I am currently trying to prove that if $f:X\rightarrow Y$ is a closed continuous map between topological spaces such that $f^{-1}(y)$ is compact for all $y\in Y$ then f is proper. In this sense $f$ is proper if for every compact set $K\subset Y$ we have that $f^{-1}(K)$ is compact in $X$.

I am really quite lost on this one as I feel like I don't know enough about $X$ or $Y$ to really do anything. In previous problems/proofs I was working on like this I usually had that $Y$ or $X$ was locally compact or Hausdorff.

Since I didn't really know what I should do, I tried writing the pre image $f^{-1}(K)$ as:

$$f^{-1}(K)=\bigcup_{y\in K}f^{-1}(y)$$

However this doesn't really seem useful as an infinite collection of compact sets needn't be compact. I suspect continuity or the closedeness of $f$ could help but I can't see how. Any hints to get me going in the right direction would be much appreciated.


1 Answer 1


Note that $f: X \to Y$ is closed iff

For all $y \in Y$, if $O \subseteq X$ is open such that $f^{-1}[\{y\}] \subseteq O$, there exists some open $y \in U \subseteq Y$, so that $f^{-1}[U] \subseteq O$.

A sort of "reverse continuity" characteristic of closed maps. The proof is not hard. (We only need the left to right implication as a lemma).

From this: let $\mathcal{U}$ be an open cover of $f^{-1}[K]$ with $K$ compact. Use the "lemma" to define a new cover of $K$ based on $\mathcal{U}$...

  • $\begingroup$ How do I define a new cover of $K$ based on $\mathcal{U}$? If $\mathcal{U}$ is just a random collection of open sets in $X$ that happens to cover $f^{-1}(K)$ then how can I know anything about whether any of those open sets contain any $f^{-1}(y)$ or $f^{-1}(U)$ where $U$ is an open nbd of $y$. $\endgroup$
    – Chris
    Jan 19, 2022 at 22:41
  • $\begingroup$ @ChristopherQuinnLaFondJr. For every $y \in K$, $f^{-1}[\{y\}] \subseteq f^{-1}[K]$ and so is covered by finitely many $U \in \mathcal{U}$. Use their union in the lemma. $\endgroup$ Jan 19, 2022 at 22:43
  • $\begingroup$ I don’t follow what you say here $\endgroup$
    – Chris
    Jan 20, 2022 at 0:19
  • $\begingroup$ Isn’t what u wrote true for any map between top spaces? Like $f^{-1}(y)\subset f^{-1}(K)$ is true for every map if $y\in K$ so I don’t see how that helps us $\endgroup$
    – Chris
    Jan 20, 2022 at 2:52
  • $\begingroup$ @ChristopherQuinnLaFondJr. You have an open cover of the compact set $f^{-1}[\{y\}]$. What can you say now? $\endgroup$ Jan 20, 2022 at 5:50

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