# Showing $\mathscr{P}_n=\langle\zeta, \tau, \pi, \xi\rangle$.

This is Exercise 1.9.13 of Howie's "Fundamentals of Semigroup Theory".

## The Details:

Definition: The partial transformation semigroup $\mathscr{P}_n$ is the set of all partial maps from $\{1, 2, \dots , n\}$ to itself, together with composition of partial transformations.

Let $\xi$ be the partial map $$\begin{pmatrix} 2 & 3 & \dots & n \\ 2 & 3 & \dots & n \end{pmatrix}$$ and let $\zeta$, $\tau$, $\pi$ be as defined in this question.

Definition: For each subset $Y$ of $X=\{1, 2, \dots , n\}$, define $\xi_{Y}$ by $1_{X\setminus Y}$; thus $\xi=\xi_{\{1\}}$.

## Some Lemmas:

Lemma 1: We have $(1i)\xi(1i)=\xi_{\{i\}}$ for $i\in X$ and $$\xi_Y=\prod_{i\in Y}\xi_{\{i\}}.$$

Proof: This is basic computation and the recognition that $$1_{X\setminus Y}=\prod_{i\in Y}1_{X\setminus \{i\}}.$$ $\square$

Lemma 2: For each $\alpha$ in $\mathscr{P}_n$, let $\hat{\alpha}$, the completion of $\alpha$, be given by $$x\hat{\alpha}=\begin{cases} x\alpha &: x\in \operatorname{dom} \alpha \\ x &: \text{ otherwise.}\end{cases}$$ Then, for $Y=X\setminus \operatorname{dom} \alpha$, $$\alpha=\xi_Y\hat{\alpha}.$$

Proof: Let $x\in X$. Then \begin{align} x(\xi_Y\hat{\alpha})&=(x 1_{X\setminus Y})\hat{\alpha} \\ &=(x 1_{X\setminus (X\setminus \operatorname{dom} \alpha)})\hat{\alpha} \\ &=(x 1_{\operatorname{dom} \alpha})\hat{\alpha} \\ &=x\alpha. \end{align} $\square$

## The Question:

Deduce (from Lemma 1 and Lemma 2) that $\mathscr{P}_n=\langle\zeta, \tau, \pi, \xi\rangle$.

## Thoughts:

I'm stuck.

Each $\alpha\in\mathscr{P}_n$ can be written as $$\alpha=\left(\prod_{i\in X\setminus \operatorname{dom}\alpha}(1i)\xi(1i)\right)\hat{\alpha}$$ and $\hat{\alpha}\in\mathscr{T}_n$.
Hence $\mathscr{P}_n=\langle \zeta, \tau, \pi, \xi\rangle$.