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\ihead{Errata to ``The Hopf algebra of finite topologies ...''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{The Hopf algebra of finite topologies and }$T$\textbf{-partitions}

\textit{Loic Foissy and Claudia Malvenuto}

version of 4 October 2014 (arXiv preprint arXiv:1407.0476v2)

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

\begin{itemize}
\item \textbf{Various places:} You use the two different \textquotedblleft
spellings\textquotedblright\ \textquotedblleft$T$-partition\textquotedblright%
\ (with a mathmode \textquotedblleft$T$\textquotedblright) and
\textquotedblleft T-partition\textquotedblright\ (with a textmode
\textquotedblleft T\textquotedblright) synonymously. It would be better if you
keep to one of them. I, personally, prefer the latter, since the former
(falsely) suggests that it depends on some object called \textquotedblleft%
$T$\textquotedblright\ (which does not exist and is a red herring; the actual
argument is the \textquotedblleft$\mathcal{T}$\textquotedblright\ in
\textquotedblleft T-partition of $\mathcal{T}$\textquotedblright).

\item \textbf{Page 2:} Replace \textquotedblleft Retenauer\textquotedblright%
\ by \textquotedblleft Reutenauer\textquotedblright.

\item \textbf{Page 2:} \textquotedblleft in this the present
text\textquotedblright\ $\rightarrow$ \textquotedblleft in the present
text\textquotedblright.

\item \textbf{Page 2:} \textquotedblleft is totally ordered by the
refinement\textquotedblright\ $\rightarrow$ \textquotedblleft is partially
ordered by refinement\textquotedblright.

\item \textbf{Page 2:} I think \textquotedblleft the
coproducts\textquotedblright\ should be \textquotedblleft the
coproduct\textquotedblright, or not?

\item \textbf{Page 3:} In the definition of a \textquotedblleft strict
$T$-partition\textquotedblright, replace \textquotedblleft and $k\leq
_{\mathcal{T}}i$\textquotedblright\ by \textquotedblleft and $j\leq
_{\mathcal{T}}i$\textquotedblright.

\item \textbf{Page 3:} In the displayed equation%
\[
\Gamma_{(1,1,1)}\left(  \mathcal{T}\right)  =\sum_{{\scriptsize f}\text{
generalized }T\text{-partition of }\mathbf{T}}f,\ \ \ \ \ \ \ \ \ \ \Gamma
_{(1,0,0)}\left(  \mathcal{T}\right)  =\sum_{{\scriptsize f}\text{ strict
}T\text{-partition of }\mathbf{T}}f,
\]
replace both appearances of \textquotedblleft$\mathbf{T}$\textquotedblright%
\ by \textquotedblleft$\mathcal{T}$\textquotedblright.

\item \textbf{Page 3:} \textquotedblleft to defined\textquotedblright%
\ $\rightarrow$ \textquotedblleft to define\textquotedblright.

\item \textbf{Page 4:} Somewhere here it would be good to point out that
$\max\varnothing$ is to be understood as $0$. (You use this convention when
you write $\max f$ for a packed word $f$.)

\item \textbf{Page 4, \S 1.1:} In the \textquotedblleft$\Delta\left(  \left(
511423\right)  \right)  $\textquotedblright\ example, replace
\textquotedblleft$\left(  1\right)  \otimes\left(  4312\right)  $%
\textquotedblright\ by \textquotedblleft$\left(  11\right)  \otimes\left(
4312\right)  $\textquotedblright.

\item \textbf{Page 4, \S 1.1:} It would be better if you use a different
letter for the involution that you call $j$; you tend to use the letter
\textquotedblleft$j$\textquotedblright\ for integers too.

\item \textbf{Page 5:} In the first displayed equation on page 5, replace
\textquotedblleft$\sigma\otimes\tau$\textquotedblright\ by \textquotedblleft%
$\sigma\otimes\sigma^{\prime}$\textquotedblright.

\item \textbf{Page 5:} Also, in the same equation, it would be good if you
define $\sigma\otimes\sigma^{\prime}$. (I assume that it means the permutation
in $\mathfrak{S}_{n+n^{\prime}}$ which sends every $k\in\left\{
1,2,\ldots,n+n^{\prime}\right\}  $ to $\left\{
\begin{array}
[c]{c}%
\sigma\left(  k\right)  ,\ \ \ \ \ \ \ \ \ \ \text{if }k\leq n;\\
\sigma^{\prime}\left(  k-n\right)  +n,\ \ \ \ \ \ \ \ \ \ \text{if }k>n
\end{array}
\right.  $.)

\item \textbf{Page 6:} \textquotedblleft we define a special
posets\textquotedblright\ should be \textquotedblleft we define a special
poset\textquotedblright.

\item \textbf{Page 6:} \textquotedblleft(isoclasses) of special
posets\textquotedblright\ should be \textquotedblleft(isoclasses of) special
posets\textquotedblright.

\item \textbf{Page 6:} In \textquotedblleft Up to a unique isomorphism, we can
assume that $P=\left[  n\right]  $ as a totally ordered set\textquotedblright,
replace \textquotedblleft$P$\textquotedblright\ by \textquotedblleft$\left(
P,\leq_{tot}\right)  $\textquotedblright.

\item \textbf{Page 6:} Your definition of $L$ is incompatible with the
definition given in the reference [11]. (Also, the way how you identify linear
extensions of special double posets with certain permutations is incompatible
with how you do it in [11].) Maybe warn the reader about this, if this is your intention?

\item \textbf{Page 7, \S 1.2:} \textquotedblleft We represent a $P$%
-partition\textquotedblright\ should be \textquotedblleft We represent a
$P$-partition $f$\textquotedblright.

\item \textbf{Page 7, Remark:} In the summation subscript \textquotedblleft$f$
$P$-partition of $w$\textquotedblright, replace \textquotedblleft%
$w$\textquotedblright\ by \textquotedblleft$\mathcal{P}$\textquotedblright.

\item \textbf{Page 7, \S 1.2:} Replace \textquotedblleft$\left\{
w\ \mid\ Pack\left(  w\right)  =\sigma\right\}  $\textquotedblright\ by
\textquotedblleft$\left\{  w\ \mid\ Std\left(  w\right)  =\sigma\right\}
$\textquotedblright.

\item \textbf{Page 8, \S 2.1:} You write: \textquotedblleft Moreover, the open
sets of $\mathcal{T}$ are the ideals of $\leq_{\mathcal{T}}$\textquotedblright%
. This was a neat exercise to prove, but maybe a reference to a proof in the
literature would not hurt?

\item \textbf{Page 8:} In \textquotedblleft Moreover, $\leq_{T_{\leq}}=\leq$,
and $\mathcal{T}_{\leq_{\mathcal{T}}}=\mathcal{T}$\textquotedblright, replace
\textquotedblleft$T$\textquotedblright\ by \textquotedblleft$\mathcal{T}%
$\textquotedblright.

\item \textbf{Page 8:} Replace \textquotedblleft$\overline{i}\leq
_{\mathcal{T}}\overline{j}$ if $i\leq j$\textquotedblright\ by
\textquotedblleft$\overline{i}\leq_{\mathcal{T}}\overline{j}$ if
$i\leq_{\mathcal{T}}j$\textquotedblright\ (or generalize this definition of
quotient posets to an arbitrary preorder).

\item \textbf{Page 9, Definition 2:} I think \textquotedblleft
which\textquotedblright\ should be \textquotedblleft whose\textquotedblright%
\ (both in (1) and in (2)).

