# Introduction to inverse semigroups

Mark V. Lawson  Department of Mathematics and the Maxwell Institute for
Mathematical Sciences
Heriot-Watt University
Riccarton
Edinburgh EH14 4AS
Scotland
<EMAIL_ADDRESS>
###### Abstract.

This is an account of the theory of inverse semigroups, assuming only that the
reader knows the basics of semigroup theory.

##  1\. Introduction: a little history

Inverse semigroups111The early development of inverse semigroup theory ocurred
at a time of acute tensions between East and West. See the book by Christopher
Hollings [35] for an in-depth analysis. were introduced in the 1950’s by
Wagner [115] in the Soviet Union222The book [36] contains more information on
Wagner and his approach to mathematics. See also Wagner’s obituary by Boris
Schein [105]., by Preston [98] in the UK333The connection between Preston’s
introduction of what he termed ‘inverse semi-groups’ and the theory of
pseudogroups of transformations is the result of a conversation I had with
Preston in Australia in the late 1980’s., and by Ehresmann [20] in France444It
is hard to single out a single paper of Ehresmann as the starting point for
his approach to inverse semigroups — which was via a class of ordered
groupoids — since he stayed faithful to his interest in pseudogroups as the
way in which local structures were defined. In any event, Ehresmann’s papers
in this area reflect what might be called ‘an evolving corpus of work’. See
[21]. as algebraic analogues of pseudogroups of transformations. Inverse
semigroups can be viewed as generalizations of groups. Whereas group theory is
based on the notion of a symmetry — that is, a structure-preserving bijection
— inverse semigroup theory is based on the notion of a partial symmetry — that
is, a structure-preserving partial bijection. The passage from bijections to
partial bijections introduces a host of new algebraic phenomena.

In writing this chapter, I have assumed the reader is familiar with the basics
of semigroup theory such as could be gleaned from the first few chapters of
Howie [37].555On page 128, Howie quotes Clifford and Preston as saying that
inverse semigroups were the most promising class of semigroups for future
study. This prediction has turned out to be correct. Here are a few things you
should know and which I take for granted. The multiplication in a semigroup
will usually be denoted by $\cdot$ or by concatenation. If $S$ is any
semigroup then $S^{0}$ is the semigroup $S$ with an adjoined zero, and $S^{1}$
is the semigroup $S$ with an adjoined identity. If $A$ and $B$ are any subsets
of a semigroup $S$ then $AB$ is the set of all products where $a\in A$ and
$b\in B$. The study of congruences in semigroups is unavoidable since there
are not, in general, the analogues of normal subgroups as in groups or ideals
as in rings. If $\theta\colon S\rightarrow T$ is a homomorphism between
semigroups then $\mbox{\rm ker}(\theta)$ is the congruence that $\theta$
induces on $S$. On the other hand, every congruence $\rho$ on a semigroup $S$
has an associated natural homomorphism from $S$ to $S/\rho$ which maps $s$ to
$\rho(s)$. If both $S$ and $T$ are monoids then a monoid homomorphism maps the
identity of $S$ to the identity of $T$; if $S$ and $T$ are both semigroups
with zero, then a homomorphism of such semigroups maps the zero to the zero.
If the semigroups $S$ and $T$ are isomorphic then we write $S\cong T$. A
subset $I\subseteq S$ is said to be a right ideal if $IS\subseteq I$ and a
left ideal if $SI\subseteq I$. It is said to be an ideal if $SI\subseteq I$
and $IS\subseteq I$. If $a\in S$ then we write $a$ instead of $\\{a\\}$. The
right ideal $aS^{1}$ is called the principal right ideal generated by $a$,
whereas the left ideal $S^{1}a$ is called the principal left ideal generated
by $a$. We call $S^{1}aS^{1}$ the principal ideal generated by $a$. Our use of
$S^{1}$ is just a trick to ensure that $a$ belongs to the ideal. Define
$a\,\mathscr{R}\,b$ if $aS^{1}=bS^{1}$ and $a\,\mathscr{L}\,b$ if
$S^{1}a=S^{1}b$. Recall that
$\mathscr{L}\circ\mathscr{R}=\mathscr{R}\circ\mathscr{L}$. Put
$\mathscr{D}=\mathscr{L}\circ\mathscr{R}$, another equivalence relation. Put
$\mathscr{H}=\mathscr{L}\cap\mathscr{R}$. Define $a\,\mathscr{J}\,b$ if
$S^{1}aS^{1}=S^{1}bS^{1}$. These are the familiar Green’s relations. As usual,
if $\mathscr{G}$ is one of Green’s relations, then the $\mathscr{G}$-class
that contains the element $a$ is denoted by $G_{a}$. Although ideals are
useful in semigroup theory, the connection between ideals and congruences is
weaker for semigroups than it is for rings. If $\rho$ is a congruence on a
semigroup with zero $S$, then the set $I=\rho(0)$ is an ideal of $S$; however,
examples show that the congruence is not determined by this ideal.
Nevertheless, ideals can be used to construct some congruences on semigroups.
Let $I$ be an ideal in the semigroup $S$. Define a relation $\rho_{I}$ on $S$
by:

$(s,t)\in\rho_{I}\Leftrightarrow\mbox{either }s,t\in I\mbox{ or }s=t.$

Then $\rho_{I}$ is a congruence. The quotient semigroup $S/\rho_{I}$ is
isomorphic to the set $(S\setminus I)\cup\\{0\\}$ (we may assume that $0\notin
S\setminus I$) equipped with the following product: if $s,t\in S\setminus I$
then their product is $st$ if $st\in S\setminus I$, all other products are
defined to be $0$. Such quotients are called Rees quotients.

There are currently two books entirely devoted to inverse semigroup theory:
Petrich’s [97] and mine [51]. Petrich’s book is pretty comprehensive up to
1984 and is still a useful reference. Its only drawback is the poor index
which makes finding particular topics a bit of a chore. My book is less
ambitious; its goal is to motivate the study of inverse semigroups by
concentrating on concrete examples and was completed in 1998.

In writing this chapter, I have drawn mainly upon my own book [51] and some
notes I wrote [63] for a workshop at the University of Ottawa in 2010, but I
have taken the opportunity to radically rethink what I wrote there in the
light of subsequent research.

Any results below that are stated but not proved should be treated as
exercises.

Acknowledgements. I would like to thank Victoria Gould, Bernard Rybolowicz and
Aidan Sims for reading the first version of this chapter and giving me good
notes.

##  2\. Motivation: pseudogroups of transformations

Good mathematics has to be based on good examples. For example, finite
semigroup theory grew out of the theory of finite-state automata since a
finite semigroup can be associated with every finite-state automaton.
Similarly, inverse semigroup theory grew out of the theory of pseudogroups of
transformations, which goes back to the works of Sophus Lie and Henri Cartan.
If you have ever taken an introductory course in differential geometry, then
you have been exposed to pseudogroups of transformations even if that phrase
was never once uttered. So, there I shall begin. However, for those of you who
have not taken such a course, I shall quickly move on to the collection of all
partial bijections on a set, jettisoning any topology or differential
structure, and this will take us to the core motivation for studying inverse
semigroups.

Let $X$ be a nice space (for example, $X$ should be Hausdorff but this will
play no role in what follows). Let $\mathbb{R}^{n}$ be the usual space of real
$n$-vectors. We shall coordinatize $X$ using $\mathbb{R}^{n}$ but we shall
only do so locally not globally. Thus we assume that for each point $x\in X$
there exists an open set $x\in U$ and a homeomorphism $\phi\colon U\rightarrow
V$, called a chart, where $V$ is an open subset of $\mathbb{R}^{n}$. A chart
gives one way of coordinatizing elements of $U$ by means of elements of $V$.
However, it is quite possible that $x$ lies in another chart. In other words,
there could well exist another open subset of $X$, let’s say $U^{\prime}$,
which also contains $x$ and for which there is a homeomorphism $\psi\colon
U^{\prime}\rightarrow V^{\prime}$ where $V^{\prime}$ is also an open subset of
$\mathbb{R}^{n}$. This map gives another way of coordinatizing elements of
$U^{\prime}$ by means of elements of $V^{\prime}$. However, for any element in
$y\in U\cap U^{\prime}$, there are now two possible co-ordinate
representations: $\phi(y)$ and $\psi(y)$. How are they to be related? This
depends on the nature of the map $\psi\phi^{-1}\colon\phi(U\cap
U^{\prime})\rightarrow\psi(U\cap U^{\prime})$, called the change of co-
ordinates map, which connects them. This map is a homeomorphism from the open
subset $\phi(U\cap U^{\prime})$ of $\mathbb{R}^{n}$ to the open subset
$\psi(U\cap U^{\prime})$ of $\mathbb{R}^{n}$. If the map $\psi\phi^{-1}$ is
always assumed to be smooth then we say that we have defined on $X$ the
structure of a differential manifold [106].

We now dispense with $X$ and concentrate instead on the set
$\Gamma(\mathbb{R}^{n})$ of all smooth maps between the open subsets of
$\mathbb{R}^{n}$. This is just the set of all possible changes of co-ordinates
and is our first example of a pseudogroup. Usually, it is the set $X$ which is
the centre of attention, but we have shifted focus to the set of allowable co-
ordinate transformations; if we change the nature of those, then we change the
nature of the structure we are defining on $X$.

More generally, let $X$ be any topological space with topology $\tau$. Denote
the set of all homeomorphisms between open subsets of $X$ by
$\mathcal{I}(X,\tau)$. If $\tau$ is the usual topology on $\mathbb{R}^{n}$
then $\Gamma(\mathbb{R}^{n})$ is a subset of
$\mathcal{I}(\mathbb{R}^{n},\tau)$ closed under certain operations. The sets
$\mathcal{I}(X,\tau)$ are also examples of pseudogroups. The terminology
‘pseudogroup’ comes from the fact that, just like groups of transformations,
the sets $\mathcal{I}(X,\tau)$ are closed under the composition of partial
functions — don’t forget the empty function! — and are also closed under
‘inverses’ where we interpret inverses as inverses of partial functions. The
set $\Gamma(\mathbb{R}^{n})$ is likewise closed under composition of partial
functions and said inverses. There is a further property that is needed to
qualify as a pseudogroup. It is this. If $\\{\phi_{i}\colon i\in I\\}$ is any
subset of $\mathcal{I}(X,\tau)$ such that $\bigcup_{i\in I}\phi_{i}$ is a
partial bijection then, in fact, $\bigcup_{i\in
I}\phi_{i}\in\mathcal{I}(X,\tau)$. This is a completeness property. We shall
call semigroups of the form $\mathcal{I}(X,\tau)$ full pseudogroups of
transformations.

For modern applications of pseudogroups, see [16, 80].

To make our lives easier, let’s now take the non-empty set $X$ with the
discrete topology. We denote by $\mathcal{I}(X)$ the set of all bijections
between subsets of $X$. If $f\colon A\rightarrow B$, where $A,B\subseteq X$,
then we call $A$ the domain (of definition) of $f$ and $B$ the range of $f$.
We write $\mbox{\rm dom}(f)=A$ and $\mbox{\rm ran}(f)=B$. The set
$\mathcal{I}(X)$ is closed under the composition of partial functions and so
is a semigroup. It is, in fact, a monoid since the identity function defined
on $X$ belongs to $\mathcal{I}(X)$. By analogy with the symmetric group, it is
called the symmetric inverse monoid. What we mean by the term ‘inverse monoid’
will be explained in the next section. Clearly, the structure of the semigroup
$\mathcal{I}(X)$ depends only on the cardinality of $X$. Thus, if $X$ is a
finite set with $n$ elements, it makes sense to write $\mathcal{I}_{n}$
instead of $\mathcal{I}(X)$.

There are a number of structures we could look at in the semigroups
$\mathcal{I}(X)$ but we begin with the following which turns out to be the
right place to start. Let $f\in\mathcal{I}(X)$. Consider the following two
equations in $\mathcal{I}(X)$:

$f=fgf\text{ and }g=gfg.$

These have the unique solution $g=f^{-1}$. This result is the basis for the
definition of an inverse semigroup in the next section.

##  3\. Basic inverse semigroup theory

A semigroup $S$ is said to be inverse if for each $s\in S$ there exists a
unique element, denoted by $s^{-1}$, such that the following two equations are
satisfied:

$s=ss^{-1}s\text{ and }s^{-1}=s^{-1}ss^{-1}.$

######  Example 3.1.

The semigroups $\mathcal{I}(X)$ really are inverse monoids. If
$f\in\mathcal{I}(X)$ then $f^{-1}$ is the unique solution of the equations
$f=fgf$ and $g=gfg$ where $g$ is the unknown.

Observe that if $s^{-1}$ is the inverse of $s$, then $s$ is the inverse of
$s^{-1}$. We have therefore proved the following.

######  Lemma 3.2.

Let $S$ be an inverse semigroup and let $s\in S$. Then $(s^{-1})^{-1}=s$.

The elements $s^{-1}s$ and $ss^{-1}$ are idempotents, where an idempotent in a
semigroup is an element $e$ such that $e^{2}=e$. For example, $s^{-1}s$ is an
idempotent because

$(s^{-1}s)^{2}=(s^{-1}s)(s^{-1}s)=(s^{-1}ss^{-1})s=s^{-1}s.$

If $e$ is an idempotent in an inverse semigroup $S$, then $e=ee=eee$. We have
therefore proved the following.

######  Lemma 3.3.

Let $S$ be an inverse semigroup and let $e\in S$ be an idempotent. Then
$e^{-1}=e$.

It follows that every idempotent in an inverse semigroup is of the form
$s^{-1}s$ for some elements $s$ or, equivalently, of the form $ss^{-1}$, for
some element $s$.

Idempotents play an important role in inverse semigroup theory. For that
reason we introduce some special notation. If $S$ is an inverse semigroup we
denote its set of idempotents by $\mathsf{E}(S)$. Two special idempotents are
the identity element, if it exists, usually denoted by $1$, and the zero
element, if it exists, usually denoted by $0$. An inverse semigroup with
identity is called an inverse monoid and an inverse semigroup with zero is
called an inverse semigroup with zero.666There is no one term in English for a
‘semigroup with zero’. It is important to be aware right at the beginning that
even if $S$ is an inverse monoid, it is not true in general that
$s^{-1}s=1=ss^{-1}$.

######  Example 3.4.

The idempotents in $\mathcal{I}(X)$ are precisely the identity functions
defined on the subsets of $X$. Thus if $A\subseteq X$ then a typical
idempotent is $1_{A}$ which is the identity function on the set $A$. The
identity is $1_{X}$ and the zero is $1_{\varnothing}$.

The difference between monoids and semigroups is sometimes viewed as trivial.
It isn’t. For example, a commutative unital $C^{\ast}$-algebra is constructed
from a compact space whereas a commutative non-unital $C^{\ast}$-algebra is
constructed from a locally compact space.

If $S$ is any monoid then we can look at its group of units which we denote by
$\mathsf{U}(S)$. If $e$ is any idempotent in a semigroup $S$ then $eSe$ is a
monoid. I call these local monoids though you will find the misleading term
‘local submonoid’ in the literature. The group of units of $eSe$ is denoted by
$H_{e}$.

######  Example 3.5.

The units of $\mathcal{I}(X)$ are the bijective functions and these form a
group $\mathcal{S}(X)$, called the symmetric group on $X$.

######  Lemma 3.6.

Let $S$ be an inverse semigroup and let $e$ be an idempotent in $S$. Then
$eSe$ is an inverse monoid.

###### Proof.

It is clear that $eSe$ is a monoid. We prove that it is inverse. Let $a\in
eSe$. Then $a=aa^{-1}a$ and $a^{-1}=a^{-1}aa^{-1}$ in $S$. You can check that
$ea^{-1}e$ is also an inverse of $a$ since $ea=a=ae$. So, by uniqueness, we
have that $a^{-1}=ea^{-1}e$. We have therefore proved that if $a\in eSe$ then
$a^{-1}\in eSe$. It follows that $eSe$ is an inverse monoid. ∎

The idempotents $s^{-1}s$ and $ss^{-1}$ are so important, that we have some
special notation for them:

$\mathbf{d}(s)=s^{-1}s\text{ and }\mathbf{r}(s)=ss^{-1}.$

Observe that $a\mathbf{d}(a)=a$ and $\mathbf{r}(a)a=a$. If the inverse
semigroup has a zero then $a=0$ if and only if $\mathbf{d}(a)=0$
(respectively, $\mathbf{r}(a)=0$.) You can easily check that in an inverse
semigroup $a\,\mathscr{R}\,b$ if and only if $\mathbf{r}(a)=\mathbf{r}(b)$ and
$a\,\mathscr{L}\,b$ if and only if $\mathbf{d}(a)=\mathbf{d}(b)$. For this
reason, the explicit use of Green’s relations in inverse semigroups is not
very common.

An inverse subsemigroup of an inverse semigroup is a subsemigroup that is also
closed under inverses. If $S$ is an inverse subsemigroup of $T$ and
$\mathsf{E}(S)=\mathsf{E}(T)$ we say that $S$ is a wide777You will often see
the term ‘full’ used to mean ‘wide’. I prefer the term ‘wide’ because, as we
shall see, it fits in well with groupoid theory, whereas ‘full’ has
categorical connotations which are not what we want. inverse subsemigroup of
$T$.

######  Example 3.7.

The inverse semigroup $\Gamma(\mathbb{R}^{n})$ is, in fact, a wide inverse
submonoid of $\mathcal{I}(\mathbb{R}^{n},\tau)$ where $\tau$ is the usual
topology on $\mathbb{R}^{n}$.

It is now time for our first example. We can, in fact, express this as a
lemma.

######  Lemma 3.8.

  1. (1)

A group is an inverse semigroup having a unique idempotent.

  2. (2)

An inverse semigroup having a unique idempotent is a group.

###### Proof.

(1) Let $G$ be a group. Then for any element $g$ in $G$ we have that
$g^{-1}g=1=gg^{-1}$. Multiply the first equation on the left by $g$ to get
$gg^{-1}g=g$ and multiply the second equation on the left by $g^{-1}$ to get
$g^{-1}gg^{-1}=g^{-1}$. On the other hand, suppose that $gxg=g$ and $xgx=x$.
Multiply the first equation by $g^{-1}$ on the left and by $g^{-1}$ on the
right. This gives us $x=g^{-1}$ which also satisfies the second equation. It
follows that every group is an inverse semigroup. Let $g$ be an idempotent in
$G$. Then $gg=g$. Multiply the left-hand side of this equation by $g^{-1}$ to
get $g=1$, and $1$ is clearly an idempotent. Thus the only idempotent in a
group is the identity.

