# On topological McAlister semigroups

Serhii Bardyla  Universität Wien, Institut für Mathematik, Kurt Gödel Research
Center, Wien, Austria<EMAIL_ADDRESS>
###### Abstract.

In this paper we consider McAlister semigroups over arbitrary cardinals and
investigate their algebraic and topological properties. We show that the group
of automorphisms of a McAlister semigroup $\mathcal{M}_{\lambda}$ is
isomorphic to the direct product $Sym(\lambda){\times}\mathbb{Z}_{2}$, where
$Sym(\lambda)$ is the group of permutations of the cardinal $\lambda$. This
fact correlates with the result of Mashevitzky, Schein and Zhitomirski which
states that the group of automorphisms of the free inverse semigroup over a
cardinal $\lambda$ is isomorphic to the wreath product of $Sym(\lambda)$ and
$\mathbb{Z}_{2}$. Each McAlister semigroup admits a compact semigroup
topology. Consequently, the Green’s relations $\mathscr{D}$ and $\mathscr{J}$
coincide in McAlister semigroups. The latter fact complements results of
Lawson. We showed that each non-zero element of a Hausdorff semitopological
McAlister semigroup is isolated. This fact is an analogue of the result of
Mesyan, Mitchell, Morayne and Péresse, who proved that each non-zero element
of Hausdorff topological polycyclic monoid is isolated. Also, it follows that
the free inverse semigroup over a singleton admits only the discrete Hausdorff
shift-continuous topology. We proved that a Hausdorff locally compact
semitopological semigroup $\mathcal{M}_{1}$ is either compact or discrete.
This fact is similar to the result of Gutik, who showed that a Hausdorff
locally compact semitopological polycyclic monoid $\mathcal{P}_{1}$ is either
compact or discrete. However, this dichotomy does not hold for the semigroup
$\mathcal{M}_{2}$. Moreover, $\mathcal{M}_{2}$ admits continuum many different
Hausdorff locally compact inverse semigroup topologies.

###### Key words and phrases:

McAlister semigroups, semitopological semigroup, locally compact semigroup,
automorphism

###### 2020 Mathematics Subject Classification:

22A15, 20M18

The author was supported by the Austrian Science Fund FWF (Grant M 2967).

##  1\. Introduction and Preliminaries

We shall follow the terminology of [9, 17]. All topological spaces are assumed
to be Hausdorff. A semigroup $S$ is called an inverse semigroup if for each
element $a\in S$ there exists a unique inverse element $a^{-1}\in S$ such that
$aa^{-1}a=a$ and $a^{-1}aa^{-1}=a^{-1}$. The map $inv:S\rightarrow S$,
$a\rightarrow a^{-1}$ is called an inversion.

A topologized semigroup is a topological space endowed with a semigroup
operation. A topologized semigroup $S$ is called

  * •

semitopological, if for each element $x\in S$ the shifts
$l_{x}(s):s\rightarrow xs$ and $r_{x}(s):s\rightarrow sx$ are continuous;

  * •

topological, if the semigroup operation is continuous in $S$;

  * •

topological inverse, if $S$ is an inverse semigroup, and semigroup operation
together with inversion are continuous in $S$.

A topology $\tau$ on a semigroup $S$ is called

  * •

shift-continuous, if $(S,\tau)$ is a semitopological semigroup;

  * •

semigroup, if $(S,\tau)$ is a topological semigroup;

  * •

inverse semigroup, if $(S,\tau)$ is a topological inverse semigroup.

A finite sequence of elements of a nonempty set $A$ is called a word over $A$.
The empty word is denoted by $\varepsilon$. For a non-zero cardinal $\lambda$
by $F_{\lambda}$ we denote the free monoid over a set $A$ of cardinality
$\lambda$. Note that $F_{\lambda}$ can be represented as a set of all words
over the set $A$ endowed with the semigroup operation of concatenation. More
precisely, the concatenation of two words $a=a_{1}\ldots a_{n}\in F_{\lambda}$
and $b=b_{1}\ldots b_{m}\in F_{\lambda}$ is the word $ab=a_{1}\ldots
a_{n}b_{1}\ldots b_{m}\in F_{\lambda}$. We agree that $\varepsilon
a=a\varepsilon=a$ for each $a\in F_{\lambda}$, that is $\varepsilon$ is the
unit of $F_{\lambda}$. By $|a|$ we denote the length of a word $a\in
F_{\lambda}\setminus\\{\varepsilon\\}$. We assume that $|\varepsilon|=0$. If
$a=a_{1}\ldots a_{n}\in F_{\lambda}$, then let $a^{t}=a_{n}\ldots a_{1}$.
Also, we agree that $a^{t}=a$ for any $a\in F_{\lambda}$ such that $|a|\leq
1$.

For a non-zero cardinal $\lambda$ the polycyclic monoid
$\mathcal{P}_{\lambda}$ is the set
$F_{\lambda}{\times}F_{\lambda}\sqcup\\{0\\}$ endowed with the following
semigroup operation:

$\begin{split}&(a,b)\cdot(c,d)=\left\\{\begin{array}[]{ccl}(c_{1}a,d),&\hbox{if~{}~{}}c=c_{1}b&\hbox{for
some~{}}c_{1}\in F_{\lambda};\\\ (a,b_{1}d),&\hbox{if~{}~{}}b=b_{1}c&\hbox{for
some~{}}b_{1}\in F_{\lambda};\\\ 0,&\hbox{otherwise;}&\end{array}\right.\\\
&\hbox{~{}and~{}}\quad(a,b)\cdot 0=0\cdot(a,b)=0\cdot 0=0.\end{split}$

Polycyclic monoids over finite cardinals were introduced by Nivat and Perrot
in [26]. Basic algebraic properties of polycyclic monoids are described in
Chapter 9.3 from [17]. In [5] Gutik and the author investigated algebraic
properties of polycyclic monoids over arbitrary cardinals. It turns out that
they share many common properties with the classical polycyclic monoids.
Nowadays polycyclic monoids are well-studied algebraic objects (see [15, 19])
and have fruitful applications. In particular, they are universal objects in
the class of graph inverse semigroups. Namely, each graph inverse semigroup
$G(E)$ over a directed graph $E$ is a subsemigroup of the polycyclic monoid
$\mathcal{P}_{|G(E)|}$ [3]. Also, polycyclic monoids are useful in a
construction of Thompson groups (see [18, 21]).

It is easy to see that the set
$I=\mathcal{P}_{\lambda}{\times}\\{0\\}\cup\\{0\\}{\times}\mathcal{P}_{\lambda}$
is a two-sided ideal of the direct product
$\mathcal{P}_{\lambda}{\times}\mathcal{P}_{\lambda}$. Let
$\mathcal{S}_{\lambda}$ denotes the Rees quotient semigroup
$(\mathcal{P}_{\lambda}{\times}\mathcal{P}_{\lambda})/I$. By $0$ we naturally
denote the point $q(I)\in\mathcal{S}_{\lambda}$, where $q$ is the quotient
map. The subsemigroup
$\mathcal{M}_{\lambda}=\\{((a,b),(c,d))\mid\varepsilon\neq
ac^{t}=bd^{t}\\}\cup\\{0\\}$ of $\mathcal{S}_{\lambda}$ is called the
McAlister semigroup. The semigroups $\mathcal{M}_{k}$, $k\in\mathbb{N}$ were
introduced by McAlister in [24]. McAlister semigroups are closely related to
tiling semigroups constructed from a one-dimensional tiling. Also, they can be
considered as a generalization of the free inverse semigroup with one
generator, because the semigroup $\mathcal{M}_{1}$ is isomorphic to the free
inverse semigroup over a singleton with adjoined zero [20]. For alternative
representations (which do not use polycyclic monoids) and other algebraic
properties of McAlister semigroups see Chapter 9.4 from [17] or papers of
Lawson and McAlister [20, 24].

Let us note that polycyclic monoids as well as McAlister semigroups are
closely related to the well-known bicyclic monoid $\mathcal{B}$. In
particular, the polycyclic monoid $\mathcal{P}_{1}$ is isomorphic to the
bicyclic monoid with adjoined zero. Topological bicyclic monoid was
investigated by Eberhart and Selden [8]. They showed that the bicyclic
semigroup admits only the discrete semigroup topology, and if ${\mathcal{B}}$
is a dense proper subsemigroup of a topological semigroup $S$, then
$I=S\setminus{\mathcal{B}}$ is a two-sided ideal of $S$. Bertman and West [6]
showed that the bicyclic monoid admits only the discrete shift-continuous
topology. Compact topological semigroups cannot contain an isomorphic copy of
the bicyclic monoid [1]. Embedding of the bicyclic monoid into compact-like
topological semigroups was discussed in [2, 11, 13].

Topological polycyclic monoids and their subsemigroups were investigated by
Mesyan, Mitchell, Morayne and Péresse [25]. In particular, they showed that
each non-zero element of a topological graph inverse semigroup is isolated and
a locally compact topological graph inverse semigroup over a finite graph $E$
is discrete. Since a polycyclic monoid $\mathcal{P}_{\lambda}$ is isomorphic
to the graph inverse semigroup over the graph $E$ which contains one vertex
and $\lambda$ many loops, their results imply that each non-zero element of a
topological polycyclic monoid is isolated and locally compact topological
polycyclic monoids $\mathcal{P}_{k}$, $k\in\mathbb{N}$ are discrete. Gutik and
the author in [5] showed that each non-zero element of a semitopological
polycyclic monoid is isolated and for every non-zero cardinal $\lambda$ a
locally compact topological polycyclic monoid $\mathcal{P}_{\lambda}$ is
discrete. Graph inverse semigroups which admit only the discrete locally
compact semigroup topology were characterized in [4].

In this paper we investigate algebraic and topological properties of McAlister
semigroups.

##  2\. Algebraic properties of McAlister semigroups

######  Lemma 2.1.

For each non-zero element $x\in\mathcal{M}_{\lambda}$ there exist words
$u,v,w\in F_{\lambda}$ such that $x=((u,uv),(wv^{t},w))$ or
$x=((uv,u),(w,wv^{t}))$.

###### Proof.

Let $((a,b),(c,d))\in\mathcal{M}_{\lambda}\setminus\\{0\\}$, where
$a=a_{1}\ldots a_{n}$, $b=b_{1}\ldots b_{m}$, $c=c_{1}\ldots c_{k}$,
$d=d_{1}\ldots d_{p}$. Recall that $ac^{t}=bd^{t}\neq\varepsilon$ witnessing
that $a_{1}\ldots a_{n}c_{k}\ldots c_{1}=b_{1}\ldots b_{m}d_{p}\ldots d_{1}$.
Three cases are possible:

  1. (1)

$n>m$ and $k<p$;

  2. (2)

$n<m$ and $k>p$;

  3. (3)

$n=m$ and $k=p$.

1) It follows that
$a_{1}=b_{1},\ldots,a_{m}=b_{m},a_{m+1}=d_{p},\ldots,a_{n}=d_{p-n+m+1},c_{k}=d_{p-n+m},\ldots,c_{1}=d_{1}$.
Put $v\equiv a_{m+1}\ldots a_{n}$, $u\equiv a_{1}\ldots a_{m}$ and $w\equiv
c_{1}\ldots c_{k}$. Then $((a,b),(c,d))=((uv,u),(w,wv^{t}))$.

2) It follows that
$b_{1}=a_{1},\ldots,b_{n}=a_{n},b_{n+1}=c_{k},\ldots,b_{m}=c_{k-m+n+1},d_{p}=c_{k-m+n},\ldots,d_{1}=c_{1}$.
Put $v\equiv b_{n+1}\ldots b_{m}$, $u\equiv a_{1}\ldots a_{n}$ and $w\equiv
d_{1}\ldots d_{p}$. Then $((a,b),(c,d))=((u,uv),(wv^{t},w))$.