\item \textbf{Page 11, \S 2.3:} When you define $Std\left(  \mathcal{T}%
\right)  $ and $\mathcal{T}_{\mid Y}$, it would be useful to point out how
these definitions translate into the language of preorders. Namely:

\begin{itemize}
\item If $X$ is a finite totally ordered set of cardinality $n$, if
$\mathcal{T}$ is a topology on $X$, and if $\phi$ is the unique increasing
bijection from $X$ to $\left[  n\right]  $, then two elements $i$ and $j$ of
$\left[  n\right]  $ satisfy $i\leq_{Std\left(  \mathcal{T}\right)  }j$ if and
only if $\phi^{-1}\left(  i\right)  \leq_{\mathcal{T}}\phi^{-1}\left(
j\right)  $.

\item If $X$ is a finite set, if $\mathcal{T}$ is a topology on $X$, and if
$Y\subseteq X$, then two elements $i$ and $j$ of $Y$ satisfy $i\leq
_{\mathcal{T}_{\mid Y}}j$ if and only if $i\leq_{\mathcal{T}}j$.
\end{itemize}

\item \textbf{Page 11, Proposition 6:} Replace \textquotedblleft$n\geq
1$\textquotedblright\ by \textquotedblleft$n\geq0$\textquotedblright.

\item \textbf{Page 11, proof of Proposition 6:} At the beginning of this
proof, I would suggest adding the observation that if $X$ is a finite totally
ordered set and if $\mathcal{T}$ is a topology on $X$, then%
\[
\Delta\left(  Std\left(  \mathcal{T}\right)  \right)  =\sum_{O\in\mathcal{T}%
}Std\left(  \mathcal{T}_{\mid X\setminus O}\right)  \otimes Std\left(
\mathcal{T}_{\mid O}\right)  .
\]
This formula generalizes the formula that you use to define $\Delta$, and it
can be easily derived from the latter (since the open sets of $\mathcal{T}$
and the open sets of $Std\left(  \mathcal{T}\right)  $ are in an obvious
1-to-1 correspondence). You later tacitly use this formula when you compute
$\left(  \Delta\otimes Id\right)  \circ\Delta\left(  \mathcal{T}\right)  $ and
$\left(  Id\otimes\Delta\right)  \circ\Delta\left(  \mathcal{T}\right)  $.

\item \textbf{Page 12:} It would be helpful to explain the notations
$O.O^{\prime}$ (meaning $O\sqcup O^{\prime}\left(  +n\right)  $) and
$O\downarrow\left[  n^{\prime}\right]  $ (meaning $O\sqcup\left[  n^{\prime
}\right]  \left(  +n\right)  $), where $O\in\mathcal{T}$ and $O^{\prime}%
\in\mathcal{T}^{\prime}$ for two topologies $\mathcal{T}\in\mathbf{T}_{n}$ and
$\mathcal{T}^{\prime}\in\mathbf{T}_{n^{\prime}}$.

\item \textbf{Page 12:} On the second line of the computation that shows
$\Delta\left(  \mathcal{T}.\mathcal{T}^{\prime}\right)  =\Delta\left(
\mathcal{T}\right)  .\Delta\left(  \mathcal{T}^{\prime}\right)  $, replace
\textquotedblleft$\mathcal{T}_{\left[  n^{\prime}\right]  \setminus O^{\prime
}}^{\prime}$\textquotedblright\ by \textquotedblleft$\mathcal{T}_{\mid\left[
n^{\prime}\right]  \setminus O^{\prime}}^{\prime}$\textquotedblright, and also
replace \textquotedblleft$\mathcal{T}\mid_{O^{\prime}}$\textquotedblright\ by
\textquotedblleft$\mathcal{T}^{\prime}\mid_{O^{\prime}}$\textquotedblright.

\item \textbf{Page 12:} On the third line of the same computation, again
replace \textquotedblleft$\mathcal{T}\mid_{O^{\prime}}$\textquotedblright\ by
\textquotedblleft$\mathcal{T}^{\prime}\mid_{O^{\prime}}$\textquotedblright.

\item \textbf{Page 12:} On the second-to-last line of the last computation of
this page, replace \textquotedblleft$\Delta\left(  \mathcal{T}\right)
-1\otimes\mathcal{T}^{\prime}$\textquotedblright\ by \textquotedblleft%
$\Delta\left(  \mathcal{T}^{\prime}\right)  -1\otimes\mathcal{T}^{\prime}%
$\textquotedblright.

\item \textbf{Page 12:} On the last line of the last computation of this page,
replace \textquotedblleft$\left(  \mathcal{T}\otimes1\right)  \downarrow
\Delta\left(  \mathcal{T}\right)  $\textquotedblright\ by \textquotedblleft%
$\left(  \mathcal{T}\otimes1\right)  \downarrow\Delta\left(  \mathcal{T}%
^{\prime}\right)  $\textquotedblright.

\item \textbf{Page 13, \S 2.4:} Here is one observation that you tacitly use
in some of your arguments: If $X$ is a finite totally ordered set with total
ordering $\leq_{tot}$, and if $\mathcal{T}$ is a topology on $X$ such that
$\mathcal{T}$ is $T_{0}$, then
\begin{equation}
Std\left(  \mathcal{T}\right)  \cong\left(  X,\leq_{\mathcal{T}},\leq
_{tot}\right)  \label{p13.2.4.a}%
\end{equation}
as special posets.

\item \textbf{Page 14, proof of Proposition 8:} Replace \textquotedblleft%
$\sim_{\mathcal{T}}^{\prime}$\textquotedblright\ by \textquotedblleft%
$\sim_{\mathcal{T}^{\prime}}$\textquotedblright\ (on the 4th line of page 14).

\item \textbf{Page 14, proof of Proposition 8:} Replace \textquotedblleft If
$\mathcal{T}\in\mathbf{T}_{n}$, $n\geq1$\textquotedblright\ by
\textquotedblleft If $\mathcal{T}\in\mathbf{T}_{n}$, $n\geq0$%
\textquotedblright.

\item \textbf{Page 14, proof of Proposition 8:} Replace \textquotedblleft%
$Std\left(  O/\sim_{\mathcal{T}}\right)  $\textquotedblright\ by
\textquotedblleft$O/\sim_{\mathcal{T}}$\textquotedblright.

\item \textbf{Page 14, proof of Proposition 8:} Remove the words
\textquotedblleft If $\mathcal{T}$ has $k$ equivalence
classes\textquotedblright\ (as you never get to use $k$).

\item \textbf{Page 14, proof of Proposition 8:} In the last displayed equation
of this proof, you rewrite $\left(  \left[  n\right]  \setminus O\right)
/\sim_{\mathcal{T}}$ as $\overline{Std\left(  \mathcal{T}_{\mid\left[
n\right]  \setminus O}\right)  }$, and rewrite $O/\sim_{\mathcal{T}}$ as
$\overline{Std\left(  \mathcal{T}_{\mid O}\right)  }$. It would be good if you
would add some explanation of why this rewriting is possible. The reason is
that whenever $Z$ is a subset of $\left[  n\right]  $ which is a union of
equivalence classes of $\mathcal{T}$\ \ \ \ \footnote{In particular, $Z$ can
be any open set of $\mathcal{T}$ or any closed set of $\mathcal{T}$.}, we have
$Z/\sim_{\mathcal{T}}\cong\overline{Std\left(  \mathcal{T}_{\mid Z}\right)  }$
as special posets. This fact is straightforward to check (using
(\ref{p13.2.4.a})), but in my opinion is worth explicitly stating and
explicitly referencing when you use it.

