\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}%
\usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage}
\usepackage[all,cmtip]{xy}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{framed}
\usepackage{amsmath}
\usepackage{comment}
\usepackage{color}
\usepackage{hyperref}
\usepackage[sc]{mathpazo}
\usepackage[T1]{fontenc}
\usepackage{amsthm}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{LastRevised=Monday, January 02, 2023 00:04:59}
%TCIDATA{SuppressPackageManagement}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\theoremstyle{definition}
\newtheorem{theo}{Theorem}[section]
\newenvironment{theorem}[1][]
{\begin{theo}[#1]\begin{leftbar}}
{\end{leftbar}\end{theo}}
\newtheorem{lem}[theo]{Lemma}
\newenvironment{lemma}[1][]
{\begin{lem}[#1]\begin{leftbar}}
{\end{leftbar}\end{lem}}
\newtheorem{prop}[theo]{Proposition}
\newenvironment{proposition}[1][]
{\begin{prop}[#1]\begin{leftbar}}
{\end{leftbar}\end{prop}}
\newtheorem{defi}[theo]{Definition}
\newenvironment{definition}[1][]
{\begin{defi}[#1]\begin{leftbar}}
{\end{leftbar}\end{defi}}
\newtheorem{remk}[theo]{Remark}
\newenvironment{remark}[1][]
{\begin{remk}[#1]\begin{leftbar}}
{\end{leftbar}\end{remk}}
\newtheorem{coro}[theo]{Corollary}
\newenvironment{corollary}[1][]
{\begin{coro}[#1]\begin{leftbar}}
{\end{leftbar}\end{coro}}
\newtheorem{conv}[theo]{Convention}
\newenvironment{condition}[1][]
{\begin{conv}[#1]\begin{leftbar}}
{\end{leftbar}\end{conv}}
\newtheorem{quest}[theo]{Question}
\newenvironment{algorithm}[1][]
{\begin{quest}[#1]\begin{leftbar}}
{\end{leftbar}\end{quest}}
\newtheorem{warn}[theo]{Warning}
\newenvironment{conclusion}[1][]
{\begin{warn}[#1]\begin{leftbar}}
{\end{leftbar}\end{warn}}
\newtheorem{conj}[theo]{Conjecture}
\newenvironment{conjecture}[1][]
{\begin{conj}[#1]\begin{leftbar}}
{\end{leftbar}\end{conj}}
\newtheorem{exmp}[theo]{Example}
\newenvironment{example}[1][]
{\begin{exmp}[#1]\begin{leftbar}}
{\end{leftbar}\end{exmp}}
\iffalse
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\fi
\newenvironment{verlong}{}{}
\newenvironment{vershort}{}{}
\newenvironment{noncompile}{}{}
\excludecomment{verlong}
\includecomment{vershort}
\excludecomment{noncompile}
\newcommand{\kk}{\mathbf{k}}
\newcommand{\id}{\operatorname{id}}
\newcommand{\ev}{\operatorname{ev}}
\newcommand{\Comp}{\operatorname{Comp}}
\newcommand{\bk}{\mathbf{k}}
\newcommand{\Nplus}{\mathbb{N}_{+}}
\newcommand{\NN}{\mathbb{N}}
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\DeclareSymbolFont{bbold}{U}{bbold}{m}{n}
\DeclareSymbolFontAlphabet{\mathbbold}{bbold}
\setlength\textheight{22.5cm}
\setlength\textwidth{15cm}
\ihead{Errata to ``Discrete Morse theory and the cohomology ring''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Discrete Morse theory and the cohomology ring}

\textit{Robin Forman}

\url{https://math.rice.edu/~forman/product.ps}

version of 2000

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

\setcounter{section}{6}

\section{Errata and addenda}

The following list contains some corrections and comments to Robin Forman's
paper \textquotedblleft Discrete Morse theory and the cohomology
ring\textquotedblright. I refer to the preprint version of 2000 of this paper
(available from \url{https://math.rice.edu/~forman/product.ps} ), but some of
the errors listed below are also contained in the published
version\footnote{\href{https://www.ams.org/journals/tran/2002-354-12/S0002-9947-02-03041-6/}{Transactions
of the American Mathematical Society \textbf{354}, issue 12, pp. 5063--5085.}%
}. The latter error are marked with an $\spadesuit$ sign.