3) It follows that $a_{i}=b_{i}$ for any $i\leq n$ and $c_{i}=d_{i}$ for any
$i\leq k$. Put $v\equiv\varepsilon$, $u\equiv a$ and $w\equiv c$. Then
$((a,b),(c,d))=((a,a),(c,c))=((a\varepsilon,a),(c,c\varepsilon^{t}))=((uv,u),(w,wv^{t}))$.
∎

For an inverse semigroup $S$ the Green’s relations $\mathscr{L}$,
$\mathscr{R}$, $\mathscr{H}$, $\mathscr{D}$ and $\mathscr{J}$ are defined as
follows:

$a\mathscr{L}b$ | if and only if |  $a^{-1}a=b^{-1}b$;
---|---|---
$a\mathscr{R}b$ | if and only if |  $aa^{-1}=bb^{-1}$;
$a\mathscr{J}b$ | if and only if |  $SaS=SbS$;
|  $\mathscr{H}=\mathscr{L}\cap\mathscr{R}$; |
|  $\mathscr{D}=\mathscr{L}{\circ}\mathscr{R}=\mathscr{R}{\circ}\mathscr{L}$. |

The Green’s relations $\mathscr{L},\mathscr{R},\mathscr{H}$ and $\mathscr{D}$
on McAlister semigroups $\mathcal{M}_{n}$, $n\in\mathbb{N}$ were described by
Lawson in [20]. The following lemma complements mentioned above results.

######  Lemma 2.2.

Let $((a_{1},b_{1}),(c_{1},d_{1}))$ and $((a_{2},b_{2}),(c_{2},d_{2}))$ be
arbitrary non-zero elements of a semigroup $\mathcal{M}_{\lambda}$. Then the
following conditions hold:

  1. (1)

$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{L}((a_{2},b_{2}),(c_{2},d_{2}))$ iff
$b_{1}=b_{2}$ and $d_{1}=d_{2}$;

  2. (2)

$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{R}((a_{2},b_{2}),(c_{2},d_{2}))$ iff
$a_{1}=a_{2}$ and $c_{1}=c_{2}$;

  3. (3)

$\mathscr{H}$-classes are singletons;

  4. (4)

$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{D}((a_{2},b_{2}),(c_{2},d_{2}))$ iff
$a_{1}c_{1}^{t}=a_{2}c_{2}^{t}$;

  5. (5)

the relations $\mathscr{D}$ and $\mathscr{J}$ coincide on
$\mathcal{M}_{\lambda}$.

###### Proof.

Note that for any $((a,b),(c,d))\in\mathcal{M}_{\lambda}\setminus\\{0\\}$,
$((a,b),(c,d))^{-1}=((b,a),(d,c))$.

1) Observe that

$((a_{1},b_{1}),(c_{1},d_{1}))^{-1}\cdot((a_{1},b_{1}),(c_{1},d_{1}))=((b_{1},a_{1}),(d_{1},c_{1}))\cdot((a_{1},b_{1}),(c_{1},d_{1}))=((b_{1},b_{1}),(d_{1},d_{1})).$

Similarly it can be checked that
$((a_{2},b_{2}),(c_{2},d_{2}))^{-1}\cdot((a_{2},b_{2}),(c_{2},d_{2}))=((b_{2},b_{2}),(d_{2},d_{2}))$.
By the definition of the relation $\mathscr{L}$,

$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{L}((a_{2},b_{2}),(c_{2},d_{2}))\qquad\hbox{iff}\qquad((b_{1},b_{1}),(d_{1},d_{1}))=((b_{2},b_{2}),(d_{2},d_{2})).$

The latter equality is true iff $b_{1}=b_{2}$ and $d_{1}=d_{2}$. Hence
condition 1 holds.

2) Observe that

$((a_{1},b_{1}),(c_{1},d_{1}))\cdot((a_{1},b_{1}),(c_{1},d_{1}))^{-1}=((a_{1},b_{1}),(c_{1},d_{1}))\cdot((b_{1},a_{1}),(d_{1},c_{1}))=((a_{1},a_{1}),(c_{1},c_{1})).$

Similarly it can be checked that
$((a_{2},b_{2}),(c_{2},d_{2}))\cdot((a_{2},b_{2}),(c_{2},d_{2}))^{-1}=((a_{2},a_{2}),(c_{2},c_{2}))$.
By the definition of the relation $\mathscr{R}$,

$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{R}((a_{2},b_{2}),(c_{2},d_{2}))\qquad\hbox{iff}\qquad((a_{1},a_{1}),(c_{1},c_{1}))=((a_{2},a_{2}),(c_{2},c_{2})).$

The latter equality is true iff $a_{1}=a_{2}$ and $c_{1}=c_{2}$. Hence
condition 2 holds.

3) By the definition of the relation $\mathscr{H}$,
$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{H}((a_{2},b_{2}),(c_{2},d_{2}))$ iff

$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{L}((a_{2},b_{2}),(c_{2},d_{2}))\quad\hbox{
and
}\quad((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{R}((a_{2},b_{2}),(c_{2},d_{2})).$

At this point condition 3 follows from conditions 1 and 2.

4) ($\Rightarrow$) Since $\mathscr{D}=\mathscr{L}{\circ}\mathscr{R}$ there
exists an element
$((a_{3},b_{3}),(c_{3},d_{3}))\in\mathcal{M}_{\lambda}\setminus\\{0\\}$ such
that

$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{L}((a_{3},b_{3}),(c_{3},d_{3}))\quad\hbox{
and
}\quad((a_{3},b_{3}),(c_{3},d_{3}))\mathscr{R}((a_{2},b_{2}),(c_{2},d_{2})).$

Conditions 1 and 2 imply that $b_{1}=b_{3}$, $d_{1}=d_{3}$ and $a_{2}=a_{3}$,
$c_{2}=c_{3}$. Then

$a_{1}c_{1}^{t}=b_{1}d_{1}^{t}=b_{3}d_{3}^{t}=a_{3}c_{3}^{t}=a_{2}c_{2}^{t}.$

($\Leftarrow$) Since $a_{2}c_{2}^{t}=a_{1}c_{1}^{t}=b_{1}d_{1}^{t}$, we have
that $((a_{2},b_{1}),(c_{2},d_{1}))\in\mathcal{M}_{\lambda}\setminus\\{0\\}$.
By conditions 1 and 2,

$((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{L}((a_{2},b_{1}),(c_{2},d_{1}))\qquad\hbox{and}\qquad((a_{2},b_{1}),(c_{2},d_{1}))\mathscr{R}((a_{2},b_{2}),(c_{2},d_{2})).$

Hence $((a_{1},b_{1}),(c_{1},d_{1}))\mathscr{D}((a_{2},b_{2}),(c_{2},d_{2}))$.

5) By Proposition 2.19 from [12], the relations $\mathscr{D}$ and
$\mathscr{J}$ coincide on each compact topological semigroup $S$. So, it is
sufficient to construct a compact semigroup topology on
$\mathcal{M}_{\lambda}$. It will be done in Theorem 3.3 after some preparatory
work made before. ∎

Observe that to each $\mathscr{D}$-class
$D\subset\mathcal{M}_{\lambda}\setminus\\{0\\}$ corresponds a certain word
$p\in F_{\lambda}\setminus\\{\varepsilon\\}$. Namely, fix any
$((a,b),(c,d))\in D$ and put $p=ac^{t}$. By Lemma 2.2, the word $p$ does not
depend on the choice of $((a,b),(c,d))\in D$. Further, by $D_{p}$ we will
denote the $\mathscr{D}$-class in $\mathcal{M}_{\lambda}$ which contains an
element $((a,b),(c,d))$ with $ac^{t}=p$. The following lemma is an analogue of
Proposition 8 from [20] but for McAlister semigroups over arbitrary cardinals.

######  Lemma 2.3.

For each $p\in F_{\lambda}\setminus\\{\varepsilon\\}$ the following statements
hold:

  * •

$|D_{p}|=(|p|+1)^{2}$;

  * •

if $e\leq f$ for some idempotents $e,f\in D_{p}$, then $e=f$.

###### Proof.

Observe that there are $|p|+1$ possibilities to represent the word
$p\in\mathcal{F}_{\lambda}\setminus\\{\varepsilon\\}$ as $ac^{t}$, where
$a,c\in F_{\lambda}$. It follows that there are $(|p|+1)^{2}$ possibilities to
construct an element $((a,b),(c,d))$ such that $ac^{t}=bd^{t}=p$. Thus the set
$D_{p}$ contains $(|p|+1)^{2}$ elements. Note that idempotents of
$\mathcal{M}_{\lambda}\setminus\\{0\\}$ have the form $((a,a),(c,c))$, where
$ac^{t}\neq\varepsilon$. Fix any two idempotents
$((a,a),(c,c))\leq((b,b),(d,d))$ in $D_{p}$. It follows that
$((a,a),(c,c))\cdot((b,b),(d,d))=((a,a),(c,c))$. The definition of the
semigroup operation in $\mathcal{M}_{\lambda}$ implies that there exists
$a_{1},c_{1}\in F_{\lambda}$ such that $a=a_{1}b$ and $c=c_{1}d$. Since
$bd^{t}=p=ac^{t}=a_{1}bd^{t}c_{1}^{t}$ we deduce that
$a_{1}=c_{1}=\varepsilon$. Hence $((a,a),(c,c))=((b,b),(d,d))$. ∎

######  Proposition 2.4.

Fix any words $q,p\in F_{\lambda}\setminus\\{\varepsilon\\}$ and elements
$((a,b),(c,d))\in D_{p}$, $((x,y),(u,v))\in D_{q}$. If ${\bf
z}=((a,b),(c,d))\cdot((x,y),(u,v))\neq 0$, then there exist words
$s_{1},r_{1},s_{2},r_{2}\in F_{\lambda}$ such that $s_{1}pr_{1}=s_{2}qr_{2}$
and ${\bf z}\in D_{s_{1}pr_{1}}=D_{s_{2}qr_{2}}$.

###### Proof.

Recall that $p=ac^{t}=bd^{t}$ and $q=xu^{t}=yv^{t}$. Since ${\bf z}\neq 0$ one
of the following cases holds.

  1. (1)

$x=wb$ and $u=zd$ for some $w,z\in F_{\lambda}$;

  2. (2)

$x=wb$ and $d=zu$ for some $w,z\in F_{\lambda}$;

  3. (3)

$b=wx$ and $d=zu$ for some $w,z\in F_{\lambda}$;

  4. (4)

$b=wx$ and $u=zd$ for some $w,z\in F_{\lambda}$.

1) Observe that

${\bf z}=((a,b),(c,d))\cdot((wb,y),(zd,v))=((wa,y),(zc,v)).$

It follows that $wac^{t}z^{t}=yv^{t}$ which implies that $wpz^{t}=q$. Hence
${\bf z}\in D_{q}=D_{wpz^{t}}$. Put
$s_{1}=w,r_{1}=z^{t},s_{2}=r_{2}=\varepsilon$.

2) Observe that

${\bf z}=((a,b),(c,zu))\cdot((wb,y),(u,v))=((wa,y),(c,zv)).$

It follows that $wac^{t}=yv^{t}z^{t}$ which implies that $wp=qz^{t}$. Hence
${\bf z}\in D_{wp}=D_{qz^{t}}$. Put
$s_{1}=w,r_{1}=s_{2}=\varepsilon,r_{2}=z^{t}$.

3) Observe that

${\bf z}=((a,wx),(c,zu))\cdot((x,y),(u,v))=((a,wy),(c,zv)).$

It follows that $ac^{t}=wyv^{t}z^{t}$ which implies that $p=wqz^{t}$. Hence
${\bf z}\in D_{p}=D_{wqz^{t}}$. Put
$s_{1}=r_{1}=\varepsilon,s_{2}=w,r_{2}=z^{t}$.

4)Observe that

${\bf z}=((a,wx),(c,d))\cdot((x,y),(zd,v))=((a,wy),(zc,v)).$

It follows that $ac^{t}z^{t}=wyv^{t}$ which implies that $pz^{t}=wq$. Hence
${\bf z}\in D_{pz^{t}}=D_{wq}$. Put
$s_{1}=r_{2}=\varepsilon,r_{1}=z^{t},s_{2}=w$. ∎

######  Lemma 2.5.

For any $a,b\in\mathcal{M}_{\lambda}\setminus\\{0\\}$ the set $\\{x:ax=b$, or
$xa=b\\}$ is finite.

