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

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!

• 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.
Commented Aug 2, 2011 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
Commented Aug 2, 2011 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 $U_x=\{x\}$ Commented Aug 2, 2011 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
Commented Aug 2, 2011 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
Commented Aug 2, 2011 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 $$V' \cap V_p = f_{|U_p}(U' \cap U_p)$$ 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.

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