I have only read Sections 1 and 2 of the paper completely; the corrections to
the other sections thus are likely to be less than comprehensive.

\begin{itemize}
\item \textbf{page 2:} In the complex $\mathcal{M}^{\ast}$, the first arrow
should be \textquotedblleft$\leftarrow$\textquotedblright\ instead of
\textquotedblleft$\rightarrow$\textquotedblright.

\item[$\spadesuit$] \textbf{page 12, \S 1:\footnote{This is on page 5071 of
the published version.}} In \textquotedblleft sign chosen so that
$\left\langle a,\partial V\left(  b\right)  \right\rangle =-1$%
\textquotedblright, replace \textquotedblleft$V\left(  b\right)
$\textquotedblright\ by \textquotedblleft$V\left(  a\right)  $%
\textquotedblright.

\item \textbf{page 12, \S 1:} \textquotedblleft if $a$ for all simplices
$a$\textquotedblright\ $\rightarrow$ \textquotedblleft for all simplices
$a$\textquotedblright.

\item[$\spadesuit$] \textbf{page 12, \S 1:\footnote{This is on page 5071 of
the published version.}} Near the bottom of this page, you claim that every
simplex $a$ of $M$ satisfies exactly one of the ofllowing:

(i) $a$ is the smaller simplex in one $V$-pair\footnote{By \textquotedblleft%
$V$-pair\textquotedblright, I mean a pair of simplices that belongs to $V$.};

(ii) $a$ is the larger simplex in one $V$-pair;

(iii) $a$ is critical.

This is correct, but should perhaps be justified. The nontrivial part of the
proof is showing that (i) and (ii) cannot hold at the same time, i.e., that a
simplex $a^{\left(  p\right)  }\in M$ cannot be both the smaller simplex in
one $V$-pair $\left\{  a^{\left(  p\right)  }<b^{\left(  p+1\right)
}\right\}  $ and the larger simplex in another $V$-pair $\left\{  c^{\left(
p-1\right)  }<a^{\left(  p\right)  }\right\}  $ at the same time.

So let me show this: Assume the contrary. Thus, there exists a simplex
$a^{\left(  p\right)  }\in M$ that is both the smaller simplex in one $V$-pair
$\left\{  a^{\left(  p\right)  }<b^{\left(  p+1\right)  }\right\}  $ and the
larger simplex in another $V$-pair $\left\{  c^{\left(  p-1\right)
}<a^{\left(  p\right)  }\right\}  $ at the same time. Consider this simplex
$a^{\left(  p\right)  }$ and these two pairs. Thus, $\left\{  a^{\left(
p\right)  }<b^{\left(  p+1\right)  }\right\}  \in V$ and $\left\{  c^{\left(
p-1\right)  }<a^{\left(  p\right)  }\right\}  \in V$.

Since a simplex of dimension $k$ is just a $\left(  k+1\right)  $-element set,
we see from $a^{\left(  p\right)  }<b^{\left(  p+1\right)  }$ that the set $b$
contains $a$ as a subset but its size is just $1$ larger than the size of $a$.
Therefore, $b=a\cup\left\{  x\right\}  $ for some element $x\notin a$.
Consider this $x$. Similarly, from $c^{\left(  p-1\right)  }<a^{\left(
p\right)  }$, we see that $a=c\cup\left\{  y\right\}  $ for some element
$y\notin c$. Consider this $y$. Note that $a\subseteq b$ (since $a^{\left(
p\right)  }<b^{\left(  p+1\right)  }$) and $y\in\left\{  y\right\}  \subseteq
c\cup\left\{  y\right\}  =a$.

Now, define a simplex $d:=b\setminus\left\{  y\right\}  $. Then, $d\subseteq
b$, so that $d\in M$ (since $b\in M$, but $M$ is a simplicial complex).
Moreover, $y\in a\subseteq b$, so that the size of the set $b\setminus\left\{
y\right\}  $ is $1$ smaller than the size of $b$. In other words, the size of
the set $d$ is $1$ smaller than the size of $b$ (since $d=b\setminus\left\{
y\right\}  $). Hence, the simplex $d$ has dimension $p$ (since $b$ has
dimension $p+1$). Thus, we can write $d$ as $d^{\left(  p\right)  }$.

However, $a=c\cup\left\{  y\right\}  $, thus $c=a\setminus\left\{  y\right\}
$ (since $y\notin c$). Hence, $c=\underbrace{a}_{\subseteq b}\setminus\left\{
y\right\}  \subseteq b\setminus\left\{  y\right\}  =d$. In other words,
$c^{\left(  p-1\right)  }\subseteq d^{\left(  p\right)  }$ (since
$c=c^{\left(  p-1\right)  }$ and $d=d^{\left(  p\right)  }$). Therefore,
$c^{\left(  p-1\right)  }<d^{\left(  p\right)  }$. In other words, $d^{\left(
p\right)  }>c^{\left(  p-1\right)  }$.

