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Is a bijective local homeomorphism a global homeomorphism? What about diffeomorphisms?

I don't know if it's true this property, I'm not sure. If someone can prove it I would be very grateful, and if not I would welcome a counterexample because I can not think. Thank you very much. At worst, if not true, someone knows a sufficient condition to fulfill what I want? Thank you very much!

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    @Theo: yes, I think you agreed with me, or did you see something wrong?2011-08-02

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Here's a very detailed proof.

Let's say we have a continuous map $f:X \to Y$ of topological spaces of which we know:

  • $f$ is a local homeomorphism, that is for every $p \in X$ exist the open subsets $U \subseteq X$, $V \subseteq Y$ with $p \in U$ and such that $f_{|U}:U \to V$ is a homeomorphism
  • $f$ is bijective, that is there is an inverse map $f^{-1}:Y \to X$

In order to prove that $f$ is a homeomorphism we need to prove that $f^{-1}$ is continuous.

So, let $U' \subseteq X$ an open set and $V' = (f^{-1})^{-1}(U') = f(U')$. For each $p \in V'$ let $U_p$, $V_p$ as above (i.e. $f_{|U_p}: U_p \to V_p$ is homeomorphism), then $ V' \cap V_p = f_{|U_p}(U' \cap U_p) $ is open because $f_{|U_p}$ is an homeomorphism (and therefore an open map). Furthermore $V'= \cup_{p \in V'} V' \cap V_p$ is open, as union of open sets.

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    Thanks for clarifying @Theo, I've overlooked this fact.2011-08-02