and thus $\alpha\left(  x\right)
-x\in D$.

Thus, for every $x\in C$, we have%
\begin{align*}
\alpha^{L+1}\left(  x\right)   &  =\alpha\left(  \underbrace{\alpha^{L}\left(
x\right)  }_{=x+L\left(  \alpha\left(  x\right)  -x\right)  }\right)
=\alpha\left(  x+L\left(  \alpha\left(  x\right)  -x\right)  \right) \\
&  =\underbrace{\alpha\left(  x\right)  }_{=x+\left(  \alpha\left(  x\right)
-x\right)  }+L\underbrace{\alpha\left(  \alpha\left(  x\right)  -x\right)
}_{\substack{=\alpha\left(  x\right)  -x\\\text{(by (\ref{6.5.1}) (applied to
}\alpha\left(  x\right)  -x\text{ instead of }x\text{),}\\\text{because
}\alpha\left(  x\right)  -x\in D\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\alpha\text{ is }K\text{-linear}\right) \\
&  =x+\underbrace{\left(  \alpha\left(  x\right)  -x\right)  +L\left(
\alpha\left(  x\right)  -x\right)  }_{=\left(  L+1\right)  \left(
\alpha\left(  x\right)  -x\right)  }=x+\left(  L+1\right)  \left(
\alpha\left(  x\right)  -x\right)  .
\end{align*}
In other words, (\ref{6.5.pf.1}) is proven for $\ell=L+1$. This completes the
induction step. Thus, the induction proof of (\ref{6.5.pf.1}) is complete.

Now, applying (\ref{6.5.pf.1}) to $\ell=p$, we obtain $\alpha^{p}\left(
x\right)  =x+p\left(  \alpha\left(  x\right)  -x\right)  $ for every $x\in C$.
Thus, every $x\in C$ satisfies%
\[
\alpha^{p}\left(  x\right)  =x+\underbrace{p\left(  \alpha\left(  x\right)
-x\right)  }_{\substack{=0\\\text{(since }K\text{ has characteristic
}p\text{)}}}=x=\operatorname*{id}\nolimits_{C}\left(  x\right)  .
\]
In other words, $\alpha^{p}=\operatorname*{id}\nolimits_{C}$. This proves
Lemma 6.5.
\end{proof}

\begin{quote}
\textbf{Lemma 6.6.} Let $K$ be a field of characteristic $p$, where $p$ is a
prime. Let $k$ be an integer not divisible by $p$. Let $C$ be a $K$-vector
space, and let $D$ be a $K$-vector subspace of $C$. Let $\beta:C\rightarrow C$
be a $K$-linear map such that%
\begin{equation}
\left(  \beta^{p-1}\left(  x\right)  =x\ \ \ \ \ \ \ \ \ \ \text{for every
}x\in D\right)  \label{6.6.1}%
\end{equation}
and%
\begin{equation}
\left(  \beta\left(  x\right)  \equiv kx\operatorname{mod}%
D\ \ \ \ \ \ \ \ \ \ \text{for every }x\in C\right)  . \label{6.6.2}%
\end{equation}
Then, $\left(  \beta^{p-1}\right)  ^{p}=\operatorname*{id}\nolimits_{C}$.
\end{quote}

\begin{proof}
[Proof of Lemma 6.6.]We claim that for every $\ell\in\mathbb{N}$, we have%
\begin{equation}
\beta^{\ell}\left(  x\right)  \equiv k^{\ell}x\operatorname{mod}%
D\ \ \ \ \ \ \ \ \ \ \text{for every }x\in C. \label{6.6.pf.1}%
\end{equation}


\textit{Proof of (\ref{6.6.pf.1}).} We are going to prove (\ref{6.6.pf.1}) by
induction over $\ell$:

\textit{Induction base:} We have $\underbrace{\beta^{0}}_{=\operatorname*{id}%
}\left(  x\right)  =\operatorname*{id}\left(  x\right)  =x$ and
$\underbrace{k^{0}}_{=1}x=x$ for every $x\in C$. Thus, $\beta^{0}\left(
x\right)  =x=k^{0}x$ for every $x\in C$. In other words, (\ref{6.6.pf.1})
holds for $\ell=0$. This completes the induction base.

\textit{Induction step:} Let $L\in\mathbb{N}$. Assume that (\ref{6.6.pf.1})
holds for $\ell=L$. We now must prove (\ref{6.6.pf.1}) for $\ell=L+1$.

Since (\ref{6.6.pf.1}) holds for $\ell=L$, we have
\begin{equation}
\beta^{L}\left(  x\right)  \equiv k^{L}x\operatorname{mod}%
D\ \ \ \ \ \ \ \ \ \ \text{for every }x\in C. \label{6.6.pf.2}%
\end{equation}
Thus, for every $x\in C$, we have%
\begin{align*}
\beta^{L+1}\left(  x\right)   &  =\beta^{L}\left(  \beta\left(  x\right)
\right)  \equiv k^{L}\underbrace{\beta\left(  x\right)  }_{\substack{\equiv
kx\operatorname{mod}D\\\text{(by (\ref{6.6.2}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{6.6.pf.2}), applied to }\beta\left(  x\right)  \text{ instead
of }x\right) \\
&  \equiv\underbrace{k^{L}k}_{=k^{L+1}}x=k^{L+1}x\operatorname{mod}D.
\end{align*}
In other words, (\ref{6.6.pf.1}) is proven for $\ell=L+1$. This completes the
induction step. Thus, the induction proof of (\ref{6.6.pf.1}) is complete.

Now notice that, by Fermat's Little Theorem, we have $k^{p-1}\equiv
1\operatorname{mod}p$ (since $p$ is a prime, and $k$ is an integer not
divisible by $p$). Hence, $k^{p-1}-1$ is divisible by $p$, so that $\left(
k^{p-1}-1\right)  x=0$ for every $x\in C$ (since $K$ has characteristic $p$).
On the other hand, applying (\ref{6.6.pf.1}) to $\ell=p-1$, we obtain
$\beta^{p-1}\left(  x\right)  \equiv k^{p-1}x\operatorname{mod}D$ for every
$x\in C$. Thus, every $x\in C$ satisfies%
\[
\beta^{p-1}\left(  x\right)  \equiv k^{p-1}x=x+\underbrace{\left(
k^{p-1}x-x\right)  }_{=\left(  k^{p-1}-1\right)  x=0}=x\operatorname{mod}D.
\]


Combined with (\ref{6.6.1}), this yields that we can apply Lemma 6.5 to
$\alpha=\beta^{p-1}$. This yields $\left(  \beta^{p-1}\right)  ^{p}%
=\operatorname*{id}\nolimits_{C}$, and thus Lemma 6.6 is proven.
\end{proof}

The next lemma we need is a basic property of connected graded bialgebras:

\begin{quote}
\textbf{Lemma 1.17.} Let $H$ be a connected graded bialgebra over a field $K$
(not necessarily of characteristic $p$). Let $n$ be a positive integer. Let
$x\in H_{n}$. Then,%
\[
\Delta\left(  x\right)  \in x\otimes1+1\otimes x+\sum_{k=1}^{n-1}H_{k}\otimes
H_{n-k}.
\]



\end{quote}

Note that this Lemma 1.17 is a slight rewriting of Proposition II.1.1 of [M],
but let us give a proof for the sake of completeness.

\begin{proof}
[Proof of Lemma 1.17.]In the following, $\operatorname*{id}$ will always
denote the identity map $\operatorname*{id}\nolimits_{H}$ of $H$.

