text:
\section{Introduction}

In this article we develop a novel tool to study the fixed point property of finite $T_0$--spaces (or equivalently finite posets) which consists in analysing a new finite $T_0$--space which is constructed from the one whose fixed point property we want to study. More specifically, given a finite $T_0$--space $X$ we construct a new finite $T_0$--space $\mathscr{C}(X)$ whose elements are connected subspaces of $X$ and such that each continuous map $f\colon X\to X$ induces a continuous map $\mathscr{C}(f)\colon \mathscr{C}(X)\to \mathscr{C}(X)$. This assignment is not functorial but defines an endofunctor of the homotopy category of finite $T_0$--spaces.

We show with several examples that this tool is useful to study the fixed point property of finite $T_0$--spaces and we develop many properties of this construction so as to gain a better understanding of it. Among them, we prove that our construction preserves homotopy type and we study how it changes if we remove beat points of the given space.

\section{Preliminaries}

\subsection*{Finite topological spaces}

Let $X$ be a finite topological space. For each $x\in X$ the \emph{minimal open set} that contains $x$ is defined as the intersection of all the open subsets of $X$ that contain $x$ and is denoted by $U_x$ (or by $U_x^X$ if we need to make explicit the topological space in which it is considered). We denote by $F_x$ the intersection of all closed sets that contain $x$. We also denote $\widehat U_x = U_x-\{x\}$ and $\widehat F_x = F_x-\{x\}$. The set $\{U_x\mid x\in X\}$ is a basis for the topology of $X$. We can define a preorder in $X$ by $x_1\leq x_2$ if and only if $U_{x_1}\subseteq U_{x_2}$. Moreover, this defines a bijective correspondence between topologies and preorders in a given finite set, under which $T_0$--topologies correspond to partial orders \cite{alexandroff1937diskrete}. This correspondence allows us to consider a finite $T_0$--space as a finite poset, and a finite poset as a finite $T_0$--space. In addition, a map between finite $T_0$--spaces is continuous if and only if it is order-preserving.

Let $X$ be a finite topological space with associated preorder $\leq$. We define $X^{\textup{op}}$ as the finite space with underlying set $X$ and associated preorder given by $x\leq^{\textup{op}}y$ if and only if $y\leq x$ for all $x,y\in X$.

Let $X$ and $Y$ be finite $T_0$--spaces and let $f,g\colon X\to Y$ be continuous maps. Then $f$ and $g$ are homotopic (denoted by $f \simeq g$) if and only if there exist $n\in \mathbb{N}$ and continuous maps $f_0, f_1, \ldots, f_n\colon X \to Y$ such that $f = f_0\leq f_1\geq \ldots \leq f_n = g$ \cite[Corollary 1.2.6]{barmak2011algebraic}.

As a consequence, if $f$ and $g$ are such that for each $x\in X$, $f(x)$ and $g(x)$ are comparable, then
$f\simeq g$ (since both $f$ and $g$ are comparable with the continuous map $h=\max\{f,g\}$).

Let $X$ be a finite $T_0$--space and let $x\in X$. We say that $x$ is a \emph{down beat point} of $X$ if $\widehat U_x$ has a maximum element. We say that $x$ is an \emph{up beat point} of $X$ is $\widehat F_x$ has a minimum element. If $x$ is a down beat point or an up beat point, we say that $x$ is a \emph{beat point} of $X$. It is well known that if $X$ is a finite $T_0$--space and $x_0\in X$ is a beat point, then $X-\{x_0\}$ is a strong deformation retract of $X$ \cite[Proposition 1.3.4]{barmak2011algebraic}.

\subsection*{bp--retracts}

We recall the following definition from \cite{cianci2019combinatorial}.

\begin{definition}[{\cite[Definition 3.1]{cianci2019combinatorial}}]
Let $X$ be a finite $T_0$--space and let $A\subseteq X$. We say that $A$ is a \emph{dbp--retract} (resp.\ \emph{ubp--retract}) of $X$ if $A$ can be obtained from $X$ by successively removing down beat points (resp.\ up beat points), that is, if there exist $n\in\mathbb{N}_0$ and a sequence $X=X_0\supseteq X_1\supseteq\cdots\supseteq X_n=A$ of subspaces of $X$ such that, for all $i\in\{1,\ldots,n\}$, the space $X_i$ is obtained from $X_{i-1}$ by removing a single down beat point (resp.\ up beat point) of $X_{i-1}$.
\end{definition}

Clearly, if $X$ is a finite $T_0$--space and $A$ is a dbp--retract of $X$ then the minimal elements of $X$ belong to $A$ since they can not be down beat points of any subspace of $X$. Similarly, if $X$ is a finite $T_0$--space and $A$ is an ubp--retract of $X$ then the maximal elements of $X$ belong to $A$.

Observe also that if $X$ is a finite $T_0$--space and $A\subseteq X$ then $A$ is an ubp--retract of $X$ if and only if $A^\textup{op}$ is a dbp--retract of $X^\textup{op}$.

The following results of \cite{cianci2019combinatorial} give characterizations of dbp--retracts which will be needed in this work. Clearly, similar results hold for ubp--retracts.

\begin{theo}[{\cite[Theorem 3.5]{cianci2019combinatorial}}]
\label{theo_dbpr_equivalences}
Let $X$ be a finite $T_0$--space, let $A$ be a subspace of $X$ and let $i\colon A\to X$ be the inclusion map. Then, the following propositions are equivalent:
\begin{enumerate}[(1)]
\item $A$ is a dbp--retract of $X$.
\item There exists a continuous function $f\colon X\to X$ such that $f\leq\textup{id}_X$, $f^{2}=f$ and $f(X)=A$.
\item There exists a unique continuous function $f\colon X\to X$ such that $f\leq\textup{id}_X$, $f^{2}=f$ and $f(X)=A$.
\item There exists a retraction $r\colon X\to A$ of $i$ such that $ir\leq\textup{id}_X$.
\item There exists a unique retraction $r\colon X\to A$ of $i$ such that $ir\leq\textup{id}_X$.
\end{enumerate}
\end{theo}

\begin{theo}[{\cite[Theorem 3.7]{cianci2019combinatorial}}]
\label{theo_Ux_cap_A}
Let $X$ be a finite $T_0$--space and let $A$ be a subspace of $X$. Then, $A$ is a dbp--retract of $X$ if and only if $U^{X}_x\cap A$ has a maximum for every $x\in X$. Equivalently, $A$ is a dbp--retract of $X$ if and only if $U^{X}_x\cap A$ has a maximum for every $x\in X-A$.
\end{theo}

From the proof of theorem 3.7 of \cite{cianci2019combinatorial} it follows that if $A$ is a dbp-retract of a finite  $T_0$--space $X$ and $r\colon X\to A$ is the corresponding retraction, then $r(x)=\max(U^X_x\cap A)$ for all $x\in X$.

\subsection*{Fixed point property}

Let $X$ be a topological space.
We say that a continuous map $f\colon X\to X$ has a \emph{fixed point} if there exists $x\in X$ such that $f(x)=x$.
We say that $X$ has the \emph{fixed point property{}} if all continuous maps $f\colon X\to X$ have a fixed point. It is easy to see that if a topological space fails to be $T_0$ or connected, then it does not have the fixed point propery. Thus, we will only consider connected finite $T_0$--spaces in this article.

It is known that if $X$ is a finite $T_0$--space, $f\colon X \to X$ is a continuous map and there exists $x\in X$ such that $f(x)$ and  $x$ are comparable, then $f$ has a fixed point. From this fact we deduce the following result.

\begin{prop}
\label{Uhat}
 Let $X$ be a finite $T_0$--space  and let $f\colon X\to X$ be a continuous map. If there exists $x \in X$ such that $\widehat{U}_x$ (respectively $\widehat{F}_x$) has the fixed point property, and $f(\widehat{U}_x) \subseteq U_x$ (respectively $f(\widehat{F}_x) \subseteq F_x$), then $f$ has a fixed point.
\end{prop}

\begin{proof}
Suppose that $f(\widehat{U}_x) \subseteq U_x$. If there exists $y \in \widehat{U}_x$ such that $f(y) = x$, then $f$ has a fixed point. Otherwise, $f(\widehat{U}_x) \subseteq \widehat{U}_x$ and hence $f$ has a fixed point.
\end{proof}

We recall the following result.

\begin{prop}{\cite[Corollary 10.1.4]{barmak2011algebraic}}
\label{prop_XheY_ppf}
 Let $X$ and $Y$ be homotopy equivalent finite $T_0$--spaces. If $X$ has the fixed point property, then $Y$ also has the fixed point property.
\end{prop}

It follows from the previous proposition that if $X$ is a finite $T_0$--space with the fixed point property and $x$ is a beat point of $X$, then $X-\{x\}$ also has the fixed point property. As a further consequence, if $X$ is a contractible finite $T_0$--space, then it has the fixed point property. Therefore, if a finite $T_0$--space has a maximum or minimum element, then it is contractible and hence it has the fixed point property.

Let $X$ be a finite $T_0$--space and let $n\in\mathbb{N}$ such that $n\geq 2$. We say that $X$ is a \emph{$2n$--crown} if $X$ is homeomorphic to the finite $T_0$--space given by the following Hasse diagram
\begin{center}
\begin{tikzpicture}[y=1.5cm]
\tikzstyle{every node}=[font=\scriptsize]
\foreach \x in {1,2} \draw (\x,0) node(c\x)[inner sep=2pt]{$\bullet$} node[below=1]{$\x$};
\foreach \x in {3} \draw (\x,0) node(c\x)[inner sep=2pt]{$\cdots$};
\foreach \x in {4} \draw (\x,0) node(c\x)[inner sep=2pt]{$\bullet$} node[below=1]{${n-1}$};
\foreach \x in {5} \draw (\x,0) node(c\x)[inner sep=2pt]{$\bullet$} node[below=1]{$n$};

\foreach \x in {1,2} \draw (\x,1) node(b\x)[inner sep=2pt]{$\bullet$} node[above=1]{$n+\x$};
\foreach \x in {3} \draw (\x,1) node(b\x)[inner sep=2pt]{$\cdots$};
\foreach \x in {4} \draw (\x,1) node(b\x)[inner sep=2pt]{$\bullet$} node[above=1]{$2n-1$};
\foreach \x in {5} \draw (\x,1) node(b\x)[inner sep=2pt]{$\bullet$} node[above=1]{$2n$};

\foreach \x in {1,2} \draw (b1)--(c\x);
\foreach \x in {1} \draw (b2)--(c\x);
\draw[shorten >=1cm,shorten <=0cm] (b2)--(c3);
\draw[shorten >=1cm,shorten <=0cm] (c2)--(b3);
\draw[shorten >=1cm,shorten <=0cm] (b4)--(c3);
\draw[shorten >=1cm,shorten <=0cm] (c4)--(b3);
\foreach \x in {5} \draw (b4)--(c\x);
\foreach \x in {4,5} \draw (b5)--(c\x);
\end{tikzpicture}
\end{center}
For each element $a$ in a $2n$--crown $X$, we have that $X-\{a\}$ is contractible. From this fact we obtain the following proposition.

\begin{prop} \label{prop_crown}
Let $n\in\mathbb{N}$ such that $n\geq 2$ and let $X$ be a $2n$--crown. Let $f\colon X\to X$ be a continuous map. If $f$ is not bijective then $f$ has a fixed point.
\end{prop}

\section{The $\mathscr{C}$ construction }
\label{section-C_constr}

Let $S$ be a set. The power set of $X$ will be denoted by $\mathcal{P}(S)$ and the set $\mathcal{P}(S)-\{\varnothing \}$ will be denoted by $\mathcal{P}_{\neq\varnothing}(S)$.

\begin{definition}
Let $X$ be a finite $T_0$--space. Let $A$ be a non-empty subset of $X$. We define
\begin{displaymath}
U^X_A = \bigcap_{a\in A} U^X_a  \qquad \textnormal{and} \qquad F^X_A = \bigcap_{a\in A} F^X_a
\end{displaymath}
\end{definition}

When there is no risk of confusion, the superindices $X$ of the previous definition will be omitted and thus $U^X_A$ and $F^X_A$ will be denoted by $U_A$ and $F_A$ respectively.

Note that if $X$ is a finite $T_0$--space and $A$ is a non-empty subset of $X$, then $U_A$ is an open subset of $X$ and $F_A$ is a closed subset of $X$.

From now on, if $X$ is a topological space, $\mathcal{C}(X)$ will denote the set of connected components of $X$.

\begin{definition}
 Let $X$ be a finite $T_0$--space. We define
 \begin{align*}
  \mathscr{U}_0(X) &= \{U_A\subseteq X \mid A\in \mathcal{P}_{\neq\varnothing}(\mxl[X]) \wedge U_A\neq \varnothing \}\textup{ and}\\
  \mathscr{F}_0(X) &= \{F_A\subseteq X \mid A\in \mathcal{P}_{\neq\varnothing}(\mnl[X]) \wedge F_A\neq \varnothing \}.
 \end{align*}

We also define
\begin{align*}
 \mathscr{U}(X) &= \{C\subseteq X \mid C\in\mathcal{C}(U) \textup{ for some } U\in \mathscr{U}_0(X)\} \textup{ and}\\
 \mathscr{F}(X) &= \{C\subseteq X \mid C\in\mathcal{C}(F) \textup{ for some } F\in \mathscr{F}_0(X) \}.
 \end{align*}
\end{definition}

Note that $\mathscr{U}(X)$ and $\mathscr{F}(X)$ are finite sets.

\begin{rem}
\label{rem_cc_of_U_F_are_open_closed}
Let $X$ be a finite $T_0$--space.

(1) Clearly, the elements of $\mathscr{F}(X)$ are closed subsets of $X$. In addition, since finite topological spaces are locally connected it follows that the elements of $\mathscr{U}(X)$ are open subsets of $X$.

(2) For all $a\in \mxl[X]$, $U_a\in \mathscr{U}(X)$ since $U_a$ is connected. In a similar way, for all $b\in \mnl[X]$, $F_b\in \mathscr{F}(X)$.
\end{rem}

The following proposition states that the sets $\mathscr{U}(X)$ and $\mathscr{F}(X)$ of the previous definition are disjoint, except in a trivial case.

\begin{prop}
\label{prop_UF_disjuntos}
 Let $X$ be a finite $T_0$--space.
 \begin{enumerate}[(a)]
  \item If $X$ is connected and $\mathscr{U}(X)\cap \mathscr{F}(X)\neq\varnothing$, then $X$ has a maximum element and a minimum element.
  \item If $X$ has a maximum element then $\mathscr{U}(X) = \{X\}$.
  \item If $X$ has a minimum element then $\mathscr{F}(X) = \{X\}$.
  \item If $X$ is connected and $\mathscr{U}(X)\cap \mathscr{F}(X)\neq\varnothing$, then $\mathscr{U}(X) = \mathscr{F}(X) = \{X\}$.
 \end{enumerate}
\end{prop}

\begin{proof}\

 (\emph{a}) Let $C\in\mathscr{U}(X)\cap \mathscr{F}(X)$. Then $C$ is non-empty, open and closed in $X$ by remark \ref{rem_cc_of_U_F_are_open_closed}. Since $X$ is connected, $C=X$.

 Since $C\in\mathscr{U}(X)$, there exists $U\in \mathscr{U}_0(X)$ such that $C$ is a connected component of $U$. Hence $U=X$. Since $X$ is finite and $X=U$ has at most one maximal element of $X$, then $X$ has a unique maximal element, that is, $X$ has a maximum element.

 In a similar way, it follows that $X$ has a minimum element.

 (\emph{b}) If $a=\max X$, then $\mathscr{U}_0(X) = \{U_a\}=\{X\}$ and hence $\mathscr{U}(X)=\{X\}$.

 (\emph{c}) Analogous to the proof of (\emph{b}).

 (\emph{d}) It follows immediately from (\emph{a}), (\emph{b}) and (\emph{c}).
\end{proof}

Observe that in the previous proposition, the hypothesis that $X$ is connected is necessary for parts (\emph{a}) and (\emph{d}). Indeed, consider the discrete space of two points $X=\{0,1\}$. Clearly $F_0=\{0\}=U_0$. Hence $\{0\}\in \mathscr{U}(X)\cap \mathscr{F}(X)$, but $X$ has neither a maximum element nor a minimum element. Therefore item ($a$) does not hold. In addition, it is easy to verify that $\mathscr{U}(X)=\mathscr{F}(X)=\{\{0\},\{1\}\}$, and thus item ($d$) does not hold.

\begin{definition}
\label{defi_orden_en_C(X)}
Let $X$ be a finite $T_0$--space. We define the relation $\leq$ in $\mathscr{U}(X)\cup \mathscr{F}(X)$ as follows.