\item \textbf{Page 14, commutative diagram:} I would rather not use the label
\textquotedblleft$\iota$\textquotedblright\ for the inclusion map
$\mathbf{H}_{\mathbf{SP}}\rightarrow\mathbf{H}_{\mathbf{T}}$, given that
$\iota$ already means something different.

\item \textbf{Page 15, Definition 9:} Replace \textquotedblleft$\mathcal{T}%
\in\mathbf{T}_{l}$\textquotedblright\ by \textquotedblleft$\mathcal{T}%
^{\prime}\in\mathbf{T}_{l}$\textquotedblright.

\item \textbf{Page 15, Proposition 10:} I think the claim that
\textquotedblleft$\iota$ is an isometry for this pairing\textquotedblright\ is
wrong, unless I incorrectly understood the definition of $\iota$. For a
counterexample, let $\mathcal{T}=%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%& 1 \ar@{-}[dl] \ar@{-}[dr] & \\
%2 & & 3
%}}}%
%BeginExpansion
\xymatrix{
& 1 \ar@{-}[dl] \ar@{-}[dr] & \\
2 & & 3
}%
%EndExpansion
$ and $\mathcal{T}^{\prime}=%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%& 2 \ar@{-}[dl] \ar@{-}[dr] & \\
%1 & & 3
%}}}%
%BeginExpansion
\xymatrix{
& 2 \ar@{-}[dl] \ar@{-}[dr] & \\
1 & & 3
}%
%EndExpansion
$ (where both $\mathcal{T}$ and $\mathcal{T}^{\prime}$ are $T_{0}$, and I draw
them as the Hasse diagrams of their posets). Then, $Pic\left(  \mathcal{T}%
,\mathcal{T}^{\prime}\right)  $ has one element (namely, the map sending
$1,2,3$ to $3,1,2$, respectively), but $Pic\left(  \iota\left(  \mathcal{T}%
\right)  ,\iota\left(  \mathcal{T}^{\prime}\right)  \right)  $ is empty (since
$i\left(  \mathcal{T}\right)  =%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%2 & & 3 \\
%& 1 \ar@{-}[ul] \ar@{-}[ur] &
%}}}%
%BeginExpansion
\xymatrix{
2 & & 3 \\
& 1 \ar@{-}[ul] \ar@{-}[ur] &
}%
%EndExpansion
$ and $\iota\left(  \mathcal{T}^{\prime}\right)  =%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%1 & & 3 \\
%& 2 \ar@{-}[ul] \ar@{-}[ur] &
%}}}%
%BeginExpansion
\xymatrix{
1 & & 3 \\
& 2 \ar@{-}[ul] \ar@{-}[ur] &
}%
%EndExpansion
$).

On the other hand, if you change the definition of $\iota$ so that $\iota$
also reverses the labelling of the ground set, then this new $\iota$ is an
isometry for the pairing (in fact, then you have $Pic\left(  \mathcal{T}%
,\mathcal{T}^{\prime}\right)  =Pic\left(  \iota\left(  \mathcal{T}\right)
,\iota\left(  \mathcal{T}^{\prime}\right)  \right)  $ as sets), but then
Proposition 6 (3) no longer holds.

\item \textbf{Page 15, proof of Proposition 10:} It would be helpful to
observe that $n=n_{1}+n_{2}$ (directly after \textquotedblleft Let $f\in
Pic\left(  \mathcal{T}_{1}.\mathcal{T}_{2},\mathcal{T}\right)  $%
\textquotedblright).

\item \textbf{Page 15, proof of Proposition 10:} You write: \textquotedblleft
Moreover, by restriction, $Std\left(  f_{\mid\left[  n_{1}\right]  }\right)  $
is a picture between $\mathcal{T}_{1}$ and $Std\left(  \mathcal{T}%
_{\mid\left[  n\right]  \setminus O_{f}}\right)  $ and $Std\left(
f_{\mid\left[  n_{1}+n_{2}\right]  \setminus\left[  n_{2}\right]  }\right)  $
is a picture between $\mathcal{T}_{2}$ and $Std\left(  \mathcal{T}_{\mid
O}\right)  $\textquotedblright, replace \textquotedblleft$\mathcal{T}_{\mid
O}$\textquotedblright\ by \textquotedblleft$\mathcal{T}_{\mid O_{f}}%
$\textquotedblright. Also, I think you should define what \textquotedblleft%
$Std\left(  f_{\mid\left[  n_{1}\right]  }\right)  $\textquotedblright\ and
\textquotedblleft$Std\left(  f_{\mid\left[  n_{1}+n_{2}\right]  \setminus
\left[  n_{2}\right]  }\right)  $\textquotedblright\ mean. The definition that
you are using and omitting seems to be the following one:

Let $X_{1}$ be a finite totally ordered set of cardinality $m_{1}$. Let
$\phi_{1}$ be the unique increasing bijection from $X_{1}$ to $\left[
m_{1}\right]  $. Let $\mathcal{S}_{1}$ be a topology on $X_{1}$.

Let $X_{2}$ be a finite totally ordered set of cardinality $m_{2}$. Let
$\phi_{2}$ be the unique increasing bijection from $X_{2}$ to $\left[
m_{2}\right]  $. Let $\mathcal{S}_{2}$ be a topology on $X_{2}$.

Let $f$ be a map $X_{1}\rightarrow X_{2}$. Then, $Std\left(  f\right)  $ is
defined to be the map $\phi_{2}\circ f\circ\phi_{1}^{-1}:\left[  m_{1}\right]
\rightarrow\left[  m_{2}\right]  $.

\item \textbf{Page 16, proof of Proposition 10:} In \textquotedblleft$\sharp
Pic\left(  \mathcal{T}_{1},Std\left(  \mathcal{T}_{\mid\left[  n\right]
\setminus O}\right)  \right)  \times\sharp Pic\left(  \mathcal{T}%
_{2},Std\left(  \mathcal{T}_{\mid O}\right)  \right)  $\textquotedblright,
replace the \textquotedblleft$\times$\textquotedblright\ sign by a
\textquotedblleft$\cdot$\textquotedblright.

\item \textbf{Page 16, proof of Proposition 10:} Replace \textquotedblleft%
$\left\langle Pic\left(  \mathcal{T}_{2},Std(\mathcal{T}_{\mid O}\right)
\right\rangle $\textquotedblright\ by \textquotedblleft$\left\langle
\mathcal{T}_{2},Std\left(  \mathcal{T}_{\mid O}\right)  \right\rangle
$\textquotedblright.

\item \textbf{Page 16, \S 3.1:} It would make more sense if you define your
partial order not just on $\mathbf{T}_{n}$, but more generally on the set of
all topologies on $X$ whenever $X$ is a finite set. This more general notion
is used in the first step of your proof of Theorem 12 (3).

\item \textbf{Page 18, proof of Theorem 12 (1):} Replace \textquotedblleft%
$I_{2}=I\cap\left(  \left[  k+l\right]  \setminus\left[  l\right]  \right)
\left(  -k\right)  $\textquotedblright\ by \textquotedblleft$I_{2}=\left(
I\cap\left(  \left[  k+l\right]  \setminus\left[  l\right]  \right)  \right)
\left(  -k\right)  $\textquotedblright.

