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, 2017 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, 2017 at 10:19
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    $\begingroup$ 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, 2017 at 10:24
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    $\begingroup$ In your example small open nbhds of 0 don't map to open sets in $S^1$. $\endgroup$ Dec 24, 2017 at 10:25
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    $\begingroup$ @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, 2017 at 10:41

1 Answer 1


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.


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