Given $C_1,C_2\in \mathscr{U}(X)\cup \mathscr{F}(X)$, we will say that $C_1\leq C_2$ if and only if one of the following conditions holds:
\begin{itemize}
 \item $C_1, C_2\in \mathscr{U}(X)$ and $C_1\subseteq C_2$,
 \item $C_1, C_2\in \mathscr{F}(X)$ and $C_1\supseteq C_2$,
 \item $C_1\in \mathscr{F}(X)$, $C_2\in \mathscr{U}(X)$ and $C_1\cap C_2 \neq \varnothing$.
\end{itemize}
\end{definition}

It is not difficult to prove that the previously defined relation $\leq$ is an order relation in $\mathscr{U}(X)\cup\mathscr{F}(X)$.

\begin{definition}
Let $X$ be a finite $T_0$--space.
\begin{itemize}
\item We define $\mathscr{U}(X)$ as the finite $T_0$--space associated to the poset $(\mathscr{U}(X),\subseteq)$.
\item We define $\mathscr{F}(X)$ as the finite $T_0$--space associated to the poset $(\mathscr{F}(X),\supseteq)$.
\item We define $\mathscr{C}(X)$ as the finite $T_0$--space associated to the poset $(\mathscr{U}(X)\cup\mathscr{F}(X),\leq)$, where $\leq$ is the order defined in \ref{defi_orden_en_C(X)}.
\end{itemize}
\end{definition}

Observe that $\mathscr{U}(X)$ and $\mathscr{F}(X)$ are subspaces of $\mathscr{C}(X)$.

The following is a simple example of the $\mathscr{C}$ construction.

\begin{ex} \label{ex_easy}
Let $X$ be the finite $T_0$--space given by the following Hasse diagram.
\begin{center}
\begin{tikzpicture}
\tikzstyle{every node}=[font=\scriptsize]
\foreach \x in {0,1} \draw (\x,0) node(\x){$\bullet$} node[below=1]{\x};
\foreach \x in {2,3,4} \draw (\x-2.5,1) node(\x){$\bullet$} node[above=1]{\x};
\foreach \x in {2,3,4} \draw (0)--(\x);
\foreach \x in {3,4} \draw (1)--(\x);
\end{tikzpicture}
\end{center}
Then the Hasse diagram of $\mathscr{C}(X)$ is
\begin{center}
\begin{tikzpicture}[x=1.5cm, y=1.5cm]
\path node (U2) at (0,3) [CCC] {2\\0}
node (U3) at (1,3) [CCC] {3\\0 1}
node (U4) at (2,3) [CCC] {4\\0 1}

node (U34a) at (0.5,2) [CCC] {0}
node (U34b) at (1.5,2) [CCC] {1}

node (F01a) at (0.7,1) [CCC] {3}
node (F01b) at (1.7,1) [CCC] {4}

node (F0) at (0.7,0) [CCC] {2 3 4\\0}
node (F1) at (1.7,0) [CCC] {3 4\\1};

\draw [-] (U34a) to (U3) to (U34b) to (U4) to (U34a) to (U2);
\draw [-] (F0) to (F01a) to (F1) to (F01b) to (F0);
\draw[-,bend left=20] (F0) to (U34a);
\draw[-,bend left=20] (F1) to (U34b);
\draw[-,bend right=20] (F01a) to (U3);
\draw[-,bend right=20] (F01b) to (U4);
\end{tikzpicture}
\end{center}
\end{ex}

The following proposition describes the maximal and minimal elements of the poset $\mathscr{C}(X)$ for any given finite $T_0$--space $X$.

\begin{prop}
 \label{prop_mnl_mxl_de_C(X)}
 Let $X$ be a finite $T_0$--space. Then \[\mxl[\mathscr{C}(X)] = \{U_a \mid a\in \mxl[X] \}\quad \textup{and}\quad \mnl[\mathscr{C}(X)] = \{F_a \mid a\in \mnl[X] \}.\]
\end{prop}

\begin{proof}
Let $a\in \mxl[X]$. Since $U_a$ is connected, $U_a\in \mathscr{C}(X)$. Let $C\in \mathscr{U}(X)$ such that $C\geq U_a$. There exists a non-empty subset $A\subseteq \mxl[X]$ such that $C$ is a connected component of $U_A$. Then $U_a\subseteq C\subseteq U_A$. It follows that $a\leq x$ for all $x\in A$, and thus $a = x$ for all $x\in A$. Hence $A=\{a\}$ and $U_a = U_A = C$. Therefore, $U_a\in \mxl[\mathscr{C}(X)]$.

Now suppose that $C\in \mxl[\mathscr{C}(X)]$. Suppose, in addition, that $C\in \mathscr{F}(X)$ and let $x\in C$. Since $X$ is finite, there exists $b\in \mxl[X]$ such that $x\leq b$. Hence $b\in C$ since $C$ is closed. Thus, $C\cap U_b\neq \varnothing$ and then $C<U_b$ which contradicts the maximality of $C$.

Thus $C\in \mathscr{U}(X)$. Then there exists a non-empty subset $A\subseteq \mxl[X]$ such that $C$ is a connected component of $U_A$. Let $a\in A$. Then $C\subseteq U_a$, that is, $C\leq U_a$. Hence $C=U_a$.

The second identity can be proved in a similar way.
\end{proof}

The following proposition shows that the construction $\mathscr{F}$ can be interpreted as a dual version of $\mathscr{U}$.

\begin{prop}
\label{prop_C_op}
Let $X$ be a finite $T_0$--space.
\begin{enumerate}[(a)]
\item For all $A\subseteq \mnl[X]$, $F_A^X = \left(U_A^{X^{\textup{op}}}\right)^{\textup{op}}$.
\item $\mathscr{F}(X) = \mathscr{U}(X^{\textup{op}})$ as sets.
\item $\mathscr{C}(X^{\textup{op}}) =\left(\mathscr{C}(X) \right)^{\textup{op}}$. In particular, $\mathscr{F}(X) = (\mathscr{U}(X^{\textup{op}}))^{\textup{op}}$ as posets with the corresponding partial orders induced from $\mathscr{C}(X)$ and $\mathscr{C}(X^{\textup{op}})$ respectively.
\end{enumerate}
\end{prop}

\begin{proof}\

(\emph{a}) Let $A\subseteq \mnl[X]$. Note that for all $a\in A$, $F_a = \left(U_a^{X^{\textup{op}}} \right)^{\textup{op}}$. Hence
\[
F_A^X =\bigcap_{a\in A} F_a = \bigcap_{a\in A}\left(U_a^{X^{\textup{op}}} \right)^{\textup{op}}= \left(\bigcap_{a\in A} U_a^{X^{\textup{op}}}\right)^{\textup{op}} = \left(U_A^{X^{\textup{op}}}\right)^{\textup{op}}.
\]

(\emph{b}) Follows from (\emph{a}).

($c$) It follows from ($b$) that $\mathscr{U}(X^\textup{op})\cup \mathscr{F}(X^\textup{op}) = \mathscr{F}(X)\cup \mathscr{U}(X)$. Then the underlying sets of $\mathscr{C}(X^{\textup{op}})$ and $\left(\mathscr{C}(X) \right)^{\textup{op}}$ coincide. Let $\leq$ denote the partial order of $\mathscr{C}(X)$ and let $\preceq$ denote the partial order of $\mathscr{C}(X^{\textup{op}})$. Let $C_1,C_2\in \mathscr{C}(X^{\textup{op}})$.
\begin{itemize}
\item If $C_1,C_2\in \mathscr{U}(X^{\textup{op}})=\mathscr{F}(X)$, then $C_1\preceq C_2$ if and only if  $C_1\subseteq C_2$ if and only if $C_2\leq C_1$ if and only if $C_1\leq^{\textup{op}} C_2$.
\item If $C_1,C_2\in \mathscr{F}(X^{\textup{op}})=\mathscr{U}(X)$, then $C_1\preceq C_2$ if and only if  $C_1\supseteq C_2$ if and only if $C_2\leq C_1$ if and only if $C_1\leq^{\textup{op}} C_2$.
\item If $C_1\in \mathscr{F}(X^{\textup{op}})$ and $C_2\in \mathscr{U}(X^{\textup{op}})$, then $C_1\preceq C_2$ if and only if  $C_1\cap C_2\neq \varnothing$ if and only if $C_2\leq C_1$ if and only if $C_1\leq^{\textup{op}} C_2$.
\end{itemize}
Hence, the partial orders $\preceq$ and $\leq^{\textup{op}}$ coincide.
\end{proof}

\begin{definition}
Let $X$ be a finite $T_0$--space and let $B\subseteq X$ be a subset. We define
\begin{displaymath}
\ub B=\{a\in\mxl[X]\mid B\subseteq U_a \}  \qquad  \textnormal{and} \qquad \lb B=\{a\in\mnl[X]\mid B\subseteq F_a \}.
\end{displaymath}
\end{definition}

\begin{rem}
Let $X$ be a finite $T_0$--space and let $B\subseteq X$ be a subset.

If $\ub B \neq \varnothing$ then $B\subseteq U_{\ub B}$. If, in addition, $B$ is a connected subspace of $X$ then there exists a unique connected component $C_0$ of $U_{\ub B}$ such that $B\subseteq C_0$.

Similarly, if $\lb B \neq \varnothing$ then $B\subseteq F_{\lb B}$, and if, in addition, $B$ is a connected subspace of $X$ then there exists a unique connected component $C_0$ of $F_{\lb B}$ such that $B\subseteq C_0$.
\end{rem}

\begin{lemma}
\label{lema-para_conexo_minimal}
 Let $X$ be a finite $T_0$--space and let $B\subseteq X$ be a connected subspace. Then:
 \begin{enumerate}[(a)]
  \item $\ub B \neq \varnothing$ if and only if $\{C\in\mathscr{U}(X)\mid B\subseteq C \} \neq\varnothing$.
  \item $\lb B \neq\varnothing$ if and only if $\{C\in\mathscr{F}(X)\mid B\subseteq C \} \neq\varnothing$.
 \end{enumerate}
\end{lemma}

\begin{proof}
 We will prove only item (\emph{a}). The proof of item (\emph{b}) is similar.

 ($\Rightarrow$) Let $a_0\in \ub B$. Then $U_{a_0}\in\{C\in\mathscr{U}(X)\mid B\subseteq C \}$.

 ($\Leftarrow$) Let $C_0\in\mathscr{U}(X)$ such that $B\subseteq C_0$ and let $A\subseteq \mxl[X]$ be a non-empty subset such that $C_0$ is a connected component of $U_A$. Let $a_0\in A$. Then $B\subseteq U_A\subseteq U_{a_0}$. Hence $a_0\in\ub B$.
\end{proof}

The following proposition states a key property of the $\mathscr{C}$ construction.

\begin{prop}
\label{prop_conexo_minimal}
Let $X$ be a finite $T_0$--space and let $B\subseteq X$ be a connected non-empty subspace.
\begin{enumerate}[(a)]
\item If $\{C\in \mathscr{U}(X) \mid B\subseteq C \}$ is non-empty, then it has a minimum element, which is the connected component of $U_{\ub B}$ that contains $B$.
\item If $\{C\in \mathscr{F}(X) \mid B\subseteq C \}$ is non-empty, then it has a maximum element, which is the connected component of $F_{\lb B}$ that contains $B$.
\end{enumerate}
\end{prop}

\begin{proof}
By the previous lemma, $\ub B\neq\varnothing$. Let $C_0$ be the unique connected component of $U_{\ub B}$ such that $B\subseteq C_0$.

Let $C\in \mathscr{U}(X)$ such that $B\subseteq C$. There exists a non-empty subset $A\subseteq \mxl[X]$ such that $C$ is a connected component of $U_{A}$. Hence $B\subseteq U_a$ for all $a\in A$. Then, $A\subseteq \ub B$ and thus $U_{\ub B}\subseteq U_{A}$. Hence $C_0\subseteq U_{A}$.

Since $C$ is a connected component of $U_{A}$ and $\varnothing\neq B\subseteq C_0 \cap C$, it follows that $C_0\subseteq C$, that is, $C_0\leq C$. Hence $C_0 = \min \{C\in \mathscr{U}(X) \mid B\subseteq C\}$.

The second part can be proved in a similar way.
\end{proof}

The following corollary follows immediately from the previous proposition.

\begin{coro} \label{coro_cc_of_U_Bsharp_F_Bflat}
Let $X$ be a finite $T_0$--space.
\begin{enumerate}[(a)]
\item Let $B\in\mathscr{U}(X)$. Then $B$ is a connected component of $U_{\ub B}$.
\item Let $B\in\mathscr{F}(X)$. Then $B$ is a connected component of $F_{\lb B}$.
\end{enumerate}
\end{coro}

Proposition \ref{prop_conexo_minimal} allows us to give the following definition.

\begin{definition}
Let $X$ be a finite $T_0$--space and let $x\in X$. We define
\begin{displaymath}
{\bf C}_\U(x)=\min \{C\in \mathscr{U}(X) \mid x\in C \}
\end{displaymath}
and
\begin{displaymath}
{\bf C}_\F(x)=\max \{C\in \mathscr{F}(X) \mid x\in C \}.
\end{displaymath}
\end{definition}

\begin{prop}
Let $X$ be a finite $T_0$--space.
\begin{enumerate}[(a)]
\item The set $\mathscr{U}(X)$ is a basis for a topology in $X$ which is coarser than the topology of $X$.
\item Let $\mathcal{T}$ be the topology generated by $\mathscr{U}(X)$. The Kolmogorov quotient of $(X,\mathcal{T})$ is homeomorphic to a subspace of the finite $T_0$--space associated to the poset $\mathscr{U}(X)$.
\end{enumerate}
\end{prop}

\begin{proof}\

($a$) Since $U_a\in\mathscr{U}(X)$ for all $a\in\mxl[X]$ it follows that $\displaystyle\bigcup_{C\in\mathscr{U}(X)}C=X$. Now let $C_1,C_2\in\mathscr{U}(X)$ and let $x\in C_1\cap C_2$. Then $x\in{\bf C}_\U(x)\subseteq C_1 \cap C_2$ by \ref{prop_conexo_minimal}. Thus, $\mathscr{U}(X)$ is a basis for a topology in $X$ which, by \ref{rem_cc_of_U_F_are_open_closed}, is clearly coarser than the topology of $X$.

($b$) Let $K$ be the Kolmogorov quotient of $(X,\mathcal{T})$ and let $q\colon X\to K$ be the quotient map. Let $f\colon X\to \mathscr{U}(X)$ be defined by
$f(x)={\bf C}_\U(x)$. We will prove that $f$ is order-preserving. Let $a,b\in X$ be such that $a\leq b$ then $a\in{\bf C}_\U(b)$ since ${\bf C}_\U(b)$ is an open subset of $X$ which contains $b$. Thus ${\bf C}_\U(a)\subseteq {\bf C}_\U(b)$. Hence, $f$ is order-preserving, and therefore, it is a continuous map.

Now, let $c,d\in X$ be such that $q(c)=q(d)$. Thus, $\{C\in U(X) \mid q(c)\in C\}=\{C\in U(X) \mid q(d)\in C\}$ since $\mathscr{U}(X)$ is a basis for the topology $\mathcal{T}$. Hence, $f(c)={\bf C}_\U(c)={\bf C}_\U(d)=f(d)$. Therefore, there exists a continuous map $\overline{f}\colon K\to \mathscr{U}(X)$ such that $\overline{f}q=f$.