\item \textbf{Page 18, proof of Theorem 12 (1):} You have not defined what
$\left(  I\cap\left(  \left[  k+l\right]  \setminus\left[  l\right]  \right)
\right)  \left(  -k\right)  $ means. (To fix this, add \textquotedblleft The
set $O\left(  -n\right)  $ is the set $\left\{  k-n\ \mid\ k\in O\right\}
$\textquotedblright\ immediately after the sentence \textquotedblleft The set
$O\left(  +n\right)  $ is the set $\left\{  k+n\ \mid\ k\in O\right\}
$\textquotedblright\ on page 9.)

\item \textbf{Page 18, proof of Theorem 12 (1):} Replace \textquotedblleft%
$I_{1}\sqcup I_{2}\left[  k\right]  $\textquotedblright\ by \textquotedblleft%
$I_{1}\sqcup I_{2}\left(  +k\right)  $\textquotedblright.

\item \textbf{Page 18, proof of Theorem 12 (2):} Replace \textquotedblleft any
open sets\textquotedblright\ by \textquotedblleft any open
set\textquotedblright.

\item \textbf{Page 19, proof of Theorem 12 (3):} I personally find the
equality sign between%
\[
\sum_{O\in\mathcal{T}}\sum_{\substack{\mathcal{T}^{\prime}\leq\mathcal{T}%
,\ O\in\mathcal{T}^{\prime},\\\left(  \left[  n\right]  \setminus O\right)
<_{\mathcal{T}^{\prime}}O}}R_{Std\left(  \mathcal{T}_{\mid\left[  n\right]
\setminus O}^{\prime}\right)  }\otimes R_{Std\left(  \mathcal{T}_{\mid
O}^{\prime}\right)  }%
\]
and%
\[
\sum_{O\in\mathcal{T}}\sum_{\substack{\mathcal{S}\leq Std\left(
\mathcal{T}_{\mid\left[  n\right]  \setminus O}\right)  ,\\\mathcal{S}%
^{\prime}\leq Std\left(  \mathcal{T}_{\mid O}\right)  }}R_{\mathcal{S}}\otimes
R_{\mathcal{S}^{\prime}}%
\]
a bit underexplained. Here is how I would explain why these terms are equal:

First, we need a lemma:

\textbf{Lemma A.} Let $O\in\mathcal{T}$, $\mathcal{S}^{\prime}\leq
\mathcal{T}_{\mid O}$ and $\mathcal{S}\leq\mathcal{T}_{\mid\left[  n\right]
\setminus O}$.

\textbf{(a)} There exists a unique topology $\mathcal{T}^{\prime}\in
\mathbf{T}_{n}$ such that $\mathcal{S}^{\prime}=\mathcal{T}_{\mid O}^{\prime}%
$, $\mathcal{S}=\mathcal{T}_{\mid\left[  n\right]  \setminus O}^{\prime}$ and
$\left(  \left[  n\right]  \setminus O\right)  <_{\mathcal{T}^{\prime}}O$.

\textbf{(b)} This topology $\mathcal{T}^{\prime}$ is given by $\mathcal{T}%
^{\prime}=\left\{  \Omega\cup O\ \mid\ \Omega\in\mathcal{S}\right\}
\cup\mathcal{S}^{\prime}$.

\textbf{(c)} This topology $\mathcal{T}^{\prime}$ further satisfies
$\mathcal{T}^{\prime}\leq\mathcal{T}$ and $O\in\mathcal{T}^{\prime}$.

\textit{Proof of Lemma A.} Parts \textbf{(a)} and \textbf{(b)} of Lemma A are
merely the result of your First step, but with $\mathcal{S}$ and
$\mathcal{S}^{\prime}$ renamed as $\mathcal{S}^{\prime}$ and $\mathcal{S}$.

\textbf{(c)} From \textbf{(b)}, we have $\mathcal{T}^{\prime}=\left\{
\Omega\cup O\ \mid\ \Omega\in\mathcal{S}\right\}  \cup\mathcal{S}^{\prime}$.
Thus, $O\in\mathcal{T}^{\prime}$. It remains to prove that $\mathcal{T}%
^{\prime}\leq\mathcal{T}$. In other words, we need to prove that
$I\in\mathcal{T}$ for every $I\in\mathcal{T}^{\prime}$. Indeed, let
$I\in\mathcal{T}^{\prime}$. Then, $I\in\mathcal{T}^{\prime}=\left\{
\Omega\cup O\ \mid\ \Omega\in\mathcal{S}\right\}  \cup\mathcal{S}^{\prime}$,
so that either $I\in\left\{  \Omega\cup O\ \mid\ \Omega\in\mathcal{S}\right\}
$ or $I\in\mathcal{S}^{\prime}$. In the first case, we have $I=\Omega\cup O$
for some $\Omega\in\mathcal{S}$; now, this $\Omega$ satisfies $\Omega
\in\mathcal{S}\subseteq\mathcal{T}_{\mid\left[  n\right]  \setminus O}$ and
thus $\Omega\cup O\in\mathcal{T}$, so that $I=\Omega\cup O\in\mathcal{T}$. In
the second case, $I\in\mathcal{S}^{\prime}\subseteq\mathcal{T}_{\mid
O}\subseteq\mathcal{T}$ (since $O$ is open in $\mathcal{T}$). Hence, we have
proven $I\in\mathcal{T}$ in both cases. Thus, $\mathcal{T}^{\prime}%
\leq\mathcal{T}$ is proven, and thus the proof of Lemma A \textbf{(c)} is complete.

As a consequence of Lemma A, we have:

\textbf{Lemma B.} Let $O\in\mathcal{T}$, $\mathcal{S}^{\prime}\leq
\mathcal{T}_{\mid O}$ and $\mathcal{S}\leq\mathcal{T}_{\mid\left[  n\right]
\setminus O}$. Then, there exists a unique topology $\mathcal{T}^{\prime}%
\in\mathbf{T}_{n}$ such that $\mathcal{T}^{\prime}\leq\mathcal{T}$%
,$\ O\in\mathcal{T}^{\prime}$, $\left(  \left[  n\right]  \setminus O\right)
<_{\mathcal{T}^{\prime}}O$, $\mathcal{S}=\mathcal{T}_{\mid\left[  n\right]
\setminus O}^{\prime}$ and $\mathcal{S}^{\prime}=\mathcal{T}_{\mid O}^{\prime
}$.

\textit{Proof of Lemma B.} The uniqueness follows from Lemma A \textbf{(a)},
while the existence follows from parts \textbf{(a)} and \textbf{(c)} of Lemma A.

