Let $f:(X,d) \to (Y,d')$ an inverse continuous map , X ,Y are metric spaces.
Show that $f$ is homeomorphism.
My approach:
$f$ is inverse $\implies f $ is injective and onto.
Then , $f$ is continuous $\implies f^{-1} $ is continuous.
This is all what I have to prove ? the problem looks very simple to me and I am not entirly sure about it.Thanks.
The statement isn't true. Consider $i: \mathbb{R} \mapsto \mathbb{R}$ given by $i (x) = x$ (identity map), where the domain $\mathbb{R}$ is taken under the discrete metric $d (x, y) = \begin{cases} 0 & x = y \\ 1 & x \ne y \end{cases}$, and the range $\mathbb{R}$ is taken under the usual metric $d' (x, y) = |x - y|$. The map $i$ is clearly bijective. It is also continuous: all subsets of $\mathbb{R}$, indeed, all pre-images under $i$, are open in the discrete topology. However, $i$ is not open: $i (\{ 0 \}) = \{ 0 \}$, which is not open in the usual metric topology on $\mathbb{R}$.