We will prove now that if $k,l\in K$ are such that $\overline{f}(k)\leq \overline{f}(l)$ then $k\leq l$. Let $k,l\in K$ be such that $\overline{f}(k)\leq \overline{f}(l)$. Let $x,y\in X$ be such that $k=q(x)$ and $l=q(y)$. Then $f(x)=\overline{f}(k)\leq \overline{f}(l)=f(y)$. Hence ${\bf C}_\U(x)\subseteq {\bf C}_\U(y)$. Then, for all $C\in \mathscr{U}(X)$ such that $y\in C$ we obtain that $x\in{\bf C}_\U(x)\subseteq {\bf C}_\U(y)\subseteq C$, and thus $x\in C$. Therefore $x\leq y$. Hence, $k=q(x)\leq q(y)=l$.
\end{proof}

\begin{definition} \label{def_Uf_Ff_Cf}
Let $X$ and $Y$ be finite $T_0$--spaces and let $f\colon X\to Y$ be a continuous map. We define
\begin{displaymath}
\mathscr{U}(f)\colon \mathscr{U}(X)\to \mathscr{U}(Y) \qquad \textnormal{by} \qquad \mathscr{U}(f)(C)= \min \{D\in\mathscr{U}(Y) \mid f(C)\subseteq D \},
\end{displaymath}
and
\begin{displaymath}
\mathscr{F}(f)\colon \mathscr{F}(X)\to \mathscr{F}(Y) \qquad \textnormal{by} \qquad \mathscr{F}(f)(C)= \max \{D\in\mathscr{F}(Y) \mid f(C)\subseteq D \}.
\end{displaymath}
If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$, we define
\begin{displaymath}
\mathscr{C}(f)\colon \mathscr{C}(X)\to \mathscr{C}(Y) \qquad \textnormal{by} \qquad \mathscr{C}(f)(C)=
\begin{cases}
\mathscr{U}(f)(C) & \textup{if } C\in\mathscr{U}(X), \\
\mathscr{F}(f)(C) & \textup{if } C\in\mathscr{F}(X).
\end{cases}
\end{displaymath}
\end{definition}

We will prove now that $\mathscr{U}(f)$ is well defined. Let $C\in\mathscr{U}(X)$. Then there exists $a\in \mxl[X]$ such that $C\subseteq U_a$. Let $b\in\mxl[Y]$ such that $f(a)\leq b$. Then $f(C)\subseteq U_{f(a)}\subseteq U_b$. Hence $\ub{f(C)}\neq \varnothing$. Then $\{D\in\mathscr{U}(Y) \mid f(C)\subseteq D \}$ is non-empty by \ref{lema-para_conexo_minimal}, and thus it has a minimum element by \ref{prop_conexo_minimal}.

In a similar way it can be proved that $\mathscr{F}(f)$ is well defined and thus $\mathscr{C}(f)$ is also well defined.

\medskip

Note that the map $\mathscr{C}(f)$ might not be well defined without the assumption $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$. Note also that if $\mathscr{U}(X)\cap\mathscr{F}(X)\neq\varnothing$ then either $X$ is not connected or it has minimum and maximum elements by \ref{prop_UF_disjuntos}.

\begin{rem} \label{rem_def_de_Uf_and_Ff}
Let $X$ and $Y$ be finite $T_0$--spaces and let $f\colon X\to Y$ be a continuous map.

Observe that, if $C\in\mathscr{U}(X)$, then, by \ref{prop_conexo_minimal}, $\mathscr{U}(f)(C)$ is the connected component of $U_{\ub {f(C)}}$ that contains $f(C)$.

In a similar way, if $C\in\mathscr{F}(X)$, then $\mathscr{F}(f)(C)$ is the connected component of $F_{\lb {f(C)}}$ that contains $f(C)$.
\end{rem}

We will show in the following proposition that the maps defined in \ref{def_Uf_Ff_Cf} are continuous.

\begin{prop}
Let $X$ and $Y$ be finite $T_0$--spaces and let $f\colon X\to Y$ be a continuous map. Then the maps $\mathscr{U}(f)$ and $\mathscr{F}(f)$ are continuous. If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ then the map $\mathscr{C}(f)\colon \mathscr{C}(X)\to \mathscr{C}(Y)$ is continuous.
\end{prop}

\begin{proof}
Let $C_1, C_2\in \mathscr{C}(X)$ such that $C_1\leq C_2$.
\begin{itemize}
 \item If $C_1,C_2\in \mathscr{U}(X)$, then $C_1\subseteq C_2$. Hence $f(C_1)\subseteq f(C_2)\subseteq \mathscr{U}(f)(C_2)$. Then $\mathscr{U}(f)(C_1)\subseteq \mathscr{U}(f)(C_2)$ by definition of $\mathscr{U}(f)$. Thus, $\mathscr{U}(f)(C_1)\leq \mathscr{U}(f)(C_2)$.
 \item If $C_1,C_2\in \mathscr{F}(X)$, then $C_2\subseteq C_1$ and $f(C_2)\subseteq f(C_1)\subseteq \mathscr{F}(f)(C_1)$. Hence $\mathscr{F}(f)(C_2)\subseteq \mathscr{F}(f)(C_1)$ by definition of $\mathscr{F}(f)$. Thus, $\mathscr{F}(f)(C_1)\leq \mathscr{F}(f)(C_2)$.
 \item If $C_1\in \mathscr{F}(X)$ and $C_2\in \mathscr{U}(X)$, then $C_1\cap C_2\neq \varnothing$. Hence $\varnothing \neq f(C_1)\cap f(C_2)\subseteq \mathscr{C}(f)(C_1)\cap \mathscr{C}(f)(C_2)$. Thus, $\mathscr{C}(f)(C_1)\leq \mathscr{C}(f)(C_2)$.
\end{itemize}
The result follows.
\end{proof}

\begin{prop} \label{prop_Uf_homeo}
Let $X$ and $Y$ be finite $T_0$--spaces and let $f\colon X\to Y$ be a homeomorphism. Then
\begin{enumerate}[(a)]
\item for all $C\in \mathscr{U}(X)$, $\mathscr{U}(f)(C)=f(C)$,
\item for all $C\in \mathscr{F}(X)$, $\mathscr{F}(f)(C)=f(C)$,
\item $\mathscr{U}(f)$ and $\mathscr{F}(f)$ are homeomorphisms.
\end{enumerate}
\end{prop}

\begin{proof}\

($a$) Since $f$ is a homeomorphism, $f(\mxl[X])=\mxl[Y]$ and $f(U_A)=U_{f(A)}$ for each non-empty subset $A\subseteq \mxl[X]$. Let $C\in \mathscr{U}(X)$. Then $C$ is a connected component of $U_{\ub C}$ and hence $f(C)$ is a connected component of $U_{f(\ub C)}$. Thus $\mathscr{U}(f)(C)=f(C)$.

($b$) Analogous to ($a$).

($c$) Let $C\in \mathscr{U}(X)$. Then applying ($a$) we obtain that $\mathscr{U}(f^{-1})\mathscr{U}(f)(C)=\mathscr{U}(f^{-1})(f(C))=f^{-1}(f(C))=C$. Thus $\mathscr{U}(f^{-1})\mathscr{U}(f) = \textup{id}_{\mathscr{U}(X)}$. Similarly $\mathscr{U}(f)\mathscr{U}(f^{-1}) = \textup{id}_{\mathscr{U}(Y)}$.

The proof for $\mathscr{F}(f)$ follows similarly.
\end{proof}

In the following example we will see that the constructions $\mathscr{U}$, $\mathscr{F}$ and $\mathscr{C}$ are not functorial.

\begin{ex}
Let $X$ be the finite $T_0$--space whose underlying set is $\{0,1,2,3,4\}$ and whose topology is generated by the basis $\{\{0\},\{0,1\},\{2\},\{0,1,2,3\},\{0,1,2,4\}\}$. The Hase diagrams of $X$ and $\mathscr{U}(X)$ are shown below.
\smallskip
\begin{center}
\begin{tikzpicture}[baseline=-1.25cm]
\tikzstyle{every node}=[font=\scriptsize]
\draw (0,0) node(0){$\bullet$} node[below=1]{0};
\draw (0,1) node (1){$\bullet$} node[left=1]{1};
\draw (1,1) node (2){$\bullet$} node[right=1]{2};
\foreach \x in {3,4} \draw (\x-3,2) node(\x){$\bullet$} node[above=1]{\x};

\draw (0)--(1);
\foreach \x in {4,3} \draw (1)--(\x);
\foreach \x in {4,3} \draw (2)--(\x);
\draw (0.5,-0.9) node(X){\normalsize $X$};
\end{tikzpicture}
\hspace*{2cm}
\begin{tikzpicture}[x=2.3cm, y=2cm]
\path node (U3) at (0,1) [CC] {3\\1 2\\0}
 node (U4) at (1,1) [CC] {4\\1 2\\0}
 node (U1) at (0,0) [CC] {1\\0}
 node (U2) at (1,0) [CC] {2};

\draw (U1)--(U4)--(U2)--(U3)--(U1);
\draw (0.5,-0.5) node(UX){$\mathscr{U}(X)$};
\end{tikzpicture}
\end{center}

Let $f\colon X\to X$ be the map defined by
\begin{center}
\begin{tabular}{c | c c c c c}
$x$ & 0 & 1 & 2 & 3& 4\\[-2pt]
\hline
$f(x)$& 2 & 3 & 3 & 3& 3
\end{tabular}
\end{center}
Clearly $f$ is a continuous map. Let $g\colon X\to X$ be the constant map with value 0. Then $fg$ is the constant map with value 2. It follows that $\mathscr{U}(g)$ is the constant map with value $\{0,1\}$ and that $\mathscr{U}(fg)$ is the constant map with value $\{2\}$. Then
\[\mathscr{U}(f)\mathscr{U}(g)(\{2\})=\mathscr{U}(f)(\{0,1\})= U_3 \neq \{2\} = \mathscr{U}(fg)(\{2\}).
\]
Hence, $\mathscr{U}(fg)\neq\mathscr{U}(f)\mathscr{U}(g)$.

This implies that $\mathscr{C}(fg)\neq\mathscr{C}(f)\mathscr{C}(g)$, and the fact that $\mathscr{F}$ is not functorial can be deduced from this example considering opposite spaces.
\end{ex}

\begin{prop}
Let $X$, $Y$ and $Z$ be finite $T_0$--spaces and let $g\colon X\to Y$ and $f\colon Y\to Z$ be continuous maps. Then $\mathscr{U}(fg) \leq \mathscr{U}(f)\mathscr{U}(g)$ and $\mathscr{F}(fg) \geq \mathscr{F}(f)\mathscr{F}(g)$. If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ and $\mathscr{U}(Y)\cap\mathscr{F}(Y)=\varnothing$ then $\mathscr{C}(fg) \simeq \mathscr{C}(f)\mathscr{C}(g)$.
\end{prop}

\begin{proof}
Let $C\in \mathscr{U}(X)$. Then
\[fg(C)\subseteq f(\mathscr{U}(g)(C))\subseteq \mathscr{U}(f)(\mathscr{U}(g)(C)).
\]
Hence, $\mathscr{U}(fg)(C)\subseteq\mathscr{U}(f)(\mathscr{U}(g)(C))$. Therefore, $\mathscr{U}(fg) \leq \mathscr{U}(f)\mathscr{U}(g)$. In a similar way it can be proved that $\mathscr{F}(fg) \geq \mathscr{F}(f)\mathscr{F}(g)$.

If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ and $\mathscr{U}(Y)\cap\mathscr{F}(Y)=\varnothing$ then the maps $\mathscr{C}(fg)$ and $\mathscr{C}(f)\mathscr{C}(g)$ are defined. Since $\mathscr{U}(fg) \leq \mathscr{U}(f)\mathscr{U}(g)$ and $\mathscr{F}(fg) \geq \mathscr{F}(f)\mathscr{F}(g)$ we obtain that, for all $C\in\mathscr{C}(X)$, the elements $\mathscr{C}(fg)(C)$ and $\mathscr{C}(f)\mathscr{C}(g)(C)$ are comparable. Thus, the last assertion follows.
\end{proof}

\begin{prop} \label{prop_f_leq_g_Uf_leq_Ug}
Let $X$ and $Y$ be finite $T_0$--spaces and let $f,g\colon X\to Y$ be continuous maps.
\begin{enumerate}[(a)]
\item If $f\leq g$, then $\mathscr{U}(f)\leq \mathscr{U}(g)$ and $\mathscr{F}(f)\leq \mathscr{F}(g)$.
\item If $f\leq g$ and $\mathscr{U}(X)\cap \mathscr{F}(X)=\varnothing$, then $\mathscr{C}(f)\leq \mathscr{C}(g)$.
\item If $f\simeq g$, then $\mathscr{U}(f)\simeq \mathscr{U}(g)$ and $\mathscr{F}(f)\simeq \mathscr{F}(g)$.
\item If $f\simeq g$ and $\mathscr{U}(X)\cap \mathscr{F}(X)=\varnothing$, then $\mathscr{C}(f)\simeq \mathscr{C}(g)$.
\end{enumerate}
\end{prop}

\begin{proof}

We will prove ($a$). Item ($b$) follows from ($a$), while items ($c$) and ($d$) follow from ($a$) and ($b$).

Let $C\in \mathscr{U}(X)$. If $x\in C$ then $f(x)\leq g(x)\in g(C)\subseteq \mathscr{U}(g)(C)$, and since $\mathscr{U}(g)(C)$ is an open subset of $Y$ it follows that $f(x)\in \mathscr{U}(g)(C)$. Therefore, $f(C)\subseteq \mathscr{U}(g)(C)$. Hence $\mathscr{U}(f)(C)\leq \mathscr{U}(g)(C)$.

Now let $C\in \mathscr{F}(X)$. If $x\in C$ then $g(x)\geq f(x)\in f(C)\subseteq \mathscr{F}(f)(C)$, and since $\mathscr{F}(f)(C)$ is a closed subset of $Y$ it follows that $g(x)\in \mathscr{F}(f)(C)$. Therefore, $g(C)\subseteq \mathscr{F}(f)(C)$. Hence $\mathscr{F}(f)(C)\leq \mathscr{F}(g)(C)$.
\end{proof}

\begin{coro}
Let $X$ and $Y$ be finite $T_0$--spaces and let $f\colon X\to Y$ be a homotopy equivalence. Then $\mathscr{U}(f)$ and $\mathscr{F}(f)$ are homotopy equivalences. If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ and $\mathscr{U}(Y)\cap\mathscr{F}(Y)=\varnothing$ then $\mathscr{C}(f)$ is a homotopy equivalence.
\end{coro}

\section{Fixed points}

The following proposition and its corollaries relate the $\mathscr{C}$ construction to the fixed point property.

\begin{prop}
\label{prop_pf_con_ppf_implica_f_pf_general}
Let $X$ be a finite $T_0$--space and let $f\colon X\to X$ be a continuous map. Let $C\in \mathscr{C}(X)$ be a subspace of $X$ with the fixed point property.
\begin{enumerate}[(a)]
\item If $C\in \mathscr{U}(X)$ and $\mathscr{U}(f)(C)\leq C$ then $f$ has a fixed point.
\item If $C\in \mathscr{F}(X)$ and $\mathscr{F}(f)(C)\geq C$ then $f$ has a fixed point.
\item If $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ and $\mathscr{C}(f)(C)=C$, then $f$ has a fixed point.
\end{enumerate}
\end{prop}

\begin{proof}
Under the assumptions of item ($a$) we obtain that $f(C)\subseteq \mathscr{U}(f)(C) \subseteq C$. Hence the map $f$ has a fixed point. Item ($b$) can be proved in a similar way and item ($c$) follows immediately from ($a$) and ($b$).
\end{proof}

\begin{coro}
Let $X$ be a finite $T_0$--space.
\begin{itemize}
\item If the space $\mathscr{U}(X)$ (resp. $\mathscr{F}(X)$) has the fixed point property and every $C\in\mathscr{U}(X)$ (resp. every $C\in\mathscr{F}(X)$) has the fixed point property then $X$ has the fixed point property.
\item If $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$, the space $\mathscr{C}(X)$ has the fixed point property and every $C\in\mathscr{C}(X)$ has the fixed point property then $X$ has the fixed point property.
\end{itemize}
\end{coro}

\begin{proof}
We will prove the result in the case that $\mathscr{U}(X)$ has the fixed property. Let $f\colon X \to X$ be a continuous map. Then the map $\mathscr{U}(f)\colon\mathscr{U}(X)\to\mathscr{U}(X)$ has a fixed point. Thus $f$ has a fixed point by the previous proposition.
\end{proof}

\begin{coro}
Let $X$ be a finite $T_0$--space  and let $f\colon X\to X$ be a continuous map.
\begin{enumerate}[(a)]
\item If there exist $a\in \mxl[X]$ and $C\in \mathscr{U}(X)$ such that $U_a\geq C$ and $\mathscr{U}(f)(C) = U_a$, then $f$ has a fixed point.
\item If there exist $a\in \mnl[X]$ and $C\in \mathscr{F}(X)$ such that $F_a\leq C$ and $\mathscr{F}(f)(C) = F_a$, then $f$ has a fixed point.
\end{enumerate}
\end{coro}

\begin{proof}
From the hypotheses of item ($a$) we obtain that $\mathscr{U}(f)(U_a)\geq \mathscr{U}(f)(C) = U_a$. And since $U_a\in \mxl[\mathscr{C}(X)]$, $\mathscr{U}(f)(U_a) = U_a$. Then $f$ has a fixed point by proposition \ref{prop_pf_con_ppf_implica_f_pf_general}.

