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\def \HH{\text{\bsf HHeyt}}
\def \Gra{\text{\bsf Gra}}
\def \PS{\text{\bsf PSp}}
\def \SS{\text{\bsf SymGraph}_S}
\def \HP{\text{\bsf HPSp}}
\def \CH{\text{\bsf CHeyt}}
\def \Set{\text{\bsf Set}}
\def \Top{\text{\bsf Top}}
\def \Frm{\text{\bsf Frm}}
\def \Met{\text{\bsf Metr}}
\def \End{\operatorname{End}}
\def \Aut{\operatorname{Aut}}
\def \Idp{\operatorname{Idp}}
\def \Fix{\operatorname{Fix}}
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\titul
{Finite commutative monoids of open maps}

\bigskip\bigskip
\heading
A. Pultr\footnotemark"*" and J. Sichler\footnote"*"
{The authors gratefully acknowledge the support of the
National Science and Engineering Research Council of 
Canada and of the Grant Agency of the Czech 
Republic under Grant 201/96/0119.}
\endheading

 \vskip7mm               \eightpoint
 \roster
\item""{\bf Abstract :}
A finite commutative semigroup is isomorphic to the 
semigroup of all open continuous mappings of a 
Hausdorff space into itself, resp.  of all complete 
Heyting endomorphisms of a Hausdorff Heyting 
algebra if and only if it is a product of a group and 
several groups augmented by zero. 
Characterization of this type does not 
apply in the infinite case.
\medskip

\noindent {\bf 1991 Mathematics Subject Classification :}
Primary: 20M30. Secondary: 06D20, 54D10, 54C10
\medskip

\noindent {\bf Keywords :}
endomorphism, open continuous map, 
complete Heyting homomorphism, strong graph 
homomorphism, Hausdorff property

      \endroster \tenpoint
\vskip10mm


  
%\address
%\flushpar
%\parindent=0pt
%Applied Mathematics, MFF KU\newline
%Malostransk\'e n\'am. 25\newline
%118 00 Praha 1\newline
%Czech Republic
%\endaddress
%\email
%pultr\@kam.ms.mff.cuni.cz
%\endemail
%\address
%\flushpar
%\parindent=0pt
%Mathematics\newline
%University of Manitoba\newline
%Winnipeg, Manitoba\newline
%Canada R3T 2N2
%\endaddress
%\email
%sichler\@cc.umanitoba.ca
%\endemail
%\endtopmatter




In 1966, Paalman de Miranda proved that a certain 
finite monoid cannot be represented as the monoid 
$\frak O\frak p(X)$ of all open continuous mappings of any 
Hausdorff space $X$ into itself \cite{9}.  Her 
discovery interestingly contrasts with these two facts:
\roster
\item"--"
every monoid is isomorphic to the monoid of all 
{\it quasi-open\/} continuous mappings of a {\it metric\/} space
into itself (this, together with similar results, can 
be found in \cite{15}, see also \cite{14,16}),
\item"--"
every monoid is isomorphic to the monoid $\frak O\frak p(X)$ for a 
$T_1$-space $X$, see \cite{14,16} and, as observed in \cite{13}, 
even for a space $X$ that is both $T_1$ and sober.
\endroster

In \cite{13}, we have extended Paalman de Miranda's 
result to the considerably wider context of Hausdorff 
Heyting algebras and their complete Heyting 
homomorphisms (the word `wider' is somewhat 
imprecise, for there exist marginal cases of 
Hausdorff spaces whose corresponding Heyting 
algebra is {\it not\/} Hausdorff, see Isbell \cite{5}).  In 
addition, we found that, apart from cyclic groups, 
only a single one-generated finite monoid is 
representable in either category.  


In the present paper we describe all finite 
commutative monoids isomorphic to the monoid $\frak O\frak p(X)$ 
for some Hausdorff space $X$, or to the monoid of all 
complete Heyting endomorphisms of some Hausdorff 
Heyting algebra in the sense of Isbell \cite{5}.  It 
turns out that these monoids are products
$$G_0\times G^0_1\times\cdots\times G^0_n$$
in which $G_0$ is a group and each monoid 
$G^0_j$ (if any) is a group augmented by a 
zero.  This also characterizes finite commutative 
monoids formed by strong endomorphisms of 
symmetric graphs.  These three categories strongly 
contrast their respective non-Hausdorf or 
non-symmentric counterparts:  the categories of 
\roster
\item"--"
$T_1$-spaces and open maps,
\item"--"
complete Heyting algebras and complete Heyting 
homomorphisms,
\item"--"
directed graphs and strong homomorphisms
\endroster
are algebraically universal (see \cite{14,16}, \cite{13}, 
\cite{14}), and hence every monoid is representable 
as an endomorphism monoid in each of these 
wider categories.

We also show that the situation changes for infinite 
endomorphism monoids:  for instance, the additive monoid 
of natural numbers is representable, and so are all its 
finite or countably infinite powers.  

\medskip
Only basic notions of category theory, those found in the 
initial chapters of \cite{1} or \cite{8}, are needed here.  
The few facts on frames not explained below can be 
found in \cite{7} or in \cite{13}.  A topological 
construction presented in Section~4 is based on a 
construction from Chapter~VI of \cite{14}; to aid the 
reader and to avoid unnecessary repetition, we refer to 
\cite{14} in considerable detail and adopt the notation 
used there.  

