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Here's a problem I've had a hard time with

If $f: M\rightarrow N$ is a cover map and $M$ is a m-manifold, will $N$ also be a m-manifold? A manifold is a space locally Euclidean space that is Hausdorff and Second Countable.

What if $N$ were a n-manifold, will $M$ also be a $n$-manifold?

Since I'm very new to topology, a heuristic explanation would be enough. I have some idea as to what the answers would be but I am not completely sure.

mathemagician
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    If $f$ is a cover map, there exist open sets $U\subset M$, $V\subset N$, which satisfies $f:U\rightarrow V$ is homeomorphism. Because homeomorphic manifold must have same dimensions, $\dim M=\dim U=\dim V=\dim N$. – gaoxinge Oct 06 '14 at 13:18

2 Answers2

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Suppose $f\colon M\to N$ is a covering map. If $N$ is a (topological) $n$-manifold, then $M$ is too. The converse, however, is true only with the additional assumption that $N$ is Hausdorff.

Here's a counterexample in which $N$ is not Hausdorff. Let $M = \mathbb R^2\smallsetminus \{(0,0)\}$, and let $N$ be the quotient space of $M$ by the equivalence relation generated by $$ (x,y) \sim (2^n x, 2^{-n}y), \qquad n\in\mathbb Z. $$ Then the quotient map $f\colon M\to N$ is a covering map, but $N$ is not Hausdorff, so it's not a manifold.

To see that $N$ is not Hausdorff, it's useful to look at the equivalence classes of $(1,0)$ and $(0,1)$.

Jack Lee
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  • Do you know of a reference for the converse (when $N$ is Hausdorff)? Also, my understanding is that the converse must read "If $M$ is a smooth or topological $n$-manifold and $N$ is Hausdorff, then $N$ is a topological $n$-manifold." That is, smoothness of $M$ doesn't imply that $N$ has a smooth structure at all, much less one for which the covering map becomes smooth. (See this MSE question: http://math.stackexchange.com/questions/349693/a-covering-map-from-a-differentiable-manifold) – Jason DeVito - on hiatus Oct 06 '14 at 18:11
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    @Jason, you're right about smoothness -- I was a little sloppy in my original post. I've deleted the reference to smoothness. But if you're interested, one correct converse in the smooth category is "if $M$ is a smooth manifold, the covering automorphism group acts smoothly on $M$, and $N$ is Hausdorff, then $N$ is a smooth manifold." – Jack Lee Oct 06 '14 at 18:22
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    @Jason, I don't know offhand of a reference where the converse is proved in detail. But in the topological category, it's not hard to put together a proof. First, because $f$ is a local homeomorphism, $N$ is locally Euclidean. Second, because $f$ is an open map, it takes a countable basis for $M$ to a countable basis for $N$, so $N$ is second-countable. You can find more details in my Introduction to Topological Manifolds. – Jack Lee Oct 06 '14 at 18:24
  • Both posts make perfect sense. Thank you. – Jason DeVito - on hiatus Oct 06 '14 at 18:27
  • Will it always be true that if $N$ is a manifold then $M$ is too? – mathemagician Oct 07 '14 at 01:39
  • @mathemagician: Yes, if $N$ is a topological manifold, then $M$ is too. (In the smooth category, if $N$ is a smooth manifold and $f$ is a smooth local diffeomorphism, then $M$ is also a smooth manifold.) – Jack Lee Oct 07 '14 at 04:40
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Being a manifold is at least partly a local property. This is of course given, since a covering map is locally a homeomorphism and hence looks locally as the Euclidean space! Write this down explicitely to understand it, please. But of course this observation yields the dimension of $M$ which is defined locally!

If $M$ would not be necessary connected, it might fail to be a manifold since you could take an uncountable amount of copies of $M$ to cover $N$. This would fail to be second countable. But reasonable covering spaces of manifolds are manifolds again, since they correspond to a countable indexed subgroup of $\pi_1(N)$ (since $N$ is a manifold) and hence you can pull a countable basis back via the covering map.

Note that the Hausdorff is preserved by coverings. (as well the locally euclidean property as mentioned above, which is trivial being a local property). Indeed, take two points on $M$ and observe whether a local trivial neighborhood of both seperates them, otherwise they will be in different sheets. I leave the details to you.

Make sure you prove the first and last part for yourself in detail, it will be a great help.

Daniel Valenzuela
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