###### Proof.

Recall that $\mathcal{S}_{\lambda}$ is the Rees quotient semigroup
$(\mathcal{P}_{\lambda}{\times}\mathcal{P}_{\lambda})/J$, where
$J=\\{0\\}{\times}\mathcal{P}_{\lambda}\cup\mathcal{P}_{\lambda}{\times}\\{0\\}$.
The point $q(J)\in\mathcal{S}_{\lambda}$, where $q$ is the quotient map is
naturally denoted by $0$. Fix any non-zero elements
$a,b\in\mathcal{S}_{\lambda}$. The definition of $\mathcal{S}_{\lambda}$
ensures that $a=((c,d),(e,f))$, $b=((p,q),(r,s))$ and
$\\{(a,b),(c,d),(p,q),(r,s)\\}\subset\mathcal{P}_{\lambda}\setminus\\{0\\}$.
By Lemma 1 from [25], the sets

$A=\\{x\in\mathcal{P}_{\lambda}:(c,d)\cdot x=(p,q),\hbox{ or
}x\cdot(c,d)=(p,q)\\}\hbox{ and }$ $B=\\{x\in\mathcal{P}_{\lambda}:(e,f)\cdot
x=(r,s),\hbox{ or }x\cdot(e,f)=(r,s)\\}$

are finite. Then the set

$\\{x\in\mathcal{S}_{\lambda}:ax=b,\hbox{ or }xa=b\\}\subset A{\times}B$

is finite. Recall that the semigroup $\mathcal{M}_{\lambda}$ is a subsemigroup
of $\mathcal{S}_{\lambda}$. It follows that for any
$a,b\in\mathcal{M}_{\lambda}\setminus\\{0\\}$ the set
$\\{x\in\mathcal{M}_{\lambda}:ax=b$, or $xa=b\\}$ is contained in the finite
set $\\{x\in\mathcal{S}_{\lambda}:ax=b,\hbox{ or }xa=b\\}$, witnessing that it
is finite as well. ∎

Groups of automorphism of free inverse semigroups and related structures were
investigated by Schein and coauthors in [22, 23, 27]. Since McAlister
semigroups generalize the free inverse semigroup over a singleton, the problem
of describing the group $\mathrm{Aut}(\mathcal{M}_{\lambda})$ of automorphisms
of the McAlister semigroup $\mathcal{M}_{\lambda}$ arises naturally.

For this we are going to introduce two kinds of automorphisms of
$\mathcal{M}_{\lambda}$. Assume that $F_{\lambda}$ is the free monoid over the
set $\\{a_{\alpha}:\alpha\in\lambda\\}$. Let $f:\lambda\rightarrow\lambda$ be
a bijection. Note that $f$ generates the automorphism $h$ of $F_{\lambda}$
defined by $h(a_{\alpha_{1}}\ldots a_{\alpha_{n}})=a_{f(\alpha_{1})}\ldots
a_{f(\alpha_{n})}$ and $h(\varepsilon)=\varepsilon$. The automorphism $h$
generates two maps
$\phi_{f},\psi_{f}:\mathcal{M}_{\lambda}\rightarrow\mathcal{M}_{\lambda}$
defined by the formulae:

$\phi_{f}[((a,b),(c,d))]=((h(a),h(b)),(h(c),h(d))),\quad\hbox{ and
}\quad\phi_{f}(0)=0;$
$\psi_{f}[((a,b),(c,d))]=((h(c),h(d)),(h(a),h(b))),\quad\hbox{ and
}\quad\psi_{f}(0)=0.$

Let $g$ be an automorphism of $F_{\lambda}$ and fix any
$a=a_{\alpha_{1}}\ldots a_{\alpha_{n}}\in F_{\lambda}$. Then the following
holds:

$g(a)^{t}=[g(a_{\alpha_{1}})\ldots
g(a_{\alpha_{n}})]^{t}=g(a_{\alpha_{n}})\ldots g(a_{\alpha_{1}})=g(a^{t}).$

Therefore, if $ac^{t}=bd^{t}$, then

$h(a)h(c)^{t}=h(a)h(c^{t})=h(ac^{t})=h(bd^{t})=h(b)h(d^{t})=h(b)h(d)^{t}.$

It follows that

$h(c)h(a)^{t}=[h(a)h(c)^{t}]^{t}=[h(b)h(d)^{t}]^{t}=h(d)h(b)^{t}.$

Hence the maps $\phi_{f}$ and $\psi_{f}$ are defined correctly.

Words $a,b\in F_{\lambda}$ are called suffix-incomparable if neither $a=cb$
nor $b=ca$ for any $c\in F_{\lambda}$. Note that for any non-zero elements
$(a,b)$ and $(c,d)$ of the polycyclic monoid $\mathcal{P}_{\lambda}$,
$(a,b)\cdot(c,d)=0$ iff the words $b$ and $c$ are suffix-incomparable.

The next theorem describes the set $\mathrm{Aut}(\mathcal{M}_{\lambda})$ of
automorphisms of the semigroup $\mathcal{M}_{\lambda}$.

######  Proposition 2.6.

$\mathrm{Aut}(\mathcal{M}_{\lambda})=\\{\phi_{f}:f$ is a bijection of
$\lambda\\}\cup\\{\psi_{f}:f$ is a bijection of $\lambda\\}$.

###### Proof.

Let us check that for a fixed bijection $f$ the map $\phi_{f}$ is an
automorphism. The injectivity of $\phi_{f}$ follows from the injectivity of
$h$. To prove that the map $\phi_{f}$ is surjective, fix any non-zero point
$((a,b),(c,d))\in\mathcal{M}_{\lambda}$. Clearly,
$\phi_{f}[((h^{-1}(a),h^{-1}(b)),(h^{-1}(c),h^{-1}(d)))]=((a,b),(c,d))$. It
remains to check that

$((h^{-1}(a),h^{-1}(b)),(h^{-1}(c),h^{-1}(d)))\in\mathcal{M}_{\lambda}.$

Since $h^{-1}$ is an automorphism of $F_{\lambda}$ the arguments above imply
that $h^{-1}(p)^{t}=h^{-1}(p^{t})$ for any $p\in F_{\lambda}$. It follows that

$h^{-1}(a)h^{-1}(c)^{t}=h^{-1}(a)h^{-1}(c^{t})=h^{-1}(ac^{t})=h^{-1}(bd^{t})=h^{-1}(b)h^{-1}(d^{t})=h^{-1}(b)h^{-1}(d)^{t},$

witnessing that
$((h^{-1}(a),h^{-1}(b)),(h^{-1}(c),h^{-1}(d)))\in\mathcal{M}_{\lambda}$. Hence
the map $\phi_{f}$ is a bijection.

Let us check that $\phi_{f}$ is a homomorphism. Fix any elements
$((a,b),(c,d))$ and $((x,y),(u,v))$ of $\mathcal{M}_{\lambda}$ and put ${\bf
z}=((a,b),(c,d))\cdot((x,y),(u,v))$. Two cases are possible:

  1. (1)

${\bf z}=0$;

  2. (2)

${\bf z}\neq 0$.

In case 1 we have that the words $b$ and $x$ are suffix-incomparable, or the
words $d$ and $u$ are suffix-incomparable. It is straightforward to check that
suffix-incomparability is preserved by automorphisms of $F_{\lambda}$. Thus,
the words $h(b)$ and $h(x)$ are suffix-incomparable, or the words $h(d)$ and
$h(u)$ are suffix-incomparable. Hence

$\phi_{f}[((a,b),(c,d))]\cdot\phi_{f}[((x,y),(u,v))]=((h(a),h(b)),(h(c),h(d)))\cdot((h(x),h(y)),(h(u),h(v)))=0=\phi_{f}({\bf
z}).$

Second case has four subcases:

  * (2.1)

$x=wb$ and $u=zd$ for some $w,z\in F_{\lambda}$;

  * (2.2)

$x=wb$ and $d=zu$ for some $w,z\in F_{\lambda}$;

  * (2.3)

$b=wx$ and $d=zu$ for some $w,z\in F_{\lambda}$;

  * (2.4)

$b=wx$ and $u=zd$ for some $w,z\in F_{\lambda}$.

2.1) Observe that

${\bf z}=((a,b),(c,d))\cdot((wb,y),(zd,v))=((wa,y),(zc,v)).$

Then

$\begin{split}\phi_{f}[((a,b),(c,d))]\cdot\phi_{f}[((x,y),(u,v))]=((h(a),h(b)),(h(c),h(d)))\cdot((h(wb),h(y)),(h(zd),h(v)))=\\\
((h(a),h(b)),(h(c),h(d)))\cdot((h(w)h(b),h(y)),(h(z)h(d),h(v)))=\\\
((h(wa),h(y)),(h(zc),h(v)))=\phi_{f}({\bf z}).\end{split}$

2.2) Observe that

${\bf z}=((a,b),(c,zu))\cdot((wb,y),(u,v))=((wa,y),(c,zv)).$

Then

$\begin{split}\phi_{f}[((a,b),(c,d))]\cdot\phi_{f}[((x,y),(u,v))]=((h(a),h(b)),(h(c),h(zu)))\cdot((h(wb),h(y)),(h(u),h(v)))=\\\
=((h(a),h(b)),(h(c),h(z)h(u)))\cdot((h(w)h(b),h(y)),(h(u),h(v)))=\\\
((h(wa),h(y)),(h(c),h(zv)))=\phi_{f}({\bf z}).\end{split}$

2.3) Observe that

${\bf z}=((a,wx),(c,zu))\cdot((x,y),(u,v))=((a,wy),(c,zv)).$

Then

$\begin{split}\phi_{f}[((a,b),(c,d))]\cdot\phi_{f}[((x,y),(u,v))]=((h(a),h(wx)),(h(c),h(zu)))\cdot((h(x),h(y)),(h(u),h(v)))=\\\
=((h(a),h(w)h(x)),(h(c),h(z)h(u)))\cdot((h(x),h(y)),(h(u),h(v)))=\\\
((h(a),h(wy)),(h(c),h(zv)))=\phi_{f}({\bf z}).\end{split}$

2.4) Observe that

${\bf z}=((a,wx),(c,d))\cdot((x,y),(zd,v))=((a,wy),(zc,v)).$

Then

$\begin{split}\phi_{f}[((a,b),(c,d))]\cdot\phi_{f}[((x,y),(u,v))]=((h(a),h(wx)),(h(c),h(d)))\cdot((h(x),h(y)),(h(zd),h(v)))=\\\
=((h(a),h(w)h(x)),(h(c),h(d)))\cdot((h(x),h(y)),(h(z)h(d),h(v)))=\\\
((h(a),h(wy)),(h(zc),h(v)))=\phi_{f}({\bf z}).\end{split}$

Hence for each bijection $f:\lambda\rightarrow\lambda$ the map $\phi_{f}$ is
an automorphism of $\mathcal{M}_{\lambda}$.