\vskip10mm

\heading
1. Preliminaries
\endheading

\noindent
{\bf 1.1.} A  {\it concrete category\/}  is, as usual, a category $
\Cal C$ 
equipped with a faithful functor $\vert\;\vert :\Cal C@>>>\Set$ into the 
category $\Set$ of all sets and mappings. We say that 
a finite coproduct 
$$\{j_i:X_i@>>>X\mid i=1,\dots,n\}$$
in $\Cal C$ is a {\it concrete sum\/} in a concrete category $(\Cal C
,\vert\;\vert )$ 
provided $\vert X\vert$ is a disjoint union $\vert X_1\vert\cup\dots
\cup\vert X_n\vert$ and 
$\vert j_i\vert :\vert X_i\vert@>>>\vert X\vert$ are the corresponding inclusion maps.  
If this is the case, we write 
$$X=X_1\cup\dots\cup X_n=\bigcup X_i.$$
A $\Cal C-$object $Y$ is called a {\it summand\/} of a $\Cal C$-object $
X$ if 
$X=Y\cup Z$ is a concrete sum for some $Z$.  
Then, of course, the $\Cal C$-object $Z$ is a summand of $X$ as 
well and $\vert Z\vert =\vert X\smallsetminus Y\vert$.

A concrete category has {\it tame summands}, and is 
called a $TS$-{\it category}, if it has these three 
properties:
\roster
\item"(TS1)" 
for any two summands $X_1$ and $X_2$ of $X$ there exists a 
summand $Y$ of $X$ with $\vert Y\vert =\vert X_1\vert\cap\vert X_
2\vert$,
\item"(TS2)"
all preimages of summands under $\Cal C$-morphisms are 
summands,
\item"(TS3)"
for any two concrete sums $X=\bigcup X_i$ and $X=\bigcup Y_j$ 
there exists a 
common refinement, that is, a concrete sum 
$X=\bigcup Z_{ij}$ with $\vert Z_{ij}\vert =\vert X_i\vert\cap\vert 
Y_j\vert$.
\endroster
There is no danger of confusion in writing 
$Y=X_1\cap X_2$ rather than $\vert Y\vert =\vert X_1\vert\cap\vert 
X_2\vert$ in (TS1).  By 
(TS1), for any two summands $X_1$ and $X_2$ of $X$, there 
is a summand $Z$ of $X$ with $\vert Z\vert =\vert X_1\vert\smallsetminus
\vert X_2\vert$, and we 
will simply write $Z=X_1\smallsetminus X_2$ here again.

There are numerous examples of $TS$-categories. The 
category 
$$\Top_{2,o}$$
of Hausdorff spaces and their open continuous maps 
is a $TS$-category and, together with categories 
related to it, will be of particular interest here.  

\proclaim{1.2.  Proposition} Let $X_1,\dots,X_n$ be 
summands (not necessarily disjoint) of an object $X$ in 
a $TS$-category $\Cal C$, and let $\vert X\vert =\vert\bigcup X_i
\vert$.  If $f:\vert X\vert@>>>\vert Y\vert$ 
is a mapping such that each $f\restriction\vert X_i\vert$ carries 
a $\Cal C$-morphism $X_i@>>>Y$, then $f$ carries a $\Cal C$-morphism 
$X@>>>Y$. 
\endproclaim

\demo{Proof} Use the common refinement of the $n$  
concrete sums $X=X_i\cup (X\smallsetminus X_i)$.
\qed\enddemo

\medskip
\noindent
{\bf 1.3.}  Let $(X,\le )$ be a poset.  We say that a subset 
$Y\subseteq X$ is {\it decreasing\/} if $x\le y\in Y$ implies $x\in 
Y$, and 
{\it increasing\/} if $x\ge y\in Y$ implies $x\in Y$.  Recall that an 
ordered topological space $X=(X,\tau ,\le )$ is a {\it Priestley 
space\/} if $(X,\le )$ is a poset and $(X,\tau )$ is a compact 
topological space, and if for any closed decreasing 
set $Y\subset X$ and any $x\in X\smallsetminus Y$ there exists a clopen
decreasing set $B$ such that $Y\subseteq B$ and $x\notin B$.  Any 
continuous, order preserving mapping $f:X@>>>X'$ 
between Priestley spaces $X$, $X'$ is called a {\it Priestley 
map.\/}  In what follows, 
$$\PS$$
will denote the category of all Priestley spaces and 
all Priestley maps.  

\proclaim{Observation} A subset $Y$ of a Priestley 
space $X$ is a summand of $X$ if and only if it is 
increasing, decreasing and clopen in $X$.  Hence $\PS$ is 
a $TS$-category.  \qed\endproclaim

\medskip
\noindent
{\bf 1.4.}  Recall that a {\it frame\/} is any complete lattice $
L$ in 
which $(\bigvee S)\wedge a=\bigvee \{s\wedge a\mid s\in S\}$ for any $
S\subseteq L$ and 
$a\in L$, and that a frame homomorphism $h:L@>>>L'$ is 
any mapping preserving finite meets and arbitrary 
joins.  A typical instance of a frame is the lattice 
$\frak O(X)$ of all open sets of a topological space $X$; and if 
$f:X@>>>X'$ is a continuous map into a space $X'$, then 
the inverse-image map $f^{-1}=\frak O(f):\frak O(X')@>>>\frak O(X
)$ is a 
frame homomorphism.  

For the so-called {\it sober\/} spaces, that is, spaces 
satisfying a certain condition weaker than the 
Hausdorff separation axiom, the mapping $f\mapsto \frak O(f)$ is 
an invertible contravariant correspondence of 
continuous maps $f:X@>>>X'$ to frame homomorphisms 
$\frak O(f):\frak O(X')@>>>\frak O(X)$.  Moreover, any sober space $
X$ can 
be reconstructed from its frame $\frak O(X)$, and this is 
why a frame can be viewed as a natural 
generalization of a (sober) space. 

A frame $L$ is called {\it Hausdorff\/} (in the sense of 
Isbell, see \cite{5,7}) if the codiagonal map $L\oplus L@>>>L$ 
of the free product $L\oplus L$ into $L$ satisfies a certain 
condition, applied in \cite{13}.  The actual form of 
this condition is secondary, for we shall use only its 
consequence proved in \cite{13}, and use it only once.  
It should be noted that the frame $\frak O(X)$ of a 
Hausdorff space $X$ is not necessarily Hausdorff in 
the sense of Isbell, but that $\frak O(X)$ is Hausdorff 
whenever $X$ is regular.  

