Are all right inverses of a given function, injective?

Question

Suppose we have a function P, I want to show that all right-inverses of P are injective. I know that if a function has a left-inverse, then it bound to be injective. Isn't it correct that all right-inverses of P have P as left-inverse, thus are injective?

Answer 1

This results from general result on sets:

Let $X,Y,Z\;$ be sets and $f:X\longrightarrow Y, \;g:Y\longrightarrow Z\;$ two maps. Then:

1. If $\;g\circ f$ is injective, $f$ is injective.

2. If $\;g\circ f$ is surjective, $g$ is surjective.

Proofs thereof by contrapositive.

Answer 2

You have exactly the right idea! You simply need to formalize it.

Suppose that $P$ is a function and that $g$ is a right inverse of $P$. Suppose now that $x,y$ are elements of the domain of $g$ such that $g(x)=g(y).$ To show that $g$ is injective, we must show that $x=y.$ The only properties we haven't yet used is the right inverse property and the definition of functions. Since $P\circ g$ is well-defined, then $g(x)=g(y)$ is an element of the domain of $P.$ Thus, since $P$ is a function and $g(x)=g(y),$ then $$P\bigl(g(x)\bigr)=P\bigl(g(y)\bigr),$$ meaning $$(P\circ g)(x)=(P\circ g)(y).$$ What do you know about $P\circ g$ that can get you the rest of the way?