Theorem: Let $f:X\rightarrow Y$ be a continuous bijection, $X$ is locally compact, and $Y$ is Hausdorff. Then $f$ is a homeomorphism.

So let $U\subset X$ be closed, it's enough to show $f(U)$ is closed in $Y$.

Since $X$ is locally compact, for each $x\in U$, there exists a compact neighborhood $U_x$, and $U_x\cap U$ is also a neighborhood of $x$. Now $U=U\cap(\cup_{x\in U}U_x)$ where each $U_x$ is a compact subset of $X$. But how do I show that $f(U)$ is compact in $Y$?

Now suppose that I've shown that $f(U)$ is compact in the Hausdorff space $Y$, I can show that for each $y$ not in $f(U)$, there exists an open neighborhood of $y$ that does not intersect $f(U)$. But how does that lead to $f(U)$ being closed?

  • 1
    $\begingroup$ As the counterexample of @G.Sassatelli shows this is false. On the other hand, there is a version of this theorem which allows $X$ to be noncompact, but an additional hypothesis is required, namely that $f$ is a proper map, meaning that for every compact subspace $C \subset Y$ the subspace $f^{-1}(C) \subset X$ is compact. $\endgroup$
    – Lee Mosher
    Oct 18, 2017 at 13:00
  • $\begingroup$ As for your last paragraph: if you could show that for each $y\in Y\setminus f(U),$ there is an open neighborhood of $y$ disjoint from $f(U),$ then what you'd have shown is that $Y\setminus f(U)$ is open, which would show that $f(U)$ is closed. $\endgroup$ Oct 18, 2017 at 13:16

2 Answers 2


The theorem you are trying to prove is false. Consider $X=[0,1)$, $Y=S^1$ and $f(x)=(\cos (2\pi x),\sin (2\pi x))$. This is a continuous bijection of locally compact metric spaces, but it is not a homeomorphism.

As a fact to keep in mind, it is true that a continuous bijective map $\Bbb R^m\to \Bbb R^m$ is a homeomorphism, but it requires more than general topology to be proved.

Added: The same theorem with "$X$ compact" instead of "locally compact" is true, and perhaps this was the intention of the assignment.

  • $\begingroup$ But it was presented as an exercise. Do you think it holds if I strengthen the condition to $X$ is compact? Are there any weaker conditions for which it holds? $\endgroup$ Oct 18, 2017 at 12:00
  • $\begingroup$ Yes, for $X$ compact it holds. And, quite likely, this was the intention of the assignment. $\endgroup$
    – user228113
    Oct 18, 2017 at 12:05
  • $\begingroup$ Shouldn't be $X=[0,2\pi [$ instead? Or then $(\cos(2\pi x),\sin (2\pi x)) $? $\endgroup$
    – Ivo Terek
    Oct 18, 2017 at 16:34
  • 1
    $\begingroup$ @IvoTerek The latter. My original intention was to write $e^{2\pi i x}$, but then I decided to switch notation. While doing it, I lost the coefficient. $\endgroup$
    – user228113
    Oct 18, 2017 at 17:14

It is true if $X$ is compact. Take $F\subseteq X$ closed. Then $F$ is compact, because $X$ is. By continuity, $f[F]$ is compact. Since $Y$ is Hausdorff, $f[F]$ is closed.


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