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\ihead{Errata to pre-Lie algebra survey, version 18 December 2009}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{A short survey on pre-Lie algebras}

\textit{Dominique Manchon}

\url{https://lmbp.uca.fr/~manchon/biblio/ESI-prelie2009.pdf}

version of 18 December 2009

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the paper \textquotedblleft A short
survey on pre-Lie algebras\textquotedblright\ by the numbers under which they
appear in this paper (specifically, in its version of 18 December 2009). All
page numbers appearing below are actual page numbers on the top headers of the
pages, not the page numbers of the PDF file.

I have read only parts of the paper, so this list of errors is likely incomplete.

\appendix


\section{Errata}

\begin{itemize}
\item \textbf{Page 3, \S 1.1:} In the whole Section 1.1, you appear to make
the assumption that $\operatorname*{char}k=0$.

\item \textbf{Page 3, \S 1.1:} The word \textquotedblleft
compatible\textquotedblright\ (in \textquotedblleft with a compatible
decreasing filtration\textquotedblright) is confusing: It creates the
impression that there is an additional \textquotedblleft
compatibility\textquotedblright\ condition on the filtration. I think it is
really just an abbreviation for the requirement that \textquotedblleft%
$A_{p}\vartriangleright A_{q}\subset A_{p+q}$\textquotedblright; in this case,
it is probably best to remove this word (since you already make the
requirement that \textquotedblleft$A_{p}\vartriangleright A_{q}\subset
A_{p+q}$\textquotedblright\ explicitly).

\item \textbf{Page 3, \S 1.1:} In \textquotedblleft$A=A_{1}\supset
A_{2}\subset A_{3}\supset\cdots$\textquotedblright, replace the
\textquotedblleft$\subset$\textquotedblright\ sign by a \textquotedblleft%
$\supset$\textquotedblright\ sign.

\item \textbf{Page 3, \S 1.1:} Replace \textquotedblleft reduces to $\left\{
0\right\}  $\textquotedblright\ by \textquotedblleft is $\left\{  0\right\}
$\textquotedblright.

\item \textbf{Page 3, equation (9):} Replace \textquotedblleft$\sum
\limits_{i\geq0}B_{i}L_{\Omega\left(  a\right)  }^{i}a$\textquotedblright\ by
\textquotedblleft$\sum\limits_{i\geq0}\dfrac{B_{i}}{i!}L_{\Omega\left(
a\right)  }^{i}a$\textquotedblright.

\item \textbf{Page 5, equation (17):} It should be said here that the map
$L_{a}$ is extended to the fictitious unit $\mathbf{1}$ by $L_{a}%
\mathbf{1}:=a$.

\item \textbf{Page 6, \S 1.3:} Replace \textquotedblleft$V=\bigoplus
\limits_{n\geq0}$\textquotedblright\ by \textquotedblleft$V=\bigoplus
\limits_{n\geq1}V_{n}$\textquotedblright. (The direct sum should start at
$n=1$ in order for the symmetric algebra $S\left(  V\right)  $ to have
finite-dimensional homogeneous components, which in turn is necessary to
ensure that its graded dual is a Hopf algebra.)

\item \textbf{Page 6, \S 2.1:} In the definition of an augmented operad,
replace \textquotedblleft for any $a\in\mathcal{P}_{k},\ b\in\mathcal{P}%
_{l},\ a\in\mathcal{P}_{m}$\textquotedblright\ by \textquotedblleft for any
$a\in\mathcal{P}_{k},\ b\in\mathcal{P}_{l},\ c\in\mathcal{P}_{m}%
$\textquotedblright.

\item \textbf{Page 7, \S 2.1:} On the first line of page 7, replace
\textquotedblleft\textsl{free }$\mathcal{O}$\textsl{-algebra}%
\textquotedblright\ by \textquotedblleft\textsl{free }$\mathcal{P}%
$\textsl{-algebra}\textquotedblright.

\item \textbf{Page 7, equation (27):} Replace \textquotedblleft$:$%
\textquotedblright\ by \textquotedblleft$:=$\textquotedblright.