Note that for each semigroup $S$ the map $g:S^{2}\rightarrow S^{2}$,
$g[(x,y)]=(y,x)$ is an automorphism. Moreover, if $J$ is a two sided ideal of
$S^{2}$ such that $(a,b)\in J$ iff $(b,a)\in J$, then $g$ generates an
automorphism $g^{*}$ of the Rees quotient semigroup $S^{2}/J$ defined by
$g^{*}(a,b)=g(a,b)$ if $(a,b)\notin J$ and $g^{*}(0)=0$, where by $0$ we
denote the image of $J$ under the quotient map. If $X$ is a subsemigroup of
$S^{2}/J$ such that $(a,b)\in X$ iff $(b,a)\in X$, then the restriction
$g^{*}|_{X}$ is an automorphism of $X$. Applying the above arguments to the
polycyclic monoid $\mathcal{P}_{\lambda}$ we get that the map $\psi_{id}$ is
an automorphism of the semigroup $\mathcal{M}_{\lambda}$, where
$id:\lambda\rightarrow\lambda$ is the identity map. Observe that
$\psi_{f}=\phi_{f}\circ\psi_{id}=\psi_{id}\circ\phi_{f}$ for any bijection $f$
on $\lambda$, witnessing that $\psi_{f}$ is an automorphism of
$\mathcal{M}_{\lambda}$. Hence

$\\{\phi_{f}:f\hbox{ is a bijection of }\lambda\\}\cup\\{\psi_{f}:f\hbox{ is a
bijection of }\lambda\\}\subset\mathrm{Aut}(\mathcal{M}_{\lambda}).$

To show the converse inclusion fix any automorphism $\phi$ of the McAlister
semigroup $\mathcal{M}_{\lambda}$. Let

$G_{1}=\\{((\varepsilon,a_{\alpha}),(a_{\alpha},\varepsilon)):\alpha\in\lambda\\}\quad\hbox{
and }\quad
G_{2}=\\{((a_{\alpha},\varepsilon),(\varepsilon,a_{\alpha})):\alpha\in\lambda\\}.$

######  Claim 1.

The set $G_{1}\cup G_{2}\cup\\{0\\}$ generates $\mathcal{M}_{\lambda}$

###### Proof.

By $\langle G_{1}\cup G_{2}\cup\\{0\\}\rangle$ we denote the subsemigroup of
$\mathcal{M}_{\lambda}$ which is generated by $G_{1}\cup G_{2}\cup\\{0\\}$.
Observe that for each word $u=u_{1}\ldots u_{n}\in
F_{\lambda}\setminus\\{\varepsilon\\}$,

$((\varepsilon,u),(u^{t},\varepsilon))=((\varepsilon,u_{1}),(u_{1},\varepsilon))\cdot((\varepsilon,u_{2}),(u_{2},\varepsilon))\cdots((\varepsilon,u_{n}),(u_{n},\varepsilon))\quad\hbox{and}$
$((u,\varepsilon),(\varepsilon,u^{t}))=((u_{n},\varepsilon),(\varepsilon,u_{n}))\cdot((u_{n-1},\varepsilon),(\varepsilon,u_{n-1}))\cdots((u_{1},\varepsilon),(\varepsilon,u_{1})).$

Hence
$\\{((\varepsilon,u),(u^{t},\varepsilon)),((u,\varepsilon),(\varepsilon,u^{t}))\\}\subset\langle
G_{1}\cup G_{2}\cup\\{0\\}\rangle$ for each $u\in
F_{\lambda}\setminus\\{\varepsilon\\}$. Fix any non-zero element
$g\in\mathcal{M}_{\lambda}$. By Lemma 2.1, we can represent $g$ as
$((a,aw),(bw^{t},b))$ or $((aw,a),(b,bw^{t}))$ for some $a,b,w\in
F_{\lambda}$. Assume that $g=((a,aw),(bw^{t},b))$. Let
$x=((\varepsilon,wb^{t}),(bw^{t},\varepsilon))$,
$y=((awb^{t},\varepsilon),(\varepsilon,bw^{t}a^{t}))$ and
$z=((\varepsilon,aw),(w^{t}a^{t},\varepsilon))$. By the preceding arguments,
$\\{x,y,z\\}\subset\langle G_{1}\cup G_{2}\cup\\{0\\}\rangle$. It remains to
calculate that

$\begin{split}xyz=((\varepsilon,wb^{t}),(bw^{t},\varepsilon))\cdot((awb^{t},\varepsilon),(\varepsilon,bw^{t}a^{t}))\cdot((\varepsilon,aw),(w^{t}a^{t},\varepsilon))=\\\
((a,\varepsilon),(bw^{t},bw^{t}a^{t}))\cdot((\varepsilon,aw),(w^{t}a^{t},\varepsilon))=((a,aw),(bw^{t},b))=g.\end{split}$

If $g=((aw,a),(b,bw^{t}))$, then let
$x=((aw,\varepsilon),(\varepsilon,w^{t}a^{t}))$,
$y=((\varepsilon,awb^{t}),(bw^{t}a^{t},\varepsilon))$ and
$z=((wb^{t},\varepsilon),(\varepsilon,bw^{t}))$. Clearly,
$\\{x,y,z\\}\subset\langle G_{1}\cup G_{2}\cup\\{0\\}\rangle$. It remains to
calculate that

$\begin{split}xyz=((aw,\varepsilon),(\varepsilon,w^{t}a^{t}))\cdot((\varepsilon,awb^{t}),(bw^{t}a^{t},\varepsilon))\cdot((wb^{t},\varepsilon),(\varepsilon,bw^{t}))=\\\
((aw,awb^{t}),(b,\varepsilon))\cdot((wb^{t},\varepsilon),(\varepsilon,bw^{t}))=((aw,a),(b,bw^{t}))=g.\end{split}$

Thus $g\in\langle G_{1}\cup G_{2}\cup\\{0\\}\rangle$. Hence
$\mathcal{M}_{\lambda}=\langle G_{1}\cup G_{2}\cup\\{0\\}\rangle$. ∎

Let us note that in the previous claim we add $0$ to the set of generators of
$\mathcal{M}_{\lambda}$ only to establish the case $\lambda=1$. If $\lambda>1$
this is not necessary.

######  Claim 2.

$\phi(G_{1}\cup G_{2})=G_{1}\cup G_{2}$.

###### Proof.

Let $M$ be the set of all maximal elements of the semilattice
$E(\mathcal{M}_{\lambda})$. It is easy to see that $e\in M$ iff
$e=((\varepsilon,\varepsilon),(a_{\alpha},a_{\alpha}))$ or
$e=((a_{\alpha},a_{\alpha}),(\varepsilon,\varepsilon))$ for some
$\alpha\in\lambda$. Since the restriction of $\phi$ on the semilattice
$E(\mathcal{M}_{\lambda})$ is an automorphism of $E(\mathcal{M}_{\lambda})$ we
deduce that $\phi|_{M}$ is a bijection of $M$.

First we shall show that $\phi(G_{1})\subset G_{1}\cup G_{2}$. Let us note
that

$G_{1}\cup
G_{2}=\\{((a,b),(c,d))\in\mathcal{M}_{\lambda}:|ac^{t}|=1\\}\setminus M.$

Fix any $g=((\varepsilon,a_{\alpha}),(a_{\alpha},\varepsilon))\in G_{1}$ and
let $((a,b),(c,d))=\phi(g)$. To derive a contradiction, assume that
$|ac^{t}|=|bd^{t}|>1$. Then,

$\phi[((\varepsilon,\varepsilon),(a_{\alpha},a_{\alpha}))]=\phi(gg^{-1})=\phi(g)\cdot\phi(g)^{-1}=((a,b),(c,d))\cdot((b,a),(d,c))=((a,a),(c,c))\notin
M.$

But this contradicts to the fact $\phi(M)=M$. The obtained contradiction
implies that $|ac^{t}|=|bd^{t}|=1$. Since $\phi$ is an automorphism and
$g\notin E(\mathcal{M}_{\lambda})$ we deduce that $\phi(g)\notin
E(\mathcal{M}_{\lambda})$. Consequently, $\phi(g)\in G_{1}\cup G_{2}$,
witnessing that $\phi(G_{1})\subset G_{1}\cup G_{2}$. Similarly, it can be
showed that $\phi(G_{2})\subset G_{1}\cup G_{2}$. It follows that
$\phi(G_{1}\cup G_{2})\subset G_{1}\cup G_{2}$. Since $\phi^{-1}$ is an
automorphism of $\mathcal{M}_{\lambda}$, the same arguments imply that
$\phi^{-1}(G_{1}\cup G_{2})\subset G_{1}\cup G_{2}$. Then $G_{1}\cup
G_{2}=\phi(\phi^{-1}(G_{1}\cup G_{2}))\subset\phi(G_{1}\cup G_{2})$. Hence
$\phi(G_{1}\cup G_{2})=G_{1}\cup G_{2}$. ∎

######  Claim 3.

$\phi(G_{1})=G_{1}$ and $\phi(G_{2})=G_{2}$, or $\phi(G_{1})=G_{2}$ and
$\phi(G_{2})=G_{1}$.

###### Proof.

Two cases are possible:

  1. (1)

$\phi(G_{1})\subset G_{1}$;

  2. (2)

$\phi(G_{1})\cap G_{2}\neq\emptyset$.

1\. If $\phi(G_{1})=G_{1}$, then
$\phi(G_{2})=\phi(G_{1}^{-1})=\phi(G_{1})^{-1}=G_{1}^{-1}=G_{2}$ and we are
done. Assuming that $G_{1}\setminus\phi(G_{1})\neq\emptyset$ and taking into
account that $\phi(G_{1}\cup G_{2})=G_{1}\cup G_{2}$, we obtain that there
exists $s=((a_{\alpha},\varepsilon),(\varepsilon,a_{\alpha}))\in G_{2}$ such
that $\phi(s)=((\varepsilon,a_{\beta}),(a_{\beta},\varepsilon))\in G_{1}$.
Since $\phi(G_{1}\cup G_{2})=G_{1}\cup G_{2}$, there exists
$t=((a_{\gamma},\varepsilon),(\varepsilon,a_{\gamma}))\in G_{2}$ such that
$\phi(t)=s$. Note that $\alpha\neq\beta$, or equivalently $s\neq\phi(s)^{-1}$,
because $s\in G_{2}$ and

$\phi(s)^{-1}=\phi(s^{-1})=\phi[((\varepsilon,a_{\alpha}),(a_{\alpha},\varepsilon))]\in\phi(G_{1})\subset
G_{1}.$

Also note that
$st=((a_{\gamma}a_{\alpha},\varepsilon),(\varepsilon,a_{\alpha}a_{\gamma}))\neq
0$ which implies that $\phi(st)\neq 0$ as well. On the other hand

$\phi(st)=\phi(s)\cdot\phi(t)=((\varepsilon,a_{\beta}),(a_{\beta},\varepsilon))\cdot((a_{\alpha},\varepsilon),(\varepsilon,a_{\alpha}))=0.$

The obtained contradiction implies that if case 1 holds, then
$\phi(G_{1})=G_{1}$ and $\phi(G_{2})=G_{2}$.

2\. If $G_{2}\subset\phi(G_{1})$, then
$\phi(G_{2})=\phi(G_{1}^{-1})=\phi(G_{1})^{-1}\supset G_{2}^{-1}=G_{1}$. Since
$\phi$ is a bijection and $\phi(G_{1}\cup G_{2})=G_{1}\cup G_{2}$ we get that
$\phi(G_{1})=G_{2}$ and $\phi(G_{2})=G_{1}$. Assume that
$G_{2}\setminus\phi(G_{1})\neq\emptyset$. Since $\phi(G_{1})\cap
G_{2}\neq\emptyset$, there exists
$s=((\varepsilon,a_{\alpha}),(a_{\alpha},\varepsilon))\in G_{1}$ such that
$\phi(s)=((a_{\beta},\varepsilon),(\varepsilon,a_{\beta}))\in G_{2}$. Since
$\phi(G_{1}\cup G_{2})=G_{1}\cup G_{2}$ there exists
$t=((a_{\gamma},\varepsilon),(\varepsilon,a_{\gamma}))\in G_{2}$ such that
$\phi(t)=((a_{\delta},\varepsilon),(\varepsilon,a_{\delta}))\in
G_{2}\setminus\phi(G_{1})$. Note that $\gamma\neq\alpha$, or equivalently
$t\neq s^{-1}$, because $\phi(t)\in G_{2}$ and

$\phi(s^{-1})=\phi(s)^{-1}=((\varepsilon,a_{\beta}),(a_{\beta},\varepsilon))\in
G_{1}.$

Then

$0=\phi(0)=\phi(st)=\phi(s)\cdot\phi(t)=((a_{\beta},\varepsilon),(\varepsilon,a_{\beta}))\cdot((a_{\delta},\varepsilon),(\varepsilon,a_{\delta}))=((a_{\delta}a_{\beta},\varepsilon),(\varepsilon,a_{\beta}a_{\delta}))\neq
0.$

The obtained contradiction implies that if case 2 holds, then
$\phi(G_{1})=G_{2}$ and $\phi(G_{2})=G_{1}$. ∎

So, we either have $\phi(G_{1})=G_{1}$ and $\phi(G_{2})=G_{2}$, or
$\phi(G_{1})=G_{2}$ and $\phi(G_{2})=G_{1}$. In the first case $\phi$
generates a bijection $f:\lambda\rightarrow\lambda$ defined by:
$f(\alpha)=\beta$ iff
$\phi[((\varepsilon,a_{\alpha}),(a_{\alpha},\varepsilon))]=((\varepsilon,a_{\beta}),(a_{\beta},\varepsilon))$.
At this point it is straightforward to check that the restriction of $\phi$ on
$G_{1}\cup G_{2}\cup\\{0\\}$ coincides with the restriction of $\phi_{f}$ on
$G_{1}\cup G_{2}\cup\\{0\\}$. Recall that by Claim 1 the set $G_{1}\cup
G_{2}\cup\\{0\\}$ generates $\mathcal{M}_{\lambda}$. Since both $\phi$ and
$\phi_{f}$ are homomorphisms we get that $\phi=\phi_{f}$.