(2) Let $S$ be an inverse semigroup with exactly one idempotent. Call this
idempotent $e$. For any element $a\in S$, we already know that $a^{-1}a$ and
$aa^{-1}$ are idempotents. It follows that $a^{-1}a=e=aa^{-1}$ for any element
$a\in S$. On the other hand, $e$ is the identity. To see why, observe that
$ea=(aa^{-1})a=a$. Similarly, $ae=a$. We have therefore proved that $S$ is a
group with identity $e$. ∎

The most important fact about inverse semigroups, and one that is not at all
obvious from the definition, is that if $e$ and $f$ are any idempotents in an
inverse semigroup $S$ then $ef=fe$. We say that the idempotents commute. For
the benefit of ring theorists, observe that we are not saying that the
idempotents are central. Inverse semigroups in which the idempotents are
central are called Clifford semigroups.

######  Proposition 3.9.

Idempotents commute in an inverse semigroup.

###### Proof.

This is an example of a proof that is elementary but not easy. Let $e$ and $f$
be idempotents in the inverse semigroup $S$. Put $x=(ef)^{-1}$. We prove first
that $fxe$ is an idempotent:

$(fxe)^{2}=(fxe)(fxe)=f(x(ef)x)e=fxe.$

Next, we claim that $fxe$ is the inverse of $ef$:

$ef(fxe)ef=ef^{2}xe^{2}f=efxef=ef$

and

$(fxe)ef(fxe)=fxe^{2}f^{2}xe=(fxe)^{2}=fxe,$

by what we proved above. It follows that

$(ef)^{-1}=fxe.$

We have therefore proved that $(ef)^{-1}$ is an idempotent. Now, we apply
Lemma 3.2 and Lemma 3.3 to deduce that $ef$ is an idempotent. We have
therefore proved that in an inverse semigroup the product of any two
idempotents is itself an idempotent. It follows that both $ef$ and $fe$ are
idempotents. Next, we show that $fe$ is the inverse of $ef$:

$(ef)(fe)(ef)=ef^{2}e^{2}f=efef=ef$

and

$(fe)(ef)(fe)=fe^{2}f^{2}e=fefe=fe.$

We have therefore proved that $(ef)^{-1}=fe$. But $ef$ is an idempotent and
so, by Lemma 3.3, it is its own inverse. We have therefore proved that
$ef=fe$, as required. ∎

######  Example 3.10.

In the symmetric inverse monoid, the product of the idempotents $1_{A}$ and
$1_{B}$ is the idempotent $1_{A\cap B}$. Because the intersection operation on
two sets is commutative, we see why the idempotents commute in this special
case.

Because we know that idempotents commute, we can now prove the following:

######  Lemma 3.11.

Let $S$ be an inverse semigroup. Then $(st)^{-1}=t^{-1}s^{-1}$.

###### Proof.

We prove that $t^{-1}s^{-1}$ is the inverse of $st$. We have that

$st(t^{-1}s^{-1})st=s(tt^{-1})(s^{-1}s)t=s(s^{-1}s)(tt^{-1})t=st$

where we have used the fact that idempotents in an inverse semigroup commute.
Similarly,

$t^{-1}s^{-1}(st)t^{-1}s^{-1}=t^{-1}s^{-1}.$

By virtue of the fact that inverses are unique, we deduce that
$(st)^{-1}=t^{-1}s^{-1}$. ∎

We also see that the analogue of conjugation holds. We shall use the term
‘conjugation’ below.

######  Lemma 3.12.

If $e$ is an idempotent of an inverse semigroup and $s$ is any element of that
inverse semigroup then $ses^{-1}$ is an idempotent.

###### Proof.

This follows by a direct calculation

$(ses^{-1})^{2}=ses^{-1}ses^{-1}=se^{2}(s^{-1}s)s^{-1}=ses^{-1},$

where we have again used the fact that idempotents commute. ∎

Let $e$ and $f$ be idempotents in an inverse semigroup. Then we may define a
relation on them by $e\leq f$ precisely when $e=ef$. Because idempotents
commute, it is not hard to show that this defines a partial order on the set
of idempotents. We can now characterize Clifford semigroups.

######  Lemma 3.13.

Let $S$ be an inverse semigroup. Then $S$ is a Clifford semigroup if and only
if $\mathbf{d}(s)=\mathbf{r}(s)$ for every $s\in S$.

###### Proof.

We use the partial order we have defined on the idempotents of an inverse
semigroup. Let $S$ be a Clifford semigroup and let $s\in S$. Since the
idempotents are central $s=s(s^{-1}s)=(s^{-1}s)s$. Multiply the last equality
on the right by $s^{-1}$ to deduce that $\mathbf{r}(s)\leq\mathbf{d}(s)$. It
now follows by symmetry that $\mathbf{d}(s)=\mathbf{r}(s)$.

Suppose now that $\mathbf{d}(s)=\mathbf{r}(s)$ for all $s\in S$. Let $e$ be
any idempotent and $s$ arbitrary. We shall prove that $se=es$. By assumption,
$\mathbf{d}(es)=\mathbf{r}(es)$. That is $(es)^{-1}es=es(es)^{-1}$. We now use
Lemma 3.11 and Lemma 3.3 and the fact that idempotents commute to get
$s^{-1}es=ss^{-1}e$. Because of our assumption, $ss^{-1}=s^{-1}s$. Thus
$s^{-1}es=s^{-1}se$. Multiplying on the left by $s$, and using the fact that
idempotents commute and our assumption, gives $es=se$, as required. ∎

An inverse semigroup $S$ is said to be a union of groups if
$S=\bigcup_{e\in\mathsf{E}(S)}H_{e}$. By Lemma 3.13, we may now deduce the
following

######  Lemma 3.14.

Let $S$ be an inverse semigroup. Then $S$ is a Clifford semigroup if and only
if it is a union of groups.

###### Proof.

Let $S$ be a Clifford semigroup and let $s\in S$. Put
$e=\mathbf{d}(s)=\mathbf{r}(s)$. Then $s\in eSe$. But, in fact, $s$ belongs to
the group of units of $eSe$ which is $H_{e}$. Thus $s\in H_{e}$ and so we have
proved that every Clifford semigroup is a union of groups.

Conversely, suppose that $S$ is an inverse semigroup which is a union of
groups. Then for each $s\in S$, we have that $s\in H_{e}$ for some idempotent
$e$. We have that $\mathbf{d}(s)=e=\mathbf{r}(s)$. It follows that $S$
Clifford. ∎

I have included Lemma 3.14 because group theorists sometimes tend to see
inverse semigroups as unions of groups. These are, in fact, a very special
class of inverse semigroups. Clifford semigroups can be described explicitly
as ‘strong semilattices of groups’ or, what amounts to the same thing, as
presheaves of groups over meet semilattices. This applies to, in particular,
abelian inverse semigroups. For strong semilattices of groups, see [37,
Chapter IV, Section 2].

We can obtain another characterization of Clifford semigroups which is
analogous to part of [30, part (c) of Theorem 3.2] but at the price of an
extra assumption on the semilattice of idempotents. We say that a meet
semilattice is $0$-disjunctive if for all $0\neq f<e$ there exists $0\neq
g\leq e$ such that $fg=0$. Thus, semilattices which are $0$-disjunctive have a
weak notion of complement. A non-zero element $a$ of an inverse semigroup is
said to be an infinitesimal if $a^{2}=0$.

######  Lemma 3.15.

Let $S$ be an inverse semigroup, the semilattice of idempotents of which is
$0$-disjunctive. Then $S$ is a Clifford semigroup if and only if $S$ contains
no infinitesimals.

###### Proof.

Suppose that $S$ is a Clifford semigroup. Let $a$ be an infinitesimal. Then
$a^{2}=0$ thus $aa=0$. It follows that $\mathbf{d}(a)\mathbf{r}(a)=0$. But, in
a Clifford semigroup $\mathbf{d}(a)=\mathbf{r}(a)$ by Lemma 3.13. It follows
that $\mathbf{d}(a)=0$ and so $a=0$ which is a contradiction. Thus there are
no infinitesimals.

To prove the converse, suppose that there are no infinitesmials. Let $a\in S$
be arbitrary and non-zero. We shall prove that $\mathbf{d}(a)=\mathbf{r}(a)$.
Suppose not. Put $e=\mathbf{d}(a)\mathbf{r}(a)$. If $e=0$ then
$a^{-1}aaa^{-1}=0$ and so $a^{2}=0$. But this means that $a$ is an
infinitesimal. It follows that $e\neq 0$. There are now two cases. Suppose,
first, that $e=\mathbf{d}(a)$. Then $\mathbf{d}(a)<\mathbf{r}(a)$. By
assumption, there exists $0<f\leq\mathbf{r}(a)$ such that $f\mathbf{d}(a)=0$.
Put $b=fa$. If $b=0$ then it is easy to see that $f=0$, which is a
contradiction. It follows that $b\neq 0$. You can easily check that $b$ is an
infinitesimal. This is a contradiction. We can now deal with the second case
where $e<\mathbf{d}(a)$. By assumption, there is a non-zero idempotent $f$
such that $ef=0$ and $f\leq\mathbf{d}(a)$. Put $b=af$. If $b=0$ then
$(af)^{-1}af=0$ which implies that $f=0$. It follows that $b\neq 0$. However,
$b^{2}=afaf=a\mathbf{d}(a)f\mathbf{r}(a)af=aefaf=0$. But this contradicts the
assumption that there are no infinitesimals. It follows that
$\mathbf{d}(a)=\mathbf{r}(a)$ for all $a\in S$. Thus by Lemma 3.13, we have
shown that $S$ is a Clifford semigroup. ∎

Although the idempotents in an inverse semigroup are not in general central,
we do have the following result, which tells us how idempotents ‘pass through’
non-idempotent elements.

######  Lemma 3.16.

Let $S$ be an inverse semigroup in which $e$ is an idempotent and $s$ is any
element.

  1. (1)

$es=sf$ for some idempotent $f$.

  2. (2)

$se=gs$ for some idempotent $g$.

###### Proof.

We prove only (1) since the proof of (2) is similar. We have that
$es=ess^{-1}s=e(ss^{-1})s$. Now we use the fact that idempotents commute to
get $es=(ss^{-1})es$. But this is equal to $s(s^{-1}es)$ and now we use Lemma
3.12 to deduce that $f=s^{-1}es$ is an idempotent. We have therefore proved
that $es=sf$. ∎

We have described one extreme example of inverse semigroup. We now describe
another. By a meet semilattice, we mean a partially ordered set $(P,\leq)$
with the property that every pair of elements $a,b\in P$ has a greatest lower
bound, denoted by $a\wedge b$. Thus, $a\wedge b\leq a,b$ and if $c\leq a,b$
then $c\leq a\wedge b$. We call $a\wedge b$ the meet of $a$ and $b$.888We
shall usually denote the meet by $\cdot$ or concatenation. The proofs of the
following are immediate from the definitions.

######  Lemma 3.17.

Let $(P,\leq)$ be a meet semilattice and let $a,b,c\in P$. Then

  1. (1)

$a\wedge(b\wedge c)=(a\wedge b)\wedge c$.

  2. (2)

$a\wedge b=b\wedge a$.

  3. (3)

$a\wedge a=a$

A semigroup is called a band if every element is an idempotent. Observe that
we have proved in Lemma 3.17 that if $(P,\leq)$ is a meet semilattice then
$(P,\wedge)$ is a commutative band.

######  Lemma 3.18.

Let $S$ be a commutative band. Define $a\leq b$ by $a=ab$. Then $(S,\leq)$ is
a partially ordered set and, in fact, a meet semilattice in which $a\wedge
b=ab$.

###### Proof.

We first check that $\leq$ is a partial order. By assumption, each element of
$S$ is an idempotent. It follows that $a\leq a$. Suppose that $a\leq b$ and
$b\leq a$. Then $a=ab$ and $b=ba$. But the semigroup is commutative and so
$a=b$. Finally, suppose that $a\leq b$ and $b\leq c$. Then $a=ab$ and $b=bc$.
It follows that $a=abc=ac$. Thus $a\leq c$. We have verified that $(S,\leq)$
is a partially ordered set. We now show that $a\wedge b=ab$. Observe first
that $(ab)a=a^{2}b=ab$, using commutativity and the fact that every element is
idempotent. Also, $(ab)b=ab^{2}=ab$. We have show that $ab\leq a,b$. Suppose,
now that $c\leq a,b$. Then $c=ca=cb$, and so $c(ab)=cb=c$. This shows that
$c\leq ab$ and thus proves that $ab=a\wedge b$. ∎

As a result of Lemma 3.17 and Lemma 3.18, we can think about meet semilattices
in two equivalent ways: either as partially ordered sets in which each pair of
elements has a greatest lower bound or as commutative bands.

Observe that if $S$ is any inverse semigroup then, since the idempotents
commute, $\mathsf{E}(S)$ is a subsemigroup of $S$. It follows that
$\mathsf{E}(S)$ is always a commutative band and so it becomes a meet
semilattice when we define $e\wedge f=ef$. For this reason, it is usual to
refer to $\mathsf{E}(S)$ as the semilattice of idempotents of $S$.

We now have the following companion to Lemma 3.8.

######  Lemma 3.19.

  1. (1)

A meet semilattice is an inverse semigroup with respect to the meet operation
in which every element is an idempotent.

  2. (2)

An inverse semigroup in which every element is an idempotent is a meet
semilattice.

###### Proof.

The proof of (1) is immediate by Lemma 3.17. (2) Suppose that $S$ is an
inverse semigroup in which every element is an idempotent. Then
$S=\mathsf{E}(S)$. The result now follows by Lemma 3.18, since $\mathsf{E}(S)$
is a commutative band. ∎

Our next result is often useful in showing that a semigroup is or is not
inverse. We need a definition first. A semigroup $S$ is said to be regular if
for each $a\in S$ there exists an element $b$ such that $a=aba$ and $b=bab$.
The element $b$ is called an inverse of $a$. Observe that both $ab$ and $ba$
are idempotents. In showing that a semigroup is regular, it is enough to check
that for each element $a$ there is an element $x$ such that $a=axa$. The
reason is that $b=xax$ is then an inverse of $a$. Inverse semigroups are the
regular semigroups in which every element has a unique inverse.

######  Proposition 3.20.

A regular semigroup is inverse if and only if its idempotents commute.

###### Proof.

Let $S$ be a regular semigroup in which the idempotents commute and let $u$
and $v$ be inverses of $x$. Then

$u=uxu=u(xvx)u=(ux)(vx)u,$

where both $ux$ and $vx$ are idempotents. Thus, since idempotents commute, we
have that

$u=(vx)(ux)u=vxu=(vxv)xu=v(xv)(xu).$

Again, $xv$ and $xu$ are idempotents and so

$u=v(xu)(xv)=v(xux)v=vxv=v.$

Hence $u=v$. The converse follows by Proposition 3.9. ∎

######  Example 3.21.

As an example of Proposition 3.20 in action, observe that the full
transformation monoid $\mathcal{T}(X)$, of all functions from $X$ to itself
where $X$ has at least two elements, is regular but not inverse. We show first
that it is regular. Let $f\in\mathcal{I}(X)$. There are two cases. Suppose
first that $f$ is surjective; for each $y\in X$ choose $x_{y}\in f^{-1}(y)$.
Define $g\colon X\rightarrow X$ by $g(y)=x_{y}$. That is, $g$ maps an element
to one of its pre-images under $f$. We calculate $fgf$. Let $x\in X$ and
$y=f(x)$. Then

$(fgf)(x)=(fg)(f(x))=f(g(y))=f(x_{y})=y.$

It follows that $f=fgf$. Suppose now that $f$ is not surjective. Choose any
element $x_{0}\in X$. For each element $y$ in the image of $f$ choose
$x_{y}\in f^{-1}(y)$. Now define $g\colon X\rightarrow X$ as follows: if $y$
is in the image of $f$, then define $g(y)=x_{y}$, and if $y$ is not in the
image of $f$, then define $g(y)=x_{0}$. A similar calculation to the one above
shows that $f=fgf$. We have therefore shown that the full transformation
monoid is regular. Choose two distinct elements from $X$ that we call $x_{1}$
and $x_{2}$. Define two functions $c_{1},c_{2}\colon X\rightarrow X$ where
$c_{1}$ is the constant function to $x_{1}$ and $c_{2}$ is the constant
function to $x_{2}$. The functions $c_{1}$ and $c_{2}$ are distinct and both
are idempotents. But $c_{1}c_{2}=c_{1}$ whereas $c_{2}c_{1}=c_{2}$. We have
shown that the idempotents do not commute. Thus $\mathcal{T}(X)$ is regular
but not inverse.

We now consider homomorphisms between inverse semigroups. Our first result is
actually slightly more general.

######  Lemma 3.22.

Let $S$ be an inverse semigroup and let $T$ be any semigroup. Suppose that
$\theta\colon S\rightarrow T$ is a semigroup homomorphism. Then $\theta(S)$,
the image of $S$, is an inverse semigroup.

###### Proof.

We shall use Proposition 3.20 to prove that $\theta(S)$ is inverse. We show
first that it is regular. Let $\theta(s)\in\theta(S)$. Then, since $s\in S$
inverse, there is an element $s^{-1}\in S$ such that $s=ss^{-1}s$ and
$s^{-1}=s^{-1}ss^{-1}$. It follows that
$\theta(s)=\theta(s)\theta(s^{-1})\theta(s)$ and
$\theta(s^{-1})=\theta(s^{-1})\theta(s)\theta(s^{-1})$. This shows that
$\theta(S)$ is regular.

We now prove that the idempotents in $\theta(S)$ commute. To begin with, we
have to show that idempotents in the image come from idempotents in the
domain.999This is an instance of what is often termed Lallement’s Lemma. Let
$\theta(s)$ be an idempotent in $\theta(S)$. We shall prove that there is an
idempotent $e$ in $S$ such that $\theta(e)=\theta(s)$. Consider the element
$e=ss^{-2}s=(ss^{-1})(s^{-1}s)$. It is the product of two idempotents in $S$
and so it is an idempotent. In particular, $e$ is an idempotent in $S$. We
calculate $\theta(e)$. We have that

$\theta(e)=\theta(ss^{-2}s)=\theta(s)\theta(s^{-2})\theta(s).$

But $\theta(s)=\theta(s)^{2}=\theta(s^{2})$. We therefore have that

$\theta(e)=\theta(s^{2}s^{-2}s^{2})=\theta(s^{2})=\theta(s)^{2}=\theta(s).$

Now, let $\theta(s)$ and $\theta(t)$ be idempotents in $\theta(S)$. By our
result above, there are idempotents $e$ and $f$ in $S$ such that
$\theta(e)=\theta(s)$ and $\theta(f)=\theta(t)$. It follows that
$\theta(ef)=\theta(s)\theta(t)$. But idempotents in inverse semigroups commute
and so $ef=fe$. We have therefore shown that
$\theta(s)\theta(t)=\theta(t)\theta(s)$. Thus, the idempotents in $\theta(S)$
commute. We have therefore proved that the image of $\theta$ is an inverse
semigroup by Proposition 3.20. ∎

Homomorphisms of inverse semigroups are just semigroup homomorphisms.
Isomorphisms of inverse semigroups are just semigroup isomorphisms.

