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?
Best Answer
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.