Preservation of Local Compactness

[Euge]

Let $f : X \to Y$ be a surjective continuous closed map of topological spaces such that every fiber $f^{-1}(y)$ is compact. Show that $Y$ is locally compact if $X$ is locally compact.

 

[WWGD]

Doesnt this have to see with Saturated subsets of quotient maps, IIRC?

 

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Doesnt this have to see with Saturated subsets of quotient maps, IIRC?

I don't understand the question, could you please clarify?

 

[WWGD]

I don't understand the question, could you please clarify?

Apologies, I remember the concept of saturated subsets playing a role in proofs when dealing with fibrations and coverings. Let me try to make the argument more precise.

 

[Office_Shredder]

I spent a long time on this before noticing you said f was a closed map! It doesn't seem like this has gotten a lot of traction, but it's really just chaining all the topology definitions together.

Spoiler

given $y\in Y$, we need to find $y\in U\subset K$ with $K$ compact and $U$ open.

Since $f$ is subjective, there exists $x\in X$ such that $f(x)=y$. $X$ is locally compact so there exists $x\in V\subset L$ with $V$ open and $L$ is compact.
$y\in f(V)\subset f(L)$. Since $f$ is continuous, $f(L)$ is compact. But $f(V)$ may not be open. The rest of the proof fixes this issue.

$X-V$ is closed, and $f$ is a closed map, so $f(X-V)$ is closed. So $Y-f(X-V)$ is an open set, and it's a subset of $f(L)$: if $z\in Y-f(X-V)$, we know that $f(a)=z$ for some $a$, and $f(a)\notin f(X-V)$ means $a\in V$ which we know is contained in $L$. So we're close. But if $f$ is many to 1, this set may be too small, e.g. $y$ might be in $f(X-V)$.

The final correction uses the compactness of $$f^{-1}(y)$. Let $x_\alpha$ be the pre image of $y$, indexed by some set $A$ for notational convenience (the index set can literally be the pre image). For each $\alpha$, $x_\alpha \in V_\alpha \subset L_\alpha$ with each $V_\alpha$ open and $L_\alpha$ compact. The $V_\alpha$s are an open cover of $f^{-1}(y)$, so there is a finite subcover, which is indexed by a finite set $M\subset A$. $X-\bigcup_M V_\alpha$ is a closed set such that $y\notin f(X-\bigcup_M V_\alpha)$. So $y\in Y-f(X-\bigcup_M V_\alpha)$ which is an open set, that is contained in $f(\bigcup_M L_\alpha)$, by similar argument in the one index case. But $\bigcup_M L_\alpha$ is a finite union of compact sets so is compact, and then applying $f$ gives a compact subset of $Y$.