\item \textbf{Page 8, \S 2.3:} \textquotedblleft Here the notation $V<W$ means
that $x<y$ for any vertex $x$ of $v$ and any vertex $y$ of $w$ such that $x$
and $y$ are comparable\textquotedblright\ is technically correct, but somewhat
confusing because the sum in (31) has the condition $W<V$ rather than $V<W$.
It would be better to replace this by \textquotedblleft Here the notation
$W<V$ means that $x<y$ for any vertex $x$ of $w$ and any vertex $y$ of $v$
such that $x$ and $y$ are comparable\textquotedblright.

\item \textbf{Page 8, \S 2.3:} In the last sentence of \S 2.3, replace
\textquotedblleft$\phi\left(  t\right)  $\textquotedblright\ by
\textquotedblleft$\Phi\left(  t\right)  $\textquotedblright.

\item \textbf{Page 10, equation (41):} Replace \textquotedblleft%
$s\rightarrow_{i}v$\textquotedblright\ by \textquotedblleft$s\rightarrow_{v}%
t$\textquotedblright.

\item \textbf{Page 11, Proposition 3.2:} Replace \textquotedblleft%
$\alpha\left(  \bullet=1\right)  $\textquotedblright\ by \textquotedblleft%
$\alpha\left(  \bullet\right)  =1$\textquotedblright.

\item \textbf{Page 11, after Proposition 3.2:} \textquotedblleft The condition
$\alpha\left(  \bullet\right)  $ is in fact dropped\textquotedblright\ should
be \textquotedblleft The condition $\alpha\left(  \bullet\right)  =1$ is in
fact dropped\textquotedblright.

In the same sentence \textquotedblleft\lbrack4, Propsition
15]\textquotedblright\ should be \textquotedblleft\lbrack4, Proposition
16]\textquotedblright\ (at least if we refer to arXiv version 4 of [4]; the
referenced version 2 doesn't seem to include this proposition at all).

\item \textbf{Page 12, reference [2]:} The correct pages are 521--534, not 512--534.

\item \textbf{Page 13, reference [16]:} The arXiv ID is
\texttt{arXiv:0910.2166}, not \texttt{arXiv:math/0910.2166}.

\item \textbf{Page 13, reference [17]:} The page range for this paper is 295--316.
\end{itemize}

\section{Addenda}

\subsection{Page 3, \S 1.1}

In \S 1.1 on page 3, you write that \textquotedblleft The application $W$ is
clearly a bijection\textquotedblright. It took me a long time to see why this
is true! I don't consider this argument simple enough as to be left to the
reader. Let me give this argument in more detail. (This was partly written by GPT-5.5.)

First, I state a useful lemma:\footnote{Lemma \ref{lem.banach-cons} is
\textbf{not} concerned with the setting of \S 1.1; instead, it introduces a
more general setting. In particular, the $A$ in Lemma \ref{lem.banach-cons} is
not necessarily a Lie algebra, and not necessarily a vector space, but just an
abelian group.}

\begin{lemma}
\label{lem.banach-cons}Let $A$ be any abelian group (written additively). Let
$A=A_{1}\supset A_{2}\supset A_{3}\supset\cdots$ be any decreasing filtration
of $A$ (by subgroups) such that $\bigcap\limits_{j\geq1}A_{j}=0$. Let
$W:A\rightarrow A$ be any map (not necessarily a group homomorphism). Assume
that for every $n\geq1$ and every $a\in A$ and $b\in A$, the following holds:
If $a\equiv b\operatorname{mod}A_{n}$, then%
\begin{equation}
W\left(  a\right)  -a\equiv W\left(  b\right)  -b\operatorname{mod}A_{n+1}.
\label{eq.lem.banach-cons.ass}%
\end{equation}
Then:

\begin{enumerate}
\item[\textbf{(a)} ] The map $W$ is injective.

