Proving conditions for a covering map to be a homeomorphism

[PsychonautQQ]

Homework Statement

Let p: E-->B be a covering map. Let E be path connected and B be simply connected. Prove that p is a homeomorphism.

Homework Equations

The Attempt at a Solution

I'm really struggling with this. Can anyone give me any insights? B is simply connected so any two paths with the same end points are homotopic and it's fundamental group is trivial. Help please?

 

[andrewkirk]

I don't know whether this will lead anywhere, but it seems to me the only difference between a covering map and a homeomorphism is that a covering map is not injective, as both are continuous, open, surjective and have local continuous inverses.

So we might try a proof by contradiction: assume there are two points e1 and e2 in E that map via p to the same point b in B and try to derive a contradiction, eg that E is not path-connected or that B is not simply-connected, or that p is not continuous or not open. I'd be tempted to use the path-connectedness of E and think about paths between e1 and e2. Their images under p will be loops in B and, since B is simply-connected, they can be homotopically shrink to the constant map with image {b}. What happens to the pre-images of those loops as we use a homotopy to shrink them in B? I have the feeling something will break in E.

 

[fresh_42]

Does covering map mean something more than the usual bundle projection?

 

[PsychonautQQ]

A covering map is a continuous surjective map p: E--B such that for every point in B there exists a neighborhood U such that it's preimage is the disjoint union of open sets in E such that each 'slice' is homeomorphic to U

 

[fresh_42]

A covering map is a continuous surjective map p: E--B such that for every point in B there exists a neighborhood U such that it's preimage is the disjoint union of open sets in E such that each 'slice' is homeomorphic to U

Yes. And what prevents us from taking usual Euclidean spaces as $E$ and $B$ and an ordinary linear projection $p\,$? How are $E$ and $B$ supposed to be homeomorphic?

 

[PsychonautQQ]

Hm, I'm not completely sure what you're getting at, but i'll expand on what I was thinking and maybe it will relate.

Somehow the fact that B is simply connected and E is path connected means that the preimage of p will be a bijection; normally the preimage of a covering map contains multiple 'slices'. The thing is I think of simply connected strictly in terms of paths, ie any two paths in a simple connected space with the same end/initial points are path homotopic. So when I'm trying to show that p^-1 is a bijection by showing how it acts on an arbitrary neighborhood U, i feel like I'm missing something, like I need to get paths involved somehow.

Back to what you said though, I'm not completely familiar with what a linear projection of Euclidan spaces is exactly, I mean I'm sure I have an idea... I guess I think of Euclidian space as being R^2 or R^3, and a linear projection would be maybe like a sphere projecting to a circle... Hot damn maybe I do more about those words than I thought hahaha. Anyway,if what I'm saying isn't complete bolderdash then for a linear projection to be a homeomorphism we'd need the preimage and image to be the same thing? in my example at least... Like the ball B^2 in R^3 is projected to R^2 would be a homeomorphism I presume.

Hope I said something relevant in these paragraphs.

 

[fresh_42]

My objection is, $p \, : \, \mathbb{R}^3 \longrightarrow \mathbb{R}^2$ defines a smooth projection of a Riemannian manifold $\mathbb{R}^3=\mathbb{R}\times \mathbb{R}^2$ above the basis plane with real lines as fibers. All vector spaces are path and simple connected. I admit it is a rather trivial fiber bundle, but nevertheless an allowed one. So where do you see a homeomorphism?

 

[PsychonautQQ]

Yeah, so I don't know all those words yet, but Uhm, I see a homeomorphism I guess if we restrict the map to a plane in R^3, say we fix the Z value... then p(x,y,c) = (x,y) where c is a constant is a homeomorphism

 

[fresh_42]

Yeah, so I don't know all those words yet, but Uhm, I see a homeomorphism I guess if we restrict the map to a plane in R^3, say we fix the Z value... then p(x,y,c) = (x,y) where c is a constant is a homeomorphism

Yes. This is called a section, the slice you mentioned above. I simply wanted to point out, that connectedness properties cannot change the dimensional difference between the total space $E$ and the basis space $B$, which would be needed for both to be bijectively related.

 

[andrewkirk]

@PsychonautQQ D'oh, I just realized! Doesn't this theorem follow as a fairly easy corollary to the one you just completed here? How many elements are there in the fundamental group of B? How might we use that together with the earlier theorem to work out how many elements there are in the pre-image under p of a point in B?

 

[PsychonautQQ]

Yes. This is called a section, the slice you mentioned above. I simply wanted to point out, that connectedness properties cannot change the dimensional difference between the total space $E$ and the basis space $B$, which would be needed for both to be bijectively related.

Oh, thanks for the great insight, I really appreciate it :D.

But doesn't this insight contradict what I'm trying to prove? I'm trying to show that a covering map p: E-->B Is a homeomorphism and thus bijective if B is simply connected and E is path connected, so the connectedness property has something to do with the mapping being a bijection, right?

 

[PsychonautQQ]

@PsychonautQQ D'oh, I just realized! Doesn't this theorem follow as a fairly easy corollary to the one you just completed here? How many elements are there in the fundamental group of B? How might we use that together with the earlier theorem to work out how many elements there are in the pre-image under p of a point in B?

Because B is simply connected the fundamental group is trivial, thus is consists of one equivalence class where the elements of the equivalence class are all path homotopic to a constant map. So since every loop in the fundamental group is path homotopic, if we take two loops in the fundamental group of B and lift it to a path in E, these lifting's will also be path homotopic and end at the same point by uniqueness of path homotopy liftings.. Is this all correct?

I still am missing something, but now I can smell the bijection I think. I just need to relate these path liftings to the preimage of the covering map.

let f and g be lifted to paths r and s, then p*r=f and p*s=g, hmm...

 

[andrewkirk]

Because B is simply connected the fundamental group is trivial

That's a good start, but you haven't gone on to use the result of that other theorem I linked. Using that theorem, we don't even need to think about homotopies. That theorem says that, for any point $b_0$ in B, there's a bijection between a subset of the fundamental group of B and $p^{-1}(b_0)$. So what can we say about the size of $p^{-1}(b_0)$, what does that tell us about $p^{-1}(B)$ and what sort of map does that tell us that $p$ is?