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Suppose $f:X \rightarrow Y$ is a perfect map between topological spaces $X$ and $Y$, i.e. $f$ is continuous, surjective and closed. Also suppose that the fibers of $f$ are compact. If $Y$ is locally compact, will $X$ be as well?

I define locally compact as follows: A space $X$ is locally compact if for each $x \in X$ there exists a compact subset $C$ that contains a neighborhood of $x$.

Initially I thought this would be the case, however, I can't really seem to find a good proof. This led me to believe that it would only work for Hausdorff spaces, but I'm not sure if that's true either...

I'm totally stuck and don't know how to proceed from here.

Geigercounter
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  • There are lots of ways one can start the proof and which fact/condition to use where is also bit hard. See https://math.stackexchange.com/q/4453577/861687. – user264745 May 22 '22 at 20:22
  • @user264745 I have already proved that. My question is about the converse: if $Y$ is locally compact, will $X$ be? – Geigercounter May 22 '22 at 20:30
  • Okay. We can actually prove a stronger result. See https://math.stackexchange.com/q/4450307/861687. compact $\Rightarrow$ locally compact. – user264745 May 22 '22 at 20:34
  • @user264745 great! thanks! however a space can be locally compact while not being compact. Does it still hold then? That's where I am stuck. I don't know how to prove it when we don't have the compact condition. For example $\mathbb{R}$ with the Euclidean topology is locally compact but not compact. – Geigercounter May 22 '22 at 20:36
  • What a coincidence! I’m stuck at this same reasoning, look debate between Nico and me in this post. Maybe you can clarify some stuff. – user264745 May 22 '22 at 20:41
  • See in this post. Henno Brandsma used above reasoning(of course in different context) in comment section. Counter defence for that reasoning is urysohn lemma don’t generalize to regular space. I think your claim may not be correct because James Dugundji didn’t show this(what you’re trying to prove) result. – user264745 May 22 '22 at 20:51
  • @user264745 I was getting suspicious that I didn't find a lot of questions about the converse result. There are some questions that ask for the equivalence, but no satisfactory answers (https://math.stackexchange.com/questions/4237232/x-is-locally-compact-iff-y-is-locally-compact-if-there-is-a-perfect-map-f and https://math.stackexchange.com/questions/4272683/perfect-map-and-locally-compact-spaces?noredirect=1&lq=1) Note that they might use different definitions for local compactness (i.e. assuming Hausdorff) – Geigercounter May 22 '22 at 20:59
  • Indeed, in above link you provided, Hausdorff is assumed. Without that condition your hypothesis don’t hold. – user264745 May 22 '22 at 21:10
  • @user264745 So do you know a counterexample for when it isn't Hausdorff? and do you know of a proof for the case that it is Hausdorff? The answers is the linked questions only have really vague and unhelpful hints imo, I wasn't able to figure out a proof based on those hints – Geigercounter May 22 '22 at 21:40
  • No. I don’t know concrete counterexample(I avoid this kind of problem). If Hausdorff is assumed, then first link you provided is solution, try to follow hint and use Hausdorff condition to show $\overline{f(O)}$ is compact, because it’s closed. – user264745 May 22 '22 at 21:45
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    @user264745 too bad for the counterexample... Thanks anyway! One question though: you're talking about the "If"-part in the hint right? I find that it isn't really much of a hint since it just repeats the definitions. Would be able to clarify the hint a little bit? – Geigercounter May 22 '22 at 21:51
  • @user264745 any updates on the proof? I haven't been able to find it. However I think I'm close to getting a counter example – Geigercounter May 24 '22 at 08:57
  • See https://math.stackexchange.com/q/4457432/861687, my above logic is wrong. I haven’t yet tryed to prove it but soon(few hours) I will try and tell you my progress. – user264745 May 24 '22 at 09:01
  • That hint contains typo/error. $f(O_{f(x)})$ have no meaning! Image of codomain, that is absurd. The whole hint don’t make any sense, IMO. Can you double check my claim in comment section of this post? – user264745 May 24 '22 at 11:28
  • I thought if means if..then.. statement and only if means opposite direction but I googled it. I was wrong. Still that hint makes no sense. – user264745 May 24 '22 at 11:37
  • @user264745 that's what I meant by the hints being not helpful. I'm going to let this problem rest for a bit and come back in a while, maybe I'll make some progress then – Geigercounter May 24 '22 at 18:43

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