\medskip
\noindent
{\bf 1.5. }
The distributive law of 1.4 implies that any frame $L$ 
is also a complete Heyting algebra, and that the 
Heyting operation $\to$ is uniquely determined by $L$.  
Moreover, 
\roster
\item""
for Hausdorff spaces $X$ and $X'$, complete 
Heyting homomorphisms $\frak O(X')@>>>\frak O(X)$ correspond to 
those continuous maps $f:X@>>>X'$ which are open, see 
\cite{2,6}.  
\endroster
In what follows, 
$$\HH$$
will denote the category of all Heyting algebras that 
are Hausdorff in the sense of Isbell, and of all their 
complete Heyting homomorphisms.  


\medskip
\noindent
{\bf 1.6. }
We now recall the celebrated {\it Priestley duality}, a 
contravariant natural isofunctor 
$$\Cal D:\PS@>>>\text{\bsf D}$$
of the category $\PS$ of Priestley spaces onto the 
category {\bsf D} of distributive (0,1)-lattices and 
their (0,1)-homomorphisms.  For any Priestley space 
$X$, the elements of $\Cal D(X)$ are represented by clopen 
decreasing subsets of $X$ and, for any Priestley map 
$f:Y@>>>X$, the homomorphism $\Cal D(f):\Cal D(X)@>>>\Cal D(Y)$ is the 
restriction of the inverse image map $f^{-1}$ to the 
collection of these representing sets.

In what follows, the image of the category $\HH$ under 
the inverse $\Cal D^{-1}$ of $\Cal D$ will be denoted as 
$$\HP$$
and referred to as the category of all $HP$-{\it spaces\/} and 
$HP$-{\it maps}.  It is clear that $\HP$ is not a full 
subcategory of $\PS$.
  
\vskip10mm

\heading
2. Hausdorff categories
\endheading
\subheading{2.1. Definition }
 We say that a
$TS$-category $\Cal C$ is {\it Hausdorff\/} if, for any $\Cal C$-object $
X$, 
the image 
$$h[X]=\{x\mid h(x)=x\}$$
of any endomorphism $h:X@>>>X$ with $h^2=h$ is a 
summand of $X$.  

\proclaim{2.2.  Proposition} The category $\Top_{2,o}$ is 
Hausdorff.  \endproclaim

\demo{Proof} If $h:X@>>>X$ is continuous and open, 
then its image $h[X]$ is open.  Since $h[X]=\{x\mid h(x)=x\}$ 
and $X$ is Hausdorff, the set $h[X]$ is also closed, and it 
is clear that any clopen subset of $X$ is a summand of 
$X$ in $\Top_{2,o}$.  \qed\enddemo

\proclaim{2.3.  Proposition} The category $\HP$ is 
Hausdorff.  \endproclaim

\demo{Proof} We already know that $\HP$ is a 
$TS$-category. Let $X$ be a Hausdorff Priestley space, 
and let $h=h^2\in\End X$.  Since $h$ is the Priestley dual 
of a complete Heyting endomorphism, the image 
$Y=h[X]$ is a closed subspace of $X$ that is also 
decreasing, see \cite{12}.  Furthermore, the $\HP$-map 
$h$ has a decomposition $X@>{h_1}>>Y@>{h_2}>>X$ in which the 
Priestley map $h_1$ is surjective, and the inclusion map 
$h_2$ is such that $\Cal D(h_2)$ is a complete surjective 
Heyting homomorphism.  In particular, $\Cal D(h_2)$ is a 
surjective frame homomorphism, and hence $\Cal D(Y)$ is 
Hausdorff, see \cite{7}. Therefore $Y\in\HP$.

Next we note that $h_2$ is an equalizer of $h$ and $\text{id}_X$.  
Indeed, $h\circ h_2=\text{id}_X\circ h_2$ follows from 
$h\circ h_2\circ h_1=h\circ h=h=h_2\circ h_1$ because $h_1$ is surjective.  
On the other hand, if $h\circ k=k$, then $h_2\circ (h_1\circ k)=k$, 
and the factorizing map $h_1\circ k$ is unique because $h_2$ is 
one-to-one.  This implies that the surjective 
complete Heyting homomorphism $\Cal D(h_2):\Cal D(X)@>>>\Cal D(Y)$ is 
a coequalizer of $\Cal D(h)$ and $\text{id}_{\Cal D(X)}$.  But then, 
by Lemma 4.4 of \cite{13}, there exists a 
complemented $a\in \Cal D(X)$ such that, for any $b,c\in \Cal D(X
)$, 
$$\Cal D(h_2)(b)=\Cal D(h_2)(c)\text{ }\Leftrightarrow\text{ }b\wedge 
a=c\wedge a.$$
Therefore $\Cal D(h_2)(b)=1_{\Cal D(Y)}$ if and only if $b\ge a$ in $
\Cal D(X)$.  
Simultaneously, since $h_2$ is the inclusion map of $Y$ 
into $X$, it follows that $h^{-1}_2(b)=b\cap Y$ for any clopen 
decreasing $b\subseteq X$.  In particular, 
$Y=1_{\Cal D(Y)}=\Cal D(h_2)(a)=h^{-1}_2(a)=a\cap Y$, and hence $
Y\subseteq a$.  
If there exists an $x\in a\smallsetminus Y$ then, by 1.6,
there exists a clopen decreasing set 
$b\subseteq X$ with $x\notin b$ and $Y\subseteq b$.  But then 
$\Cal D(h_2)(b)=h^{-1}_2(b)=Y=1_{\Cal D(Y)}$ while $b\not \ge a$ -- a contradiction.  
Therefore $Y=a$.  Since the element $a$ is 
complemented in $\Cal D(X)$, the set $Y=h[X]$ is increasing, 
decreasing and clopen in $X$, and this is obviously true 
also for the set $X\smallsetminus Y$.  Therefore $X$ is a summand of
$X$, see 1.3. \qed
\enddemo