######  Lemma 3.23.

Let $\theta\colon S\rightarrow T$ be a homomorphism between inverse
semigroups. Then:

  1. (1)

$\theta(s^{-1})=\theta(s)^{-1}$ for all $s\in S$.

  2. (2)

If $e$ is an idempotent then $\theta(e)$ is an idempotent.

  3. (3)

If $\theta(s)$ is an idempotent then there is an idempotent $e$ in $S$ such
that $\theta(s)=\theta(e)$.

  4. (4)

$\theta(S)$ is an inverse subsemigroup of $T$.

  5. (5)

If $U$ is an inverse subsemigroup of $T$ then $\theta^{-1}(U)$ is an inverse
subsemigroup of $S$.

###### Proof.

(1) Clearly, $\theta(s)\theta(s^{-1})\theta(s)=\theta(s)$ and
$\theta(s^{-1})\theta(s)\theta(s^{-1})=\theta(s^{-1})$. Thus, by uniqueness of
inverses, we have that $\theta(s^{-1})=\theta(s)^{-1}$.

(2) $\theta(e)^{2}=\theta(e)\theta(e)=\theta(e^{2})=\theta(e)$.

(3) If $\theta(s)^{2}=\theta(s)$, then
$\theta(s^{-1}s)=\theta(s^{-1})\theta(s)=\theta(s)^{-1}\theta(s)=\theta(s)^{2}=\theta(s)$,
where we have used the fact that every idempotent in an inverse semigroup is
its own inverse.

(4) This is immediate by Lemma 3.22.

(5) Straightforward. ∎

Let $\theta\colon S\rightarrow T$ be a homomorphism of inverse semigroups.
Then the restriction map,
$(\theta\mid\mathsf{E}(S))\colon\mathsf{E}(S)\rightarrow\mathsf{E}(T)$, is
well-defined since homomorphisms map idempotents to idempotents by part (2) of
Lemma 3.23 The homomorphism $\theta$ is said to be idempotent-separating if
$(\theta\mid\mathsf{E}(S))$ is actually injective. We say that a congruence is
idempotent-separating if its associated natural homomorphism is idempotent-
separating. We say that $S$ is an idempotent-separating cover of $T$ if
$\theta$ is both surjective and idempotent-separating.

It remains to be shown that inverse semigroups really do encode partial
bijections. Given an inverse semigroup $S$ our goal is therefore to construct
an injective homomorphism $\lambda\colon S\rightarrow\mathcal{I}(X)$ for some
set $X$. We shall do just this by generalizing the famililar Cayley’s Theorem
from group theory. As a starting point, we follow the group theory and take as
our underlying set $X$ the set $S$ itself. We now have to associate an element
of $\mathcal{I}(S)$ with each element $a\in S$. This is delivered by the
following lemma.

######  Lemma 3.24.

Let $S$ be an inverse semigroup and let $a\in S$. Then
$\lambda_{a}\colon\mathbf{d}(a)S\rightarrow\mathbf{r}(a)S$ defined by by
$\lambda_{a}(x)=ax$ is a well-defined partial bijection.

###### Proof.

This is well-defined because $aS=aa^{-1}S$ as the following set inclusions
show

$aS=aa^{-1}aS\subseteq aa^{-1}S\subseteq aS.$

Also $\lambda_{a^{-1}}\colon\mathbf{r}(a)S\rightarrow\mathbf{d}(a)S$,
$\lambda_{a^{-1}}\lambda_{a}$ is the identity on $\mathbf{d}(a)S$, and
$\lambda_{a}\lambda_{a^{-1}}$ is the identity on $\mathbf{r}(a)S$. Thus
$\lambda_{a}$ is a bijection and $\lambda_{a}^{-1}=\lambda_{a^{-1}}$. ∎

We now define a function $\lambda\colon S\rightarrow\mathcal{I}(S)$ by
$\lambda(a)=\lambda_{a}$. The following theorem completes the circle of ideas
begun in Section 2 and justifies the claim that inverse semigroups are the
abstract theory of partial bijections.

######  Theorem 3.25 (Wagner-Preston representation theorem).

Every inverse semigroup can be embedded in a symmetric inverse monoid.

###### Proof.

We prove first that $\lambda$ is a semigroup homomorphism. To that end, we
prove that $\lambda_{a}\lambda_{b}=\lambda_{ab}$. If $e$ and $f$ are any
idempotents of an inverse semigroup $S$ then

$eS\cap fS=efS.$

Thus

$\mbox{dom}(\lambda_{a})\cap\mbox{im}(\lambda_{b})=a^{-1}aS\cap
bb^{-1}S=a^{-1}abb^{-1}S.$

Hence

$\mbox{dom}(\lambda_{a}\lambda_{b})=\lambda_{b}^{-1}(a^{-1}abb^{-1}S)=b^{-1}a^{-1}aS=b^{-1}a^{-1}abS$

where we use the following subset inclusions

$b^{-1}a^{-1}aS=b^{-1}bb^{-1}a^{-1}aS=b^{-1}a^{-1}abb^{-1}S\subseteq
b^{-1}a^{-1}abS\subseteq b^{-1}a^{-1}aS.$

Thus $\mbox{dom}(\lambda_{a}\lambda_{b})=\mbox{dom}(\lambda_{ab})$. It is
immediate from the definitions that $\lambda_{a}\lambda_{b}$ and
$\lambda_{ab}$ have the same effect on elements, and so $\lambda$ is a
homomorphism. It remains to prove that $\lambda$ is injective. Suppose that
$\lambda_{a}=\lambda_{b}$. Then $a=ba^{-1}a$ and $b=ab^{-1}b$. Observe that
$ab^{-1}b=(ba^{-1}a)b^{-1}b=b(b^{-1}b)a^{-1}a=ba^{-1}a=a$, where we have used
the fact that idempotents commute. It follows that $a=b$. ∎

We conclude this section by stating a deeper result that we shall not prove.
Let $S$ be a finite semigroup. Then for each $s\in S$ there exists some power
$s^{n}$ which is an idempotent. Let $\theta\colon S\rightarrow T$ be a
homomorphism between finite semigroups. Suppose that $\theta(s)$ is an
idempotent in $T$. Then for some natural number $n\geq 1$, the element $s^{n}$
is also an idempotent. But $\theta(s^{n})=\theta(s)^{n}=\theta(s)$. Thus, for
finite semigroups, if $\theta(s)$ is an idempotent then there is an idempotent
$e$ such that $\theta(e)=\theta(s)$. If $S$ is an inverse semigroup then every
subsemigroup (notice: I did not say inverse subsemigroup on purpose) has
commuting idempotents. We say that a semigroup $T$ divides a semigroup $S$ if
there is a subsemigroup $S^{\prime}$ of $S$ such that $T$ is a homomorphic
image of $S^{\prime}$. Suppose now that $S$ is a finite inverse semigroup.
Then every subsemigroup $S^{\prime}$ of $S$ is a finite semigroup that has
commuting idempotents. Thus every homomorphic image of $S^{\prime}$ is a
finite semigroup with commuting idempotents. We have therefore proved the
following:

######  Lemma 3.26.

Every semigroup that divides a finite inverse semigroup has commuting
idempotents.

The following theorem was proved by Chris Ash [3] and is deep. It is the
converse of the above lemma.

######  Theorem 3.27 (Ash).

Every finite semigroup with commuting idempotents divides a finite inverse
semigroup.

This is a beautiful result, and a stunning piece of mathematics. It uses
finite combinatorics, in the guise of Ramsey Theory, as part of the proof. See
[49] for the connections between combinatorics and semigroup theory.

##  4\. The natural partial order

Partial bijections can be compared with each other: we can say that one
partial bijection is the restriction of another. This leads to a partial order
on the set of partial bijections of a set. It might be thought that this
partial order has to be imposed on an inverse semigroup but, remarkably, this
partial order can be defined in purely algebraic terms. It is therefore called
the natural partial order. It will follow that every inverse semigroup is, in
fact, a partially ordered semigroup with respect to this order.

On an inverse semigroup, define $s\leq t$ iff $s=ts^{-1}s$. This looks one-
sided because the idempotent appears on the right-hand side. It is here that
we invoke Lemma 3.16.

######  Lemma 4.1.

In an inverse semigroup, the following are equivalent:

  1. (1)

$s\leq t$.

  2. (2)

$s=te$ for some idempotent $e$.

  3. (3)

$s=ft$ for some idempotent $f$.

  4. (4)

$s=ss^{-1}t$.

###### Proof.

(1)$\Rightarrow$(2). This is immediate.

(2)$\Rightarrow$(3). This is immediate by Lemma 3.16.

(3)$\Rightarrow$(4). Suppose that $s=ft$. Then $fs=s$ and so
$fss^{-1}=ss^{-1}$. It follows that $s=ss^{-1}t$.

(4)$\Rightarrow$(1). Suppose that $s=ss^{-1}t$. By Lemma 3.16, we know that
$s=ti$ for some idempotent $i$. Observe that $si=s$ and so $s^{-1}si=s^{-1}s$.
It readily follows that $s=ts^{-1}s$ giving $s\leq t$. ∎

Let $S$ be a semigroup equipped with a partial order $\leq$. We say that $S$
is a partially ordered semigroup if $a\leq b$ and $c\leq d$ imply that $ac\leq
bd$. We may now establish the main properties of the relation we have defined.

######  Lemma 4.2.

Let $S$ be an inverse semigroup.

  1. (1)

The relation $\leq$ is a partial order.

  2. (2)

If $s\leq t$ then $s^{-1}\leq t^{-1}$.

  3. (3)

The semigroup $S$ is partially ordered with respect to $\leq$.

  4. (4)

If $e$ and $f$ are idempotents then $e\leq f$ if and only if $e=ef=fe$.

  5. (5)

$s\leq e$, where $e$ is an idempotent, implies that $s$ is an idempotent.

  6. (6)

Let $T$ be an inverse semigroup and let $\theta\colon S\rightarrow T$ be a
homomorphism. Then $a\leq b$ in $S$ implies that $\theta(a)\leq\theta(b)$ in
$T$.

###### Proof.

(1) Observe that $a\leq a$ since $a=a\mathbf{d}(a)$. Suppose that $a\leq b$
and $b\leq a$. Then $a=b\mathbf{d}(a)$ and $b=a\mathbf{d}(b)$. Then
$b\mathbf{d}(a)=b$ from which it follows that $a=b$. Suppose that $a\leq b$
and $b\leq c$. Using Lemma 4.1, it is easy to show that $a\leq c$.

(2) This is immediate using the definition and Lemma 4.1.

(3) This is immediate using the definition, Lemma 4.1 and Lemma 3.16.

(4) Straightforward.

(5) This is immediate from the definition and the fact that the product of
idempotents is an idempotent.

(6) This is immediate from the fact that the natural partial order is
algebraically defined, Lemma 4.1 and the fact that idempotents are mapped to
idempotents by homomorphisms. ∎

Part (1) of Lemma 4.2 leads us to dub $\leq$ the natural partial order on $S$.
If a partial order is studied in relation to an inverse semigroup, then it is
always this one.

Part (2) of Lemma 4.2 needs to be highlighted since readers familiar with
lattice-ordered groups might have been expecting something different.

Part (4) of Lemma 4.2 tells us that when the natural partial order is
restricted to the semilattice of idempotents we get back the usual ordering on
the idempotents that we have already defined.

######  Example 4.3.

The natural partial order in the symmetric inverse monoids $\mathcal{I}(X)$ is
precisely the usual restriction order on partial functions. Observe that by
the Wagner-Preston representation theorem Theorem 3.25, we have that $a\leq b$
if and only if $\lambda_{a}\subseteq\lambda_{b}$.

The natural partial order is used to define a class of inverse monoids. We say
that an inverse monoid $S$ is factorizable if for each element $s\in S$ there
exists a unit $g$ such that $s\leq g$,

######  Example 4.4.

The symmetric inverse monoids $\mathcal{I}(X)$ are factorizable if $X$ is
finite. Let $f\in\mathcal{I}(X)$. Then not only do $\mbox{dom}(f)$ and
$\mbox{ran}(f)$ have the same cardinality but so too do the sets
$X\setminus\mbox{dom}(f)$ and $X\setminus\mbox{ran}(f)$. Choose any bijection
$g$ from $X\setminus\mbox{dom}(f)$ to $X\setminus\mbox{ran}(f)$. Then $f\cup
g$ is a bijection and so is an element of the group of units of
$\mathcal{I}(X)$. However, there are many choices for $g$ and so there are
many units that extend $f$. It is precisely this lack of uniqueness in the
unit that implies that we cannot reduce inverse monoid theory to group theory.

Our next result tells us that the partial order encodes how far from being a
group an inverse semigroup is.

######  Lemma 4.5.

An inverse semigroup is a group if and only if the natural partial order is
the equality relation.

###### Proof.

Let $S$ be an inverse semigroup in which the natural partial order is
equality. If $e$ and $f$ are any idempotents then $ef\leq e,f$ and so $e=f$.
It follows that there is exactly one idempotent. We deduce that $S$ is a group
by Lemma 3.8. The proof of the converse is immediate. ∎

The natural partial order is very important in studying inverse semigroups.
For this reason, it is appropriate here to introduce some terminology and
notation from the theory of partially ordered sets.

In any partially ordered set $(X,\leq)$, a subset $Y\subseteq X$ is said to be
an order ideal if $x\leq y\in Y$ implies that $x\in Y$. More generally, if $Y$
is any subset of $X$ then define

$Y^{\downarrow}=\\{x\in X\colon x\leq y\text{ for some }y\in Y\\}.$

This is the order ideal generated by $Y$. If $y\in X$ then we denote
$\\{y\\}^{\downarrow}$ by $y^{\downarrow}$ and call it the principal order
ideal generated by $y$. If $Y$ is any subset of $X$, define

$Y^{\uparrow}=\\{x\in X\colon x\geq y\text{ for some }y\in Y\\}.$

If $Y=\\{y\\}$ we denote $\\{y\\}^{\uparrow}$ by $y^{\uparrow}$. If $P$ and
$Q$ are partially ordered sets then a function $\theta\colon X\rightarrow Y$
is said to be isotone if $x\leq y$ in $X$ implies that
$\theta(x)\leq\theta(y)$ in $Y$. An order-isomorphism between two partially
ordered sets is a bijective isotone function whose inverse is also isotone.

Part (5) of Lemma 4.2 tells us that the semilattice of idempotents of an
inverse semigroup $S$ is an order ideal in $S$ with respect to the natural
partial order.

Part (6) of Lemma 4.2 tells us that homomorphisms between inverse semigroups
are isotone.

Idempotents and non-idempotents are closely related in an inverse semigroup.

######  Lemma 4.6.

Let $S$ be an inverse semigroup. Then there is an order-isomorphism between
the set $a^{\downarrow}$ and the set $\mathbf{d}(a)^{\downarrow}$ (and,
likewise, with the set $\mathbf{r}(a)^{\downarrow}$).

###### Proof.

Define a map from $a^{\downarrow}$ to $\mathbf{d}(a)^{\downarrow}$ by
$b\mapsto\mathbf{d}(b)$. This is well-defined since if $b\leq a$ then
$\mathbf{d}(b)\leq\mathbf{d}(a)$. It is isotone since if $b_{2}\leq b_{1}\leq
a$ then $\mathbf{d}(b_{2})\leq\mathbf{d}(b_{1})\leq\mathbf{d}(a)$. From the
definition of the natural partial order, it is immediate that this map is
injective. We define a map from $\mathbf{d}(a)^{\downarrow}$ to
$a^{\downarrow}$ by $e\mapsto ae$. It is routine to check that this is well-
defined and isotone. These two maps are mutually inverse. It follows that the
partially ordered sets are order-isomorphic. ∎

Looking below an idempotent we see only idempotents, but what happens if we
look up? The answer is that we don’t necessarily see only idempotents. The
symmetric inverse monoid is an example. This leads us to the following
definition. An inverse semigroup $S$ is said to be $E$-unitary if $e\leq s$,
where $e$ is an idempotent, implies that $s$ is an idempotent. An inverse
semigroup with zero $S$ is said to be $E^{\ast}$-unitary101010The term
$0$-$E$-unitary is also used. if $0\neq e\leq s$ where $e$ is an idempotent
implies that $s$ is an idempotent.

######  Lemma 4.7.

Let $S$ be an inverse semigroup with zero. Then it is $E$-unitary if and only
if it is a meet semilattice.

###### Proof.

Suppose that $S$ is $E$-unitary. If $a\in S$ then $0\leq a$. But $0$ is an
idempotent, and so $a$ is an idempotent. The proof of the converse is
imediate. ∎

The above lemma explains why we have made two definitions above, depending on
whether the inverse semigroup does not or does have a zero. This bifurcation
between inverse semigroups-without-zero and inverse semigroups-with-zero
permeates the subject.

There was a time when the study of $E$-unitary inverse semigroups was the
centre of attention. The two papers by Don McAlister [84, 85] are probably the
most significant in that they describe the structure of $E$-unitary inverse
semigroups in terms of simpler building blocks and relate them to arbitrary
inverse semigroups. The theory of $E^{\ast}$-unitary inverse semigroups has
also been pursued. There are both interesting examples of such inverse
semigroups and the analogue of McAlister’s theory can be developed in the case
of the so-called strongly $E^{\ast}$-unitary inverse semigroups. See [11, 52,
53], for example.

The following lemma tells us that in an inverse semigroup, there is a
relationship between two elements that have a common upper bound.

######  Lemma 4.8.

Let $S$ be an inverse semigroup and suppose that $a,b\leq c$. then $a^{-1}b$
and $ab^{-1}$ are idempotents.

###### Proof.

By part (2) of Lemma 4.2, we have that $a^{-1},b^{-1}\leq c^{-1}$. By part (3)
of Lemma 4.2, we have that $a^{-1}b\leq c^{-1}c$ and $ab^{-1}\leq cc^{-1}$. It
follows by part (5) of Lemma 4.2, that both $a^{-1}b$ and $ab^{-1}$ are
idempotents. ∎

Lemma 4.8 leads us to the following definition. Let $a,b\in S$, an inverse
semigroup. Define $a\sim b$ if and only if $a^{-1}b,ab^{-1}\in\mathsf{E}(S)$.
This is called the compatibility relation. If $a\sim b$ and if their least
upper bound exists, we denote it by $a\vee b$ and call it the join of $a$ and
$b$. A subset of an inverse semigroup is said to be compatible if the elements
are pairwise compatible. For example, for each element $a\in S$ in an inverse
semigroup, the set $a^{\downarrow}$ is compatible.

