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\ihead{Remark on polyhedral cones}
\ohead{\today}
\begin{document}

\title{A remark on polyhedral cones from packed words and from finite topologies}
\author{Darij Grinberg}
\date{\today\ (unfinished!)}
\maketitle
\tableofcontents

\section{The main theorem}

The purpose of this little note is to prove \cite[Theorem 5.2]{MNT} using the
machinery of \cite{FM}.

I shall use the notations of \cite{FM} (except that I write
$\operatorname*{WQSym}$ instead of $\mathbf{WQSym}$). Here is a brief overview
of these notations:

\begin{itemize}
\item We fix a field $\mathbb{K}$.

\item We let $\mathbb{N}=\left\{  0,1,2,\ldots\right\}  $ and $\mathbb{N}%
_{>0}=\left\{  1,2,3,\ldots\right\}  $.

\item For each $n\in\mathbb{N}$, we let $\left[  n\right]  $ denote the set
$\left\{  1,2,\ldots,n\right\}  $. In particular, $\left[  0\right]
=\varnothing$.

\item A \textit{word} means a $n$-tuple of positive integers for some
$n\in\mathbb{N}$. In this case, the $n$ is called the \textit{length} of the
word. A word $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $ is identified with
the map $\left[  n\right]  \rightarrow\mathbb{N}_{>0},\ i\mapsto w_{i}$.

\item A word $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $ is said to be
\textit{packed} if and only if $\left\{  w_{1},w_{2},\ldots,w_{n}\right\}
=\left[  k\right]  $ for some $k\in\mathbb{N}$. In this case, the $k$ is
denoted by $\max w$. (Note that $k$ is the largest entry of $w$ if $w$ is nonempty.)

For example, the word $\left(  3,1,2,1,3\right)  $ is packed (with
$\max\left(  3,1,2,1,3\right)  =3$), and so is the empty word $\left(
{}\right)  $ (with $\max\left(  {}\right)  =0$); but the word $\left(
3,1,3\right)  $ is not packed.

\item If $w$ is any word, then the \textit{packing} of $w$ is the packed word
$\operatorname*{Pack}w$ obtained by replacing the smallest number that appears
in $w$ by $1$ (as often as it appears), replacing the second-smallest number
that appears in $w$ by $2$ (as often as it appears), and so on. More formally,
it can be defined as follows: Write $w$ as $w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  $. Let $W=\left\{  w_{1},w_{2},\ldots,w_{n}\right\}  $ be the
set of all entries of $w$, and let $m=\left\vert W\right\vert $. Let $\phi$ be
the unique increasing bijection from $W$ to $\left[  m\right]  $. Then,
$\operatorname*{Pack}w$ is defined to be the word $\left(  \phi\left(
w_{1}\right)  ,\phi\left(  w_{2}\right)  ,\ldots,\phi\left(  w_{n}\right)
\right)  $.

For example,%
\[
\operatorname*{Pack}\left(  4,1,7,2,4,1\right)  =\left(  3,1,4,2,3,1\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \operatorname*{Pack}\left(
4,2\right)  =\left(  2,1\right)  .
\]
Also, $\operatorname*{Pack}w=w$ for any packed word $w$.

\item We let $\operatorname*{WQSym}$ denote the free $\mathbb{K}$-vector space
with basis $\left(  w\right)  _{w\text{ is a packed word}}$. We define a
$\mathbb{K}$-bilinear operation $.$ (you're reading right: our symbol for this
operation is a period) on this vector space $\operatorname*{WQSym}$ by setting%
\[
f.g=\sum_{\substack{h=\left(  h_{1},h_{2},\ldots,h_{n+m}\right)  \text{ is a
packed word of length }n+m;\\\operatorname*{Pack}\left(  h_{1},h_{2}%
,\ldots,h_{n}\right)  =f\text{ and }\operatorname*{Pack}\left(  h_{n+1}%
,h_{n+2},\ldots,h_{n+m}\right)  =g}}h
\]
for any two packed words $f$ and $g$, where $n$ and $m$ are the lengths of $f$
and $g$. Equipping $\operatorname*{WQSym}$ with this operation $.$ as
multiplication, we obtain a $\mathbb{K}$-algebra with unity $\left(
{}\right)  $ (the empty word). When we refer to the $\mathbb{K}$-algebra
$\operatorname*{WQSym}$ below, we shall always understand it to be equipped
with this $\mathbb{K}$-algebra structure.

For example, in $\operatorname*{WQSym}$, we have%
\[
\left(  1,1\right)  .\left(  2,1\right)  =\left(  1,1,2,1\right)  +\left(
2,2,2,1\right)  +\left(  1,1,3,2\right)  +\left(  2,2,3,1\right)  +\left(
3,3,2,1\right)  .
\]


The $\mathbb{K}$-algebra $\operatorname*{WQSym}$ has various further
structures -- such as a Hopf algebra structure, and an embedding into the ring
of noncommutative formal power series (see \cite[\S 4.3.2]{MNT}, where
$\operatorname*{WQSym}$ is constructed via this embedding, and where the image
of a packed word $u$ under this embedding is denoted by $\mathbf{M}_{u}$). We
won't need this extra structure.
\end{itemize}

Let me add a few more definitions.\footnote{A \textit{set composition} of a
set $X$ means a tuple $\left(  X_{1},X_{2},\ldots,X_{k}\right)  $ of disjoint
nonempty subsets of $X$ such that $X_{1}\cup X_{2}\cup\cdots\cup X_{k}=X$.}

\begin{definition}
Let $n\in\mathbb{N}$. Let $u$ be a packed word of length $n$. Let $r=\max u$.
Define $B_{i}=u^{-1}\left(  \left\{  i\right\}  \right)  $ for every
$i\in\left[  r\right]  $. (Thus, $\left(  B_{1},B_{2},\ldots,B_{r}\right)  $
is a set composition of $\left[  n\right]  $; it is what is called the
\textquotedblleft set composition of $\left[  n\right]  $ encoded by
$u$\textquotedblright\ in \cite{MNT}.) Now, we define a polyhedral cone
$K_{u}$ in $\mathbb{R}^{n}$ by%
\[
K_{u}=\left\{  \left(  x_{1},x_{2},\ldots,x_{n}\right)  \in\mathbb{R}%
^{n}\ \mid\ \sum_{j=1}^{k}\sum_{i\in B_{j}}x_{i}\geq
0\ \ \ \ \ \ \ \ \ \ \text{for all }k=1,2,\ldots,r\right\}  .
\]

\end{definition}

\begin{definition}
For any two sets $X$ and $Y$, let $\operatorname*{Map}\left(  X,Y\right)  $
denote the set of all maps from $X$ to $Y$. Define a $\mathbb{K}$-vector space
$\mathfrak{M}$ by $\mathfrak{M}=\bigoplus_{n\geq0}\operatorname*{Map}\left(
\mathbb{R}^{n},\mathbb{K}\right)  $ (where each $\operatorname*{Map}\left(
\mathbb{R}^{n},\mathbb{K}\right)  $ becomes a $\mathbb{K}$-vector space by
pointwise addition and multiplication with scalars). We make $\mathfrak{M}$
into a $\mathbb{K}$-algebra, whose multiplication is defined as follows: For
any $n\in\mathbb{N}$, any $m\in\mathbb{N}$, any $f\in\operatorname*{Map}%
\left(  \mathbb{R}^{n},\mathbb{K}\right)  $ and $g\in\operatorname*{Map}%
\left(  \mathbb{R}^{m},\mathbb{K}\right)  $, we define $fg$ to be the element
of $\operatorname*{Map}\left(  \mathbb{R}^{n+m},\mathbb{K}\right)  $ which
sends every $\left(  x_{1},x_{2},\ldots,x_{n+m}\right)  \in\mathbb{R}^{n+m}$
to $f\left(  x_{1},x_{2},\ldots,x_{n}\right)  g\left(  x_{n+1},x_{n+2}%
,\ldots,x_{n+m}\right)  $.
\end{definition}

\begin{definition}
For every $n\in\mathbb{N}$ and any subset $S$ of $\mathbb{R}^{n}$, we define a
map $\underline{1}_{S}\in\operatorname*{Map}\left(  \mathbb{R}^{n}%
,\mathbb{K}\right)  \subseteq\mathfrak{M}$ as the indicator function of $S$
(that is, the map which sends every $s\in S$ to $1$ and every $s\in
\mathbb{R}^{n}\setminus S$ to $0$).
\end{definition}

Our goal is to show:

\begin{theorem}
\label{thm.MNT5.2}The map%
\begin{align*}
\alpha:\operatorname*{WQSym}  &  \rightarrow\mathfrak{M},\\
u  &  \mapsto\left(  -1\right)  ^{\max u}\underline{1}_{K_{u}}%
\end{align*}
is a $\mathbb{K}$-algebra homomorphism.
\end{theorem}

This is a stronger version of \cite[Theorem 5.2]{MNT}\footnote{Notice that
\cite[Theorem 5.2]{MNT} talks not about our map $\alpha:\operatorname*{WQSym}%
\rightarrow\mathfrak{M}$, but rather about a map $\mathcal{P}\rightarrow
\operatorname*{WQSym}$ where $\mathcal{P}$ is a certain subquotient of
$\mathfrak{M}$ (namely, the subalgebra of $\mathfrak{M}$ generated by
$\underline{1}_{K_{u}}$, taken modulo functions with measure-zero support).
These two maps are \textquotedblleft in some sense\textquotedblright\ inverse
(allowing us to derive \cite[Theorem 5.2]{MNT} from Theorem \ref{thm.MNT5.2}).
I find Theorem \ref{thm.MNT5.2} the more natural statement.
\par
Notice that \cite{MNT} denotes by $\left(  \mathbf{M}_{u}\right)  _{u\text{ is
a packed word}}$ the basis of $\operatorname*{WQSym}$ that we call $\left(
u\right)  _{u\text{ is a packed word}}$.}, and a particular case of
\cite[Theorem 8.1]{MNT}\footnote{At least, I suspect so -- I have not checked
all the details. I also suspect that the whole \cite[Theorem 8.1]{MNT} can be
obtained in a similar way as we prove Theorem \ref{thm.MNT5.2} below.}.

\section{The proof}

We shall prove Theorem \ref{thm.MNT5.2} using a detour via the algebra
$\mathbf{H}_{\mathbf{T}}$ defined in \cite[Chapter 2]{FM}. We shall use the
following notations from \cite[Chapter 2]{FM}:

\begin{itemize}
\item If $X$ is a set, then a \textit{topology} on $X$ is defined to be a
family $\mathcal{T}$ of subsets of $X$ that satisfies the following three properties:

\begin{itemize}
\item We have $\varnothing\in\mathcal{T}$ and $X\in\mathcal{T}$.

\item The union of any number of sets in $\mathcal{T}$ is again a set in
$\mathcal{T}$.

\item The intersection of any finite number of sets in $\mathcal{T}$ is again
a set in $\mathcal{T}$.
\end{itemize}

We will only use this concept in the case when $X$ is finite; in this case,
the difference between \textquotedblleft any number of sets in $\mathcal{T}%
$\textquotedblright\ and \textquotedblleft any finite number of sets in
$\mathcal{T}$\textquotedblright\ is immaterial (since $\mathcal{T}$ itself
must be finite), and therefore unions and intersections play symmetric roles
in the notion of a topology on $X$.

\begin{itemize}
\item If $\mathcal{T}$ is a topology on $X$, then the sets belonging to
$\mathcal{T}$ are called the \textit{open sets} of $\mathcal{T}$. The
complements of these open sets (inside $X$) are called the \textit{closed
sets} of $\mathcal{T}$.
\end{itemize}

\item If $X$ is a set, then a \textit{preorder} on $X$ is defined to be a
binary relation $\preccurlyeq$ on $X$ that is reflexive and transitive (but,
unlike a partial order, doesn't need to be antisymmetric). Both partial orders
and equivalence relations are preorders.

\item If $X$ is a set, and if $\preccurlyeq$ is a preorder on $X$, then an
\textit{ideal} of $\left(  X,\preccurlyeq\right)  $ means a subset $I$ of $X$
that has the following property:

\begin{itemize}
\item If $i\in I$ and $j\in X$ satisfy $i\preccurlyeq j$, then $j\in I$.
\end{itemize}

\item If $X$ is a finite set, then there is a canonical bijection between
$\left\{  \text{topologies on }X\right\}  $ and $\left\{  \text{preorders on
}X\right\}  $. This bijection (sometimes called the \textit{Alexandrov
correspondence}) proceeds as follows:

\begin{itemize}
\item If $\preccurlyeq$ is a preorder on $X$, then we can define a topology
$\mathcal{T}_{\preccurlyeq}$ on $X$ by%
\[
\mathcal{T}_{\preccurlyeq}=\left\{  \text{ideals of }\left(  X,\preccurlyeq
\right)  \right\}  .
\]
We shall denote this topology $\mathcal{T}_{\preccurlyeq}$ as the
\textit{topology corresponding to }$\preccurlyeq$.

\item If $\mathcal{T}$ is a topology on $X$, then we can define five binary
relations $\leq_{\mathcal{T}}$, $\geq_{\mathcal{T}}$ and $\sim_{\mathcal{T}}$
on $X$ by setting%
\[
\left(  a\leq_{\mathcal{T}}b\right)  \ \Longleftrightarrow\ \left(  \text{each
}I\in\mathcal{T}\text{ satisfying }a\in I\text{ satisfies }b\in I\right)  ;
\]%
\[
\left(  a\geq_{\mathcal{T}}b\right)  \ \Longleftrightarrow\ \left(  \text{each
}I\in\mathcal{T}\text{ satisfying }b\in I\text{ satisfies }a\in I\right)  ;
\]%
\[
\left(  a\sim_{\mathcal{T}}b\right)  \ \Longleftrightarrow\ \left(  \text{each
}I\in\mathcal{T}\text{ satisfies the equivalence }\left(  a\in I\right)
\Longleftrightarrow\left(  b\in I\right)  \right)  ;
\]%
\[
\left(  a<_{\mathcal{T}}b\right)  \ \Longleftrightarrow\ \left(
a\leq_{\mathcal{T}}b\text{ but not }a\geq_{\mathcal{T}}b\right)
\ \Longleftrightarrow\ \left(  a\leq_{\mathcal{T}}b\text{ but not }%
a\sim_{\mathcal{T}}b\right)  ;
\]%
\[
\left(  a>_{\mathcal{T}}b\right)  \ \Longleftrightarrow\ \left(
a\geq_{\mathcal{T}}b\text{ but not }a\leq_{\mathcal{T}}b\right)
\ \Longleftrightarrow\ \left(  a\geq_{\mathcal{T}}b\text{ but not }%
a\sim_{\mathcal{T}}b\right)  .
\]
The three binary relations $\leq_{\mathcal{T}}$, $\geq_{\mathcal{T}}$ and
$\sim_{\mathcal{T}}$ are preorders on $X$, and the relation $\sim
_{\mathcal{T}}$ is an equivalence relation (whence the quotient set
$X/\sim_{\mathcal{T}}$ is well-defined). The relations $<_{\mathcal{T}}$ and
$>_{\mathcal{T}}$ are strict partial orders. We shall refer to the relation
$\leq_{\mathcal{T}}$ as the \textit{preorder corresponding to }$\mathcal{T}$.
\end{itemize}

These assignments of a topology to a preorder and vice versa are mutually
inverse: If $\preccurlyeq$ is a preorder on $X$, then $\leq_{\mathcal{T}%
_{\preccurlyeq}}$ is precisely $\preccurlyeq$. Conversely, if $\mathcal{T}$ is
a topology on $X$, then $\mathcal{T}_{\leq_{\mathcal{T}}}$ is precisely
$\mathcal{T}$.