The second item follows in a similar way.
\end{proof}

\begin{prop}
Let $X$ be a finite $T_0$--space and let $f\colon X\to X$ be a continuous map. If $f$ has a fixed point, then the maps
$\mathscr{U}(f)$ and $\mathscr{F}(f)$ have fixed points.
\end{prop}

\begin{proof}
Let $x\in X$ be such that $f(x) = x$. Let $C_0=\min\{C\in\mathscr{U}(X) \mid x\in C\}$. Note that this minimum element exists by proposition \ref{prop_conexo_minimal}. It follows that $x = f(x)\in f(C_0)\subseteq \mathscr{U}(f)(C_0)$. Then $C_0\leq \mathscr{U}(f)(C_0)$. Hence, $\mathscr{U}(f)$ has a fixed point.

In a similar way it can be proved that the map $\mathscr{F}(f)$ has a fixed point.
\end{proof}

The following example shows that the converse of the previous proposition does not hold.

\begin{ex} \label{ex_2}
Let $X$ be the finite $T_0$--space whose underlying set is $\{0,1,2,3,4,5\}$ and whose topology is generated by the basis $\{\{0\},\{1\},\{0,1,2\},\{0,1,3\},\{0,1,2,3,4\},\{0,1,2,3,5\}\}$. The Hase diagrams of $X$ and $\mathscr{U}(X)$ are shown below.

\begin{center}
\begin{tikzpicture}
\tikzstyle{every node}=[font=\scriptsize]
\foreach \x in {0,1} \draw (\x,0) node(\x){$\bullet$} node[below=1]{\x};
\draw (0,1) node (2){$\bullet$} node[left=1]{2};
\draw (1,1) node (3){$\bullet$} node[right=1]{3};
\foreach \x in {4,5} \draw (\x-4,2) node(\x){$\bullet$} node[above=1]{\x};

\foreach \x in {2,3} \draw (0)--(\x);
\foreach \x in {2,3} \draw (1)--(\x);
\foreach \x in {4,5} \draw (2)--(\x);
\foreach \x in {4,5} \draw (3)--(\x);
\draw (-1,2) node(X){\normalsize $X$};
\end{tikzpicture}
\hspace*{2cm}
\begin{tikzpicture}[x=2.3cm, y=2cm]
\path node (U4) at (0,1) [CC] {4\\2 3\\0 1}
 node (U5) at (1,1) [CC] {5\\2 3\\ 0 1}
 node (U45) at (0.5,0) [CC] {2 3\\0 1};

\draw [-] (U45) to (U4);
\draw [-] (U45) to (U5);
\draw (-0.8,1) node(UX){$\mathscr{U}(X)$};
\end{tikzpicture}
\end{center}

Let $f\colon X\to X$ be the continuous map defined by
\begin{center}
\begin{tabular}{c | c c c c c c }
$x$ & 0 & 1 & 2 & 3& 4 & 5 \\[-2pt]
\hline
$f(x)$& 1 & 0 & 3 & 2& 5 & 4
\end{tabular}
\end{center}
Clearly $\mathscr{U}(X)$ has the fixed point property and hence the map $\mathscr{U}(f)$ has a fixed point. In a similar way, $\mathscr{F}(f)$ has a fixed point. However, the map $f$ does not have fixed points.
\end{ex}

\section{Examples}

In this section we give several examples of application which show that the tools of section \ref{section-C_constr} are useful to study the fixed point property of finite posets.

First we give the following lemmas.

\begin{lemma}
\label{lema_para_Q4_Q5}
 Let $X$ the finite $T_0$--space given by the following Hasse diagram.
 \[
 \begin{tikzpicture}
 \tikzstyle{every node}=[font=\scriptsize]

  \foreach \x in {0,1,2} \draw (\x,0) node(\x){$\bullet$} node[below=1]{\x};
  \foreach \x in {3,4,5} \draw (\x-3,1) node(\x){$\bullet$} node[right=1]{\x};
  \foreach \x in {6,7,8} \draw (\x-6,2) node(\x){$\bullet$} node[above=1]{\x};

  \foreach \x in {3,4} \draw (0)--(\x);
  \foreach \x in {3,5} \draw (1)--(\x);
  \foreach \x in {4,5} \draw (2)--(\x);

  \foreach \x in {6,7} \draw (3)--(\x);
  \foreach \x in {6,8} \draw (4)--(\x);
  \foreach \x in {7,8} \draw (5)--(\x);
 \end{tikzpicture}
 \]

If $f\colon X\to X$ is a continuous map without fixed points, then $f(\{3,4,5\})=\{3,4,5\}$.
\end{lemma}

\begin{proof}
We will prove first that $f(\{3,4,5\})\subseteq X-\mxl[X]$. Let $x\in \{3,4,5\}$ and suppose that $f(x)\in \mxl[X]$. Then $f(F_x)=\{f(x)\}$ and hence there exist $a,b\in \mxl[X]$ such that $a\neq b$ and $f(a)=f(b)=f(x)$. Observe that $\{0,1,2,3,4,5\}\subseteq U_a \cup U_b$. Hence
\[f(\widehat{U}_{f(x)})\subseteq \{0,1,2,3,4,5\}\subseteq f(U_a \cup U_b) \subseteq U_{f(x)}.
 \]
Since $\widehat{U}_{f(x)}$ is contractible, $f$ has a fixed point by \ref{Uhat} which entails a contradiction.

Since $f(\{3,4,5\})\subseteq X-\mxl[X]$, it follows that $f(X-\mxl[X])\subseteq X-\mxl[X]$. And since $f$ does not have fixed points, $f$ is bijective in this subspace by \ref{prop_crown}. In particular $f(\{3,4,5\})=\{3,4,5\}$.
\end{proof}

\begin{lemma} \label{lema_para_Q7}
  Let $X$ the finite $T_0$--space given by the following Hasse diagram.
 \[
 \begin{tikzpicture}
 \tikzstyle{every node}=[font=\scriptsize]
  \foreach \x in {0,1,2} \draw (\x,0) node(\x){$\bullet$} node[below=1]{\x};
  \foreach \x in {3,4,5} \draw (\x-3,1) node(\x){$\bullet$} node[right=1]{\x};
  \foreach \x in {6,7,8} \draw (\x-6,2) node(\x){$\bullet$} node[above=1]{\x};

   \foreach \x in {3,4,5} \draw (0)--(\x);
   \foreach \x in {3,4,5} \draw (1)--(\x);
   \foreach \x in {3,5} \draw (2)--(\x);

   \foreach \x in {6,7} \draw (3)--(\x);
   \foreach \x in {6,8} \draw (4)--(\x);
   \foreach \x in {7,8} \draw (5)--(\x);
 \end{tikzpicture}
 \]

 Let $f\colon X\to X$ a continuous map without fixed points. Then $f(\{3,4,5\}) = \{3,4,5\}$.
\end{lemma}

\begin{proof}
Let $x\in \{0,1\}$. Since $f$ does not have fixed points, $f(x)$ is not comparable with $x$. Hence $f(x)\in \{0,1,2\}-\{x\}$. In addition, note that $\widehat F_2\subseteq F_x$. Then, if $f(x) = 2$, it follows that $f(\widehat F_2)\subseteq f(F_x) \subseteq F_2$, with $\widehat F_2$ contractible. Hence by \ref{Uhat}, $f$ has a fixed point, which entails a contradiction. Then $f(0)=1$ and $f(1)=0$. It follows that $f(F_{0,1})\subseteq F_{0,1}$. In addition, as $f$ does not have fixed points, the restriction $f|\colon F_{0,1} \to F_{0,1}$ is bijective by \ref{prop_crown}. In particular, $f(\{3,4,5\}) = \{3,4,5\}$.
\end{proof}

Now we are ready to give the examples.

\begin{figure}
\begin{subfigure}[b]{0.25\textwidth}
\centering
\begin{tikzpicture}
\tikzstyle{every node}=[font=\scriptsize]
\foreach \x in {0,1,2} \draw (\x,0) node(\x){$\bullet$} node[below=1]{\x};
\foreach \x in {3,4,5} \draw (\x-3,1) node(\x){$\bullet$} node[right=1]{\x};
\foreach \x in {6,7} \draw (\x-5.5,2) node(\x){$\bullet$} node[right=1]{\x};
\foreach \x in {8,9,10} \draw (\x-8,3) node(\x){$\bullet$} node[above=1]{\x};

\foreach \x in {3,4} \draw (0)--(\x);
\foreach \x in {3,5} \draw (1)--(\x);
\foreach \x in {4,5} \draw (2)--(\x);
\foreach \x in {8,10} \draw (3)--(\x);
\foreach \x in {6,7} \draw (4)--(\x);
\foreach \x in {6,7} \draw (5)--(\x);
\foreach \x in {8,9} \draw (6)--(\x);
\foreach \x in {9,10} \draw (7)--(\x);
\end{tikzpicture}
\caption{$P^{3323}$}
\label{fig-Q8}
\end{subfigure}
\begin{subfigure}[b]{0.35\textwidth}
\centering
\begin{tikzpicture}
\tikzstyle{every node}=[font=\scriptsize]
\foreach \x in {0,1,2} \draw (\x,-0.2) node(\x){$\bullet$} node[below=1]{\x};
\foreach \x in {3,4,5,6,7} \draw (\x-4,1) node(\x){$\bullet$} node[right=1]{\x};
\foreach \x in {8,9,10} \draw (\x-8,2.2) node(\x){$\bullet$} node[above=1]{\x};

\foreach \x in {3,4,5} \draw (0)--(\x);
\foreach \x in {3,4,6,7} \draw (1)--(\x);
\foreach \x in {5,6,7} \draw (2)--(\x);

\foreach \x in {3,4,6,7} \draw (8)--(\x);
\foreach \x in {3,5,7} \draw (9)--(\x);
\foreach \x in {4,5,6} \draw (10)--(\x);
\end{tikzpicture}
\caption{$P^{343}_1$}
\label{fig-Q6}
\end{subfigure}
\begin{subfigure}[b]{0.35\textwidth}
\begin{tikzpicture}
\tikzstyle{every node}=[font=\scriptsize]
\foreach \x in {0,1,2} \draw (\x,-0.2) node(\x){$\bullet$} node[below=1]{\x};
\foreach \x in {3,4,5,6,7} \draw (\x-4,1) node(\x){$\bullet$} node[right=1]{\x};
\foreach \x in {8,9,10} \draw (\x-8,2.2) node(\x){$\bullet$} node[above=1]{\x};

\foreach \x in {3,4,5} \draw (0)--(\x);
\foreach \x in {3,4,6,7} \draw (1)--(\x);
\foreach \x in {5,6,7} \draw (2)--(\x);

\foreach \x in {3,4,6} \draw (8)--(\x);
\foreach \x in {3,5,6,7} \draw (9)--(\x);
\foreach \x in {4,5,7} \draw (10)--(\x);
\end{tikzpicture}
\caption{$P^{343}_2$}
\label{fig-Q7}
\end{subfigure}
\caption{}
\end{figure}

\begin{ex}
\label{ejm_Q8}
We will prove that the space $P^{3323}$ of \cite[Fig.1]{Schroder1993fpp11} has the fixed point property. Its Hasse diagram is given in figure \ref{fig-Q8}. The Hasse diagrams of $\mathscr{U}(P^{3323})$ and $\mathscr{F}(P^{3323})$ are shown in figure \ref{fig-CQ8} and the shaded elements of these diagrams are contractible subspaces of $P^{3323}$.

\begin{figure}
\begin{subfigure}[b]{0.48\textwidth}
\centering
\begin{tikzpicture}[x=2.3cm, y=2cm]
\path node[CCC] (U8) at (0,6) { 8\\ 6\\ 3 4 5\\ 0 1 2}
node[CCC] (U9) at (1,6) { 9\\ 6 7\\ 4 5\\ 0 1 2}
node (U10) at (2,6) [CCC] { 10\\ 7\\ 3 4 5\\ 0 1 2}

node (U89) at (0,5) [CCC] { 6\\ 4 5\\ 0 1 2}
node (U810) at (1,5) [CC] { 3 4 5\\ 0 1 2}
node (U910) at (2,5) [CCC] { 7\\ 4 5\\ 0 1 2}

node (U8910) at (1,4) [CCC] { 4 5\\ 0 1 2};

\draw [-] (U8) to (U810) to (U10) to (U910) to (U9) to (U89) to (U8);
\draw [-] (U8910) to (U89);
\draw [-] (U8910) to (U810);
\draw [-] (U8910) to (U910);
\end{tikzpicture}
\caption{$\mathscr{U}(P^{3323})$}
\label{fig-UQ8}
\end{subfigure}
\begin{subfigure}[b]{0.48\textwidth}
\centering
\begin{tikzpicture}[x=2.3cm, y=2cm]
\path node (F012) at (1,3) [CCC] { 8 9 10\\ 6 7}

node (F01) at (0,2) [CC] { 8 9 10\\ 6 7\\ 3}
node (F02) at (1,2) [CCC] { 8 9 10\\ 6 7\\ 4}
node (F12) at (2,2) [CCC] { 8 9 10\\ 6 7\\ 5}

node (F0) at (0,1) [CCC] { 8 9 10\\ 6 7\\ 3 4\\ 0}
node (F1) at (1,1) [CCC] { 8 9 10\\ 6 7\\ 3 5\\ 1}
node (F2) at (2,1) [CCC] { 8 9 10\\ 6 7\\ 4 5\\ 2};

\draw[-] (F012) to (F01);
\draw[-] (F012) to (F02);
\draw[-] (F012) to (F12);
\draw[-] (F0) to (F02) to (F2) to (F12) to (F1) to (F01) to (F0);

\end{tikzpicture}
\caption{$\mathscr{F}(P^{3323})$}
\label{fig-FQ8}
\end{subfigure}
\caption{\!}
\label{fig-CQ8}
\end{figure}

Suppose that there exists a continuous map $f\colon P^{3323}\to P^{3323}$ without fixed points.

Let $C_1=\{0,1,2,3,4,5\}\in \mathscr{U}(X)$. Since the poset $\mathscr{U}(P^{3323})$ has a minimum element, the map $\mathscr{U}(f)\colon \mathscr{U}(P^{3323})\to \mathscr{U}(P^{3323})$ has a fixed point, which must be $C_1$ by \ref{prop_pf_con_ppf_implica_f_pf_general}. Thus, $f(C_1)\subseteq \mathscr{U}(f)(C_1)=C_1$.

Let $C_2=\{3,6,7,8,9,10\}\in \mathscr{F}(P^{3323})$. Proceeding as in the previous paragraph we obtain that $C_2$ is a fixed point of $\mathscr{F}(f)$. Thus, $f(C_2)\subseteq \mathscr{F}(f)(C_2)=C_2$.

Hence, $f(C_1\cap C_2)\subseteq C_1\cap C_2$ and hence $3$ is a fixed point of $f$, which entails a contradiction.