In the second case $\phi$ generates a bijection $f:\lambda\rightarrow\lambda$
defined by: $f(\alpha)=\beta$ iff
$\phi[((\varepsilon,a_{\alpha}),(a_{\alpha},\varepsilon))]=((a_{\beta},\varepsilon),(\varepsilon,a_{\beta}))$.
At this point it is straightforward to check that the restriction of $\phi$ on
$G_{1}\cup G_{2}\cup\\{0\\}$ coincides with the restriction of $\psi_{f}$ on
$G_{1}\cup G_{2}\cup\\{0\\}$, witnessing that $\phi=\psi_{f}$.

Hence $\mathrm{Aut}(\mathcal{M}_{\lambda})=\\{\phi_{f}:f$ is a bijection of
$\lambda\\}\cup\\{\psi_{f}:f$ is a bijection of $\lambda\\}$. ∎

By $Sym(\lambda)$ we denote the group of permutations of the cardinal
$\lambda$, and by $\mathbb{Z}_{2}$ the two-element group $\\{1,-1\\}$. Due to
Mashevitzky, Schein and Zhitomirski [23], the group of automorphisms of the
free inverse semigroup $\mathbf{F}_{X}$ over a set $X$ of cardinality
$\lambda$ is isomorphic to the wreath product of $Sym(\lambda)$ and
$\mathbb{Z}_{2}$. In particular, if the set $X$ is singleton, then the group
$\mathrm{Aut}(\mathbf{F}_{X})$ is isomorphic to $\mathbb{Z}_{2}$. The next
theorem indicates that McAlister semigroups are related not only to the free
inverse semigroup over a singleton, but also to arbitrary free inverse
semigroups.

######  Theorem 2.7.

The group $\mathrm{Aut}(\mathcal{M}_{\lambda})$ is isomorphic to the direct
product $Sym(\lambda){\times}\mathbb{Z}_{2}$.

###### Proof.

The isomorphism $i:\mathrm{Aut}(\mathcal{M}_{\lambda})\rightarrow
Sym(\lambda){\times}\mathbb{Z}_{2}$ is defined by the following formulae:

$i(\phi_{f})=(f,1)\quad\hbox{ and }\quad i(\psi_{f})=(f,-1),\quad\hbox{ for
each bijection }f\in Sym(\lambda).$

Let us sketch why the map $i$ is an isomorphism. By Proposition 2.6, $i$ is a
bijection. Recall that the set $G_{1}\cup G_{2}\cup\\{0\\}$ generates
$\mathcal{M}_{\lambda}$ (see Claim 1 in the proof of Proposition 2.6). Observe
that

$(\phi_{f}\circ\phi_{g})|_{G_{1}\cup G_{2}\cup\\{0\\}}=\phi_{f\circ
g}|_{G_{1}\cup G_{2}\cup\\{0\\}}.$

Since $\phi_{f}\circ\phi_{g}$ as well as $\phi_{f\circ g}$ are automorphisms
of $\mathcal{M}_{\lambda}$ we deduce that $\phi_{f}\circ\phi_{g}=\phi_{f\circ
g}$. It follows that

$i(\phi_{f})i(\phi_{g})=(f,1)(g,1)=(f\circ g,1)=i(\phi_{f\circ
g})=i(\phi_{f}\circ\phi_{g}).$

Observe that $(\psi_{f}\circ\psi_{g})|_{G_{1}\cup
G_{2}\cup\\{0\\}}=\phi_{f\circ g}|_{G_{1}\cup G_{2}\cup\\{0\\}}$. It follows
that $\psi_{f}\circ\psi_{g}=\phi_{f\circ g}$. Then

$i(\psi_{f})i(\psi_{g})=(f,-1)(g,-1)=(f\circ g,1)=i(\phi_{f\circ
g})=i(\psi_{f}\circ\psi_{g}).$

Repeating the same arguments, one can check that
$\psi_{f}\circ\phi_{g}=\psi_{f\circ g}$ and
$\phi_{f}\circ\psi_{g}=\psi_{f\circ g}$. Then

$i(\psi_{f})i(\phi_{g})=(f,-1)(g,1)=(f\circ g,-1)=i(\psi_{f\circ
g})=i(\psi_{f}\circ\phi_{g})\qquad\hbox{and}$
$i(\phi_{f})i(\psi_{g})=(f,1)(g,-1)=(f\circ g,-1)=i(\psi_{f\circ
g})=i(\phi_{f}\circ\psi_{g}).$

Hence the map $i$ is an isomorphism. ∎

Since the semigroup $\mathcal{M}_{1}$ is isomorphic to the free inverse
semigroup over a singleton with adjoined zero, Theorem 2.7 implies the
following known result.

######  Corollary 2.8.

The group of automorphisms of the free inverse semigroup over a singleton is
isomorphic to $\mathbb{Z}_{2}$.

##  3\. Topological properties of McAlister semigroups

A word $a$ of $F_{\lambda}$ is called a suffix of a word $b\in F_{\lambda}$ if
there exists $c\in F_{\lambda}$ such that $b=ca$.

######  Theorem 3.1.

Each non-zero element of a semitopological semigroup $\mathcal{M}_{\lambda}$
is isolated.

###### Proof.

First, assume that $\lambda=1$ and the free monoid $F_{1}$ is taken over the
singleton $\\{a\\}$. Fix any non-zero element
$((a^{n},a^{m}),(a^{k},a^{l}))\in\mathcal{M}_{1}$. Put
$t=\max\\{n,m,k,l\\}+1$. Note that the elements
$x=((a^{t},a^{t}),(\varepsilon,\varepsilon))$ and
$y=((\varepsilon,\varepsilon),(a^{t},a^{t}))$ are idempotents in
$\mathcal{M}_{1}$. Since shifts are continuous on the Hausdorff space
$\mathcal{M}_{1}$, the retracts

$R_{1}=x\cdot\mathcal{M}_{1},\qquad R_{2}=\mathcal{M}_{1}\cdot x,\qquad
R_{3}=y\cdot\mathcal{M}_{1}\quad\hbox{ and }\quad R_{4}=\mathcal{M}_{1}\cdot
y$

are closed subsets of $\mathcal{M}_{1}$. It is easy to check that the
mentioned above retracts do not contain the point
$((a^{n},a^{m}),(a^{k},a^{l}))$. It follows that the set
$U=\mathcal{M}_{1}\setminus(\cup_{i=1}^{4}R_{i})$ is an open neighborhood of
$((a^{n},a^{m}),(a^{k},a^{l}))$. Let us show that the set $U$ is finite.
Assume that $z=((a^{p},a^{q}),(a^{r},a^{s}))\in U$. Since $z\notin R_{1}$ we
deduce that $p<t$, because otherwise $z=xz\in x\cdot\mathcal{M}_{1}=R_{1}$.
Since $z\notin R_{2}$ we get that $q<t$, because otherwise
$z=zx\in\mathcal{M}_{1}\cdot x=R_{2}$. Similarly one can check that the
assumption $z\notin R_{3}$ provides that $r<t$. Finally, we deduce that $s<t$
from the condition $z\notin R_{4}$. It follows that
$U\subset\\{((a^{p},a^{q}),(a^{r},a^{s}))\in\mathcal{M}_{1}:\max\\{p,q,r,s\\}<t\\}$,
witnessing that the set $U$ is finite. The Hausdorffness of $\mathcal{M}_{1}$
implies that the point $((a^{n},a^{m}),(a^{k},a^{l}))$ is isolated.

Assuming that $\lambda\geq 2$, fix any non-zero point $((x,y),(u,v))$ of the
McAlister semigroup $\mathcal{M}_{\lambda}$ and two distinct words $z_{1}$ and
$z_{2}$ in $F_{\lambda}$ such that $|z_{1}|=|z_{2}|=1$. Observe that the
following statements hold:

  1. (1)

$((x,y),(u,v))\cdot((z_{1}y,z_{1}y),(z_{1}v,z_{1}v))=((z_{1}x,z_{1}y),(z_{1}u,z_{1}v))\neq
0$;

  2. (2)

$((x,y),(u,v))\cdot((z_{2}y,z_{2}y),(z_{2}v,z_{2}v))=((z_{2}x,z_{2}y),(z_{2}u,z_{2}v))\neq
0$;

  3. (3)

$((z_{1}x,z_{1}x),(z_{1}u,z_{1}u))\cdot((x,y),(u,v))=((z_{1}x,z_{1}y),(z_{1}u,z_{1}v))\neq
0$;

  4. (4)

$((z_{2}x,z_{2}x),(z_{2}u,z_{2}u))\cdot((x,y),(u,v))=((z_{2}x,z_{2}y),(z_{2}u,z_{2}v))\neq
0.$

Since $\mathcal{M}_{\lambda}$ is a semitopological semigroup, there exist open
neighborhoods $U_{1}$, $U_{2}$, $U_{3}$, $U_{4}$ of $((x,y),(u,v))$ such that

$U_{1}\cdot((z_{1}y,z_{1}y),(z_{1}v,z_{1}v))\cup
U_{2}\cdot((z_{2}y,z_{2}y),(z_{2}v,z_{2}v))\subset\mathcal{M}_{\lambda}\setminus\\{0\\}\hbox{
and}$ $((z_{1}x,z_{1}x),(z_{1}u,z_{1}u))\cdot
U_{3}\cup((z_{2}x,z_{2}x),(z_{2}u,z_{2}u))\cdot
U_{4}\subset\mathcal{M}_{\lambda}\setminus\\{0\\}.$

Let $U=\cap_{i=1}^{4}U_{i}$ and fix any non-zero element $((a,b),(c,d))\in U$.
By the choice of $U$, the following statements hold:

  1. (1)

$((a,b),(c,d))\cdot((z_{1}y,z_{1}y),(z_{1}v,z_{1}v))\neq 0$;

  2. (2)

$((a,b),(c,d))\cdot((z_{2}y,z_{2}y),(z_{2}v,z_{2}v))\neq 0$;

  3. (3)

$((z_{1}x,z_{1}x),(z_{1}u,z_{1}u))\cdot((a,b),(c,d))\neq 0$;

  4. (4)

$((z_{2}x,z_{2}x),(z_{2}u,z_{2}u))\cdot((a,b),(c,d))\neq 0$.

The above statements imply that the following conditions hold:

  1. (1)

$[b$ is a suffix of $z_{1}y$, or $z_{1}y$ is a suffix of $b]$ and $[d$ is a
suffix of $z_{1}v$, or $z_{1}v$ is a suffix of $d]$;

  2. (2)

$[b$ is a suffix of $z_{2}y$, or $z_{2}y$ is a suffix of $b]$ and $[d$ is a
suffix of $z_{2}v$, or $z_{2}v$ is a suffix of $d]$;

  3. (3)

$[a$ is a suffix of $z_{1}x$, or $z_{1}x$ is a suffix of $a]$ and $[c$ is a
suffix of $z_{1}u$, or $z_{1}u$ is a suffix of $c]$;

  4. (4)

$[a$ is a suffix of $z_{2}x$, or $z_{2}x$ is a suffix of $a]$ and $[c$ is a
suffix of $z_{2}u$, or $z_{2}u$ is a suffix of $c]$.

