Problem: Let $f:X \rightarrow Y$ and $g:Y \rightarrow Z$ where $X,Y$ and $Z$ are metric spaces. If $g \circ f$ is homeomorphism and $f$ is surjective prove that $f$ and $g$ are homeomorphism.
The problem I could not solve but I prove that $f,g$ are bijective. So, if $f$ and $g$ are continuous then $f^{-1}$ and $g^{-1}$ are continuous because :
$$ (g \circ f)^{-1} = f^{-1} \circ g^{-1} \implies f^{-1}=(g \circ f)^{-1}\circ g \wedge g^{-1}=f \circ (g\circ f)^{-1} $$
My question is whether the result can be proven without assuming that $f$ and $g$ are continuous. If so, I would like help because I have not been able to prove it. Thanks!
The claim is false without the assumption that $f$ and $g$ are continuous. Here is a counterexample:
Let $f,g :\mathbb{R} \to \mathbb{R}$ be defined as $$f(x)=g(x)= \left\{ \begin{array}{lc} x &\mbox{ if } x \in \mathbb{Q} \\ -x&\mbox{ if } x \notin \mathbb{Q} \end{array} \right.$$
Then, $f$ is surjective and $g \circ f(x)=x \forall x \in \mathbb R$ is a homemorphism. But $f,g$ are not.