\item[\textbf{(b)}] Assume that $A$ is complete with respect to the filtration
$A=A_{1}\supset A_{2}\supset A_{3}\supset\cdots$. Then, the map $W$ is a bijection.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.banach-cons}.]\textbf{(a)} Let $a$ and $b$ be two
elements of $A$ such that $W\left(  a\right)  =W\left(  b\right)  $. We shall
show that%
\begin{equation}
a-b\in A_{m}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\left\{  1,2,3,\ldots
\right\}  . \label{pf.p3.l1.0.a.1}%
\end{equation}


\begin{proof}
[Proof of (\ref{pf.p3.l1.0.a.1}).]We shall prove (\ref{pf.p3.l1.0.a.1}) by
induction on $m$:

\textit{Induction base:} We have $a-b\in A=A_{1}$. In other words,
(\ref{pf.p3.l1.0.a.1}) holds for $m=1$. This completes the induction base.

\textit{Induction step:} Let $M\in\left\{  1,2,3,\ldots\right\}  $ be such
that (\ref{pf.p3.l1.0.a.1}) holds for $m=M$. We must show that
(\ref{pf.p3.l1.0.a.1}) holds for $m=M+1$ as well.

We have assumed that (\ref{pf.p3.l1.0.a.1}) holds for $m=M$. In other words,
$a-b\in A_{M}$. In other words, $a\equiv b\operatorname{mod}A_{M}$. Hence,
(\ref{eq.lem.banach-cons.ass}) (applied to $n=M$) yields $W\left(  a\right)
-a\equiv W\left(  b\right)  -b\operatorname{mod}A_{M+1}$. Hence,%
\[
a=\underbrace{W\left(  a\right)  }_{=W\left(  b\right)  }-\underbrace{\left(
W\left(  a\right)  -a\right)  }_{\equiv W\left(  b\right)
-b\operatorname{mod}A_{M+1}}\equiv W\left(  b\right)  -\left(  W\left(
b\right)  -b\right)  =b\operatorname{mod}A_{M+1}.
\]
In other words, $a-b\in A_{M+1}$. That is, (\ref{pf.p3.l1.0.a.1}) holds for
$m=M+1$. This completes the induction step. Thus, (\ref{pf.p3.l1.0.a.1}) is proved.
\end{proof}

Now, (\ref{pf.p3.l1.0.a.1}) shows that $a-b\in A_{m}$ for each $m\in\left\{
1,2,3,\ldots\right\}  $. Thus,
\[
a-b\in\bigcap_{m\geq1}A_{m}=\bigcap\limits_{j\geq1}A_{j}=0.
\]
That is, $a-b=0$, so that $a=b$.

Forget that we fixed $a$ and $b$. We thus have shown that if $a$ and $b$ are
two elements of $A$ such that $W\left(  a\right)  =W\left(  b\right)  $, then
$a=b$. In other words, the map $W$ is injective. This proves Lemma
\ref{lem.banach-cons} \textbf{(a)}. \medskip

\textbf{(b)} We assumed that $A$ is complete. In other words, every Cauchy
sequence in $A$ has a limit. Here, a \emph{Cauchy sequence} means an infinite
sequence $\left(  a_{n}\right)  _{n\geq1}$ of elements of $A$ such that for
each $N\geq1$, there exists some $n\geq1$ such that%
\begin{equation}
\text{all integers }m_{1},m_{2}\geq n\text{ satisfy }a_{m_{1}}-a_{m_{2}}\in
A_{N}.\label{pf.p3.l1.0.b.1}%
\end{equation}
Furthermore, a \emph{limit} of such a sequence $\left(  a_{n}\right)
_{n\geq1}$ means an element $x\in A$ such that for each $N\geq1$, there exists
some $n\geq1$ such that
\begin{equation}
\text{all }m\geq n\text{ satisfy }x-a_{m}\in A_{N}.\label{pf.p3.l1.0.b.lim}%
\end{equation}


Now, let $u\in A$. We shall construct an $x\in A$ satisfying $W\left(
x\right)  =u$. This will prove that $W$ is surjective, and thus bijective
(since Lemma \ref{lem.banach-cons} \textbf{(a)} yields that $W$ is injective).