Note that words $z_{1}y$ and $z_{2}y$ are different and $|z_{1}y|=|z_{2}y|$.
Then the following cases are not possible:

  * •

$z_{1}y$ is a suffix of $b$ and $z_{2}y$ is a suffix of $b$;

  * •

$z_{1}y$ is a suffix of $b$ and $b$ is a suffix of $z_{2}y$;

  * •

$b$ is a suffix of $z_{1}y$ and $z_{2}y$ is a suffix of $b$.

Hence conditions 1 and 2 imply that $b$ is a suffix of $z_{1}y$ and $z_{2}y$,
witnessing that $b$ is a suffix of $y$.

Since the words $z_{1}v$ and $z_{2}v$ are different and $|z_{1}v|=|z_{2}v|$,
the following cases are impossible:

  * •

$z_{1}v$ is a suffix of $d$ and $z_{2}v$ is a suffix of $d$;

  * •

$z_{1}v$ is a suffix of $d$ and $d$ is a suffix of $z_{2}v$;

  * •

$d$ is a suffix of $z_{1}v$ and $z_{2}v$ is a suffix of $d$.

Hence conditions 1 and 2 provide that $d$ is a suffix of $v$. Using similar
arguments, one can check that conditions 3 and 4 ensure that $a$ is a suffix
of $x$ and $c$ is a suffix of $u$. It remains to observe that for a fixed
words $x,y,u,v\in F_{\lambda}$ there are only finitely many elements
$((a,b),(c,d))\in\mathcal{M}_{\lambda}$ such that $a$ is a suffix of $x$, $b$
is a suffix of $y$, $c$ is a suffix $u$ and $d$ is a suffix of $v$. Hence the
neighborhood $U$ is finite, witnessing that the point $((x,y),(u,v))$ is
isolated. ∎

######  Corollary 3.2.

A semitopological free inverse semigroup over a singleton is discrete.

A word $x\in F_{\lambda}$ is called a subword of a word $y\in F_{\lambda}$ if
there exist $a,b\in F_{\lambda}$ such that $y=axb$.

Despite the fact that compact topological semigroups cannot contain neither
bicyclic monoid nor polycyclic monoids [1, 5], the next theorem shows that
McAlister semigroups admit compact inverse semigroup topology.

######  Theorem 3.3.

There exists the unique topology $\tau$ on $\mathcal{M}_{\lambda}$ such that
$(\mathcal{M}_{\lambda},\tau)$ is a compact topological inverse semigroup.

###### Proof.

The topology $\tau$ is defined as follows:

  * •

each non-zero element of $\mathcal{M}_{\lambda}$ is isolated;

  * •

the family $\\{U_{A}:A$ is a finite subset of
$F_{\lambda}\setminus\\{\varepsilon\\}\\}$ forms an open neighborhood base at
$0$, where $U_{A}=\mathcal{M}_{\lambda}\setminus(\cup_{a\in A}D_{a})$.

Obviously, the space $(\mathcal{M}_{\lambda},\tau)$ is Hausdorff. Since each
$\mathscr{D}$-class in $\mathcal{M}_{\lambda}$ is finite (recall Lemma 2.3),
the space $(\mathcal{M}_{\lambda},\tau)$ is compact. Since $D_{a}^{-1}=D_{a}$
for each $a\in F_{\lambda}\setminus\\{\varepsilon\\}$, we have that
$U_{A}^{-1}=U_{A}$ for each finite subset $A\subset
F_{\lambda}\setminus\\{\varepsilon\\}$. It follows that the inversion is
continuous in $(\mathcal{M}_{\lambda},\tau)$. To prove that
$(\mathcal{M}_{\lambda},\tau)$ is a topological semigroup it is sufficient to
show that for each basic open neighborhood $U_{A}$ of $0$ and
$x\in\mathcal{M}_{\lambda}\setminus\\{0\\}$ there exists a basic open
neighborhood $U_{B}$ of $0$ such that $x\cdot U_{B}\cup U_{B}\cdot x\cup
U_{B}\cdot U_{B}\subset U_{A}$. So, fix any basic open neighborhood $U_{A}$ of
$0$, where $A=\\{a_{1},\ldots,a_{n}\\}\subset
F_{\lambda}\setminus\\{\varepsilon\\}$. Let

$B=\\{b:\hbox{ there exists }i\leq n\hbox{ such that }b\hbox{ is a subword of
}a_{i}\\}\setminus\\{\varepsilon\\}.$

We claim that $U_{B}$ is a two-sided ideal of $\mathcal{M}_{\lambda}$. Indeed,
fix any $x\in\mathcal{M}_{\lambda}$ and $y\in D_{q}\subset U_{B}$. Note that
$q\notin B$. If $xy\neq 0$, then Proposition 2.4 implies that $xy\in D_{sqr}$
where $s,r\in F_{\lambda}$. By the choice of $B$, $sqr\in F_{\lambda}\setminus
B$. Thus $xy\in D_{sqr}\subset U_{B}$. Similarly, one can check that $yx\in
U_{B}$. Hence $U_{B}$ is a two-sided ideal of $\mathcal{M}_{\lambda}$. At this
point it is easy to see that $x\cdot U_{B}\cup U_{B}\cdot x\cup U_{B}\cdot
U_{B}\subset U_{B}\subset U_{A}$, witnessing that
$(\mathcal{M}_{\lambda},\tau)$ is a topological inverse semigroup. The
uniqueness of the topology $\tau$ follows from Theorem 3.1. ∎

Eberhart and Selden [8] showed that if the bicyclic monoid ${\mathcal{B}}$ is
a dense proper subsemigroup of a topological semigroup $S$, then
$I=S\setminus{\mathcal{B}}$ is a two-sided ideal of the semigroup $S$. Mesyan,
Mitchell, Morayne and Péresse [25] showed that if a polycyclic monoid
$\mathcal{P}_{\lambda}$ is a dense proper subsemigroup of a topological
semigroup $S$, then $I=(S\setminus{\mathcal{P}_{\lambda}})\cup\\{0\\}$ is a
two-sided ideal of the semigroup $S$. Note that the above statements still
hold if $S$ is a semitopological semigroup [5]. The following lemma shows that
McAlsiter semigroups also possess this property.

######  Lemma 3.4.

Let $S$ be a semitopological semigroup which contains $\mathcal{M}_{\lambda}$
as a dense proper subsemigroup. Then
$(S\setminus\mathcal{M}_{\lambda})\cup\\{0\\}$ is a two-sided ideal in $S$.

###### Proof.

Since $\mathcal{M}_{\lambda}$ is dense in $S$ and $S$ is a semitopological
semigroup, we obtain that $0s=s0=0$ for any $s\in S$. Let us show that
$\mathcal{M}_{\lambda}\cdot(S\setminus\mathcal{M}_{\lambda})\subset(S\setminus\mathcal{M}_{\lambda})\cup\\{0\\}$.
Assuming the contrary, fix $s\in S\setminus\mathcal{M}_{\lambda}$ and
$x\in\mathcal{M}_{\lambda}\setminus\\{0\\}$ such that
$y=xs\in\mathcal{M}_{\lambda}\setminus\\{0\\}$. By Theorem 3.1, $y$ is an
isolated point in $\mathcal{M}_{\lambda}$. Since $\mathcal{M}_{\lambda}$ is
dense in $S$ we get that $y$ is isolated in $S$. The continuity of left shifts
in $S$ implies that there exists an open neighborhood $U$ of $s$ such that
$x\cdot U=\\{y\\}$. Since $s$ is an accumulation point of the set
$\mathcal{M}_{\lambda}$, we have that the set $U\cap\mathcal{M}_{\lambda}$ is
infinite. But this contradicts Lemma 2.5. The obtained contradiction implies
that
$\mathcal{M}_{\lambda}\cdot(S\setminus\mathcal{M}_{\lambda})\subset(S\setminus\mathcal{M}_{\lambda})\cup\\{0\\}$.

Now let us show that $(S\setminus\mathcal{M}_{\lambda})\cup\\{0\\}$ is a left
ideal in $S$. At this point it is sufficient to prove that
$st\in(S\setminus\mathcal{M}_{\lambda})\cup\\{0\\}$ for any $s,t\in
S\setminus\mathcal{M}_{\lambda}$. To derive a contradiction, assume that there
exist elements $s,t\in S\setminus\mathcal{M}_{\lambda}$ such that
$y=st\in\mathcal{M_{\lambda}}\setminus\\{0\\}$. Recall that the point $y$ is
isolated in $S$. Since $S$ is a semitopological semigroup, there exists an
open neighborhood $U$ of $s$ such that $U\cdot t=\\{y\\}$. The density of
$\mathcal{M}_{\lambda}$ in $S$ yields that the open set $U$ contains a point
$x\in\mathcal{M}_{\lambda}\setminus\\{0\\}$. Then
$xt=y\in\mathcal{M}_{\lambda}\setminus\\{0\\}$ which contradicts to the
inclusion
$\mathcal{M}_{\lambda}\cdot(S\setminus\mathcal{M}_{\lambda})\subset(S\setminus\mathcal{M}_{\lambda})\cup\\{0\\}$.
Hence $(S\setminus\mathcal{M}_{\lambda})\cup\\{0\\}$ is a left ideal in $S$.
Similarly it can be showed that $(S\setminus\mathcal{M}_{\lambda})\cup\\{0\\}$
is a right ideal in $S$. ∎

##  4\. Locally compact semitopological McAlister semigroups

Locally compact topological graph inverse semigroups were investigated by
Mesyan, Mitchell, Morayne and Péresse in [25]. They showed that a locally
compact topological graph inverse semigroup over a finite graph $E$ is
discrete. This result implies that a locally compact topological polycyclic
monoids $\mathcal{P}_{k}$, $k\in\mathbb{N}$ are discrete. Gutik and the author
[5] showed that for every non-zero cardinal $\lambda$ a locally compact
topological polycyclic monoid $\mathcal{P}_{\lambda}$ is discrete. Graph
inverse semigroups which admit only the discrete locally compact semigroup
topology were characterized in [4]. In [10] Gutik showed that a locally
compact semitopological bicyclic monoid with adjoined zero (i.e., polycyclic
monoid $\mathcal{P}_{1}$) is either compact or discrete. It turns out that
this kind of dichotomy also holds for some generalizations of the bicyclic
monoid, including all polycyclic monoids.

######  Proposition 4.1.

Assume that $\mathcal{M}_{k}$ is a locally compact semitopological semigroup
for some positive integer $k$. Let $U$ be a compact open neighborhood of $0$
and $A_{U}=\\{a\in F_{\lambda}:D_{a}\cap U\neq\emptyset\\}$. Then
$D_{a}\subset U$ for all but finitely many $a\in A_{U}$.

###### Proof.

Assume that $\mathcal{M}_{k}$ is a locally compact semitopological semigroup.
Theorem 3.1 implies that all non-zero points in $\mathcal{M}_{k}$ are
isolated. Consequently, each open neighborhood of $0$ is closed. Let $U$ be
any compact neighborhood of $0$. The compactness of $U$ implies that for each
open neighborhood $V\subset U$ of $0$ the set $U\setminus V$ is finite.

