A covering space of CW complex has an induced CW complex structure.

Question

Let $X$ be a $CW$ complex, and let $q : E \rightarrow X$ be a covering map. Prove that $E$ has a $CW$ decomposition for which each cell is mapped homeomorphically by $q$ onto a cell of $X$.
Hint: If $A \subseteq X$ is a locally path-connected subset, then the restriction of $q$ to each component of $q^{-1}(A)$ is a covering map onto its image.

This problem is in Lee's Toplogical Manifolds on pages 303.
We can define characteristic maps $\tilde \Phi$ of $E$ as follows.
Let $\Phi : D \rightarrow X$ be a characteristic map for a cell $e$ of $X$. Since $D$ is simply connected, for each fiber of an element in $e$ there exist a unique lifting $\tilde \Phi : D \rightarrow E$. I have checked that the set of $\tilde \Phi (Int D)$ form a cell decomposition. And this cell decomposition has Weak topology. But I am unable to show that this has the property of closure finiteness. Let $\Phi(Int D) = \tilde e$. $q (\bar {\tilde e}) = \bar e$ intersects finitely many cells of $X$ since $X$ is a $CW$ complex. But I cannot prove that $\bar {\tilde e}$ intersects finitely many cells of $E$. I'd like to have some hints.

Answer 1

Given a space $E$ equipped with a cell decomposition for which it has the weak topology and the boundary of every cell is contained in the union of the lower-dimensional cells, closure-finiteness is equivalent to the statement that for each $n$, the $n$-skeleton $E^n\subseteq E$ (i.e., the union of the cells of dimension $\leq n$) has the weak topology. Indeed, (a paraphrase of) this latter condition is often taken as the definition of a CW-complex instead of closure-finiteness (for instance, this is the definition in Hatcher's Algebraic Topology; he proves the equivalence with the closure-finite definition as Proposition A.2 in the Appendix).

Given that you have already shown any covering space of a CW-complex has the weak topology, this is now easy: the $n$-skeleton $E^n$ is a covering space of the $n$-skeleton $X^n$ of $X$, and $X^n$ is also a CW-complex.

More directly, given that each $E^n$ has the weak topology, you can prove closure-finiteness by induction on the dimension of the cells: if you know that closure-finiteness holds for cells of dimension $\leq n$, then you know that $E^n$ is a CW-complex. It follows that the attaching map $\partial D^{n+1}\to E^n$ of any $(n+1)$-cell intersects only finitely many cells of $E^n$, since any compact subset of a CW-complex is contained in finitely many cells.