We have $x\in H_{n}$ and thus%
\begin{align*}
\Delta\left(  x\right)   &  \in\Delta\left(  H_{n}\right)  \subseteq\sum
_{k=0}^{n}H_{k}\otimes H_{n-k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }H\text{
is a graded coalgebra}\right) \\
&  =H_{0}\otimes\underbrace{H_{n-0}}_{=H_{n}}+\sum\limits_{k=1}^{n-1}%
H_{k}\otimes H_{n-k}+H_{n}\otimes\underbrace{H_{n-n}}_{=H_{0}}\\
&  =H_{0}\otimes H_{n}+\sum\limits_{k=1}^{n-1}H_{k}\otimes H_{n-k}%
+H_{n}\otimes H_{0}.
\end{align*}
Hence, there exist some $u\in H_{0}\otimes H_{n}$, $v\in\sum\limits_{k=1}%
^{n-1}H_{k}\otimes H_{n-k}$ and $w\in H_{n}\otimes H_{0}$ such that
$\Delta\left(  x\right)  =u+v+w$. Consider these $u$, $v$ and $w$.

We now will prove the two equalities $u=x\otimes1$ and $w=1\otimes x$. The
proofs of these equalities are analogous (they only differ in the order of the
tensorands), so it will be enough to prove $w=1\otimes x$ only.

According to my definition of a connected graded bialgebra (see my remark
about ``Page 1070, fifth line of this page''), the map $\epsilon\mid_{H_{0}%
}:H_{0}\rightarrow K$ is an isomorphism (since $H$ is connected). Thus, the
map $\operatorname*{id}\otimes\left(  \epsilon\mid_{H_{0}}\right)  :H\otimes
H_{0}\rightarrow H\otimes K$ is an isomorphism.

Adding the relations $u\in H_{0}\otimes H_{n}$ and $v\in\sum\limits_{k=1}%
^{n-1}H_{k}\otimes H_{n-k}$, we obtain%
\[
u+v\in H_{0}\otimes H_{n}+\sum\limits_{k=1}^{n-1}H_{k}\otimes H_{n-k}%
=\sum\limits_{k=0}^{n-1}H_{k}\otimes H_{n-k},
\]
so that%
\begin{align*}
\left(  \operatorname*{id}\otimes\epsilon\right)  \left(  u+v\right)   &
\in\left(  \operatorname*{id}\otimes\epsilon\right)  \left(  \sum
\limits_{k=0}^{n-1}H_{k}\otimes H_{n-k}\right)  =\sum\limits_{k=0}%
^{n-1}\underbrace{\left(  \operatorname*{id}\otimes\epsilon\right)  \left(
H_{k}\otimes H_{n-k}\right)  }_{=\operatorname*{id}\left(  H_{k}\right)
\otimes\epsilon\left(  H_{n-k}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{id}\otimes
\epsilon\text{ is }K\text{-linear}\right) \\
&  =\sum\limits_{k=0}^{n-1}\operatorname*{id}\left(  H_{k}\right)
\otimes\underbrace{\epsilon\left(  H_{n-k}\right)  }%
_{\substack{=0\\\text{(since }n-k\geq1\text{ (because }k\leq n-1\text{), and
since}\\H\text{ is a graded coalgebra)}}}=\sum\limits_{k=0}^{n-1}%
\operatorname*{id}\left(  H_{k}\right)  \otimes0=0.
\end{align*}
In other words, $\left(  \operatorname*{id}\otimes\epsilon\right)  \left(
u+v\right)  =0$.

Let $\operatorname*{kan}\nolimits_{1}$ be the canonical isomorphism
$H\rightarrow H\otimes K$ which sends every $x\in H$ to $x\otimes1$. By the
axioms of a coalgebra, $\left(  \operatorname*{id}\otimes\epsilon\right)
\circ\Delta=\operatorname*{kan}\nolimits_{1}$ (since $H$ is a coalgebra).
Hence,%
\begin{align*}
\left(  \left(  \operatorname*{id}\otimes\epsilon\right)  \circ\Delta\right)
\left(  x\right)   &  =\operatorname*{kan}\nolimits_{1}\left(  x\right)
=\underbrace{x}_{=\operatorname*{id}\left(  x\right)  }\otimes\underbrace{1}%
_{\substack{=\epsilon\left(  1\right)  =\left(  \epsilon\mid_{H_{0}}\right)
\left(  1\right)  \\\text{(since }1\in H_{0}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{kan}%
\nolimits_{1}\right) \\
&  =\operatorname*{id}\left(  x\right)  \otimes\left(  \epsilon\mid_{H_{0}%
}\right)  \left(  1\right)  =\left(  \operatorname*{id}\otimes\left(
\epsilon\mid_{H_{0}}\right)  \right)  \left(  x\otimes1\right)  .
\end{align*}
Compared with%
\begin{align*}
\left(  \left(  \operatorname*{id}\otimes\epsilon\right)  \circ\Delta\right)
\left(  x\right)   &  =\left(  \operatorname*{id}\otimes\epsilon\right)
\left(  \underbrace{\Delta\left(  x\right)  }_{=u+v+w}\right)  =\left(
\operatorname*{id}\otimes\epsilon\right)  \left(  u+v+w\right) \\
&  =\underbrace{\left(  \operatorname*{id}\otimes\epsilon\right)  \left(
u+v\right)  }_{=0}+\left(  \operatorname*{id}\otimes\epsilon\right)  \left(
w\right)  =\left(  \operatorname*{id}\otimes\epsilon\right)  \left(  w\right)
\\
&  =\underbrace{\left(  \left(  \operatorname*{id}\otimes\epsilon\right)
\mid_{H\otimes H_{0}}\right)  }_{=\operatorname*{id}\otimes\left(
\epsilon\mid_{H_{0}}\right)  }\left(  w\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }w\in\underbrace{H_{n}}_{\subseteq H}\otimes H_{0}\subseteq
H\otimes H_{0}\right) \\
&  =\left(  \operatorname*{id}\otimes\left(  \epsilon\mid_{H_{0}}\right)
\right)  \left(  w\right)  ,
\end{align*}
this yields $\left(  \operatorname*{id}\otimes\left(  \epsilon\mid_{H_{0}%
}\right)  \right)  \left(  w\right)  =\left(  \operatorname*{id}\otimes\left(
\epsilon\mid_{H_{0}}\right)  \right)  \left(  x\otimes1\right)  $. Since
$\operatorname*{id}\otimes\left(  \epsilon\mid_{H_{0}}\right)  $ is injective
(because $\operatorname*{id}\otimes\left(  \epsilon\mid_{H_{0}}\right)  $ is
an isomorphism), this yields $w=x\otimes1$. As we have said, the proof of
$u=1\otimes x$ is similar to the proof of $w=x\otimes1$ that we just did
(except we have to switch the order of the tensorands). Thus, we now have%
\[
\Delta\left(  x\right)  =u+v+w=\underbrace{w}_{=x\otimes1}+\underbrace{u}%
_{=1\otimes x}+\underbrace{v}_{\in\sum\limits_{k=1}^{n-1}H_{k}\otimes H_{n-k}%
}\in x\otimes1+1\otimes x+\sum_{k=1}^{n-1}H_{k}\otimes H_{n-k}.
\]
This proves Lemma 1.17.
\end{proof}

\begin{quote}
\textbf{Corollary 1.18.} Let $H$ be a connected graded bialgebra over a field
$K$ (not necessarily of characteristic $p$). Let $\ell\in\mathbb{N}$.