We define a sequence $\left(  a_{n}\right)  _{n\geq1}$ of elements of $A$
recursively by%
\begin{align}
a_{1}  &  =u\ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.p3.l1.0.b.2bas}\\
a_{n}  &  =a_{n-1}-W\left(  a_{n-1}\right)  +u\ \ \ \ \ \ \ \ \ \ \text{for
each }n\geq2. \label{pf.p3.l1.0.b.2rec}%
\end{align}
Then, we claim that
\begin{equation}
W\left(  a_{n}\right)  \equiv u\operatorname{mod}A_{n}%
\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq1. \label{pf.p3.l1.0.b.3}%
\end{equation}


\begin{proof}
[Proof of (\ref{pf.p3.l1.0.b.3}).]We shall prove (\ref{pf.p3.l1.0.b.3}) by
induction on $n$:

\textit{Base case:} For $n=1$, the claim (\ref{pf.p3.l1.0.b.3}) is true, since
$W\left(  a_{1}\right)  -u\in A=A_{1}$ and thus $W\left(  a_{1}\right)  \equiv
u\operatorname{mod}A_{1}$.

\textit{Induction step:} Let $M\in\left\{  1,2,3,\ldots\right\}  $ be such
that (\ref{pf.p3.l1.0.b.3}) holds for $n=M$. We must show that
(\ref{pf.p3.l1.0.b.3}) holds for $n=M+1$ as well.

We have assumed that (\ref{pf.p3.l1.0.b.3}) holds for $n=M$. In other words,
$W\left(  a_{M}\right)  \equiv u\operatorname{mod}A_{M}$. Hence, $u-W\left(
a_{M}\right)  \in A_{M}$. But (\ref{pf.p3.l1.0.b.2rec}) (applied to $n=M+1$)
yields
\[
a_{M+1}=a_{M}-W\left(  a_{M}\right)  +u=a_{M}+\left(  u-W\left(  a_{M}\right)
\right)  .
\]
Thus, $a_{M+1}-a_{M}=u-W\left(  a_{M}\right)  \in A_{M}$. In other words,
$a_{M+1}\equiv a_{M}\operatorname{mod}A_{M}$. Hence,
(\ref{eq.lem.banach-cons.ass}) (applied to $n=M$, $a=a_{M+1}$ and $b=a_{M}$)
yields
\[
W\left(  a_{M+1}\right)  -a_{M+1}\equiv W\left(  a_{M}\right)  -a_{M}%
\operatorname{mod}A_{M+1}.
\]
Therefore,
\begin{align*}
W\left(  a_{M+1}\right)   &  =\underbrace{a_{M+1}}_{=a_{M}+\left(  u-W\left(
a_{M}\right)  \right)  }+\underbrace{\left(  W\left(  a_{M+1}\right)
-a_{M+1}\right)  }_{\equiv W\left(  a_{M}\right)  -a_{M}\operatorname{mod}%
A_{M+1}}\\
&  \equiv\left(  a_{M}+\left(  u-W\left(  a_{M}\right)  \right)  \right)
+\left(  W\left(  a_{M}\right)  -a_{M}\right) \\
&  =u\operatorname{mod}A_{M+1}.
\end{align*}
In other words, (\ref{pf.p3.l1.0.b.3}) holds for $n=M+1$. This completes the
induction step. Thus, (\ref{pf.p3.l1.0.b.3}) is proved.
\end{proof}

Next, we claim that the sequence $\left(  a_{n}\right)  _{n\geq1}$ is a Cauchy sequence.

\begin{proof}
[Proof.]Let $N\geq1$. We must find some $n\geq1$ such that
(\ref{pf.p3.l1.0.b.1}) holds.

We claim that $n=N$ works. Indeed, let $m_{1},m_{2}\geq N$ be two integers. We
must show that $a_{m_{1}}-a_{m_{2}}\in A_{N}$.

