Given that $(g \circ f)$ is invertible, can we conclude that $f$ and $g$ are invertible?

Question

Given that $(g \circ f)$ is invertible, can we conclude that $f$ and $g$ are invertible?

I have previously proved that if $f$ and $g$ are invertible, then $g \circ f$ is invertible, however I am not sure if the converse of this statement is true.

Answer 1

No, for example for any set $X$ with at least two points and $x\in X$, the obvious maps $f:\{x\} \to X$ and $g:X \to \{x\}$ satisfy $g\circ f=Id_{\{x\}}$ but it is easy to see that neither $f$ nor $g$ are invertible.

Answer 2

The simple example by Nitrogen already tells you the answer is no. But you can say some things: if $g \circ f$ is injective, $f$ is injective: otherwise $x \neq y$ in the domain of $f$ exist with $f(x) = f(y)$ but then also $(g \circ f)(x) = g(f(x)) = g(f(y)) = (g \circ f)(y)$ contradicting the injectivity of $g \circ f$.

Also, if $g \circ f$ is surjective, so is $g$: let $z$ be in the codomain of $g$ (which is also the codomain of $g \circ f$). Some $x$ in the domain of $f$ exists such that $(g \circ f)(x) = z = g(f(x))$. But then $f(x)$ witnesses that $g$ assumes $z$ as a value (and $f(x)$ is in the domain of $g$).

So if $g \circ f$ is bijective, $f$ is injective and $g$ is surjective, which is true in the simple counterexample with $x \in X, |X| \ge 2$: $f: \{x\} \rightarrow X, f(x) = x$ and $g: X \rightarrow \{x\}, \forall p \in X g(p) = x$.