# 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.) – user39082 Feb 14 '15 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 $(N)$)? – Michael Albanese Mar 16 '16 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. – Ronnie Brown Mar 19 '16 at 23:03