Also, recall that $d\subseteq b$. In other words, $d^{\left(  p\right)
}\subseteq b^{\left(  p+1\right)  }$. Hence, $d^{\left(  p\right)
}<b^{\left(  p+1\right)  }$.

We shall now show that $a=d$.

Recall that $V$ is the set of all pairs $\left\{  u^{\left(  k\right)
}<v^{\left(  k+1\right)  }\right\}  $ of simplices in $M$ satisfying $f\left(
u\right)  \geq f\left(  v\right)  $. Hence, we have $f\left(  c\right)  \geq
f\left(  a\right)  $ (since $\left\{  c^{\left(  p-1\right)  }<a^{\left(
p\right)  }\right\}  \in V$) and $f\left(  a\right)  \geq f\left(  b\right)  $
(since $\left\{  a^{\left(  p\right)  }<b^{\left(  p+1\right)  }\right\}  \in
V$). In other words, we have $f\left(  a\right)  \leq f\left(  c\right)  $ and
$f\left(  b\right)  \leq f\left(  a\right)  $.

Now, we are in one of the following two cases:

\textit{Case 1:} We have $f\left(  d\right)  \leq f\left(  c\right)  $.

\textit{Case 2:} We have $f\left(  d\right)  >f\left(  c\right)  $.

Let us first consider Case 1. In this case, we have $f\left(  d\right)  \leq
f\left(  c\right)  $.

Since $f$ is a discrete Morse function, the set%
\[
\left\{  v^{\left(  p\right)  }>c^{\left(  p-1\right)  }\ \mid\ f\left(
v\right)  \leq f\left(  c\right)  \right\}
\]
has size $\leq1$ (by the definition of a discrete Morse function). Hence, any
two elements of this set must be equal. Since both simplices $a^{\left(
p\right)  }$ and $d^{\left(  p\right)  }$ belong to this set (because
$a^{\left(  p\right)  }>c^{\left(  p-1\right)  }$ and $f\left(  a\right)  \leq
f\left(  c\right)  $ and $d^{\left(  p\right)  }>c^{\left(  p-1\right)  }$ and
$f\left(  d\right)  \leq f\left(  c\right)  $), we thus conclude that these
two simplices $a^{\left(  p\right)  }$ and $d^{\left(  p\right)  }$ are equal.
In other words, $a=d$. Thus, we have proved $a=d$ in Case 1.

Let us now consider Case 2. In this case, we have $f\left(  d\right)
>f\left(  c\right)  $. Hence, $f\left(  d\right)  >f\left(  c\right)  \geq
f\left(  a\right)  \geq f\left(  b\right)  $, so that $f\left(  d\right)  \geq
f\left(  b\right)  $.

Since $f$ is a discrete Morse function, the set%
\[
\left\{  v^{\left(  p\right)  }<b^{\left(  p+1\right)  }\ \mid\ f\left(
v\right)  \geq f\left(  b\right)  \right\}
\]
has size $\leq1$ (by the definition of a discrete Morse function). Hence, any
two elements of this set must be equal. Since both simplices $a^{\left(
p\right)  }$ and $d^{\left(  p\right)  }$ belong to this set (because
$a^{\left(  p\right)  }<b^{\left(  p+1\right)  }$ and $f\left(  a\right)  \geq
f\left(  b\right)  $ and $d^{\left(  p\right)  }<b^{\left(  p+1\right)  }$ and
$f\left(  d\right)  \geq f\left(  b\right)  $), we thus conclude that these
two simplices $a^{\left(  p\right)  }$ and $d^{\left(  p\right)  }$ are equal.
In other words, $a=d$. Thus, we have proved $a=d$ in Case 2.

We now have shown that $a=d$ in both Cases 1 and 2. Thus, $a=d$ always holds.
However, $y\notin d$ (since $d=b\setminus\left\{  y\right\}  $). But this
contradicts $y\in a=d$. This contradiction shows that our assumption was
false, qed.

\item[$\spadesuit$] \textbf{page 13, definition of }$m\left(  s\right)
$\textbf{ for an upper gradient step:\footnote{This is on page 5071 of the
published version.}} After \textquotedblleft$m\left(  s\right)  =-\left\langle
a_{0},\partial b_{0}\right\rangle \left\langle \partial b_{0},a_{1}%
\right\rangle $\textquotedblright, I would add \textquotedblleft$=\left\langle
\partial Va_{0},a_{1}\right\rangle $\textquotedblright\ (since this is tacitly
being used later on).

