I‘m trying to solve the following problem:
Let $S$ be a set. Let $p:S\to S$ be a projector, i.e, $p\circ p = p$.
Let $G \subset Map(S,S)$ a subset closed under composition, i.e. if $f,g \in G$, then $f\circ g \in G$ and such that $(G, \circ)$ is a group.
- Show that $G$ contains one and only one projector.
I think the uniqueness of this projector can be shown taking $p, p'$ and assuming both are projectors in G.
But how do I prove it's existence in G? Is this $p$ the neutral element of $G$? If so, how do I show/see this?
So with the hint in the comment, it is clear that $p$ is the neutral element. Since we have that $p$ holds $e\circ e = e$, thus $p = e$
- Let $p(S) = T$, $f\in G$ and $q = p|_{S-T}: S-T \to T$. Show that $f(S)=T$ and that $f|_{T}$ is a bijection. What is the inverse of $f$ in $G$?
My attempt:
Since $f\circ p = f$, we have that $f(p(S)) = f(T) \overset{!}= T$. Is this just by definition of $f$ and $G$? Is then $f^{-1}=f$, the inverse of $f$, since $f(f(T))= f(T) = T = p(S)$?
This is really confusing me, any hint or advice would be appreciated.