What is the univesal cover of 2-sphere for which the two poles are connected to each other with a line?

Question

How should i compute the universal cover of 2-sphere for which the two poles are connected to each other with a line? I think it should be $ S^2 $ but it looks like a ruined doughnut and the universal cover for doughnut is $ R^2$

Answer 1

You are correct about the space looking like a ruined donut.

The universal covering space of this is infinitely many spheres (in both directions), such that the North pole of the i$^{\text{th}}$ sphere is glued to the South pole of the (i+1)$^{\text{st}}$ sphere. The covering map just maps each sphere down.

Edit: As Roy says, this space is only homotopy equivalent to the universal cover. Inserting a line segment between each sphere gives the correct answer.

Answer 2

do you know the universal covering space of a circle? it is obtained by cutting the circle apart to make an interval, and then joining together infinitely many intervals end to end. Your example is homotopy equivalent to what you get by contracting the line to a point, hence to a space obtained by identifying the two poles of a sphere, hence a torus with a loop shrunk to a point. This is a little like a fat circle. thus the universal covering space is obtained as Max Nagy says, by separating this space at the point where the poles are joined, and then sticking infinitely many of them together end to end, or pole to pole. but technically, since your space had a line joinging the two poles, in the universal cover we should introduce also a line joining every consecutive pair of poles. So I think the answer is what Max Nagy said but with this extra family of line segments added in. This is homotopy equivalent to Max Nagy's answer.

Answer 3

The sphere with two points identified with a line is homotopy equivalent to the sphere with two points identified, which is in turn $S^2 \vee S^1$. You can see the latter claim by placing the attaching line outside of the sphere and shrinking it down to a loop.

With this, you can construct the universal cover as a cayley graph free on two generators, but each node is a sphere.

Answer 4

Rather than going through homotopy equivalent spaces and then "undoing" the homotopy equivalence to get the actual covering space, here's a way to see what's going on directly, i.e., working with homeomorphisms. For purposes of visualization, imagine the original space as an ordinary sphere $S$ sitting in 3-dimensional Euclidean space, and imagine the line $L$ joining the two poles as being drawn outside the sphere (where it's easier for me to see). Imagine also a meridian $M$ of the sphere, an arc in the sphere joining the two poles. Notice that $L\cup M$ is topologically a circle, which I'll call $C$ for short. So your space can be described as the result of taking a circle $C$ and gluing it to a sphere $S$ by identifying an arc $M$ on the sphere with an arc of the circle.

Gluing a 2-sphere to a (nice) space along a contractible set doesn't affect the fundamental group, so to construct the universal cover, the main task is to get the universal cover of $C$; the $S$ part will be easy to handle later.

The universal cover of a circle $C$ is the real line $\mathbb R$, and the arc $M$ in $C$ corresponds, under the covering map, to an infinite sequence of closed intervals $M_n\subseteq\mathbb R$ ($n\in\mathbb Z$) tending to infinity in both directions.

Each $M_n$ needs to have a copy of $S$ attached to it. These copies are literally homeomorphic to the sphere $S$ because $S$ has trivial fundamental group so it can be covered only by disjoint unions of homeomorphic copies of itself. So we take a sequence of copies $S_n$ of $S$, and we attach each of them to $\mathbb R$ at the corresponding $M_n$, identifying $M_n$ with a meridian of $S_n$. That's the universal covering we want.

In my description of the universal cover, I mentioned the line $\mathbb R$ first and then glued the spheres to it. One could equivalently describe the same space by staring with the infinite sequence of 2-spheres and gluing a line to them along disjoint intervals tending to infinity. Furthermore, since the glued intervals are already in the spheres, all that's really added when we attach the line is the segments of $\mathbb R$ between consecutive $M_n$'s. These segments join the north pole of one sphere $S_n$ to the south pole of the next, $S_{n+1}$. So my description is equivalent to the one in Max Nagy's answer.

Incidentally, after doing such a computation, I find it useful to check that I've really got at least a local homeomorphism. The original space had points (the poles) where it's not locally Euclidean; a little neighborhood looks like a disk with a whisker sticking out of the middle. So the universal cover had better also have such points. It does, namely the ends of the $M_n$'s. This sort of "sanity check" has saved me from errors more than once.