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Let $p: C \to X$ is a covering map. Suppose that $C$ is a differentiable manifold. Is X - differentiable manifold?

More precisely, I am interested in the case where $C$ is Submanifold of Lie algebra, $p$ is the exponential map, and $X= Im \, p$.

Alex-omsk
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    There is some categorical weirdness here. Strictly speaking $p$ is only continuous (plus rest of definition of covering map) and not smooth because a priori we don't know $X$ has a differentiable structure. Is the question, can we endow $X$ with a smooth structure? Or is the question is there a smooth structure so that $p$ is a smooth covering map? Maybe this amounts to the same question? – Matt Apr 03 '13 at 02:49

2 Answers2

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Apparently it is not the case that $X$ must be a differentiable manifold. In

D. Ruberman. Invariant Knots of free involutions of $S^4$, Top. Appl. 18 (1984), 217-224

Ruberman shows, among other things, the existence of a topological manifold $X$ which is homotopy equivalent but not homeomorphic to $\mathbb{R}P^4$. Further, $X$ is not smoothable.

On the other hand, the universal cover $C$ of $X$ will be a compact simply connected topological manifold with $\pi_2(C) = \pi_2(X) = 0$ (since $X$ is homotopy equivalent to $\mathbb{R}P^4$ and $\pi_2(\mathbb{R}P^4) = 0$). By Freedman's classification, this implies $C$ is homeomorphic to $S^4$. In particular, $C$ can be given the structure of a smooth manifold.

As far as you "more precisely", I'm not sure what you mean. Generally, $\exp$ is not a covering map.

Jason DeVito - on hiatus
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$\color{red}{\text{Beware:}}$ I am quite unsure what I had in mind when I wrote this. Right now I cannot see how to deal with Neal's objection in the comment! I'll leave this here to record the gap.

Yes, there is a canonical smooth structure on $X$ characterized by the fact that the map $p$ becomes, with respect to it, a local diffeomorphism.

You can construct its charts by composing charts out of $C$ with local sections of $p$.

Mariano Suárez-Álvarez
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    I think there is some ambiguity here in the choice of local sections of $p$ when checking that transitions are smooth. What if the deck transformations are not diffeomorphisms? – Neal Apr 03 '13 at 04:31
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    Dear Mariano, don't you just mean "by composing charts out of $C$ with restrictions of $p$" ? (The maps you mention cannot be composed) – Georges Elencwajg Apr 03 '13 at 12:08
  • For example, if there were an unsmoothable $K(\pi,1)$ topological manifold, its universal cover would be diffeomorphic to Euclidean space. – Neal Apr 04 '13 at 01:39
  • Mariano: Since both of our answers can't be correct, would you mind taking a look at mine for any possible errors? I admit I'm somewhat outside of my comfort area with my answer, so the mistake likely lies on my end - I just can't see it. Also, how do you answer Neal's (first) objection? Thank you! – Jason DeVito - on hiatus Apr 18 '13 at 03:29
  • I honestly do not recall what I was thinking, exactly :-/ I'll add a warning! – Mariano Suárez-Álvarez Apr 18 '13 at 03:41