Inverse semigroup

Remarks

Equivalently, $S$ is regular and any two idempotents of $S$ commute. In an inverse semigroup, $(a^{-1})^{-1} = a$ and $(ab)^{-1} = b^{-1} a^{-1}$.

Example: the symmetric inverse monoid $I_X$ on a set $X$, the set of all bijections $f \colon A \to B$ between subsets $A, B \subseteq X$. The product $fg$ is the composite $f \circ g$, defined on the points $x$ with $x$ in the domain of $g$ and $g(x)$ in the domain of $f$. For $f \colon A \to B$, the inverse bijection $f^{-1} \colon B \to A$ satisfies $f f^{-1} f = f$ and $f^{-1} f f^{-1} = f^{-1}$, and it is the only element that does. Every group is also an inverse semigroup.

Non-example: the left zero semigroup $\set{e, f}$ with $xy = x$ for all $x, y$. Here $exe = e$ and $xex = x$ for both $x = e$ and $x = f$, so $e$ has two such partners and uniqueness fails. Accordingly, the idempotents $e$ and $f$ do not commute: $ef = e$ but $fe = f$.