Now that Lemma B is proven, we have%
\begin{align*}
&  \sum_{O\in\mathcal{T}}\sum_{\substack{\mathcal{T}^{\prime}\leq
\mathcal{T},\ O\in\mathcal{T}^{\prime},\\\left(  \left[  n\right]  \setminus
O\right)  <_{\mathcal{T}^{\prime}}O}}R_{Std\left(  \mathcal{T}_{\mid\left[
n\right]  \setminus O}^{\prime}\right)  }\otimes R_{Std\left(  \mathcal{T}%
_{\mid O}^{\prime}\right)  }\\
&  =\sum_{O\in\mathcal{T}}\sum_{\substack{\mathcal{S}\leq\mathcal{T}%
_{\mid\left[  n\right]  \setminus O},\\\mathcal{S}^{\prime}\leq\mathcal{T}%
_{\mid O}}}\underbrace{\sum_{\substack{\mathcal{T}^{\prime}\leq\mathcal{T}%
,\ O\in\mathcal{T}^{\prime},\\\left(  \left[  n\right]  \setminus O\right)
<_{\mathcal{T}^{\prime}}O,\\\mathcal{S}=\mathcal{T}_{\mid\left[  n\right]
\setminus O}^{\prime},\ \mathcal{S}^{\prime}=\mathcal{T}_{\mid O}^{\prime}%
}}R_{Std\left(  \mathcal{S}\right)  }\otimes R_{Std\left(  \mathcal{S}%
^{\prime}\right)  }}_{\substack{\text{this sum has precisely one
term}\\\text{(due to Lemma B), and thus equals}\\R_{Std\left(  \mathcal{S}%
\right)  }\otimes R_{Std\left(  \mathcal{S}^{\prime}\right)  }}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathcal{T}_{\mid\left[  n\right]
\setminus O}^{\prime}\leq\mathcal{T}_{\mid\left[  n\right]  \setminus O}\text{
and }\mathcal{T}_{\mid O}^{\prime}\leq\mathcal{T}_{\mid O}\text{ for every
}\mathcal{T}^{\prime}\leq\mathcal{T}\right) \\
&  =\sum_{O\in\mathcal{T}}\sum_{\substack{\mathcal{S}\leq\mathcal{T}%
_{\mid\left[  n\right]  \setminus O},\\\mathcal{S}^{\prime}\leq\mathcal{T}%
_{\mid O}}}R_{Std\left(  \mathcal{S}\right)  }\otimes R_{Std\left(
\mathcal{S}^{\prime}\right)  }\\
&  =\sum_{O\in\mathcal{T}}\sum_{\substack{\mathcal{S}\leq Std\left(
\mathcal{T}_{\mid\left[  n\right]  \setminus O}\right)  ,\\\mathcal{S}%
^{\prime}\leq Std\left(  \mathcal{T}_{\mid O}\right)  }}R_{\mathcal{S}}\otimes
R_{\mathcal{S}^{\prime}}%
\end{align*}
(here, we substituted $Std\left(  \mathcal{S}\right)  $ and $Std\left(
\mathcal{S}^{\prime}\right)  $ for $\mathcal{S}$ and $\mathcal{S}^{\prime}$ in
the sum).

\item \textbf{Page 19, proof of Theorem 12 (3):} In \textquotedblleft We used
the first step for the third equality\textquotedblright, replace
\textquotedblleft third\textquotedblright\ by \textquotedblleft
fourth\textquotedblright\ (or by a larger number if you add in intermediate steps).

\item \textbf{Page 19, Definition 13:} In (2), replace \textquotedblleft Let
$f$ a generalized\textquotedblright\ by \textquotedblleft Let $f$ be a
generalized\textquotedblright.

\item \textbf{Page 19, Definition 13:} In (2), replace \textquotedblleft%
$i,j\in\left[  n\right]  $\textquotedblright\ by \textquotedblleft%
$i,j,k\in\left[  n\right]  $\textquotedblright.

\item \textbf{Page 19, Definition 13:} Again, I'd recommend generalizing this
definition from the case $\mathcal{T}\in\mathbf{T}_{n}$ to the case when
$\mathcal{T}$ is a topology on a finite set $X$ equipped with a total
order.\footnote{This generality is used on page 21 (in notation like
\textquotedblleft$\mathcal{P}\left(  \mathcal{T}_{\mid O}\right)
$\textquotedblright).} I think the generalized definition will look like this:

\textbf{Definition 13'.} Let $X$ be a finite totally ordered set. Let
$\mathcal{T}$ be a topology on $X$.

\begin{enumerate}
\item[(1)] A \textit{generalized T-partition} of $\mathcal{T}$ is a surjective
map $f:X\rightarrow\left[  p\right]  $ such that if $i\leq_{\mathcal{T}}j$ in
$X$, then $f\left(  i\right)  \leq f\left(  j\right)  $ in $\left[  p\right]
$. If $f$ is a generalized T-partition of $\mathcal{T}$, we shall represent it
by the packed word $f\left(  x_{1}\right)  f\left(  x_{2}\right)  \ldots
f\left(  x_{n}\right)  $, where $\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ is
the list of the elements of $X$ in increasing order (with respect to the total
order on $X$).

\item[(2)] Let $f$ be a generalized T-partition of $\mathcal{T}$. We shall say
that $f$ is a \textit{(strict) T-partition} if for all $i,j\in X$:

\begin{itemize}
\item $i<_{\mathcal{T}}j$ and $i>j$ implies that $f\left(  i\right)  <f\left(
j\right)  $ in $\left[  p\right]  $.

\item If $i<j<k$, $i\sim_{\mathcal{T}}k$ and $f\left(  i\right)  =f\left(
j\right)  =f\left(  k\right)  $, then $i\sim_{\mathcal{T}}j$ and
$j\sim_{\mathcal{T}}k$.
\end{itemize}

\item[(3)] The set of generalized T-partitions of $\mathcal{T}$ is denoted by
$\mathcal{P}\left(  \mathcal{T}\right)  $; the set of (strict) T-partitions of
$\mathcal{T}$ is denoted by $\mathcal{P}_{s}\left(  \mathcal{T}\right)  $.

\item[(4)] If $f\in\mathcal{P}\left(  \mathcal{T}\right)  $, we put:%
\begin{align*}
\ell_{1}\left(  f\right)   &  =\sharp\left\{  \left(  i,j\right)  \in
X^{2}\ \mid\ i<_{\mathcal{T}}j,\ i<j,\text{ and }f\left(  i\right)  =f\left(
j\right)  \right\}  ,\\
\ell_{2}\left(  f\right)   &  =\sharp\left\{  \left(  i,j\right)  \in
X^{2}\ \mid\ i<_{\mathcal{T}}j,\ i>j,\text{ and }f\left(  i\right)  =f\left(
j\right)  \right\}  ,\\
\ell_{3}\left(  f\right)   &  =\sharp\left\{  \left(  i,j,k\right)  \in
X^{3}\ \mid\ i<j<k,\text{\ }i\sim_{\mathcal{T}}k,\ \right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  i\not \sim _{\mathcal{T}}j,\text{ }%
j\not \sim _{\mathcal{T}}k\text{ and }f\left(  i\right)  =f\left(  j\right)
=f\left(  k\right)  \right\}  .
\end{align*}

\end{enumerate}

It is useful to notice that if $\mathcal{T}$ and $\mathcal{S}$ are two
topologies on finite totally ordered sets $X$ and $Y$, and if $\phi
:X\rightarrow Y$ is an isomorphism of totally ordered sets which is, at the
same time, a homeomorphism between $\mathcal{T}$ and $\mathcal{S}$, then the
generalized T-partitions of $X$ are in 1-to-1 correspondence with the
generalized T-partitions of $Y$, and the same holds for strict T-partitions.
If we represent these (generalized and strict) T-partitions by packed words,
then we actually have $\mathcal{P}\left(  \mathcal{T}\right)  =\mathcal{P}%
\left(  \mathcal{S}\right)  $ and $\mathcal{P}_{s}\left(  \mathcal{T}\right)
=\mathcal{P}_{s}\left(  \mathcal{S}\right)  $. As a consequence, if $X$ is a
finite totally ordered set of cardinality $n$, and if $\mathcal{T}$ is a
topology on $X$, then the unique increasing bijection $\phi:X\rightarrow
\left[  n\right]  $ is an isomorphism of totally ordered sets which is, at the
same time, a homeomorphism between $\mathcal{T}$ and $Std\left(
\mathcal{T}\right)  $. Thus, if we represent (generalized and strict)
T-partitions by packed words, then we have $\mathcal{P}\left(  \mathcal{T}%
\right)  =\mathcal{P}\left(  Std\left(  \mathcal{T}\right)  \right)  $ and
$\mathcal{P}_{s}\left(  \mathcal{T}\right)  =\mathcal{P}_{s}\left(  Std\left(
\mathcal{T}\right)  \right)  $. You are tacitly using this rather often.