Therefore, the space $P^{3323}$ has the fixed point property.
\end{ex}

\begin{ex}
We will prove that the space $P^{343}_1$ given in \cite[Fig.1]{Schroder1993fpp11} has the fixed point property. Its Hasse diagram is given in figure \ref{fig-Q6}. The Hasse diagram of $\mathscr{C}(P^{343}_1)$ is shown in figure \ref{fig-CQ6}.

\begin{figure}
\begin{subfigure}[c]{0.48\textwidth}
\centering
\begin{tikzpicture}[x=2.3cm, y=2cm]
\path node[CCC] (U8) at (0,6)  { 8\\ 3 4 6 7\\ 0 1 2}
node (U9) at (1,6) [CCC] { 9\\ 3 5 7\\ 0 1 2}
node (U10) at (2,6) [CCC] { 10\\ 4 5 6\\ 0 1 2}

node (U89) at (0,5) [CCC] { 3 7\\ 0 1 2}
node (U810) at (1,5) [CCC] { 4 6\\ 0 1 2}
node (U910a) at (2,5) [CCC] { 5\\ 0 2}

node (U8910a) at (0.2,4) [CCC] { 0}
node (U8910b) at (1,4) [CCC] { 1}
node (U8910c) at (1.8,4) [CCC] { 2};

\draw [-] (U8) to (U810) to (U10) to (U910a) to (U9) to (U89) to (U8);

\foreach \x in {U89, U810, U910a} \draw (U8910a)--(\x);
\foreach \x in {U89, U810} \draw (U8910b)--(\x);
\foreach \x in {U89, U910a, U810} \draw (U8910c)--(\x);

\path node (F012a) at (0.2,3) [CCC] { 8}
node (F012b) at (1,3) [CCC] { 9}
node (F012c) at (1.8,3) [CCC] { 10}

node (F01) at (0,2) [CCC] { 8 9 10\\ 3 4}
node (F02a) at (1,2) [CCC] { 9 10\\ 5}
node (F12) at (2,2) [CCC] { 8 9 10\\ 6 7}

node (F0) at (0,1) [CCC] { 8 9 10\\ 3 4 5\\ 0}
node (F1) at (1,1) [CCC] { 8 9 10\\ 3 4 6 7\\ 1}
node (F2) at (2,1) [CCC] { 8 9 10\\ 5 6 7\\ 2};

\foreach \x in {F01, F12} \draw (F012a)--(\x);
\foreach \x in {F01, F02a, F12} \draw (F012b)--(\x);
\foreach \x in {F01, F02a, F12} \draw (F012c)--(\x);
\draw[-] (F0) to (F02a) to (F2) to (F12) to (F1) to (F01) to (F0);

\draw[-, red] (U810) to (F01) to (U89) to (F12) to (U810);
\draw[-, red] (F02a) to (U910a);
\draw[-, blue, bend left] (U8) to (F012a);
\draw[-, blue, bend left] (U9) to (F012b);
\draw[-, blue, bend left] (U10) to (F012c);

\draw[-, blue, bend left] (F0) to (U8910a);
\draw[-, blue, bend left] (F1) to (U8910b);
\draw[-, blue, bend left] (F2) to (U8910c);
\end{tikzpicture}
\caption{$\mathscr{C}(P^{343}_1)$}
\label{fig-CQ6}
\end{subfigure}
\begin{subfigure}[c]{0.48\textwidth}
\centering
\begin{tikzpicture}[x=2.3cm, y=2cm]
\path node (U8) at (0,6) [CCC] { 8\\ 3 4 6\\ 0 1 2}
node (U9) at (1,6) [CCC] { 9 \\ 3 5 6 7\\ 0 1 2}
node (U10) at (2,6) [CCC] { 10\\ 4 5 7\\ 0 1 2}

node (U89) at (0,5) [CCC, label=left:$A$] { 3 6\\ 0 1 2}
node (U810a) at (1,5) [CCC, label=left:$B$] { 4\\ 0 1}
node (U910) at (2,5) [CCC, label=left:$C$] { 5 7\\ 0 1 2}

node (U8910a) at (0.2,4) [CCC] { 0}
node (U8910b) at (1,4) [CCC] { 1}
node (U8910c) at (1.8,4) [CCC] { 2};

\draw [-] (U8) to (U810a) to (U10) to (U910) to (U9) to (U89) to (U8);

\foreach \x in {U89, U810a, U910} \draw (U8910a)--(\x);
\foreach \x in {U89, U810a, U910} \draw (U8910b)--(\x);
\foreach \x in {U89, U910} \draw (U8910c)--(\x);

\path node (F012a) at (0.2,3) [CCC] {8}
node (F012b) at (1,3) [CCC] {9}
node (F012c) at (1.8,3) [CCC] {10}

node (F01) at (0,2) [shape=circle, draw, fill=black!10, label=right:$A'$, align=center, font=\scriptsize] {8 9 10\\3 4}
node (F02a) at (1,2) [shape=circle, draw, fill=black!10, label=right:$B'$, align=center, font=\scriptsize] {9 10\\5}
node (F12) at (2,2) [shape=circle, draw, fill=black!10, label=right:$C'$,  align=center, font=\scriptsize] {8 9 10\\6 7}

node (F0) at (0,1) [CCC] {8 9 10\\3 4 5\\0}
node (F1) at (1,1) [CCC] {8 9 10\\3 4 6 7\\1}
node (F2) at (2,1) [CCC] {8 9 10\\5 6 7\\2};

\foreach \x in {F01, F12} \draw (F012a)--(\x);
\foreach \x in {F01, F02a, F12} \draw (F012b)--(\x);
\foreach \x in {F01, F02a, F12} \draw (F012c)--(\x);
\draw[-] (F0) to (F02a) to (F2) to (F12) to (F1) to (F01) to (F0);

\draw[-, red] (U810a) to (F01) to (U89) to (F12) to (U910) to (F02a);
\draw[-, blue, bend left] (U8) to (F012a);
\draw[-, blue, bend left] (U9) to (F012b);
\draw[-, blue, bend left] (U10) to (F012c);

\draw[-, blue, bend left] (F0) to (U8910a);
\draw[-, blue, bend left] (F1) to (U8910b);
\draw[-, blue, bend left] (F2) to (U8910c);
\end{tikzpicture}
\caption{$\mathscr{C}(P^{343}_2)$}
\label{fig-CQ7}
\end{subfigure}
\caption{}
\end{figure}

We will prove first that $\mathscr{C}(P^{343}_1)$ is homotopy equivalent to $P^{3323}$. Note that the minimal elements of $\mathscr{U}(P^{343}_1)$ are down beat points of $\mathscr{C}(P^{343}_1)$ and that the maximal elements of $\mathscr{F}(P^{343}_1)$ are up beat points of $\mathscr{C}(P^{343}_1)$. Then $\mathscr{C}(P^{343}_1)$ is homotopy equivalent to the finite $T_0$--space that is given by the Hasse diagram of figure \ref{fig-step-1}. Now observe that the element labeled $a$ in this figure is a down beat point, and by removing it we obtain the space given in figure \ref{fig-step-2}, which is homeomorphic to $P^{3323}$. Therefore $\mathscr{C}(P^{343}_1)$ is homotopy equivalent to $P^{3323}$.

\begin{figure}
\begin{subfigure}[c]{0.4\textwidth}
\centering
\begin{tikzpicture}
 \tikzstyle{every node}=[font=\scriptsize]
  \foreach \x in {0,1,2} \draw (\x,0) node(\x){$\bullet$};
  \foreach \x in {3,4,5} \draw (\x-3,1) node(\x){$\bullet$};

  \foreach \x in {9,10} \draw (\x-9,3) node(\x){$\bullet$};
  \draw (2,3) node(11){$\bullet$} node[right=1]{{\normalsize $a$}};
  \foreach \x in {12,13,14} \draw (\x-12,4) node(\x){$\bullet$};

  \foreach \x in {3,4} \draw (0)--(\x);
  \foreach \x in {3,5} \draw (1)--(\x);
  \foreach \x in {4,5} \draw (2)--(\x);

  \foreach \x in {12,13} \draw (9)--(\x);
  \foreach \x in {12,14} \draw (10)--(\x);
  \foreach \x in {13,14} \draw (11)--(\x);

  \draw[-, red] (3) to (9) to (5) to (10) to (3);
  \draw[-, red] (4) to (11);
\end{tikzpicture}
\caption{}
\label{fig-step-1}
\end{subfigure}
\begin{subfigure}[c]{0.4\textwidth}
\centering
\begin{tikzpicture}
  \tikzstyle{every node}=[font=\scriptsize]
  \foreach \x in {0,1,2} \draw (\x,0) node(\x){$\bullet$};
  \foreach \x in {3,4,5} \draw (\x-3,1) node(\x){$\bullet$};

  \foreach \x in {9,10} \draw (\x-9,3) node(\x){$\bullet$};
  \foreach \x in {12,13,14} \draw (\x-12,4) node(\x){$\bullet$};

  \foreach \x in {3,4} \draw (0)--(\x);
  \foreach \x in {3,5} \draw (1)--(\x);
  \foreach \x in {4,5} \draw (2)--(\x);

  \foreach \x in {12,13} \draw (9)--(\x);
  \foreach \x in {12,14} \draw (10)--(\x);

  \draw[-, red] (3) to (9) to (5) to (10) to (3);
  \draw[-,bend right] (4) to (13);
  \draw[-,bend right] (4) to (14);
\end{tikzpicture}
\caption{}
\label{fig-step-2}
\end{subfigure}
\caption{}
\end{figure}

Now, let $f\colon P^{343}_1\to P^{343}_1$ be a continuous map. In example \ref{ejm_Q8} we proved that $P^{3323}$ has the fixed point property. Hence, by \ref{prop_XheY_ppf}, $\mathscr{C}(P^{343}_1)$ has the fixed point property. Thus, the map $\mathscr{C}(f)$ has a fixed point. And since all the elements of $\mathscr{C}(P^{343}_1)$ are contractible subspaces of $P^{343}_1$, applying proposition \ref{prop_pf_con_ppf_implica_f_pf_general}, we obtain that $f$ has a fixed point. Therefore $P^{343}_1$ has the fixed point property.
\end{ex}

\begin{ex}
We will prove that the space $P^{343}_2$ given in \cite[Fig.1]{Schroder1993fpp11} has the fixed point property. Its Hasse diagram is given in figure \ref{fig-Q7}. The Hasse diagram of $\mathscr{C}(P^{343}_2)$ is shown in figure \ref{fig-CQ7}. Let $A$, $B$, $C$, $A'$, $B'$ and $C'$ be the sets indicated in that figure.

Suppose that $f\colon P^{343}_2\to P^{343}_2$ is a continuous map without fixed points. Since all the elements of $\mathscr{U}(P^{343}_2)$ are contractible subspaces of $P^{343}_2$, then the map $\mathscr{U}(f)$ does not have fixed points by \ref{prop_pf_con_ppf_implica_f_pf_general}. By lemma \ref{lema_para_Q7}, $\mathscr{U}(f)(\{A,B,C\})\subseteq \{A,B,C\}$. In a similar way, $\mathscr{F}(f)(\{A',B',C'\})\subseteq \{A',B',C'\}$. Consider the subspace $M = \{A,B,C,A',B',C'\}\subseteq \mathscr{C}(P^{343}_2)$. It follows that $\mathscr{C}(f)(M)\subseteq M$. Since $M$ is contractible, the map $\mathscr{C}(f)$ has fixed points and then, from \ref{prop_pf_con_ppf_implica_f_pf_general} we obtain that $f$ has fixed points, which entails a contradiction.

Therefore, $P^{343}_2$ has the fixed point property.
\end{ex}

\begin{ex}
Let $n,k\in\mathbb{N}$ such that $n\geq 4$ and $2\leq k \leq n-1$. Let $X_{n,k}$ the finite $T_0$--space whose underlying set is $\{a_1,a_2,a_3,b_1,b_2,\ldots,b_n,c_1,c_2,\ldots,c_n\}$ and whose minimal open sets are
\begin{align*}
U_{a_1}&=\{a_1\}\cup \{b_j\mid 1\leq j\leq n-1\}\cup \{c_j\mid 1\leq j\leq n\},\\
U_{a_2}&=\{a_2\}\cup \{b_j\mid 1\leq j\leq n \land j\neq n-1 \}\cup \{c_j\mid 1\leq j\leq n\},\\
U_{a_3}&=\{a_3\}\cup \{b_j\mid k\leq j\leq n\}\cup \{c_j\mid k-1\leq j\leq n\},\\
U_{b_1}&=\{b_1,c_1,c_2\},\\
U_{b_j}&=\{b_j,c_{j-1},c_{j+1}\}, \textup{ for $2\leq j\leq n-1$},\\
U_{b_n}&=\{b_n,c_{n-1},c_n\},\\
U_{c_j}&=\{c_j\}, \textup{ for $1\leq j\leq n$}.
\end{align*}

The Hasse diagram of $X_{n,k}$ is
\begin{center}
\begin{tikzpicture}[y=1.5cm]
\tikzstyle{every node}=[font=\scriptsize]
\foreach \x in {1,2} \draw (\x,0) node(c\x)[inner sep=2pt]{$\bullet$} node[below=1]{$c_\x$};
\foreach \x in {3,5} \draw (\x,0) node(c\x)[inner sep=2pt]{$\cdots$};
\foreach \x in {4} \draw (\x,0) node(c\x)[inner sep=2pt]{$\bullet$} node[below=1]{$c_k$};
\foreach \x in {6} \draw (\x,0) node(c\x)[inner sep=2pt]{$\bullet$} node[below=1]{$c_{n-2}$};
\foreach \x in {7} \draw (\x,0) node(c\x)[inner sep=2pt]{$\bullet$} node[below=1]{$c_{n-1}$};
\foreach \x in {8} \draw (\x,0) node(c\x)[inner sep=2pt]{$\bullet$} node[below=1]{$c_n$};

\foreach \x in {1,2} \draw (\x,1) node(b\x)[inner sep=2pt]{$\bullet$} node[right=1]{$b_\x$};
\foreach \x in {3,5} \draw (\x,1) node(b\x)[inner sep=2pt]{$\cdots$};
\foreach \x in {4} \draw (\x,1) node(b\x)[inner sep=2pt]{$\bullet$} node[right=1]{$b_k$};
\foreach \x in {6} \draw (\x,1) node(b\x)[inner sep=2pt]{$\bullet$} node[right=1]{$b_{n-2}$};
\foreach \x in {7} \draw (\x,1) node(b\x)[inner sep=2pt]{$\bullet$} node[right=1]{$b_{n-1}$};
\foreach \x in {8} \draw (\x,1) node(b\x)[inner sep=2pt]{$\bullet$} node[right=1]{$b_n$};

\foreach \x in {1,2,3} \draw (1.75*\x+1,2) node(a\x)[inner sep=2pt]{$\bullet$} node[above=1]{$a_\x$};

\foreach \x in {1,2,4,6,7} \draw (a1)--(b\x);
\foreach \x in {1,2,4,6,8} \draw (a2)--(b\x);
\foreach \x in {4,6,7,8} \draw (a3)--(b\x);
\foreach \x in {1,2} \draw (b1)--(c\x);
\foreach \x in {1} \draw (b2)--(c\x);
\draw[shorten >=1cm,shorten <=0cm] (b2)--(c3);
\draw[shorten >=1cm,shorten <=0cm] (c2)--(b3);
\foreach \x in {3,5} \draw[shorten >=1cm,shorten <=0cm] (b4)--(c\x);
\foreach \x in {3,5} \draw[shorten >=1cm,shorten <=0cm] (c4)--(b\x);
\draw[shorten >=1cm,shorten <=0cm] (b6)--(c5);
\draw[shorten >=1cm,shorten <=0cm] (c6)--(b5);
\foreach \x in {7} \draw (b6)--(c\x);
\foreach \x in {6,8} \draw (b7)--(c\x);
\foreach \x in {7,8} \draw (b8)--(c\x);
\end{tikzpicture}
\end{center}

Observe that the posets $X_{4,3}$ and $X_{4,2}$ are isomorphic to the posets $P^{443}_4$ and $P^{443}_5$ of \cite[Fig.1]{Schroder1993fpp11}.