\textbf{(a)} We have $\Psi^{\ell}\left(  1\right)  =1$ and $\left(  \Psi
^{\ell}-I\right)  \left(  H_{0}\right)  =0$.

\textbf{(b)} Let $n$ be a positive integer. Let $x\in H_{n}$. Then,%
\begin{equation}
\Psi^{\ell}\left(  x\right)  \equiv\ell x\operatorname{mod}\left(  \sum
_{i=1}^{n-1}H_{i}H_{n-i}\right)  \cap H_{n}. \label{1.18}%
\end{equation}



\end{quote}

\begin{proof}
[Proof of Corollary 1.18.]\textbf{(a)} There is a fact that every $K$-linear
map $f:H\rightarrow H$ satisfying $f\left(  1\right)  =1$ must also satisfy
$f^{\ast\nu}\left(  1\right)  =1$ for every $\nu\in\mathbb{N}$.

This fact is very easy to prove by induction over $\ell$ (the details are left
to the reader). Applying this fact to $f=I$ and $\nu=\ell$, we conclude that
$I^{\ast\ell}\left(  1\right)  =1$.

Since $\Psi^{\ell}$ was defined by $\Psi^{\ell}=I^{\ast\ell}$, we have
$\Psi^{\ell}\left(  1\right)  =I^{\ast\ell}\left(  1\right)  =1$.

On the other hand, since $H$ is connected, the map $\epsilon\mid_{H_{0}}%
:H_{0}\rightarrow K$ is an isomorphism.

Consider the family $\left(  1\right)  $ of vectors in the $K$-vector space
$H_{0}$. The image of this family under the isomorphism $\epsilon\mid_{H_{0}}$
is $\left(  \underbrace{\left(  \epsilon\mid_{H_{0}}\right)  \left(  1\right)
}_{=\epsilon\left(  1\right)  =1}\right)  =\left(  1\right)  $, and this is a
basis of the $K$-vector space $K$. Hence, the family $\left(  1\right)  $ is a
basis of the $K$-vector space $H_{0}$ (because every family of vectors in a
vector space whose image under some isomorphism is a basis must itself be a
basis). Hence, $H_{0}=K\cdot1$ (where $1$ denotes the unity of the $K$-algebra
$H$). Thus,%
\begin{align*}
\left(  \Psi^{\ell}-I\right)  \left(  H_{0}\right)   &  =\left(  \Psi^{\ell
}-I\right)  \left(  K\cdot1\right)  =K\cdot\underbrace{\left(  \Psi^{\ell
}-I\right)  \left(  1\right)  }_{=\Psi^{\ell}\left(  1\right)  -I\left(
1\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Psi^{\ell}-I\text{ is
}K\text{-linear}\right) \\
&  =K\cdot\left(  \underbrace{\Psi^{\ell}\left(  1\right)  }_{\substack{=1}%
}-\underbrace{I\left(  1\right)  }_{=1}\right)  =K\cdot\underbrace{\left(
1-1\right)  }_{=0}=0.
\end{align*}


This proves Corollary 1.18 \textbf{(a)}.

\textbf{(b)} Note first that $x\in H_{n}$ yields $\epsilon\left(  x\right)
\in\epsilon\left(  H_{n}\right)  =0$ (since $H$ is a graded coalgebra and
$n>0$), so that $\epsilon\left(  x\right)  =0$.

We are going to prove (\ref{1.18}) by induction over $\ell$:

\textit{Induction base:} We have
\[
\underbrace{\Psi^{0}}_{=\eta\circ\epsilon}\left(  x\right)  =\left(  \eta
\circ\epsilon\right)  \left(  x\right)  =\eta\underbrace{\left(
\epsilon\left(  x\right)  \right)  }_{=0}=\eta\left(  0\right)  =0\equiv
0x\operatorname{mod}\left(  \sum_{i=1}^{n-1}H_{i}H_{n-i}\right)  \cap H_{n}.
\]
Thus, (\ref{1.18}) holds for $\ell=0$. This completes the induction base.

\textit{Induction step:} Let $L\in\mathbb{N}$ be arbitrary. Assume that
(\ref{1.18}) holds for $\ell=L$. We now must show that (\ref{1.18}) holds for
$\ell=L+1$.

Since $\Psi^{L}$ and $\Psi^{L+1}$ are graded maps, we have $\Psi^{L}\left(
H_{n}\right)  \subseteq H_{n}$ and $\Psi^{L+1}\left(  H_{n}\right)  \subseteq
H_{n}$.

By the definition of $\Psi^{L}$, we have $\Psi^{L}=I^{\ast L}$. Also,
Corollary 1.18 \textbf{(a)} (applied to $\ell=L$) yields $\Psi^{L}\left(
1\right)  =1$ and $\left(  \Psi^{L}-I\right)  \left(  H_{0}\right)  =0$.