\item For each $n\in\mathbb{N}$, we let $\mathbf{T}_{n}$ denote the set of all
topologies on the set $\left[  n\right]  =\left\{  1,2,\ldots,n\right\}  $.

\item We let $\mathbf{T}$ denote the set $\bigsqcup\limits_{n\in\mathbb{N}%
}\mathbf{T}_{n}$.

\item If $f$ is a packed word of length $n\in\mathbb{N}$, then we define a
preorder $\leq_{f}$ on the set $\left[  n\right]  $ by setting%
\[
\left(  a\leq_{f}b\right)  \ \Longleftrightarrow\ \left(  f\left(  a\right)
\leq f\left(  b\right)  \right)  .
\]
Furthermore, if $f$ is a packed word of length $n\in\mathbb{N}$, then we let
$\mathcal{T}_{f}$ be the topology $\mathcal{T}_{\leq_{f}}$ corresponding to
this preorder $\leq_{f}$. The closed sets of this topology $\mathcal{T}_{f}$
are the sets $f^{-1}\left(  \left\{  1,2,\ldots,i\right\}  \right)  $ for
$i\in\left\{  0,1,\ldots,\max f\right\}  $.

\item If $P\subseteq\mathbb{N}$ and $n\in\mathbb{N}$, then $P\left(
+n\right)  $ shall denote the set $\left\{  k+n\mid k\in P\right\}  $. (In
other words, $P\left(  +n\right)  $ is the set $P$ shifted right by $n$ units
on the number line.)

\item If $\mathcal{T}\in\mathbf{T}_{n}$ and $\mathcal{S}\in\mathbf{T}_{m}$ are
two topologies (on the sets $\left[  n\right]  $ and $\left[  m\right]  $,
respectively) for some $n\in\mathbb{N}$ and $m\in\mathbb{N}$, then we define a
topology $\mathcal{T}.\mathcal{S}\in\mathbf{T}_{n+m}$ on the set $\left[
n+m\right]  $ by
\[
\mathcal{T}.\mathcal{S}=\left\{  O\cup\left(  P\left(  +n\right)  \right)
\ \mid\ O\in\mathcal{T}\text{ and }P\in\mathcal{S}\right\}  .
\]
Thus, we have defined a binary operation $.$ on $\mathbf{T}$. This binary
operation $.$ is associative (by \cite[Proposition 3]{FM}), and the topology
$\left\{  \varnothing\right\}  \in\mathbf{T}_{0}$ is its neutral element.

\item We let $\mathbf{H}_{\mathbf{T}}$ be the free $\mathbb{K}$-vector space
with basis $\mathbf{T}$. We equip $\mathbf{H}_{\mathbf{T}}$ with a
multiplication $.$ that linearly extends the operation $.$ on $\mathbf{T}$
(that is, the restriction of the multiplication $\mathbf{H}_{\mathbf{T}}$ to
the basis $\mathbf{T}$ should be the operation $.$ on $\mathbf{T}$). Thus,
$\mathbf{H}_{\mathbf{T}}$ becomes a $\mathbb{K}$-algebra with unity $\left\{
\varnothing\right\}  \in\mathbf{T}_{0}$.
\end{itemize}

The $\mathbb{K}$-algebra $\mathbf{H}_{\mathbf{T}}$ also has the structure of a
Hopf algebra, but we shall not need it, so we don't define it here.

We shall also use the following notation from \cite[Chapter 4]{FM}:

\begin{itemize}
\item If $X$ is a set, and if $\mathcal{T}$ is a topology on $X$, then we set%
\begin{align*}
\mathcal{P}\left(  \mathcal{T}\right)   &  =\bigsqcup_{p\in\mathbb{N}}\left\{
\text{surjective maps }f:X\rightarrow\left[  p\right]  \text{ such that every
}c\in X\text{ and }d\in X\right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  \text{satisfying }%
c\leq_{\mathcal{T}}d\text{ satisfy }f\left(  x\right)  \leq f\left(  d\right)
\right\}  .
\end{align*}
Thus, if $X=\left[  n\right]  $ for some $n\in\mathbb{N}$, then all elements
of $\mathcal{P}\left(  \mathcal{T}\right)  $ are packed words of length $n$.
\end{itemize}

Next, we define a polyhedral cone for every $\mathcal{T}\in\mathbf{T}$:

\begin{definition}
Let $n\in\mathbb{N}$ and $\mathcal{T}\in\mathbf{T}_{n}$ (that is, let
$\mathcal{T}$ be a topology on the set $\left[  n\right]  =\left\{
1,2,\ldots,n\right\}  $). Then, we define a polyhedral cone $K_{\mathcal{T}}$
in $\mathbb{R}^{n}$ by%
\[
K_{\mathcal{T}}=\left\{  \left(  x_{1},x_{2},\ldots,x_{n}\right)
\in\mathbb{R}^{n}\ \mid\ \sum_{i\in C}x_{i}\geq0\ \ \ \ \ \ \ \ \ \ \text{for
all closed sets }C\text{ of }\mathcal{T}\right\}  .
\]

\end{definition}

The following follows from the definitions:

\begin{remark}
\label{rmk.KTf}Let $u$ be a packed word. Then, $K_{u}=K_{\mathcal{T}_{u}}$,
where $\mathcal{T}_{u}$ is as defined in \cite[\S 2.1]{FM}.
\end{remark}

Let us define a few more things:

\begin{definition}
\label{def.U}Let $X$ be a finite totally ordered set, and let $\mathcal{T}$ be
a topology on $X$. We define $\mathcal{U}\left(  \mathcal{T}\right)  $ to be
the set of all $f\in\mathcal{P}\left(  \mathcal{T}\right)  $ having the
property that any two elements $i$ and $j$ of $X$ satisfying $i<_{\mathcal{T}%
}j$ must satisfy $f\left(  i\right)  <f\left(  j\right)  $. Notice that
$\mathcal{U}\left(  \mathcal{T}\right)  \subseteq\mathcal{P}\left(
\mathcal{T}\right)  $. (We can call the elements of $\mathcal{U}\left(
\mathcal{T}\right)  $ \textquotedblleft strictly increasing packed
words\textquotedblright\ for $\mathcal{T}$.) (It can also be shown that
$\mathcal{L}\left(  \mathcal{T}\right)  \subseteq\mathcal{U}\left(  T\right)
$, where $\mathcal{L}\left(  T\right)  $ is as defined in \cite[Definition
15]{FM}.)
\end{definition}

\begin{definition}
We define a $\mathbb{K}$-linear map $U:\mathbf{H}_{\mathbf{T}}\rightarrow
\operatorname*{WQSym}$ by%
\[
U\left(  \mathcal{T}\right)  =\sum_{f\in\mathcal{U}\left(  \mathcal{T}\right)
}f\ \ \ \ \ \ \ \ \ \ \text{for every }\mathcal{T}\in\mathbf{T}.
\]

\end{definition}

\begin{remark}
This map $U$ is easily seen to be the map $\Gamma_{\left(  0,0,1\right)  }$ in
the notation of \cite[Proposition 14]{FM}. Thus, $U$ is a surjective Hopf
algebra homomorphism.
\end{remark}

Now, here is a rather trivial fact:

\begin{proposition}
\label{prop.MNT5.2top}The map%
\begin{align*}
\beta:\mathbf{H}_{\mathbf{T}}  &  \rightarrow\mathfrak{M},\\
\mathcal{T}  &  \mapsto\left(  -1\right)  ^{\left\vert \left[  n\right]
/\sim_{\mathcal{T}}\right\vert }\underline{1}_{K_{\mathcal{T}}}%
\end{align*}
is a $\mathbb{K}$-algebra homomorphism from $\mathbf{H}_{\mathbf{T}}=\left(
\mathbf{H}_{\mathbf{T}},.\right)  $ to $\mathfrak{M}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.MNT5.2top} (sketched).]The proof boils down to
the observation that if $n\in\mathbb{N}$, $m\in\mathbb{N}$, $\mathcal{T}%
\in\mathbf{T}_{n}$ and $\mathcal{S}\in\mathbf{T}_{m}$, then%
\begin{align*}
K_{\mathcal{T}.\mathcal{S}}  &  =\left\{  \left(  x_{1},x_{2},\ldots
,x_{n+m}\right)  \in\mathbb{R}^{n+m}\ \mid\ \left(  x_{1},x_{2},\ldots
,x_{n}\right)  \in K_{\mathcal{T}}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }\left(  x_{n+1},x_{n+2}%
,\ldots,x_{n+m}\right)  \in K_{\mathcal{S}}\right\}  .
\end{align*}

\end{proof}

Now, we claim:

\begin{theorem}
\label{thm.commdiag}The diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\mathbf{H}_{\mathbf{T}} \ar[r]^-{U} \ar[rd]_{\beta} & \operatorname{WQSym}
%\ar[d]^{\alpha} \\
%& \mathfrak{M}
%}}}%
%BeginExpansion
\xymatrix{
\mathbf{H}_{\mathbf{T}} \ar[r]^-{U} \ar[rd]_{\beta} & \operatorname{WQSym}
\ar[d]^{\alpha} \\
& \mathfrak{M}
}%
%EndExpansion
\]
commutes. That is, we have $\beta=\alpha\circ U$.
\end{theorem}

Before we prove this, we introduce some more notations.

\begin{definition}
We define a $\mathbb{K}$-linear map $Z:\mathbf{H}_{\mathbf{T}}\rightarrow
\mathbf{H}_{\mathbf{T}}$ by%
\[
Z\left(  \mathcal{T}\right)  =\left(  -1\right)  ^{\left\vert \left[
n\right]  /\sim_{\mathcal{T}}\right\vert }\mathcal{T}%
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{ and }\mathcal{T}%
\in\mathbf{T}_{n}.
\]
It is easy to see that $Z$ is an involutive Hopf algebra isomorphism.
\end{definition}

\begin{definition}
\label{def.insert-into-top}Let $X$ be a finite totally ordered set, and let
$\mathcal{T}$ be a topology on $X$. Let $a$ and $b$ be two elements of $X$. We
define three new topologies $\mathcal{T}\looparrowleft\left(  a\leq b\right)
$, $\mathcal{T}\looparrowleft\left(  a\geq b\right)  $ and $\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  $ on $X$ as follows:%
\begin{align*}
\mathcal{T}  &  \looparrowleft\left(  a\leq b\right)  =\left\{  O\in
\mathcal{T}\ \mid\ \left(  a\in O\ \Longrightarrow\ b\in O\right)  \right\}
;\\
\mathcal{T}  &  \looparrowleft\left(  a\geq b\right)  =\left\{  O\in
\mathcal{T}\ \mid\ \left(  b\in O\ \Longrightarrow\ a\in O\right)  \right\}
;\\
\mathcal{T}  &  \looparrowleft\left(  a\sim b\right)  =\left\{  O\in
\mathcal{T}\ \mid\ \left(  a\in O\ \Longleftrightarrow\ b\in O\right)
\right\}  .
\end{align*}
(It is easy to check that these are actually topologies. Of course,
$\mathcal{T}\looparrowleft\left(  a\geq b\right)  =\mathcal{T}\looparrowleft
\left(  b\leq a\right)  $.)
\end{definition}

Here comes a collection of simple properties of these three new topologies:

\begin{lemma}
\label{lem.insert-into-top}Let $X$ be a finite totally ordered set, and let
$\mathcal{T}$ be a topology on $X$. Let $a$ and $b$ be two elements of $X$.

\textbf{(a)} We have%
\begin{align}
\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  \cap\left(
\mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)   &  =\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\label{eq.lem.insert-into-top.1}\\
\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  \cup\left(
\mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)   &  =\mathcal{T}.
\label{eq.lem.insert-into-top.2}%
\end{align}


\textbf{(b)} We have%
\[
\mathcal{T}\looparrowleft\left(  a\sim b\right)  =\left(  \mathcal{T}%
\looparrowleft\left(  a\leq b\right)  \right)  \looparrowleft\left(  a\geq
b\right)  =\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)
\looparrowleft\left(  a\leq b\right)  .
\]


\textbf{(c)} If $a\leq_{\mathcal{T}}b$, then $\mathcal{T}\looparrowleft\left(
a\leq b\right)  =\mathcal{T}$ and $\mathcal{T}\looparrowleft\left(  a\sim
b\right)  =\mathcal{T}\looparrowleft\left(  a\geq b\right)  $.

\textbf{(d)} If $b\leq_{\mathcal{T}}a$, then $\mathcal{T}\looparrowleft\left(
a\geq b\right)  =\mathcal{T}$ and $\mathcal{T}\looparrowleft\left(  a\sim
b\right)  =\mathcal{T}\looparrowleft\left(  a\leq b\right)  $.

\textbf{(e)} If $c$ and $d$ are two elements of $X$, then $c\leq
_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$ holds if and only if%
\[
\left(  c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}}a\text{ and
}b\leq_{\mathcal{T}}d\right)  \right)  .
\]


\textbf{(f)} If $c$ and $d$ are two elements of $X$, then $c\leq
_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }d$ holds if and only if%
\[
\left(  c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}}b\text{ and
}a\leq_{\mathcal{T}}d\right)  \right)  .
\]


\textbf{(g)} If $c$ and $d$ are two elements of $X$, then $c\leq
_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$ holds if and only if%
\[
\left(  c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}}a\text{ and
}b\leq_{\mathcal{T}}d\right)  \text{ or }\left(  c\leq_{\mathcal{T}}b\text{
and }a\leq_{\mathcal{T}}d\right)  \right)  .
\]


\textbf{(h)} If $c$ and $d$ are two elements of $X$, then $c\leq
_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$ holds if and only if%
\[
\left(  c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d\text{ or
}c\leq_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }d\right)  .
\]


\textbf{(i)} If $c$ and $d$ are two elements of $X$, then $c\leq_{\mathcal{T}%
}d$ holds if and only if%
\[
\left(  c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d\text{ and
}c\leq_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }d\right)  .
\]


\textbf{(j)} If $c$ and $d$ are two elements of $X$, then $c\sim
_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$ holds if and only if%
\[
\left(  c\sim_{\mathcal{T}}d\text{ or }\left(  b\leq_{\mathcal{T}}%
c\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}d\leq_{\mathcal{T}%
}a\right)  \right)  .
\]


\textbf{(k)} If $c$ and $d$ are two elements of $X$, and if we have neither
$a\leq_{\mathcal{T}}b$ nor $b\leq_{\mathcal{T}}a$, then $c\sim_{\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }d$ holds if and only if%
\[
\left(  c\sim_{\mathcal{T}}d\text{ or }\left(  c\sim_{\mathcal{T}}a\text{ and
}d\sim_{\mathcal{T}}b\right)  \text{ or }\left(  c\sim_{\mathcal{T}}b\text{
and }d\sim_{\mathcal{T}}a\right)  \right)  .
\]


\textbf{(l)} We have%
\begin{align*}
\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
\cap\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)
\right)   &  =\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\sim
b\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
\cup\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)
\right)   &  =\mathcal{P}\left(  \mathcal{T}\right)  .
\end{align*}


\textbf{(m)} Assume that neither $a\leq_{\mathcal{T}}b$ nor $b\leq
_{\mathcal{T}}a$. Then, the three sets $\mathcal{U}\left(  \mathcal{T}%
\looparrowleft\left(  a\leq b\right)  \right)  $, $\mathcal{U}\left(
\mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)  $ and $\mathcal{U}%
\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  $ are
disjoint, and their union is $\mathcal{U}\left(  \mathcal{T}\right)  $.

