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Let $p:Y\to X$ be a covering map. If $X$ has a CW complex structure, then we can give a CW complex structure on $Y$ so that $p$ becomes a cellular map, by lifting the characteristic maps (cf. Euler characteristic of covering space of CW complex). I am curious about the converse. Suppose $Y$ is a CW complex. Then can we give $X$ a CW complex structure?

user302934
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The action of $\mathbb{Z}$ on $\mathbb{R}^2\setminus\{0\}$ generated by $(x,y)\mapsto(2x,y/2)$ induces a covering map $\mathbb{R}^2\setminus\{0\}\rightarrow(\mathbb{R}^2\setminus\{0\})/\mathbb{Z}$ whose base is not Hausdorff, hence cannot be a CW-complex. (This is Exercise 1.3.25 from Hatcher's Algebraic Topology).

Thorgott
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  • Is there a counterexample in the case where both $X$ and $Y$ are compact? – user302934 Dec 07 '23 at 03:07
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    I don't know any, it might be a difficult question. In case $X$ is connected and $Y$ is compact, $X$ will be compact Hausdorff (so also paracompact, normal), will be locally contractible (this always holds), so the usual topological properties for CW-complexes hold. Proving a space does not admit a CW-structure is hard if you don't have such an easy reason to appeal to. Nonetheless, I would be highly surprised if there weren't counter-examples. – Thorgott Dec 07 '23 at 03:56