By the definition of $\Psi^{L+1}$, we have%
\[
\Psi^{L+1}=I^{\ast\left(  L+1\right)  }=I\ast\underbrace{I^{\ast L}}%
_{=\Psi^{L}}=I\ast\Psi^{L}=\mu\circ\left(  I\otimes\Psi^{L}\right)
\circ\Delta
\]
(by the definition of convolution). Thus,%
\begin{align*}
\Psi^{L+1}\left(  x\right)   &  =\left(  \mu\circ\left(  I\otimes\Psi
^{L}\right)  \circ\Delta\right)  \left(  x\right)  =\mu\left(  \left(
I\otimes\Psi^{L}\right)  \underbrace{\left(  \Delta\left(  x\right)  \right)
}_{\substack{\in x\otimes1+1\otimes x+\sum\limits_{k=1}^{n-1}H_{k}\otimes
H_{n-k}\\\text{(by Lemma 1.17)}}}\right) \\
&  \in\mu\underbrace{\left(  \left(  I\otimes\Psi^{L}\right)  \left(
x\otimes1+1\otimes x+\sum\limits_{k=1}^{n-1}H_{k}\otimes H_{n-k}\right)
\right)  }_{\substack{=\left(  I\otimes\Psi^{L}\right)  \left(  x\otimes
1\right)  +\left(  I\otimes\Psi^{L}\right)  \left(  1\otimes x\right)
+\sum\limits_{k=1}^{n-1}\left(  I\otimes\Psi^{L}\right)  \left(  H_{k}\otimes
H_{n-k}\right)  \\\text{(since }I\otimes\Psi^{L}\text{ is }K\text{-linear)}%
}}\\
&  =\mu\left(  \underbrace{\left(  I\otimes\Psi^{L}\right)  \left(
x\otimes1\right)  }_{=I\left(  x\right)  \otimes\Psi^{L}\left(  1\right)
}+\underbrace{\left(  I\otimes\Psi^{L}\right)  \left(  1\otimes x\right)
}_{=I\left(  1\right)  \otimes\Psi^{L}\left(  x\right)  }+\sum\limits_{k=1}%
^{n-1}\underbrace{\left(  I\otimes\Psi^{L}\right)  \left(  H_{k}\otimes
H_{n-k}\right)  }_{=I\left(  H_{k}\right)  \otimes\Psi^{L}\left(
H_{n-k}\right)  }\right) \\
&  =\mu\left(  I\left(  x\right)  \otimes\Psi^{L}\left(  1\right)  +I\left(
1\right)  \otimes\Psi^{L}\left(  x\right)  +\sum\limits_{k=1}^{n-1}I\left(
H_{k}\right)  \otimes\Psi^{L}\left(  H_{n-k}\right)  \right) \\
&  =\underbrace{I\left(  x\right)  }_{=x}\cdot\underbrace{\Psi^{L}\left(
1\right)  }_{=1}+\underbrace{I\left(  1\right)  }_{=1}\cdot\Psi^{L}\left(
x\right)  +\sum\limits_{k=1}^{n-1}\underbrace{I\left(  H_{k}\right)  }%
_{=H_{k}}\cdot\underbrace{\Psi^{L}\left(  H_{n-k}\right)  }%
_{\substack{\subseteq H_{n-k}\\\text{(since }\Psi^{L}\text{ is a graded map)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu\text{ is the multiplication
map}\right) \\
&  \subseteq x+\Psi^{L}\left(  x\right)  +\sum\limits_{k=1}^{n-1}H_{k}%
H_{n-k}=x+\Psi^{L}\left(  x\right)  +\sum\limits_{i=1}^{n-1}H_{i}H_{n-i}%
\end{align*}
(here, we renamed the index $k$ as $i$ in the sum), so that $\Psi^{L+1}\left(
x\right)  -x-\Psi^{L}\left(  x\right)  \in\sum\limits_{i=1}^{n-1}H_{i}H_{n-i}%
$. Combined with $\Psi^{L+1}\left(  x\right)  -x-\Psi^{L}\left(  x\right)  \in
H_{n}$ (this is because $x\in H_{n}$ and thus $\Psi^{L}\left(  x\right)
\in\Psi^{L}\left(  H_{n}\right)  \subseteq H_{n}$ and $\Psi^{L+1}\left(
x\right)  \in\Psi^{L+1}\left(  H_{n}\right)  \subseteq H_{n}$), this yields%
\[
\Psi^{L+1}\left(  x\right)  -x-\Psi^{L}\left(  x\right)  \in\left(
\sum\limits_{i=1}^{n-1}H_{i}H_{n-i}\right)  \cap H_{n}.
\]
In other words,%
\[
\Psi^{L+1}\equiv x+\underbrace{\Psi^{L}\left(  x\right)  }_{\substack{\equiv
Lx\operatorname{mod}\left(  \sum\limits_{i=1}^{n-1}H_{i}H_{n-i}\right)  \cap
H_{n}\\\text{(because (\ref{1.18}) holds for }\ell=L\text{)}}}\equiv
x+Lx=\left(  L+1\right)  x\operatorname{mod}\left(  \sum\limits_{i=1}%
^{n-1}H_{i}H_{n-i}\right)  \cap H_{n}.
\]
In other words, (\ref{1.18}) holds for $\ell=L+1$. Thus, the induction step is
done. The induction proof of (\ref{1.18}) is therefore complete.

So now we know that (\ref{1.18}) holds for every $\ell$. In other words,
Corollary 1.18 \textbf{(b)} is proven.
\end{proof}

Finally, a consequence of Proposition 1.4:

\begin{quote}
\textbf{Corollary 1.19.} Let $H$ be a bialgebra, a graded bialgebra or a Hopf
algebra. Assume that $H$ is commutative or cocommutative. Then the
characteristic operations (defined in Definition 1.2) satisfy $\left(
\Psi^{k}\right)  ^{s}=\Psi^{k^{s}}$ (where $\left(  \Psi^{k}\right)  ^{s}$
means $\underbrace{\Psi^{k}\circ\Psi^{k}\circ...\circ\Psi^{k}}_{s\text{
times}}$) for all $k\in\mathbb{N}$ and $s\in\mathbb{N}$.
\end{quote}

\begin{proof}
[Proof of Corollary 1.19.]Fix some $k\in\mathbb{N}$. We will prove $\left(
\Psi^{k}\right)  ^{s}=\Psi^{k^{s}}$ by induction over $s$:

\textit{Induction base:} We have $\left(  \Psi^{k}\right)  ^{0}=I=I^{\ast
1}=\Psi^{1}$ (because $\Psi^{1}$ was defined as $I^{\ast1}$) and $\Psi^{k^{0}%
}=\Psi^{1}$. Thus, $\left(  \Psi^{k}\right)  ^{0}=\Psi^{1}=\Psi^{k^{0}}$. In
other words, $\left(  \Psi^{k}\right)  ^{s}=\Psi^{k^{s}}$ holds for $s=0$.
This completes the induction base.

\textit{Induction step:} Let $S\in\mathbb{N}$. Assume that $\left(  \Psi
^{k}\right)  ^{s}=\Psi^{k^{s}}$ holds for $s=S$. We must then prove that
$\left(  \Psi^{k}\right)  ^{s}=\Psi^{k^{s}}$ also holds for $s=S+1$.

Since $\left(  \Psi^{k}\right)  ^{s}=\Psi^{k^{s}}$ holds for $s=S$, we have
$\left(  \Psi^{k}\right)  ^{S}=\Psi^{k^{S}}$. Applying Proposition 1.4 to
$l=k^{S}$, we get $\Psi^{k}\circ\Psi^{k^{S}}=\Psi^{k\cdot k^{S}}=\Psi
^{k^{S+1}}$ (since $k\cdot k^{S}=k^{S+1}$). Hence, $\left(  \Psi^{k}\right)
^{S+1}=\Psi^{k}\circ\underbrace{\left(  \Psi^{k}\right)  ^{S}}_{=\Psi^{k^{S}}%
}=\Psi^{k}\circ\Psi^{k^{S}}=\Psi^{k^{S+1}}$. In other words, $\left(  \Psi
^{k}\right)  ^{s}=\Psi^{k^{s}}$ is proven to hold for $s=S+1$. This completes
the induction step. Thus, the induction proof of $\left(  \Psi^{k}\right)
^{s}=\Psi^{k^{s}}$ is complete. In other words, Corollary 1.19 is proven.
\end{proof}

Now to the actual proof of Lemma 6.3:

\begin{proof}
[Proof of Lemma 6.3.]We WLOG assume that $H$ is a graded bialgebra. (The case
when $H$ is a Hopf algebra is analogous.)