In an inverse semigroup with zero, there is a refinement of the compatibility
relation which is important. In such a semigroup, a pair of idempotents $e$
and $f$ is said to be orthogonal, denoted by $e\perp f$, if and only if
$ef=0$. We define an arbitrary pair of elements $a$ and $b$ to be orthogonal,
denoted by $a\perp b$, precisely when $\mathbf{d}(a)\perp\mathbf{d}(b)$ and
$\mathbf{r}(a)\perp\mathbf{r}(b)$ You can easily check that $a\perp b$
precisely when $a^{-1}b=0=ab^{-1}$. We call $\perp$ the orthogonality
relation. Observe that $a\perp b$ implies that $a\sim b$. If an orthogonal
subset has a least upper bound then it is said to have an orthogonal join.

######  Example 4.9.

Let $f,g\in\mathcal{I}(Y)$. Then $f\sim g$ if and only if $f\cup g$ is a
partial bijection, and $f\perp g$ if and only if the domain of $f$ is disjoint
from the domain of $g$ and the range of $f$ is disjoint from the range of $g$.

######  Lemma 4.10.

Let $S$ be an inverse semigroup with zero. If $a\perp b$ and $c\in S$ then
$ac\perp bc$ and $ca\perp cb$.

###### Proof.

We prove that $ac\perp bc$; the proof of the other case is similar. It is
routine to check that $\mathbf{d}(ac)\mathbf{d}(bc)=0$. The result now follows
by symmetry. ∎

The compatibility relation is reflexive and symmetric but not, in general,
transitive, as the symmetric inverse monoid shows. However, we do have an
exact criterion for when the compatibility relation is transitive.

######  Proposition 4.11.

The compatibility relation is transitive if and only if the semigroup is
$E$-unitary.

###### Proof.

Suppose that $\sim$ is transitive. Let $e\leq s$, where $e$ is an idempotent.
Then $se^{-1}$ is an idempotent because $e=se=se^{-1}$, and $s^{-1}e$ is an
idempotent because $s^{-1}e\leq s^{-1}s$. Thus $s\sim e$. Clearly $e\sim
s^{-1}s$, and so, by our assumption that the compatibility relation is
transitive, we have that $s\sim s^{-1}s$. But $s(s^{-1}s)^{-1}=s$, so that $s$
is an idempotent.

Conversely, suppose that $S$ is $E$-unitary and that $s\sim t$ and $t\sim u$.
Clearly $(s^{-1}t)(t^{-1}u)$ is an idempotent and

$(s^{-1}t)(t^{-1}u)=s^{-1}(tt^{-1})u\leq s^{-1}u.$

But $S$ is $E$-unitary and so $s^{-1}u$ is an idempotent. Similarly, $su^{-1}$
is an idempotent. Hence $s\sim u$. ∎

There is a connection between compatible elements and certain kinds of meets.
We shall examine meets in greater generality later on in this section.

######  Lemma 4.12.

Let $S$ be an inverse semigroup.

  1. (1)

$s\sim t$ if and only if $s\wedge t$ exists and $\mathbf{d}(s\wedge
t)=\mathbf{d}(s)\wedge\mathbf{d}(t)$ and $\mathbf{r}(s\wedge
t)=\mathbf{r}(s)\wedge\mathbf{r}(t)$.

  2. (2)

If $s\sim t$ then

$s\wedge t=t\mathbf{d}(s)=s\mathbf{d}(t)=\mathbf{r}(s)t=\mathbf{r}(t)s.$

###### Proof.

(1) We prove first that $st^{-1}$ is an idempotent if and only if the greatest
lower bound $s\wedge t$ of $s$ and $t$ exists and $(s\wedge t)^{-1}(s\wedge
t)=s^{-1}st^{-1}t$. The full result then follows by the dual argument. Suppose
that $st^{-1}$ is an idempotent. Put $z=st^{-1}t$. Then $z\leq s$ and $z\leq
t$, since $st^{-1}$ is an idempotent. Let $w\leq s,t$. Then $w^{-1}w\leq
t^{-1}t$ and so $w\leq st^{-1}t=z$. Hence $z=s\wedge t$. Also

$z^{-1}z=(st^{-1}t)^{-1}(st^{-1}t)=t^{-1}ts^{-1}st^{-1}t=s^{-1}st^{-1}t.$

Conversely, suppose that $s\wedge t$ exists and $(s\wedge t)^{-1}(s\wedge
t)=s^{-1}st^{-1}t$. Put $z=s\wedge t$. Then $z=sz^{-1}z$ and $z=tz^{-1}z$.
Thus $sz^{-1}z=tz^{-1}z$, and so $st^{-1}t=ts^{-1}s$. Hence
$st^{-1}=ts^{-1}st^{-1}$, which is an idempotent.

(2) We shall prove that $s\wedge t=t\mathbf{d}(s)$ since the other equalities
follow by symmetry. Observe that $t\mathbf{d}(s)=ts^{-1}s\leq s,t$ since
$ts^{-1}=(st^{-1})^{-1}$ is an idempotent. Suppose that $x\leq s,t$. Then
$x=xx^{-1}x\leq ts^{-1}s$. It follows that $s\wedge t=t\mathbf{d}(s)$. ∎

The following result is useful since it enables us to deduce that two elements
are equal from apparently weaker conditions.

######  Lemma 4.13.

Let $S$ be an inverse semigroup. If $a\sim b$ and
$\mathbf{d}(a)=\mathbf{d}(b)$ (respectively, $\mathbf{r}(a)=\mathbf{r}(b)$)
then $a=b$.

###### Proof.

Suppose that $a\sim b$ then by Lemma 4.12 the meet $a\wedge b$ exists and
$\mathbf{d}(a\wedge b)=\mathbf{d}(a)\mathbf{d}(b)$. By assumption,
$\mathbf{d}(a)=\mathbf{d}(b)$. Thus $a\wedge b\leq a$ and $\mathbf{d}(a\wedge
b)=\mathbf{d}(a)$. It follows that $a\wedge b=a$. Similarly, $a\wedge b=b$. We
have therefore proved that $a=b$. ∎

Inverse semigroups generalize groups, but we can also construct groups from
inverse semigroups. The idea is this. Groups are abstract versions of groups
of bijections, and bijections can be constructed by glueing together
compatible sets of partial bijections. Thus, we could construct groups from
inverse semigroups by glueing together suitable compatible subsets. We show
first how to construct groups from arbitrary inverse semigroups. The
motivation for how to do this comes from Lemma 4.5, which tells us that groups
are those inverse semigroups in which the natural partial order is equality,
and part (6) of Lemma 4.2, which tells us that homomorphisms between inverse
semigroups are isotone. These two results lead us to make the following
definition. On an inverse semigroup $S$, with $s,t\in S$, define the relation
$\sigma$ by $s\,\sigma\,t$ if and only if there is an element $u$ such that
$u\leq s,t$.

######  Theorem 4.14.

Let $S$ be an inverse semigroup.

  1. (1)

$\sigma$ is a congruence on $S$.

  2. (2)

$S/\sigma$ is a group.

  3. (3)

If $\rho$ is any congruence on $S$ such that $S/\rho$ is a group then
$\sigma\subseteq\rho$.

###### Proof.

(1) We begin by showing that $\sigma$ is an equivalence relation. Reflexivity
and symmetry are immediate. To prove transitivity, let $(a,b),(b,c)\in\sigma$.
Then there exist elements $u,v\in S$ such that $u\leq a,b$ and $v\leq b,c$.
Thus $u,v\leq b$. The set $b^{\downarrow}$ is a compatible subset and so
$u\wedge v$ exists by Lemma 4.12. But $u\wedge v\leq a,c$ and so
$(a,c)\in\sigma$. The fact that $\sigma$ is a congruence follows from the fact
that the natural partial order is compatible with the multiplication.

(2) Clearly, all idempotents are contained in a single $\sigma$-class
(possibly with non-idempotent elements). Consequently, $S/\sigma$ is an
inverse semigroup with a single idempotent. Thus $S/\sigma$ is a group by
Lemma 4.5.

(3) Let $\rho$ be any congruence such that $S/\rho$ is a group. Let
$(a,b)\in\sigma$. Then $z\leq a,b$, for some $z$, by definition. Hence
$\rho(z)\leq\rho(a),\rho(b)$ since homomomorphisms between inverse semigroups
are isotone. But $S/\rho$ is a group and so its natural partial order is
equality. Hence $\rho(a)=\rho(b)$. ∎

In the light of Theorem 4.14, it is natural to call the congruence $\sigma$
the minimum group congruence and the group $S/\sigma$ the maximum group image
of $S$.

A homomorphism $\theta\colon S\rightarrow T$ is said to be idempotent-pure if
$\theta(a)$ an idempotent implies that $a$ is an idempotent.

######  Lemma 4.15.

Let $S$ be an inverse semigroup.

  1. (1)

$\sim\,\subseteq\sigma$.

  2. (2)

The congruence $\rho$ is idempotent-pure if and only if
$\rho\,\subseteq\,\sim$.

###### Proof.

(1) Suppose that $a\sim b$. Then $a\wedge b$ exists by Lemma 4.12. It follows
that $a\,\sigma\,b$.

(2) Suppose that $\rho$ is idempotent-pure and that $a\,\rho\,b$. Then
$ab^{-1}\,\rho\,bb^{-1}$. But $\rho$ is idempotent-pure and so $ab^{-1}$ is an
idempotent. Similarly, $a^{-1}b$ is an idempotent. We have proved that $a\sim
b$. Conversely, suppose that $\rho$ is a congruence such that
$\rho\,\subseteq\,\sim$ and $a\,\rho\,e$, where $e$ is an idempotent. Observe
that $a^{-1}\,\rho\,e$. Thus $aa^{-1}\,\rho\,e$. It follows that
$aa^{-1}\,\rho\,a$. Thus $(aa^{-1})a$ is an idempotent and so $a$ is an
idempotent. ∎

Inverse semigroups and their homomorphisms form a category (of structures).
The category of groups and their homomorphisms is a subcategory. The
properties of the minimum group congruence lead naturally to the following
result on the category of inverse semigroups.

######  Proposition 4.16.

The category of groups is a reflective subcategory of the category of inverse
semigroups.

###### Proof.

Let $S$ be an inverse semigroup and let $\sigma^{\natural}\colon S\rightarrow
S/\sigma$ be the associated natural homomorphism. Let $\theta\colon
S\rightarrow G$ be a homomorphism to a group $G$. Then $\mbox{\rm
ker}(\theta)$ is a group congruence on $S$ and so $\sigma\subseteq\mbox{\rm
ker}(\theta)$ by Theorem 4.14. Thus by standard semigroup theory, there is a
unique homomorphism $\theta^{\ast}$ from $S/\sigma$ to $G$ such that
$\theta=\theta^{\ast}\sigma^{\natural}$. ∎

It follows by standard category theory, such as [78, Chapter IV, Section 3],
that there is a functor from the category of inverse semigroups to the
category of groups which takes each inverse semigroup $S$ to $S/\sigma$ and if
$\theta\colon S\rightarrow T$ is a homomorphism of inverse semigroups then the
function $\psi\colon S/\sigma\rightarrow T/\sigma$ defined by
$\psi(\sigma(s))=\sigma(\theta(s))$ is the corresponding group homomorphism
(this can be checked directly).

There is another characterization of $E$-unitary inverse semigroups which is
interesting in this context.

######  Lemma 4.17.

Let $S$ be an inverse semigroup. Then the following conditions are equivalent:

  1. (1)

$S$ is $E$-unitary.

  2. (2)

$\sim\,=\sigma$.

  3. (3)

$\sigma$ is idempotent-pure.

  4. (4)

$\sigma(e)=\mathsf{E}(S)$ for any idempotent $e$.

###### Proof.

(1)$\Rightarrow$(2). By part (1) of Lemma 4.15, the compatibility relation is
contained in $\sigma$. Let $(a,b)\in\sigma$. Then $z\leq a,b$ for some $z$. It
follows that $z^{-1}z\leq a^{-1}b$ and $zz^{-1}\leq ab^{-1}$. But $S$ is
$E$-unitary and so $a^{-1}b$ and $ab^{-1}$ are both idempotents. Hence $a\sim
b$.

(2)$\Rightarrow$(3). By part (2) of Lemma 4.15, a congruence is idempotent
pure precisely when it is contained in the compatibility relation.

(3) $\Rightarrow$ (4). This is immediate from the definition of an idempotent-
pure congruence.

(4) $\Rightarrow(1)$ Suppose that $e\leq a$ where $e$ is an idempotent. Then
$(e,a)\in\sigma$. But by our assumption, the element $a$ is an idempotent. ∎

For inverse semigroups with zero the minimum group congruence is not very
interesting since the group degenerates to the trivial group. The following
example shows one concrete way to deal with this issue.

######  Example 4.18.

Let $G$ be a group and let $S$ be the inverse monoid of all isomorphisms
between the subgroups of $G$. The semilattice of idempotents is isomorphic to
the partially ordered set of subgroups of $G$. The trivial group is a subgroup
of every group. It follows that the isomorphism from the trivial subgroup to
itself is the zero of $S$. Suppose now that $G$ is infinite. Consider the
inverse subsemigroup $\Omega(G)$ of $S$ which consists of all the isomorphisms
between the subgroups of $G$ of finite index. The group
$\mbox{Comm}(G)=\Omega(G)/\sigma$ is called the abstract commensurator of $G$
[7]. See also [90]. The elements of $\mbox{Comm}(G)$ are ‘hidden symmetries’
to use the terminology of Farb and Weinberger [23].

The above example shows that we may form groups from inverse semigroups by
looking at the ‘large’ elements of the inverse semigroup (and so excluding the
zero). Here is another example. Let $S$ be an inverse semigroup. A non-zero
idempotent $e$ of an inverse semigroup $S$ is said to be essential if $ef\neq
0$ for all non-zero idempotents $f$ of $S$. An element $s$ is said to be
essential if both $s^{-1}s$ and $ss^{-1}$ are essential. Essential elements
were first defined in [9] and applied to construct groups in [55, 56]. Denote
by $S^{e}$ the set of all essential elements of $S$. We regard the elements of
$S^{e}$ as being ‘large’ elements of $S$. We call $S^{e}$ the essential part
of $S$.

######  Lemma 4.19.

Let $S$ be an inverse semigroup. Then, if non-empty, $S^{e}$ is an inverse
subsemigroup of $S$.

###### Proof.

It is clear that $S^{e}$ is closed under inverse. Let $a,b\in S^{e}$. We prove
that $\mathbf{d}(ab)$ is essential; the proof that $\mathbf{r}(ab)$ is
essential is similar. Observe, first, that $b^{-1}a^{-1}ab\neq 0$ since both
$bb^{-1}$ and $a^{-1}a$ are essential. Let $e$ be a non-zero idempotent. We
calculate $\mathbf{d}(ab)e$. Observe that $beb^{-1}\neq 0$ and is an
idempotent, since it is the conjugate of an idempotent. Thus
$a^{-1}abeb^{-1}\neq 0$. It follows that $a^{-1}abe\neq 0$ and so
$b^{-1}a^{-1}abe\neq 0$. ∎

We expect the groups $S^{e}/\sigma$ to be interesting, which indeed they are
if we choose $S$ carefully. See [56, 71, 73] for applications of the essential
part of an inverse semigroup in constructing groups.

We have seen that there is a precondition that must be satisfied in order that
a pair of elements have a join. The same is not true for meets. If every pair
of elements of an inverse semigroup has meets we say that it is a meet-
semigroup. Such semigroups were first studied by Leech [75]. There is an
alternative way to characterize inverse meet-semigroups which is often useful.

######  Lemma 4.20.

Let $S$ be an inverse semigroup.

  1. (1)

$S$ has all binary meets if and only if for each element $a\in S$ there is an
idempotent, denoted by $\phi(a)$, such that $\phi(a)\leq a$ and if $e\leq a$,
where $e$ is an idempotent, then $e\leq\phi(a)$. Thus, $\phi(a)$ is the
largest idempotent below $a$.

  2. (2)

The map $\phi\colon S\rightarrow\mathsf{E}(S)$, defined in part (1) above, is
an order-preserving idempotent function with $\mathsf{E}(S)$ as its fixed-
point set such that $\phi(ae)=\phi(a)e$ and $\phi(ea)=e\phi(a)$ for all
$e\in\mathsf{E}(S)$.

###### Proof.

(1) Suppose first that $S$ has all binary meets. For each $a\in S$ define
$\phi(a)=a\wedge\mathbf{d}(a)$. Clearly, $\phi(a)$ is an idempotent (because
it is beneath an idempotent). Let $e$ be an idempotent such that $e\leq a$.
Then $e\leq\mathbf{d}(a)$ and so $e\leq\phi(a)$. We have therefore proved that
if all binary meet exist, then the function $\phi$ exists. Conversely, suppose
that such a function $\phi$ exists. Let $a,b\in S$ and consider the element
$\phi(ab^{-1})b$. Clearly, $\phi(ab^{-1})b\leq b$. But, by definition,
$\phi(ab^{-1})\leq ab^{-1}$ and so $\phi(ab^{-1})b\leq ab^{-1}b\leq a$. Let
$c\leq a,b$. Then $cc^{-1}\leq ab^{-1}$ and so $cc^{-1}\leq\phi(ab^{-1})$. Now
$c\leq b$ and so $c=(cc^{-1})c\leq\phi(ab^{-1})b$. It follows that $a\wedge
b=\phi(ab^{-1})b$.

(2) It is easy to prove that $\phi$ is order-preserving. We prove that
$\phi(ae)=\phi(a)e$ for all $e\in\mathsf{E}(S)$. By definition $\phi(a)\leq
a$. Thus $\phi(a)e\leq ae$. But $\phi(a)e$ is an idempotent. It follows that
$\phi(a)e\leq\phi(ae)$. We have that $\phi(ae)\leq ae\leq a$. But $\phi(ae)$
is an idempotent. It follows that $\phi(ae)\leq\phi(a)$ and so
$\phi(ae)e\leq\phi(a)e$. But $\phi(ae)\leq ae$, and so $\phi(ae)e=\phi(ae)$.
Thus $\phi(ae)\leq\phi(a)e$. We have therefore proved that
$\phi(ae)=\phi(a)e$. ∎

The function $\phi$ is called a fixed-point operator.

