This entry is about semigroups with a unary operator $i$ such that $s \cdot i(s) \cdot s = s$ and $i(s) \cdot s \cdot i(s) = i(s)$. For the semigroup with a two-sided inverse, see invertible semigroup
An inverse semigroup is a semigroup $S$ (a set with an associative binary operation) such that for every element $s\in S$, there exists a unique “inverse” $s^*\in S$ such that $s s^* s = s$ and $s^* s s^* = s^*$. It is evident from this that $s^{\ast\ast} = s$.
Needless to say, a group is an inverse semigroup. More to the point however:
This inverse semigroup plays a role in the theory similar to that of permutation groups in the theory of groups. It is also paradigmatic of the general philosophy that
Groups describe global symmetries, while inverse semigroups describe local symmetries.
Other examples include:
If $X$ is a topological space, let $\Gamma(X)\subseteq I(X)$ consist of the homeomorphisms between open subsets of $X$. Then $\Gamma(X)$ is a pseudogroup of transformations on $X$ (a general pseudogroup of transformations is a sub-inverse-semigroup of $\Gamma(X)$).
If $L$ is a meet-semilattice, then $L$ is an inverse semigroup under the meet operation.
Lots to say here: the meet-semilattice of idempotents, the connection with ordered groupoids, various representation theorems.
For any $x$ in an inverse semigroup, $x x^\ast$ and $x^\ast x$ are idempotent. If $e$ is idempotent, then $e^\ast = e$.
The proof is trivial.
In an inverse semigroup, the product of any two idempotents $e, f$ is idempotent, and any two idempotents commute.
One easily checks that $(e f)^\ast = f(e f)^\ast e$, and that $f(e f)^\ast e$ is an idempotent. So $(e f)^\ast$ is idempotent; as a result, $(e f)^\ast = e f$. Thus $e f$ and similarly $f e$ are idempotent. Next we have
since $e, f, e f, f e$ are all idempotent, and so $f e = (e f)^\ast = e f$, which completes the proof.
Thus the idempotents in an inverse semigroup form a subsemigroup which is commutative and idempotent. Such a structure is the same as a meet-semilattice except for the fact that there might not have an empty meet or top element; that is, we define an order $\leq$ on idempotents by $e \leq f$ if and only if $e = e f$, whence multiplication of idempotents becomes the binary meet.
For any two elements $x, y$ in an inverse semigroup, $(x y)^\ast = y^\ast x^\ast$.
Since the idempotents $x^\ast x, y y^\ast$ commute, we have
and similarly $y^\ast x^\ast (x y)y^\ast x^\ast = y^\ast y y^\ast x^\ast x x^\ast = y^\ast x^\ast$, which is all we need.
For elements $x, y$ in an inverse semigroup, the following are equivalent:
We show $3. \Rightarrow 2.$; a similar proof shows $1. \Rightarrow 4.$ Clearly then we have $3. \Rightarrow 2. \Rightarrow 1. \Rightarrow 4. \Rightarrow 3.$
Given an idempotent $f$ such that $x = y f$, we have
which gives $3. \Rightarrow 2.$
A preorder $\leq$ is defined on an inverse semigroup by saying $x \leq y$ if any of the four conditions of Proposition is satisfied; transitivity follows by equivalence to 1. and closure of idempotents under multiplication. When restricted to idempotents, this preorder coincides with the meet-semilattice order.
If $a \leq b$ and $x \leq y$ in an inverse semigroup, then $a x \leq b y$ and $x^\ast \leq y^\ast$.
Writing $a = e b$ for some idempotent $e$, we have $a x = e (b x)$ and so $a x \leq b x$. Similarly $b x \leq b y$, so $a x \leq b y$ by transitivity. This gives $a x \leq b y$. If $x = e y$ for an idempotent $e$, then $x^\ast = (e y)^\ast = y^\ast e^\ast$; this gives $x^\ast \leq y^\ast$,
The preorder $\leq$ on an inverse semigroup is a partial order, i.e., if $x \leq y$ and $y \leq x$, then $x = y$.
From $x \leq y$ we derive $x^\ast \leq y^\ast$ and $x x^\ast \leq y y^\ast$, and similarly from $y \leq x$ we derive $y y^\ast \leq x x^\ast$. Thus $x x^\ast = y y^\ast$ since the preorder on idempotents is a meet-semilattice, which is a partial order. Then from $x \leq y$ we derive $x = x x^\ast y = y y^\ast y = y$.
Thus an inverse semigroup is naturally regarded as an internal semigroup in the category of posets (equivalently, a finite-product preserving functor from the Lawvere theory of semigroups to Pos).
In this section, an ordered groupoid means an internal groupoid in the finitely complete category of posets Pos. For any finitely complete category $C$, we observe that the forgetful functor $Gpd(C) \to SemiCat(C)$, taking an internal groupoid in $C$ to the underlying semicategory (remembering only composition of morphisms, forgetting presence of inverses and identity morphisms), has a right adjoint which takes a semicategory to the core groupoid of the category of idempotents attached to a semicategory (see here for details). (This observation is formulated in finite limit logic, and thus by a Yoneda lemma argument, its validity reduces to that of the observation in the special case $C = Set$.)