We WLOG assume that $H$ is commutative. (The case when $H$ is cocommutative
can be obtained from the case when $H$ is commutative by dualization using
Proposition 3.9.)

Fix some $k\in\mathbb{Z}$ such that $k\not \equiv 0\operatorname{mod}p$. Thus,
$k$ is not divisible by $p$. Thus, by Fermat's Little Theorem, $p\mid
k^{p-1}-1$ (since $p$ is prime), so that
\begin{equation}
\left(  k^{p-1}-1\right)  \left(  x\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for
every }x\in H \label{6.3.pf.0}%
\end{equation}
(since $K$ has characteristic $p$).

We must prove that every positive integer $n$ satisfies%
\begin{equation}
\rho_{n}\left(  \left(  \Psi^{k^{p^{n-1}}}\right)  ^{p-1}\right)  =\rho
_{n}\left(  I\right)  . \label{6.3.pf.1}%
\end{equation}


In fact, we will prove (\ref{6.3.pf.1}) by induction over $n$:

\textit{Induction base:} We have $p^{1-1}=p^{0}=1$, so that $\Psi^{k^{p^{1-1}%
}}=\Psi^{k^{1}}=\Psi^{k}$, so that $\left(  \Psi^{k^{p^{1-1}}}\right)
^{p-1}=\left(  \Psi^{k}\right)  ^{p-1}=\Psi^{k^{p-1}}$ (by Corollary 1.19,
applied to $s=p-1$).

Every $x\in H_{1}$ satisfies%
\[
\Psi^{k^{p-1}}\left(  x\right)  \equiv k^{p-1}x\operatorname{mod}\left(
\sum_{i=1}^{1-1}H_{i}H_{1-i}\right)  \cap H_{1}%
\]
(by Corollary 1.18 \textbf{(b)}, applied to $\ell=k^{p-1}$ and $n=1$). Since
$\underbrace{\left(  \sum\limits_{i=1}^{1-1}H_{i}H_{1-i}\right)  }_{=\left(
\text{empty sum}\right)  =0}\cap H_{1}=0\cap H_{1}=0$, this becomes
$\Psi^{k^{p-1}}\left(  x\right)  \equiv k^{p-1}x\operatorname{mod}0$. Hence,
every $x\in H_{1}$ satisfies $\Psi^{k^{p-1}}\left(  x\right)  \equiv
k^{p-1}x\operatorname{mod}0$, so that%
\[
\Psi^{k^{p-1}}\left(  x\right)  =k^{p-1}x=x+\underbrace{\left(  k^{p-1}%
x-x\right)  }_{\substack{=\left(  k^{p-1}-1\right)  x=0\\\text{(by
(\ref{6.3.pf.0}))}}}=x.
\]
Thus, every $x\in H_{1}$ satisfies $\left(  \Psi^{k^{p-1}}-I\right)  \left(
x\right)  =\underbrace{\Psi^{k^{p-1}}\left(  x\right)  }_{=x}%
-\underbrace{I\left(  x\right)  }_{=x}=x-x=0$ and thus $x\in
\operatorname*{Ker}\left(  \Psi^{k^{p-1}}-I\right)  $. Hence, $H_{1}%
\subseteq\operatorname*{Ker}\left(  \Psi^{k^{p-1}}-I\right)  $. In other
words, $\left(  \Psi^{k^{p-1}}-I\right)  \left(  H_{1}\right)  =0$.

On the other hand, Corollary 1.18 \textbf{(a)} (applied to $\ell=k^{p-1}$)
yields $\Psi^{k^{p-1}}\left(  1\right)  =1$ and $\left(  \Psi^{k^{p-1}%
}-I\right)  \left(  H_{0}\right)  =0$.

Now, since $\rho_{1}\left(  \Psi^{k^{p-1}}-I\right)  $ is the restriction of
the map $\Psi^{k^{p-1}}-I$ to $\bigoplus\limits_{i=0}^{1}H_{i}$, we have
\begin{align*}
\operatorname{Im}\left(  \rho_{1}\left(  \Psi^{k^{p-1}}-I\right)  \right)   &
=\left(  \Psi^{k^{p-1}}-I\right)  \underbrace{\left(  \bigoplus\limits_{i=0}%
^{1}H_{i}\right)  }_{\substack{=H_{0}\oplus H_{1}=H_{0}+H_{1}\\\text{(since
direct sums are sums)}}}=\left(  \Psi^{k^{p-1}}-I\right)  \left(  H_{0}%
+H_{1}\right) \\
&  =\underbrace{\left(  \Psi^{k^{p-1}}-I\right)  \left(  H_{0}\right)  }%
_{=0}+\underbrace{\left(  \Psi^{k^{p-1}}-I\right)  \left(  H_{1}\right)
}_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Psi^{k^{p-1}}-I\text{ is
}K\text{-linear}\right) \\
&  =0+0=0,
\end{align*}
so that $\rho_{1}\left(  \Psi^{k^{p-1}}-I\right)  =0$. Thus,%
\begin{align*}
0  &  =\rho_{1}\left(  \Psi^{k^{p-1}}-I\right)  =\rho_{1}\underbrace{\left(
\Psi^{k^{p-1}}\right)  }_{=\left(  \Psi^{k^{p^{1-1}}}\right)  ^{p-1}}-\rho
_{1}\left(  I\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho_{1}\text{
is }K\text{-linear}\right) \\
&  =\rho_{1}\left(  \left(  \Psi^{k^{p^{1-1}}}\right)  ^{p-1}\right)
-\rho_{1}\left(  I\right)  ,
\end{align*}
so that $\rho_{1}\left(  \left(  \Psi^{k^{p^{1-1}}}\right)  ^{p-1}\right)
=\rho_{1}\left(  I\right)  $. In other words, (\ref{6.3.pf.1}) holds for
$n=1$. This completes the induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that
(\ref{6.3.pf.1}) holds for $n=N$. We must then prove that (\ref{6.3.pf.1})
also holds for $n=N+1$.