Since $A_{N}$ is an abelian group, this is equivalent to showing that
$a_{m_{2}}-a_{m_{1}}\in A_{N}$. Thus, by symmetry, we WLOG assume that
$m_{1}\geq m_{2}$. For each $i\geq N$, we have%
\[
a_{i+1}=a_{i}-W\left(  a_{i}\right)  +u\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.p3.l1.0.b.2rec}), applied to }n=i+1\right)
\]
and thus
\begin{align}
a_{i+1}-a_{i}  &  =u-W\left(  a_{i}\right)  \in A_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since (\ref{pf.p3.l1.0.b.3}) yields
}W\left(  a_{i}\right)  \equiv u\operatorname{mod}A_{i}\right) \nonumber\\
&  \subset A_{N} \label{pf.p3.l1.0.b.cauchy.pdf.1}%
\end{align}
(by $A_{1}\supset A_{2}\supset A_{3}\supset\cdots$, since $i\geq N$). Hence,
by the telescope principle,
\begin{align*}
a_{m_{1}}-a_{m_{2}}  &  =\sum_{i=m_{2}}^{m_{1}-1}\underbrace{\left(
a_{i+1}-a_{i}\right)  }_{\substack{\in A_{N}\\\text{(by
(\ref{pf.p3.l1.0.b.cauchy.pdf.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}m_{1}\geq m_{2}\right) \\
&  \in\sum_{i=m_{2}}^{m_{1}-1}A_{N}\subset A_{N},
\end{align*}
since $A_{N}$ is a subgroup of $A$. Thus, (\ref{pf.p3.l1.0.b.1}) holds for
$n=N$. This proves that $\left(  a_{n}\right)  _{n\geq1}$ is a Cauchy sequence.
\end{proof}

Since $A$ is complete, the Cauchy sequence $\left(  a_{n}\right)  _{n\geq1}$
has a limit. Let $x$ be this limit. We shall show that $W\left(  x\right)  =u$.

Indeed, let $N\geq1$. Since $x$ is the limit of the sequence $\left(
a_{n}\right)  _{n\geq1}$, there exists some $n\geq1$ such that
(\ref{pf.p3.l1.0.b.lim}) holds. Consider this $n$. Set $m:=\max\left\{
n,N\right\}  $. Then, $m\geq n$ and $m\geq N$. Hence, $x-a_{m}\in A_{N}$ (by
(\ref{pf.p3.l1.0.b.lim})). From $A_{1}\supset A_{2}\supset A_{3}\supset\cdots
$, we obtain $A_{N+1}\subset A_{N}$ and $A_{m}\subset A_{N}$ (since $m\geq N$).

From $x-a_{m}\in A_{N}$, we obtain $x\equiv a_{m}\operatorname{mod}A_{N}$.
Hence, (\ref{eq.lem.banach-cons.ass}) (applied to $N$, $x$ and $a_{m}$ instead
of $n$, $a$ and $b$) yields
\[
W\left(  x\right)  -x\equiv W\left(  a_{m}\right)  -a_{m}\operatorname{mod}%
A_{N+1}.
\]
Since $A_{N+1}\subset A_{N}$, this entails%
\[
W\left(  x\right)  -x\equiv W\left(  a_{m}\right)  -a_{m}\operatorname{mod}%
A_{N}.
\]
Adding this congruence to the congruence $x\equiv a_{m}\operatorname{mod}%
A_{N}$, we obtain
\[
W\left(  x\right)  \equiv W\left(  a_{m}\right)  \operatorname{mod}A_{N}.
\]
On the other hand, (\ref{pf.p3.l1.0.b.3}) yields $W\left(  a_{m}\right)
\equiv u\operatorname{mod}A_{m}$, so that $W\left(  a_{m}\right)  \equiv
u\operatorname{mod}A_{N}$ (since $A_{m}\subset A_{N}$). Thus,%
\[
W\left(  x\right)  \equiv W\left(  a_{m}\right)  \equiv u\operatorname{mod}%
A_{N},
\]
so that $W\left(  x\right)  -u\in A_{N}$.

Forget that we fixed $N$. We thus have proved that $W\left(  x\right)  -u\in
A_{N}$ for each $N\geq1$. Hence,
\[
W\left(  x\right)  -u\in\bigcap_{N\geq1}A_{N}=\bigcap\limits_{j\geq1}A_{j}=0.
\]
Thus, $W\left(  x\right)  -u=0$, so that $W\left(  x\right)  =u$.

