Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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
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. – gary Aug 2 '11 at 21:00

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.

share|cite|improve this answer
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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.