\textbf{(n)} Assume that neither $a\leq_{\mathcal{T}}b$ nor $b\leq
_{\mathcal{T}}a$. Then,%
\begin{align*}
\left\vert X/\sim_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}\right\vert  &  =\left\vert X/\sim_{\mathcal{T}\looparrowleft\left(  a\geq
b\right)  }\right\vert =\left\vert X/\sim_{\mathcal{T}}\right\vert
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left\vert X/\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)
}\right\vert  &  =\left\vert X/\sim_{\mathcal{T}}\right\vert -1.
\end{align*}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.insert-into-top} (sketched).]Parts \textbf{(a)} and
\textbf{(b)} are straightforward to check.

\textbf{(c)} Assume that $a\leq_{\mathcal{T}}b$. Then, every $O\in\mathcal{T}$
satisfies $\left(  a\in\mathcal{T}\ \Longrightarrow\ b\in\mathcal{T}\right)
$. Hence, $\mathcal{T}\looparrowleft\left(  a\leq b\right)  =\mathcal{T}$ by
the definition of $\mathcal{T}\looparrowleft\left(  a\leq b\right)  $. From
Lemma \ref{lem.insert-into-top} \textbf{(b)}, we have $\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  =\underbrace{\left(  \mathcal{T}%
\looparrowleft\left(  a\leq b\right)  \right)  }_{=\mathcal{T}}\looparrowleft
\left(  a\geq b\right)  =\mathcal{T}\looparrowleft\left(  a\geq b\right)  $.
Thus, Lemma \ref{lem.insert-into-top} \textbf{(c)} is proven.

\textbf{(d)} The proof of part \textbf{(d)} is similar to that of \textbf{(c)}.

\textbf{(e)} $\Longleftarrow:$ Assume that $\left(  c\leq_{\mathcal{T}}d\text{
or }\left(  c\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}d\right)
\right)  $. We need to check that $c\leq_{\mathcal{T}\looparrowleft\left(
a\leq b\right)  }d$ holds. In other words, we need to check that every
$O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  $ satisfying $c\in O$
satisfies $d\in O$. So let us fix an $O\in\mathcal{T}\looparrowleft\left(
a\leq b\right)  $ satisfying $c\in O$. We must prove that $d\in O$.

We have $O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  \subseteq
\mathcal{T}$ (by the definition of $\mathcal{T}\looparrowleft\left(  a\leq
b\right)  $). Thus, if $c\leq_{\mathcal{T}}d$, then $d\in O$. Hence, for the
rest of this proof, we WLOG assume that we don't have $c\leq_{\mathcal{T}}d$.
Thus, by assumption, we have $c\leq_{\mathcal{T}}a$ and $b\leq_{\mathcal{T}}%
d$. Therefore, $a\in O$ (since $c\in O$ and $c\leq_{\mathcal{T}}a$). But
$O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  $, and therefore
$\left(  a\in O\ \Longrightarrow\ b\in O\right)  $ (by the definition of
$\mathcal{T}\looparrowleft\left(  a\leq b\right)  $), so that $b\in O$ (since
$a\in O$), and thus $d\in O$ (since $b\leq_{\mathcal{T}}d$). This completes
the proof of the $\Longleftarrow$ direction of Lemma \ref{lem.insert-into-top}
\textbf{(e)}.

$\Longrightarrow:$ Assume that $c\leq_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }d$ holds. We need to check that \newline$\left(  c\leq
_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}}a\text{ and }%
b\leq_{\mathcal{T}}d\right)  \right)  $. We can WLOG assume that we don't have
$c\leq_{\mathcal{T}}d$. Then, we must prove that $\left(  c\leq_{\mathcal{T}%
}a\text{ and }b\leq_{\mathcal{T}}d\right)  $.

We don't have $c\leq_{\mathcal{T}}d$. Hence, there exists a $Q\in\mathcal{T}$
such that $c\in Q$ but $d\notin Q$. Consider this $Q$. If we had $\left(  a\in
Q\ \Longrightarrow\ b\in Q\right)  $, then $Q$ would belong to $\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  $, which would yield $d\in Q$ (since
$c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$ and $c\in Q$),
which would contradict $d\notin Q$. Hence, we cannot have $\left(  a\in
Q\ \Longrightarrow\ b\in Q\right)  $. Thus, $a\in Q$ and $b\notin Q$.

Let $O\in\mathcal{T}$ be such that $c\in O$. We shall prove that $a\in O$.
Indeed, assume the contrary. Then, $a\notin O$. Thus, $a\notin Q\cap O$, so
that $\left(  a\in Q\cap O\ \Longrightarrow\ b\in Q\cap O\right)  $. Since
$Q\cap O\in\mathcal{T}$ (because $Q\in\mathcal{T}$ and $O\in\mathcal{T}$),
this yields $Q\cap O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  $.
Since we also have $c\in Q\cap O$ (since $c\in Q$ and $c\in O$), this yields
$d\in Q\cap O$ (since $c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}d$), so that $d\in Q\cap O\subseteq Q$, which contradicts $d\notin Q$. This
contradiction proves that our assumption was wrong. Hence, $a\in O$ is proven.
Forget now that we fixed $O$. Thus we have shown that $a\in O$ for every
$O\in\mathcal{T}$ which satisfies $c\in O$. In other words, $c\leq
_{\mathcal{T}}a$.

Let $O\in\mathcal{T}$ be such that $b\in O$. We shall prove that $d\in O$.
Indeed, assume the contrary. Then, $d\notin O$. Thus, $d\notin Q\cup O$ (since
$d\notin Q$ and $d\notin O$). But $b\in O\subseteq Q\cup O$, so that $\left(
a\in Q\cup O\ \Longrightarrow\ b\in Q\cup O\right)  $. Since $Q\cup
O\in\mathcal{T}$ (because $Q\in\mathcal{T}$ and $O\in\mathcal{T}$), this
yields $Q\cup O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  $. Since we
also have $c\in Q\cup O$ (since $c\in Q$), this yields $d\in Q\cup O$ (since
$c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$), which
contradicts $d\notin Q\cup O$. This contradiction proves that our assumption
was wrong. Hence, $d\in O$ is proven. Forget now that we fixed $O$. Thus we
have shown that $d\in O$ for every $O\in\mathcal{T}$ which satisfies $b\in O$.
In other words, $b\leq_{\mathcal{T}}d$.

We thus have shown that $\left(  c\leq_{\mathcal{T}}a\text{ and }%
b\leq_{\mathcal{T}}d\right)  $. This completes the proof of the
$\Longrightarrow$ direction of Lemma \ref{lem.insert-into-top} \textbf{(e)}.

\textbf{(f)} The proof of part \textbf{(f)} is analogous to that of
\textbf{(e)}.

\textbf{(g)} Let $c$ and $d$ be two elements of $X$. Then, we have the
following logical equivalence:%
\begin{align*}
&  \ \left(  c\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d\right)
\\
&  \Longleftrightarrow\ \left(  c\leq_{\left(  \mathcal{T}\looparrowleft
\left(  a\leq b\right)  \right)  \looparrowleft\left(  a\geq b\right)
}d\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.insert-into-top} \textbf{(b)}}\right) \\
&  \Longleftrightarrow\ \left(  c\leq_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }d\text{ or }\left(  c\leq_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }b\text{ and }a\leq_{\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  }d\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.insert-into-top}
\textbf{(f)}}\right) \\
&  \Longleftrightarrow\ \left(  \left(  c\leq_{\mathcal{T}}d\text{ or }\left(
c\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}d\right)  \right)  \text{
or }\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left(  \left(  c\leq_{\mathcal{T}}b\text{ or
}\left(  c\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}b\right)  \right)
\text{ and }\left(  a\leq_{\mathcal{T}}d\text{ or }\left(  a\leq_{\mathcal{T}%
}a\text{ and }b\leq_{\mathcal{T}}d\right)  \right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.insert-into-top}
\textbf{(e)}, applied to each of the three inequalities}\right) \\
&  \Longleftrightarrow\ \left(  c\leq_{\mathcal{T}}d\text{ or }\left(
c\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}d\right)  \text{ or
}\left(  c\leq_{\mathcal{T}}b\text{ and }a\leq_{\mathcal{T}}d\right)  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{after simplifying using the transitivity
and reflexivity of }\leq_{\mathcal{T}}\right)  .
\end{align*}
This proves Lemma \ref{lem.insert-into-top} \textbf{(g)}.

\textbf{(h)} This is just a rewriting of Lemma \ref{lem.insert-into-top}
\textbf{(g)} using parts \textbf{(e)} and \textbf{(f)}.

\textbf{(i)} $\Longrightarrow:$ This is clear.

$\Longleftarrow:$ Assume that $\left(  c\leq_{\mathcal{T}\looparrowleft\left(
a\leq b\right)  }d\text{ and }c\leq_{\mathcal{T}\looparrowleft\left(  a\geq
b\right)  }d\right)  $. We need to show that $c\leq_{\mathcal{T}}d$. Indeed,
assume the contrary.

We have $c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$. Thus,
Lemma \ref{lem.insert-into-top} \textbf{(e)} yields that we must have $\left(
c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}}a\text{ and }%
b\leq_{\mathcal{T}}d\right)  \right)  $. Since we assumed that $c\leq
_{\mathcal{T}}d$ does not hold, this yields $\left(  c\leq_{\mathcal{T}%
}a\text{ and }b\leq_{\mathcal{T}}d\right)  $. Similarly, $\left(
c\leq_{\mathcal{T}}b\text{ and }a\leq_{\mathcal{T}}d\right)  $. Thus,
$c\leq_{\mathcal{T}}b\leq_{\mathcal{T}}d$, which contradicts our assumption
that not $c\leq_{\mathcal{T}}d$. This contradiction completes the proof.

\textbf{(j)} We have $c\sim_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}d$ if and only if $\left(  c\leq_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }d\text{ and }d\leq_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }c\right)  $. We can rewrite each of the two statements
$c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$ and
$d\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }c$ using Lemma
\ref{lem.insert-into-top} \textbf{(e)}, and then simplify the result; we end
up with Lemma \ref{lem.insert-into-top} \textbf{(j)}.

\textbf{(k)} Let $c$ and $d$ be two elements of $X$. Assume that we have
neither $a\leq_{\mathcal{T}}b$ nor $b\leq_{\mathcal{T}}a$. We have
$c\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$ if and only if
$\left(  c\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d\text{ and
}d\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }c\right)  $. We can
rewrite each of the two statements $c\leq_{\mathcal{T}\looparrowleft\left(
a\sim b\right)  }d$ and $d\leq_{\mathcal{T}\looparrowleft\left(  a\sim
b\right)  }c$ using Lemma \ref{lem.insert-into-top} \textbf{(g)}, and then
simplify the result (a disjunction with $9$ cases, of which many can be ruled
out due to the assumption that neither $a\leq_{\mathcal{T}}b$ nor
$b\leq_{\mathcal{T}}a$); we end up with Lemma \ref{lem.insert-into-top}
\textbf{(k)}.

\textbf{(l)} \textit{Proof of }$\mathcal{P}\left(  \mathcal{T}\looparrowleft
\left(  a\leq b\right)  \right)  \cap\mathcal{P}\left(  \mathcal{T}%
\looparrowleft\left(  a\geq b\right)  \right)  =\mathcal{P}\left(
\mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  $\textit{:} Whenever
$f$ is a surjective map $X\rightarrow\left[  p\right]  $ for some
$p\in\mathbb{N}$, we have the following logical equivalence:%
\begin{align*}
&  \ \left(  f\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  \cap\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(
a\geq b\right)  \right)  \right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  f\in\mathcal{P}\left(
\mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  \right)
}_{\Longleftrightarrow\left(  \text{every }c\in X\text{ and }d\in X\text{
satisfying }c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d\text{
satisfy }f\left(  c\right)  \leq f\left(  d\right)  \right)  }\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \wedge\text{ }\underbrace{\left(
f\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  b\leq a\right)
\right)  \right)  }_{\Longleftrightarrow\left(  \text{every }c\in X\text{ and
}d\in X\text{ satisfying }c\leq_{\mathcal{T}\looparrowleft\left(  b\leq
a\right)  }d\text{ satisfy }f\left(  c\right)  \leq f\left(  d\right)
\right)  }\right) \\
&  \Longleftrightarrow\ \left(  \left(  \text{every }c\in X\text{ and }d\in
X\text{ satisfying }c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}d\text{ satisfy }f\left(  c\right)  \leq f\left(  d\right)  \right)  \right.
\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \wedge\text{ }\left(  \text{every }c\in X\text{
and }d\in X\text{ satisfying }c\leq_{\mathcal{T}\looparrowleft\left(  b\leq
a\right)  }d\text{ satisfy }f\left(  c\right)  \leq f\left(  d\right)
\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{every }c\in X\text{ and }d\in X\text{
satisfying }\underbrace{\left(  c\leq_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }d\text{ or }c\leq_{\mathcal{T}\looparrowleft\left(  a\geq b\right)
}d\right)  }_{\substack{\Longleftrightarrow\left(  c\leq_{\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }d\right)  \\\text{(by Lemma
\ref{lem.insert-into-top} \textbf{(h)})}}}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{ satisfy }f\left(  c\right)  \leq
f\left(  d\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{every }c\in X\text{ and }d\in X\text{
satisfying }c\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d\text{
satisfy }f\left(  c\right)  \leq f\left(  d\right)  \right) \\
&  \Longleftrightarrow\ \left(  f\in\mathcal{P}\left(  \mathcal{T}%
\looparrowleft\left(  a\sim b\right)  \right)  \right)  .
\end{align*}
Thus, $\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)
\right)  \cap\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\geq
b\right)  \right)  =\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\sim
b\right)  \right)  $ is proven.

