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This is a question from an old exam paper (Question 21G); regular covering maps were not lectured this year so I'm having a bit of trouble with it. (The material is not examinable this year, so this question is purely out of interest.)

Firstly, what precisely is a "connected covering map" $p : X \to Y$? I imagine it means $X$ is at least connected, but the definition of "regular covering map" I found requires $X$ to be locally path-connected as well. In any case, my intuition tells me that the claims are probably false if we don't have some kind of path-connectedness.

I think I have a proof that a 2-to-1 connected covering map is regular: fix $x \in X$, let $y = p(x)$ and let $\tilde{x} \ne x$ be the other preimage of $y$ under $p$. There is an open neighbourhood $U \subset X$ of $x$ such that $p(U)$ is evenly covered and by taking $U$ small enough we may assume $U$ is path-connected. Let $\tilde{U}$ be the homeomorphic copy about $\tilde{x}$. Then, for every $x' \in U$, there is a path $u : I \to U$ from $x$ to $x'$, and we can uniquely lift $p \circ u$ to a path $\tilde{u} : I \to \tilde{U}$ from $\tilde{x}$ to some $\tilde{x}'$. This defines a homeomorphism $f : U \to \tilde{U}$, and we can extend to a well-defined homeomorphism $X \to X$ since $X$ is connected and each fibre has exactly 2 points. By construction this $f$ is the deck transformation which demonstrates that $p$ is regular.

However, the proof that $p$ is a regular cover when $\pi_1 (Y, y)$ is abelian eludes me for the moment. I imagine this has something to do with Galois coverings or somesuch, but the lecturer this year mostly focused on universal covers. I'm looking for a complete proof of this claim, but hints would also be appreciated.

I can construct two triple covers of $S_1 \vee S_1$, and I believe they're not homeomorphic. Both are $S_1 \vee S_1$ with a single circle glued to it at two points; one is glued to both halves of $S_1 \vee S_1$, and the other is glued to the same half twice. I think the former is a regular cover, and the latter is not. Am I correct?

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Usually coverings are discussed with the hypothesis that the base space $Y$ is connected, locally path-connected and semi-locally simply connected (which is the usual hypothesis one uses to prove the universal cover exists) –  Mariano Suárez-Alvarez May 6 '11 at 21:41
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To prove anything interesting about covering spaces you need to assume X is connected and locally path-connected as well. Otherwise you can't get uniqueness anywhere. –  JSchlather May 6 '11 at 22:34
    
Okay I believe that a covering map $p: X \rightarrow Y$ is normal if $p_\ast(\pi_1(X,x))$ is a normal subgroup of $\pi_1(Y,p(x))$. Which reduces a,b to group theory questions. –  JSchlather May 6 '11 at 22:37
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There is a close relation between the group of covering transformations and the image $H = p_\ast \pi_1(X,x)$ of the fundamental group of $X$ in $\pi_1(Y,y)$ (where $y = p(x)$). One of these remarkable facts - which can be found in Munkres' "Topology" - states that $H$ is a normal subgroup of $\pi_1(Y,y)$ iff the covering is regular. –  Sam May 6 '11 at 22:51

1 Answer 1

So I'm going to do this first question in terms of group theory. Let $H= p_\ast \pi_1(X,x)$ and $G=\pi_1(Y,y)$. As Sam mentioned one definition of a covering map being regular is if $H$ is normal in $G$. We'll go off of that definition.

If $p$ is a $2$-$1$ map then $p$ is regular.

Observe that since that the index of $H$ in $G$ is equal to the cardinality of the fiber, which in this case is $2$. We have that the index of $H$ in $G$ is 2 and it's a basic result of group theory that if a subgroup has index $2$ then it is normal. So $H$ is normal in $G$ as desired.

If $\pi_1(Y,y)$ is abelian then $p$ is regular.

This is immediate because every subgroup of an abelian group is normal.

For the second part, I think the easiest example of a normal covering of $S^1\vee S^1$ is to take three points and to join each of them with two line segments. For a non normal covering, draw a triangle and then at one vertex draw a circle. In between the other vertex draw two lines, this is a covering that is not normal.

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Thank you, but the definition of regular covering map I was working with is the one where the group of deck transformations acts transitively on the fibres. –  Zhen Lin May 7 '11 at 19:37
    
@Zhen: That these conditions are equivalent when $Y$ satisfies the usual properties is not so hard to prove. See, for example, proposition 1.39 in Hatcher's book. –  Dylan Moreland Jul 6 '11 at 5:52
    
Hatcher also has a lot of covering spaces of $S^1\vee S^1$ starting on page 58. –  Joe Johnson 126 Jul 6 '11 at 6:10

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