Questions about Covering maps, manifolds, compactness

[PsychonautQQ]

Suppose $p: C-->X$ is a covering map.

a) If $C$ is an n-manifold and $X$ is Hausdorff, show that $X$ is an n-manifold.
b) If $X$ is an n-manifold, show that $C$is an n-manifold
c) suppose that $X$ is a compact manifold. Show that $C$ is compact if and only if p is a finite sheeted covering.Thoughts:

a) I will have to show that $X$ is second countable and has a basis of n-euclidean balls. To show that it has a basis of n-euclidean balls, I could take the pre image of any basis element of $X$ and then use any component of the preimage which will be homeomorphic to both a euclidean n-ball and the basis element in $X$. To show that $X$ is second countable I don't know what I would do yet.

b) The key here is using the fact that slices of the preimage of $p$ will map homeomorphically to their source neighborhood in $X$. Perhaps if I can show that the preimage of basis elements of $X$ is surjective onto $C$ the proof will not be far away.

c) I could suppose $C$ is compact and then proceed by contradiction: Take the preimage of any neighborhood in $X$ and use this infinite preimage to construct an open covering of $C$ with no finite sub covering. I am hesitant about this strategy because I haven't seen how I can use the fact that $X$ is a compact manifold.

Any feedback is appreciated as always! Thanks PF!

 

[andrewkirk]

For (a) and (b) we need to prove the second-countability and locally Euclidean properties. For (b) we also need to prove C is Hausdorff. So, five things to prove overall between the two sub-questions. I found three of them not too hard.

For (a) a natural suggestion for a countable basis for X would be $\mathscr B^X=\{f(B)\ :\ B\in\mathscr B^C\}$ where $\mathscr B^C$ is a countable basis for C. Then take an open set U in X and show it can be written as a union of elements of $\mathscr B^X$.

My proof of that uses the axiom of choice to first express X as a union of open sets each of which is homeomorphic to a stack of homeomorphic images in C. From there on it is easy. I am pondering whether there's a version that doesn't use Choice.

For (b) the hard bit will be showing second-countability. The trouble with just considering pre-images of basis elements of X is that, if we split them up into their connected components, there may be uncountably many since, as I recall, there is nothing in the definition of covering space to say that the stack of slices above an open set in X can't be uncountable. But if we don't split them up then we may lose Hausdorffness of C since we may no longer be able to separate different slices by open sets.

So I currently have no answer for second-countability in (b) and only a Choice-tainted answer for second-countability in (a). But I can prove the other bits. Let me know if you want any hints for them.

I haven't thought about (c) yet. 'If and only if' questions make me feel tired, because one has to prove BOTH directions.

 

[math771]

a) To show that X is second countable, we can use the fact that C is an n-manifold and thus has a countable basis of n-euclidean balls. Then, since p is a covering map, for any x in X, there exists an open neighborhood U of x that is evenly covered by p. By the definition of a covering map, this means that the preimage of U under p is a disjoint union of open sets in C, each of which is homeomorphic to U. Since C has a countable basis of n-euclidean balls, the preimage of U can be written as a countable union of n-euclidean balls, which forms a basis for X. Thus, X is second countable.

b) To show that C is an n-manifold, we can use the fact that X is an n-manifold and p is a covering map. Since X is an n-manifold, for any x in X, there exists an open neighborhood U of x that is homeomorphic to an open subset of R^n. Since p is a covering map, the preimage of U under p is a disjoint union of open sets in C, each of which is homeomorphic to U. Thus, for any x in C, there exists an open neighborhood V of x that is homeomorphic to an open subset of R^n. Therefore, C is an n-manifold.

c) Suppose p is a finite sheeted covering. This means that for any x in X, the preimage of x under p has a finite number of points. Since X is compact, any open covering of X has a finite subcovering. Therefore, if we take an open covering of C and map it to X using p, the finite number of points in the preimage of each point in X will give us a finite subcovering of C. Thus, C is compact.

Now suppose p is not a finite sheeted covering. This means that there exists a point x in X whose preimage under p is infinite. Since X is compact, any open covering of X must have a finite subcovering. However, if we take an open covering of C and map it to X using p, the infinite preimage of x will prevent us from finding a finite subcovering of C. Therefore, C is not compact.