We will prove that the spaces $X_{n,k}$ have the fixed point property for all $n,k$ as above. To this end we will compute $\mathscr{U}(X_{n,k})$. Let
\begin{align*}
A &=\{b_j\mid 1\leq j \leq n-2\} \cup \{c_j\mid 1\leq j \leq n-1\}, \\
B &=\{b_j \mid j\geq k \ \land\ j\not\equiv n \modu{2} \} \cup \{c_j \mid j\geq k-1 \ \land\ j\equiv n \modu{2} \}, \\
C &=\{b_j \mid j\geq k \ \land\ j\equiv n \modu{2} \} \cup \{c_j \mid j\geq k-1 \ \land\ j\not\equiv n \modu{2} \}\cup\{c_n\}, \\
D &=\{b_j \mid k\leq j\geq n-3 \ \land\ j\not\equiv n \modu{2} \} \cup \{c_j \mid k-1\leq j \leq n-2 \ \land\ j\equiv n \modu{2} \}, \\
E &=\{b_j \mid k\leq j\leq n-2 \ \land\ j\equiv n \modu{2} \} \cup \{c_j \mid k-1\leq j\leq n-1 \ \land\ j\not\equiv n \modu{2} \}, \\
F &= \{c_n\}, \textnormal{ and } \\
S &= X_{n,k}-\mxl[X_{n,k}].
\end{align*}
Observe that the connected components of $U_{a_1a_2}$ are $A$ and $F$, the connected components of $U_{a_1a_3}$ are $B$ and $E$, the connected components of $U_{a_2a_3}$ are $C$ and $D$ and that the connected components of $U_{a_1a_2a_3}$ are $D$, $E$ and $F$. Hence $\mathscr{U}(X_{n,k})$ is the finite space given by the following Hasse diagram
\begin{center}
\begin{tikzpicture}[x=1cm, y=1.2cm]
\tikzstyle{every node}=[font=\scriptsize]
\draw (0,2) node (Ua1){$\bullet$} node[above=1]{$U_{a_1}$};
\draw (1,2) node (Ua2){$\bullet$} node[above=1]{$U_{a_2}$};
\draw (2,2) node (Ua3){$\bullet$} node[above=1]{$U_{a_3}$};

\draw (0,1) node (A){$\bullet$} node[right=1]{$A$};
\draw (1,1) node (B){$\bullet$} node[right=1]{$B$};
\draw (2,1) node (C){$\bullet$} node[right=1]{$C$};

\draw (0,0) node (D){$\bullet$} node[below=1]{$D$};
\draw (1,0) node (E){$\bullet$} node[below=1]{$E$};
\draw (2,0) node (F){$\bullet$} node[below=1]{$F$};

\draw [-] (Ua1) to (B) to (Ua3) to (C) to (Ua2) to (A) to (Ua1);
\draw [-] (A) to (E) to (C) to (F) to (B) to (D) to (A);
\end{tikzpicture}
\end{center}

Suppose that $f\colon X_{n,k}\to X_{n,k}$ is a continuous map without fixed points. Since all the elements of $\mathscr{U}(X_{n,k})$ are contractible subspaces of $X_{n,k}$, then $\mathscr{U}(f)$ does not have fixed points by \ref{prop_pf_con_ppf_implica_f_pf_general}. Thus, by \ref{lema_para_Q4_Q5}, $\mathscr{U}(f)(\{A,B,C\})= \{A,B,C\}$. Since $S=A\cup B\cup C$, we have that $f(S)=f(A\cup B\cup C) = f(A)\cup f(B)\cup f(C) \subseteq A\cup B\cup C = S$.

We will prove now that the restriction $f|\colon S\to S$ is not bijective. Note that
\begin{align*}
\#A = 2n-3 > n-k +2 \geq \#B
\end{align*}
where the first inequality holds since $n+k >5$. Thus, the sets $A$, $B$ and $C$ do not all have the same cardinality and since $\mathscr{U}(f)$ does not have fixed points and $\mathscr{U}(f)(\{A,B,C\})= \{A,B,C\}$ it follows that there exists $T\in \{A,B,C\}$ such that $\# \mathscr{U}(f)(T)< \# T$. Hence $\# f(T)< \# T$ and then $f$ is not bijective.

Finally, since $f|\colon S\to S$ is not bijective, it has a fixed point by \ref{prop_crown} which entails a contradiction.
\end{ex}

\section{Beat points}

\begin{rem}
Note that the space $X$ of example \ref{ex_2} is not contractible but $\mathscr{U}(X)$ is. Thus $X$ and $\mathscr{U}(X)$ do not have the same homotopy type.
\end{rem}

\begin{prop} \label{prop_C_cap_A}
Let $X$ be a finite $T_0$--space and let $A$ be a bp-retract of $X$. Let $r\colon X\to A$ be the corresponding retraction.
\begin{enumerate}[(a)]
\item If $C\in\mathscr{C}(X)$, then $r(C)=C\cap A$.
\item If $\mxl[X] \subseteq A$ and $C\in\mathscr{U}(X)$, then $C\cap A\in\mathscr{U}(A)$.
\item If $\mnl[X] \subseteq A$ and $C\in\mathscr{F}(X)$, then $C\cap A\in\mathscr{F}(A)$.
\end{enumerate}
\end{prop}

\begin{proof}\

($a$) Suppose that $A$ is a dbp-retract of $X$. Clearly, $C\cap A \subseteq r(C)$ and $r(C)\subseteq A$. It remains to prove that $r(C)\subseteq C$.

Suppose first that $C\in\mathscr{U}(X)$. Let $z\in C$. Since $r(z)\leq z$ and $C$ is an open subset of $X$ it follows that $r(z)\in C$.

Suppose now that $C\in\mathscr{F}(X)$. Let $z\in C$. Note that $z\geq y$ for all $y\in\lb C$. Then $r(z)\geq r(y)=y$ for all $y\in\lb C$, that is, $r(z)\in F_{\lb C}$. Since $r(z)\leq z$ it follows that $C\cup\{r(z)\}$ is a connected subspace of $F_{\lb C}$. Thus $C\cup\{r(z)\}\subseteq C$ and hence $r(z)\in C$.

($b$) Note that $\mxl[A]=\mxl[X]$ and that $C\cap A$ is connected and non-empty by item ($a$). Since $C\in\mathscr{U}(X)$, $C$ is a connected component of $U^X_{\ub C}$. Then $C\cap A\subseteq U^X_{\ub C}\cap A=U^{A}_{\ub C}$.

Thus, there exists a connected component $C_0$ of $U^{A}_{\ub C}$ such that $C\cap A\subseteq C_0$. Then $C_0 \subseteq U^{X}_{\ub C}$ and since $\varnothing\neq C\cap A\subseteq C\cap C_0$ and $C$ is a connected component of $U^X_{\ub C}$ we obtain that $C_0\subseteq C$.
Hence, $C\cap A\subseteq C_0 \subseteq C\cap A$. Therefore $C\cap A= C_0 \in \mathscr{U}(A)$.

($c$) Analogous to that of item ($b$).
\end{proof}

\begin{prop} \label{prop_U_beat_points}
Let $X$ be a finite $T_0$--space and let $A$ be a bp-retract of $X$. Let $r\colon X\to A$ be the corresponding retraction and let $i\colon A \to X$ be the inclusion map.
\begin{enumerate}[(a)]
\item If $\mxl[X]\subseteq A$ then $\mathscr{U}(r)(C)=r(C)$ for all $C\in\mathscr{U}(X)$ and $\mathscr{U}(r)\mathscr{U}(i)=\textup{id}_{\mathscr{U}(A)}$.
\item If $A$ is an ubp-retract of $X$ then $\mathscr{U}(r)$ is a homeomorphism.
\item If $\mxl[X] \subseteq A$ and $A$ is a dbp-retract of $X$ then $\mathscr{U}(i)\mathscr{U}(r)\leq \textup{id}_{\mathscr{U}(X)}$.
\end{enumerate}
\end{prop}

\begin{proof}\

($a$) Let $C\in\mathscr{U}(X)$. Then $r(C)=C\cap A\in \mathscr{U}(A)$ by \ref{prop_C_cap_A}. It follows that $\mathscr{U}(r)(C)=r(C)$.

Let $D\in \mathscr{U}(A)$. Then
\begin{align*}
D &= ri(D)\subseteq r(\mathscr{U}(i)(D)) \subseteq \mathscr{U}(r)(\mathscr{U}(i)(D)) = \mathscr{U}(i)(D)\cap A \subseteq U^X_{\ub D} \cap A = U^{A}_{\ub D}
\end{align*}
by \ref{rem_def_de_Uf_and_Ff}. Since $D$ is a connected component of $U^{A}_{\ub D}$ by \ref{coro_cc_of_U_Bsharp_F_Bflat} and $\mathscr{U}(r)\mathscr{U}(i)(D)$ is connected, it follows that $\mathscr{U}(r)\mathscr{U}(i)(D)=D$.

($b$) Since $A$ is an ubp-retract then $\mxl[X]\subseteq A$ and hence $\mathscr{U}(r)\mathscr{U}(i)=\textup{id}_{\mathscr{U}(A)}$ by item ($a$). Let $C\in\mathscr{U}(X)$. Then
\begin{align*}
\mathscr{U}(i)\mathscr{U}(r)(C)=\mathscr{U}(i)(r(C))=\mathscr{U}(ir)(C).
\end{align*}
Note that $ir(C)=C\cap A$ and since $C\in\mathscr{U}(X)$ it follows that $\mathscr{U}(ir)(C) \subseteq C$.

Let $z\in C$. Since $r(z)\geq z$, $r(z)\in ir(C)\subseteq \mathscr{U}(ir)(C)$ and $\mathscr{U}(ir)(C)$ is an open subset of $X$, we obtain that $z\in\mathscr{U}(ir)(C)$. Hence, $\mathscr{U}(ir)(C) = C$.

Therefore, $\mathscr{U}(i)\mathscr{U}(r)(C)=C$.

($c$) Let $C\in\mathscr{U}(X)$. Then
\begin{align*}
\mathscr{U}(i)\mathscr{U}(r)(C)=\mathscr{U}(i)(r(C))=\mathscr{U}(ir)(C) \leq \mathscr{U}(\textup{id}_X)(C)=C
\end{align*}
by \ref{prop_f_leq_g_Uf_leq_Ug}.
\end{proof}

Clearly, an analogous result holds for $\mathscr{F}$.

\begin{prop} \label{prop_F_beat_points}
Let $X$ be a finite $T_0$--space and let $A$ be a bp-retract of $X$. Let $r\colon X\to A$ be the corresponding retraction and let $i\colon A \to X$ be the inclusion map.
\begin{enumerate}[(a)]
\item If $\mnl[X]\subseteq A$ then $\mathscr{F}(r)(C)=r(C)$ for all $C\in\mathscr{F}(X)$ and $\mathscr{F}(r)\mathscr{F}(i)=\textup{id}_{\mathscr{F}(A)}$.
\item If $A$ is a dbp-retract of $X$ then $\mathscr{F}(r)$ is a homeomorphism.
\item If $\mnl[X]\subseteq A$ and $A$ is an ubp-retract of $X$ then $\mathscr{F}(i)\mathscr{F}(r)\geq \textup{id}_{\mathscr{F}(X)}$.
\end{enumerate}
\end{prop}

\begin{proof}
Similar to the proof of \ref{prop_U_beat_points}.
\end{proof}

\begin{rem}
Let $X$ be a finite $T_0$--space and let $A$ be a dbp-retract of $X$ such that $\mxl[X]\subseteq A$. Let $r$ and $i$ be as in \ref{prop_U_beat_points}. Then $\mathscr{U}(r)\mathscr{U}(i)=\textup{id}_{\mathscr{U}(A)}$ and $\mathscr{U}(i)\mathscr{U}(r)\leq \textup{id}_{\mathscr{U}(X)}$. Thus, $\mathscr{U}(i)$ is an embedding, and applying \ref{theo_dbpr_equivalences} it follows that $\mathscr{U}(i)(\mathscr{U}(A))$ is a dbp-retract of $\mathscr{U}(X)$, that is, $\mathscr{U}(i)(\mathscr{U}(A))$ can be obtained from $\mathscr{U}(X)$ by removing down beat points. Observe that $\mathscr{U}(i)(\mathscr{U}(A))$ is homeomorphic to $\mathscr{U}(A)$.
\end{rem}

The following examples show that several of the hypotheses of the previous propositions are necessary.

\begin{ex}
Let $X$ be the finite $T_0$--space given in figure \ref{fig_ex_3-X}. Note that 4 is a down beat point of $X$. The spaces $\mathscr{U}(X)$, $X-\{4\}$ and $\mathscr{U}(X-\{4\})$ are given in figures \ref{fig_ex_3-UX}, \ref{fig_ex_3b-X} and \ref{fig_ex_3b-UX}, respectively.

Observe that $\{0,2,4\}\in \mathscr{U}(X)$ but $\{0,2\}\notin \mathscr{U}(X-\{4\})$. Thus, \ref{prop_C_cap_A} might not hold if $\mxl[X]\not\subseteq A$.

Let $i\colon X-\{4\}\to X$ be the inclusion map and let $r\colon X\to X-\{4\}$ be the retraction. Note that $\mathscr{U}(r)(\{0,2,4\})=\{0,1,2,3\}\neq r(\{0,2,4\})$ and that $\mathscr{U}(i)\mathscr{U}(r)(\{0,2,4\})=\{0,1,2,3\}\not\leq\{0,2,4\}$. Hence, items ($a$) and ($c$) of proposition \ref{prop_U_beat_points} might not hold if $\mxl[X]\not\subseteq A$.

\begin{figure}
\begin{subfigure}[c]{0.24\textwidth}
\centering
\begin{tikzpicture}
\tikzstyle{every node}=[font=\scriptsize]
\draw (0.5,0) node(0){$\bullet$} node[below=1]{0};
\foreach \x in {1,2} \draw (\x-1,1) node(\x){$\bullet$} node[left=1]{\x};
\foreach \x in {3,4} \draw (\x-3,2) node(\x){$\bullet$} node[left=1]{\x};

\foreach \x in {1,2} \draw (0)--(\x);
\foreach \x in {3} \draw (1)--(\x);
\foreach \x in {3,4} \draw (2)--(\x);
\end{tikzpicture}
\caption{$X$}
\label{fig_ex_3-X}
\end{subfigure}
\begin{subfigure}[c]{0.24\textwidth}
\centering
\begin{tikzpicture}[x=2.1cm, y=1.7cm]
\path node (U3) at (0,1) [CC] {3\\1 2\\0}
 node (U4) at (1,1) [CC] {4\\2\\0}
 node (U34) at (0.5,0) [CC] {2\\0};

\draw [-] (U34) to (U3);
\draw [-] (U34) to (U4);
\end{tikzpicture}
\caption{$\mathscr{U}(X)$}
\label{fig_ex_3-UX}
\end{subfigure}
\begin{subfigure}[c]{0.24\textwidth}
\centering
\begin{tikzpicture}
\tikzstyle{every node}=[font=\scriptsize]
\draw (0.5,0) node(0){$\bullet$} node[below=1]{0};
\foreach \x in {1,2} \draw (\x-1,1) node(\x){$\bullet$} node[left=1]{\x};
\foreach \x in {3} \draw (\x-3,2) node(\x){$\bullet$} node[left=1]{\x};

\foreach \x in {1,2} \draw (0)--(\x);
\foreach \x in {3} \draw (1)--(\x);
\foreach \x in {3} \draw (2)--(\x);
\end{tikzpicture}
\caption{$X-\{4\}$}
\label{fig_ex_3b-X}
\end{subfigure}
\begin{subfigure}[c]{0.22\textwidth}
\centering
\begin{tikzpicture}[x=2.3cm, y=2cm]
\path node (U3) at (0,1) [CC] {3\\1 2\\0};

\end{tikzpicture}
\caption{$\mathscr{U}(X-\{4\})$}
\label{fig_ex_3b-UX}
\end{subfigure}
\caption{ }
\end{figure}
\end{ex}

\begin{ex}
Let $X$ be the finite $T_0$--space given in figure \ref{fig_ex_4-X}. Note that 1 is a down beat point of $X$. The spaces $\mathscr{U}(X)$, $X-\{1\}$ and $\mathscr{U}(X-\{1\})$ are given in figures \ref{fig_ex_4-UX}, \ref{fig_ex_4b-X} and \ref{fig_ex_4b-UX}, respectively.

