# Confusions about proof regarding proper maps

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.

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 $$\mathcal{U}$$...

• 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 $y$. Jan 19, 2022 at 22:41
• @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. Jan 19, 2022 at 22:43
• I don’t follow what you say here Jan 20, 2022 at 0:19
• 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 Jan 20, 2022 at 2:52
• @ChristopherQuinnLaFondJr. You have an open cover of the compact set $f^{-1}[\{y\}]$. What can you say now? Jan 20, 2022 at 5:50