\item \textbf{Page 20, Proposition 14:} Remove the spurious closing
parenthesis in \textquotedblleft$j\circ\Gamma_{\left(  q_{1},q_{2}%
,q_{3}\right)  })$\textquotedblright.

\item \textbf{Page 20, Remarks (2):} This is not very precise. It would be
more correct to say that if $\mathcal{T}\in\mathbf{T}$ is $T_{0}$, then a
strict T-partition of $\mathcal{T}$ is the same as a packed word which is, at
the same time, a $\left(  P,\operatorname*{id}\right)  $-partition in
Stanley's sense [16], where $P$ is the poset associated to $\mathcal{T}$ and
$\operatorname*{id}:\left[  n\right]  \rightarrow\left[  n\right]  $ is the
identity map (which we use as a labelling). Unlike you, Stanley does not
require his $\left(  P,\omega\right)  $-partitions to be packed, and he uses
the word \textquotedblleft$P$-partition\textquotedblright\ (without the
$\omega$) for the particular case when the labelling $\omega$ is natural
(which the identity map not always is!).

\item \textbf{Page 21:} Replace \textquotedblleft Then $f=h_{\mid O}=h_{\mid
O^{\prime}}=f^{\prime}$ and $g=Pack\left(  h_{\mid\left[  n\right]  \setminus
O}\right)  =Pack\left(  h_{\mid\left[  n\right]  \setminus O}\right)
=g^{\prime}$\textquotedblright\ by \textquotedblleft Then $f=h_{\mid\left[
n\right]  \setminus O}=h_{\mid\left[  n\right]  \setminus O^{\prime}%
}=f^{\prime}$ and $g=h_{\mid O}\left[  -\max\left(  f\right)  \right]
=h_{\mid O^{\prime}}\left[  -\max\left(  f^{\prime}\right)  \right]
=g^{\prime}$ (where, for any set $S$, any function $\varphi:S\rightarrow
\mathbb{Z}$ and any integer $z$, we let $\varphi\left[  -z\right]  $ be the
function $S\rightarrow\mathbb{Z},\ i\mapsto\varphi\left(  i\right)
-z$)\textquotedblright. (I would rather not use the \textquotedblleft%
$Pack$\textquotedblright\ operator here, as it can mean different things; we
do \textbf{not} want to \textquotedblleft pack\textquotedblright\ the domain.)

\item \textbf{Page 21:} Replace \textquotedblleft$O=h^{-1}\left(  \left\{
k+1,\ldots\max\left(  h\right)  \right\}  \right)  $\textquotedblright\ by
\textquotedblleft$O=h^{-1}\left(  \left\{  k+1,\ldots,\max\left(  h\right)
\right\}  \right)  $\textquotedblright.

\item \textbf{Page 21:} Replace \textquotedblleft$g=Pack\left(  h_{\mid
O}\right)  $\textquotedblright\ by \textquotedblleft$g=h_{\mid O}\left[
-k\right]  $\textquotedblright.

\item \textbf{Pages 22-23, \S 4.2:} Again, it would help to generalize the
definition of $\mathcal{L}\left(  \mathcal{T}\right)  $ to the case when
$\mathcal{T}$ is a topology on an arbitrary finite totally ordered set rather
than on $\left[  n\right]  $. This generalization is similar to the
generalization I suggested for Definition 13, and you are already using it on
page 24 when you write things like \textquotedblleft$\mathcal{L}\left(
\mathcal{T}_{\mid O}\right)  $\textquotedblright.

\item \textbf{Page 22, proof of Proposition 14:} Replace \textquotedblleft%
$\mathcal{T}_{w}$\textquotedblright\ by \textquotedblleft$\mathcal{T}_{f}%
$\textquotedblright.

\item \textbf{Page 22, proof of Proposition 14:} On the same line, replace
\textquotedblleft$\Gamma_{q}\left(  T_{f}\right)  $\textquotedblright\ by
\textquotedblleft$\Gamma_{q}\left(  \mathcal{T}_{f}\right)  $%
\textquotedblright.

\item \textbf{Page 22, proof of Proposition 14:} In the last paragraph of this
proof, replace \textquotedblleft$\left(  q_{2},q_{2},q_{3}\right)
$\textquotedblright\ by \textquotedblleft$\left(  q_{2},q_{1},q_{3}\right)
$\textquotedblright\ (twice).

\item \textbf{Page 22, proof of Proposition 14:} In the last paragraph of this
proof, replace \textquotedblleft$q_{2}^{\ell_{2}\left(  g\right)  }q_{1}%
^{\ell_{1}\left(  g\right)  }q_{2}^{\ell_{2}\left(  g\right)  }$%
\textquotedblright\ by \textquotedblleft$q_{2}^{\ell_{2}\left(  g\right)
}q_{1}^{\ell_{1}\left(  g\right)  }q_{3}^{\ell_{3}\left(  g\right)  }%
$\textquotedblright.

\item \textbf{Page 23, Remark:} Replace \textquotedblleft$\mathcal{L}\left(
T\right)  \subseteq P\left(  T\right)  $\textquotedblright\ by
\textquotedblleft$\mathcal{L}\left(  T\right)  \subseteq\mathcal{P}\left(
T\right)  $\textquotedblright.

\item \textbf{Page 23, Proposition 16:} Replace \textquotedblleft$f\left(
1\right)  ,\ldots f\left(  n\right)  $\textquotedblright\ by \textquotedblleft%
$f\left(  1\right)  ,\ldots,f\left(  n\right)  $\textquotedblright.

\item \textbf{Page 23, proof of Proposition 16:} Replace \textquotedblleft%
$f\left(  1\right)  ,\ldots f\left(  n\right)  $\textquotedblright\ by
\textquotedblleft$f\left(  1\right)  ,\ldots,f\left(  n\right)  $%
\textquotedblright\ (twice).

\item \textbf{Page 23, proof of Proposition 16:} On the second line of the
displayed computation, replace \textquotedblleft${f}\,\shuffle\,{f^{\prime}}%
$\textquotedblright\ by \textquotedblleft$f^{\prime}\shuffle f^{\prime\prime}%
$\textquotedblright.

\item \textbf{Page 24, proof of Proposition 16:} \textquotedblleft are union
of\textquotedblright\ $\rightarrow$ \textquotedblleft are unions
of\textquotedblright.

\item \textbf{Page 24, proof of Proposition 16:} Replace \textquotedblleft of
both $\mathbf{H}_{\mathcal{T}}$ and\textquotedblright\ by \textquotedblleft of
both $\mathbf{H}_{\mathbf{T}}$ and\textquotedblright.