It remains to prove $\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  \cup\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(
a\geq b\right)  \right)  =\mathcal{P}\left(  \mathcal{T}\right)  $. We shall
achieve this by proving both inclusions separately:

\textit{Proof of }$\mathcal{P}\left(  \mathcal{T}\right)  \subseteq
\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
\cup\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)
\right)  $\textit{:} Let $f\in\mathcal{P}\left(  \mathcal{T}\right)  $. We
must prove that $f\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  \cup\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(
a\geq b\right)  \right)  $.

We WLOG assume that $f\left(  a\right)  \leq f\left(  b\right)  $. We shall
now show that $f\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  $. This will yield that $f\in\mathcal{P}\left(
\mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  \cup\mathcal{P}%
\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)  $, and thus
complete this proof of $\mathcal{P}\left(  \mathcal{T}\right)  \subseteq
\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
\cup\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)
\right)  $.

Let $c\in X$ and $d\in X$ be such that $c\leq_{\mathcal{T}\looparrowleft
\left(  a\leq b\right)  }d$. In order to prove that $f\in\mathcal{P}\left(
\mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  $, we must now show
that $f\left(  c\right)  \leq f\left(  d\right)  $.

We have $c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$. Due to
Lemma \ref{lem.insert-into-top} \textbf{(e)}, this yields that \newline%
$\left(  c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}}a\text{ and
}b\leq_{\mathcal{T}}d\right)  \right)  $. In the first of these cases,
$f\left(  c\right)  \leq f\left(  d\right)  $ follows immediately from
$f\in\mathcal{P}\left(  \mathcal{T}\right)  $; thus, let us assume that we are
in the second case. Thus, $c\leq_{\mathcal{T}}a$ and $b\leq_{\mathcal{T}}d$.
From $f\in\mathcal{P}\left(  \mathcal{T}\right)  $, we thus obtain $f\left(
c\right)  \leq f\left(  a\right)  $ and $f\left(  b\right)  \leq f\left(
d\right)  $. Hence, $f\left(  c\right)  \leq f\left(  a\right)  \leq f\left(
b\right)  \leq f\left(  d\right)  $, qed.

\textit{Proof of }$\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  \cup\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(
a\geq b\right)  \right)  \subseteq\mathcal{P}\left(  \mathcal{T}\right)
$\textit{:} We now need to show that $\mathcal{P}\left(  \mathcal{T}%
\looparrowleft\left(  a\leq b\right)  \right)  \cup\mathcal{P}\left(
\mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)  \subseteq
\mathcal{P}\left(  \mathcal{T}\right)  $. To do so, it is clearly enough to
prove $\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)
\right)  \subseteq\mathcal{P}\left(  \mathcal{T}\right)  $ and $\mathcal{P}%
\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)
\subseteq\mathcal{P}\left(  \mathcal{T}\right)  $. We shall only show the
first of these two relations, as the second is analogous. Let $f\in
\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
$. Then, every $c\in X$ and $d\in X$ satisfying $c\leq_{\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }d$ satisfy $f\left(  c\right)  \leq
f\left(  d\right)  $. Hence, every $c\in X$ and $d\in X$ satisfying
$c\leq_{\mathcal{T}}d$ satisfy $f\left(  c\right)  \leq f\left(  d\right)  $
(since every $c\in X$ and $d\in X$ satisfying $c\leq_{\mathcal{T}}d$ satisfy
$c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$ (due to Lemma
\ref{lem.insert-into-top} \textbf{(e)})). In other words, $f\in\mathcal{P}%
\left(  \mathcal{T}\right)  $. Since this is proven for every $f\in
\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
$, we thus conclude that $\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(
a\leq b\right)  \right)  \subseteq\mathcal{P}\left(  \mathcal{T}\right)  $.

The proof of Lemma \ref{lem.insert-into-top} \textbf{(l)} is thus complete.

\textbf{(m)} It is clearly enough to prove the three equalities%
\begin{align}
\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
&  =\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(
a\right)  <f\left(  b\right)  \right\}  ;\label{pf.lem.insert-into-top.k.1}\\
\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)
&  =\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(
a\right)  =f\left(  b\right)  \right\}  ;\label{pf.lem.insert-into-top.k.2}\\
\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)
&  =\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(
a\right)  >f\left(  b\right)  \right\}  . \label{pf.lem.insert-into-top.k.3}%
\end{align}
We shall only check the first two of these three equalities (since the third
one is analogous to the first).

Let us first check that $a<_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}b$. Indeed, it is clear from the definition of $\mathcal{T}\looparrowleft
\left(  a\leq b\right)  $ that $a\leq_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }b$. Thus, in order to prove that $a<_{\mathcal{T}\looparrowleft
\left(  a\leq b\right)  }b$, we must only show that we don't have
$b\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }a$. To achieve this,
we assume the contrary. Lemma \ref{lem.insert-into-top} \textbf{(e)} (applied
to $c=b$ and $d=a$) thus yields that $\left(  b\leq_{\mathcal{T}}a\text{ or
}\left(  b\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}a\right)  \right)
$. In either of these cases, we must have $b\leq_{\mathcal{T}}a$, which
contradicts the assumption that neither $a\leq_{\mathcal{T}}b$ nor
$b\leq_{\mathcal{T}}a$. So $a<_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }b$ is proven.

Next, we are going to prove (\ref{pf.lem.insert-into-top.k.1}) by showing its
two inclusions separately:

\textit{Proof of }$\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  \subseteq\left\{  f\in\mathcal{U}\left(  \mathcal{T}%
\right)  \ \mid\ f\left(  a\right)  <f\left(  b\right)  \right\}  $\textit{:}
Let $g\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)
\right)  $. Thus, $g\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(
a\leq b\right)  \right)  $, and every two elements $i$ and $j$ of $X$
satisfying $i<_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }j$ must
satisfy $g\left(  i\right)  <g\left(  j\right)  $. Applying the latter fact to
$i=a$ and $j=b$, we obtain $g\left(  a\right)  <g\left(  b\right)  $ (since
$a<_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }b$).

Moreover, $g\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  \subseteq\mathcal{P}\left(  \mathcal{T}\looparrowleft
\left(  a\leq b\right)  \right)  \cup\mathcal{P}\left(  \mathcal{T}%
\looparrowleft\left(  a\geq b\right)  \right)  =\mathcal{P}\left(
\mathcal{T}\right)  $ (by Lemma \ref{lem.insert-into-top} \textbf{(l)}).

Let now $i$ and $j$ be any two elements of $X$ satisfying $i<_{\mathcal{T}}j$.
We shall show that $g\left(  i\right)  <g\left(  j\right)  $.

Indeed, $i<_{\mathcal{T}}j$, thus $i\leq_{\mathcal{T}}j$ and therefore
$i\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }j$ (due to Lemma
\ref{lem.insert-into-top} \textbf{(e)}). Assume (for the sake of
contradiction) that $j\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}i$. Then, $i\sim_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }j$, and
thus (by Lemma \ref{lem.insert-into-top} \textbf{(j)}, applied to $c=i$ and
$d=j$) we have $\left(  i\sim_{\mathcal{T}}j\text{ or }\left(  b\leq
_{\mathcal{T}}i\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}%
j\leq_{\mathcal{T}}a\right)  \right)  $. But neither of these two cases can
occur (since $i<_{\mathcal{T}}j$ precludes $i\sim_{\mathcal{T}}j$, and since
$b\leq_{\mathcal{T}}i\leq_{\mathcal{T}}a$ contradicts our assumption that not
$b\leq_{\mathcal{T}}a$). Hence, we have our contradiction. Thus, our
assumption (that $j\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }i$)
was false. We therefore have $i\leq_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }j$ but not $j\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}i$. In other words, $i<_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }%
j$. Thus, $g\left(  i\right)  <g\left(  j\right)  $ (since $g\in
\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  $).

Now, let us forget that we fixed $i$ and $j$. We thus have shown that any two
elements $i$ and $j$ of $X$ satisfying $i<_{\mathcal{T}}j$ satisfy $g\left(
i\right)  <g\left(  j\right)  $. In other words, $g\in\mathcal{U}\left(
\mathcal{T}\right)  $ (since we already know that $g\in\mathcal{P}\left(
\mathcal{T}\right)  $). Thus, $g$ is an element of $\mathcal{U}\left(
\mathcal{T}\right)  $ and satisfies $g\left(  a\right)  <g\left(  b\right)  $.
In other words, $g\in\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)
\ \mid\ f\left(  a\right)  <f\left(  b\right)  \right\}  $. Since this is
proven for every $g\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(
a\leq b\right)  \right)  $, we thus conclude that \newline$\mathcal{U}\left(
\mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  \subseteq\left\{
f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(  a\right)
<f\left(  b\right)  \right\}  $.

\textit{Proof of }$\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)
\ \mid\ f\left(  a\right)  <f\left(  b\right)  \right\}  \subseteq
\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
$\textit{:} Let \newline$g\in\left\{  f\in\mathcal{U}\left(  \mathcal{T}%
\right)  \ \mid\ f\left(  a\right)  <f\left(  b\right)  \right\}  $. Then,
$g\in\mathcal{U}\left(  \mathcal{T}\right)  $ and $g\left(  a\right)
<g\left(  b\right)  $. From $g\in\mathcal{U}\left(  \mathcal{T}\right)  $, we
obtain $g\in\mathcal{P}\left(  \mathcal{T}\right)  $.

Let now $c\in X$ and $d\in X$ be such that $c\leq_{\mathcal{T}\looparrowleft
\left(  a\leq b\right)  }d$. We now aim to show that $g\left(  c\right)  \leq
g\left(  d\right)  $.

Indeed, from $c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$, we
obtain $\left(  c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}%
}a\text{ and }b\leq_{\mathcal{T}}d\right)  \right)  $ (by Lemma
\ref{lem.insert-into-top} \textbf{(e)}). In the first of these two cases, we
obtain $g\left(  c\right)  \leq g\left(  d\right)  $ immediately (since
$g\in\mathcal{P}\left(  \mathcal{T}\right)  $), while in the second case we
obtain%
\begin{align*}
g\left(  c\right)   &  \leq g\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }c\leq_{\mathcal{T}}a\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right) \\
&  <g\left(  b\right)  \leq g\left(  d\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }b\leq_{\mathcal{T}}d\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right)  .
\end{align*}
Thus, $g\left(  c\right)  \leq g\left(  d\right)  $ is proven in either case.

Now, let us forget that we fixed $c$ and $d$. We thus have proven that
$g\left(  c\right)  \leq g\left(  d\right)  $ for any $c\in X$ and $d\in X$
satisfying $c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$. In
other words, $g\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  $.

Now, let $c\in X$ and $d\in X$ be such that $c<_{\mathcal{T}\looparrowleft
\left(  a\leq b\right)  }d$. We now aim to show that $g\left(  c\right)
<g\left(  d\right)  $.

Indeed, from $c<_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$, we
obtain $c\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$, and thus
\newline$\left(  c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}%
}a\text{ and }b\leq_{\mathcal{T}}d\right)  \right)  $ (by Lemma
\ref{lem.insert-into-top} \textbf{(e)}). In the second of these two cases, we
have%
\begin{align*}
g\left(  c\right)   &  \leq g\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }c\leq_{\mathcal{T}}a\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right) \\
&  <g\left(  b\right)  \leq g\left(  d\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }b\leq_{\mathcal{T}}d\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right)  .
\end{align*}
Thus, $g\left(  c\right)  <g\left(  d\right)  $ is proven in the second case.
We thus WLOG assume that we are in the first case. That is, we have
$c\leq_{\mathcal{T}}d$. If $c<_{\mathcal{T}}d$, then we can immediately
conclude that $g\left(  c\right)  <g\left(  d\right)  $ (since $g\in
\mathcal{U}\left(  \mathcal{T}\right)  $). Hence, we WLOG assume that we don't
have $c<_{\mathcal{T}}d$. Thus, $c\sim_{\mathcal{T}}d$ (since $c\leq
_{\mathcal{T}}d$), so that $d\leq_{\mathcal{T}}c$. Hence, $\left(
d\leq_{\mathcal{T}}c\text{ or }\left(  d\leq_{\mathcal{T}}a\text{ and }%
b\leq_{\mathcal{T}}c\right)  \right)  $, so that Lemma
\ref{lem.insert-into-top} \textbf{(e)} (applied to $d$ and $c$ instead of $c$
and $d$) yields $d\leq_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }c$.
But this contradicts $c<_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }%
d$. Thus, we have obtained a contradiction, and our proof of $g\left(
c\right)  <g\left(  d\right)  $ is complete.

Now, let us forget that we fixed $c$ and $d$. We thus have proven that
$g\left(  c\right)  <g\left(  d\right)  $ for any $c\in X$ and $d\in X$
satisfying $c<_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$. In other
words, $g\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  $ (since $g\in\mathcal{P}\left(  \mathcal{T}\looparrowleft
\left(  a\leq b\right)  \right)  $). Since this is proven for every
$g\in\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(
a\right)  <f\left(  b\right)  \right\}  $, we thus conclude that $\left\{
f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(  a\right)
<f\left(  b\right)  \right\}  \subseteq\mathcal{U}\left(  \mathcal{T}%
\looparrowleft\left(  a\leq b\right)  \right)  $.

Combining $\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)
\right)  \subseteq\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)
\ \mid\ f\left(  a\right)  <f\left(  b\right)  \right\}  $ with \newline%
$\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(
a\right)  <f\left(  b\right)  \right\}  \subseteq\mathcal{U}\left(
\mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  $, we obtain
(\ref{pf.lem.insert-into-top.k.1}).

Let us next check that $a\sim_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }b$. Indeed, it is clear from the definition of $\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  $ that $a\leq_{\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }b$ and that $b\leq_{\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }a$. Combining these, we obtain
$a\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }b$.

Next, we are going to prove (\ref{pf.lem.insert-into-top.k.2}) by showing its
two inclusions separately:

\textit{Proof of }$\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\sim
b\right)  \right)  \subseteq\left\{  f\in\mathcal{U}\left(  \mathcal{T}%
\right)  \ \mid\ f\left(  a\right)  =f\left(  b\right)  \right\}  $\textit{:}
Let $g\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)
\right)  $. Thus, $g\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(
a\sim b\right)  \right)  $, and every two elements $i$ and $j$ of $X$
satisfying $i<_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }j$ must
satisfy $g\left(  i\right)  <g\left(  j\right)  $. We have $a\sim
_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }b$ and $g\in
\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)
$; thus, $g\left(  a\right)  =g\left(  b\right)  $.