Enumerate $A_{U}=\\{a_{n}:n\in\mathbb{N}\\}$. To derive a contradiction,
assume that there exists an infinite subset $B\subset\mathbb{N}$ such that
$D_{a_{n}}\setminus U\neq\emptyset$ for each $n\in B$. Let us imagine
$\mathscr{D}$-classes $D_{a_{n}},n\in B$ using the classical “egg-box” picture
[7, Chapter 2]. Then $D_{a_{n}}$ can be associated with the set
$S_{n}=\\{(i,j):i,j\leq|a_{n}|+1\\}$ (recall Lemma 2.3). If for some set $A$
(in our case the role of the set $A$ will play $U$) the sets $S_{n}\setminus
A$ and $S_{n}\cap A$ are nonempty, then there exists an element $(i,j)\in
A\cap S_{n}$ such that at least one of its “neighbors” is not in $A$. That is

$(i-1,j)\in S_{n}\setminus A,\quad\hbox{ or }\quad(i,j-1)\in S_{n}\setminus
A,\quad\hbox{ or }\quad(i+1,j)\in S_{n}\setminus A,\quad\hbox{ or
}\quad(i,j+1)\in S_{n}\setminus A.$

Note that the notion of “neighbor” in $D_{a_{n}}$ is a bit different. For an
element

$((b_{1}\ldots b_{n},c_{1}\ldots c_{m}),(d_{1}\ldots d_{k},e_{1}\ldots
e_{p}))\in D_{a_{n}}$

the set of its “neighbors” consists of the following four elements:

  * •

$((b_{1}\ldots b_{n-1},c_{1}\ldots c_{m}),(d_{1}\ldots d_{k}b_{n},e_{1}\ldots
e_{p}))$;

  * •

$((b_{1}\ldots b_{n},c_{1}\ldots c_{m-1}),(d_{1}\ldots d_{k},e_{1}\ldots
e_{p}c_{m}))$;

  * •

$((b_{1}\ldots b_{n}d_{k},c_{1}\ldots c_{m}),(d_{1}\ldots d_{k-1},e_{1}\ldots
e_{p}))$;

  * •

$((b_{1}\ldots b_{n},c_{1}\ldots c_{m}e_{p}),(d_{1}\ldots d_{k},e_{1}\ldots
e_{p-1}))$.

Hence if we formalize the preceeding arguments, then we obtain that for each
$n\in B$ there exist words $b_{n},c_{n},d_{n},e_{n}\in F_{k}$ and a word
$x_{n}\in F_{k}$ with $|x_{n}|=1$ such that at least one of the following
cases hold:

  1. (1)

$((b_{n}x_{n},c_{n}),(d_{n},e_{n}))\in U\cap D_{a_{n}}\quad$ and
$\quad((b_{n},c_{n}),(d_{n}x_{n},e_{n}))\in D_{a_{n}}\setminus U;$

  2. (2)

$((b_{n},c_{n}x_{n}),(d_{n},e_{n}))\in U\cap D_{a_{n}}\quad$ and
$\quad((b_{n},c_{n}),(d_{n},e_{n}x_{n}))\in D_{a_{n}}\setminus U;$

  3. (3)

$((b_{n},c_{n}),(d_{n}x_{n},e_{n}))\in U\cap D_{a_{n}}\quad$ and
$\quad((b_{n}x_{n},c_{n}),(d_{n},e_{n}))\in D_{a_{n}}\setminus U;$

  4. (4)

$((b_{n},c_{n}),(d_{n},e_{n}x_{n}))\in U\cap D_{a_{n}}\quad$ and
$\quad((b_{n},c_{n}x_{n}),(d_{n},e_{n}))\in D_{a_{n}}\setminus U.$

The Pigeonhole principle implies that there exist an infinite subset $C\subset
B$ and $i\in\\{1,2,3,4\\}$ such that for each $n\in C$ case $i$ holds. Since
the free monoid $F_{k}$ contains precisely $k$ words of length $1$, using one
more time the Pigeonhole principle we can find an infinite subset $D\subset C$
such that the words $x_{n}$ and $x_{m}$ coincide, whenever $n,m\in D$. For
each $n\in D$ let us denote $x_{n}$ simply by $x$. At this point we have four
possibilities:

$(i=1)$ In this case for each $n\in D$

$((b_{n}x,c_{n}),(d_{n},e_{n}))\in U\cap D_{a_{n}}\quad\hbox{ and
}\quad((b_{n},c_{n}),(d_{n}x,e_{n}))\notin U\cap D_{a_{n}}.$

Consider the product $((\varepsilon,x),(x,\varepsilon))\cdot 0=0$. Since the
semigroup $\mathcal{M}_{k}$ is semitopological, there exists an open
neighborhood $V\subset U$ of $0$ such that
$((\varepsilon,x),(x,\varepsilon))\cdot V\subset U$. Observe that for each
$n\in D$

$((\varepsilon,x),(x,\varepsilon))\cdot((b_{n}x,c_{n}),(d_{n},e_{n}))=((b_{n},c_{n}),(d_{n}x,e_{n}))\notin
U.$

It follows that $((b_{n}x,c_{n}),(d_{n},e_{n}))\notin V$ for each $n\in D$.
Since the elements $((b_{n}x,c_{n}),(d_{n},e_{n}))$, $n\in D$ belong to
different $\mathscr{D}$-classes, we get that the set
$T=\\{((b_{n}x,c_{n}),(d_{n},e_{n})):n\in D\\}$ is infinite. Moreover, $T$ is
contained in $U\setminus V$ which contradicts to the fact that the set
$U\setminus V$ is finite.

$(i=2)$ In this case for each $n\in D$

$((b_{n},c_{n}x),(d_{n},e_{n}))\in U\cap D_{a_{n}}\quad\hbox{ and
}\quad((b_{n},c_{n}),(d_{n},e_{n}x))\notin U\cap D_{a_{n}}.$

Consider the product $0\cdot((x,\varepsilon),(\varepsilon,x))=0$. There exists
an open neighborhood $V\subset U$ of $0$ such that
$V\cdot((x,\varepsilon),(\varepsilon,x))\subset U$. Observe that for each
$n\in D$

$((b_{n},c_{n}x),(d_{n},e_{n}))\cdot((x,\varepsilon),(\varepsilon,x))=((b_{n},c_{n}),(d_{n},e_{n}x))\notin
U.$

Hence the infinite set $\\{((b_{n},c_{n}x),(d_{n},e_{n})):n\in D\\}$ is
contained in $U\setminus V$ which implies a contradiction.

$(i=3)$ In this case for each $n\in D$

$((b_{n},c_{n}),(d_{n}x,e_{n}))\in U\cap D_{a_{n}}\quad\hbox{ and
}\quad((b_{n}x,c_{n}),(d_{n},e_{n}))\notin U\cap D_{a_{n}}.$

Consider the product $((x,\varepsilon),(\varepsilon,x))\cdot 0=0$. There
exists an open neighborhood $V\subset U$ of $0$ such that
$((x,\varepsilon),(\varepsilon,x))\cdot V\subset U$. Observe that for each
$n\in D$

$((x,\varepsilon),(\varepsilon,x))\cdot((b_{n},c_{n}),(d_{n}x,e_{n}))=((b_{n}x,c_{n}),(d_{n},e_{n}))\notin
U.$

Hence the infinite set $\\{((b_{n},c_{n}),(d_{n}x,e_{n})):n\in D\\}$ is
contained in $U\setminus V$ which implies a contradiction.

$(i=4)$ In this case for each $n\in D$

$((b_{n},c_{n}),(d_{n},e_{n}x))\in U\cap D_{a_{n}}\quad\hbox{ and
}\quad((b_{n},c_{n}x),(d_{n},e_{n}))\notin U\cap D_{a_{n}}.$

Consider the product $0\cdot((\varepsilon,x),(x,\varepsilon))=0$. There exists
an open neighborhood $V\subset U$ of $0$ such that
$V\cdot((\varepsilon,x),(x,\varepsilon))\subset U$. Observe that for each
$n\in D$

$((b_{n},c_{n}),(d_{n},e_{n}x))\cdot((\varepsilon,x),(x,\varepsilon))=((b_{n},c_{n}x),(d_{n},e_{n}))\notin
U.$

Hence the infinite set $\\{((b_{n},c_{n}),(d_{n},e_{n}x)):n\in D\\}$ is
contained in $U\setminus V$ which implies a contradiction.

The obtained contradiction completes the proof of the proposition. ∎

######  Theorem 4.2.

A locally compact semitopological semigroup $\mathcal{M}_{1}$ is either
compact or discrete.

###### Proof.

If we assume that the free monoid $F_{1}$ is taken over the singleton
$\\{a\\}$, then it is easy to see that non-zero $\mathscr{D}$-classes in
$\mathcal{M}_{1}$ have the form $D_{a^{n}}$, $n\in\mathbb{N}$. Assuming that
$\mathcal{M}_{1}$ is not discrete, fix any compact infinite open neighborhood
$U$ of $0$. Let $A=\\{n\in\mathbb{N}:D_{a^{n}}\cap U\neq\emptyset\\}$. Since
$\mathscr{D}$-classes are finite (see Lemma 2.3), the set $A$ is infinite. To
derive a contradiction, assume that the set $B=\mathbb{N}\setminus A$ is
infinite as well. In this case there exists an infinite subset
$\\{n_{k}:k\in\mathbb{N}\\}\subset A$ such that
$\\{n_{k}+1:k\in\mathbb{N}\\}\subset B$. Consider the product
$((a,a),(\varepsilon,\varepsilon))\cdot 0=0$. The continuity of left shifts in
$\mathcal{M}_{1}$ implies that there exists an open neighborhood $V\subset U$
of $0$ such that $((a,a),(\varepsilon,\varepsilon))\cdot V\subset U$.
Proposition 4.1 ensures that $D_{a^{n_{k}}}\subset U$ for all but finitely
many $k\in\mathbb{N}$. Since the set $U\setminus V$ is finite, there exists
$k\in\mathbb{N}$ such that $D_{a^{n_{k}}}\subset V$, in particular,
$((\varepsilon,\varepsilon),(a^{n_{k}},a^{n_{k}}))\in V$. It follows that

$((a,a),(\varepsilon,\varepsilon))\cdot((\varepsilon,\varepsilon),(a^{n_{k}},a^{n_{k}}))=((a,a),(a^{n_{k}},a^{n_{k}}))\in
D_{a^{n_{k}+1}}\subset\mathcal{M}_{1}\setminus U.$

The obtained contradiction implies that the set $B=\mathbb{N}\setminus A$ is
finite. By Proposition 4.1, all but finitely many $\mathscr{D}$-classes are
contained in $U$. Lemma 2.3 implies that the set $\mathcal{M}_{1}\setminus U$
is finite. Hence the semitopological semigroup $\mathcal{M}_{1}$ is compact. ∎

Since the free inverse semigroup over a singleton with adjoined zero is
isomorphic to the semigroup $\mathcal{M}_{1}$, Theorem 4.2 implies the
following:

######  Corollary 4.3.

A locally compact semitopological free inverse semigroup over a singleton with
adjoined zero is either compact or discrete.

######  Lemma 4.4.

Each two-sided ideal $I\subset F_{\lambda}$ generates a locally compact
inverse semigroup topology $\tau_{I}$ on $\mathcal{M}_{\lambda}$.