\noindent
{\bf 2.4. }
Recall that a pair $(X,R)$ is a directed graph if $X$ is a 
set and $R\subseteq X\times X$ is a binary relation on $X$, and that 
a mapping $f:X@>>>Y$ is a graph homomorphism from 
$(X,R)$ to a directed graph $(Y,S)$ if $(f(x),f(x'))\in S$ for 
every $(x,x')\in R$.  If $\circ$ denotes also the composition of 
relations, then the condition defining a graph 
homomorphism is equivalent to the inclusion 
$$f\circ R\subseteq S\circ f.$$
A map $f:X@>>>Y$ is called a {\it strong homomorphism \/}
from $(X,R)$ to $(Y,S)$ if 
$$f\circ R=S\circ f.$$
The category of all directed graphs and their strong 
homomorphisms is algebraically universal, and hence 
every monoid is isomorphic to the monoid of all 
strong endomorphisms of some directed graph, see 
\cite{14}.  

As we will see,
this is no longer true for strong homomorphisms of 
{\it symmetric\/} directed graphs, that is, graphs $(X,R)$ 
satisfying $(x,x')\in R$ if and only if $(x',x)\in R$ (see 3.5 
and 4.6 below).  Thus, in a sense, the graph 
symmetry plays a role similar to the Hausdorff 
topological axiom. 

\proclaim{Observation} The category
$$\SS$$
of all symmetric directed graphs and their strong 
homomorphisms is Hausdorff.
\endproclaim

\demo{Proof} It is easy to verify that $Y\subseteq X$ is a 
summand of a symmetric directed graph $(X,R)$ 
exactly when 
$$xRy\text{ and }y\in Y\text{ imply }x\in Y,\tag s$$
and that $\SS$ is a $TS$-category.  Let $f$ be any (not 
necessarily idempotent) strong endomorphism of 
$(X,R)$.  To show that $f[X]$ is a factor, suppose that 
$xRy$ and $y\in f[X]$.  Then $y=f(t)$ for some $t\in X$, and 
hence $x(R\circ f)t$.  But then $x(f\circ R)t$ because $f$ is 
strong, and hence $x=f(u)$ for some $u$ with $uRt$.  
Therefore $f[X]$ satisfies (s), and hence it is a 
summand in $(X,R)$.  \qed\enddemo

\noindent
{\bf 2.5.}
Our main interest is the behaviour of endomorphism 
monoids in $\HH$ or $\HP$, and now we return to these 
categories.   Since the latter category does not 
contain all of $\Top_{2,o}$ and because proofs for both 
categories are essentially identical, we work in the 
more general setting of Hausdorff categories that 
covers both of them.  Moreover, this notion applies 
to other categories, such as $\SS$.
  
\vskip10mm

\heading
3. Endomorphisms in Hausdorff categories
\endheading

\noindent
{\bf 3.1.}
With the exception of Theorem 3.6 and its 
corollary, the symbol $X$ will always denote an object 
in a Hausdorff $TS$-category.  In what follows,
\roster
\item""
$\End X$ is the endomorphism monoid of $X$,
\item""
$\Aut X$ is the automorphism group of $X$, and 
\item""
$\Idp X=\{f\in\End X\mid f^2=f\}$.
\endroster

\proclaim{3.2. Proposition} Let $\End X$ be finite and 
commutative. Then every $f\in\End X$ has a 
decomposition
$$f=g\circ h\text{ with }g\in\Aut X\text{ and }h\in\Idp X,$$
in which the idempotent $h$ is uniquely determined.
\endproclaim

\demo{Proof} Since $\End X$ is finite, there is a least 
integer $n\ge 0$ such that $f^{n+k}=f^n$ for some $k>0$.  

If $n=0$, then $f\in\Aut X$ and we are done.

Suppose that $n>0$. Then
$$X\supset f[X]\supset\dots\supset f^n[X]=f^{n+i}[X]\text{ for all }
i>0.$$
For any integer $kr\ge n$ we thus have $f^{kr}[X]=f^n[X]$.  
From $f^{2kr}=f^{kr}$ it then follows that $F=f^n[X]$ is a 
summand.  But then the preimage $f^{-1}(F)\supseteq F$ of $F$ is 
also a summand, and hence the map $\varphi :X@>>>X$ defined 
by 
$$\varphi (x)=\cases f(x)&\text{for }x\in f^{-1}(F),\\
x&\text{for }x\notin f^{-1}(F)\endcases $$
is an endomorphism of $X$. For any $x\in f^{-2}(F)\smallsetminus f^{-1}
(F)$ 
we then have 
$$\varphi f(x)=f^2(x)\in F\text{ and }f\varphi (x)=f(x)\notin F,$$
which is impossible. From $f^{-2}(F)\supseteq f^{-1}(F)$ it then 
follows that 
$$f^{-2}(F)=f^{-1}(F)$$
and hence, by induction,
$$X=f^{-n}(F)=f^{-1}(F).$$
But then $f[X]=F$ and, consequently,
$$f^{k+1}=f.$$
Therefore the restriction of $f^k$ to $F$ is the identity 
on $F$, and hence the mapping $g:X@>>>X$ given by
$$g(x)=\cases f(x)&\text{for }x\in F,\\
x&\text{for }x\notin F\endcases $$
belongs to $\Aut X$. Furthermore, 
$$\text{if }x\notin F\text{ then }gf(x)=fg(x)=f(x).\tag 3.2.1$$
Next we define $h\in\End X$ by 
$$h(x)=\cases x&\text{for }x\in F,\\
f(x)&\text{for }x\notin F.\endcases $$
Since $h^2(x)=hf(x)=f(x)=h(x)$ for $x\notin F$, we have 
$h\in\Idp X$.  If $x\notin F$, then $gh(x)=gf(x)=f(x)$ by 
\thetag{3.2.1} and, if $x\in F$, then $gh(x)=g(x)=f(x)$ 
again.  Therefore $f=g\circ h$ as claimed.  