\item \textbf{Page 24:} Here you have used the surjectivity of $L$ to prove
that $\left(  \mathbf{WQSym},\shuffle,\Delta\right)  $ is a Hopf algebra. In
your place, I would add a remark that the same argument (but with $L$ replaced
by $\Gamma_{\left(  1,0,0\right)  }$) can be showed that $\mathbf{WQSym}$ with
its standard structure (that is, $\left(  \mathbf{WQSym},.,\Delta\right)  $)
is a Hopf algebra. This fact is, of course, commonplace, but there is virtue
in having a readable proof that does not use the alphabet doubling trick. (My
problem with the alphabet doubling trick is the scarcity of authors bothering
to explain it in a way that is formally correct.)

\item \textbf{Page 25, proof of Lemma 18:} Replace \textquotedblleft As
$i\sim_{\mathcal{T}}k$\textquotedblright\ by \textquotedblleft As
$i\sim_{\mathcal{T}_{f}}k$\textquotedblright.

\item \textbf{Page 26, proof of Proposition 19, First step:} You write:
\textquotedblleft This is the generalization of lemma 18\textquotedblright. I
think you are only generalizing the $\Longrightarrow$ part of Lemma 18 here.

\item \textbf{Page 26, proof of Proposition 19, First step:} Replace
\textquotedblleft If $i\not \sim _{\mathcal{T}}j$,\textquotedblright\ by
\textquotedblleft Assume that $i\not \sim _{\mathcal{T}}j$.\textquotedblright%
\ (because you are carrying this assumption through the next few sentences).

\item \textbf{Page 26, proof of Proposition 19, Second step:} When you say
\textquotedblleft necessarily the $C_{p,r}$ are intervals\textquotedblright, I
think you mean \textquotedblleft no element of $C_{p,s}$ is between two
elements of $C_{p,r}$ for any $s\neq r$\textquotedblright\ (so, the $C_{p,r}$
are intersections of intervals with $g^{-1}\left(  \left\{  p\right\}
\right)  $, although I don't think this perspective is helpful).

\item \textbf{Page 26, proof of Proposition 19, Second step:} In the Unicity
proof, replace \textquotedblleft are all distincts\textquotedblright\ by
\textquotedblleft are all distinct\textquotedblright.

\item \textbf{Page 26, proof of Proposition 19, Second step:} In the Unicity
proof, it would be better to replace \textquotedblleft If $p<q$ and
$c_{q,s}<c_{p,r}$\textquotedblright\ by \textquotedblleft If $p<q$ and
$c_{q,s}\leq c_{p,r}$\textquotedblright. (While this is equivalent, it
simplifies the argument.)

\item \textbf{Page 26, proof of Proposition 19, Second step:} In the Existence
proof, replace \textquotedblleft if $x\in C_{p,r}$\textquotedblright\ by
\textquotedblleft if $i\in C_{p,r}$\textquotedblright.

\item \textbf{Page 26, proof of Proposition 19, Second step:} In the Existence
proof, replace \textquotedblleft$p=g\left(  x\right)  \leq g\left(  y\right)
=q$\textquotedblright\ by \textquotedblleft$p=g\left(  i\right)  \leq g\left(
j\right)  =q$\textquotedblright.

\item \textbf{Page 26, proof of Proposition 19, Second step:} In the Existence
proof, replace \textquotedblleft$f\left(  x\right)  <f\left(  y\right)
$\textquotedblright\ by \textquotedblleft$f\left(  i\right)  <f\left(
j\right)  $\textquotedblright.

\item \textbf{Page 27, proof of Proposition 19, Second step:} In the Existence
proof, replace \textquotedblleft$p=g\left(  x\right)  \leq g\left(  y\right)
=q$\textquotedblright\ by \textquotedblleft$p=g\left(  i\right)  \leq g\left(
j\right)  =q$\textquotedblright.

\item \textbf{Page 27, proof of Proposition 19, Third step:} Replace
\textquotedblleft$x^{\prime},y^{\prime}\in\mathbf{H}_{\mathcal{T}}%
$\textquotedblright\ by \textquotedblleft$x^{\prime},y^{\prime}\in
\mathbf{H}_{\mathbf{T}}$\textquotedblright.

\item \textbf{Page 28, proof of Corollary 28:} Replace \textquotedblleft This
is the first step of the proof of proposition 19\textquotedblright\ by
\textquotedblleft The first equality sign follows from the first and the
second step of Proposition 19; the second equality sign follows from Lemma
18\textquotedblright.

\item \textbf{Page 28, Remark:} \textquotedblleft for all packed
word\textquotedblright\ $\rightarrow$ \textquotedblleft for any packed
word\textquotedblright.

\item \textbf{Page 29, \S 4.4:} You write: \textquotedblleft a T-partition of
$\mathcal{T}$ is a P-partition of the poset associated to $\mathcal{T}$, in
Stanley's sense [16]\textquotedblright. This is not precise; see my comment on
page 20, Remarks (2) for how to correct it.

\item \textbf{Page 29, \S 4.4:} In the second commutative diagram, I think the
\textquotedblleft$\theta_{\left(  1,0,0\right)  }$\textquotedblright\ arrow
should be \textquotedblleft$\varphi_{\left(  1,0,0\right)  }$%
\textquotedblright.

\item \textbf{Page 30, proof of Lemma 24, part 1:} Replace \textquotedblleft
and $j>i$\textquotedblright\ by \textquotedblleft and $i>j$\textquotedblright.

\item \textbf{Page 30, proof of Lemma 24, part 2:} After \textquotedblleft
Hence, if $c_{p}=c_{q}$\textquotedblright, add \textquotedblleft and
$p<q$\textquotedblright.

\item \textbf{Page 30, proof of Lemma 24, part 2:} Before \textquotedblleft By
definition of the standardization\textquotedblright, add \textquotedblleft
Then\textquotedblright\ (to make clear that the $p<q$ and $c_{p}\neq c_{q}$
hypotheses still apply).

\item \textbf{Page 30, proof of Lemma 25:} Replace \textquotedblleft%
$\sigma^{-1}$ is increasing on $i_{p},\ldots,j_{p}$\textquotedblright\ by
\textquotedblleft$\sigma^{-1}$ is increasing on $\left\{  i_{p},\ldots
,j_{p}\right\}  $\textquotedblright.

\item \textbf{Page 30, proof of Lemma 25:} Replace \textquotedblleft%
$\sigma^{-1}\left(  i\right)  =k$ and $\sigma^{-i}\left(  j\right)
=l$\textquotedblright\ by \textquotedblleft$\sigma\left(  i\right)  =k$ and
$\sigma\left(  j\right)  =l$\textquotedblright.

\item \textbf{Page 30, proof of Lemma 25:} On the last line of page 30, you
write \textquotedblleft$k<l$\textquotedblright. I would add some more details
on why this is true: We have $i=\sigma^{-1}\left(  k\right)  $ and
$j=\sigma^{-1}\left(  l\right)  $. Now, $f\left(  \underbrace{\sigma
^{-1}\left(  k\right)  }_{=i}\right)  =f\left(  i\right)  =p<q=f\left(
\underbrace{j}_{=\sigma^{-1}\left(  l\right)  }\right)  =f\left(  \sigma
^{-1}\left(  l\right)  \right)  $. Recalling our construction of $f$, we see
that this yields $k<l$ (since $f\left(  \sigma^{-1}\left(  g\right)  \right)
$ weakly increases with $g$).

