Connected covering spaces of the 2-sphere with a diameter and a disk.

Question

Consider the unit sphere $S^2 \subset \mathbb R^3$, the $xy$-plane unit disk $D = \{(x, y, 0) \in \mathbb{R}^3 \mid x^2 + y^2 \leq 1\}$, and the line segment $\ell = \{(0, 0, z) \in \mathbb R^3: |z| \leq 1 \}$, and let $X = S^2 \cup D \cup \ell$.

I want to find the possible connected 2-fold covers of $X$. There is a similar thread on StackExchange where there is not a disk involved. I feel as though I understand the solution there relatively well. I'll spell it out in case the link breaks:

A 2-sphere with a diameter (and no disk) is homotopy equivalent to $S^1 \vee S^2$ and the double cover must be homotopy equivalent to $S^2 \vee S^1 \vee S^2$. We can apply further homotopies to arrive at a covering space given by two spheres $S_1, S_2$ and two line segments $\ell_1$ and $\ell_2$, where $\ell_1$ connects the north pole of $S_1$ to the south pole of $S_2$, and $\ell_2$ connects the north pole of $S_2$ to the south pole of $S_1$.

This is fine and all, but I'm having a really difficult time when we add the inner disk. I have found that $X$ is homotopy equivalent to $S^1 \vee S^1 \vee S^2 \vee S^2$ by contracting the inner disk to get a wedge of 2-spheres with line segment going through the north and south poles of each 2-sphere; this space is then homotopy equivalent to $S^1 \vee S^1 \vee S^2\vee S^2$.

Taking the 2-fold cover of $S^1 \vee S^1 \vee S^2 \vee S^2$, I get $(\bigvee_{i=1}^3 S^1) \vee (\bigvee_{j=1}^4 S^2)$. More clearly, I get $S^1 \vee S^1 \vee S^1$ with two copies of $S^2$ at each intersection of the $S^1$'s.

How do I then apply a homotopy to get something which covers $X$? Everything I have tried turns out not to be a cover.

Answer 1

I believe I have solved my own question. Here is the covering space I found; basically, one needs to move the equators outside of the spheres while still intersecting the diameters. Here is a picture. I don't know if there is another connected 2-sheeted covered which is not homeomorphic to this.

Answer 2

Note that in your solution, there exists a pair of points with neighbourhoods homeomorphic to an open interval, which when removed, leave behind two disjoint pieces, each with non-trivial second homology group. Namely pick one point from each of the arcs joining the "warted spheres".

The following solution has no such pair of points, so is not homeomorphic to your example.

It is easy to see that there are no further examples up to homeomorphism, by following through on the comment by @MoisheKohan:

Let $a\in\pi_1(X)$ be represented by the loop starting at the origin, going up along the interval to the north pole, going along the sphere to the equator, and returning to the origin along the disk. Similarly let $b\in\pi_1(X)$ be represented by the loop starting at the origin, going down along the interval to the south pole, going along the sphere to the equator, and returning to the origin along the disk.

Then $\pi_1(X)$ is freely generated by $a,b$ and the double covers of $X$ correspond to the three epimorphisms $$\pi_1(X)\to \mathbb{Z}/2\mathbb{Z}.$$

The epimorphism mapping $a\mapsto 1, b\mapsto 0$ corresponds to the cover where lifting a based loop homotopic to $a$ takes you to a different point to where you started the lift (the other point covering the origin), whilst lifting a based loop homotopic to $b$ returns you to your starting point.

This cover is the one you drew in your solution.

By symmetry of $a$ and $b$, the cover corresponding to the epimorphism mapping $a\mapsto 0, b\mapsto 1$ is homeomorphic to the first solution. Here by a homeomorphism of covers we mean a pair of homeomorphisms - one between covering spaces and one between bases (not necessarily the identity), which commute with the covering maps.

The solution drawn above corresponds to the final epimorphism, mapping $a\mapsto 1, b\mapsto 1$. Here following the lift of either $a$ or $b$ takes you to a new point.

Thus there are precisely two non-homeomorphic double covers of $X$.