Lifting local compactness to a covering space

Question

(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.)?

Answer 1

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 $B$).