Finally, suppose that $f=g_1h_1=g_2h_2$ with $g_i\in\Aut X$ 
and $h_i\in\Idp X$.  Then $g_1h_1=g_2h_2h_2=g_1h_1h_2$ and hence 
$h_1=h_1h_2$.  By symmetry $h_2=h_2h_1$, and then $h_1=h_2$ 
follows from $h_1h_2=h_2h_1$.  \qed\enddemo

\noindent
{\bf 3.3. }
For any $f\in\End X$ define 
$$\Fix(f)=\{x\in X\mid f(x)=x\}\text{ and }\Mov(f)=\{x\in X\mid f
(x)\ne x\}.$$
Since $X$ belongs to a Hausdorff category, for any 
$h\in\Idp X$, 
$$\Fix(h)=h[X]\text{ and }\Mov(h)\text{ are summands of }X.\tag 3.3.1$$

\proclaim{3.4.  Lemma} Let $\End X$ be commutative.  
Then for any $h,k\in\Idp X$ and $f\in\End X$ 
$$h(\Mov(h))\subseteq\Fix(f),\text{ and}\tag 3.4.1$$
$$\text{if }x\in\Mov(h)\cap\Mov(k)\text{ then }h(x)=k(x).\tag 3.4.2$$
\endproclaim

\demo{Proof} By \thetag{3.3.1}, there is an 
endomorphism
$$\varphi (x)=\cases x&\text{for }x\in\Mov(h),\\
f(x)&\text{for }x\in\Fix(h).\endcases $$
For $x\in\Mov(h)$ we thus have 
$fh(x)=\varphi h(x)=h\varphi (x)=h(x)$, and this proves 
\thetag{3.4.1}. Suppose that $x\in\Mov(h)\cap\Mov(k)$. 
Then $h(x)\in\Fix(k)$ and $k(x)\in\Fix(h)$, by \thetag{3.4.1},
and hence $h(x)=kh(x)=hk(x)=k(x)$. \qed
\enddemo

\proclaim{3.5. Theorem} Let $X$ be an object of a 
Hausdorff category, and let $\End X$ be finite and 
commutative. Then there is an $n\ge 0$ such that 
$$\End X\cong G_0\times G^0_1\times\dots\times G^0_n$$
where $G_0$ is a group and $G^0_1,\dots,G^0_n$ are groups 
augmented by an external zero.
\endproclaim

\demo{Proof} The monoid $\Idp X$ is a semilattice. 
For every $h\in\Idp X$ we write $\mu (h)=\Mov(h)$.  
By \thetag{3.4.1}, for any $k\in\Idp X$ we have 
$\Mov(h)\cup\Mov(k)\subseteq\Mov(hk)$ and, since the reverse 
inclusion is obvious, we conclude that 
$$\mu (hk)=\mu (h)\cup\mu (k)\text{ for any }h,k\in\Idp X.$$
Secondly, if $h,k\in\Idp X$ and $\mu (h)=\mu (k)$ then, by 
\thetag{3.4.2}, for every $x\in\Mov(h)=\Mov(k)$ we 
have $h(x)=k(x)$, and then $h=k$ follows.  Therefore $\mu$ 
is an injective homomorphism onto a join semilattice 
$(J,\cup )$ of subsets of $X$, and $\emptyset =\mu (\text{id}_X)\in 
J$.

If $S$ is a summand of $X$ and $h\in\Idp X$, then 
$\Mov(h)\cap S$ is a summand of $X$ as well, see 
\thetag{3.3.1}, and hence the mapping $h^{(S)}:X@>>>X$ 
given by 
$$h^{(S)}(x)=\cases h(x)&\text{for }x\in\Mov(h)\cap S,\\
x&\text{for }x\notin\Mov(h)\cap S\endcases $$
is an idempotent endomorphism with 
$\Mov(h^{(S)})=\Mov(h)\cap S$.  Since $S=\Mov(k)$ and 
$S=X\smallsetminus\Mov(k)$ are summands of $X$, see \thetag{3.3.1},
we conclude that $\mu$ maps $\Idp X$ isomorphically onto 
the join reduct $(J,\cup )$ of a finite Boolean algebra $J$ of 
subsets of $X$.  If $J=\{\emptyset \}$, then $\Idp X=\{\text{id}_
X\}$ and hence 
$\End X=\Aut X$.  From now on, we assume that $J$ is 
non-trivial, and let $h_1,\dots,h_n\in\Idp X$ denote those 
endomorphisms for which $\{\mu (h_i)\mid i=1,\dots,n\}$ is the 
set of atoms of $J$.  Since any finite Boolean algebra 
$J$ with more than two elements is atomistic, for 
every $h\in\Idp X$ there is a (possibly empty) unique 
subset $A\subseteq \{1,\dots,n\}$ such that $h$ is the composite of 
the `atomic' idempotents $h_i$ with $i\in A$, and we write 
$h=h_A$.  Thus the sets 
$$M_i=\Mov(h_i)\text{ with }i=1,\dots,n$$
are nonvoid, pairwise disjoint minimal summands of 
$X$, and 
$$\Mov(h_A)=\bigcup \{M_i\mid i\in A\}.$$
From \thetag{3.4.2} it now follows that 
$$h_A\restriction M_i=\cases h_i\restriction M_i&\text{for }i\in 
A,\\
\text{id}_{M_i}&\text{for }i\notin A.\endcases \tag 3.5.1$$
Finally, write $M_0=X\smallsetminus (M_1\cup\dots\cup M_n)$.  Then $M_
0$ is 
a nonvoid summand of $X$ because of \thetag{TS3}, and 
hence 
$$X=M_0\cup M_1\cup\dots\cup M_n\text{ is a concrete sum}.\tag 3.5.2$$
It is clear that 
$$M_0=\bigcap \{\Fix(h)\mid h\in\Idp X\}.\tag 3.5.3$$