Note that $\mathscr{U}(X)$ and $\mathscr{U}(X-\{1\})$ are not homeomorphic. Hence, item ($b$) of \ref{prop_U_beat_points} might not hold if $A$ is a dbp-retract of $X$.
\begin{figure}
\begin{subfigure}[c]{0.24\textwidth}
\centering
\begin{tikzpicture}
\tikzstyle{every node}=[font=\scriptsize]
\draw (1,0) node(0){$\bullet$} node[below=1]{0};
\draw (1.5,1) node(1){$\bullet$} node[left=1]{1};
\foreach \x in {2,3,4} \draw (\x-2,2) node(\x){$\bullet$} node[left=1]{\x};

\foreach \x in {1,2} \draw (0)--(\x);
\foreach \x in {3,4} \draw (1)--(\x);
\end{tikzpicture}
\caption{$X$}
\label{fig_ex_4-X}
\end{subfigure}
\begin{subfigure}[c]{0.28\textwidth}
\centering
\begin{tikzpicture}[x=1.5cm, y=1.3cm]
\path node (U2) at (0,2) [CC] {2\\0}
node (U3) at (1,2) [CC] {3\\1\\0}
node (U4) at (2,2) [CC] {4\\1\\0}
node (U34) at (1.5,1) [CC] {1\\0}
node (U234) at (1,0) [CC] {0};

\draw [-] (U34) to (U3);
\draw [-] (U34) to (U4);
\draw [-] (U234) to (U2);
\draw [-] (U234) to (U34);
\end{tikzpicture}
\caption{$\mathscr{U}(X)$}
\label{fig_ex_4-UX}
\end{subfigure}
\begin{subfigure}[c]{0.22\textwidth}
\centering
\begin{tikzpicture}[x=0.9cm, y=0.7cm]
\tikzstyle{every node}=[font=\scriptsize]
\draw (1,0) node(0){$\bullet$} node[below=1]{0};
\foreach \x in {2,3,4} \draw (\x-2,2) node(\x){$\bullet$} node[left=1]{\x};

\foreach \x in {2,3,4} \draw (0)--(\x);
\end{tikzpicture}
\caption{$X-\{1\}$}
\label{fig_ex_4b-X}
\end{subfigure}
\begin{subfigure}[c]{0.22\textwidth}
\centering
\begin{tikzpicture}[x=1.1cm, y=0.8cm]
\path node (U2) at (0,2) [CC] {2\\0}
node (U3) at (1,2) [CC] {3\\0}
node (U4) at (2,2) [CC] {4\\0}
node (U234) at (1,0) [CC] {0};

\draw [-] (U234) to (U2);
\draw [-] (U234) to (U3);
\draw [-] (U234) to (U4);
\end{tikzpicture}
\caption{$\mathscr{U}(X-\{1\})$}
\label{fig_ex_4b-UX}
\end{subfigure}
\caption{ }
\end{figure}
\end{ex}

\begin{prop}
Let $X$ be a finite $T_0$--space and let $a\in\mxl[X]$ be a down beat point of $X$. Let $b=\max \widehat U_a$.
\begin{enumerate}[(a)]
\item If $b\notin\mxl[X-\{a\}]$ then $U_a$ is a down beat point of $\mathscr{U}(X)$ and $\mathscr{U}(X-\{a\})$ is a ubp-retract of $\mathscr{U}(X)-\{U_a\}$.
\item If $b\in\mxl[X-\{a\}]$ then $\mathscr{U}(X)$ and $\mathscr{U}(X-\{a\})$ are homeomorphic.
\end{enumerate}
\end{prop}

\begin{proof}\

($a$) Since $b\notin\mxl[X-\{a\}]$ then there exists $d\in\mxl[X]$ such that $d>b$ and $d\neq a$. Hence $U_b=U_{ad}$ and thus $U_b\in \mathscr{U}(X)$. Clearly $U_b < U_a$. We will prove that $U_b = \max \widehat U^{\mathscr{U}(X)}_{U_a}$. Let $C\in \mathscr{U}(X)$ be such that $C<U_a$. Clearly $a\notin C$ (since otherwise $U_a\subseteq C$). Thus, $C\subseteq \widehat U_a = U_b$. Therefore, $U_a$ is a down beat point of $\mathscr{U}(X)$.

Let $i\colon X-\{a\}\to X$ be the inclusion map and let $r\colon X\to X-\{a\}$ be the retraction which corresponds to the removal of the beat point $a$. Note that $\mathscr{U}(X-\{a\})\subseteq \mathscr{U}(X)-\{U_a\}$, and hence $\mathscr{U}(i)(D)=D$ for all $D\in\mathscr{U}(X-\{a\})$. Then $\mathscr{U}(i)\colon \mathscr{U}(X-\{a\})\to \mathscr{U}(X)$ is the inclusion map.

Let $D\in\mathscr{U}(X-\{a\})$. Note that $\mathscr{U}(r)\mathscr{U}(i)(D)=\mathscr{U}(r)(D)=D$ since $a\notin D$. Hence $\mathscr{U}(r)\mathscr{U}(i)=\textup{id}_{\mathscr{U}(X-\{a\})}$.

Let $C\in \mathscr{U}(X)-\{U_a\}$. Applying \ref{prop_C_cap_A} we obtain that $C=C-\{a\}=r(C)\subseteq\mathscr{U}(r)(C)=\mathscr{U}(i)\mathscr{U}(r)(C)$. From
\ref{theo_dbpr_equivalences} it follows that $\mathscr{U}(X-\{a\})$ is a ubp-retract of $\mathscr{U}(X)-\{U_a\}$.

($b$) Since $b\in\mxl[X-\{a\}]$, $b$ is an up beat point of $X$. Hence $\mathscr{U}(X)$ is homeomorphic to $\mathscr{U}(X-\{b\})$ by \ref{prop_U_beat_points}, which is homeomorphic to $\mathscr{U}(X-\{a\})$ since $X-\{b\}$ and $X-\{a\}$ are homeomorphic.
\end{proof}

\begin{prop}
Let $X$ be a finite $T_0$--space. Then $\mathscr{U}(X)$ does not have up beat points and $\mathscr{F}(X)$ does not have down beat points.
\end{prop}

\begin{proof}
Suppose that $C$ is an up beat point of $\mathscr{U}(X)$. Let $D=\min \widehat F^{\mathscr{U}(X)}_C$. Then
\begin{displaymath}
\mxl[\mathscr{U}(X)]\cap F^{\mathscr{U}(X)}_D = \mxl[\mathscr{U}(X)]\cap \widehat F^{\mathscr{U}(X)}_C = \mxl[\mathscr{U}(X)]\cap F^{\mathscr{U}(X)}_C
\end{displaymath}
Thus, by \ref{prop_mnl_mxl_de_C(X)},
\begin{align*}
\ub C &= \{a \in \mxl[X] \mid  U_a\in F^{\mathscr{U}(X)}_C\}  = \{a \in \mxl[X] \mid  U_a\in F^{\mathscr{U}(X)}_D\} = \ub D
\end{align*}
Then, by \ref{coro_cc_of_U_Bsharp_F_Bflat}, $C$ and $D$ are connected components of $U_{\ub C}$ and since $C\subseteq D$, it follows that $C=D$, which entails a contradiction.

The second part of the proposition can be proved in a similar way.
\end{proof}

\section{Properties of $\mathscr{U}(X)$}

\begin{prop}\label{prop_Xprime_ubpr}
Let $X$ be a finite $T_0$--space such that every $C\in \mathscr{U}(X)$ has a maximum element. Let $X'=\{\max C \mid C\in \mathscr{U}(X)\}$. Then $X'$ is an ubp-retract of $X$.
\end{prop}

\begin{proof}
Let $i\colon X'\to X$ be the inclusion map. For each $x\in X$ let $C(x)=\min \{C\in \mathscr{U}(X) \mid x\in C \}$. Let $r\colon X \to X'$ be defined by $r(x)=\max C(x)$.

We claim that $r$ is continuous. Indeed, let $x_1,x_2\in X$ such that $x_1\leq x_2$. Since $C(x_2)$ is an open subset of $X$ we obtain that $x_1\in C(x_2)$ and thus $C(x_1)\subseteq C(x_2)$. Hence, $r(x_1)\leq r(x_2)$.

Clearly, $ir\geq \textup{id}_X$. We will prove now that $ri=\textup{id}_{X'}$. Let $z\in X'$. Then there exists $C\in \mathscr{U}(X)$ such that $z=\max C$. Since $C(z)$ is an open subset of $X$ we obtain that $U_z\subseteq C(z) \subseteq C \subseteq U_z$. It follows that $ri(z)=z$.

Therefore, $X'$ is an ubp-retract of $X$ by \ref{theo_dbpr_equivalences}.
\end{proof}

\begin{rem}
Observe that, under the hypotheses of the previous proposition, from \ref{prop_U_beat_points} and \ref{prop_Xprime_ubpr} we obtain that $\mathscr{U}(X')\cong \mathscr{U}(X)$.
\end{rem}

\begin{prop} \label{prop_UX_iso_a_Xprime}
Let $X$ be a finite $T_0$--space such that every $C\in \mathscr{U}(X)$ has a maximum element. Let $X'=\{\max C \mid C\in \mathscr{U}(X)\}$.
Then there exists mutually inverse isomorphisms $\varphi\colon \mathscr{U}(X) \to X'$ and $\psi \colon X' \to \mathscr{U}(X)$ given by $\varphi(C)=\max C$ and $\psi(z)=U^X_z$.
\end{prop}

\begin{proof}
Observe that if $C\in \mathscr{U}(X)$ and $\max C=z$ then, as in the proof of \ref{prop_Xprime_ubpr}, we obtain that $C=U^X_z$. The result follows.
\end{proof}

\begin{lemma} \label{lemma_CC_of_UX}
Let $X$ be a finite $T_0$--space. Let $B\subseteq \mxl[\mathscr{U}(X)]$ be a non-empty subset. Let $A=\{a\in\mxl[X]\mid U^X_a\in B\}$. Then
\begin{enumerate}[(a)]
\item $\mxl[U^{\mathscr{U}(X)}_B]=\mathcal{C}(U^X_A)$, and
\item $\mathcal{C}(U^{\mathscr{U}(X)}_B)=\{U_M\mid M\in\mathcal{C}(U^X_A)\}$.
\end{enumerate}
\end{lemma}

\begin{proof}\

($a$) First note that $B=\{U^X_a \mid a \in A\}$ by \ref{prop_mnl_mxl_de_C(X)}.

Let $D\in \mxl[U^{\mathscr{U}(X)}_B]$. Then $D\subseteq U^X_A$. Since $D$ is connected, there exists a connected component $D_0\in\mathcal{C}(U^X_A)$ such that $D\subseteq D_0$. It follows that $D_0\in U^{\mathscr{U}(X)}_B$. From the maximality of $D$ we obtain that $D=D_0$ and hence $D\in \mathcal{C}(U^X_A)$.

Now, let $D\in \mathcal{C}(U^X_A)$. Then, $D\in U^{\mathscr{U}(X)}_B$. Let $C\in U^{\mathscr{U}(X)}_B$ such that $D\subseteq C$. Then $C\subseteq U^X_A$, and since $C$ is connected it follows that $C\subseteq D$. Thus, $C=D$. Therefore, $D\in \mxl[U^{\mathscr{U}(X)}_B]$.

($b$) From ($a$) we obtain that
\begin{align*}
\displaystyle U^{\mathscr{U}(X)}_B=\bigcup_{M\in \mathcal{C}(U^X_A)} U^{\mathscr{U}(X)}_M.
\end{align*}
Now, observe that if $M_1,M_2\in \mathcal{C}(U^X_A)$ are such that $M_1\neq M_2$ then $U^{\mathscr{U}(X)}_{M_1}\cap U^{\mathscr{U}(X)}_{M_2}=\varnothing$ since $M_1\cap M_2 = \varnothing$. The result follows.
\end{proof}

\begin{prop} \label{prop_UUX_isomorphic_to_UX}
Let $X$ be a finite $T_0$--space. Then there exists mutually inverse isomorphisms $\varphi\colon \mathscr{U}(\mathscr{U}(X)) \to \mathscr{U}(X)$ and $\psi \colon \mathscr{U}(X) \to \mathscr{U}(\mathscr{U}(X))$ given by $\varphi(D)=\max D$ and $\psi(C)=U^{\mathscr{U}(X)}_C$.
\end{prop}

\begin{proof}
First observe that every $C\in \mathscr{U}(\mathscr{U}(X))$ has a maximum element by item ($b$) of \ref{lemma_CC_of_UX}. Now, if $C\in\mathscr{U}(X)$ then $C$ is a connected component of $U^X_{\ub C}$ and hence $U_C$ is a connected component of $U^{\mathscr{U}(X)}_{\{U_a\mid a \in \ub C\}}$ by item ($b$) of \ref{lemma_CC_of_UX}. Therefore, $\{\max D \mid D\in \mathscr{U}(\mathscr{U}(X))\}=\mathscr{U}(X)$. The result follows from \ref{prop_UX_iso_a_Xprime}.
\end{proof}

\begin{coro}
Let $X$ be a finite $T_0$--space. Then there exists a finite $T_0$--space $Y$ such that $X\cong \mathscr{U}(Y)$ if and only if $X\cong \mathscr{U}(X)$.
\end{coro}

\begin{proof}
Follows easily from \ref{prop_Uf_homeo} and \ref{prop_UUX_isomorphic_to_UX}.
\end{proof}

\section{Other properties of the $\mathscr{C}$--construction}

In this section we will develop more results about the $\mathscr{C}$--construction. Not all of them are related to the fixed point property but are interesting on their own and permit a better understanding of this construction. In addition, these results show that the $\mathscr{C}$--construction is an interesting object of study.

In the following proposition we will see that if $X$ is a finite $T_0$--space, then $X$ is homotopy equivalent to the Grothendieck construction on a functor from $\mathscr{U}(X)$ to the category of finite posets, or equivalently to the the non-Hausdorff homotopy colimit of a $\mathscr{U}(X)$--diagram of finite posets \cite{fernandez2016homotopy}.

Recall that if $A$ is a finite poset, $\mathcal{P}$ is the category of finite posets and $F\colon A\to \mathcal{P}$ a functor, then the Grothendieck construction for $F$ is the poset $\int F$ with underlying set $\bigcup\limits_{a\in A} (\{a\}\times F(a))$ and with partial order defined as follows: if $a,b\in A$, $x\in F(a)$, $y\in F(b)$ then $(a,x)\leq (b,y)$ if and only if $a\leq b$ and $F(a\leq b)(x)\leq y$.

\begin{definition}
Let $X$ be a finite $T_0$--space. We define the functor $\mathcal{I}_X\colon \mathscr{U}(X)\to \mathcal{P}$ by $\mathcal{I}_X(C)=C$ and $\mathcal{I}_X(C\subseteq C')$ as the inclusion map $C\hookrightarrow C'$.

We also define maps $\rho_X\colon \int \mathcal{I}_X \to X$ and $\iota_X\colon X\to \int \mathcal{I}_X$ by $\rho_X(C,x) = x$ for all $(C,x)\in \int \mathcal{I}_X$ and $\iota(x)=({\bf C}_\U(x), x)$ for all $x\in X$.
\end{definition}

\begin{prop} \label{prop_X_dbpr_hocolim}
Let $X$ be a finite $T_0$--space. Then, the maps $\rho_X\colon \int \mathcal{I}_X \to X$ and $\iota_X\colon X\to \int I_X$ are continuous, $\rho_X \iota_X=\textup{id}_X$ and $\iota_X\rho_X \leq\textup{id}_{\int \mathcal{I}_X}$.

In particular, $\iota_X(X)$ is a dbp--retract of $\int \mathcal{I}_X$ and $X$ is homotopy equivalent to $\int \mathcal{I}_X$.
\end{prop}

\begin{proof}
Let $(C_1,x_1),(C_2,x_2)\in\int \mathcal{I}_X$ such that $(C_1,x_1)\leq (C_2,x_2)$. Then $C_1\subseteq C_2$ and $x_1=\textup{inc}(x_1)\leq x_2$ in $C_2$. Hence $\rho_X(C_1,x_1) = x_1\leq x_2 = \rho_X(C_2,x_2)$ in $X$. Thus, $\rho_X$ is continuous.

Now, let $x_1,x_2\in X$ such that $x_1\leq x_2$. Since $x_2\in {\bf C}_\U(x_2)$ and ${\bf C}_\U(x_2)$ is an open subset of $X$, then $x_1\in {\bf C}_\U(x_2)$. Hence, ${\bf C}_\U(x_1)\leq {\bf C}_\U(x_2)$ and thus $\iota_X(x_1)=({\bf C}_\U(x_1),x_1)\leq ({\bf C}_\U(x_2),x_2)=\iota(x_2)$. Therefore, $\iota_X$ is continuous.