\item[$\spadesuit$] \textbf{page 13, definition of }$m\left(  s\right)
$\textbf{ for a lower gradient step:\footnote{This is on page 5072 of the
published version.}} After \textquotedblleft$m\left(  s\right)  =-\left\langle
\partial a_{0},b_{0}\right\rangle \left\langle b_{0},\partial a_{1}%
\right\rangle $\textquotedblright, I would add \textquotedblleft$=\left\langle
V\partial a_{0},a_{1}\right\rangle $\textquotedblright\ (since this is tacitly
being used later on). This equality follows from the fact that (if we WLOG
assume that $b_{0}$ is oriented so that $V\left(  b_{0}\right)  =a_{1}$) we
have $\left\langle b_{0},\partial a_{1}\right\rangle =\left\langle
b_{0},\partial Vb_{0}\right\rangle =-1$ and $\left\langle \partial a_{0}%
,b_{0}\right\rangle =\left\langle V\partial a_{0},Vb_{0}\right\rangle
=\left\langle V\partial a_{0},a_{1}\right\rangle $.

\item \textbf{page 15:} The first two words on this page should be
\textquotedblleft critical simplices\textquotedblright, not \textquotedblleft
gradient paths\textquotedblright.

\item[$\spadesuit$] \textbf{page 15, Lemma 1.3 (i):\footnote{This is on page
5072 of the published version.}} Add a comma before \textquotedblleft$s_{r-1}%
$\textquotedblright.

\item[$\spadesuit$] \textbf{page 15, Lemma 1.3 (ii):\footnote{This is on page
5072 of the published version.}} Remove the word \textquotedblleft
nontrivial\textquotedblright.

\item[$\spadesuit$] \textbf{page 16:\footnote{This is on page 5073 of the
published version.}} \textquotedblleft straighforward\textquotedblright%
\ $\rightarrow$ \textquotedblleft straightforward\textquotedblright.

\item[$\spadesuit$] \textbf{page 16:\footnote{This is on page 5073 of the
published version.}} The sentence following Theorem 1.4 should be part of
Theorem 1.4 (in particular, it should be italicized).

\item[$\spadesuit$] \textbf{page 18:\footnote{This is on page 5075 of the
published version.}} You say: \textquotedblleft The general case is not much
harder.\textquotedblright. Let me elaborate on this:

We have $\Phi^{\infty}=\Phi^{N}$. Recall also that $\partial\circ\Phi
=\Phi\circ\partial$; in other words, the operator $\Phi$ commutes with
$\partial$. Hence, any power $\Phi^{i}$ of $\Phi$ also commutes with
$\partial$. In other words, we have%
\begin{equation}
\partial\circ\Phi^{i}=\Phi^{i}\circ\partial\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\mathbb{N}.\label{p18.1}%
\end{equation}


For any two operators $\alpha,\beta:C_{\ast}\left(  M,\mathbb{Z}\right)
\rightarrow C_{\ast}\left(  M,\mathbb{Z}\right)  $, we shall write
$\alpha\simeq\beta$ (and say that $\alpha$ and $\beta$ are
\textit{chain-homotopic}) if and only if there exists an operator $K:C_{\ast
}\left(  M,\mathbb{Z}\right)  \rightarrow C_{\ast+1}\left(  M,\mathbb{Z}%
\right)  $ satisfying $\beta-\alpha=\partial\circ K+K\circ\partial$. The
relation $\simeq$ is an equivalence relation (this is a fundamental result and
easy to check).

