Let $X$ be a set. A partial map on $X$ is an application defined from a subset of $X$ into $X$ . We denote by $\FFF(X)$ the set of partial map on $X$ . Given $\alpha\in\FFF(X)$ , we denote by $\domi(\alpha)$ and $\rang(\alpha)$ respectively the domain and the range of $\alpha$ , i.e. $$\domi(\alpha),\rang{\alpha}\subseteq X,\ \ \alpha:\domi(\alpha)\rightarrow X,\ \ \alpha(\domi(\alpha))=\rang(\alpha).$$ We define the composition of two partial map $\alpha,\beta\in\FFF(X)$ as the partial map $\alpha\circ\beta\in\FFF(X)$ with domain $$\domi(\alpha\circ\beta)=\beta^{-1}(\rang(\beta)\cap\domi(\alpha))=\gbra{x\in\domi(\beta)\,|\,\alpha(x)\in\domi(\beta)}$$ defined by the common rule $$\alpha\circ\beta(x)=\alpha(\beta(x)),\ \ \forall x\in\domi{(\alpha\circ\beta)}.$$ It is easily verified that the $\FFF(X)$ with the composition $\circ$ is a semigroup.

A partial map $\alpha\in\FFF(X)$ is said bijective when it is bijective as a map $\alpha:\rang(\alpha)\rightarrow\domi(\alpha)$ . It can be proved that the subset $\III(X)\subseteq\FFF(X)$ 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 $\III(X)$ .