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I was thinking today, what is the universal cover of a torus with the "donut hole" shrunk to a point?

I am certain it must include a sphere, but that can't be enough because of the point at the center of the manifold.

Is it a wedge of two spheres perhaps?

Intuitively something like this seems correct. How should I think about these types of problems in general? Geometrically, what should I look for when searching for a universal cover?

Johnny Apple
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2 Answers2

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It is the infinite chains of 2-spheres, each pair of consecutive spheres meet in one point, and consecutive spheres are disjoint. The generator of the covering group acts as the shift sending each sphere to the next one.

Moishe Kohan
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  • How do I see that this actually covers the space? – Johnny Apple May 16 '14 at 05:36
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    @JohnnyApple: I described the covering group action. Using it, you see that the quotient of the infinite chain of spheres by the action of the covering group is $S^2$ with two distinct points identified. This is exactly your torus with a meridianal circle pinched to one point, which is the complex you are asking about. – Moishe Kohan May 16 '14 at 19:04
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To elaborate on @studiosus's response: Rather than thinking of a long chain of spheres, think of a long chain of sausage links.

We can think of the usual torus $T$ by starting with a cylinder, $S^1\times [0,1]$, and identifying the top and bottom circles; i.e., $T = S^1\times [0,1]/\sim$ where $(x,0)\sim (x,1)$ for every $x\in S^1$. Now, the torus with no hole, $\Sigma$, can be obtained by identifying that glued circle to a single point. I.e., we take $S^1\times [0,1]/\sim_1$, where $(x,0)\sim_1 (y,1)$ for all $x,y\in S^1$.

But let's do this a different way: First identify all the points $(x,0)\in S^1\times [0,1]$ to a point $P$ and all the points $(y,1)\in S^1\times [0,1]$ to a different point $Q$. (Topologically, this is a sphere, but we're thinking of it as a sausage link.) Now identify the points $P$ and $Q$; topologically, this is a sphere with two points identified, but we're thinking of it as attaching the ends of the sausage link to one another.

If we think of this as a sequence of two equivalence relations we're modding out by, we have $(x,0)\sim_2 (y,0)$ and $(x,1)\sim_2 (y,1)$ for all $x,y\in S^1$, and on this quotient we then identify $P$ and $Q$. Thus, we've obtained the identification space $S^1\times [0,1]/\sim_1$, so our sausage link is in fact homeomorphic to the torus with no hole, $\Sigma$.

Now, the universal cover is the infinite chain of sausage links, namely, $$S^1\times\Bbb R/\sim_4\,, \quad\text{where } (x,n)\sim_4 (y,n) \text{ for all }x,y\in S^1 \text{ and } n\in\Bbb Z.$$

Ted Shifrin
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  • I understand everything you said until the last sentence. We have that the space itself is like one "sausage link" connected to itself at the ends. How do we go from there to an infinite connected circle of sausage links as the cover? How do I recognize this as the cover once I've identified the space as a "sausage link"? – Johnny Apple May 18 '14 at 08:24
  • OK, two comments. First, you can see that the infinite chain of sausage links is locally homeomorphic to our "attached" individual sausage link: At one of the integer points, it looks exactly like what happens when we attach $P$ and $Q$. Second, the projection map $\pi\colon S^1\times\Bbb R/\sim_4 \to S^1\times [0,1]/\sim_1$ is given by the usual map $\Bbb R\to [0,1]/(0\sim 1)$ in the second factor; in other words, mod out by the translation $z\mapsto z+n$. This is our group of deck transformations ($\Bbb Z$). – Ted Shifrin May 18 '14 at 11:22
  • I totally agree. But let's say I had everything up until the last line. How do I then go on to guess that the covering space is what you said it is? – Johnny Apple May 19 '14 at 00:21
  • Because to get a simply connected covering space you have to "unidentify" $P$ and $Q$. Then, wanting a local homeomorphism at the preimages of that point, you need (recursively) an infinite chain. After all, how do you decide that $\Bbb R$ is the universal covering space of $S^1$? – Ted Shifrin May 19 '14 at 00:27
  • Ok. So this infinite chain of sausage links is closed in on itself then (like a circle of infinite radius made up of sausage links). Why does a finite circle of sausage links not work? Is it simply because the fundamental group is $\Bbb Z$? – Johnny Apple May 19 '14 at 00:30
  • Go back to understanding the universal cover of $S^1$. That's really underlying this whole story. A finite circle will not be simply connected! – Ted Shifrin May 19 '14 at 00:32