Now, forget that we fixed $u$. We have shown that for each $u\in A$, there
exists some $x\in A$ satisfying $W\left(  x\right)  =u$. In other words, the
map $W$ is surjective. Combined with Lemma \ref{lem.banach-cons} \textbf{(a)},
this shows that $W$ is bijective. This proves Lemma \ref{lem.banach-cons}
\textbf{(b)}.
\end{proof}

We now apply Lemma \ref{lem.banach-cons} to the setting of \S 1.1. Thus, $A$
is a complete filtered left pre-Lie algebra over a field $k$ of characteristic
$0$, with filtration
\[
A=A_{1}\supset A_{2}\supset A_{3}\supset\cdots
\]
satisfying $A_{p}\vartriangleright A_{q}\subset A_{p+q}$ and $\bigcap
\limits_{j\geq1}A_{j}=0$. We need to prove that the map
\begin{align*}
W:A  &  \rightarrow A\\
a  &  \mapsto e^{L_{a}}\mathbf{1}-\mathbf{1}%
\end{align*}
is a bijection (where $\mathbf{1}$ is a fictitious unit outside of $A$, but we
have $e^{L_{a}}\mathbf{1}-\mathbf{1}\in A$ nevertheless because the constant
term of the power series $e^{x}$ is $1$).

Explicitly, each $a\in A$ satisfies $e^{L_{a}}=\sum\limits_{r\geq0}\dfrac
{1}{r!}L_{a}^{r}$ and therefore%
\begin{equation}
W\left(  a\right)  =e^{L_{a}}\mathbf{1}-\mathbf{1}=\sum\limits_{r\geq0}%
\dfrac{1}{r!}L_{a}^{r}\mathbf{1}-\mathbf{1}=\sum_{r\geq1}\dfrac{1}{r!}%
L_{a}^{r}\mathbf{1}. \label{eq.p3.Wa=sum}%
\end{equation}
For each positive integer $r$, we shall denote the element
\[
L_{a}^{r}\mathbf{1}=\underbrace{a\vartriangleright\left(  a\vartriangleright
\left(  a\vartriangleright\left(  \cdots\vartriangleright\left(
a\vartriangleright a\right)  \right)  \right)  \right)  }_{\text{with }r\text{
copies of }a}\in A
\]
by $a_{\left[  r\right]  }$. Thus, the equality (\ref{eq.p3.Wa=sum}) can be
rewritten as%
\[
W\left(  a\right)  =\sum_{r\geq1}\dfrac{1}{r!}a_{\left[  r\right]  }.
\]
(The sum converges because $a_{\left[  r\right]  }\in A_{r}$ for all $r\geq1$,
as can be easily checked by induction on $r$.) Hence, for each $a\in A$, we
have%
\begin{equation}
W\left(  a\right)  -a=\sum_{r\geq1}\dfrac{1}{r!}a_{\left[  r\right]  }%
-a=\sum_{r\geq2}\dfrac{1}{r!}a_{\left[  r\right]  } \label{eq.p3.Wminusid}%
\end{equation}
(since $\dfrac{1}{1!}a_{\left[  1\right]  }=a_{\left[  1\right]  }=a$). Note
that the definition of $a_{\left[  r\right]  }$ readily yields
\begin{equation}
a_{\left[  r+1\right]  }=L_{a}a_{\left[  r\right]  }=a\vartriangleright
a_{\left[  r\right]  } \label{eq.p3.ar+1}%
\end{equation}
for each $r\geq1$.

We shall now check that our map $W:A\rightarrow A$ satisfies the assumption
(\ref{eq.lem.banach-cons.ass}) of Lemma \ref{lem.banach-cons}. So let $n\geq1$
and $a,b\in A$ be such that $a\equiv b\operatorname{mod}A_{n}$. We must prove
that
\[
W\left(  a\right)  -a\equiv W\left(  b\right)  -b\operatorname{mod}A_{n+1}.
\]
To do so, we first observe that
\begin{equation}
a_{\left[  r\right]  }-b_{\left[  r\right]  }\in A_{n+r-1}\quad\text{for every
}r\geq1. \label{eq.p3.Pr-difference}%
\end{equation}


\begin{proof}
[Proof of (\ref{eq.p3.Pr-difference}).]Indeed, (\ref{eq.p3.Pr-difference}) is
proved by induction on $r$:

The \textit{base case} $r=1$ says that $a_{\left[  1\right]  }-b_{\left[
1\right]  }\in A_{n}$; in other words, it says that $a-b\in A_{n}$ (since
$a_{\left[  1\right]  }=a$ and $b_{\left[  1\right]  }=b$). But this is just a
restatement of our hypothesis $a\equiv b\operatorname{mod}A_{n}$. Hence the
base case is complete.