###### Proof.

We define $\tau_{I}$ as follows: each non-zero element of
$(\mathcal{M}_{\lambda},\tau_{I})$ is isolated and an open neighborhood base
at $0$ consists of the sets $U_{A}=\cup_{a\in A}{D_{a}}\cup\\{0\\}$, where
$A\subset I$, $|I\setminus A|<\omega$. Since $\mathscr{D}$-classes are finite
in $\mathcal{M}_{\lambda}$ it is easy to see that the space
$(\mathcal{M}_{\lambda},\tau_{I})$ is locally compact and Hausdorff. Note that
$U_{A}^{-1}=U_{A}$ for each $A\subset I$. Hence the inversion is continuous in
$(\mathcal{M}_{\lambda},\tau_{I})$. Fix any $A\subset I$ such that the set
$B=I\setminus A$ is finite. Enumerate $B=\\{b_{1},\ldots,b_{n}\\}$ and let

$C=\\{x\in F_{\lambda}:x\hbox{ is a subword of }b_{i}\hbox{ for some }i\leq
n\\}\setminus\\{\varepsilon\\}.$

Note that the set $C$ is finite and put $J=I\setminus C$. One can easily check
that $J$ is a two-sided ideal in $F_{\lambda}$. We claim that $U_{J}$ is a two
sided ideal in $\mathcal{M}_{\lambda}$. Indeed, fix any
$((a,b),(c,d))\in\mathcal{M}_{\lambda}$ and $((x,y),(u,v))\in D_{p}\subset
U_{J}$. Note that $p\in J$. Proposition 2.4 implies that either ${\bf
z}=((a,b),(c,d))\cdot((x,y),(u,v))=0\in U_{J}$ or there exist words $s,r\in
F_{\lambda}$ such that ${\bf z}\in D_{spr}$. Since $J$ is a two-sided ideal in
$F_{\lambda}$, the word $spr$ belongs to $J$ which implies that ${\bf z}\in
D_{spr}\subset U_{J}$. Hence $U_{J}$ is a left ideal in
$\mathcal{M}_{\lambda}$. Similarly it can be shown that $U_{J}$ is a right
ideal. Thus, for each $((a,b),(c,d))\in\mathcal{M}_{\lambda}$ we have that

$((a,b),(c,d))\cdot U_{J}\cup U_{J}\cdot((a,b),(c,d))\cup U_{J}\cdot
U_{J}\subset U_{J}\subset U_{A}.$

It follows that $(\mathcal{M}_{\lambda},\tau_{I})$ is a topological inverse
semigroup. ∎

A family $\mathcal{A}$ of infinite subsets of $\mathbb{N}$ is called almost
disjoint if $|A\cap B|<\omega$ for each $A,B\in\mathcal{A}$. By Theorem 1.3
from [16], there exists an almost disjoint family
$\mathcal{A}\subset\mathcal{P}(\mathbb{N})$ of cardinality continuum. The next
proposition shows that Theorem 4.2 cannot be generalized for the McAlister
semigroup $\mathcal{M}_{2}$.

######  Proposition 4.5.

The semigroup $\mathcal{M}_{2}$ admits continuum many distinct locally compact
inverse semigroup topologies.

###### Proof.

Assume that the free monoid $F_{2}$ is taken over the set $\\{a,b\\}$ and
consider the ideals $I_{n}=F_{2}ab^{n}aF_{2}\subset F_{2}$, $n\in\mathbb{N}$.
Note that $ab^{n}a\in I_{m}$ iff $n=m$ for each $n,m\in\mathbb{N}$. Let
$\mathcal{A}$ be any almost disjoint family of infinite subsets of
$\mathbb{N}$ of cardinality continuum. For each $A\in\mathcal{A}$ let
$J_{A}=\cup_{n\in A}I_{n}$ and note that $J_{A}$ is an ideal in $F_{2}$. We
claim that the set $J_{A}\setminus J_{B}$ is infinite for each distinct
$A,B\in\mathcal{A}$. Indeed, consider the infinite subset $C=\\{ab^{n}a:n\in
A\\}\subset J_{A}$. Recall that $ab^{n}a\in J_{B}$ iff $n\in B$. Since the
sets $A$ and $B$ are almost disjoint the set $C\setminus J_{B}\subset
J_{A}\setminus J_{B}$ is infinite. Therefore, for each distinct
$A,B\in\mathcal{A}$ the locally compact topologies $\tau_{J_{A}}$ and
$\tau_{J_{B}}$ (see Lemma 4.4) are distinct. Recall that by Proposition 2.6,
there exist only four different automorphisms of $\mathcal{M}_{2}$. The
Pigeonhole principle implies that we can find a subfamily
$\\{\tau_{\alpha}:\alpha\in\mathfrak{c}\\}\subset\\{\tau_{J_{A}}:A\in\mathcal{A}\\}$
such that for each $\alpha\neq\beta$ the locally compact topological inverse
semigroups $(\mathcal{M}_{2},\tau_{\alpha})$ and
$(\mathcal{M}_{2},\tau_{\beta})$ are not topologically isomorphic. ∎

The next lemma provides a method of constructing locally compact inverse
semigroup topologies on $\mathcal{M}_{\lambda}$, $\lambda\geq\omega$. Also, it
shows that Proposition 4.1 does not hold in the case $\lambda\geq\omega$.

######  Lemma 4.6.

For any infinite cardinal $\lambda$ there exists a locally compact inverse
semigroup topology on $\mathcal{M}_{\lambda}$ such that $0$ has an infinite
compact neighborhood $U$ intersecting each $\mathscr{D}$-class by at most one
element.

###### Proof.

Assume that the free monoid $F_{\lambda}$ is taken over the set
$\\{a_{\alpha}:\alpha\in\lambda\\}$. For each $n\in\omega\subset\lambda$ put
$U_{n}=\\{((a_{k},a_{k}),(a_{k},a_{k})):n<k<\omega\\}\cup\\{0\\}$. Let $\tau$
be the topology on $\mathcal{M}_{\lambda}$ defined as follows: each non-zero
element of $\mathcal{M}_{\lambda}$ is isolated and the family
$\\{U_{n}:n\in\omega\\}$ forms an open neighborhood base at $0$. Obviously,
the space $(\mathcal{M}_{\lambda},\tau)$ is Hausdorff and locally compact
($U_{0}$ is a compact neighborhood of $0$). Let us check that $\tau$ is an
inverse semigroup topology. Since $U_{n}^{-1}=U_{n}$ the inversion is
continuous. To prove the continuity of the multiplication in
$(\mathcal{M}_{\lambda},\tau)$ it is sufficient to consider the following
cases:

  1. (1)

$0\cdot 0=0$;

  2. (2)

$x\cdot 0=0$, where $x\in\mathcal{M}_{\lambda}\setminus\\{0\\}$;

  3. (3)

$0\cdot x=0$, where $x\in\mathcal{M}_{\lambda}\setminus\\{0\\}$.

1) Note that $U_{n}\cdot U_{n}=U_{n}$ for each $n\in\omega$. It follows that
the semigroup operation is continuous in case 1.

Fix any non-zero element $x=((a,b),(c,d))\in\mathcal{M}_{\lambda}$.

2) If $b\neq\varepsilon$, then let $p$ be the last letter of the word $b\in
F_{\lambda}$. Clearly, there exists $n\in\omega$ such that
$((p,p),(p,p))\notin U_{n}$. Then $((a,b),(c,d))\cdot U_{n}=\\{0\\}$. If
$b=\varepsilon$, then the definition of $\mathcal{M}_{\lambda}$ implies that
$d\neq\varepsilon$. Let $p$ be the last letter of the word $d\in F_{\lambda}$.
There exists $n\in\omega$ such that $((p,p),(p,p))\notin U_{n}$. Then
$((a,b),(c,d))\cdot U_{n}=\\{0\\}$, witnessing that the semigroup operation is
continuous in case 2.

3) If $a\neq\varepsilon$, let $p$ be the last letter of the word $a\in
F_{\lambda}$. Clearly, there exists $n\in\omega$ such that
$((p,p),(p,p))\notin U_{n}$. Then $U_{n}\cdot((a,b),(c,d))=\\{0\\}$. If
$a=\varepsilon$, then the definition of $\mathcal{M}_{\lambda}$ implies that
$c\neq\varepsilon$. Let $p$ be the last letter of the word $c\in F_{\lambda}$.
There exists $n\in\omega$ such that $((p,p),(p,p))\notin U_{n}$. Then
$U_{n}\cdot((a,b),(c,d))=\\{0\\}$, witnessing that the semigroup operation is
continuous in case 3.

Hence $\tau$ is an inverse semigroup topology on $\mathcal{M}_{\lambda}$.
Lemma 2.2 ensures that for each $n,m\in\omega$,
$((a_{n},a_{n}),(a_{n},a_{n}))\mathscr{D}((a_{m},a_{m}),(a_{m},a_{m}))$ iff
$n=m$. It follows that $U_{0}$ is compact neighborhood of $0$ such that
$|U_{0}\cap D_{u}|\leq 1$ for each $u\in
F_{\lambda}\setminus\\{\varepsilon\\}$. ∎

## References

  * [1] L. Anderson, R. Hunter and R. Koch, Some results on stability in semigroups, Trans. Amer. Math. Soc. 117 (1965), 521–529.
  * [2] T. Banakh, S. Dimitrova and O. Gutik, _Embedding the bicyclic semigroup into countably compact topological semigroups_ , Topology Appl. 157:18 (2010), 2803–2814.
  * [3] S. Bardyla, _On universal objects in the class of graph inverse semigroups_ , Eur. J. Math. 6 (2020), 4–13.
  * [4] S. Bardyla, _On locally compact topological graph inverse semigroups_ , Topology Appl. 267 (2019), 106873.
  * [5] S. Bardyla, O. Gutik, _On a semitopological polycyclic monoid_ , Algebra Discr. Math. 21:2 (2016), 163–183.
  * [6] M. Bertman, T. West, _Conditionally compact bicyclic semitopological semigroups_ , Proc. Roy. Irish Acad. A76:21–23 (1976), 219–226.
  * [7] A. H. Clifford, G. B. Preston, _The Algebraic Theory of Semigroups_ , Vol. I, Amer. Math. Soc. Surveys 7, 1961.
  * [8] C. Eberhart, J. Selden, _On the closure of the bicyclic semigroup_ , Trans. Amer. Math. Soc. 144 (1969), 115–126.
  * [9] R. Engelking, _General Topology_ , 2nd ed., Heldermann, Berlin, 1989.
  * [10] O. Gutik, _On the dichotomy of the locally compact semitopological bicyclic monoid with adjoined zero,_ Visn. L’viv. Univ., Ser. Mekh.-Mat. 80 (2015), 33–41.
  * [11] O. Gutik, D. Repov̌s, _On countably compact 0-simple topological inverse semigroups_ , Semigroup Forum 75:2 (2007), 464–469.
  * [12] W. Hajji, B. Steinberg, _A parametrization of the irreducible representation of a compact inverse semigroup_ , Commun. Algebra, 43 (2015), 5261–5281.
  * [13] J. Hildebrant and R. Koch, Swelling actions of $\Gamma$-compact semigroups, Semigroup Forum 33:1 (1986), 65–85.
  * [14] D. Jones, M. Lawson, _Graph inverse semigroups: Their characterization and completion_ , Journal of Algebra 409 (2014), 444–473.
  * [15] D. Jones, M. Lawson, _Strong representations of the polycyclic inverse monoids: Cycles and atoms,_ Periodica Mathematica Hungarica 64 (2012), 53–87.
  * [16] K. Kunen, _Set theory_ , Studies in Logic and the Foundations of Mathematics, vol. 102, North-Holland Publishing Company, (1980).
  * [17] M. Lawson, _Inverse Semigroups. The Theory of Partial Symmetries_ , World Scientific, Singapore, 1998.
  * [18] M. Lawson, _The Polycyclic Monoids $P_{n}$ and the Thompson Groups $V_{n,1}$_, Commun. Algebra 35:12 (2007), 4068–4087.
  * [19] M. Lawson, _Primitive partial permutation representations of the polycyclic monoids and branching function systems_ , Period. Math. Hungar. 58 (2009), 189–207.
  * [20] M. Lawson, _McAlister Semigroups_ , Journal of Algebra 202 (1998), 276–294.
  * [21] M. Lawson, _The polycyclic inverse monoids and the Thompson groups revisited_ , preprint, arXiv:2006.15338.
  * [22] G. Mashevitzky, B. Schein, _Automorphisms of the endomorphism semigroup of a free monoid or a free semigroup_ , Proc. Amer. Math. Soc. 131:6 (2003), 1655–1660.
  * [23] G. Mashevitzky, B. Schein and G. Zhitomirski, _Automorphisms of the endomorphism semigroup of a free inverse semigroup_ , Commun. Algebra 34 (2006), 3569–3584.
  * [24] D. McAlister, _Inverse semigroups which are separated over a subsemigroup_ , Trans. Amer. Math. Soc. 182 (1973), 85–117.
  * [25] Z. Mesyan, J. Mitchell, M. Morayne and Y. Péresse, _Topological graph inverse semigroups_ , Topology Appl. 208 (2016), 106–-126.
  * [26] M. Nivat, J.-F. Perrot, _Une généralisation du monoide bicyclique_ , C. R. Acad. Sci., Paris, Sér. A 271 (1970), 824–827.
  * [27] O. Poliakova, B. Schein, _A new construction for free inverse semigroups_ , Journal of Algebra 288 (2005), 20–58.