\item \textbf{Page 31, proof of Lemma 25:} I would expect some more detail on
why \textquotedblleft By definition of $f$, $i$ is the greatest element of
$C_{p}$, with $p=f\left(  i\right)  $\textquotedblright. The argument that I
have (maybe not the shortest possible) is as follows: Let $p=f\left(
i\right)  $. Recall that $\sigma\in Std\left(  f\right)  $; thus, $\sigma$ is
increasing on $C_{p}$. By definition of $f$, we have $l\leq k$ for every $l$
satisfying $f\left(  \sigma^{-1}\left(  l\right)  \right)  =f\left(
\sigma^{-1}\left(  k\right)  \right)  $ (since $f\left(  \sigma^{-1}\left(
k+1\right)  \right)  =f\left(  \sigma^{-1}\left(  k\right)  \right)  +1$).
Substituting $\sigma\left(  u\right)  $ for $l$ here, and recalling that
$f\left(  \underbrace{\sigma^{-1}\left(  k\right)  }_{=i}\right)  =f\left(
i\right)  =p$ and $k=\sigma\left(  i\right)  $, we can rewrite this as
follows: We have $\sigma\left(  u\right)  \leq\sigma\left(  i\right)  $ for
every $u$ satisfying $f\left(  u\right)  =p$. In other words, $\sigma\left(
u\right)  \leq\sigma\left(  i\right)  $ for every $u\in C_{p}$. Thus, $u\leq
i$ for every $u\in C_{p}$ (since $\sigma$ is increasing on $C_{p}$). Hence,
$i$ is the greatest element of $C_{p}$.

The argument why \textquotedblleft$j=\sigma^{-1}\left(  k+1\right)  $ is the
smallest element of $C_{p+1}$\textquotedblright\ is similar.

\item \textbf{Page 31, proof of Lemma 25:} In Case (1) of the proof of
$f\left(  \sigma^{-1}\left(  i\right)  \right)  =g\left(  \sigma^{-1}\left(
i\right)  \right)  $, I would add \textquotedblleft$=g\left(  \sigma
^{-1}\left(  i\right)  \right)  +1$\textquotedblright\ before
\textquotedblleft$=g\left(  \sigma^{-1}\left(  i+1\right)  \right)
$\textquotedblright. The same change could be made in Case (2), and a similar
change in Case (3).

\item \textbf{Page 31, proof of Lemma 25:} In the proof of $f\left(
\sigma^{-1}\left(  i\right)  \right)  =g\left(  \sigma^{-1}\left(  i\right)
\right)  $, I would define $p=f\left(  \sigma^{-1}\left(  i\right)  \right)  $
in all three Cases (1), (2) and (3), not just in Case (1) as you currently do.

\item \textbf{Page 31, proof of Theorem 23, part 1, }$\Longrightarrow$
\textbf{direction:} You write: \textquotedblleft By construction of
$\psi_{\sigma}\left(  I\right)  $\textquotedblright. But I don't see how what
comes after this follows from the construction of $\psi_{\sigma}\left(
I\right)  $. Instead, I would say it follows from $I=M\left(  f\right)  $.

\item \textbf{Page 31, proof of Theorem 23, part 1, }$\Longrightarrow$
\textbf{direction:} Replace \textquotedblleft we should have $f\left(
k^{\prime}\right)  >f\left(  l^{\prime}\right)  $\textquotedblright\ by
\textquotedblleft we should have $f\left(  k^{\prime}\right)  \geq f\left(
l^{\prime}\right)  $\textquotedblright.

\item \textbf{Page 31, proof of Theorem 23, part 1, }$\Longrightarrow$
\textbf{direction:} Do you actually use the \textquotedblleft$g\left(
k^{\prime}\right)  <g\left(  l^{\prime}\right)  $\textquotedblright%
\ observation? I don't see how.

\item \textbf{Page 31, proof of Theorem 23, part 1, }$\Longrightarrow$
\textbf{direction:} Replace \textquotedblleft So for all $k^{\prime}\in C_{q}%
$, $l^{\prime}\in C_{q+1}$, $k^{\prime}<l^{\prime}$\textquotedblright\ by
\textquotedblleft So for all $k^{\prime}\in C_{q}^{\prime}$ and $l^{\prime}\in
C_{q+1}^{\prime}$, we have $k^{\prime}\leq k<l^{\prime}$\textquotedblright%
\ (note that I added inverted commas and a $k$).

\item \textbf{Page 31, proof of Theorem 23, part 1, }$\Longleftarrow$
\textbf{direction:} Replace \textquotedblleft If $g\left(  k\right)  \leq
g\left(  l\right)  $, we put $\sigma\left(  k\right)  =i$ and $\sigma\left(
l\right)  =j$. By construction of $\psi_{\sigma}\left(  J\right)  $, $i<j$. By
construction of $\psi_{\sigma}\left(  I\right)  $, $f\left(  k\right)
=f\left(  \sigma^{-1}\left(  i\right)  \right)  =\leq f\left(  \sigma
^{-1}\left(  j\right)  \right)  =f\left(  l\right)  $.\textquotedblright%
\ (which is not completely true\footnote{I replaced the $\leq$ sign in
$g\left(  k\right)  \leq g\left(  l\right)  $ by a $<$ sign, since otherwise
your $i<j$ claim does not hold (unless we also assume $k<l$, which is not the
optimal way).}) by the following argument: \textquotedblleft If $g\left(
k\right)  <g\left(  l\right)  $, then we have $\sigma\left(  k\right)
<\sigma\left(  l\right)  $ (since $\sigma=Std\left(  g\right)  $) and thus
$f\left(  k\right)  \leq f\left(  l\right)  $ (as $\sigma=Std\left(  f\right)
$). The same conclusion holds if $g\left(  k\right)  =g\left(  l\right)  $
(this follows directly from the previous bullet point). Thus, if $g\left(
k\right)  \leq g\left(  l\right)  $, then $f\left(  k\right)  \leq f\left(
l\right)  $.\textquotedblright

\item \textbf{Page 31, proof of Theorem 23, part 1, }$\Longleftarrow$
\textbf{direction:} I don't understand how you prove that if $g\left(
k\right)  <g\left(  l\right)  $ and $k>l$, then $f\left(  k\right)  <f\left(
l\right)  $.\ \ \ \ \footnote{More precisely, I don't understand how you get
$k\leq l$. Maybe you are using the Remark from page 29, but I do not see how.}
I would show this differently:

Let $k$ and $l$ be such that $g\left(  k\right)  <g\left(  l\right)  $ and
$k>l$. As shown in the previous bullet point, we have $f\left(  k\right)  \leq
f\left(  l\right)  $ (since $g\left(  k\right)  \leq g\left(  l\right)  $). If
$f\left(  k\right)  \neq f\left(  l\right)  $, we are thus done. Hence, assume
that $f\left(  k\right)  =f\left(  l\right)  $. Then, $\sigma\left(  k\right)
>\sigma\left(  l\right)  $ (since $\sigma=Std\left(  f\right)  $ and $k>l$).
But $g\left(  k\right)  <g\left(  l\right)  $ and thus $\sigma\left(
k\right)  <\sigma\left(  l\right)  $ (since $\sigma=Std\left(  g\right)  $),
which contradicts $\sigma\left(  k\right)  >\sigma\left(  l\right)  $. This
contradiction finishes the proof.

\item \textbf{Page 31, Remark:} Replace \textquotedblleft$Pack\left(
f\right)  =\sigma$\textquotedblright\ by \textquotedblleft$Std\left(
f\right)  =\sigma$\textquotedblright.
\end{itemize}


\end{document}