For the \textit{induction step}, assume that (\ref{eq.p3.Pr-difference}) holds
for some $r\geq1$. Then, (\ref{eq.p3.ar+1}) yields
\begin{align*}
a_{\left[  r+1\right]  }-b_{\left[  r+1\right]  }  &  =a\vartriangleright
a_{\left[  r\right]  }-b\vartriangleright b_{\left[  r\right]  }\\
&  =\underbrace{a\vartriangleright a_{\left[  r\right]  }-b\vartriangleright
a_{\left[  r\right]  }}_{=\left(  a-b\right)  \vartriangleright a_{\left[
r\right]  }}+\underbrace{b\vartriangleright a_{\left[  r\right]
}-b\vartriangleright b_{\left[  r\right]  }}_{=b\vartriangleright\left(
a_{\left[  r\right]  }-b_{\left[  r\right]  }\right)  }\\
&  =\underbrace{\left(  a-b\right)  }_{\substack{\in A_{n}\\\text{(since
}a\equiv b\operatorname{mod}A_{n}\text{)}}}\vartriangleright
\underbrace{a_{\left[  r\right]  }}_{\in A_{r}}+\underbrace{b}_{\in A=A_{1}%
}\vartriangleright\underbrace{\left(  a_{\left[  r\right]  }-b_{\left[
r\right]  }\right)  }_{\substack{\in A_{n+r-1}\\\text{(by the induction
hypothesis)}}}\\
&  \in\underbrace{A_{n}\vartriangleright A_{r}}_{\subset A_{n+r}%
}+\underbrace{A_{1}\vartriangleright A_{n+r-1}}_{\subset A_{1+\left(
n+r-1\right)  }=A_{n+r}}\subset A_{n+r}+A_{n+r}=A_{n+r}.
\end{align*}
This proves (\ref{eq.p3.Pr-difference}) for $r+1$, and thus completes the
induction. Hence, (\ref{eq.p3.Pr-difference}) is proved.
\end{proof}

Note that each $r\geq2$ satisfies $n+r-1\geq n+1$ and thus%
\begin{equation}
A_{n+r-1}\subset A_{n+1} \label{eq.p3.Pr-dsub4}%
\end{equation}
(since $A_{1}\supset A_{2}\supset A_{3}\supset\cdots$).

Now, using (\ref{eq.p3.Wminusid}), we get
\begin{align*}
\left(  W\left(  a\right)  -a\right)  -\left(  W\left(  b\right)  -b\right)
&  =\sum_{r\geq2}\dfrac{1}{r!}a_{\left[  r\right]  }-\sum_{r\geq2}\dfrac
{1}{r!}b_{\left[  r\right]  }=\sum_{r\geq2}\dfrac{1}{r!}\underbrace{\left(
a_{\left[  r\right]  }-b_{\left[  r\right]  }\right)  }_{\substack{\in
A_{n+r-1}\\\text{(by (\ref{eq.p3.Pr-difference}))}}}\\
&  \in\sum_{r\geq2}\dfrac{1}{r!}\underbrace{A_{n+r-1}}_{\substack{\subset
A_{n+1}\\\text{(by (\ref{eq.p3.Pr-dsub4}))}}}\subset\sum_{r\geq2}\dfrac{1}%
{r!}A_{n+1}\subset A_{n+1}%
\end{align*}
(since $A_{n+1}$ is closed with respect to the topology defined by our
filtration). In other words,%
\[
W\left(  a\right)  -a\equiv W\left(  b\right)  -b\operatorname{mod}A_{n+1}.
\]
This proves the assumption (\ref{eq.lem.banach-cons.ass}) of Lemma
\ref{lem.banach-cons}. Hence, Lemma \ref{lem.banach-cons} \textbf{(b)} shows
that $W$ is bijective. This justifies the sentence \textquotedblleft The
application $W$ is clearly a bijection\textquotedblright.


\end{document}