Now, for any $i\in\mathbb{N}$, we have%
\begin{align*}
& \partial\circ\left(  \Phi^{i}\circ V\right)  +\left(  \Phi^{i}\circ
V\right)  \circ\partial\\
& =\underbrace{\partial\circ\Phi^{i}}_{\substack{=\Phi^{i}\circ\partial
\\\text{(by (\ref{p18.1}))}}}\circ V+\Phi^{i}\circ V\circ\partial\\
& =\Phi^{i}\circ\partial\circ V+\Phi^{i}\circ V\circ\partial=\Phi^{i}%
\circ\underbrace{\left(  \partial\circ V+V\circ\partial\right)  }%
_{\substack{=\Phi-1\\\text{(since }\Phi=1+\partial\circ V+V\circ
\partial\text{)}}}\\
& =\Phi^{i}\circ\left(  \Phi-1\right)  =\underbrace{\Phi^{i}\circ\Phi}%
_{=\Phi^{i+1}}-\underbrace{\Phi^{i}\circ1}_{=\Phi^{i}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi^{i}\text{ is a linear
map}\right)  \\
& =\Phi^{i+1}-\Phi^{i}.
\end{align*}
In other words, for any $i\in\mathbb{N}$, we have $\Phi^{i+1}-\Phi
^{i}=\partial\circ\left(  \Phi^{i}\circ V\right)  +\left(  \Phi^{i}\circ
V\right)  \circ\partial$. Hence, for any $i\in\mathbb{N}$, we have $\Phi
^{i}\simeq\Phi^{i+1}$ (since $\Phi^{i+1}-\Phi^{i}=\partial\circ K+K\circ
\partial$ for $K:=\Phi^{i}\circ V$). In other words, we have the following
chain of relations:%
\[
\Phi^{0}\simeq\Phi^{1}\simeq\Phi^{2}\simeq\Phi^{3}\simeq\cdots.
\]
Since the relation $\simeq$ is an equivalence relation, we thus find $\Phi
^{0}\simeq\Phi^{N}$. In other words, $1\simeq\Phi^{\infty}$ (since $\Phi
^{0}=1$ and $\Phi^{N}=\Phi^{\infty}$). In other words, there exists an
operator $K:C_{\ast}\left(  M,\mathbb{Z}\right)  \rightarrow C_{\ast+1}\left(
M,\mathbb{Z}\right)  $ satisfying $\Phi^{\infty}-1=\partial\circ
K+K\circ\partial$, qed.

\item \textbf{page 20, \S 3:} \textquotedblleft to be the
dual\textquotedblright\ $\rightarrow$ \textquotedblleft be the
dual\textquotedblright.

\item \textbf{page 21:} \textquotedblleft The proof easily
adapted\textquotedblright\ $\rightarrow$ \textquotedblleft The proof can be
easily adapted\textquotedblright.

\item[$\spadesuit$] \textbf{page 21:\footnote{This is on page 5077 of the
published version.}} Again, in the complex between Theorem 3.2 and Theorem
3.3, the first arrow should be \textquotedblleft$\leftarrow$\textquotedblright%
\ instead of \textquotedblleft$\rightarrow$\textquotedblright.

\item \textbf{page 23:} Remove the period at the end of the last displayed
equation on this page.

\item \textbf{page 24, Theorem 3.9:} Add \textquotedblleft
for\textquotedblright\ before \textquotedblleft$i=1,2,\ldots,k$%
\textquotedblright.

\item \textbf{page 25:} In the first displayed equation on this page, the
\textquotedblleft$\mathcal{L}$\textquotedblright\ on the right hand side
should be an \textquotedblleft$L$\textquotedblright\ (normal font, not calligraphic).

\item[$\spadesuit$] \textbf{page 25:\footnote{This is on page 5080 of the
published version.}} In the fourth displayed equation on this page, the comma
in \textquotedblleft$b_{1}^{\ast}\otimes b_{2}^{\ast},\otimes\cdots\otimes
b_{\ell}^{\ast}$\textquotedblright\ should be removed.

\item \textbf{page 27, \S 5:} \textquotedblleft Then 4.2\textquotedblright%
\ $\rightarrow$ \textquotedblleft Then Corollary 4.2\textquotedblright.

\item \textbf{page 28, Example 2:} On the right hand side of the last
displayed equation on this page, I think you are missing a factor of $B^{\ast
}$.

\item \textbf{page 29, Example 3:} Remove the period at the end of the cocomplex:

\item[$\spadesuit$] \textbf{page 29, Example 3:\footnote{This is on page 5083
of the published version.}} Replace \textquotedblleft$L_{P}$\textquotedblright%
\ by \textquotedblleft$L_{p}$\textquotedblright.

\item[$\spadesuit$] \textbf{page 30, Example 4:\footnote{This is on page 5083
of the published version.}} \textquotedblleft vertices $v$ of $G$%
\textquotedblright\ $\rightarrow$ \textquotedblleft vertices $v$ of
$M$\textquotedblright.

\item[$\spadesuit$] \textbf{page 30, Example 4:\footnote{This is on page 5083
of the published version.}} Is $C$ supposed to mean the abelian group
$\mathbb{Z}$ ?
\end{itemize}


\end{document}