Since (\ref{6.3.pf.1}) holds for $n=N$, we have%
\[
\rho_{N}\left(  \left(  \Psi^{k^{p^{N-1}}}\right)  ^{p-1}\right)  =\rho
_{N}\left(  I\right)  .
\]


Every $x\in\bigoplus\limits_{i=0}^{N}H_{i}$ satisfies $\left(  \Psi
^{k^{p^{N-1}}}\right)  ^{p-1}\left(  x\right)  =\left(  \rho_{N}\left(
\left(  \Psi^{k^{p^{N-1}}}\right)  ^{p-1}\right)  \right)  \left(  x\right)  $
(since $\rho_{N}\left(  \left(  \Psi^{k^{p^{N-1}}}\right)  ^{p-1}\right)  $ is
the restriction of the map $\left(  \Psi^{k^{p^{N-1}}}\right)  ^{p-1}$ to
$\bigoplus\limits_{i=0}^{N}H_{i}$) and $I\left(  x\right)  =\left(  \rho
_{N}\left(  I\right)  \right)  \left(  x\right)  $ (since $\rho_{N}\left(
I\right)  $ is the restriction of the map $I$ to $\bigoplus\limits_{i=0}%
^{N}H_{i}$). Thus,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{every }x\in\bigoplus\limits_{i=0}^{N}H_{i}\text{ satisfies}\\
\left(  \Psi^{k^{p^{N-1}}}\right)  ^{p-1}\left(  x\right)
=\underbrace{\left(  \rho_{N}\left(  \left(  \Psi^{k^{p^{N-1}}}\right)
^{p-1}\right)  \right)  }_{=\rho_{N}\left(  I\right)  }\left(  x\right)
=\left(  \rho_{N}\left(  I\right)  \right)  \left(  x\right)  =I\left(
x\right)  =x.
\end{array}
\right)  . \label{6.3.pf.2a}%
\end{equation}


Let $E$ be the $K$-vector subspace $\bigoplus\limits_{i=0}^{N}H_{i}$ of $H$.
Then, (\ref{6.3.pf.2a}) rewrites as follows:%
\begin{equation}
\text{Every }x\in E\text{ satisfies }\left(  \Psi^{k^{p^{N-1}}}\right)
^{p-1}\left(  x\right)  =x. \label{6.3.pf.2b}%
\end{equation}
On the other hand,
\begin{equation}
\text{every }j\in\left\{  0,1,...,N\right\}  \text{ satisfies }H_{j}\subseteq
E \label{6.3.pf.2c}%
\end{equation}
(because for every $j\in\left\{  0,1,...,N\right\}  $, the space $H_{j}$ is an
addend of the direct sum $\bigoplus\limits_{i=0}^{N}H_{i}$, and thus is
contained in $\bigoplus\limits_{i=0}^{N}H_{i}=E$).

We are now going to show that%
\begin{equation}
\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)  \left(  H_{n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\left\{  0,1,...,N+1\right\}  .
\label{6.3.pf.3}%
\end{equation}


\textit{Proof of (\ref{6.3.pf.3}).} Let $n\in\left\{  0,1,...,N+1\right\}  $
be arbitrary. Then, $n\leq N+1$, so that $n-1\leq N$.

Corollary 1.18 \textbf{(a)} (applied to $\ell=k^{p^{N}\left(  p-1\right)  }$)
yields $\Psi^{k^{p^{N}\left(  p-1\right)  }}\left(  1\right)  =1$ and
\newline$\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)  \left(
H_{0}\right)  =0$. Hence, (\ref{6.3.pf.3}) is already proven when $n=0$. Thus,
for the rest of the proof, we can WLOG assume that $n\neq0$. Assume this.
Then, $n$ is a positive integer.

Since $\Psi^{k^{p^{N-1}}}$ is a graded map, it satisfies $\Psi^{k^{p^{N-1}}%
}\left(  H_{n}\right)  \subseteq H_{n}$. Hence, it restricts to a $K$-linear
map $\beta:H_{n}\rightarrow H_{n}$ which satisfies $\left(  \beta\left(
x\right)  =\Psi^{k^{p^{N-1}}}\left(  x\right)  \text{ for every }x\in
H_{n}\right)  $. Consider this $\beta$. Then,%
\begin{equation}
\beta^{\ell}\left(  x\right)  =\left(  \Psi^{k^{p^{N-1}}}\right)  ^{\ell
}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{N}%
\text{ and }x\in H_{n}. \label{6.3.pf.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{6.3.pf.5}).} We will prove (\ref{6.3.pf.5})
by induction over $\ell$:
\par
\textit{Induction base:} We have $\underbrace{\beta^{0}}_{=\operatorname*{id}%
\nolimits_{H_{n}}}\left(  x\right)  =\operatorname*{id}\nolimits_{H_{n}%
}\left(  x\right)  =x=\underbrace{\operatorname*{id}}_{=\left(  \Psi
^{k^{p^{N-1}}}\right)  ^{0}}\left(  x\right)  =\left(  \Psi^{k^{p^{N-1}}%
}\right)  ^{0}\left(  x\right)  $. Thus, (\ref{6.3.pf.5}) holds for $\ell=0$.
The induction base is now complete.
\par
\textit{Induction step:} Let $L\in\mathbb{N}$. Assume that (\ref{6.3.pf.5})
holds for $\ell=L$. Now we must show that (\ref{6.3.pf.5}) also holds for
$\ell=L+1$.
\par
Let $x\in H_{n}$. We can apply (\ref{6.3.pf.5}) to $L$ and $\beta\left(
x\right)  $ instead of $\ell$ and $x$ (since we assumed that (\ref{6.3.pf.5})
holds for $\ell=L$). This gives us $\beta^{L}\left(  \beta\left(  x\right)
\right)  =\left(  \Psi^{k^{p^{N-1}}}\right)  ^{L}\left(  \beta\left(
x\right)  \right)  $. Thus,%
\[
\beta^{L+1}\left(  x\right)  =\beta^{L}\left(  \beta\left(  x\right)  \right)
=\left(  \Psi^{k^{p^{N-1}}}\right)  ^{L}\underbrace{\left(  \beta\left(
x\right)  \right)  }_{=\Psi^{k^{p^{N-1}}}\left(  x\right)  }=\left(
\Psi^{k^{p^{N-1}}}\right)  ^{L}\left(  \Psi^{k^{p^{N-1}}}\left(  x\right)
\right)  =\left(  \Psi^{k^{p^{N-1}}}\right)  ^{L+1}\left(  x\right)  .
\]
Thus, we have proven that $\beta^{L+1}\left(  x\right)  =\left(
\Psi^{k^{p^{N-1}}}\right)  ^{L+1}\left(  x\right)  $ for every $x\in H_{n}$.
In other words, (\ref{6.3.pf.5}) holds for $\ell=L+1$. This completes the
induction step. Thus, the induction proof of (\ref{6.3.pf.5}) is complete.}

Let $C$ be the $K$-vector space $H_{n}$. Let $D$ be the $K$-vector subspace
$\left(  \sum\limits_{i=1}^{n-1}H_{i}H_{n-i}\right)  \cap H_{n}$ of $C$. Since
$C=H_{n}$, the map $\beta$ is a map $C\rightarrow C$ (since $\beta$ is a map
$H_{n}\rightarrow H_{n}$).

