Let $X,~Y$ be two metric spaces, $E\subseteq X$, $f:E\to Y$. I have a question that is lingering in my mind long. If $f$ is continuous and one-to-one, then does $f^{-1}:f(E)\to E$ continuous? I think the answer is true, is it? Is there counterexample? However, I read several books, some of them just prove that if $E$ is compact, then it is so. Why don't they more generally discuss it?
PS: The definition of continuity is: $f:E\to Y$ is called continuous iff the preimage $f^{-1}(U)$ is open in $E$ for all open set $U$ in $Y$.
It is not true in general. Just take the identity from $\mathbb R$ endowed with the discrete metric into $\mathbb R$ endowed with the usual metric. It is continuous, but its inverse isn't.