######  Example 4.21.

We can deduce right away that the symmetric inverse monoids are meet-monoids
because if $f\in\mathcal{I}(X)$ then we can define the idempotent $1_{A}$
where $A$ is the set of all points that $f$ fixes (and if $f$ does not fix any
points then $A=\varnothing$.)

The $E^{\ast}$-unitary semigroups also enjoy a property that is more
significant than it looks. The following was first noted by [76].

######  Proposition 4.22.

An $E^{\ast}$-unitary inverse semigroup has meets of all pairs of elements.

###### Proof.

Let $s$ and $t$ be any pair of elements. Suppose that there exists a non-zero
element $u$ such that $u\leq s,t$. Then $uu^{-1}\leq st^{-1}$ and $uu^{-1}$ is
a non-zero idempotent. Thus $st^{-1}$ is an idempotent. Similarly $s^{-1}t$ is
an idempotent. It follows that $s\wedge t$ exists by Lemma 4.12. If the only
element below $s$ and $t$ is $0$ then $s\wedge t=0$. ∎

To conclude this section, we state another deep result about finite inverse
monoids. We need a definition first. An inverse monoid $S$ is said to be
$F$-inverse if every $\sigma$-class contains a greatest element.

######  Lemma 4.23.

Every $F$-inverse monoid is $E$-unitary.

###### Proof.

Let $S$ be an $F$-inverse monoid. Suppose that $e\leq a$. Then
$\sigma(e)=\sigma(a)$. Each $\sigma$-class contains a maximum element. Since
we are working in a monoid, $\sigma(e)=\sigma(1)$. Let the maximum element in
$\sigma(1)$ be $x$. Then $1\leq x$. It follows that $1=x1=x$. Thus the maximum
element of $\sigma(1)$ is $1$ itself. Thus $a\leq 1$ and so $a$ is an
idempotent. ∎

You can find out a lot more about $F$-inverse monoids, here [6]. In [5,
Theorem 2.7], the following deep theorem is proved and the background to it
explained.

######  Theorem 4.24.

Every finite inverse monoid has a finite $F$-inverse monoid cover.

The implications of this theorem for finite inverse monoid theory are
explained in [50], but it is enough to say that this is another example of a
beautiful piece of mathematics and is related to Ash’s deep result Theorem
3.27.

##  5\. Inverse semigroups as non-commutative lattices

It is only a slight exaggeration to say that in the four decades after inverse
semigroups were introduced the main focus of researchers was on the purely
algebraic properties of inverse semigroups.111111This is a tendancy rather
than an absolute. For example, the papers [103] and [104] deal with the
natural partial order. Subsequently, the properties of the natural partial
order have come to the fore. In this section, we shall study inverse
semigroups with respect to this natural partial order. From this point of
view, inverse semigroups can themselves be regarded as ‘non-commutative meet
semilattices’. However, we shall find it fruitful to regard certain inverse
semigroups as being ‘non-commutative lattices’ of various kinds, always with
the proviso that, because of Lemma 4.8, the join will not always be defined.
Meets of idempotents will always usually be denoted by concatenation.
Specifically, we shall take as our points of departure the various classes of
lattice (each with top $1$ and bottom $0$): a frames are the complete
infinitely distributive lattices; distributive lattices have all binary meets
and binary joins with binary meets distributing over binary joins and vice-
versa; Boolean algebras are those distibutive lattices in which each element
$x$ has a complement $\bar{x}$ such that $x\bar{x}=0$ and $x\vee\bar{x}=1$.
Observe that homomorphisms of distributive lattices automatically preserve
complements. Boolean algebras are particularly important so it will be useful
to have a way of defining them which involves only products and complements.
For the standard axioms for a Boolean algebra see [61, Chapter 2]. The
following lemma contains some axioms due to Frink [26]. If you have trouble
proving that this really is a Boolean algebra, see [93].

######  Lemma 5.1.

Consider the following structure $(B,\cdot,a\mapsto\bar{a},0)$ where
$(B,\cdot)$ is a commutative band and $ab=a$ if and only if $a\bar{b}=0$.
Define $a+b=\overline{(\bar{a}\cdot\bar{b})}$. Then $(B,\cdot,+,0,\bar{0})$ is
a Boolean algebra.

We call these Frink’s axioms for a Boolean algebra.

We begin with frames. The lattice of open sets of a topological space is an
example of a frame The theory of frames can be viewed as the theory of
topological spaces in which the open sets, and not the points, are taken as
primary. As well as being an interesting theory in its own right [38] with
important applications, it is also a key ingredient in topos theory [79].
Johnstone discusses the origins of frame theory in his book on this subject
[38, Chapter II]. One sentence is significant for the goals of this section.
He writes on page 76:

> It was Ehresmann …and his student Bénabou …who first took the decisive step
> in regarding complete Heyting algebras as ‘generalized topological spaces’.

However, Johnstone does not say why Ehresmann was led to his frame-theoretic
viewpoint of topological spaces. In fact, it was Ehresmann’s paper [20], which
we have cited above as being one of the origins of inverse semigroup theory,
which led to the theory of frames. Ehresmann was interested in pseudogroups of
transformations. Amongst those are the full transformation pseudogroups
$\mathcal{I}(X,\tau)$ of homeomorphisms between the open subsets of $\tau$.
The idempotents of such pseudogroups are the identity functions defined on the
open subsets of $\tau$. Of course, these form frames. More generally, we
define a pseudogroup to be an inverse semigroup in which all compatible
subsets have joins and in which multiplication distributes over such joins.
The semilattice of idempotents of a pseudogroup is a frame and so we can
regard pseudogroups as being non-commutative generalizations of frames. Using
this language, we can say that Schein [104] proved that associated with every
inverse semigroup is a universal pseudogroup. Pseudogroups themselves are the
subject of [102].

We can continue in this vein, but consider instead only finite joins. We say
that an inverse monoid is distributive if it has all finite joins and
multiplication distributes over such joins. The semilattice of idempotents of
a distributive inverse monoid is a distributive lattice. We can therefore
regard distributive inverse monoids as being non-commutative distributive
lattices. Distributive inverse monoids were first studied in [42].

An inverse monoid is said to be Boolean if it is distributive and its
semilattice of idempotents is in fact a Boolean algebra. We can therefore
regard Boolean inverse monoids as being non-commutative Boolean algebras.
Boolean inverse monoids have shown themselves to be particularly interesting.

The relationships between these various classes of inverse semigroup is
described in [66]. Boolean inverse monoids were introduced in [59], and the
fact that associated with every inverse semigroup is a universal Boolean
inverse semigroup is proved in [62]. Associated with every Boolean inverse
monoid is its type monoid: this is always an abelian refinement monoid [47].
The type monoid is then studied in great detail in [117]. The following table
summarizes this whole approach to inverse semigroups:

Commutative | Non-commutative
---|---
Meet semilattice | Inverse semigroup
Frame | Pseudogroup
Distributive lattice | Distributive inverse monoid
Boolean algebra | Boolean inverse monoid

In the remainder of this section, we shall be particularly interested in
Boolean inverse monoids, but we shall prove some results in greater
generality. Our first results hold in any inverse semigroup

######  Lemma 5.2.

Let $S$ be an inverse semigroup.

  1. (1)

Here, we shall need the fact that the inverse semigroup has a zero. Suppose
that $a,b\leq c$. If $\mathbf{d}(a)\perp\mathbf{d}(b)$ then
$\mathbf{r}(a)\perp\mathbf{r}(b)$.

  2. (2)

If $a\wedge b$ exists then $c(a\wedge b)=ca\wedge cb$ and $(a\wedge
b)c=ac\wedge bc$.

###### Proof.

(1) We are given that $a=c\mathbf{d}(a)$ and $b=c\mathbf{d}(b)$. Using these
equations and the fact that idempotents commute, it is easy to check that
$\mathbf{r}(a)\mathbf{r}(b)=0$.

(2) We are given that $a\wedge b$ exists. We prove that $ca\wedge cb$ exists
and that $(a\wedge b)c=ac\wedge bc$. The proof of the other case is similar.
Observe that $c(a\wedge b)\leq ca,cb$. Now let $y\leq ca,cb$. Then
$c^{-1}y\leq c^{-1}ca,c^{-1}cb$ and so $c^{-1}y\leq a,b$. It follows that
$c^{-1}y\leq a\wedge b$. Thus $cc^{-1}y\leq c(a\wedge b)$. But $y\leq ca$ and
so $y=ca\mathbf{d}(y)$. It follows that $cc^{-1}y=y$. Thus $y\leq ca,cb$
implies that $y\leq c(a\wedge b)$. It follows that $c(a\wedge b)=ca\wedge cb$
∎

We now specialize to distributive inverse monoids. Result (2) below is what
remains of one of the distributive laws.

######  Lemma 5.3.

Let $S$ be a distributive inverse monoid.

  1. (1)

Suppose that $a\sim b$. Then $\mathbf{d}{(a\vee
b)}=\mathbf{d}(a)\vee\mathbf{d}(b)$ and $\mathbf{r}{(a\vee
b)}=\mathbf{r}(a)\vee\mathbf{r}(b)$

  2. (2)

Suppose that both $a\vee b$ and $c\wedge(a\vee b)$ exist. Then both $c\wedge
a$ and $c\wedge b$ exist, the join $(c\wedge a)\vee(c\wedge b)$ exists and
$c\wedge(a\vee b)=(c\wedge a)\vee(c\wedge b)$.

###### Proof.

(1) We prove the first result; the proof of the second is similar. We
calculate $(a\vee b)^{-1}(a\vee b)$. We first use the fact that $a\mapsto
a^{-1}$ is an order-isomorphism. It follows that $(a\vee b)^{-1}(a\vee
b)=(a^{-1}\vee b^{-1})(a\vee b)$. Now multiply out to get

$a^{-1}a\vee a^{-1}b\vee b^{-1}a\vee b^{-1}b.$

But $a\sim b$ and so both $a^{-1}b$ and $b^{-1}a$ are idempotents. Morover,
$a^{-1}b\leq a^{-1}a$ and $b^{-1}a\leq b^{-1}b$. The result now follows.

(2) Let $x\leq c,a$. Then $x\leq c\wedge(a\vee b)$. It follows that
$x\mathbf{d}(a)\leq(c\wedge(a\vee b))\mathbf{d}(a)$. But $x\leq a$ implies
that $x\mathbf{d}(a)=x$. It follows that $x\leq(c\wedge(a\vee
b))\mathbf{d}(a)$. On the other hand, $(c\wedge(a\vee b))\mathbf{d}(a)\leq
c,a$. It follows that

$c\wedge a=(c\wedge(a\vee b))\mathbf{d}(a).$

Similarly,

$c\wedge b=(c\wedge(a\vee b))\mathbf{d}(b).$

Observe that since $c\wedge a,c\wedge b\leq a\vee b$ it follows that $(c\wedge
a)\sim(c\wedge b)$. Thus $(c\wedge a)\vee(c\wedge b)$ exists. Observe that
$(c\wedge a)\vee(c\wedge b)\leq c\wedge a,c\wedge b$. It follows that
$(c\wedge a)\vee(c\wedge b)\leq c,a\vee b$. Thus,

$(c\wedge a)\vee(c\wedge b)\leq c\wedge(a\vee b).$

Let $x\leq a\vee b,c$. Then $x=(a\vee b)\mathbf{d}(x)$. From $x\leq a\vee b$
we get that $x\mathbf{d}(a)\leq(a\vee b)\mathbf{d}(a)$. Now, $(a\vee
b)\mathbf{d}(a)=a$. It follows that $x\mathbf{d}(a)\leq a$. But $x\leq c$ and
so $x\mathbf{d}(a)\leq c\mathbf{d}(a)$. Thus we have proved that
$x\mathbf{d}(a)\leq a,c\mathbf{d}(a)$. We now apply Lemma 5.2, and deduce that
$(a\wedge c)\mathbf{d}(a)=a\wedge c\mathbf{d}(c)$. Thus $x\mathbf{d}(a)\leq
a\wedge c\mathbf{d}(c)$. Similarly, $x\mathbf{d}(b)\leq c\mathbf{d}(b)\wedge
b$. It follows that

$x\mathbf{d}(a)\vee x\mathbf{d}(b)\leq(c\mathbf{d}(a)\wedge
a)\vee(c\mathbf{d}(b)\wedge b)$

and so

$x\mathbf{d}(a\vee b)\leq(c\mathbf{d}(a)\wedge a)\vee(c\mathbf{d}(b)\wedge
b).$

By the above, we deduce that

$x\leq(c\mathbf{d}(a)\wedge a)\vee(c\mathbf{d}(b)\wedge b).$

But $(c\mathbf{d}(a)\wedge a)\vee(c\mathbf{d}(b)\wedge b)\leq(c\wedge
a)\vee(c\wedge b)$ and the result follows. ∎

By induction, the second result above can be generalized to any number of
joins.

We now specialize to Boolean inverse monoids. Let $S$ be a Boolean inverse
monoid. If $y\leq x$, define

$x\setminus y=x\overline{\mathbf{d}(y)}.$

######  Lemma 5.4.

Let $S$ be a Boolean inverse monoid.

  1. (1)

$\mathbf{d}(x\setminus y)=\mathbf{d}(x)\overline{\mathbf{d}(y)}$.

  2. (2)

If $y\leq x$ then $y\perp(x\setminus y)$ and $x=y\vee(x\setminus y)$.

  3. (3)

Suppose that $a\leq x$ is such that $a\perp y$ and $x=y\vee a$ then
$a=x\setminus y$.

  4. (4)

$\mathbf{r}(x\setminus y)=\mathbf{r}(x)\overline{\mathbf{r}(y)}$.

###### Proof.

(1) Straightforward from the definition.

(2) By definition $(x\setminus y)\leq x$, and from part (1), we have that
$\mathbf{d}(x\setminus y)=\mathbf{d}(x)\overline{\mathbf{d}(y)}$. It follows
that $\mathbf{d}(y)\perp\mathbf{d}(x\setminus y)$. We now apply Lemma 5.2, to
deduce that $y\perp(x\setminus y)$. Clearly, $y\vee(x\setminus y)\leq x$. But
$\mathbf{d}(y\vee(x\setminus
y))=\mathbf{d}(y)\vee\mathbf{d}(x)\overline{\mathbf{d}(y)})=\mathbf{d}(x)$. It
follows that $x=y\vee(x\setminus y)$.

(3) Observe that $\mathbf{d}(x)=\mathbf{d}(y)\vee\mathbf{d}(a)$ and
$\mathbf{d}(a)\perp\mathbf{d}(y)$. It follows that
$\mathbf{d}(a)=\mathbf{d}(x)\overline{\mathbf{d}(y)}$. But
$a=x\mathbf{d}(a)=x\overline{\mathbf{d}(y)}=x\setminus y$.

(4) This follows by part (3) above. ∎

A morphism between distributive inverse monoids is a homomorphism of monoids
with zero that preserves binary joins.

In working with Boolean inverse monoids, it is often much easier to calculate
with orthogonal joins than arbitrary ones. The following result provides
circumstances under which we lose nothing by doing this.

######  Lemma 5.5.

Let $S$ be an inverse semigroup.

  1. (1)

We make the following assumptions about $S$: the semilattice of idempotents of
$S$ forms a Boolean algebra under the natural partial order; all finite
orthogonal joins exist; multiplication distributes over finite orthoginal
joins. Then $S$ is a Boolean inverse monoid.

  2. (2)

Let $S$ and $T$ be Boolean inverse monoids. Let $\theta\colon S\rightarrow T$
be a monoid homomorphism of monoid with zero that preserves orthogonal joins.
Then $\theta$ preserves all binary compatible joins.

###### Proof.

(1) We have to prove that all binary joins exist and that multiplication
distributes over such joins. Let $e,f\in\mathsf{E}(S)$. Then using the
distributive law, we have that $e\bar{f}\vee f=e\vee f$. But $e\bar{f}$ and
$f$ are orthogonal. It follows that $a(e\vee f)=a(e\bar{f}\vee
f)=ae\bar{f}\vee af$. Observe that $(ae\bar{f}\vee af)\mathbf{d}(ae)=ae$. It
follows that $ae,af\leq ae\bar{f}\vee af$. On the other hand, if $ae,af\leq x$
then $ae\bar{f}\vee af\leq x$. We have therefore proved that

$a(e\vee f)=ae\vee af$

where $e$ and $f$ are any idempotents. Suppose, now, that $a\sim b$. Then
$a\wedge b$ exists by Lemma 4.12. We have that

$(a\setminus a\wedge b)\vee b=a\overline{\mathbf{d}(a)\mathbf{d}(b)}\vee b$

since, in this case, $\mathbf{d}(a\wedge b)=\mathbf{d}(a)\mathbf{d}(b)$ by
Lemma 4.12. But
$\overline{\mathbf{d}(a)\mathbf{d}(b)}=\overline{\mathbf{d}(a)}\vee\overline{\mathbf{d}(b)}$.
We now use our result above to deduce that

$(a\setminus a\wedge b)\vee b=a\overline{\mathbf{d}(b)}\vee b,$

an orthogonal join. Observe that $(a\overline{\mathbf{d}(b)}\vee
b)\mathbf{d}(a)=a$, where we use the fact that $b\mathbf{d}(a)=a\mathbf{d}(b)$
by Lemma 4.12, our result above, and a little Boolean algebra. We therefore
have that $a,b\leq a\overline{\mathbf{d}(b)}\vee b$. Suppose that $a,b\leq x$.
Then it is routine to check that $a\overline{\mathbf{d}(b)}\vee b\leq x$. We
have therefore proved that $a\vee b$ exists and is equal to the orthogonal
join $a\overline{\mathbf{d}(b)}\vee b$.

It remains to show that multiplication distributes over binary joins. Suppose
that $a\sim b$. It can be proved directly that that $ca\sim cb$. We prove that

$c(a\vee b)=ca\vee cb.$

We have that

$c(a\vee b)=c(a\overline{\mathbf{d}(b)}\vee b)=ca\overline{\mathbf{d}(b)}\vee
cb.$

Thus

$ca\leq ca\overline{\mathbf{d}(b)}\vee cb.$

Now,

$(ca\overline{\mathbf{d}(b)}\vee
cb)\mathbf{d}(ca)=ca\overline{\mathbf{d}(b)}\vee cb\mathbf{d}(ca).$

We now use the fact that $ca\sim cb$ and Lemma 4.12 to get

$ca\overline{\mathbf{d}(b)}\vee
cb\mathbf{d}(ca)=ca\overline{\mathbf{d}(b)}\vee ca\mathbf{d}(cb)\leq ca.$

It follows that

$ca=(ca\overline{\mathbf{d}(b)}\vee cb)\mathbf{d}(ca).$

We deduce that $ca,cb\leq ca\overline{\mathbf{d}(b)}\vee cb$. Suppose that
$ca,cb\leq x$. Then it is routine to check that
$ca\overline{\mathbf{d}(b)}\vee cb\leq x$. Whence $ca\vee
cb=ca\overline{\mathbf{d}(b)}\vee cb$ and so $c(a\vee b)=ca\vee cb$.