Let $f\in\End X$ and $h\in\Idp X$. If $x\in\Mov(h)$, then 
$fh(x)=h(x)$ by \thetag{3.4.1}, while, for any $x\in\Fix(h)$, 
we obviously have $fh(x)=f(x)$. In particular, 
$$\text{if }g\in\Aut X\text{ and }h\in\Idp X,\text{ then }gh(x)=\cases 
h(x)&\text{for }x\in\Mov(h),\\
g(x)&\text{for }x\in\Fix(h).\endcases \tag 3.5.4$$
If $g\in\Aut X$ and $h=h_i$ for some 
$i=1,\dots,n$, then $h_ig(x)=gh_i(x)=h_i(x)$ for every 
$x\in M_i=\Mov(h_i)$, that is,
$$h_ig\restriction M_i=h_i\restriction M_i\text{ for any }g\in\Aut 
X\text{ and }i=1,\dots,n.\tag 3.5.5$$
Since $g$ is invertible, we have $h_ig(x)=gh_i(x)\ne g(x)$ 
for every $x\in M_i$.  Thus $g[M_i]\subseteq M_i$ for every 
$i=1,\dots,n$, and it follows that 
$$g[M_j]=M_j\text{ for any }g\in\Aut X\text{ and }j=0,1,\dots,n\tag 3.5.6$$

Now let $f\in\End X$ be arbitrary.  Then $f=g\circ h$ for 
some $g\in\Aut X$ and a uniquely determined $h\in\Idp X$, 
by Proposition 3.2.  Since $h=h_A$ for a unique 
$A\subseteq \{1,\dots,n\}$, from \thetag{3.5.4} and \thetag{3.5.1} 
it then follows that
$$f(x)=\cases h_j(x)&\text{if }x\in M_j\text{ and }j\in A,\\
g(x)&\text{if }x\in M_j\text{ and }j\in \{0,1,\dots,n\}\smallsetminus
A.\endcases \tag 3.5.7$$
For $j=0,1,\dots,n$, 
denote $f_j=f\restriction M_j$.  Then $f_j=h_j$ for $j\in A$ and 
$f_j\in\Aut M_j=G_j$ for all $j\in \{0,1,\dots,n\}\smallsetminus A$.  Since $
h_i$ 
is an external zero of $G_i$ for all $i=1,\dots,n$, see 
\thetag{3.5.5}, this shows that 
$$\varphi (f)=(f_0,f_1,\dots,f_n)$$
is a well-defined mapping of $\End X$ into the Cartesian 
product
$$P=G_0\times G^0_1\times\dots\times G^0_n$$
of the group $G_0$ and monoids $G^0_i=G_i\cup \{0_i\}$ obtained 
from the groups $G_i$ by the addition of an external zero $0_i$ for 
each $i=1,\dots n$.  Using \thetag{3.5.7}, it is easy to 
verify that $\varphi :\End X@>>>P$ is a monoid homomorphism.  

Conversely, let $(f_0,f_1,\dots,f_n)\in P$. Then $f_0\in\Aut M_0$ 
and $f_i\in\Aut M_i\cup \{0_i\}$, and the 
mapping 
$f=\psi (f_0,f_1,\dots,f_n)$ given by 
$$f(x)=\cases f_0(x)&\text{for }x\in M_0\\
f_i(x)&\text{for }x\in M_i\text{ with }f_i\ne 0_i,\\
h_i(x)&\text{for }x\in M_i\text{ with }f_i=0_i\endcases $$
belongs to $\End X$ because of \thetag{3.5.2}.  It is clear 
that the mapping $\psi$ is the inverse of the 
homomorphism $\varphi$.  \qed\enddemo


\proclaim{Corollary 3.6} Let $X$ be an object of a 
Hausdorff category.  Let $\End X$ be finite and 
commutative, and let $\Aut X$ be trivial.  Then $\End X$ is 
isomorphic to a free join semillatice with zero, or, 
equivalently, to a semilattice $(2^n,\cup )$.
\qed\endproclaim

\vskip10mm

\heading
4. Monoid representations
\endheading

In this section, we shall use certain spaces and 
graphs constructed in \cite{14} as building blocks for 
representations of the finite commutative monoids 
described by Theorem 3.5 as endomorphism monoids 
in the categories $\Top_{2,o}$ (and hence also in $\HH$ and 
$\PS$) and $\SS$.  First we turn to the topological 
construction.  

\medskip
\noindent
{\bf 4.1.}
For $j=0,1,\dots,$ let $H^j$ be non-trivial metric 
continua with distinguished points $a^j_0,a^j_1,a^j_2\in H^j$ 
constructed in Section~6 of Chapter~VI in \cite{14}.  
We divide up these continua into two
families 
$$\Cal H^i=\{H^{2k+i}\mid k=1,2,\dots\}\text{ with }i=0,1,$$
and denote 
$$H^j_0=H^j\smallsetminus \{a^j_0,a^j_1,a^j_2\}$$
Further, we recall the metric spaces $\Cal M(X,R,\varphi )$ 
amalgamated from these metric continua $H^j$ elsewhere 
in Chapter~VI of \cite{14}, and consider only those 
spaces $\Cal M(X,R,\varphi )$ such that
$$\text{only }H^j\in \Cal H^1\text{ occur in }\Cal M(X,R,\varphi 
).\tag $*$ $$
Thus we begin with 
\roster
\item"--"
non-trivial metric continua $H^j$ with 
distinguished points $a^j_0,a^j_1,a^j_2\in H^j$ for $j=0,1,\dots$,
\item"--"
a countably infinite collection $\Cal P_1$ of metric spaces 
$P=\Cal M(X,R,\varphi )$ subject to ($*$) above.  
\endroster
The proof of Lemma 6.9 in Section~6 of Chapter~VI in 
\cite{14} shows that $\Cal P_1$ has these properties:
\roster
\item"(4.1.1)"
if $H^j$ is embedded into a space $P\in \Cal P_1$ in such a way that 
$H^j_0$ is open, then every continuous map $f:H^k_0@>>>P$ 
with $f[H^k_0]\cap H^j_0\ne\emptyset$ is either a constant or else $
j=k$ 
and $f$ is the embedding,
\item"(4.1.2)"
if $H^j\in \Cal H^0$ and $P\in \Cal P_1$ then any continuous $f:H^
j@>>>P$ is constant.  
\endroster

