I know that if $f : X \rightarrow Y$ is a continuous bijection from a compact space $X$ to a Hausdorff space $Y$, then $f$ is an homeomorphism.

So I was thinking that if we relax the assumption $X$ compact to $X$ locally compact, it should be true as well. Using the above result, $f$ is an homeomorphism if we restrict it to a compact neighborhood of $X$. Since we can find a compact neighborhood around every point of $X$, $f$ should be a local homeomorphism. But a bijective local homeomorphism is a global homeomorphism, so that would be it.

Yet, if I'm not mistaken, the map $f : [0, 2\pi[ \rightarrow S^1, f(\theta) = e^{i\theta}$ is a counterexample. What is wrong with my reasoning ?

  • $\begingroup$ @Wojowu Thank you for your answer. I was wondering then, what I'm not understanding about this answer : math.stackexchange.com/a/55148/272494 $\endgroup$ – Desura Dec 24 '17 at 10:14
  • $\begingroup$ @Wojowu I mean, doesn't the answer prove that a bijective local homeomorphism is a global homeomorphism ? $\endgroup$ – Desura Dec 24 '17 at 10:19
  • 1
    $\begingroup$ Ah, sorry, I had a wrong definition of being a local homeomorphism in mind. It is indeed true that a bijective local homeomorphism is a global homeomorphism. However, your map $f$ needn't be a local homeomorphism - in the definition, we need the map to be a homeomorphism from some open subset of $X$ to an open subset of $Y$. A compact neighbourhood is definitely not going to be an open set. $\endgroup$ – Wojowu Dec 24 '17 at 10:24
  • 2
    $\begingroup$ In your example small open nbhds of 0 don't map to open sets in $S^1$. $\endgroup$ – Eero Hakavuori Dec 24 '17 at 10:25
  • 1
    $\begingroup$ @Wojowu By definition of a neighborhood, there must be an open set involved. So if $X$ is locally compact, for every $x \in X$, there exists $U$ an open set, $V$ a compact set, such that $x \in U \subset V \subset X$. So if $f$ is an homeomorphism on $V$, the restriction to $U$ is an homeomorphism as well. But I found the mistake I think. It is an homeomorphism in the subspace topology of $U$ induced by $V$, which is not the same as the subspace topology induced by $X$ since $V$ is not open, as you said. So that's not a local homeomorphism, as you were thinking. Thanks. $\endgroup$ – Desura Dec 24 '17 at 10:41

A simpler example is just the identity from a discrete $\mathbb{R}$ to the usual $\mathbb{R}$, both are locally compact metrisable, any map from a discrete space is continuous. The compact neighbourhoods in the first space are only the finite ones, so there we do have that there is a local homeomorphism (the finite sets are homeomorphic in both spaces), but they're not neighbourhoods in the image.

Likewise, for your example, the compact neighbourhoods $[0,r]$ of $0$ in $[0,2\pi)$ do have compact homeomorphic images in $S^1$ but these are not neighbourhoods of $f(0)$ in $S^1$ any more.

To go from something like a local homeomorphism to global one, we need a stronger condition. Something like: for every $x \in X$ and every compact neighbourhood $C$ of $x$ in $X$, we must have that $f[C]$ is a (necessarily homeomorphic) neighbourhood of $f(x)$ as well. And as we saw, this is by no means garantueed from just being continuous and bijective.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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