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!

  • $\begingroup$ 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? $\endgroup$
    – t.b.
    Commented Aug 2, 2011 at 14:12
  • $\begingroup$ you have a continuous bijection with a continuous inverse, i.e. a homeomorphism (so it seems, your question could use some editing) $\endgroup$
    – yoyo
    Commented Aug 2, 2011 at 15:11
  • 1
    $\begingroup$ 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 $U_x=\{x\}$ $\endgroup$ Commented Aug 2, 2011 at 16:05
  • $\begingroup$ @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 :) ) $\endgroup$
    – Andy
    Commented Aug 2, 2011 at 18:04
  • 2
    $\begingroup$ 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. $\endgroup$
    – gary
    Commented Aug 2, 2011 at 21:00

1 Answer 1


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.

  • $\begingroup$ For the case of a local diffeomorphism just note that the inverse function theorem shows that the inverse is smooth. $\endgroup$
    – t.b.
    Commented Aug 2, 2011 at 16:17
  • $\begingroup$ $V'\cap V_p$ 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.) $\endgroup$ Commented Aug 2, 2011 at 16:24
  • $\begingroup$ @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. $\endgroup$
    – t.b.
    Commented Aug 2, 2011 at 17:10
  • $\begingroup$ Thanks for clarifying @Theo, I've overlooked this fact. $\endgroup$ Commented Aug 2, 2011 at 18:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .