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Problem: Let $X$ be a connected CW-complex, and $Y$ be a connected topological space. Suppose $p: X\to Y$ is a covering map. Does there exist a CW-structure on $Y$? More generally, is $Y$ homotopically equivalent to a CW-complex?

Note that the other direction is clear: Any covering of a connected CW-complex can always be given a CW-structure, lifting the characteristics map of cells of base space such that the covering map is a cellular map.

I believe that the answer to the above problem is no, but I have no counterexample.

$\bullet$ Notice that $Y$ is locally path-connected as the covering map is a local homeomorphism, hence $Y$ is path-connected also. So, we can not consider spaces $\{0\}\cup\left\{\frac{1}{n}:n\in\Bbb N\right\}$ or Topologist Sine Curve as $Y$. Notice that both $\{0\}\cup\left\{\frac{1}{n}:n\in\Bbb N\right\}$ or Topologist Sine curve are not homotopically equivalent to a CW-complex.

$\bullet$ Similarly, we can not consider the Hawaiian Earring(this is not semi-locally simply connected) as $Y$: The connected CW-complex $X$ has the universal cover so that $X$ is semi-locally simply connected, but the property "semi-locally simply connected" is preserved under a local homeomorphism.

So, I am run out of examples. Any help will be appreciated. Thanks in advance.

Sumanta
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1 Answers1

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Andrew Ranicki conjectured (or at least asked) if a compact, 4 manifold has a CW structure, if and only if, it is smoothable.

As far as I know this is open, but assuming the conjecture is true the answer to your question is that a CW complex may cover a non CW complex. Let $M$ be a nonsmoothable 4-manifold with infinite fundamental group, these are known to exist. Let $\widetilde{M}$ denote its universal cover. $\widetilde{M}$ is noncompact, and it is known that noncompact 4-manifolds are triangulated. Hence, $\widetilde{M}$ has a CW structure and it covers $M$ which is not a CW complex, if Ranicki's conjecture is true.

In this case, the base of this cover is homotopy equivalent to a CW complex since all compact manifolds are homotopy equivalent to CW complexes.

Connor Malin
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  • Thanks for sharing all these. – Sumanta Jan 13 '21 at 02:33
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    @SumantaDas No need to give the check mark; I've not given a complete answer. I suspect there is a nonmanifold counterexample, but as you point out it is difficult because one must find a global obstruction to being a CW complex which is very hard to do. – Connor Malin Jan 13 '21 at 03:50