Furthermore, we recall that for any monoid $S$ there 
exists a space $P(S)\in \Cal P$$_1$ such that all non-constant 
continuous maps $f:P(S)@>>>P(S)$ form a monoid 
isomorphic to $S$.  Thus, in particular, 
\roster
\item"(4.1.3)"
for any group $G$ there exists some $P(G)\in \Cal P_1$ such 
that the monoid of all open continuous mappings 
$f:P(G)@>>>P(G)$ is isomorphic to $G$.
\endroster

\medskip
\noindent
{\bf 4.2. Construction.}
For a metric space $P$ and a non-trivial metric 
continuum $H^j$, we define a space 
$$P\star H^j$$
on the Cartesian product $P\times H^j$ by requiring that 
\roster
\item"--"
the neighbourhoods of $(x,a^j_0)$ are the sets containing 
$(U\times \{a^j_0\})\times (\{x\}\times V)$, where $U$ is a neighbourhood of $
x$ 
in $P$ and $V$ is a neighbourhood of $a^j_0$ in $H^j$, while
\item"--"
the neighbourhoods of any $(x,y)$ with $y\ne a^j_0$ are the 
sets containing $\{x\}\times V$, where $V$ is a neighbourhood of 
$y$ in $H^j$.  
\endroster
Having renamed each $(x,a^j_0)$ as $x$, we observe that $P$ 
is a subspace of $P\star H^j$. The space $P\star H^j$ may thus 
be visualized as the space obtained by attaching 
separate copies of $H^j$ to members of $P$ by their 
distinguished element $a^j_0$. 

The claim below is immediate.

\proclaim{Lemma} The mapping 
$g\times\text{id}:P\star H^j@>>>Q\star H^j$ is continuous and open for 
any open continuous $g:P@>>>Q$.  \qed\endproclaim

\proclaim{4.3.  Lemma} Let $Q$ be such that every 
continuous mapping of $H^k$ to $Q$ is constant, and let a 
mapping 
$$f:P\star H^k@>>>Q\star H^j$$
be open continuous. Then $k=j$ and $f=g\times\text{id}$ for 
some open continuous $g:P@>>>Q$.
\endproclaim

\demo{Proof} Let $x\in P$.  Since $f$ is open, the set 
$f[\{x\}\times H^k]$ is not a singleton, and hence $f[\{x\}\times 
H^k]\not \subseteq Q$ 
by the hypothesis.  But then \thetag{4.1.1} implies 
that $k=j$ and there is a unique $y\in Q$ such that 
$f(x,z)=(y,z)$ for all $z\in H^k$.  This gives rise to a 
unique mapping $g:P@>>>Q$ for which $f=g\times\text{id}$.  It is 
routine to verify that $g$ is open.  \qed\enddemo

\medskip
\noindent
{\bf 4.4.}
Recall that, for any space $P$, the symbol $\frak O\frak p(P)$ 
denotes the monoid of all open continuous selfmaps 
of $P$.

\proclaim{Lemma} Let $\{P_a\mid a\in A\}$ be a countable 
collection of metric spaces from $\Cal P_1$.  Then there 
exists a metric space $P$ such that 
$$\frak O\frak p(P)\cong\prod \{\frak O\frak p(P_a\mid a\in A\}.$$
\endproclaim
\demo{Proof} Select distinct metrizable continua 
$H^{j(a)}\in \Cal H^0$ and then apply \thetag{4.1.2}, Lemma 4.3 
and Lemma 4.2.  \qed\enddemo

\medskip
\noindent
{\bf 4.5.}  Now we turn to the category $\SS$ of strong 
homomorphisms od symmetric directed graphs.  

Let $A$ be an arbitrary set, and let $Z=(\{0\},\{(0,0)\})$ 
denote the singleton graph in the alg-universal 
category $\Gra_c$ of all connected directed graphs and 
their homomorphisms.  From Theorem 3 of \cite{4} 
and from the results of Chapter~II of \cite{14} it 
follows that for every $a\in A$ there is a full 
embedding $\Phi_a$ of $\Gra_c$ into its full subcategory 
formed by connected {\it symmetric\/} graphs such that 
\roster
\item"(4.5.1)" 
the homomorphism $\Phi_a:(X,R)@>>>\Phi_a(Z)$ belongs to $\SS$ 
for any $(X,R)$, 
\item"(4.5.2)" 
there is no homomorphism $\Phi_a(X,R)@>>>\Phi_{a'}(X',R')$ 
for distinct $a,a'\in A$.  
\endroster
Since $\Gra_c$ is an alg-universal category, 
for every group $G$ there exists a graph 
$(X,R)$ with $\End(X,R)\cong G$.  But then $\End\Phi_a(X,R)\cong 
G$ 
as well and, since every $g\in\End\Phi_a(X,R)$ is invertible, 
the monoid of all strong endomorphisms of $\Phi_a(X,R)$ is 
isomorphic to the group $G$. Therefore 
\roster
\item"(4.5.3)"
for any $a\in A$, every group is isomorphic to the 
monoid of all strong endomorphisms of some $\Phi_a(X,R)$.  
\endroster

\medskip
\noindent
{\bf 4.6.}
The category $\Met_o$ of all metric spaces and all their 
open continuous maps is a subcategory of $\Top_{2,o}$.  
Since metric spaces are regular, the contravariant 
functor $\frak D$ from 1.4 maps $\Met_o$ onto a full 
subcategory of $\HH$.  Accordingly, the positive, 
constructive claim of the theorem below for 
categories other than $\SS$ will be established through 
a construction of appropriate metric spaces.  

