symmetric inverse semigroup

Let $X$ be a set. A partial map on $X$ is an application defined from a subset of $X$ into $X$. We denote by $\mathfrak{F}(X)$ the set of partial map on $X$. Given $\alpha\in\mathfrak{F}(X)$, we denote by $\mathrm{dom}(\alpha)$ and $\mathrm{ran}(\alpha)$ respectively the domain and the range of $\alpha$, i.e.

 $\mathrm{dom}(\alpha),\mathrm{ran}{\alpha}\subseteq X,\ \ \alpha:\mathrm{dom}(% \alpha)\rightarrow X,\ \ \alpha(\mathrm{dom}(\alpha))=\mathrm{ran}(\alpha).$

We define the composition of two partial map $\alpha,\beta\in\mathfrak{F}(X)$ as the partial map $\alpha\circ\beta\in\mathfrak{F}(X)$ with domain

 $\mathrm{dom}(\alpha\circ\beta)=\beta^{-1}(\mathrm{ran}(\beta)\cap\mathrm{dom}(% \alpha))=\left\{x\in\mathrm{dom}(\beta)\,|\,\alpha(x)\in\mathrm{dom}(\beta)\right\}$

defined by the common rule

 $\alpha\circ\beta(x)=\alpha(\beta(x)),\ \ \forall x\in\mathrm{dom}{(\alpha\circ% \beta)}.$

It is easily verified that the $\mathfrak{F}(X)$ with the composition $\circ$ is a semigroup.

A partial map $\alpha\in\mathfrak{F}(X)$ is said bijective when it is bijective as a map $\alpha:\mathrm{ran}(\alpha)\rightarrow\mathrm{dom}(\alpha)$. It can be proved that the subset $\mathfrak{I}(X)\subseteq\mathfrak{F}(X)$ of the partial bijective maps on $X$ is an inverse semigroup (with the composition $\circ$), that is called symmetric inverse semigroup on $X$. Note that the symmetric group on $X$ is a subgroup of $\mathfrak{I}(X)$.

Title symmetric inverse semigroup SymmetricInverseSemigroup 2013-03-22 16:11:14 2013-03-22 16:11:14 Mazzu (14365) Mazzu (14365) 6 Mazzu (14365) Definition msc 20M18 partial map composition of partial maps symmetric inverse semigroup