Moreover,%
\begin{align*}
g  &  \in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)
\right)  =\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)
\right)  \cap\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\geq
b\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.insert-into-top}
\textbf{(l)}}\right) \\
&  \subseteq\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  \subseteq\mathcal{P}\left(  \mathcal{T}\looparrowleft
\left(  a\leq b\right)  \right)  \cup\mathcal{P}\left(  \mathcal{T}%
\looparrowleft\left(  a\geq b\right)  \right)  =\mathcal{P}\left(
\mathcal{T}\right)
\end{align*}
(by Lemma \ref{lem.insert-into-top} \textbf{(l)}).

Now, let $i$ and $j$ be any two elements of $X$ satisfying $i<_{\mathcal{T}}%
j$. We shall show that $g\left(  i\right)  <g\left(  j\right)  $.

Indeed, $i<_{\mathcal{T}}j$, thus $i\leq_{\mathcal{T}}j$ and therefore
$i\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }j$ (due to Lemma
\ref{lem.insert-into-top} \textbf{(g)}). Assume (for the sake of
contradiction) that $j\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)
}i$. Then, $i\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }j$, and
thus (by Lemma \ref{lem.insert-into-top} \textbf{(k)}, applied to $c=i$ and
$d=j$) we have $\left(  i\sim_{\mathcal{T}}j\text{ or }\left(  i\sim
_{\mathcal{T}}a\text{ and }j\sim_{\mathcal{T}}b\right)  \text{ or }\left(
i\sim_{\mathcal{T}}b\text{ and }j\sim_{\mathcal{T}}a\right)  \right)  $. But
neither of these three cases can occur\footnote{Indeed, the first case
($i\sim_{\mathcal{T}}j$) is precluded by the fact that $i<_{\mathcal{T}}j$.
The second case ($i\sim_{\mathcal{T}}a$ and $j\sim_{\mathcal{T}}b$) cannot
occur since it would lead to $a\sim_{\mathcal{T}}i\leq_{\mathcal{T}}%
j\sim_{\mathcal{T}}b$, which would contradict the assumption that we have
neither $a\leq_{\mathcal{T}}b$ nor $b\leq_{\mathcal{T}}a$. The third case
($i\sim_{\mathcal{T}}b$ and $j\sim_{\mathcal{T}}a$) cannot occur for a similar
reason.}. Hence, we have our contradiction. Thus, our assumption (that
$j\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }i$) was false. We
therefore have $i\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }j$
but not $j\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }i$. In other
words, $i<_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }j$. Thus,
$g\left(  i\right)  <g\left(  j\right)  $ (since $g\in\mathcal{U}\left(
\mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  $).

Now, let us forget that we fixed $i$ and $j$. We thus have shown that any two
elements $i$ and $j$ of $X$ satisfying $i<_{\mathcal{T}}j$ satisfy $g\left(
i\right)  <g\left(  j\right)  $. In other words, $g\in\mathcal{U}\left(
\mathcal{T}\right)  $ (since we already know that $g\in\mathcal{P}\left(
\mathcal{T}\right)  $). Thus, $g$ is an element of $\mathcal{U}\left(
\mathcal{T}\right)  $ and satisfies $g\left(  a\right)  =g\left(  b\right)  $.
In other words, $g\in\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)
\ \mid\ f\left(  a\right)  =f\left(  b\right)  \right\}  $. Since this is
proven for every $g\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(
a\sim b\right)  \right)  $, we thus conclude that \newline$\mathcal{U}\left(
\mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  \subseteq\left\{
f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(  a\right)
=f\left(  b\right)  \right\}  $.

\textit{Proof of }$\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)
\ \mid\ f\left(  a\right)  =f\left(  b\right)  \right\}  \subseteq
\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)
$\textit{:} Let \newline$g\in\left\{  f\in\mathcal{U}\left(  \mathcal{T}%
\right)  \ \mid\ f\left(  a\right)  =f\left(  b\right)  \right\}  $. Then,
$g\in\mathcal{U}\left(  \mathcal{T}\right)  $ and $g\left(  a\right)
=g\left(  b\right)  $. From $g\in\mathcal{U}\left(  \mathcal{T}\right)  $, we
obtain $g\in\mathcal{P}\left(  \mathcal{T}\right)  $.

Let now $c\in X$ and $d\in X$ be such that $c\leq_{\mathcal{T}\looparrowleft
\left(  a\sim b\right)  }d$. We now aim to show that $g\left(  c\right)  \leq
g\left(  d\right)  $.

Indeed, from $c\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$, we
obtain \newline$\left(  c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq
_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}d\right)  \text{ or }\left(
c\leq_{\mathcal{T}}b\text{ and }a\leq_{\mathcal{T}}d\right)  \right)  $ (by
Lemma \ref{lem.insert-into-top} \textbf{(g)}). In the first of these three
cases, we obtain $g\left(  c\right)  \leq g\left(  d\right)  $ immediately
(since $g\in\mathcal{P}\left(  \mathcal{T}\right)  $). In the second case, we
obtain%
\begin{align*}
g\left(  c\right)   &  \leq g\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }c\leq_{\mathcal{T}}a\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right) \\
&  =g\left(  b\right)  \leq g\left(  d\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }b\leq_{\mathcal{T}}d\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right)  .
\end{align*}
In the third case, we obtain%
\begin{align*}
g\left(  c\right)   &  \leq g\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }c\leq_{\mathcal{T}}b\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right) \\
&  =g\left(  a\right)  \leq g\left(  d\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }a\leq_{\mathcal{T}}d\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right)  .
\end{align*}
Thus, $g\left(  c\right)  \leq g\left(  d\right)  $ is proven in either case.

Now, let us forget that we fixed $c$ and $d$. We thus have proven that
$g\left(  c\right)  \leq g\left(  d\right)  $ for any $c\in X$ and $d\in X$
satisfying $c\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$. In
other words, $g\in\mathcal{P}\left(  \mathcal{T}\looparrowleft\left(  a\sim
b\right)  \right)  $.

Now, let $c\in X$ and $d\in X$ be such that $c<_{\mathcal{T}\looparrowleft
\left(  a\sim b\right)  }d$. We now aim to show that $g\left(  c\right)
<g\left(  d\right)  $.

Indeed, from $c<_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$, we
obtain $c\leq_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$, and thus
\newline$\left(  c\leq_{\mathcal{T}}d\text{ or }\left(  c\leq_{\mathcal{T}%
}a\text{ and }b\leq_{\mathcal{T}}d\right)  \text{ or }\left(  c\leq
_{\mathcal{T}}b\text{ and }a\leq_{\mathcal{T}}d\right)  \right)  $ (by Lemma
\ref{lem.insert-into-top} \textbf{(g)}). We study these three cases separately:

\begin{itemize}
\item Assume that we are in the first case, i.e., we have $c\leq_{\mathcal{T}%
}d$. Then, $c<_{\mathcal{T}}d$ (since otherwise, we would have $d\leq
_{\mathcal{T}}c$, and therefore $d\leq_{\mathcal{T}\looparrowleft\left(  a\sim
b\right)  }c$ (by Lemma \ref{lem.insert-into-top} \textbf{(g)}), which would
contradict $c<_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$). Hence,
$g\left(  c\right)  <g\left(  d\right)  $ (since $g\in\mathcal{U}\left(
\mathcal{T}\right)  $).

\item Assume that we are in the second case, i.e., we have $\left(
c\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}d\right)  $. Then,%
\begin{align*}
g\left(  c\right)   &  \leq g\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }c\leq_{\mathcal{T}}a\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right) \\
&  =g\left(  b\right)  \leq g\left(  d\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }b\leq_{\mathcal{T}}d\text{ and }g\in\mathcal{P}\left(
\mathcal{T}\right)  \right)  .
\end{align*}
If at least one of the strict inequalities $c<_{\mathcal{T}}a$ or
$b<_{\mathcal{T}}d$ holds, then we can strengthen this to a strict inequality
$g\left(  c\right)  <g\left(  d\right)  $ (because $g\in\mathcal{U}\left(
\mathcal{T}\right)  $), and thus be done. Hence, we WLOG assume that none of
the inequalities $c<_{\mathcal{T}}a$ or $b<_{\mathcal{T}}d$ holds. Thus,
$c\sim_{\mathcal{T}}a$ and $b\sim_{\mathcal{T}}d$. Hence, $c\sim
_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }a$ and $b\sim
_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$ (by Lemma
\ref{lem.insert-into-top} \textbf{(k)}), so that $c\sim_{\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }a\sim_{\mathcal{T}\looparrowleft\left(
a\sim b\right)  }b\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$,
which contradicts $c<_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$.
Hence, we are done in the second case as well.

\item The third case is similar to the second case.
\end{itemize}

Thus, our proof of $g\left(  c\right)  <g\left(  d\right)  $ is complete in
each case.

Now, let us forget that we fixed $c$ and $d$. We thus have proven that
$g\left(  c\right)  <g\left(  d\right)  $ for any $c\in X$ and $d\in X$
satisfying $c<_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }d$. In other
words, $g\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\sim
b\right)  \right)  $ (since $g\in\mathcal{P}\left(  \mathcal{T}\looparrowleft
\left(  a\sim b\right)  \right)  $). Since this is proven for every
$g\in\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(
a\right)  =f\left(  b\right)  \right\}  $, we thus conclude that $\left\{
f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(  a\right)
=f\left(  b\right)  \right\}  \subseteq\mathcal{U}\left(  \mathcal{T}%
\looparrowleft\left(  a\sim b\right)  \right)  $.

Combining $\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)
\right)  \subseteq\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)
\ \mid\ f\left(  a\right)  =f\left(  b\right)  \right\}  $ with \newline%
$\left\{  f\in\mathcal{U}\left(  \mathcal{T}\right)  \ \mid\ f\left(
a\right)  =f\left(  b\right)  \right\}  \subseteq\mathcal{U}\left(
\mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  $, we obtain
(\ref{pf.lem.insert-into-top.k.2}).

Now, our proof of Lemma \ref{lem.insert-into-top} \textbf{(m)} is complete.

\textbf{(n)} If $c$ and $d$ are two elements of $X$, then $c\sim
_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }d$ holds if and only if%
\[
\left(  c\sim_{\mathcal{T}}d\text{ or }\left(  b\leq_{\mathcal{T}}%
c\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}d\leq_{\mathcal{T}%
}a\right)  \right)
\]
(according to Lemma \ref{lem.insert-into-top} \textbf{(j)}). Since $\left(
b\leq_{\mathcal{T}}c\leq_{\mathcal{T}}a\text{ and }b\leq_{\mathcal{T}}%
d\leq_{\mathcal{T}}a\right)  $ cannot hold (because of our assumption that not
$b\leq_{\mathcal{T}}a$), this simplifies as follows: If $c$ and $d$ are two
elements of $X$, then $c\sim_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}d$ holds if and only if $c\sim_{\mathcal{T}}d$. Thus, the equivalence
relation $\sim_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }$ is
identical to $\sim_{\mathcal{T}}$. Hence, $\left\vert X/\sim_{\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }\right\vert =\left\vert X/\sim
_{\mathcal{T}}\right\vert $. Similarly, $\left\vert X/\sim_{\mathcal{T}%
\looparrowleft\left(  a\geq b\right)  }\right\vert =\left\vert X/\sim
_{\mathcal{T}}\right\vert $. Thus, \newline$\left\vert X/\sim_{\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }\right\vert =\left\vert X/\sim
_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }\right\vert =\left\vert
X/\sim_{\mathcal{T}}\right\vert $ is proven. It remains to show $\left\vert
X/\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }\right\vert
=\left\vert X/\sim_{\mathcal{T}}\right\vert -1$.

Lemma \ref{lem.insert-into-top} \textbf{(k)} yields the following: If $c$ and
$d$ are two elements of $X$, then $c\sim_{\mathcal{T}\looparrowleft\left(
a\sim b\right)  }d$ holds if and only if%
\[
\left(  c\sim_{\mathcal{T}}d\text{ or }\left(  c\sim_{\mathcal{T}}a\text{ and
}d\sim_{\mathcal{T}}b\right)  \text{ or }\left(  c\sim_{\mathcal{T}}b\text{
and }d\sim_{\mathcal{T}}a\right)  \right)  .
\]
In other words, two elements of $X$ are equivalent under the equivalence
relation $\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }$ if and
only if either they are equivalent under $\sim_{\mathcal{T}}$, or one of them
is in the $\sim_{\mathcal{T}}$-class of $a$ while the other is in the
$\sim_{\mathcal{T}}$-class of $b$. Thus, when passing from the equivalence
relation $\sim_{\mathcal{T}}$ to $\sim_{\mathcal{T}\looparrowleft\left(  a\sim
b\right)  }$, the equivalence classes of $a$ and $b$ get merged (and these two
classes used to be separate for $\sim_{\mathcal{T}}$, because of our
assumption that neither $a\leq_{\mathcal{T}}b$ nor $b\leq_{\mathcal{T}}a$),
while all other equivalence classes stay as they were. Thus, the total number
of equivalence classes decreases by $1$. In other words, $\left\vert
X/\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }\right\vert
=\left\vert X/\sim_{\mathcal{T}}\right\vert -1$. This completes the proof of
Lemma \ref{lem.insert-into-top} \textbf{(n)}.
\end{proof}

\begin{lemma}
\label{lem.indicator-parallelogram}Let $n\in\mathbb{N}$ and $\mathcal{T}%
\in\mathbf{T}_{n}$. Let $a$ and $b$ be two elements of $\left[  n\right]  $.
Then,%
\[
\underline{1}_{K_{\mathcal{T}}}=\underline{1}_{K_{\mathcal{T}\looparrowleft
\left(  a\leq b\right)  }}+\underline{1}_{K_{\mathcal{T}\looparrowleft\left(
a\geq b\right)  }}-\underline{1}_{K_{\mathcal{T}\looparrowleft\left(  a\sim
b\right)  }}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.indicator-parallelogram}.]It is clearly enough to
prove that
\begin{equation}
K_{\mathcal{T}}=K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }\cap
K_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }
\label{pf.lem.indicator-parallelogram.1}%
\end{equation}
and%
\begin{equation}
K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }=K_{\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }\cup K_{\mathcal{T}\looparrowleft
\left(  a\geq b\right)  }. \label{pf.lem.indicator-parallelogram.2}%
\end{equation}


Before we start proving these statements, let us rewrite the definition of
$K_{\mathcal{S}}$ for any topology $\mathcal{S}$ on $\left[  n\right]  $.
Namely, if $O$ is a subset of $\left[  n\right]  $, then we define a subset
$K_{O}$ of $\mathbb{R}^{n}$ by%
\[
K_{O}=\left\{  \left(  x_{1},x_{2},\ldots,x_{n}\right)  \in\mathbb{R}%
^{n}\ \mid\ \sum_{i\in\left[  n\right]  \setminus O}x_{i}\geq0\right\}  .
\]
It is now clear that any topology $\mathcal{S}$ on $\left[  n\right]  $
satisfies%
\begin{equation}
K_{\mathcal{S}}=\bigcap_{O\in\mathcal{S}}K_{O}.
\label{pf.lem.indicator-parallelogram.rewrite}%
\end{equation}
(Indeed, this is just a restatement of the definition of $K_{\mathcal{S}}$,
since the closed sets of $\mathcal{S}$ are the sets of the form $\left[
n\right]  \setminus O$ with $O$ being an open set of $\mathcal{S}$.)