(2) Suppose that $a\sim b$. Then $a\vee b=a\overline{\mathbf{d}(b)}\vee b$,
which is an orthogonal join. By assumption,

$\theta(a\vee
b)=\theta(a\overline{\mathbf{d}(b)})\vee\theta(b)=\theta(a)\theta(\overline{\mathbf{d}(b)}))\vee\theta(b).$

The result now follows since $\theta(\bar{e})=\overline{\theta(e)}$ where $e$
is any idempotent. ∎

For the remainder of this section, we shall focus on Boolean inverse monoids.
As we shall prove in Theorem 5.10, they arise naturally as soon as we try to
map inverse monoids to the multiplicative monoids of rings, such as in
representation theory. Accordingly, we begin with some results about
idempotents in rings. Recall that all our rings will be unital. If $e$ and $f$
are idempotents in a ring $R$ then we say they are orthogonal if $ef=0$
denoted by $e\perp f$. A finite set of idempotents is orthogonal if each
distinct pair of elements is orthogonal. A sum of a finite number of
orthogonal idempotents will be called an orthogonal sum. More generally, a sum
of orthogonal elements taken from an inverse submonoid will be called an
orthogonal sum.

######  Lemma 5.6.

Let $R$ be a unital ring.

  1. (1)

If $e$ is an idempotent in $R$, then $1-e$ is an idempotent in $R$; in
addition, the idempotents $e$ and $1-e$ are orthogonal.

  2. (2)

If $e$ and $f$ are orthogonal idempotents then $e+f$ is an idempotent; this
can be generalized to any finite sets of orthogonal idempotents.

  3. (3)

If $e$ and $f$ are orthogonal idemptents then

$1-(e+f)=(1-e)(1-f).$

  4. (4)

If $e$ and $f$ are commuting idempotents, then

$e+f-ef=f+e(1-f),$

an orthogonal sum.

  5. (5)

If $e$ and $f$ are commuting idempotents, then

$e\circ f=e+f-ef$

is an idempotent.

  6. (6)

If $e$ and $f$ are commuting idempotents then

$(1-e)(1-f)=1-e\circ f.$

  7. (7)

If $e,f,g$ is a set of commuting idempotents, then $e$ commutes with $f\circ
g$ and $g$ commutes with $e\circ f$ and

$(e\circ f)\circ g=e\circ(f\circ g).$

  8. (8)

If $e_{1},\ldots,e_{m}$ are orthogonal idempotents, then

$1-\left(\sum_{i=1}^{m}e_{i}\right)=\prod_{i=1}^{m}(1-e_{i}).$

  9. (9)

Let $e_{1},\ldots,e_{m}$ be a set of commuting idempotents. Define

$[e_{1},\ldots,e_{m}]=(\ldots((e_{1}\circ e_{2})\circ e_{3})\ldots)\circ
e_{m}.$

Then

$\prod_{i=1}^{m}(1-e_{i})=1-[e_{1},\ldots,e_{m}].$

  10. (10)

Let $e_{1},\ldots,e_{m}$ be a set of commuting idempotents. Then

$[e_{1},\ldots,e_{m}]$

is defined to be

$e_{1}(1-e_{2})\ldots(1-e_{m})+e_{2}(1-e_{3})\ldots(1-e_{m})+\ldots+e_{m-1}(1-e_{m})+e_{m},$

a sum of orthogonal idempotents.

###### Proof.

The proofs of (1)–(7) are straightforward.

(8) By part (3) and induction.

(9) By part (6) and induction.

(10) By part (4) and induction. ∎

We begin with a lemma which is a special case of the result we want to prove.

######  Lemma 5.7.

Let $E\subseteq R$, where $R$ is a unital ring. Suppose that $0,1\in E$ and
$E$ is a commutative band under the induced multiplication from $R$. Then
there is a subset $E\subseteq B\subseteq R$ such that $B$ is a Boolean algebra
with meet given by the product in the ring.

###### Proof.

If $E$ is any semilattice in $R$, define

$E^{\prime}=\\{e(1-e_{1})\ldots(1-e_{m})\colon e,e_{1},\ldots,e_{m}\in
E\\}\cup E.$

Then it is easy to check that this is a commutative band. If $E$ is any
semilattice in $R$, define $E^{\perp}$ to be the set of all sums of finite
sets of orthogonal elements of $E$. Then it is easy to check that this is a
commutative band. Put $B=(E^{\prime})^{\perp}$. This is a commutative band. To
prove that $B$ is a Boolean algebra, we use Lemma 5.1. We claim that if $e\in
B$ then $1-e\in B$. An element of $B$ is an orthogonal join of idempotents
that belong to $E^{\prime}$. Observe that if $e_{1},\ldots,e_{m}$ are any
orthogonal idempotents of $B$ then
$1-\left(\sum_{i=1}^{m}e_{i}\right)=\prod_{i=1}^{m}(1-e_{i})$ by part (8) of
Lemma 5.6 where each $e_{i}\in E^{\prime}$. If $e\in E$ then $1-e\in
E^{\prime}$. Thus we need only concentrate on idempotents of the form

$e(1-f_{1})\ldots(1-f_{n})\in E^{\prime}$

where $f_{1},\ldots,f_{n}\in E$. Then

$1-e(1-f_{1})\ldots(1-f_{n})=(1-e)+e[f_{1},\ldots,f_{n}],$

using part (9) of Lemma 5.6, which is a sum of orthogonal idempotents by part
(10) of Lemma 5.6. We have therefore proved that $B$ is closed under taking
complements. Let $e,f\in B$ and suppose that $ef=e$. Then $e(1-f)=e-ef=e-e=0$.
On the other hand, suppose that $e(1-f)=0$. Then $e=ef$. We have that $B$
satisfies Frink’s axioms and so $B$ is a Boolean algebra. ∎

To help us prove our main result, it is useful to have the following two
lemmas. The first says that to check if an inverse monoid is Boolean it is
enough to prove that we have orthogonal joins.

######  Lemma 5.8.

Let $S\subseteq R$, where $R$ is a unital ring. Suppose that $0,1\in S$ and
that $S$ is an inverse monoid under the induced multiplication from $R$. In
addition, we suppose that $S$ is closed under sums of pairs of orthogonal
elements and, if $e\in S$ is an idempotent, then $1-e\in S$. Then $S$ is a
Boolean inverse monoid.

###### Proof.

It is immediate that $\mathsf{E}(S)$ satisfies Frink’s axioms and so is a
Boolean algebra. Let $e$ and $f$ be orthogonal idempotents. Then $e+f=e\vee
f$; to see why observe that $(e+f)e=e$ and $(e+f)f=f$. Thus $e,f\leq e+f$. On
the other hand, suppose that $e,f\leq i$. Then $(e+f)i=e+f$. Thus $e+f\leq i$.
It follows that $e+f=e\vee f$. Let $a,b\in S$ be arbitrary orthogonal
elements. They are, in particular, compatible. We prove that $a+b=a\vee b$ and
the result then follows by Lemma 5.5. Observe that

$(a+b)\mathbf{d}(a)=a\mathbf{d}(a)+b\mathbf{d}(b)=a\mathbf{d}(a)+a\mathbf{d}(b)=a\mathbf{d}(a)=a$

where we have used Lemma 4.12. Thus $a\leq a+b$. Similarly $b\leq a+b$.
Suppose that $a,b\leq x$. Then
$a+b=x\mathbf{d}(a)+x\mathbf{d}(b)=x(\mathbf{d}(a)\vee\mathbf{d}(b))$. Thus
$a+b\leq x$. We have proved that $a+b=a\vee b$ if $a$ and $b$ are orthogonal.
∎

Our second result deals with orthogonal sums.

######  Lemma 5.9.

Let $S\subseteq R$, where $R$ is a unital ring. Suppose that $0,1\in S$ and
that $S$ is an inverse monoid under the induced multiplication from $R$.
Define $S^{\perp}$ to be the sums of all finite orthogonal sets of $S$. Then
$S^{\perp}$ is an inverse monoid and its set of idempotents is
$\mathsf{E}(S)^{\perp}$.

###### Proof.

Let $a,b\in S^{\perp}$. Then the fact that $ab\in S^{\perp}$ follows by Lemma
4.10. Observe that if $a=a_{1}+\ldots+a_{n}$, an orthogonal sum of elements of
$S$, then defining $a^{-1}=a_{1}^{-1}+\ldots+a_{n}^{-1}$ we get that
$a=aa^{-1}a$. Thus $S^{\perp}$ is a regular semigroup where, for example,
$aa^{-1}=a_{1}a_{1}^{-1}+\ldots+a_{n}a_{n}^{-1}$. Suppose that
$a=a_{1}+\ldots+a_{n}$ is an idempotent. Then $a^{2}=a$ and so
$a^{2}a_{1}^{-1}=a_{1}a_{1}^{-1}$. It follows that
$a_{1}^{2}a_{1}^{-1}=a_{1}a_{1}^{-1}$. We deduce that $a_{1}$ is an
idempotent. In a similar way, we may deduce that each of $a_{1},\ldots,a_{n}$
is an idempotent and so $a$ is a sum of orthogonal idempotents. We have
therefore provded that $\mathsf{E}(S^{\perp})=\mathsf{E}(S)^{\perp}$. It now
follows that $S^{\perp}$ is an inverse monoid since it is regular and the
idempotents commute. ∎

We can now prove the more general result which shows us that Boolean inverse
monoids arise naturally.

######  Theorem 5.10.

Let $S\subseteq R$, where $R$ is a unital ring. Suppose that $0,1\in S$ and
$S$ is an inverse monoid under the induced multiplication from $R$. Then there
is a subset $S\subseteq T\subseteq R$ such that $T$ is a Boolean inverse
monoid.

###### Proof.

Put $E=\mathsf{E}(S)$. Define $E^{\prime}$ as in the proof of Lemma 5.7. It is
easy to check that $E^{\prime}$ is closed under conjugation by elements of
$S$. Put $S^{\prime}=SE^{\prime}$. Then $S^{\prime}$ is a regular monoid and
the set of idempotents of $S^{\prime}$ is precisely the set $E^{\prime}$. It
follows by Proposition 3.9, that $S^{\prime}$ is also an inverse monoid.
Define $T$ to be the sums of finite orthogonal subsets of $S^{\prime}$. Then
$T$ is an inverse monoid whose set of idempotents is precisely the set
$(\mathsf{E}(S)^{\prime})^{\perp}$, using the notation of the proof of Lemma
5.7. This is proved using Lemma 5.9. It now follows by Lemma 5.8 that $T$ is a
Boolean inverse monoid. ∎

##  6\. The underlying groupoid

The product we have defined on the symmetric inverse monoid $\mathcal{I}(X)$
is not the only one, nor perhaps even the most obvious one, that we might
define. Given partial bijections $f$ and $g$, we could define $fg$ only when
the domain of $f$ is equal to the range of $g$. When we do this, we are
regarding $f$ and $g$ as being functions rather than as partial functions.
With respect to this ‘restricted product’, $\mathcal{I}(X)$ becomes a
groupoid. What we have done for the special case of the symmetric inverse
monoids, we can also do for arbitrary inverse semigroups, but first we review
the basics of groupoid theory we shall need.

Categories are usually regarded as categories of structures with their
morphisms. They can, however, also be regarded as algebraic structures in
their own right, no different from groups, rings and fields except that the
binary operation is only partially defined. We shall define categories from
this purely algebraic point of view. Let $C$ be a set equipped with a partial
binary operation which we shall denote by $\cdot$ or by concatenation. If
$x,y\in C$ and the product $x\cdot y$ is defined we write $\exists x\cdot y$.
An element $e\in C$ is called an identity if $\exists e\cdot x$ implies
$e\cdot x=x$, and $\exists x\cdot e$ implies $x\cdot e=x$. The set of
identities of $C$ is denoted $C_{o}$, where the subscript ‘o’ stands for
‘object’. The pair $(C,\cdot)$ is said to be a category if the following
axioms hold:

(C1):

$x\cdot(y\cdot z)$ exists if and only if $(x\cdot y)\cdot z$ exists, in which
case they are equal.

(C2):

$x\cdot(y\cdot z)$ exists if and only if $x\cdot y$ and $y\cdot z$ exist.

(C3):

For each $x\in C$ there exist identities $e$ and $f$ such that $\exists x\cdot
e$ and $\exists f\cdot x$.

From axiom (C3), it follows that the identities $e$ and $f$ are uniquely
determined by $x$. We write $e={\bf d}(x)$ and $f={\bf r}(x)$, where ${\bf
d}(x)$ is the domain identity and ${\bf r}(x)$ is the range identity.121212As
you will see, there is no contradiction with the notation we introduced
earlier. Observe that $\exists x\cdot y$ if and only if ${\bf d}(x)={\bf
r}(y)$. The elements of a category are called arrows. We say that the arrow
$x$ starts at $\mathbf{d}(x)$ and ends at $\mathbf{r}(x)$. If $C$ is a
category and $e$ and $f$ identities in $C$ then we put

$\mbox{\rm hom}(e,f)=\\{x\in C\colon\>{\bf d}(x)=f\mbox{ and }{\bf
r}(x)=e\\},$

the set of arrows from $f$ to $e$. Subsets of $C$ of the form $\mbox{\rm
hom}(e,f)$ are called hom-sets. We also put $\mbox{\rm end}(e)=\mbox{\rm
hom}(e,e)$, the local monoid at $e$. We define subcategories in the obvious
way. Viewed in this light, a catgeory is a monoid with many identities since
the categories with exactly one identity are precisely the monoids.

We shall not need arbitary categories but those that generalize groups. A
category $C$ is said to be a groupoid if for each $x\in C$ there is an element
$x^{-1}$ such that $x^{-1}x={\bf d}(x)$ and $xx^{-1}={\bf r}(x)$. The element
$x^{-1}$ is unique with these properties. In the case of groupoids, the local
monoids are, in fact, local groups. It can happen that a groupoid consists
entirely of its local units; in this case, we say that the groupoid is a union
of groups. A groupoid with exactly one identity is a group. Two elements $x$
and $y$ of a groupoid are said to be connected if there is an arrow starting
at ${\bf d}(x)$ and ending at ${\bf d}(y)$. This defines an equivalence
relation on the groupoid whose equivalence classes are called the connected
components131313The use of the word ‘connected’ here is unfortunate. It has
nothing to do with topology. The groupoids here are discrete. of the groupoid.
A groupoid with one connected component is said to be connected. Every
groupoid can be written as a disjoint union of connected groupoids. We say
that a groupoid is principal if for any identities $e$ and $f$ there is at
most one arrow from $e$ to $f$.

######  Lemma 6.1.

Let $G$ be a groupoid. Then $G$ is principal if and only if all its local
groups are trivial.

###### Proof.

If a groupoid is principal, it is clear that the local groups are trivial.
Suppose, now, that the local groups are trivial. We prove that the groupoid is
principal. Suppose that $x$ and $y$ are arrows that start at $e$ and end at
$f$. Then $y^{-1}x$ begins and ends at $e$. This means that, under our
assumption, $y^{-1}x=e$. We deduce that $x=y$. ∎

We define subgroupoids in the obvious way.

######  Example 6.2.

Let $X$ be a set. The set $X\times X$ becomes a groupoid when we define
$\mathbf{d}(x,y)=(y,y)$, $\mathbf{r}(x,y)=(x,x)$ and $(x,y)^{-1}=(y,x)$;
define a partial product by $(x,y)(y,z)=(x,z)$. Now, let $\sim$ be an
equivalence relation on $X$. Define $G$ to consist of those ordered pairs
$(x,y)$ where $x\sim y$. It is easy to check that $G$ is a subgroupoid of
$X\times X$. It is, in fact, a principal groupoid. If $G$ is an arbitrary
principal groupoid, then it defines an equivalence relation on the set
$X=G_{o}$. In fact, principal groupoids and equivalence relations are
different ways of defining the same thing.

At this point, category theorists should look away since we shall convert a
category into a semigroup. If $C$ is a category as we have defined it above,
then we can convert it into a semigroup with zero by adjoining a zero and
defining all undefined products to be zero. We denote the semigroup with zero
that arises by $C^{0}$.

Now we can return to inverse semigroups. Let $S$ be an arbitrary inverse
semigroup. Define the restricted product141414Sometimes referred to as the
trace product. of two elements $s$ and $t$ in $S$ to be $s\cdot t=st$ if
$s^{-1}s=tt^{-1}$ and undefined otherwise. The following result simply tells
us that what we expect to happen actually does happen.

######  Proposition 6.3.

Every inverse semigroup $S$ is a groupoid with respect to its restricted
product.

###### Proof.

We begin by showing that all idempotents of $S$ are identities of $(S,\cdot)$.
Let $e\in S$ be an idempotent and suppose that $e\cdot x$ is defined. Then
$e=xx^{-1}$ and $e\cdot x=ex$. But $ex=(xx^{-1})x=x$. Similarly, if $x\cdot e$
is defined then it is equal to $x$. We now check that the axioms (C1), (C2)
and (C3) hold.

Axiom (C1) holds: suppose that $x\cdot(y\cdot z)$ is defined. Then

$x^{-1}x=(y\cdot z)(y\cdot z)^{-1}\mbox{ and }y^{-1}y=zz^{-1}.$

But

$(y\cdot z)(y\cdot z)^{-1}=yzz^{-1}y^{-1}=yy^{-1}.$

Hence $x^{-1}x=yy^{-1}$, and so $x\cdot y$ is defined. Also
$(xy)^{-1}(xy)=y^{-1}y=zz^{-1}$. Thus $(x\cdot y)\cdot z$ is defined. It is
clear that $x\cdot(y\cdot z)$ is equal to $(x\cdot y)\cdot z$. A similar
argument shows that if $(x\cdot y)\cdot z$ exists then $x\cdot(y\cdot z)$
exists and they are equal.

Axiom (C2) holds: suppose that $x\cdot y$ and $y\cdot z$ are defined. We show
that $x\cdot(y\cdot z)$ is defined. We have that $x^{-1}x=yy^{-1}$ and
$y^{-1}y=zz^{-1}$. Now

$(yz)(yz)^{-1}=y(zz^{-1})y^{-1}=y(y^{-1}y)y^{-1}=yy^{-1}=x^{-1}x.$

Thus $x\cdot(y\cdot z)$ is defined. The proof of the converse is
straightforward.