Our goal is to apply Lemma 6.6. In order to do so, we must show that the
conditions (\ref{6.6.1}) and (\ref{6.6.2}) of Lemma 6.6 are satisfied. Let us
first prove that the condition (\ref{6.6.2}) is satisfied:

Since $k$ is not divisible by $p$, and since $p$ is prime, we have
$k^{p-1}\equiv1\operatorname{mod}p$ (by Fermat's Little Theorem). Thus,
$\left(  k^{p-1}\right)  ^{1+p+p^{2}+...+p^{N-2}}\equiv1^{1+p+p^{2}%
+...+p^{N-2}}=1\operatorname{mod}p$. Since $\left(  k^{p-1}\right)
^{1+p+p^{2}+...+p^{N-2}}=k^{\left(  p-1\right)  \left(  1+p+p^{2}%
+...+p^{N-2}\right)  }=k^{p^{N-1}-1}$ (because $\left(  p-1\right)  \left(
1+p+p^{2}+...+p^{N-2}\right)  =p^{N-1}-1$), this becomes $k^{p^{N-1}-1}%
\equiv1\operatorname{mod}p$, so that $k^{p^{N}}=k\underbrace{k^{p^{N-1}-1}%
}_{\equiv1\operatorname{mod}p}\equiv k\operatorname{mod}p$. Hence, $p\mid
k^{p^{N}}-k$. Thus, $\left(  k^{p^{N}}-k\right)  x=0$ for every $x\in C$
(since $K$ has characteristic $p$).

Now, every $x\in C$ satisfies%
\begin{align*}
\beta\left(  x\right)   &  =\Psi^{k^{p^{N-1}}}\left(  x\right) \\
&  \equiv k^{p^{N-1}}x\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary 1.18
\textbf{(b)} (applied to }\ell=k^{p^{N-1}}\text{), since }x\in C=H_{n}\right)
\\
&  =kx+\underbrace{\left(  k^{p^{N-1}}x-kx\right)  }_{=\left(  k^{p^{N}%
}-k\right)  x=0}=kx\operatorname{mod}\left(  \sum_{i=1}^{n-1}H_{i}%
H_{n-i}\right)  \cap H_{n}.
\end{align*}
Since $\left(  \sum\limits_{i=1}^{n-1}H_{i}H_{n-i}\right)  \cap H_{n}=D$, this
rewrites as follows:%
\begin{equation}
\text{Every }x\in C\text{ satisfies }\beta\left(  x\right)  \equiv
kx\operatorname{mod}D. \label{6.3.pf.4}%
\end{equation}


Now to checking the condition (\ref{6.6.1}).

By Proposition 1.4, the characteristic operations of $H$ are algebra
homomorphisms (since $H$ is commutative); in particular, $\Psi^{k^{p^{N-1}%
\left(  p-1\right)  }}$ is an algebra homomorphism. Since $\left(
\Psi^{k^{p^{N-1}}}\right)  ^{p-1}=\Psi^{k^{p^{N-1}\left(  p-1\right)  }}$ (by
Corollary 1.19, applied to $k^{p^{N-1}}$ and $p-1$ instead of $k$ and $s$),
this yields that $\left(  \Psi^{k^{p^{N-1}}}\right)  ^{p-1}$ is a $K$-algebra homomorphism.

Now, let $x\in D$ be arbitrary. (Note that we are requiring $x\in D$ now, not
only $x\in C$.) Then,%
\begin{align*}
x  &  \in D=\left(  \sum\limits_{i=1}^{n-1}H_{i}H_{n-i}\right)  \cap
H_{n}\subseteq\sum\limits_{i=1}^{n-1}\underbrace{H_{i}}_{\substack{\subseteq
E\\\text{(since }i\leq n-1\leq N\text{,}\\\text{so that }i\in\left\{
0,1,...,N\right\}  \text{,}\\\text{so that }H_{i}\subseteq E\\\text{(by
(\ref{6.3.pf.2c}), applied to }j=i\text{))}}}\ \ \underbrace{H_{n-i}%
}_{\substack{\subseteq E\\\text{(since }i\geq1\text{, thus }n-i\leq n-1\leq
N\text{,}\\\text{so that }n-i\in\left\{  0,1,...,N\right\}  \text{,}\\\text{so
that }H_{n-i}\subseteq E\\\text{(by (\ref{6.3.pf.2c}), applied to
}j=n-i\text{))}}}\\
&  \subseteq\sum\limits_{i=1}^{n-1}EE\subseteq EE\ \ \ \ \ \ \ \ \ \ \left(
\text{since }EE\text{ is a }K\text{-vector space}\right) \\
&  =\left(  \text{the set of all }K\text{-linear combinations of elements of
the form }ee^{\prime}\text{ with }e\in E\text{ and }e^{\prime}\in E\right)  .
\end{align*}
Hence, $x$ is a $K$-linear combination of elements of the form $ee^{\prime}$
with $e\in E$ and $e^{\prime}\in E$. In other words, we can write $x$ in the
form $x=\sum\limits_{i=1}^{I}\lambda_{i}e_{i}e_{i}^{\prime}$ for some
$I\in\mathbb{N}$, some elements $\lambda_{1}$, $\lambda_{2}$, $...$,
$\lambda_{I}$ of $K$, some elements $e_{1}$, $e_{2}$, $...$, $e_{I}$ of $E$,
and some elements $e_{1}^{\prime}$, $e_{2}^{\prime}$, $...$, $e_{I}^{\prime}$
of $E$. Consider this $I$, these $\lambda_{1}$, $\lambda_{2}$, $...$,
$\lambda_{I}$, these $e_{1}$, $e_{2}$, $...$, $e_{I}$, and these
$e_{1}^{\prime}$, $e_{2}^{\prime}$, $...$, $e_{I}^{\prime}$. Then,%
\begin{align*}
\beta^{p-1}\left(  x\right)   &  =\left(  \Psi^{k^{p^{N-1}}}\right)
^{p-1}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{6.3.pf.5})
(applied to }\ell=p-1\text{), since }x\in D\subseteq E=H_{n}\right) \\
&  =\left(  \Psi^{k^{p^{N-1}}}\right)  ^{p-1}\left(  \sum\limits_{i=1}%
^{I}\lambda_{i}e_{i}e_{i}^{\prime}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }x=\sum\limits_{i=1}^{I}\lambda_{i}e_{i}e_{i}^{\prime}\right) \\
&  =\sum\limits_{i=1}^{I}\lambda_{i}\underbrace{\left(  \Psi^{k^{p^{N-1}}%
}\right)  ^{p-1}\left(  e_{i}\right)  }_{\substack{=e_{i}\\\text{(by
(\ref{6.3.pf.2b}), applied to }e_{i}\text{ instead of }x\text{)}%
}}\underbrace{\left(  \Psi^{k^{p^{N-1}}}\right)  ^{p-1}\left(  e_{i}^{\prime
}\right)  }_{\substack{=e_{i}^{\prime}\\\text{(by (\ref{6.3.pf.2b}), applied
to }e_{i}^{\prime}\text{ instead of }x\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \Psi^{k^{p^{N-1}}}\right)
^{p-1}\text{ is a }K\text{-algebra homomorphism}\right) \\
&  =\sum\limits_{i=1}^{I}\lambda_{i}e_{i}e_{i}^{\prime}=x.
\end{align*}