\textit{Proof of (\ref{pf.lem.indicator-parallelogram.1}):} From
(\ref{pf.lem.indicator-parallelogram.rewrite}), we obtain $K_{\mathcal{T}%
}=\bigcap_{O\in\mathcal{T}}K_{O}$ and $K_{\mathcal{T}\looparrowleft\left(
a\leq b\right)  }=\bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }K_{O}$ and $K_{\mathcal{T}\looparrowleft\left(  a\geq b\right)
}=\bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\geq b\right)  }K_{O}$. Thus,%
\begin{align*}
\underbrace{K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }}%
_{=\bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  }K_{O}}%
\cap\underbrace{K_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }%
}_{=\bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\geq b\right)  }K_{O}}  &
=\left(  \bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  }%
K_{O}\right)  \cap\left(  \bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\geq
b\right)  }K_{O}\right) \\
&  =\bigcap_{O\in\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)
\right)  \cup\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)
}K_{O}\\
&  =\bigcap_{O\in\mathcal{T}}K_{O}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.lem.insert-into-top.2})}\right) \\
&  =K_{\mathcal{T}}.
\end{align*}
This proves (\ref{pf.lem.indicator-parallelogram.1}).

\textit{Proof of (\ref{pf.lem.indicator-parallelogram.2}):} It is easy to see
that $K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }\subseteq
K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }$%
\ \ \ \ \footnote{\textit{Proof.} Indeed, (\ref{eq.lem.insert-into-top.1})
yields $\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
\cap\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)
=\mathcal{T}\looparrowleft\left(  a\sim b\right)  $, so that $\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  \subseteq\mathcal{T}\looparrowleft
\left(  a\leq b\right)  $. Now, from
(\ref{pf.lem.indicator-parallelogram.rewrite}), we obtain $K_{\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }=\bigcap_{O\in\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }K_{O}$ and $K_{\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }=\bigcap_{O\in\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }K_{O}$. Thus,%
\begin{align*}
K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }  &  =\bigcap
_{O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  }K_{O}\subseteq
\bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\sim b\right)  }K_{O}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathcal{T}\looparrowleft\left(
a\sim b\right)  \subseteq\mathcal{T}\looparrowleft\left(  a\leq b\right)
\right) \\
&  =K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  },
\end{align*}
qed.}, and similarly $K_{\mathcal{T}\looparrowleft\left(  a\geq b\right)
}\subseteq K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }$. Combining
these two relations, we obtain $K_{\mathcal{T}\looparrowleft\left(  a\leq
b\right)  }\cup K_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }\subseteq
K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }$. Hence, in order to
prove (\ref{pf.lem.indicator-parallelogram.2}), it remains to show that
$K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }\subseteq K_{\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }\cup K_{\mathcal{T}\looparrowleft
\left(  a\geq b\right)  }$. So let us do this now.

Let $y\in K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }$. Our goal is
to show that $y\in K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }\cup
K_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }$. In fact, assume the
contrary. Then, $y\notin K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)
}$ and $y\notin K_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }$.

We have $y\notin K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }%
=\bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  }K_{O}$ (by
(\ref{pf.lem.indicator-parallelogram.rewrite})). Hence, there exists a
$P\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  $ such that $y\notin
K_{P}$. Similarly, using $y\notin K_{\mathcal{T}\looparrowleft\left(  a\geq
b\right)  }$, we can see that there exists a $Q\in\mathcal{T}\looparrowleft
\left(  a\geq b\right)  $ such that $y\notin K_{Q}$. Consider these $P$ and
$Q$.

We have $P\in\mathcal{T}\looparrowleft\left(  a\leq b\right)  =\left\{
O\in\mathcal{T}\ \mid\ \left(  a\in O\ \Longrightarrow\ b\in O\right)
\right\}  $. Thus, $P\in\mathcal{T}$ and $\left(  a\in P\ \Longrightarrow
\ b\in P\right)  $. But we do not have $\left(  b\in P\ \Longrightarrow\ a\in
P\right)  $\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then,
$\left(  b\in P\ \Longrightarrow\ a\in P\right)  $. Combining this with
$\left(  a\in P\ \Longrightarrow\ b\in P\right)  $, we obtain $\left(  a\in
P\ \Longleftrightarrow\ b\in P\right)  $. Hence, $P\in\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  $ (by the definition of $\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  $). Now, $y\in K_{\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }=\bigcap_{O\in\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }K_{O}$ (by
(\ref{pf.lem.indicator-parallelogram.rewrite})). But $\bigcap_{O\in
\mathcal{T}\looparrowleft\left(  a\sim b\right)  }K_{O}\subseteq K_{P}$ (since
$P\in\mathcal{T}\looparrowleft\left(  a\sim b\right)  $), so that $y\in
\bigcap_{O\in\mathcal{T}\looparrowleft\left(  a\sim b\right)  }K_{O}\subseteq
K_{P}$, which contradicts $y\notin K_{P}$. This contradiction proves that our
assumption was wrong, qed.}. Hence, $a\notin P$ and $b\in P$ (since $\left(
a\in P\ \Longrightarrow\ b\in P\right)  $ but not $\left(  b\in
P\ \Longrightarrow\ a\in P\right)  $).

We have thus shown that $P\in\mathcal{T}$, $a\notin P$ and $b\in P$.
Similarly, we find that $Q\in\mathcal{T}$, $b\notin Q$ and $a\in Q$. Now, it
is easy to see that $P\cap Q\in\mathcal{T}\looparrowleft\left(  a\sim
b\right)  $\ \ \ \ \footnote{\textit{Proof.} From $P\in\mathcal{T}$ and
$Q\in\mathcal{T}$, we infer that $P\cap Q\in\mathcal{T}$. Also, $a\notin P\cap
Q$ (since $a\notin P$), so that $\left(  a\in P\cap Q\ \Longrightarrow\ b\in
P\cap Q\right)  $. Moreover, $b\notin P\cap Q$ (since $b\notin Q$), and thus
$\left(  b\in P\cap Q\ \Longrightarrow\ a\in P\cap Q\right)  $. Combined with
$\left(  a\in P\cap Q\ \Longrightarrow\ b\in P\cap Q\right)  $, this yields
$\left(  a\in P\cap Q\ \Longleftrightarrow\ b\in P\cap Q\right)  $. Thus,
$P\cap Q$ is an element of $\mathcal{T}$ satisfying $\left(  a\in P\cap
Q\ \Longleftrightarrow\ b\in P\cap Q\right)  $. Hence, $P\cap Q\in\left\{
O\in\mathcal{T}\ \mid\ \left(  a\in O\ \Longleftrightarrow\ b\in O\right)
\right\}  =\mathcal{T}\looparrowleft\left(  a\sim b\right)  $, qed.} and
$P\cup Q\in\mathcal{T}\looparrowleft\left(  a\sim b\right)  $%
\ \ \ \ \footnote{\textit{Proof.} From $P\in\mathcal{T}$ and $Q\in\mathcal{T}%
$, we infer that $P\cup Q\in\mathcal{T}$. Also, $b\in P\cup Q$ (since $b\in
P$), so that $\left(  a\in P\cup Q\ \Longrightarrow\ b\in P\cup Q\right)  $.
Moreover, $a\in P\cup Q$ (since $a\in Q$), and thus $\left(  b\in P\cup
Q\ \Longrightarrow\ a\in P\cup Q\right)  $. Combined with $\left(  a\in P\cup
Q\ \Longrightarrow\ b\in P\cup Q\right)  $, this yields $\left(  a\in P\cup
Q\ \Longleftrightarrow\ b\in P\cup Q\right)  $. Thus, $P\cup Q$ is an element
of $\mathcal{T}$ satisfying $\left(  a\in P\cup Q\ \Longleftrightarrow\ b\in
P\cup Q\right)  $. Hence, $P\cup Q\in\left\{  O\in\mathcal{T}\ \mid\ \left(
a\in O\ \Longleftrightarrow\ b\in O\right)  \right\}  =\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  $, qed.}.

Let us write $y\in\mathbb{R}^{n}$ in the form $y=\left(  y_{1},y_{2}%
,\ldots,y_{n}\right)  $. We have $\left(  y_{1},y_{2},\ldots,y_{n}\right)
=y\notin K_{P}=\left\{  \left(  x_{1},x_{2},\ldots,x_{n}\right)  \in
\mathbb{R}^{n}\ \mid\ \sum_{i\in\left[  n\right]  \setminus P}x_{i}%
\geq0\right\}  $. Hence, $\sum_{i\in\left[  n\right]  \setminus P}y_{i}<0$.
Similarly, from $y\notin K_{Q}$, we obtain $\sum_{i\in\left[  n\right]
\setminus Q}y_{i}<0$.

We have
\begin{align*}
\left(  y_{1},y_{2},\ldots,y_{n}\right)   &  =y\in K_{\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }=\bigcap_{O\in\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  }K_{O}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.lem.indicator-parallelogram.rewrite})}\right) \\
&  \subseteq K_{P\cap Q}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }P\cap
Q\in\mathcal{T}\looparrowleft\left(  a\sim b\right)  \right) \\
&  =\left\{  \left(  x_{1},x_{2},\ldots,x_{n}\right)  \in\mathbb{R}^{n}%
\ \mid\ \sum_{i\in\left[  n\right]  \setminus\left(  P\cap Q\right)  }%
x_{i}\geq0\right\}  ,
\end{align*}
so that $\sum_{i\in\left[  n\right]  \setminus\left(  P\cap Q\right)  }%
y_{i}\geq0$. The same argument can be applied to $P\cup Q$ instead of $P\cap
Q$, and leads to $\sum_{i\in\left[  n\right]  \setminus\left(  P\cup Q\right)
}y_{i}\geq0$.

But any two subsets $A$ and $B$ of $\left[  n\right]  $ satisfy $\sum_{i\in
A}y_{i}+\sum_{i\in B}y_{i}=\sum_{i\in A\cup B}y_{i}+\sum_{i\in A\cap B}y_{i}$.
Applying this to $A=\left[  n\right]  \setminus P$ and $B=\left[  n\right]
\setminus Q$, we obtain%
\begin{align*}
\sum_{i\in\left[  n\right]  \setminus P}y_{i}+\sum_{i\in\left[  n\right]
\setminus Q}y_{i}  &  =\sum_{i\in\left(  \left[  n\right]  \setminus P\right)
\cup\left(  \left[  n\right]  \setminus Q\right)  }y_{i}+\sum_{i\in\left(
\left[  n\right]  \setminus P\right)  \cap\left(  \left[  n\right]  \setminus
Q\right)  }y_{i}\\
&  =\sum_{i\in\left[  n\right]  \setminus\left(  P\cap Q\right)  }y_{i}%
+\sum_{i\in\left[  n\right]  \setminus\left(  P\cup Q\right)  }y_{i}%
\end{align*}
(since $\left(  \left[  n\right]  \setminus P\right)  \cup\left(  \left[
n\right]  \setminus Q\right)  =\left[  n\right]  \setminus\left(  P\cap
Q\right)  $ and $\left(  \left[  n\right]  \setminus P\right)  \cap\left(
\left[  n\right]  \setminus Q\right)  =\left[  n\right]  \setminus\left(
P\cup Q\right)  $). Thus,%
\[
\sum_{i\in\left[  n\right]  \setminus\left(  P\cap Q\right)  }y_{i}+\sum
_{i\in\left[  n\right]  \setminus\left(  P\cup Q\right)  }y_{i}%
=\underbrace{\sum_{i\in\left[  n\right]  \setminus P}y_{i}}_{<0}%
+\underbrace{\sum_{i\in\left[  n\right]  \setminus Q}y_{i}}_{<0}<0.
\]
This contradicts%
\[
\underbrace{\sum_{i\in\left[  n\right]  \setminus\left(  P\cap Q\right)
}y_{i}}_{\geq0}+\underbrace{\sum_{i\in\left[  n\right]  \setminus\left(  P\cup
Q\right)  }y_{i}}_{\geq0}\geq0.
\]
This contradiction proves that our assumption was wrong. Hence, $y\in
K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }\cup K_{\mathcal{T}%
\looparrowleft\left(  a\geq b\right)  }$. Since we have proven this for every
$y\in K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }$, we thus conclude
that $K_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }\subseteq
K_{\mathcal{T}\looparrowleft\left(  a\leq b\right)  }\cup K_{\mathcal{T}%
\looparrowleft\left(  a\geq b\right)  }$. This finishes the proof of
(\ref{pf.lem.indicator-parallelogram.2}).

Now that both (\ref{pf.lem.indicator-parallelogram.1}) and
(\ref{pf.lem.indicator-parallelogram.2}) are proven, Lemma
\ref{lem.indicator-parallelogram} easily follows.
\end{proof}

\begin{definition}
\label{def.T-additive}Let $V$ be a $\mathbb{K}$-vector space. A $\mathbb{K}%
$-linear map $f:\mathbf{H}_{\mathbf{T}}\rightarrow V$ is said to be
$\mathbf{T}$\textit{-additive} if and only if every $n\in\mathbb{N}$, every
$\mathcal{T}\in\mathbf{T}_{n}$ and every two distinct elements $a$ and $b$ of
$\left[  n\right]  $ satisfy%
\begin{equation}
f\left(  \mathcal{T}\right)  =f\left(  \mathcal{T}\looparrowleft\left(  a\leq
b\right)  \right)  +f\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)
\right)  -f\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  .
\label{eq.lem.induction-step.cond}%
\end{equation}

\end{definition}

\begin{proposition}
\label{prop.T-additive.=}Let $V$ be a $\mathbb{K}$-vector space. Let $f$ and
$g$ be two $\mathbf{T}$-additive $\mathbb{K}$-linear maps $\mathbf{H}%
_{\mathbf{T}}\rightarrow V$. Assume that $f\left(  \mathcal{T}_{u}\right)
=g\left(  \mathcal{T}_{u}\right)  $ for every packed word $u$. Then, $f=g$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.T-additive.=}.]It is clearly enough to show
that
\begin{equation}
f\left(  \mathcal{T}\right)  =g\left(  \mathcal{T}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\mathcal{T}\in\mathbf{T}.
\label{pf.prop.T-additive.=}%
\end{equation}


For any topology $\mathcal{T}$ on a finite set $X$, we let $h\left(
\mathcal{T}\right)  $ denote the nonnegative integer $\sharp\left\{  \left(
x,y\right)  \in X^{2}\ \mid\ \text{neither }x\leq_{\mathcal{T}}y\text{ nor
}y\leq_{\mathcal{T}}x\right\}  $. We shall prove (\ref{pf.prop.T-additive.=})
by strong induction over $h\left(  \mathcal{T}\right)  $. So we fix some
$\mathcal{T}\in\mathbf{T}$, and we want to prove (\ref{pf.prop.T-additive.=}),
assuming that every $\mathcal{S}\in\mathbf{T}$ satisfying $h\left(
\mathcal{S}\right)  <h\left(  \mathcal{T}\right)  $ satisfies%
\begin{equation}
f\left(  \mathcal{S}\right)  =g\left(  \mathcal{S}\right)  .
\label{pf.prop.T-additive.=.ass}%
\end{equation}


Let $n\in\mathbb{N}$ be such that $\mathcal{T}\in\mathbf{T}_{n}$. If there
exist no two elements $a$ and $b$ of $\left[  n\right]  $ satisfying neither
$a\leq_{\mathcal{T}}b$ nor $b\leq_{\mathcal{T}}a$, then we have $\mathcal{T}%
=\mathcal{T}_{u}$ for some packed word $u$, and this $u$ satisfies $f\left(
\mathcal{T}_{u}\right)  =g\left(  \mathcal{T}_{u}\right)  $ (due to the
assumption of the proposition); thus, (\ref{pf.prop.T-additive.=}) follows
immediately (since $\mathcal{T}=\mathcal{T}_{u}$). Hence, we can WLOG assume
that such two elements $a$ and $b$ exist. Consider these two elements. Of
course, $a$ and $b$ are distinct.