Axiom (C3) holds: for each element $x$ we have that $x\cdot(x^{-1}x)$ is
defined, and we have seen that idempotents of $S$ are identities. Thus we put
${\bf d}(x)=x^{-1}x$. Similarly, we put $xx^{-1}={\bf r}(x)$. It is now clear
that $(S,\cdot)$ is a category. The fact that it is a groupoid is immediate. ∎

We call $(S,\cdot)$ the (underlying) groupoid of $S$. We can use the
underlying groupoid to reveal something of the structure of inverse
semigroups.

######  Example 6.4.

We can interpret Lemma 3.14 in terms of the structure of the underlying
groupoid. An inverse semigroup is a Clifford semigroup if and only if its
underlying groupoid is a union of groups.

In the light of Proposition 6.3, it is now natural to picture an element $a$
of an inverse semigroup as follows:

$\mathbf{d}(a)\stackrel{{\scriptstyle a}}{{\longrightarrow}}\mathbf{r}(a).$

where we now call $\mathbf{d}(a)$ the domain idempotent of $a$ and
$\mathbf{r}(a)$ is the range idempotent of $a$. The underlying groupoid
structure more or less does away with the need to deal directly with Green’s
relations. If $e$ and $f$ are idempotents we write $e\,\mathscr{D}\,f$ to mean
that there exists an element $a$ such that $\mathbf{d}(a)=e$ and
$\mathbf{r}(a)=f$. We write $a\,\mathscr{D}\,b$ to mean that
$\mathbf{d}(a)\,\mathscr{D}\,\mathbf{d}(b)$. The relation $\mathscr{D}$ really
is Green’s relation $\mathscr{D}$; by the same token $a\,\mathscr{L}\,b$ if
and only if $\mathbf{d}(a)=\mathbf{d}(b)$ and $a\,\mathscr{R}\,b$ if and only
if $\mathbf{r}(a)=\mathbf{r}(b)$. We define $a\,\mathscr{H}\,b$ if and only if
$a\,\mathscr{L}\,b$ and $a\,\mathscr{R}\,b$; in other words, $a$ and $b$
belong to the same hom-set. An inverse semigroup is said to be bisimple if its
underlying groupoid consists of one connected component. An inverse semigroup
with zero is said to be $0$-bisimple if its underlying groupoid consists of
two connected components.

In passing from an inverse semigroup to its underlying groupoid, we do lose
some information. As the following result shows, the information that is lost
is encoded by the natural partial order.

######  Lemma 6.5.

Let $S$ be an inverse semigroup. Then for any $s,t\in S$ there exist elements
$s^{\prime}\leq s$ and $t^{\prime}\leq t$ such that $st=s^{\prime}\cdot
t^{\prime}$ where the product on the right is the restricted product.

###### Proof.

Put $e=\mathbf{d}(s)\mathbf{r}(t)$ and define $s^{\prime}=se$ and
$t^{\prime}=et$. Observe that $\mathbf{d}(s^{\prime})=e$ and
$\mathbf{r}(t^{\prime})=e$ and that $st=s^{\prime}t^{\prime}$. ∎

It is possible to formalize the idea of a groupoid equipped with a partial
order in such a way that the original semigroup product can be recaptured.
This is the approach that Ehresmann took to studying inverse semigroups. See
[51, Chapter] for exactly how this is done.151515This groupoid, which is
discrete, is quite different from the one that Paterson constructs [94].

We shall now deal with the missing $\mathscr{J}$-relation. First, we need a
slight extension of Lemma 6.5.

######  Lemma 6.6.

Let $S$ be an inverse semigroup. Then $abc=a^{\prime}\cdot b^{\prime}\cdot
c^{\prime}$, which is a restricted product where $a^{\prime}\leq a$,
$b^{\prime}\leq b$ and $c^{\prime}\leq c$.

###### Proof.

Because of associativity, it doesn’t matter how we bracket. We write
$abc=(ab)c$. Put $d=ab$. Then $dc=d^{\prime}\cdot c^{\prime}$. But
$d^{\prime}=ab\mathbf{r}(c)$. Thus $(a(b\mathbf{r}(c)))\cdot c^{\prime}$. Now
write $a(b\mathbf{r}(c))=a^{\prime}\cdot b^{\prime}$ where $b^{\prime}\leq
b\mathbf{r}(c)\leq b$. We have therefore written $abc=a^{\prime}\cdot
b^{\prime}\cdot c^{\prime}$ using the fact that the multiplication in a
groupoid is associative. ∎

We can now describe the $\mathscr{J}$-relation on inverse semigroups.

######  Lemma 6.7.

Let $S$ be an inverse semigroup. Then $SaS\subseteq SbS$ if and only if
$a\,\mathscr{D}\,b^{\prime}\leq b$ for some element $b^{\prime}\in S$.

###### Proof.

Suppose, first, that $SaS\subseteq SbS$. Then $a=xby$ for some $x,y\in S$. By
Lemma 6.6, we may write $a=x^{\prime}\cdot b^{\prime}\cdot y^{\prime}$. You
can check that $a\,\mathscr{D}\,b^{\prime}$. But $b^{\prime}\leq b$. We have
therefore proved one direction. To prove the converse, suppose that
$a\,\mathscr{D}\,b^{\prime}\leq b$ for some element $b^{\prime}\in S$. Let
$\stackrel{{\scriptstyle
x}}{{\mathbf{d}(a)\longrightarrow\mathbf{d}(b^{\prime})}}$. Put
$y=b^{\prime}\cdot x\cdot a^{-1}$. Then $a=y^{-1}\cdot b^{\prime}\cdot x$.
Similarly, $b^{\prime}=y\cdot a\cdot x^{-1}$ It follows that
$SaS=Sb^{\prime}S$. Clearly, $Sb^{\prime}S\subseteq SbS$. We have therefore
shown that $SaS\subseteq SbS$. ∎

We can use the theory we have developed to generalize Lemma 3.8.

######  Lemma 6.8.

Let $S$ be an inverse semigroup with zero. Then $S$ is isomorphic to a
groupoid with a zero adjoined if and only if the natural partial order is
equality on the set $S\setminus\\{0\\}$.

###### Proof.

We prove one direction only. Let $S$ be an inverse semigroup with zero such
that the natural partial order is equality on the set $S\setminus\\{0\\}$. Let
$G$ be the underlying groupoid of $S$ with the component $\\{0\\}$ removed.
There is an obvious bijection between $S$ and $G^{0}$. We prove that this is a
homomorphism. Let $a,b\in S$ be any non-zero elements. Then
$ab=a^{\prime}\cdot b^{\prime}$ where $a^{\prime}\leq a$ and $b^{\prime}\leq
b$ by Lemma 6.5. As a result of our assumption on the natural partial order,
either $a^{\prime}=a$ and $b^{\prime}=b$, in which case the product is a
groupoid product, or at least one of $a^{\prime}$ and $b^{\prime}$ is equal to
zero — in which case $ab=0$. Thus a product in $S$ is either equal to zero or
a restricted product but not both. ∎

In addition to the underlying groupoid, we may sometimes be able to associate
another, smaller, groupoid to an inverse semigroup with zero. Let $S$ be an
inverse semigroup with zero. An element $s\in S$ is said to be an atom if
$t\leq s$ implies that $t=0$ or $t=s$.

######  Lemma 6.9.

Let $S$ be an inverse semigroup.

  1. (1)

If $x$ is an atom then $x^{-1}$ is an atom.

  2. (2)

If $x$ is an atom then both $\mathbf{d}(x)$ and $\mathbf{r}(x)$ are atoms

  3. (3)

Suppose that $\mathbf{d}(x)$ is an atom then $x$ is an atom; similarly, if
$\mathbf{r}(x)$ is an atom then $x$ is an atom.

  4. (4)

If $x$ and $y$ are atoms and the restricted product $x\cdot y$ is defined then
$x\cdot y$ is an atom.

  5. (5)

If $x$ and $y$ are distinct compatible atoms then $x\perp y$.

###### Proof.

(1) Immediate from the properties of the inverse.

(2) We prove that $\mathbf{d}(x)$ is an atom; the proof that $\mathbf{r}(x)$
is an atom is similar. Suppose that $e\leq\mathbf{d}(x)$. Then $xe\leq x$. It
follows that either $xe=0$ or $xe=x$. Suppose, first, that $xe=0$. Then
$e\mathbf{d}(x)=0$ and so $e=0$. Alternatively, if $xe=x$ then
$e\mathbf{d}(x)=\mathbf{d}(x)$ and so $e=\mathbf{d}(x)$. This shows that
$\mathbf{d}(x)$ is an atom.

(3) We prove that if $\mathbf{d}(x)$ is an atom then $x$ is an atom; the proof
of the other statement is analogous. Suppose that $y\leq x$. Then
$\mathbf{d}(y)\leq\mathbf{d}(x)$. Since $\mathbf{d}(x)$ is an atom either
$\mathbf{d}(y)=0$ or $\mathbf{d}(y)=\mathbf{d}(x)$. It follows that either
$y=0$ or $y=x$. We have therefore proved that $x$ is an atom.

(4) We are given that $x$ and $y$ are atoms and that $x\cdot y$ exists.
Observe first that $x\cdot y\neq 0$. Let $z\leq xy$. Then
$z=x(y\mathbf{d}(z))$. Now, $y\mathbf{d}(z)=0$ or $y\mathbf{d}(z)=y$, since
$y$ is an atom. If the former then $z=0$ and if the latter then $z=xy$. We
have therefore proved that $x\cdot y$ is an atom.

(5) Let $x$ be an atom. Then by part (2) above it follows that $\mathbf{d}(x)$
is an atom. Similarly, $\mathbf{d}(y)$ is an atom. If the product
$\mathbf{d}(x)\mathbf{d}(y)$ is non-zero then in fact
$\mathbf{d}(x)=\mathbf{d}(y)$. But $x\sim y$ and so $x=y$ by Lemma 4.13, which
contradicts our assumption that $x$ and $y$ are distinct. It follows that
$\mathbf{d}(x)\perp\mathbf{d}(y)$. A similar argument shows that
$\mathbf{r}(x)\perp\mathbf{r}(y)$ from which it follows that $x\perp y$. ∎

It follows by Lemma 6.9, that the set of atoms of $S$, if non-empty, forms a
groupoid, which we shall call the atomic groupoid of $S$ and denote by
$\mathsf{A}(S)$.

######  Example 6.10.

The finite symmetric inverse monoid $\mathcal{I}(X)$ has an interesting atomic
groupoid. It consists of those partial bijections the domains of which contain
exactly one element of $X$. This groupoid is isomorphic to the groupoid
$X\times X$ defined in Example 6.2.

We shall describe all finite Boolean inverse monoids in terms of groupoids.
The reader will recall that the finite Boolean algebras are isomorphic to the
powerset Boolean algebras defined on the finite set of atoms. We shall replace
finite sets by finite groupoids. We need some definitions first. If $A$ and
$B$ are subsets of a category $C$ then $AB$ is the set of all products $ab$
where $a\in A$, $b\in B$ and $\mathbf{d}(a)=\mathbf{r}(b)$. If $A$ is a subset
of a groupoid then $A^{-1}$ is the set of all $a^{-1}$ where $a\in A$. We
first show how to construct finite Boolean inverse monoids from finite
groupoids. If $G$ is a groupoid then a subset $A\subseteq G$ is said to be a
local bisection if both $AA^{-1}$ and $A^{-1}A$ consist entirely of
identities. You can check that a subset $A\subseteq G$ is a local bisection if
$a,b\in A$ and $\mathbf{d}(a)=\mathbf{d}(b)$ (respectively,
$\mathbf{r}(a)=\mathbf{r}(b)$) then $a=b$,

######  Proposition 6.11.

Let $G$ be a finite groupoid. Then $\mathsf{K}(G)$, the set of all local
bisections of $G$ under subset multiplication, is a finite Boolean inverse
monoid, the set of atoms of which forms a groupoid isomorphic to $G$.

###### Proof.

You can check that the product of two local bisections is a local bisection.
If $A$ is a local bisection, then so is $A^{-1}=\\{a^{-1}\colon a\in A\\}$ and
$A=AA^{-1}A$. The idempotents are just the subsets of $G_{o}$ and the product
of two idempotents is just the intersection of these two sets. It follows that
$\mathsf{K}(G)$ is an inverse semigroup since it is a regular semigroup with
commuting idempotents. It is a monoid with identity $G_{o}$ and has a zero
$\varnothing$. It has a Boolean algebra of idempotents. Observe that $A\leq B$
if and only if $A\subseteq B$. You can check that $A\sim B$ if and only if
$A\cup B$ is a local bisection. It is now easy to check that $\mathsf{K}(G)$
is a Boolean inverse monoid. The atoms are the singleton sets $\\{g\\}$ and
form a groupoid isomorphic to $G$. ∎

We shall now go in the opposite direction. Our first result is an immediate
consequence of finiteness.

######  Lemma 6.12.

Let $S$ be a finite Boolean inverse monoid. Then each non-zero element is
above an atom.

We now connect elements with the atoms beneath them. Let $a\in S$. Define
$\theta(a)=a^{\downarrow}\cap\mathsf{A}(S)$. Observe that
$\theta(0)=\varnothing$.

######  Lemma 6.13.

Let $S$ be a Boolean inverse semigroup. For each $a\in S$, the set $\theta(a)$
is a local bisection of the groupoid $\mathsf{A}(S)$.

###### Proof.

If $a=0$ then $\theta(a)=\varnothing$. If $a\neq 0$ then it is above at least
one atom by Lemma 6.12 and so is non-empty. Let $x,y\in\theta(a)$ such that
$\mathbf{d}(x)=\mathbf{d}(y)$. Then $x=y$. Dually, if
$\mathbf{r}(x)=\mathbf{r}(y)$ then $x=y$. ∎

If $S$ is a finite Boolean inverse monoid, then by Lemma 6.12 we may define a
function $\theta\colon S\rightarrow\mathsf{K}(\mathsf{A}(S))$ by
$\theta(0)=\varnothing$ and $\theta(a)=a^{\downarrow}\cap\mathsf{A}(S)$. It is
quite rare that we can say anything about the structure of finite semigroups
belonging to some class. Thus the following result is a pleasant surprise.

######  Theorem 6.14 (The structure of finite Boolean inverse monoids).

Let $S$ be a finite Boolean inverse monoid. Then $S$ is isomorphic to the
Boolean inverse monoid $\mathsf{K}(\mathsf{A}(S))$.

###### Proof.

It remains to show that $\theta$ (as defined above) is an isomorphism of
semigroups. First, $\theta$ is a homomorphism. Let $x$ be an atom such that
$x\leq ab$. Then $x=a(b\mathbf{d}(x))$. Thus by Lemma 6.5, we may write
$x=a^{\prime}\cdot b^{\prime}$ where $a^{\prime}\leq a$ and $b^{\prime}\leq
b$. It is easy to check that $a^{\prime}$ and $b^{\prime}$ are themselves
atoms. We have therefore proved that $\theta(ab)\subseteq\theta(a)\theta(b)$.
Conversely, let $x\in\theta(a)$ and $y\in\theta(b)$ such that the restricted
product $x\cdot y$ is defined. Then $x\cdot y=xy\leq ab$. But the restricted
product of atoms is an atom and so we have proved the first claim.

It remains to prove that $\theta$ is a bijection. We show first that
$a=\bigvee\theta(a)$. Put $b=\bigvee\theta(a)$. Then, clearly, $b\leq a$.
Suppose that $b\neq a$. It is here that we use the Boolean structure. Then
$a\setminus b\neq 0$. It follows by Lemma 6.12 that $a\setminus b$ is above an
atom $x$. But then $x\leq a$ and so $x\leq b$ (by definition). Thus $x\leq
b,a\setminus b$ which implies that $x=0$. But atoms are non-zero. It follows
that $a=\bigvee\theta(a)$ and so $\theta$ is an injection.

Now let $A\in\mathsf{K}(\mathsf{A}(S))$. We don’t lose any generality by
assuming that it is non-empty. Then $A$ is a set of compatible elements. Put
$a=\bigvee A$. Clearly, $A\subseteq\theta(a)$. Let $x$ be an atom such that
$x\leq a$. Then $x=\bigvee_{a\in A}(x\wedge a)$ by part (2) of Lemma 5.3.
Remembering that both $x$ and $a$ are atoms, it follows that $x=a$ for some
$a\in A$. We have therefore proved that $\theta(a)=A$. ∎

We now have a complete description of the finite Boolean inverse monoids: as a
result of Proposition 6.11 and Theorem 6.14, they are precisely the inverse
monoids of the form $\mathsf{K}(G)$ where $G$ is a finite groupoid. We can
apply the theory we have developed to the representation theory of arbitrary
finite inverse monoids, although a more elementary account can be found in
[111, Chapter 9]. We prove first that every finite inverse monoid $S$ can be
embedded into the Boolean inverse monoid constructed from its underlying
groupoid.

######  Lemma 6.15.

Let $S$ be a finite inverse monoid with underlying groupoid $G$.

  1. (1)

Let $a\in S$. Then $a^{\downarrow}$ is a local bisection of $G$.

  2. (2)

We have that $(ab)^{\downarrow}=a^{\downarrow}b^{\downarrow}$.

###### Proof.

(1) Let $x,y\leq a$ such that $\mathbf{d}(x)=\mathbf{d}(y)$. Then it is
immediate that $x=y$. Similar reasoning shows that if $x,y\leq a$ such that
$\mathbf{r}(x)=\mathbf{r}(y)$ then $x=y$. We have therefore shown that the set
$a^{\downarrow}$ is a local bisection of $G$.

(2) Suppose that $x\leq a$ and $y\leq b$. Then $xy\leq ab$. On the other hand,
if $c\leq ab$ then we can write $c=(\mathbf{r}(c)a)(b\mathbf{d}(c))$ and so
$x\in a^{\downarrow}b^{\downarrow}$. ∎

Let $S$ be an arbitrary inverse monoid with underlying groupoid $G$. An
element $a$ with the property that $a^{\downarrow}=\\{a\\}$, a singleton set,
will be said to be at the bottom. Define $\beta\colon
S\rightarrow\mathsf{K}(G)$ by $\beta(a)=a^{\downarrow}$. Then by Lemma 6.15,
$\beta$ is an injective homomorphism of inverse semigroups. The elements in
$1^{\downarrow}$ are precisely the idempotents, which are the identities of
$G$; the set of identities of $G$ is the monoid identity of $\mathsf{K}(G)$.
Thus the homomorphism is a monoid homomorphism. The above result, on its own,
doesn’t take us very far because we already know that every finite inverse
monoid can be embedded in a finite Boolean inverse monoid. The key point of
this embedding is that the map $\beta\colon S\rightarrow\mathsf{K}(G)$ has a
special property described by the following proposition.

