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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? –  t.b. Aug 2 '11 at 14:12
    
you have a continuous bijection with a continuous inverse, i.e. a homeomorphism (so it seems, your question could use some editing) –  yoyo Aug 2 '11 at 15:11
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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 $U_x=\{x\}$ –  Martin Sleziak Aug 2 '11 at 16:05
    
@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 :) ) –  Andy Aug 2 '11 at 18:04
    
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. –  gary Aug 2 '11 at 20:41

3 Answers 3

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 $$ V' \cap V_p = f_{|U_p}(V') $$ 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. –  t.b. Aug 2 '11 at 16:17
    
$V'\cap V_p$ 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.) –  Martin Sleziak Aug 2 '11 at 16:24
    
@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. –  t.b. Aug 2 '11 at 17:10
    
Thanks for clarifying @Theo, I've overlooked this fact. –  Martin Sleziak Aug 2 '11 at 18:01

Well, a local homeomorphism restricted to any open set (on a topological space) is a homeomorphism (I guess the image has to be open as well). So it is a continuous, open map. The only property it is missing is bijectivity, so if I understand your question, if you have a bijective local homeomorphism then it is a homeomorphism.

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By definition a local homeomorphism is a map with the property that for every point in the domain there is a neighborhood such that the restriction to that neighborhood is a homeomorphism onto its image. However, it need not be so on any open set: think of a covering map and an open set touching two sheets, for example. –  t.b. Aug 2 '11 at 15:16
    
I forgot to say that the image of that neighborhood is of course assumed to be open. Sorry about that. –  t.b. Aug 2 '11 at 17:11
    
I think in order for the restriction to any open set to be a homeomorphism, the function itself should be a global homeomorphism , e.g., by having the restriction to the open set consisting of the whole space. –  gary Aug 2 '11 at 21:08

I think the answer is no; maybe we can think of the result that a continuous bijection between X compact and Y Hausdorff is a homeomorphism; there is a reason why the result is stated as it is; if either X is not compact or Y is not Hausdorff, then the result does not always hold. I think this is a counterexample:

X=[0,1) in $\mathbb R$ ; Y=$S^1$

Then f(t):=$e^{i2\pi t}$ is a continuous bijection, and a local diffeomorphism, since $\frac{de^z}{dz}\neq 0$ , but it is not a global diffeo. , since, e.g., [0,1) is not compact, and/or , [0,1) has a 1-pt cutset, but $S^1$ does not.

EDIT: as Theo pointed out, the inverse function theorem does not apply here, since the function in question must have as a domain an open subspace of $\mathbb R^n$ , which is not the case with [0,1). Specifically, the local diffeomorphism condition is violated at p=0, which has no (subspace) 'hoods (neighborhoods) that are homeomorphic to (subspace) 'hoods of $S^1$.

Still, something nice here is that this is an example of a continuous bijection which is not a homeomorphism.

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Would you bother explaining the downvote? –  gary Aug 2 '11 at 21:27
    
Your map isn't a local diffeomorphism or a local homeomorphism by the usual conventions: no image of a neighborhood of $0$ is homeomorphic to a neighborhood of $1$ (I didn't downvote, by the way) –  t.b. Aug 2 '11 at 21:29
    
Fair enough.I guess I missed the t=0 at the chain rule; I thought $f'(t)\neq 0$, but why then do we have $\frac {df(t)}{dt}=e^{i2\pi t}i2\pi \neq 0$ –  gary Aug 2 '11 at 21:34
    
The inverse function theorem is proved for maps on open subsets of $\mathbb{R}^n$, hence for open subsets of a manifold. –  t.b. Aug 2 '11 at 21:42
    
This may sound --or actually be --just self-serving here, but, isn't there to be learnt from incorrect answers that are reasonable (as I think mine was)? Maybe one can see where things fail, or one can learn the detailed conditions for when some results hold, etc., as I did in learning the conditions of the inverse function theorem. –  gary Aug 2 '11 at 23:34

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