If $\mathcal{S}$ is any of the three posets $\mathcal{T}\looparrowleft\left(
a\leq b\right)  $, $\mathcal{T}\looparrowleft\left(  a\geq b\right)  $ and
$\mathcal{T}\looparrowleft\left(  a\sim b\right)  $, then $h\left(
\mathcal{S}\right)  <h\left(  \mathcal{T}\right)  $\ \ \ \ \footnote{This is
because $\left\{  \left(  x,y\right)  \in X^{2}\ \mid\ \text{neither }%
x\leq_{\mathcal{S}}y\text{ nor }y\leq_{\mathcal{S}}x\right\}  $ is a proper
subset of $\left\{  \left(  x,y\right)  \in X^{2}\ \mid\ \text{neither }%
x\leq_{\mathcal{T}}y\text{ nor }y\leq_{\mathcal{T}}x\right\}  $. (Proper
because $\left(  a,b\right)  $ or $\left(  b,a\right)  $ belongs to the latter
but not to the former.)}. Hence, we can apply (\ref{pf.prop.T-additive.=.ass})
to each of these three posets. We obtain
\begin{align*}
f\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)   &
=g\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  ;\\
f\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)   &
=g\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)  ;\\
f\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)   &
=g\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  .
\end{align*}
But since $f$ is $\mathbf{T}$-additive, we have
\begin{align*}
f\left(  \mathcal{T}\right)   &  =\underbrace{f\left(  \mathcal{T}%
\looparrowleft\left(  a\leq b\right)  \right)  }_{=g\left(  \mathcal{T}%
\looparrowleft\left(  a\leq b\right)  \right)  }+\underbrace{f\left(
\mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)  }_{=g\left(
\mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)  }%
-f\underbrace{\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)
\right)  }_{=g\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)
\right)  }\\
&  =g\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)
+g\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)  -g\left(
\mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  =g\left(
\mathcal{T}\right)
\end{align*}
(since $g$ is $\mathbf{T}$-additive). Thus, (\ref{pf.prop.T-additive.=}) is
proven, and the induction step is complete.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.commdiag} (sketched).]We need to show that
$\beta=\alpha\circ U$.

We notice that every topology $\mathcal{S}$ on $\left[  n\right]  $ satisfies%
\begin{align}
\left(  \beta\circ Z\right)  \left(  \mathcal{S}\right)   &  =\beta\left(
\underbrace{Z\left(  \mathcal{S}\right)  }_{\substack{=\left(  -1\right)
^{\left\vert \left[  n\right]  /\sim_{\mathcal{S}}\right\vert }\mathcal{S}%
\\\text{(by the definition of }Z\text{)}}}\right)  =\left(  -1\right)
^{\left\vert \left[  n\right]  /\sim_{\mathcal{S}}\right\vert }%
\underbrace{\beta\left(  \mathcal{S}\right)  }_{\substack{=\left(  -1\right)
^{\left\vert \left[  n\right]  /\sim_{\mathcal{S}}\right\vert }\underline{1}%
_{K_{\mathcal{S}}}\\\text{(by the definition of }\beta\text{)}}}\nonumber\\
&  =\underbrace{\left(  -1\right)  ^{\left\vert \left[  n\right]
/\sim_{\mathcal{S}}\right\vert }\left(  -1\right)  ^{\left\vert \left[
n\right]  /\sim_{\mathcal{S}}\right\vert }}_{=1}\underline{1}_{K_{\mathcal{S}%
}}\nonumber\\
&  =\underline{1}_{K_{\mathcal{S}}} \label{pf.thm.commdiag.bZ.betaZ}%
\end{align}
and%
\begin{align}
\left(  \alpha\circ U\circ Z\right)  \left(  \mathcal{S}\right)   &
=\alpha\left(  U\left(  \underbrace{Z\left(  \mathcal{S}\right)
}_{\substack{=\left(  -1\right)  ^{\left\vert \left[  n\right]  /\sim
_{\mathcal{S}}\right\vert }\mathcal{S}\\\text{(by the definition of }%
Z\text{)}}}\right)  \right) \nonumber\\
&  =\left(  -1\right)  ^{\left\vert \left[  n\right]  /\sim_{\mathcal{S}%
}\right\vert }\alpha\left(  \underbrace{U\left(  \mathcal{S}\right)
}_{\substack{=\sum_{f\in\mathcal{U}\left(  \mathcal{S}\right)  }f\\\text{(by
the definition of }U\text{)}}}\right) \nonumber\\
&  =\left(  -1\right)  ^{\left\vert \left[  n\right]  /\sim_{\mathcal{S}%
}\right\vert }\sum_{f\in\mathcal{U}\left(  \mathcal{S}\right)  }\alpha\left(
f\right)  . \label{pf.thm.commdiag.bZ.auZ}%
\end{align}


We shall now show that both maps $\beta\circ Z:\mathbf{H}_{\mathbf{T}%
}\rightarrow\operatorname*{WQSym}$ and $\alpha\circ U\circ Z:\mathbf{H}%
_{\mathbf{T}}\rightarrow\operatorname*{WQSym}$ are $\mathbf{T}$-additive.

\textit{Proof that the map }$\beta\circ Z$ \textit{is }$\mathbf{T}%
$\textit{-additive:} Let $n\in\mathbb{N}$. Let $\mathcal{T}\in\mathbf{T}_{n}$.
Let $a$ and $b$ be two distinct elements of $\left[  n\right]  $. In order to
show that $\beta\circ Z$ is $\mathbf{T}$-additive, we must prove that%
\begin{align}
&  \left(  \beta\circ Z\right)  \left(  \mathcal{T}\right) \nonumber\\
&  =\left(  \beta\circ Z\right)  \left(  \mathcal{T}\looparrowleft\left(
a\leq b\right)  \right)  +\left(  \beta\circ Z\right)  \left(  \mathcal{T}%
\looparrowleft\left(  a\geq b\right)  \right)  -\left(  \beta\circ Z\right)
\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  .
\label{pf.thm.commdiag.bZ.1}%
\end{align}


This rewrites as follows:%
\[
\underline{1}_{K_{\mathcal{T}}}=\underline{1}_{K_{\mathcal{T}\looparrowleft
\left(  a\leq b\right)  }}+\underline{1}_{K_{\mathcal{T}\looparrowleft\left(
a\geq b\right)  }}-\underline{1}_{K_{\mathcal{T}\looparrowleft\left(  a\sim
b\right)  }}%
\]
(because of (\ref{pf.thm.commdiag.bZ.betaZ})). But this is precisely the claim
of Lemma \ref{lem.indicator-parallelogram}. Hence, (\ref{pf.thm.commdiag.bZ.1}%
) is proven. We thus have shown that the map $\beta\circ Z$ is $\mathbf{T}$-additive.

\textit{Proof that the map }$\alpha\circ U\circ Z$ \textit{is }$\mathbf{T}%
$\textit{-additive:} Let $n\in\mathbb{N}$. Let $\mathcal{T}\in\mathbf{T}_{n}$.
Let $a$ and $b$ be two distinct elements of $\left[  n\right]  $. In order to
show that $\alpha\circ U\circ Z$ is $\mathbf{T}$-additive, we must prove that%
\begin{align}
&  \left(  \alpha\circ U\circ Z\right)  \left(  \mathcal{T}\right) \nonumber\\
&  =\left(  \alpha\circ U\circ Z\right)  \left(  \mathcal{T}\looparrowleft
\left(  a\leq b\right)  \right)  +\left(  \alpha\circ U\circ Z\right)  \left(
\mathcal{T}\looparrowleft\left(  a\geq b\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \alpha\circ U\circ Z\right)  \left(
\mathcal{T}\looparrowleft\left(  a\sim b\right)  \right)  .
\label{pf.thm.commdiag.auZ.1}%
\end{align}


This is rather obvious if $a\leq_{\mathcal{T}}b$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $a\leq_{\mathcal{T}}b$. Then,
Lemma \ref{lem.insert-into-top} \textbf{(c)} yields $\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  =\mathcal{T}$ and $\mathcal{T}%
\looparrowleft\left(  a\sim b\right)  =\mathcal{T}\looparrowleft\left(  a\geq
b\right)  $. Hence, (\ref{pf.thm.commdiag.auZ.1}) rewrites as%
\begin{align}
&  \left(  \alpha\circ U\circ Z\right)  \left(  \mathcal{T}\right) \nonumber\\
&  =\left(  \alpha\circ U\circ Z\right)  \left(  \mathcal{T}\right)  +\left(
\alpha\circ U\circ Z\right)  \left(  \mathcal{T}\looparrowleft\left(  a\geq
b\right)  \right)  -\left(  \alpha\circ U\circ Z\right)  \left(
\mathcal{T}\looparrowleft\left(  a\geq b\right)  \right)  .\nonumber
\end{align}
But this is obvious.}. Hence, for the rest of this proof, we WLOG assume that
we don't have $a\leq_{\mathcal{T}}b$. Similarly, we WLOG assume that we don't
have $b\leq_{\mathcal{T}}a$. Now, using (\ref{pf.thm.commdiag.bZ.auZ}), we can
rewrite the equality (\ref{pf.thm.commdiag.auZ.1}) as follows:%
\begin{align*}
&  \left(  -1\right)  ^{\left\vert \left[  n\right]  /\sim_{\mathcal{T}%
}\right\vert }\sum_{f\in\mathcal{U}\left(  \mathcal{T}\right)  }\alpha\left(
f\right) \\
&  =\left(  -1\right)  ^{\left\vert \left[  n\right]  /\sim_{\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }\right\vert }\sum_{f\in\mathcal{U}%
\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)  \right)  }%
\alpha\left(  f\right)  +\left(  -1\right)  ^{\left\vert \left[  n\right]
/\sim_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }\right\vert }%
\sum_{f\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)
\right)  }\alpha\left(  f\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -1\right)  ^{\left\vert \left[  n\right]
/\sim_{\mathcal{T}\looparrowleft\left(  a\sim b\right)  }\right\vert }%
\sum_{f\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)
\right)  }\alpha\left(  f\right)  .
\end{align*}
This can be rewritten further as%
\begin{align*}
&  \left(  -1\right)  ^{\left\vert \left[  n\right]  /\sim_{\mathcal{T}%
}\right\vert }\sum_{f\in\mathcal{U}\left(  \mathcal{T}\right)  }\alpha\left(
f\right) \\
&  =\left(  -1\right)  ^{\left\vert \left[  n\right]  /\sim_{\mathcal{T}%
}\right\vert }\sum_{f\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(
a\leq b\right)  \right)  }\alpha\left(  f\right)  +\left(  -1\right)
^{\left\vert \left[  n\right]  /\sim_{\mathcal{T}}\right\vert }\sum
_{f\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\geq b\right)
\right)  }\alpha\left(  f\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -1\right)  ^{\left\vert \left[  n\right]
/\sim_{\mathcal{T}}\right\vert -1}\sum_{f\in\mathcal{U}\left(  \mathcal{T}%
\looparrowleft\left(  a\sim b\right)  \right)  }\alpha\left(  f\right)
\end{align*}
(because Lemma \ref{lem.insert-into-top} \textbf{(n)} (applied to $X=\left[
n\right]  $) yields \newline$\left\vert \left[  n\right]  /\sim_{\mathcal{T}%
\looparrowleft\left(  a\leq b\right)  }\right\vert =\left\vert \left[
n\right]  /\sim_{\mathcal{T}\looparrowleft\left(  a\geq b\right)  }\right\vert
=\left\vert \left[  n\right]  /\sim_{\mathcal{T}}\right\vert $ and
\newline$\left\vert \left[  n\right]  /\sim_{\mathcal{T}\looparrowleft\left(
a\sim b\right)  }\right\vert =\left\vert \left[  n\right]  /\sim_{\mathcal{T}%
}\right\vert -1$). Upon cancelling $\left(  -1\right)  ^{\left\vert \left[
n\right]  /\sim_{\mathcal{T}}\right\vert }$, this simplifies to%
\[
\sum_{f\in\mathcal{U}\left(  \mathcal{T}\right)  }\alpha\left(  f\right)
=\sum_{f\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\leq b\right)
\right)  }\alpha\left(  f\right)  +\sum_{f\in\mathcal{U}\left(  \mathcal{T}%
\looparrowleft\left(  a\geq b\right)  \right)  }\alpha\left(  f\right)
+\sum_{f\in\mathcal{U}\left(  \mathcal{T}\looparrowleft\left(  a\sim b\right)
\right)  }\alpha\left(  f\right)  .
\]
But this follows immediately from Lemma \ref{lem.insert-into-top} \textbf{(m)}
(applied to $X=\left[  n\right]  $). Thus, (\ref{pf.thm.commdiag.auZ.1}) is
proven. We have thus shown that $\alpha\circ U\circ Z$ is $\mathbf{T}$-additive.