######  Proposition 6.16.

Let $S$ be a finite inverse monoid with underlying groupoid $G$ and let
$\alpha\colon S\rightarrow T$ be any monoid homomorphism to a Boolean inverse
monoid $T$. Then there is a unique morphism of Boolean inverse monoids
$\gamma\colon\mathsf{K}(G)\rightarrow T$ such that $\gamma\beta=\alpha$.

###### Proof.

Our first step is to define the function $\gamma\colon\mathsf{K}(G)\rightarrow
T$. Define $\gamma(\varnothing)=0$. If $a$ is any element of $S$ then
$\\{a\\}$ is an element of $\mathsf{K}(G)$. Let the set of elements strictly
below $a$ in $S$ be $\\{a_{1},\ldots,a_{m}\\}$. This set could well be empty
if $a$ is at the bottom — this will not cause us any problems. Since
$a_{1},\ldots,a_{m}\leq a$, these elements are pairwise compatible. It follows
that $\alpha(a_{1}),\ldots,\alpha(a_{m})$ is a compatible subset of $T$. Thus
the join $\alpha(a_{1})\vee\ldots\vee\alpha(a_{m})$ exists; if the set of
elements strictly below $a$ is empty then this join is just $0$. It is clearly
less than $\alpha(a)$ and so we may form the element
$\alpha(a)\setminus(\alpha(a_{1})\vee\ldots\vee\alpha(a_{m}))$. On the basis
of the above, define

$\gamma(\\{a\\})=\alpha(a)\setminus(\alpha(a_{1})\vee\ldots\vee\alpha(a_{m})).$

If $A$ is a local bisection which is neither empty nor a singleton set, define

$\gamma(A)=\bigvee_{a\in A}\gamma(\\{a\\}).$

This makes sense, for if $a,b\in A$, then $\\{a\\},\\{b\\}\subseteq A$ and so
$\\{a\\}\sim\\{b\\}$ in $\mathsf{K}(G)$. This completes the definition of
$\gamma$.

We show that $\gamma$ preserves binary joins. If $A\sim B$ in $\mathsf{K}(G)$
then $A\vee B=A\cup B$. From the definition of $\gamma$, it is clear that
$\gamma$ preserves binary joins. It maps the empty set to the zero by
definition.

We prove that $\gamma(a^{\downarrow})=\alpha(a)$. This proves that we have a
monoid homomorphism and that $\gamma\beta=\alpha$. Define the height of $a$ in
$S$ to be the length of a chain of maximum length from $a$. Those elements
with height zero are precisely those which are at the bottom. If $a$ has
height zero then

$\gamma(a^{\downarrow})=\alpha(a).$

We assume that we have proved that $\gamma(b^{\downarrow})=\alpha(b)$ for all
elements $b$ of height at most $n$. Let $a$ be an element of height $n+1$. Let
$b_{1},\ldots,b_{m}$ be all the elements immediately below $a$. Then, by the
induction hypothesis, we have that

$\gamma(b_{i}^{\downarrow})=\alpha(b_{i}).$

Let the elements strictly less than $a$ be $a_{1},\ldots,a_{m}$. These include
the elements $b_{1},\ldots,b_{m}$, for example, so in general each element
$a_{i}$ is beneath one of the $b_{j}$. It follows that we can write

$\alpha(a)\setminus(\alpha(a_{1})\vee\ldots\vee\alpha(a_{m}))$

as

$\alpha(a)\setminus(\alpha(b_{1}e_{1})\vee\ldots\vee\alpha(b_{n}e_{n}))$

where the idempotents $\alpha(e_{j})$ gather together by means of a join all
the idempotents that arise from showing that $a_{i}\leq b_{j}$ for various
$i$. By definition

$\gamma(a^{\downarrow})=\alpha(a)\setminus(\alpha(a_{1})\vee\ldots\vee\alpha(a_{m}))\vee\gamma(\\{a_{1}\\})\vee\ldots\vee\gamma(\\{a_{m}\\})$

but we can write

$\gamma(a^{\downarrow})=\gamma(\\{a\\})\vee\gamma(\\{a_{1},\ldots,a_{m}\\}^{\downarrow}).$

But

$\\{a_{1},\ldots,a_{m}\\}^{\downarrow}=b_{1}^{\downarrow}\cup\ldots\cup
b_{n}^{\downarrow}$

and so

$\gamma(\\{a_{1},\ldots,a_{m}\\}^{\downarrow})=\gamma(b_{1}^{\downarrow})\vee\ldots\vee\gamma(b_{n}^{\downarrow}).$

Thus

$\gamma(\\{a_{1},\ldots,a_{m}\\}^{\downarrow})=\alpha(b_{1})\vee\ldots\vee\alpha(b_{n})$

using the induction hypothesis. Thus

$\gamma(a^{\downarrow})=\alpha(a)\setminus(\alpha(a_{1})\vee\ldots\vee\alpha(a_{m}))\vee\alpha(b_{1})\vee\ldots\vee\alpha(b_{n}.)$

By our argument above

$\gamma(a^{\downarrow})=\alpha(a)\setminus(\alpha(b_{1}e_{1})\vee\ldots\vee\alpha(b_{n}e_{n}))\vee\alpha(b_{1})\vee\ldots\vee\alpha(b_{n}).$

This is equal to

$\alpha(a)\setminus(\alpha(b_{1}e_{1})\vee\ldots\vee\alpha(b_{n}e_{n}))\vee(\alpha(b_{1}e_{1})\vee\ldots\vee\alpha(b_{n}e_{n})))\vee\alpha(b_{1})\vee\ldots\vee\alpha(b_{n})$

which is just

$\alpha(a)\vee\alpha(b_{1})\vee\ldots\vee\alpha(b_{n})$

which is equal to $\alpha(a)$.

We now prove uniqueness. Suppose that
$\gamma^{\prime}\colon\mathsf{K}(S)\rightarrow T$ is a morphism of Boolean
inverse monoids such that $\gamma^{\prime}\beta=\alpha$. We show that
$\gamma^{\prime}=\gamma$. It is immediate from our assumption on
$\gamma^{\prime}$ that for all elements of height zero we have that
$\gamma(a)=\gamma^{\prime}(a)$. So, let $a\in S$ be any element which does not
have height zero. Let $\\{a_{1},\ldots,a_{m}\\}$ be the set of all elements
strictly less than $a$. Then

$\\{a_{1},\ldots,a_{m}\\}=a_{1}^{\downarrow}\cup\ldots\cup
a_{m}^{\downarrow}.$

It follows that

$\gamma^{\prime}(\\{a_{1},\ldots,a_{m}\\})=\gamma^{\prime}(a_{1}^{\downarrow})\vee\ldots\vee\gamma^{\prime}(a_{m}^{\downarrow}).$

But this is just equal to

$\gamma^{\prime}(\beta(a_{1}))\vee\ldots\vee\gamma^{\prime}(\beta(a_{m}))$

which is equal to

$\gamma(\beta(a_{1}))\cup\ldots\cup\gamma(\beta(a_{m}))$

by assumption. We have therefore proved that

$\gamma^{\prime}(\\{a_{1},\ldots,a_{m}\\})=\gamma(\\{a_{1},\ldots,a_{m}\\}).$

Observe that $\\{a\\}=a^{\downarrow}\setminus\\{a_{1},\ldots,a_{m}\\}$. Thus

$\gamma^{\prime}(\\{a\\})=\gamma^{\prime}(a^{\downarrow})\setminus\gamma^{\prime}(\\{a_{1},\ldots,a_{m}\\}).$

But

$\gamma^{\prime}(a^{\downarrow})=\gamma(a^{\downarrow})$

by assumption, and

$\gamma^{\prime}(\\{a_{1},\ldots,a_{m}\\})=\gamma(\\{a_{1},\ldots,a_{m}\\})$

by what we proved above. It follows that
$\gamma^{\prime}(\\{a\\})=\gamma(\\{a\\})$ for all elements $a\in S$. Now, let
$A$ be any non-empty local bisection. Then, since $\gamma^{\prime}$ is a
morphism of Boolean inverse monoids, we have that
$\gamma^{\prime}(A)=\bigvee_{a\in A}\gamma^{\prime}(\\{a\\})$. It follows that
$\gamma^{\prime}=\gamma$. ∎

We now apply our results to the study of the representation theory of finite
inverse monoids. The starting point is to say what we mean by the
representation theory of groupoids. Let $G$ be a finite groupoid. Then we get
a semigroup with zero $G^{0}$ by adjoining a zero; in fact, this is a special
kind of inverse semigroup by Lemma 6.8. We can therefore consider
homomorphisms $\theta\colon G^{0}\rightarrow R$ to the multiplicative monoid
of the unital ring $R$. Each identity $e\in G$ gives rise to an idempotent
$\theta(e)$ in the ring $R$. If $e$ and $f$ are distinct identities of the
groupoid $G$ then $\theta(e)$ and $\theta(f)$ are orthogonal. By assumption,
the groupoid $G$ is finite. Put $f=\sum_{e\in G_{o}}\theta(e)$. Then $f$ is an
idempotent in $R$. Consider the ring $fRf=\\{a\in R\colon faf=a\\}$. This has
identity $f$. Let $g\in G$ be an arbitary element of $G$. Let $e,e^{\prime}$
be the identities such that $a=e^{\prime}ae$. Then
$\theta(a)=\theta(e^{\prime})\theta(a)\theta(e)$. But for any identity $e$, we
have that $\theta(e)=f\theta(e)f$. It follows that $\theta(a)\in fRf$. We
therefore define a representation of a finite groupoid $G$ in a ring $R$ to be
a semigroup with zero homomorphism $\theta\colon G^{0}\rightarrow R$ such that
$1=\sum_{e\in G_{o}}\theta(e)$.

We shall also need the following definition. Let $S$ be a Boolean inverse
monoid. A monoid homorphism $\phi\colon S\rightarrow R$ to the multiplicative
monoid of the ring $R$ which maps zero to zero is said to be additive if
$\phi(a\vee b)=\phi(a)+\phi(b)$, whenever $a\perp b$. We refer the reader to
part (2) of Lemma 5.5 for the rationale for this definition.

The following theorem was proved in a different way in [111, Theorem 9.3].

######  Theorem 6.17 (Representation theory of finite inverse monoids).

Let $S$ be a finite inverse monoid with underlying groupoid $G$ and let $R$ be
a unital ring. Then there is a bijective correspondence between the set of
representations of $S$ in $R$ and the set of representations of the finite
groupoid $G$ in $R$.

###### Proof.

Let $\theta\colon S\rightarrow R$ be a monoid homomorphism to the
multiplicative monoid of the ring $R$. The image $\theta(S)$ is an inverse
submonoid of the multiplicative monoid of the ring $R$ by Lemma 3.22 and so
$\theta(S)^{0}$, the inverse monoid $\theta(S)$ with the zero of the ring $R$
adjoined, is an inverse monoid with zero. Thus by Theorem 5.10, there is a
Boolean inverse monoid $T$ such that $\theta(S)^{0}\subseteq T\subseteq R$. By
Proposition 6.16, there is therefore a unique morphism of Boolean inverse
monoids $\phi\colon\mathsf{K}(G)\rightarrow T$ such that $\phi\beta=\theta$.
We want to regard $\phi$ as a map from $\mathsf{K}(G)$ to the ring $R$. This
is an additive homomorphism. We have proved that every homomorphism
$\theta\colon S\rightarrow R$ gives rise to an additive homomorphism
$\phi\colon\mathsf{K}(G)\rightarrow R$. On the other hand, given an additive
homorphism $\phi\colon\mathsf{K}(G)\rightarrow R$, we can construct a monoid
homomorphism $\phi\beta\colon S\rightarrow R$. This leads to a bijective
correspondence between representations of $S$ in $R$ and additive
homomorphisms of $\mathsf{K}(G)$ in $R$.

We described all finite Boolean inverse monoids in Theorem 6.14. In what
follows, therefore, we may assume that $S$ is a finite Boolean inverse monoid
with atomic groupoid $G$. Let $\theta\colon S\rightarrow R$ be an additive
homomorphism of $S$. Then, by restriction, we get a semigroup homomorphism
$\theta^{\prime}$ from $G^{0}$ to the ring $R$. Since $S$ is a finite Boolean
inverse monoid, the identity of $S$ is an orthogonal join of the atomic
idempotents. It follows that the sum of the idempotents $\theta^{\prime}(e)$,
where $e\in G_{o}$, is equal to the identity of $R$. Thus, we have defined a
representation of the groupoid $G$. We now go in the opposite direction. Let
$S$ be a finite Boolean inverse monoid with atomic groupoid $G$ and suppose
that there a representation $\theta^{\prime}\colon G^{0}\rightarrow R$. We
shall now define a homomorphism $\theta$ of $S$ that extends
$\theta^{\prime}$. Let $a\in S$. Define $\theta\colon S\rightarrow R$ by
$\theta(a)=\theta^{\prime}(a_{1})+\ldots+\theta^{\prime}(a_{m})$ where
$a_{1},\ldots,a_{m}\leq a$ are all the atoms below $a$; we can assume this is
an orthogonal set by part (5) of Lemma 6.9 This is an additive homomorphism of
$S$. We therefore have a bijective correspondence between the additive
homorphisms of a Boolean inverse monoid and the representations of its atomic
groupoid.

If we put our two results together, then we have established a bijective
correspondence between representations of an inverse monoid $S$ and the
representations of its underlying groupoid $G$. ∎

##  7\. Fundamental inverse semigroups

In Section 3, we introduced the Clifford semigroups. These are the inverse
semigroups in which every element is central. Living inside every inverse
semigroup is a Clifford semigroup. For every inverse semigroup $S$, define
$\mathsf{Z}(\mathsf{E}(S))$, the centralizer of the idempotents, to be set of
all elements of $S$ which commute with every idempotent. Then
$\mathsf{Z}(\mathsf{E}(S))$ is a wide inverse subsemigroup of $S$ which is
Clifford. If $\mathsf{Z}(\mathsf{E}(S))=\mathsf{E}(S)$ we say the inverse
semigroup is fundamental. Fundamental inverse semigroups are important. In
this section, we shall study them in more detail.

To do this, we shall need the following. Define the relation $\mu$ on an
arbitrary inverse semigroup by

$(s,t)\in\mu\Leftrightarrow(\forall e\in\mathsf{E}(S))(ses^{-1}=tet^{-1}).$

It is routine to check that this is a congruence.

######  Lemma 7.1.

Let $S$ be an inverse semigroup.

  1. (1)

If $(s,t)\in\mu$ then $\mathbf{r}(s)=\mathbf{r}(t)$ and
$\mathbf{d}(s)=\mathbf{d}(t)$. It follows that $\mu\subseteq\mathscr{H}$.

  2. (2)

In the definition, we may restrict to those idempotents in
$\mathbf{d}(s)^{\downarrow}$

  3. (3)

If $(e,f)\in\mu$, where $e$ and $f$ are idempotents, then $e=f$.

###### Proof.

(1) If $(s,t)\in\mu$ then $\mathbf{r}(s)\leq\mathbf{r}(t)$. Symmetry now
delivers the answer. To prove the second claim, observe that if $(s,t)\in\mu$
then $(s^{-1},t^{-1})\in\mu$. We can now use the first claim to prove the
second claim. It follows that $\mu\subseteq\mathscr{H}$.

(2) We have used all idempotents in the definition of $\mu$ but we may
restrict to those idempotents in $\mathbf{d}(s)^{\downarrow}$ simply by
mulitplying by $\mathbf{d}(s)$.

(3) Immediate. ∎

By part (3) of Lemma 7.1, it follows that $\mu$ is an idempotent-separating
congruence. In fact, we have the following.

######  Lemma 7.2.

$\mu$ is the largest idempotent-separating congruence on $S$.

###### Proof.

Let $\rho$ be any idempotent-separating congruence on $S$ and let
$(s,t)\in\rho$. Let $e$ be any idempotent. Then $(ses^{-1},tet^{-1})\in\rho$
but $\rho$ is idempotent-separating and so $ses^{-1}=tet^{-1}$. It follows
that $(s,t)\in\mu$. Thus we have shown that $\rho\subseteq\mu$. ∎

We can now explain the connection between the congruence $\mu$ and fundamental
inverse semigroups.

######  Lemma 7.3.

Let $S$ be an inverse semigroup. Then $S$ is fundamental if and only if $\mu$
is the equality relation

###### Proof.

Suppose that $S$ is fundamental. We prove that $\mu$ is the equality relation.
Let $(s,t)\in\mu$. Then $(st^{-1},tt^{-1})\in\mu$. Let $e$ be any idempotent.
Then $(st^{-1})e(st^{-1})^{-1}=tt^{-1}ett^{-1}$. It follows that

$st^{-1}e=e\mathbf{r}(t)st^{-1}=e\mathbf{r}(s)st^{-1}=est^{-1}.$

We have therefore proved that $st^{-1}$ is central. By assumption it must be
an idempotent. It follows that $st^{-1}=tt^{-1}$. Thus $st^{-1}t=t$ but
$\mathbf{d}(t)=\mathbf{d}(s)$ and so $s=t$. To prove the connverse, suppose
that $\mu$ is the equality relation. Let $s$ commute with every idempotent.
Then $(s,ss^{-1})\in\mu$. Thus, by assumption, $s=ss^{-1}$ and so $s$ is an
idempotent. ∎

We can easily construct fundamental inverse semigroups.

######  Lemma 7.4.

Let $S$ be an inverse semigroup. Then $S/\mu$ is fundamental.

###### Proof.

Suppose that $\mu(s)$ and $\mu(t)$ are $\mu$-related in $S/\mu$. Every
idempotent in $S/\mu$ is of the form $\mu(e)$ where $e\in E(S)$. Thus

$\mu(s)\mu(e)\mu(s)^{-1}=\mu(t)\mu(e)\mu(t)^{-1}$

so that $\mu(ses^{-1})=\mu(tet^{-1})$. But both $ses^{-1}$ and $tet^{-1}$ are
idempotents, so that $ses^{-1}=tet^{-1}$ for every $e\in E(S)$. Thus
$(s,t)\in\mu$. ∎

The symmetric inverse monoid is constructed from an arbitrary set. We now show
how to construct an inverse semigroup from a meet semilattice. Let $(E,\leq)$
be a meet semilattice, and denote by $T_{E}$ be the set of all order
isomorphisms between the principal order ideals of $E$. Clearly, $T_{E}$ is a
subset of $\mathcal{I}(E)$. In fact we have the following.

######  Lemma 7.5.