I'm currently reading Topology by Munkres. I found this question online from an assignment some dude did for his topology class like back in 2017 or something like that and I'm in the process of working on it.
Suppose $X$ and $Y$ are locally compact, Hausdorff topological spaces and let $f:X \to Y$ be one-to-one, onto, continuous and proper (pre-images of compact sets are compact). Prove that $f$ is a homeomorphism.
My question: Given that the function is continuous and bijective, it should be enough to show that $f^{-1}$ is continuous.
The sets are locally compact, so for each $x \in X$ and each $y \in Y$, there is some compact subspace of $X$ and $Y$ that contains a neighbourhood of $x$ and $y$ respectively. But if the function is proper, that is to say, the pre-images of compact sets are compact and compact sets are closed by definition, wouldn't the pre-images of these closed sets also be closed? Is this not enough to conclude that $f^{-1}$ is continuous?
Thanks in advance for any tips or advice.
Best Answer
I love to use the one-point compactification in this case! :-)
If $X$ (and $Y$) is compact, the proposition is very easy to prove, because being closed and being compact in a compact Hausdorff space is just the same thing. And since the image of compact sets by a continuous function is compact, you have that the function takes closed sets to closed sets and its inverse is therefore continuous.
A locally compact Hausdorff space can be embedded in its one-point compactification: $\tilde{X} = X \cup \{\infty_X\}$ and $\tilde{Y} = Y \cup \{\infty_Y\}$. The open neighbouhoods of $\infty_Y$, for instance, are of the form $\tilde{A} = \{\infty_Y\} \cup A$, where $Y \setminus A$ is compact.
Saying that a function $f: X \rightarrow Y$ is proper, is just the same as saying it can be continuously extended to $$\tilde{f}: \tilde{X} \rightarrow \tilde{Y}$$ by setting $f(\infty_X) = \infty_Y$. To check it is continuous at $\infty_X$, just verify that $$\tilde{f}^{-1}\left(\tilde{A}\right)^c = f^{-1}\left(\tilde{A}^c\right)$$ is compact and therefore, it is a neighbouhood of $\infty_X$.
So, $\tilde{f}$ is a homeomorphism. Since $X$ and $Y$ are embedded (they have the topology induced by $\tilde{X}$ and $\tilde{Y}$, $f$ is also a homeomorphism.