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(I decided to repost this from MathSE, since the question seems to not be as easy as I had thought)

NB: In this question, local compactness is used in its weak form, i.e. in a locally compact space, every point has a compact neighbourhood. Also, T3 is the weaker condition of the pair T3/regular.

Let $p:\tilde{X}\to X$ be a covering map. Since $p$ is a local homeomorphism, $\tilde{X}$ and $X$ share the "standard" local properties: local (path) connectedness, T1, etc.

The black sheep of the family of local properties is local compactness. Of course, a suitable separation axiom (T2 for instance) makes it into a proper local property with suitable local bases and everything. Therefore, if $X$ is locally compact Hausdorff, so is $\tilde{X}$. Interestingly, if $X$ is locally compact and T3, then $\tilde{X}$ is again locally compact.

My question is, how can this fail. More precisely, given a locally compact space $X$, does there exist a non-locally compact cover $\tilde{X}$ of $X$?

I strongly believe this to be true, because the compact neighbourhoods in $X$ might be too big to be seen by the covering map, but I haven't been able to find an example. I've been trying to construct a cover of the one-point compactification of the rationals, this being the nastiest space with the required properties I could think of. I've come up with this: pick a proper neighbourhood $U$ of $\infty$ in $\mathbb{Q}^+$. Then take as a cover the space $\mathbb{Q}^+\times\{1/n;n\in\mathbb{N}\}\cup U\times\{0\}$ with the obvious projection. I think I've managed to prove this is a covering map and it looks to be not locally compact at the 0-th level, but that could just be my intuition being dead wrong.

I'd be interested in any thoughts on this. Also, somewhat less importantly, does local compactness descend from the covering space? Under what conditions do $X$ and $\tilde{X}$ share local compactness (can we in some way force the covering map to be proper, open, etc.)?

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  • $\begingroup$ Local compactness transfers through local homeomorphisms just as any other local property. $\endgroup$ Jun 7, 2011 at 20:50
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    $\begingroup$ "Every point has a compact neighborhood" really doesn't deserve the name "local compactness." "Local" compactness really should refer to "every neighborhood of a point contains a compact neighborhood" and then it would actually be a local property... $\endgroup$ Jun 7, 2011 at 20:51
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    $\begingroup$ @Fernando: this version of local compactness is not a local property. $\endgroup$ Jun 7, 2011 at 20:51
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    $\begingroup$ This version has the implication compact --> locally compact, and this has been a traditional "feature" of local compactness. Local connectedness is different in that respect. And if we add Hausdorffness it is equivalent to the "arbitarily small compact neighbourhoods" anyway. $\endgroup$ Jun 7, 2011 at 21:20
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    $\begingroup$ @Qiaochu: agreed, this is in many respects the 'bad' notion of local compactness, but it seems to be standard in some quarters. (This is the definition that Munkres gives, although he doesn't seem to be too interested in the non-Hausdorff case; of course the notions coincide when the space is Hausdorff, as Henno mentioned.) $\endgroup$
    – Todd Trimble
    Jun 7, 2011 at 21:42

1 Answer 1

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A suitable counterexample can be constructed as follows. Let $\mathbb I=\{x\in\mathbb R:0<x<1\}$ denote the open unit interval and let $B=\{0\}\cup(\omega\times\mathbb I)\cup\{1\}$ be endowed with the topology $\tau_B$ generated by the base consisting of the following sets:

$\bullet$ $\{n\}\times (a,b)$ for $0<a<b<1$ and $n\in\omega$;

$\bullet$ $\{0\}\cup(F\times (0,b))\cup((\omega\setminus F)\times (0,1)$ for a finite set $F\subset\omega$ and $b\in(0,1)$;

$\bullet$ $(F\times (a,1))\cup((\omega\setminus F)\times (0,1))\cup\{1\}$ for a finite set $F\subset\omega$ and $a \in(0,1)$;

It is easy to see that the space $B$ is $T_1$ and compact, path connected, locally path-connected, but not Hausdorff.

It can be shown that for any universal covering $p:E\to B$ over $B$ the total space $E$ is not locally compact.

The total space $E$ can be identified with the geometric realization of the Cayley graph of the free group with countably many generators (corresponding to the open unit intervals in the base space $B$).

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  • $\begingroup$ This is a very interesting answer. Thanks! I had completely forgotten about this question. $\endgroup$ Nov 29, 2017 at 15:26

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