$f:A\to B, h:=(g\circ f):A\to C$ both bijective $\implies g:B\to C$ bijective?

Question

let $f:A\to B, g:B\to C$ and $h:=(g\circ f):A\to C$ where $f,h$ are bijective. I have to prove or disprove that g is also bijective. Due to previous proofs I know that $g$ is surjective, so my working point is the injectivity of $g$.
My first idea was: Since $h:=(g\circ f)$ is bijective, $h^{-1}$ exists and is also bijective. So $h^{-1} = f^{-1}\circ g^{-1}$ and since $g^{-1}$ exists, it follows that $g$ is bijective.
I am not quite sure about my reasoning.
My second idea was: Let $a_1,a_2\in A$. $f,h$ are injecive.
So $f(a_1)=f(a_2) \implies a_1 = a_2$, $h(a_1)=h(a_2) \implies a_1=a_2 \Leftrightarrow g(f(a_1))=g(f(a_2)) \implies a_1=a_2$.
I do not see why $g(f(a_1))=g(f(a_2)) \implies a_1=a_2$ which is needed for g to be injective.
Any help is apreciated.

Answer 1

You can't conclude that $g^{-1}$ exists just because it would make $f^{-1}\circ g^{-1}$ a nice description of $h^{-1}$; you only need a right inverse of $g$ for that, and you already know that that exists because $g$ is surjective. However, we do have the fact that $g=h\circ f^{-1}$, which is bijective.