Universal covering space of CW complex has CW complex structure

We know that covering spaces of many low dimensional CW complexes such as graph (CW structures with only 0-cells and 1-cells) and all compact surfaces has CW structure. [The second fact is due to Uniformation Theorem from theory of Riemann surfaces. Basically, the only simply connected 2-manifold are the sphere, the Euclidean plane and hyperbolic plane.]

So I wonder, as put forth in the title, if this is true in general. That is, if $X$ is connected, locally path-connected and it has a CW complex structure then does its universal covering space also have a CW complex structure?

• Yes. Open cells are (embedded and) contractible, so they lift to unions of open cells. (One copy for each deck transformation, because the Universal covering Is normal.) Commented Feb 14, 2015 at 20:24

The result on the CW-topology may also be proved as a corollary of general results proved in P.I. Booth and R. Brown, Spaces of partial maps, fibred mapping spaces and the compact-open topology'', Gen. Top. Appl. 8 (1978) 181-195. This gives reasonable circumstances under which if $p: E \to B$ is a map, then the pullback functor $p^*$ from spaces over $B$ to spaces over $E$ has a right adjoint, and so preserves colimits. This immediately gives the result, and even implies more general types of result, for certain generalised inductively defined structures.
• "$(N)$ Any covering complex, $\tilde{K}$, of $K$ is a CW-complex." This presupposes that the covering is a complex. So the question becomes: is every covering of a CW complex a complex (and hence a CW-complex by $(N)$)? Commented Mar 16, 2016 at 3:36
• It is quite easy to give the cell structure on $\tilde{K}$; the more difficult part is to prove that the topology on the cover is the correct one. By the CW-structure, $K^{n+1}$ is obtained from $K^n$ by attaching $(n+1)$-cells. One wants to prove the same is true for $\tilde{K}$. For this it is helpful to know that the pullback functor $p^*$ where $p:\tilde{K}^{n+1} \to K^{n+1}$ preserves pushouts. Commented Mar 19, 2016 at 23:03