group with projector Let $S$ a set. Given $G \subset Map(S,S)$ a subset closed under composition  and containing a projector $p$, (that means $p = p\circ p$), such that $(G, \circ)$ is a group.
Now assume that $p(S) = T$ and show that for $f \in G,\space f(S) = T$.
What is the inverse of $f$?
I found that the identity element of $G$ is $p$. So $f\circ p = p$. 
So actually $f(S) = f(p(S)) = f(T)$, this would imply that S = T, since $f$ in $G$ (a group) has an inverse. But then p is also the identity map, which is not necessarily true.  So where is my mistake? And how can I show that $f(S) = T$?
The inverse of $f$ might be $f$, since $f(f(S)) = f(T) = T = p(S)$ and hence $f\circ f = p$ (We proved that $\space$ $f:T\rightarrow T$ is a bijection).
Can anyone please give me some hints?
 A: We have the semigroup $\text{Map}(S,S)$ and we are taking a sub-semigroup $G$ such that $G$ is a group. As you notice, it's not necessarily true that $G$ is a subgroup of $\text{Aut}(S)$ because for example you can take $G=\{p\}$ where $p$ is a constant map. So it's not necessarily true that given $f\in G$, the inverse of $f$ in the group coincide with the inverse function of $f$.
Given $f\in G$ let's denote $f^*$ the inverse element of $f$ in the group. To understand who is $f^*$ as a function we can consider, as you were doing, the identity $p\in G$ and define $T=p(S)$.
Now let's take the map $\varphi:G\rightarrow \text{Map}(T,T)$ given by $g\mapsto g|_T$, we have that $\varphi$ preserve compositions and also that $\varphi$ is injective because  $f|_T=g|_T\Rightarrow f\circ p= g\circ p \Rightarrow f=g$. So $G$ is isomorphic as groups to $\varphi(G):=\hat{G}$ but now the inverse elements of $\hat{G}$ coincide with the inverse functions of $\text{Map}(T,T)$ because $\varphi(p)=\text{id}_T$. Therefore, we have $f^*=\varphi^{-1}((f|_T)^{-1})$, where $(f|_T)^{-1}$ is the inverse function of $f|_T$ in $Map(T,T)$.
Now I don't know if we can go further and give an explicit way to compute $\varphi^{-1}$. But at least I don't think it's possible to do this only by knowing who is the function $p$.
