Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.