Now, it is easy to see that $\left(  \beta\circ Z\right)  \left(
\mathcal{T}_{u}\right)  =\left(  \alpha\circ U\circ Z\right)  \left(
\mathcal{T}_{u}\right)  $ for every packed word $u$%
\ \ \ \ \footnote{\textit{Proof.} Let $u$ be a packed word. Applying
(\ref{pf.thm.commdiag.bZ.betaZ}) to $\mathcal{S}=\mathcal{T}_{u}$, we obtain
$\left(  \beta\circ Z\right)  \left(  \mathcal{T}_{u}\right)  =\underline{1}%
_{K_{\mathcal{T}_{u}}}=\underline{1}_{K_{u}}$ (since Remark \ref{rmk.KTf}
yields $K_{\mathcal{T}_{u}}=K_{u}$). But applying
(\ref{pf.thm.commdiag.bZ.auZ}) to $\mathcal{S}=\mathcal{T}_{u}$ leads to%
\begin{align*}
\left(  \alpha\circ U\circ Z\right)  \left(  \mathcal{T}_{u}\right)   &
=\underbrace{\left(  -1\right)  ^{\left\vert \left[  n\right]  /\sim
_{\mathcal{T}_{u}}\right\vert }}_{\substack{=\left(  -1\right)  ^{\max
u}\\\text{(since }\left\vert \left[  n\right]  /\sim_{\mathcal{T}_{u}%
}\right\vert =\max u\text{)}}}\underbrace{\sum_{f\in\mathcal{U}\left(
\mathcal{T}_{u}\right)  }\alpha\left(  f\right)  }_{\substack{=\alpha\left(
u\right)  \\\text{(since }\mathcal{U}\left(  \mathcal{T}_{u}\right)  =\left\{
u\right\}  \text{)}}}\\
&  =\left(  -1\right)  ^{\max u}\underbrace{\alpha\left(  u\right)
}_{\substack{=\left(  -1\right)  ^{\max u}\underline{1}_{K_{u}}\\\text{(by the
definition of }\alpha\text{)}}}=\underbrace{\left(  -1\right)  ^{\max
u}\left(  -1\right)  ^{\max u}}_{=1}\underline{1}_{K_{u}}=\underline{1}%
_{K_{u}}\\
&  =\left(  \beta\circ Z\right)  \left(  \mathcal{T}_{u}\right)  ,
\end{align*}
qed.}. Hence, Proposition \ref{prop.T-additive.=} (applied to $V=\mathfrak{M}%
$, $f=\beta\circ Z$ and $g=\alpha\circ U\circ Z$) yields $\beta\circ
Z=\alpha\circ U\circ Z$. Since $Z$ is an isomorphism, we can cancel $Z$ from
this equality, and obtain $\beta=\alpha\circ U$. This proves Theorem
\ref{thm.commdiag}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.MNT5.2}.]Theorem \ref{thm.commdiag} yields
$\beta=\alpha\circ U$. Since both $\beta$ and $U$ are $\mathbb{K}$-algebra
homomorphisms, and since $U$ is surjective, this easily yields that $\alpha$
is a $\mathbb{K}$-algebra homomorphism. (Indeed, let $p\in
\operatorname*{WQSym}$ and $q\in\operatorname*{WQSym}$. Then, thanks to the
surjectivity of $U$, there exist $\mathcal{P}\in\mathbf{H}_{\mathbf{T}}$ and
$\mathcal{Q}\in\mathbf{H}_{\mathbf{T}}$ satisfying $p=U\left(  \mathcal{P}%
\right)  $ and $q=U\left(  \mathcal{Q}\right)  $. Consider these $\mathcal{P}$
and $\mathcal{Q}$. Since $U$ is a $\mathbb{K}$-algebra homomorphism, we have
$U\left(  \mathcal{P}.\mathcal{Q}\right)  =\underbrace{U\left(  \mathcal{P}%
\right)  }_{=p}\underbrace{U\left(  \mathcal{Q}\right)  }_{=q}=pq$. Now,%
\begin{align*}
&  \alpha\left(  \underbrace{p}_{=U\left(  \mathcal{P}\right)  }\right)
\cdot\alpha\left(  \underbrace{q}_{=U\left(  \mathcal{Q}\right)  }\right) \\
&  =\underbrace{\alpha\left(  U\left(  \mathcal{P}\right)  \right)
}_{=\left(  \alpha\circ U\right)  \left(  \mathcal{P}\right)  }\cdot
\underbrace{\alpha\left(  U\left(  \mathcal{Q}\right)  \right)  }_{=\left(
\alpha\circ U\right)  \left(  \mathcal{Q}\right)  }=\underbrace{\left(
\alpha\circ U\right)  }_{=\beta}\left(  \mathcal{P}\right)  \cdot
\underbrace{\left(  \alpha\circ U\right)  }_{=\beta}\left(  \mathcal{Q}\right)
\\
&  =\beta\left(  \mathcal{P}\right)  \cdot\beta\left(  \mathcal{Q}\right)
=\underbrace{\beta}_{=\alpha\circ U}\left(  \mathcal{P}.\mathcal{Q}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\beta\text{ is a }\mathbb{K}%
\text{-algebra homomorphism}\right) \\
&  =\left(  \alpha\circ U\right)  \left(  \mathcal{P}.\mathcal{Q}\right)
=\alpha\left(  \underbrace{U\left(  \mathcal{P}.\mathcal{Q}\right)  }%
_{=pq}\right)  =\alpha\left(  pq\right)  ,
\end{align*}
and this shows that $\alpha$ is a $\mathbb{K}$-algebra homomorphism.) Theorem
\ref{thm.MNT5.2} is proven.
\end{proof}

\section{Application: an alternating sum identity}

As an application of Theorem \ref{thm.MNT5.2} we can prove the following fact,
which is analogous to \cite[Corollary 4.8]{EMR}:

\begin{corollary}
\label{cor.altsum1}Let $n\in\mathbb{N}$. Let $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{n}\right)  \in\mathbb{R}^{n}$. Then,%
\[
\sum_{\substack{u\text{ is a packed word}\\\text{of length }n;\\\lambda\in
K_{u}}}\left(  -1\right)  ^{\max u}=%
\begin{cases}
\left(  -1\right)  ^{n}, & \text{if }\lambda_{1},\lambda_{2},\ldots
,\lambda_{n}\geq0;\\
0, & \text{otherwise.}%
\end{cases}
\]

\end{corollary}

This will rely on the following equality in $\operatorname*{WQSym}$:

\begin{proposition}
\label{prop.sum-pwords}Let $\zeta$ be the packed word $\left(  1\right)  $ of
length $1$. Then, in $\operatorname*{WQSym}$, we have%
\[
\zeta^{n}=\sum_{\substack{u\text{ is a packed word}\\\text{of length }n}}u.
\]

\end{proposition}

\begin{proof}
[Proof sketch.]Induction on $n$ (details are left to the reader).
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.altsum1}.]Let $\mathbb{R}_{+}$ denote the set of
all nonnegative reals. Let $\zeta\in\operatorname*{WQSym}$ be the packed word
$\left(  1\right)  $ of length $1$.

Consider the map $\alpha$ from Theorem \ref{thm.MNT5.2}. The definition of
this map $\alpha$ yields%
\[
\alpha\left(  \zeta\right)  =\underbrace{\left(  -1\right)  ^{\max\zeta}%
}_{\substack{=-1\\\text{(since }\max\zeta=1\text{)}}}\underline{1}_{K_{\zeta}%
}=-\underline{1}_{K_{\zeta}}=-\underline{1}_{\mathbb{R}_{+}}%
\]
(since the definition of $K_{\zeta}$ yields $K_{\zeta}=\mathbb{R}_{+}$).
Hence,%
\[
\left(  \alpha\left(  \zeta\right)  \right)  ^{n}=\left(  -\underline{1}%
_{\mathbb{R}_{+}}\right)  ^{n}=\left(  -1\right)  ^{n}\underbrace{\left(
\underline{1}_{\mathbb{R}_{+}}\right)  ^{n}}_{\substack{=\underline{1}%
_{\substack{\mathbb{R}_{+}^{n}}}\\\text{(this follows easily from
the}\\\text{definition of multiplication on }\mathfrak{M}\text{)}}}=\left(
-1\right)  ^{n}\underline{1}_{\substack{\mathbb{R}_{+}^{n}}}.
\]


But Proposition \ref{prop.sum-pwords} yields%
\[
\zeta^{n}=\sum_{\substack{u\text{ is a packed word}\\\text{of length }n}}u.
\]
Applying the map $\alpha$ to both sides of this equality, we obtain%
\begin{align*}
\alpha\left(  \zeta^{n}\right)   &  =\alpha\left(  \sum_{\substack{u\text{ is
a packed word}\\\text{of length }n}}u\right)  =\sum_{\substack{u\text{ is a
packed word}\\\text{of length }n}}\underbrace{\alpha\left(  u\right)
}_{\substack{=\left(  -1\right)  ^{\max u}\underline{1}_{K_{u}}\\\text{(by the
definition of }\alpha\text{)}}}\\
&  =\sum_{\substack{u\text{ is a packed word}\\\text{of length }n}}\left(
-1\right)  ^{\max u}\underline{1}_{K_{u}}.
\end{align*}
Applying both sides of this equality to $\lambda$, we obtain%
\begin{align*}
\left(  \alpha\left(  \zeta^{n}\right)  \right)  \left(  \lambda\right)   &
=\sum_{\substack{u\text{ is a packed word}\\\text{of length }n}}\left(
-1\right)  ^{\max u}\underbrace{\underline{1}_{K_{u}}\left(  \lambda\right)
}_{\substack{=%
\begin{cases}
1, & \text{if }\lambda\in K_{u};\\
0, & \text{if }\lambda\notin K_{u}%
\end{cases}
\\\text{(by the definition of }\underline{1}_{K_{u}}\text{)}}}\\
&  =\sum_{\substack{u\text{ is a packed word}\\\text{of length }n}}\left(
-1\right)  ^{\max u}%
\begin{cases}
1, & \text{if }\lambda\in K_{u};\\
0, & \text{if }\lambda\notin K_{u}%
\end{cases}
\\
&  =\sum_{\substack{u\text{ is a packed word}\\\text{of length }n;\\\lambda\in
K_{u}}}\left(  -1\right)  ^{\max u}.
\end{align*}
Hence,%
\begin{align*}
\sum_{\substack{u\text{ is a packed word}\\\text{of length }n;\\\lambda\in
K_{u}}}\left(  -1\right)  ^{\max u}  &  =\underbrace{\left(  \alpha\left(
\zeta^{n}\right)  \right)  }_{\substack{=\left(  \alpha\left(  \zeta\right)
\right)  ^{n}\\\text{(since }\alpha\text{ is a }\mathbb{K}\text{-algebra}%
\\\text{homomorphism)}}}\left(  \lambda\right)  =\underbrace{\left(
\alpha\left(  \zeta\right)  \right)  ^{n}}_{=\left(  -1\right)  ^{n}%
\underline{1}_{\substack{\mathbb{R}_{+}^{n}}}}\left(  \lambda\right) \\
&  =\left(  -1\right)  ^{n}\underbrace{\underline{1}_{\substack{\mathbb{R}%
_{+}^{n}}}\left(  \lambda\right)  }_{=%
\begin{cases}
1, & \text{if }\lambda\in\mathbb{R}_{+}^{n};\\
0, & \text{otherwise}%
\end{cases}
}=\left(  -1\right)  ^{n}%
\begin{cases}
1, & \text{if }\lambda\in\mathbb{R}_{+}^{n};\\
0, & \text{otherwise}%
\end{cases}
\\
&  =%
\begin{cases}
\left(  -1\right)  ^{n}, & \text{if }\lambda\in\mathbb{R}_{+}^{n};\\
0, & \text{otherwise}%
\end{cases}
\\
&  =%
\begin{cases}
\left(  -1\right)  ^{n}, & \text{if }\lambda_{1},\lambda_{2},\ldots
,\lambda_{n}\geq0;\\
0, & \text{otherwise}%
\end{cases}
\end{align*}
(since the condition \textquotedblleft$\lambda\in\mathbb{R}_{+}^{n}%
$\textquotedblright\ is equivalent to \textquotedblleft$\lambda_{1}%
,\lambda_{2},\ldots,\lambda_{n}\geq0$\textquotedblright). This proves
Corollary \ref{cor.altsum1}.
\end{proof}

From Corollary \ref{cor.altsum1}, we can in turn derive the precise statement
of \cite[Corollary 4.8]{EMR}:

\begin{corollary}
\label{cor.altsum2}Let $n\in\mathbb{N}$. Let $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{n}\right)  \in\mathbb{R}^{n}$. Then,%
\[
\sum_{\substack{u\text{ is a packed word}\\\text{of length }n;\\\lambda\in
K_{u}^{\circ}}}\left(  -1\right)  ^{\max u}=%
\begin{cases}
\left(  -1\right)  ^{n}, & \text{if }\lambda_{1},\lambda_{2},\ldots
,\lambda_{n}>0;\\
0, & \text{otherwise.}%
\end{cases}
\]
Here, for any packed word $u$ of length $n$, we define the subset
$K_{u}^{\circ}$ of $\mathbb{R}^{n}$ in the same way as we defined $K_{u}$, but
with the \textquotedblleft$\geq$\textquotedblright\ sign replaced by
\textquotedblleft$>$\textquotedblright.
\end{corollary}

\begin{proof}
[Proof sketch.]Pick a small $\varepsilon>0$, and let $\lambda^{\prime
}:=\left(  \lambda_{1}-\varepsilon,\lambda_{2}-\varepsilon,\ldots,\lambda
_{n}-\varepsilon\right)  $. If $\varepsilon$ has been chosen small enough
(say,
\[
0<\varepsilon<\dfrac{1}{n}\min\left\{  \sum_{i\in I}\lambda_{i}\ \mid
\ I\subseteq\left[  n\right]  \text{ satisfying }\sum_{i\in I}\lambda
_{i}>0\right\}
\]
), then any packed word $u$ of length $n$ will satisfy $\lambda\in
K_{u}^{\circ}$ if and only if it satisfies $\lambda^{\prime}\in K_{u}$, and we
will have $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}>0$ if and only if
$\lambda_{1}-\varepsilon,\lambda_{2}-\varepsilon,\ldots,\lambda_{n}%
-\varepsilon\geq0$. Hence, Corollary \ref{cor.altsum2} follows from Corollary
\ref{cor.altsum1} (applied to $\lambda^{\prime}$ and $\lambda_{i}-\varepsilon$
instead of $\lambda$ and $\lambda_{i}$).
\end{proof}

\begin{thebibliography}{9}                                                                                                %


\bibitem {FM}Loic Foissy and Claudia Malvenuto, \textit{The Hopf algebra of
finite topologies and }$T$\textit{-partitions}, arXiv preprint
arXiv:1407.0476v2.\newline\url{http://arxiv.org/abs/1407.0476v2}\newline See
\url{https://darijgrinberg.gitlab.io/algebra/topologies-errata.pdf} for
unofficial errata.

\bibitem {MNT}Fr\'{e}d\'{e}ric Menous, Jean-Christophe Novelli, Jean-Yves
Thibon, \textit{Mould calculus, polyhedral cones, and characters of
combinatorial Hopf algebras}, arXiv preprint arXiv:1109.1634v2.\newline\url{http://arxiv.org/abs/1109.1634v2}

\bibitem {EMR}Richard Ehrenborg, Sophie Morel, Margaret Readdy, \textit{Some
combinatorial identities appearing in the calculation of the cohomology of
Siegel modular varieties}, Algebraic Combinatorics \textbf{2}, issue 5 (2019),
pp. 863--878. \newline\url{http://alco.centre-mersenne.org/item/ALCO_2019__2_5_863_0}
\end{thebibliography}


\end{document}