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.


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.

  • $\begingroup$ Your direct proof is really of help to me! $\endgroup$ – Jeong Jul 28 '15 at 1:45
  • $\begingroup$ The last paragraph where you try to prove it by induction, it looks as if you were dropping the finiteness condition as in Proposition A.2. (ii) in Hatcher's book which cannot be done: consider the example given by Hatcher p.521 where $X=D^2$, all of the points in the boundary are 0-cells and the interior is the 2-cell. More concretely: The part where you say "then you know $E^n$ is a CW-complex" looks wrong, for example consider the example given. The main answer, however, looks good. $\endgroup$ – Zero Jun 19 '16 at 2:00
  • $\begingroup$ @Zero: In the context of the last paragraph, it is assumed you already know each $E^n$ has the weak topology, as in the previous paragraph. I have edited to make this explicit. $\endgroup$ – Eric Wofsey Jun 19 '16 at 3:59
  • $\begingroup$ One minor question that I'm not able to understand at the moment, why is it that the $E^n$ skeleton is a covering space of $X^n$? $\endgroup$ – Leo Lerena Apr 22 at 23:45
  • 1
    $\begingroup$ @LeoLerena: In general, if $p:A\to B$ is any covering map and $C\subseteq B$, then the restriction of $p$ to $p^{-1}(C)\to C$ is also a covering map. Here we're restricting $q:E\to X$ to $X^n$ and $q^{-1}(X^n)=E^n$. $\endgroup$ – Eric Wofsey Apr 23 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.