Now, for all $x\in X$, $\rho_X \iota_X(x)=\rho_X({\bf C}_\U(x),x)=x$. Hence $\rho_X \iota_X=\textup{id}_X$.

On the other hand, if $(C,x)\in\int \mathcal{I}_X$, then $x\in C$ and $\iota_X \rho_X(C,x)=\iota_X(x)=({\bf C}_\U(x),x)\leq (C,x)$. Hence $\iota_X \rho_X \leq \textup{id}_{\int \mathcal{I}_X}$.
\end{proof}

\begin{rem}
Note that if $X$ is a finite $T_0$--space then $\mxl[\int \mathcal{I}_X]=\{(U_a,a)\mid a\in \mxl[X]\}$.
\end{rem}

\begin{lemma}
Let $X$ be a finite $T_0$--space. For each $D\in\mathscr{U}(X)$ let $\widetilde D=\{(C,x)\in \int \mathcal{I}_X \mid C \subseteq D\}$.
\begin{enumerate}[(a)]
\item Let $B\subseteq\mxl[\int \mathcal{I}_X]$. Let $A=\{a\in\mxl[X]\mid (U_a,a)\in B\}$. Then
\begin{displaymath}
\mathcal{C}(U_{B}^{\int \mathcal{I}_X})=\{\widetilde D \mid D\in\mathcal{C}(U_A^X)\}.
\end{displaymath}
\item Let $C\in\mathscr{U}(X)$. Then $\mathscr{U}(\iota_X)(C)=\widetilde C$.
\end{enumerate}
\end{lemma}

\begin{proof}\

($a$) Note that $B=\{(U_a,a)\mid a\in A\}$ by the previous remark.

We will prove first that $U_{B}^{\int \mathcal{I}_X}=\bigcup\limits_{D\in\mathcal{C}(U_A^X)}\widetilde D$.

Let $(C,x)\in U_{B}^{\int \mathcal{I}_X}$. Then $C\subseteq U_a$ for all $a\in A$. Hence, $C\subseteq U^X_A$, and since $C$ is connected, there exists $D\in \mathcal{C}(U_A^X)$ such that $C\subseteq D$. Thus, $(C,x)\in \widetilde D$.

Now, let $D\in\mathcal{C}(U_A^X)$ and let $(C,x)\in \widetilde D$. Then $D\subseteq U_a$ for all $a\in A$. Hence $(C,x) \leq (U_a,a)$ for all $a\in A$. Thus, $(C,x) \in U_{B}^{\int \mathcal{I}_X}$.

We claim that if $D_1,D_2$ are distinct elements of $\mathcal{C}(U_A^X)$ then $\widetilde D_1\cap \widetilde D_2 = \varnothing$. Indeed, if there exists $(C,x)\in \widetilde D_1\cap \widetilde D_2$, then $C\subseteq D_1\cap D_2$ which entails a contradiction since $D_1$ and $D_2$ are different connected components of $U^X_A$.

We will prove now that $\widetilde D$ is connected for all $D\in\mathcal{C}(U_A^X)$. Let $D\in\mathcal{C}(U_A^X)$. Note that
\begin{align*}
\widetilde D &= \bigcup_{x\in D} U^{\int \mathcal{I}_X}_{(D,x)}= \bigcup_{x\in D} \left( U^{\int \mathcal{I}_X}_{(D,x)} \cup \{D\}\times D \right)
\end{align*}
Now, for each $x\in D$, $U^{\int \mathcal{I}_X}_{(D,x)} \cup \{D\}\times D$ is a connected subspace of $\int \mathcal{I}_X$ since $(D,x)\in U^{\int \mathcal{I}_X}_{(D,x)} \cap \{D\}\times D$ and $U^{\int \mathcal{I}_X}_{(D,x)}$ and $\{D\}\times D$ are connected. Therefore, $\widetilde D$ is also connected.

Finally, note that $\widetilde D$ is an open subset of $\int \mathcal{I}_X$ for all $D\in \mathcal{C}(U_A^X)$. Since $\int \mathcal{I}_X$ is a finite space, the result follows.

($b$) Let $C\in\mathscr{U}(X)$. If $x\in C$ then $\iota_X(x)=({\bf C}_\U(x),x)\in \widetilde C$ since ${\bf C}_\U(x)\subseteq C$. Thus, $\iota_X(C)\subseteq\widetilde C$.

We will prove now that $\ub{\iota_X(C)} = \ub{\widetilde C}$. Clearly, $\ub{\iota_X(C)} \supseteq \ub{\widetilde C}$ since $\iota_X(C)\subseteq\widetilde C$. Let $M\in \ub{\iota_X(C)}$. Then there exists $a\in \mxl[X]$ such that $M=(U_a,a)$. Let $x\in C$. Then $\iota_X(x)=({\bf C}_\U(x),x)\leq (U_a,a)$ and hence $x\in {\bf C}_\U(x)\subseteq U_a$. Therefore, $C\subseteq U_a$. Now, let $(C_1,x_1)\in\widetilde C$. Then $x_1\in C_1\subseteq C\subseteq U_a$. Hence, $(C_1,x_1)\leq (U_a,a)=M$. Thus, $M\in \ub{\widetilde C}$.

Now, by \ref{coro_cc_of_U_Bsharp_F_Bflat}, $\widetilde C$ is a connected component of $U^{\int \mathcal{I}_X}_{\ub{\widetilde C}}$. Since $\ub{\iota_X(C)} = \ub{\widetilde C}$ and $\iota_X(C)\subseteq\widetilde C$ we obtain that $\mathscr{U}(\iota_X)(C)=\widetilde C$ by \ref{prop_conexo_minimal}.
\end{proof}

\begin{prop}
Let $X$ be a finite $T_0$--space. Then $\mathscr{U}(\int \mathcal{I}_X)\cong \mathscr{U}(X)$. Moreover, $\mathscr{U}(\iota_X)\colon \mathscr{U}(X)\to \mathscr{U}(\int \mathcal{I}_X)$ is a homeomorphism with inverse $\mathscr{U}(\rho_X)$.
\end{prop}

\begin{proof}
Let $C\in\mathscr{U}(X)$. Then $\mathscr{U}(\iota_X)(C)=\{(D,x)\in \int \mathcal{I}_X \mid D\subseteq C\}$ by the previous lemma. Thus,
\begin{align*}
\rho_X(\mathscr{U}(\iota_X)(C))=\bigcup_{\substack{D\in\mathscr{U}(X)\\ D\subseteq C}}D = C.
\end{align*}
Hence, $\mathscr{U}(\rho_X)(\mathscr{U}(\iota_X)(C))=C$. Therefore $\mathscr{U}(\rho_X)\mathscr{U}(\iota_X)=\textup{id}_{\mathscr{U}(X)}$.

Observe that $\mathscr{U}(\iota_X)$ is surjective by the previous lemma. The result follows.
\end{proof}

\begin{definition}
Let $X$ be a finite $T_0$--space. We define $q_X\colon \int \mathcal{I}_X \to \mathscr{U}(X)$ by $q_X(C,x)=C$.
\end{definition}

\begin{prop}
Let $X$ be a finite $T_0$--space.
\begin{enumerate}
\item $q_X$ is a quotient map.
\item Let $\varphi$ and $\psi$ be the isomorphisms of proposition \ref{prop_UUX_isomorphic_to_UX}. Then the following diagram commutes
\[
\begin{tikzcd}[column sep=large, row sep=large]
\mathscr{U}(\int \mathcal{I}_X) \rar{\mathscr{U}(q_X)} \drar[swap]{\mathscr{U}(\rho_X)} &\mathscr{U}(\mathscr{U}(X)) \dar[xshift=2pt]{\varphi}\\
&\mathscr{U}(X) \uar[xshift=-2pt]{\psi}
\end{tikzcd}
\]
\end{enumerate}
\end{prop}

From \ref{prop_Xprime_ubpr} and \ref{prop_UX_iso_a_Xprime} we obtain that $\mathscr{U}(X)$ and $X$ are homotopy equivalent if every $C\in \mathscr{U}(X)$ has a maximum element. The following proposition provides a version of this result for weak equivalences.

\begin{prop}
Let $X$ be a finite $T_0$--space. Suppose that every $C\in \mathscr{U}(X)$ is weakly contractible. Then there exists a weak homotopy equivalence $X\to \mathscr{U}(X)$.
\end{prop}

\begin{proof}
Let $\mathcal{J}\colon \mathscr{U}(X)\to \mathcal{P}$ be the constant functor with value $\ast$, the one-element poset. Let $\alpha\colon \mathcal{I}_X \Rightarrow \mathcal{J}$ be the unique natural transformation. It follows that $\alpha_C$ is a weak homotopy equivalence for every $C\in \mathscr{U}(X)$ and hence, from \cite[Corollary 2.6]{fernandez2016homotopy}, we obtain that there exists a weak homotopy equivalence $\eta\colon \int \mathcal{I}_X \to \int \mathcal{J}$. Clearly $\int \mathcal{J}\cong \mathscr{U}(X)$. Besides, by \ref{prop_X_dbpr_hocolim}, $X$ is homotopy equivalent to $\int \mathcal{I}_X$. The result follows.
\end{proof}

\bibliographystyle{acm}

redpajama_set_name: RedPajamaArXiv


text:
Home [english] Region Asia Thailand Number of post-coup lèse-majest
Number of post-coup lèse-majesté arrests surges to over 100
(Bangkok, Paris) The number of individuals arrested on lèse-majesté charges since the May 2014 military coup has passed the 100 mark, FIDH and its member organizations Union for Civil Liberty (UCL) and Internet Law Reform Dialogue (iLaw) said today.
"In less than three years, the military junta has generated a surge in the number of political prisoners detained under lèse-majesté by abusing a draconian law that is inconsistent with Thailand's international obligations."
Dimitris Christopoulos, FIDH President
Article 112 of Thailand's Criminal Code (lèse-majesté) imposes jail terms for those who defame, insult, or threaten the King, the Queen, the Heir to the throne, or the Regent. Persons found guilty of violating Article 112 face prison terms of three to 15 years for each count.
The number of people who have been arrested under Article 112 of the Criminal Code has reached 105, following the arrest of six individuals on 29 April 2017. Forty-nine of them have been sentenced to prison terms of up to 30 years. To date, at least 64 individuals are either imprisoned or detained awaiting trial on lèse-majesté charges. At the time of the 22 May 2014 coup, there were six individuals behind bars under Article 112. Eighty-one of the 105 cases involved deprivation of liberty for the exercise of the right to freedom of opinion and expression. The remaining cases are related to individuals who were arrested for claiming ties to the royal family for personal gain.
"Many of those arrested are democracy activists and outspoken critics of the military regime. In some instances, they were kidnapped from their homes by military officers and interrogated in secret for several days in military camps before being formally charged. Lèse-majesté defendants are rarely granted bail, and so spend months or even years fighting their cases while in detention. All of this makes a mockery of 'justice' in Thailand's justice system."
Jon Ungpakorn, iLaw Executive Director
On 28 March 2017, following the review of the country's second periodic report under the International Covenant on Civil and Political Rights (ICCPR) in Geneva, Switzerland, the UN Human Rights Committee (CCPR), expressed concern over the "extreme sentencing practices" for those found guilty of lèse-majesté. The CCPR recommended Thailand review Article 112 to bring it into line with Article 19 of the ICCPR and reiterated that the imprisonment of persons for exercising their freedom of expression violates this provision. The CCPR also demanded the authorities release those who have been deprived of their liberty for exercising their right to freedom of expression.
"The Thai government has run out of excuses to avoid reforming lèse-majesté. Article 112 must be brought into compliance with Thailand's international obligations as demanded by numerous UN mechanisms."
Jaturong Boonyarattanasoontorn, UCL Chairman
Ms. Audrey Couprie (French, English) - Tel: +33143551412 (Paris)
Member organisations - Thailand
Internet Law Reform Dialogue
9/ 46/2 Soi Inthamara 1
Samsen Nai, Phayathai
UNION FOR CIVIL LIBERTY (UCL)
109 SUTHISARNWINICHAI ROAD
SAMSENNOK HUAYKWANG
TH-10310 BANGKOK
Release long-time lèse-majesté detainee Siraphop Kornaroot, UN body urges
Reverse the legacy of five years of military rule
More shadows than lights in political parties' human rights commitments

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Did you know that the ACL has added a new innovation to its election website for the upcoming August 21 federal poll?
ACL is offering candidates for both the House of Representatives and the Senate the exciting opportunity to speak directly to the Christian constituency about the issues that matter to them via video messages hosted at australiavotes.org.au.
The candidate videos feature of Australia Votes gives candidates the ability to share with voters their views on matters of conscience and values that go beyond party partisan politics, and to discuss issues important to their local electorate.
Several candidates have already uploaded their videos to the website, and many more are on the way. Check it all out here: http://www.australiavotes.org.au/candidate-videos/. Remember to re-check the site for more videos as August 21 draws closer.
Perhaps if there is a candidate in your electorate that you'd like to hear from but who hasn't uploaded a video yet, you could encourage them to do so. The site includes simple instructions to give all candidates an equal opportunity to share their views.


text:
ISL 2019-20: Chennaiyin FC seeks first win on home return against Mumbai City
After the thrashing at the hands of FC Goa in the first game week, Chennaiyin FC will aim to register its maiden win of the season against Mumbai City.
Shyam Vasudevan
Chennai 26 October, 2019 19:01 IST
An improved display will be expected of Lucian Goian after he struggled on his Chennaiyin debut against Mumbai City.
Two-time Indian Super League (ISL) champion Chennaiyin FC will face its first home test of the season when it hosts Mumbai City FC at the Jawaharlal Nehru Stadium here on Sunday.
The home side's campaign began in bitter fashion as it slipped to a 3-0 loss at FC Goa but head coach John Gregory, in his third season with the club, is hopeful of a winning start at home. "The ISL comes quick and you have to pick up results quickly. I made three changes against FC Goa and the boys that came on did very well. I might make one or two changes against Mumbai. I can bring in two more foreigners who did not start the FC Goa game. Tomorrow (Sunday) is a good opportunity to pick up a win," he said at the pre-match conference.
READ| ISL 2019-20: Gyan header gives NorthEast win - Talking points
Chennaiyin's defence endured a troublesome evening as it struggled to contain FC Goa's attack led by Ferran Corominas, while the attackers had a couple of decent chances. Romanian striker Dragos Firtulescu had a great chance to open the scoring when he was brought down in the box in what appeared to be a clear penalty, but much to Gregory's angst the referee saw no foul.
"It was unbelievable that penalty call (Dragos). It was a clear cut penalty. It was unbelievable that the referee and the linesman did not make the decision. If that referee made that decision in Europe, he would be demoted to the fifth tier and would never referee a game again at the top level. In India, however, he will probably referee again this weekend. The standard of refereeing needs to improve," the Englishman rued.
Gregory lambasted the refereeing decision in Chennaiyin's defeat to FC Goa earlier this week. Photo: M. Vedhan
Dragos, on the other hand, is eager to score his first goal in Chennaiyin colours. "I'm an attacking midfielder and can play anywhere in midfield. I feel better in the right but I will do what the coach asks of me...even if I have to play as a goalkeeper," he said while adding that he was eager to impress and give the fans something to cheer about on the occasion of Diwali.
Mumbai City coach Jorge Costa, whose team comes on the back of a hard-fought win over Kerala Blasters, said he was wary of the home team. "We played Chennaiyin in the pre-season and I saw their first game. That's why I say we have a very difficult game tomorrow. I still haven't decided what to do (strategy), but we will fight for three points," he said.
Chennaiyin will miss Jeje Lalpekhlua due to a long-term injury and Germanpreet Singh is on personal leave. Mumbai City will be without centre-back Mato Grgic, while midfielder Paulo Machado's participation remains doubtful.
ISL 2019-20: Gyan winner hands NorthEast United first win
Carles Cuadrat: 'Manuel Onwu has the skills to match his ambitions'
ISL 2019-20: ATK thumps listless Hyderabad - Talking points
ISL 2019-2020: ATK downs Hyderabad 5-0 for its biggest win
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