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I got stuck at some point while working on this part of an exercise from Lee's Introduction to Smooth Manifolds, 2nd edition.

The part which I am stuck on is to prove (one of the directions of proposition 4.33(c)): A topological covering map is a smooth covering map if it is a local diffeomorphism.

What I have is as follows. Let $\pi : E \to M$ be a topological covering map. Let $q\in M$. By assumption, we have some open subset $V \subset M$ and collection of disjoint open subsets $\{ U_\alpha \}$ of $E$ such that each $U_\alpha$ is homeomorphic to $V$ under $\pi$. Further, let $p_\alpha \in U_\alpha$ be such that $\pi(p_\alpha)=q$. Now, for each $\alpha$, I can find open neighbourhoods $\hat{U}_\alpha$ of $p_\alpha$ and $\hat{V}_\alpha$ of $q$ such that $\hat{U}_\alpha$ and $\hat{V}_\alpha$ are diffeomorphic. We may demand that $\hat{U}_\alpha \subset U_\alpha$.

The problem is I need one single neighbourhood of $q$ that lifts diffeomorphically, but all I have is the collection $\hat{V}_\alpha$ and taking their intersection does not necessarily give an open set.

How should I correctly approach the problem?

An alternative idea I have: Repeat the first three lines of the first proof. Now for some $\alpha '$, define $\hat{U}_{\alpha '}$ and $\hat{V}_{\alpha '}$ as above. Then lift $\hat{V}_{\alpha '}$ using $\pi^{-1}$ to all other $U_\alpha$ to obtain $\hat{U}_\alpha$. This time I know that each $\hat{U}_\alpha$ is smoothly homeomorphic to $\hat{V}_{\alpha '}$.

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  • $\begingroup$ Here wj32.org/wp/wp-content/uploads/2012/12/… the author itself uploaded his solutions. Unfortunately it explain this exercise (4.33) with "obvious". $\endgroup$ Jul 9, 2017 at 14:50
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    $\begingroup$ @TommasoSeneci: I need to point out that those solutions were not uploaded by "the author itself." The ones you linked to were written by Wen Jia Liu. I haven't ever posted solutions to the problems in my books. $\endgroup$
    – Jack Lee
    Feb 20, 2022 at 0:02

2 Answers 2

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The key is that a bijective local diffeomorphism is a diffeomorphism. So once you're finished those first three lines of your proof, use this fact, and then actually each $U_\alpha$ is diffeomorphic to $V$ under $\pi$. See here or here on math.SE for that fact.

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  • $\begingroup$ Revisiting this as I realised that this is also described later in exercise 4.39. Maybe the exercise should go before 4.33. Anyway exercise 4.38 is also used in example 4.35, so maybe this section of exercises should be moved earlier. ;) $\endgroup$
    – suncup224
    Nov 28, 2020 at 3:02
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Using the reference added in the comment and aes suggestion I found this way to prove it. Let $\pi:E\to M$ be a (topological) covering map. Take $q\in M$, there exists a neighborhood $U\subset M$ such that each connected component of $\pi^{-1}(U)$ is homeomorphic to $U$. Let $V$ be an arbitrary component. The map $\pi|_V$ is bijective (because homeomorphic) and locally diffeomorphic, therefore it is a diffeomorphism $\pi|_V:V\to U$. This shows that $\pi$ is diffeomorphic on each component and then it is a smooth covering map.

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