I want to solve this problem that I found in a qualifying topology exam:
"Let $S^1$ be the unit circle in the complex plane. How many isomporphism classes of connected covering spaces of $S^1$ exist? Construct a representant of each class."
The following theorem is taken from Hatcher's Algebraic Topology:
So, this theorem could help to find the set of basepoint-preserving isomorphism classes of path-connected covering spaces $p:(\widetilde{X},\widetilde{x}_0)\to (S^1,1)$ where $S^1$ is the unit circle and $1=(1,0)$, and we have $\pi_1(S^1,1)=\Bbb Z$, but how can we know what $\pi_1(p_*(\widetilde{X},\widetilde{x}_0))$ is if we don't know $(\widetilde{X},\widetilde{x}_0)$? Also, this theorem helps to find path-connected covering spaces, but the problem is asking for connected covering spaces, so, is this theorem useful to solve this problem or not? And finally, how can the covering spaces be represented? I guess it's by a permutation because that's the next topic in Hatcher after this theorem, but how can we do that?
Best Answer
The good news is that we do know a universal cover of $S^1$, namely $\mathbb{R}\to S^1$ given by $x\mapsto e^{ix}.$ It is easy enough to see that this is a covering map, and we know that $\mathbb{R}$ is simply connected. We also know that the fundamental group $\pi_1(S^1,1)\cong \mathbb{Z}$ acts on $\mathbb{R}$ by deck transformations, namely topological automorphisms of $\mathbb{R}$ that preserve the projection $p:\mathbb{R}\to S^1$.
Now, another important fact here is that $S^1$ is a smooth manifold, and the covering spaces of smooth manifolds are also smooth manifolds. On a smooth manifold, the topology is regular enough that connectedness is equivalent to path connectedness. So, the path connected covering spaces of $S^1$ are exactly the connected covering spaces of $S^1$. In particular, the theorem applies.
We know the subgroups of $\mathbb{Z}$, they are of the form $n\mathbb{Z}$ for $n\in \mathbb{N}$. Because $\mathbb{Z}$ is Abelian, the conjugacy classes of subgroups are in bijection with the subgroups themselves. The connected covers corresponding to each subgroup are obtained by quotienting $\mathbb{R}$ by the suitable subgroup under the $\pi_1(S^1,1)$ action. So, if we take $n\mathbb{Z}$, the quotient $\mathbb{R}/n\mathbb{Z}\cong S^1$ (topologically) and the associated covering map is $q_n: S^1\to S^1$ given by $q_n(\theta)=\theta^n$, where the elements of $S^1$ are viewed as a elements of $\mathbb{C}$.