\proclaim{Theorem} Let $\Cal C$ be any one of the 
categories $\Top_{2,o}\,$, $\HH$, $\PS$, or $\SS$, and let $S$ be a 
finite commutative monoid.  Then $S\cong\End X$ for 
some object $X\in \Cal C$ if and only if, for some $n\ge 0$, 
$$S=G_0\times G^0_1\times\dots\times G^0_n$$
where $G_0$ is a group and each $G^0_j$ is a group 
augmented by zero.  \endproclaim

\demo{Proof} To complete the proof for the three 
`topological' categories, in view of Theorem 3.5, 
we need only represent a group $G$ and any zero 
extension $G^0_j$ of a group $G_j$ by a metric space 
satisfying the hypothesis of Lemma 4.4.  To 
represent a given group $G,$ we simply use the 
metric space $P(G)\in \Cal P_1$ of \thetag{4.1.3}.  A proper 
extension $P(G_j)\cup \{p\}$ of the space $P(G_j)\in \Cal P_1$ from 
\thetag{4.1.3} by an open singleton $\{p\}$ obviously 
represents $G^0_j$ and satisfies the hypothesis of 
Lemma~4.4.  

To complete the proof, we need to represent the 
monoid $G_0\times G^0_1\times\dots\times G^0_n$ in the category $\SS$.
But this is also easy. For $i=0,1,\dots,n$, we select 
$(X_i,R_i)\in\Gra_c$ so that the strong endomorphisms of 
$\Phi_i(X_i,R_i)$ form a monoid isomorphic to $G_i$, see 
\thetag{4.5.3}, and then we form a disjoint union  
$$X=\Phi_0(X_0,R_0)\cup\bigcup_{i=1}^n\Phi_i(X_i,R_i)\cup\Phi_i(Z
).$$
Then $X$ belongs to $\SS$ and, since all graphs $\Phi_j(\dots)$ 
are connected, from \thetag{4.5.2} it follows that 
every strong endomorphism $g$ of $X$ preserves 
$\Phi_0(X_0,R_0)$ and also every $\Phi_i(X_i,R_i)\cup\Phi_i(Z)$ with 
$i=1,\dots,n$.  Therefore 
$$\End X\cong G_0\times\prod_{i=1}^n\End\big(\Phi_i(X_i,R_i)\cup\Phi_
i(Z)\big).$$
Furthermore, any endomorphism $g$ either preserves 
$\Phi_i(X_i,R_i)$ or else maps this graph to $\Phi_i(Z)$.  From 
\thetag{4.5.1} it now follows that 
$\End\big(\Phi_i(X_i,R_i)\cup\Phi_i(Z)\big)\cong G^0_i$ for all $
i=1,\dots,n$.  
Therefore $\End X\cong S$, as claimed.
\qed\enddemo

\medskip
\noindent
{\bf 4.7.}
By Theorem 4.6, all join reducts $(2^n,\cup )$ of finite 
Boolean algebras are isomorphic to $\frak O\frak p(P)$ for some 
metric space $P$.  This result can be extended to the 
infinite case.  

\proclaim{Theorem} For any set $C$, there exists a 
metric space $P$ such that $\frak O\frak p(P)$ is isomorphic to 
$(2^C,\cup )$. 
\endproclaim

\demo{Proof} Since the category of metric spaces is 
almost universal, see Chapter~VI of \cite{14}, there 
exists a colelction $\{P_c\mid c\in C\}$ of metric connected 
spaces such that a non-constant continuous 
$f:P_c@>>>P_{c'}$ exists only when $c=c'$, and then it is 
the identity mapping of $P_c$.  For the disjoint 
union 
$$P=\{p\}\cup\bigcup_{c\in C}P_c$$
with the union topology we then obviously have 
$\frak O\frak p(P)\cong (2^C,\cup ).$ \qed\enddemo

\subheading{Remark }  From 4.5 it follows that a similar
claim holds true for the Hausdorff category $\SS$ of 
symmetric graphs and their strong homomorphisms. 

\medskip
\noindent
{\bf 4.8.}
Infinite commutative monoids with no nontrivial 
invertible elements need not contain idempotents, and 
this is also true for endomorphism monoids in 
Hausdorff categories.

\proclaim{Theorem} Let $\Bbb N$ denote the additive monoid 
of all natural numbers. Then, for any $n=1,2,\dots,\omega$, 
there is a metric space $P$ with $\frak O\frak p(P)\cong \Bbb N^n$.
\endproclaim

\demo{Proof} For $H^1\in \Cal H^1$ from 4.1, denote 
$H=H^1\smallsetminus \{a^1_0\}$ and set
$$P=H\times \{0,1,\ldots \},$$
where the neighbourhoods of any $(x,k)$ with $x\ne a^1_1$ 
are all sets containing the set $U\times \{k\}$ for some 
neighbourhood $U$ of $x$ in $H$, while the neighbourhoods 
of $(a^1_1,k)$ are all sets containing the set 
$(U\times \{k\})\cup ((V\smallsetminus \{a^1_0\})\times \{k+1\})$ in which $
U$ is a 
neighbourhood of $a^1_1$ in $H$ and $V$ is a neighbourhood of 
$a^1_0$ in $H^1$. In other words, $P$ is obtained from 
$$H\cup (H^1\times \{1,2,\dots\})$$
by means of identifying each $(a^1_1,k)$ with $(a^1_0,k+1)$. 

An open map $f:P@>>>P$ is not constant, and hence, 
by \thetag{4.1.1}, it sends each $H\times \{j\}$ identically onto 
some $H\times \{k\}$, and it is not difficult to see that, in 
fact, there is some $n\ge 0$ such that $f(x,j)=(x,j+n)$ 
for all $j\ge 0$.  Since any such map $f$ is open, this 
proves that  
$$\frak O\frak p(P)\cong \Bbb N.$$
Higher powers of $\Bbb N$ can be represented because 
Lemma 4.4 applies here, see \thetag{4.1.1}.  \qed
\enddemo

\vskip10mm


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\enddocument