Now forget that we fixed $x$. We have thus shown that every $x\in D$ satisfies
$\beta^{p-1}\left(  x\right)  =x$. Combined with (\ref{6.3.pf.4}), this shows
that all conditions of Lemma 6.6 are satisfied. Hence, we can apply Lemma 6.6,
and obtain $\left(  \beta^{p-1}\right)  ^{p}=\operatorname*{id}\nolimits_{C}$.
Thus, $\operatorname*{id}\nolimits_{C}=\left(  \beta^{p-1}\right)  ^{p}%
=\beta^{\left(  p-1\right)  p}$. Hence, every $x\in C$ satisfies%
\begin{align*}
x  &  =\underbrace{\operatorname*{id}\nolimits_{C}}_{=\beta^{\left(
p-1\right)  p}}\left(  x\right)  =\beta^{\left(  p-1\right)  p}\left(
x\right)  =\left(  \Psi^{k^{p^{N-1}}}\right)  ^{\left(  p-1\right)  p}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{6.3.pf.5}), applied to
}\ell=\left(  p-1\right)  p\right) \\
&  =\Psi^{k^{p^{N-1}\left(  p-1\right)  p}}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Corollary 1.19}\\
\text{(applied to }k^{p^{N-1}}\text{ and }\left(  p-1\right)  p\text{ instead
of }k\text{ and }s\text{)}\\
\text{yields }\left(  \Psi^{k^{p^{N-1}}}\right)  ^{\left(  p-1\right)  p}%
=\Psi^{k^{p^{N-1}\left(  p-1\right)  p}}%
\end{array}
\right) \\
&  =\Psi^{k^{p^{N}\left(  p-1\right)  }}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p^{N-1}\left(  p-1\right)
p=\underbrace{p^{N-1}p}_{=p^{N}}\left(  p-1\right)  =p^{N}\left(  p-1\right)
\right)
\end{align*}
and thus%
\[
\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)  \left(  x\right)
=\underbrace{\Psi^{k^{p^{N}\left(  p-1\right)  }}\left(  x\right)  }%
_{=x}-\underbrace{I\left(  x\right)  }_{=x}=x-x=0,
\]
so that $x\in\operatorname*{Ker}\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }%
}-I\right)  $. In other words, $C\subseteq\operatorname*{Ker}\left(
\Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)  $. Hence, $\left(
\Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)  \left(  C\right)  =0$. Since
$C=H_{n}$, this becomes $\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }%
}-I\right)  \left(  H_{n}\right)  =0$. This proves (\ref{6.3.pf.3}).

Now, $\rho_{N+1}\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)  $ is
the restriction of the map $\Psi^{k^{p^{N}\left(  p-1\right)  }}-I$ to
$\bigoplus\limits_{i=0}^{N+1}H_{i}$. Hence,
\begin{align*}
&  \operatorname{Im}\left(  \rho_{N+1}\left(  \Psi^{k^{p^{N}\left(
p-1\right)  }}-I\right)  \right) \\
&  =\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)  \underbrace{\left(
\bigoplus\limits_{i=0}^{N+1}H_{i}\right)  }_{\substack{=\sum\limits_{i=0}%
^{N+1}H_{i}\\\text{(since direct sums are sums)}}}=\left(  \Psi^{k^{p^{N}%
\left(  p-1\right)  }}-I\right)  \sum\limits_{i=0}^{N+1}H_{i}\\
&  =\sum\limits_{i=0}^{N+1}\underbrace{\left(  \Psi^{k^{p^{N}\left(
p-1\right)  }}-I\right)  \left(  H_{i}\right)  }_{=0\text{ (by (\ref{6.3.pf.3}%
), applied to }n=i\text{)}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\Psi^{k^{p^{N}\left(  p-1\right)  }}-I\text{ is }K\text{-linear}\right) \\
&  =\sum\limits_{i=0}^{N+1}0=0,
\end{align*}
so that $\rho_{N+1}\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)  =0$.
Since $\rho_{N+1}\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}-I\right)
=\rho_{N+1}\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}\right)  -\rho
_{N+1}\left(  I\right)  $ (because $\rho_{N+1}$ is $K$-linear), this becomes
$\rho_{N+1}\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}\right)  -\rho
_{N+1}\left(  I\right)  =0$, so that $\rho_{N+1}\left(  \Psi^{k^{p^{N}\left(
p-1\right)  }}\right)  =\rho_{N+1}\left(  I\right)  $.

Since $\left(  \Psi^{k^{p^{N+1-1}}}\right)  ^{p-1}=\left(  \Psi^{k^{p^{N}}%
}\right)  ^{p-1}=\Psi^{k^{p^{N}\left(  p-1\right)  }}$ (by Corollary 1.19,
applied to $k^{p^{N}}$ and $p-1$ instead of $k$ and $s$), we have $\rho
_{N+1}\left(  \left(  \Psi^{k^{p^{N+1-1}}}\right)  ^{p-1}\right)  =\rho
_{N+1}\left(  \Psi^{k^{p^{N}\left(  p-1\right)  }}\right)  =\rho_{N+1}\left(
I\right)  $. Thus, (\ref{6.3.pf.1}) holds for $n=N+1$. This completes the
induction step. Thus, the induction proof of (\ref{6.3.pf.1}) is completed,
and with it the proof of Lemma 6.3.
\end{proof}

\parbreak


\textbf{Page 1086, Proposition 6.4:} I think the condition that
\textquotedblleft$k\not \equiv 0\ \left[  p\right]  ,$ $k\not \equiv
1\ \left[  p\right]  $\textquotedblright\ has to be replaced by the (stronger)
condition that $k$ be a primitive root modulo $p$. Otherwise, the sum
$H^{\left(  1\right)  }\oplus...\oplus H^{\left(  p-1\right)  }$ won't be a
well-defined direct sum anymore (since some of the addends will be equal).

\parbreak


\begin{center}
\textbf{Additional references}
\end{center}

[BF] Andrea Bonfiglioli and Roberta Fulci, \textit{A New Proof of the
Existence of Free Lie Algebras and an Application}, ISRN Geometry, Volume 2011
(2011), Article ID 126358.\newline%
\url{http://www.hindawi.com/journals/isrn.algebra/2011/247403/} 

[C] Pierre Cartier, \textit{A primer of Hopf algebras}, IHES, September
2006.\newline\url{http://www.math.osu.edu/~kerler.2/VIGRE/InvResPres-Sp07/Cartier-IHES.pdf}

[DG1] \textit{6th QEDMO 2009, Problem 4 (the Cauchy identity)}, with Solution
by Darij Grinberg.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/QEDMO6P4long.pdf}

[M] Dominique Manchon, \textit{Hopf algebras, from basics to applications to
renormalization}, Revised and updated version, may 2006,
arXiv:math/0408405v2.\newline\url{http://arxiv.org/abs/math/0408405v2}


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