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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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    I don't understand your problem. So you *have* an inverse and you wonder if it's continuous or smooth if it is locally so?2011-08-02
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    you have a continuous bijection with a continuous inverse, i.e. a homeomorphism (so it seems, your question could use some editing)2011-08-02
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    Do you mean something like: "$f:X\to Y$ is continuous, bijective and for every $x\in X$ there is a neighborhood $U_x$ such that $f|U_x: U_x \to f[U_x]$ is a homeomorphism." The answer to this question is no: Take X=discrete and Y=indiscrete topology on the same space, f=identity and $U_x=\{x\}$2011-08-02
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    @Daniel: as for the diffeomorphism, it is sufficient that it is injective, i.e. a local injective diffeomorphism is a global one. (if I remember my analysis :) )2011-08-02
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    I think this could be a sort of corollary of the inverse function theorem;given an injection, the local diffeomorphism holds in a region where the differential/Jacobian is non-zero, i.e., the IFT guarantees the existence of a local inverse for f when $J(f)\neq 0$ , but if f is injective and there is no x with $J(f)(x)= 0$ (the case of, e.g., f(x)=$x^3$ shows the two don't always coincide), I think the inverse is the global inverse inthe form given by the IFT.2011-08-02
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    Isnt the map f: [0,1)-->$S^1$: $f(t)=e^{i2\pi t}$ a counterexample? It is a continuous bijection, and the IFT tells us that it is a local diffeo. at each point, but it is not a homeomorphism (e.g., $S^1$ is compact, and [0,1) is not, or [0,1) has a single point as a cutset, and $S^1$ has no 1-pt. cutsets), let alone a diffeomorphism.2011-08-02
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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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    For the case of a local diffeomorphism just note that the inverse function theorem shows that the inverse is smooth.2011-08-02
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    $V'\cap V_p$ is open in $V_p$. This does not mean that it is open in $Y$. (See also my comment bellow the question. If I am not mistaken, it should give a counterexample to your proof.)2011-08-02
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    @Martin: There's nothing wrong here. The sets $V_p$ are assumed to be open in $Y$ which they aren't in your "counterexample". I carelessly omitted that in my comment to cduston's argument.2011-08-02
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    Thanks for clarifying @Theo, I've overlooked this fact.2011-08-02