In particular, this construction may be applied to an inverse semigroup seen as a semigroup in $Pos$:
The groupoid $Ind(S)$ attached to an inverse semigroup $S$ is the core of the category of idempotents $Idem(S)$ of $S$, which as a semigroup in $Pos$ is viewed as a one-object semicategory $B S$ in $Pos$.
In more detail: an arrow $e \to e'$ in $Idem(S)$ is a triple $(e, x, e')$ of elements in $S$, where $e, e'$ are idempotent elements and $x$ is an element such that $x e = x = e' x$. Such an arrow is invertible precisely when $e = x^\ast x$ and $e' = x x^\ast$, with inverse $x^\ast$. Thus the core consists of such arrows $x: x^\ast x \to x x^\ast$.
A key example to keep in mind is the inverse semigroup of partial bijections $\phi$ on a set, where the arrows of the corresponding groupoid are actual invertible maps $\dom(\phi) \to range(\phi)$ between subsets. In general, the object part of the associated groupoid is not just a poset, but a poset with binary meets.
The reason for the notation $Ind(S)$ is that this ordered groupoid is a so-called inductive groupoid, defined as follows:
An inductive groupoid is an internal groupoid $G$ in $Pos$ with the following additional properties:
The object part $G_0$ admits binary meets;
Given $x: e \to f$ in $G_1$ and $e' \leq e$ in $G_0$, there exists a unique $x': e' \to f$ in $G_1$ with $x' \leq x$, called the restriction $[x|_\ast e']$.
Given $x: e \to f$ in $G_1$ and $f \geq f'$ in $G_0$, there exists a unique $x': e \to f'$ in $G_1$ with $x \geq x'$, called the corestriction $[f' _\ast| x]$.
In fact conditions 2. and 3. in this definition are equivalent. A morphism of inductive groupoids $G \to G'$ is an internal functor from $G$ to $G'$ in $Pos$.
For $G$ an inductive groupoid, a tensor product $\otimes: G_1 \times G_1 \to G_1$ may be defined by the rule
where $\cdot$ indicates composition in $G$. It may be shown that $(G_1, \otimes)$ is an inverse semigroup $Inv(G)$, and the two notions are equivalent:
(Ehresmann-Schein-Nampooripad) There are canonical isomorphisms $S \to Inv(Ind(S))$ and $G \to Ind(Inv(G))$, providing an equivalent of categories $InvSemiGrp \simeq IndGpd$.
With only a subtle change in definition, the result is that one gets only groups:
Let $S$ be an inhabited semigroup with the property that for every $a \in S$ there exists a unique $x \in S$ such that $a x a = a$. Then $S$ is a group.
Since $S$ is inhabited, say by an element $b$, it has an idempotent $e$, for example $b b^\ast$. We will show that $x e = x$ for any $x$; by a similar argument $e x = x$, so that any idempotent $e$ is an identity (the identity $1$), whence the idempotents $a a^\ast$ and $a^\ast a$ equal $1$ for any $a$ and $S$ is a group.
If $a y a = a$ for unique $y$, then from $(a y a) y a = a y a = a$ it follows $y a y = y$ and hence $S$ is an inverse semigroup. The same observation means it is enough to show $(x e) x^\ast (x e) = x e$, since then also $x^\ast (x e) x^\ast = x^\ast$, which by uniqueness implies $x e = x$.
The above results on inverse semigroups apply and we derive
as was to be shown.
Every inverse semigroup $S$ can be realized as a semigroup of partial bijections on a set.
First use the conventional Cayley’s theorem to embed $S$ in $\text{End}(S)$ through the map sending $s\in S$ to the map $x\mapsto sx$. We can now consider $S$ a subset of $\text{End}(S)$.
For each map $s\in S$, $s^\ast s s^\ast=s^\ast$, so $s|_{s^\ast(S)}:s^\ast(S)\rightarrow s(S)$ has left inverse $s^\ast|_{s(S)}:s(S)\rightarrow s^\ast(S)$. Similarly, as $s s^\ast s=s$, this map is a right inverse, so $s|_{s^\ast(S)}$ is a bijection.
Consider the map $S\rightarrow I(S)$ defined by $s\mapsto (s|_{s^\ast(S)}: s^\ast(S)\rightarrow s(S))$. We now check that this map is a homomorphism. For $s,t\in S$, we want to compose the partial bijections $s|_{s^\ast(S)}$ and $t|_{t^\ast(S)}$. To do this, we first compute the overlap between the codomain of the former and the domain of the latter, which is $s(S)\cap t^\ast (S)$. The domain of $t|_{t^\ast(S)}\circ s|_{s^\ast(S)}$ is then $s^\ast (s(S)\cap t^\ast (S))=s^\ast t^\ast(S)=(ts)^\ast (S)$ as required. Finally, we check injectivity. If $s,t\in S$ and $s_{s^\ast(S)}=t_{t^\ast(S)}$, then $s^\ast(S)=t^\ast(S)$ so $ss^\ast=ts^\ast$. But then $s^\ast s s^\ast = s^\ast t s^\ast$, so $s=t$.
cohomology of inverse semi-groups?
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