1
$\begingroup$

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?

$\endgroup$
2
  • 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

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

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.