Let $p,q$ be different points of $S_{2}=\{(x,y,z)\in\mathbb{R}^{3}:x^{2}+y^{2}+z^{2}=1\}$. We consider the space $X=S_{2}/\sim$ where $\sim$ is the next relation: $x,y\in S_{2}$, $x \sim y$ if and only if $x=y$ or $x=p, y=q$ or $x = q, y = p$ ($p \neq q \in S_2$ are fixed points). I have to find the universal covering space of $X$.

It is easy to check that $\pi_{1}(X)=\mathbb{Z}$, so the cardinality of the fibers of the universal covering space is the cardinality of $\mathbb{Z}$. I have also noticed that $X$ is homeomorphic to the torus in which the central circunference has been glued into a point. Let $Y$ be this space.

I consider the universal covering space of the torus $$ \pi:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}/\mathbb{Z}^{2}\cong T^{2}, x\mapsto [x], $$ and the map $$ f:T^{2}\rightarrow Y, x\mapsto [x]. $$

I think the universal covering of $X$ have something to do with these maps, but I am not able to find it.

  • $\begingroup$ What are $p$ and $q$? The north and south poles? $\endgroup$ – Najib Idrissi Feb 4 '15 at 12:55
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    $\begingroup$ Yes, for example. I think there is no difference as long as they are different points. $\endgroup$ – Srinivasa Granujan Feb 4 '15 at 12:59
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    $\begingroup$ I think it is easier to visualize the universal cover if you consider your space as equivalent to a torus with a "collapsed" meridian. You can look at this space as an $S^{2}$ "glued" over an $S^{1}$: you can now build the universal cover thinking as the one of $S^{1}$. Take $\mathbb{R}$ (the universal cover of $S^1$) and "glue" $S^2$s over it (as the $S^2$ is "glued" over $S^1$ in the pinched torus). What you obtain is a sort of necklace of $S^{2}$s, or a tower of $S^2$ that pairwise intersect in a point. $\endgroup$ – Dario Feb 4 '15 at 13:20

Your space is homotopy equivalent to the sphere $S^2$ with a line segment (in blue in the picture) joining $p$ and $q$:

Sphere with poles joined

By continuously deforming the line segment, you can fuse the endpoints and you end up with $S^2 \vee S^1$, the wedge of a sphere and a circle at a point (I've drawn the circle outside the sphere, it doesn't change anything):

Wedge of sphere and circle

Since the sphere $S^2$ is simply connected, the universal cover of this space looks like a "necklace". It's an infinite (in both directions) number of spheres, and each sphere is linked to the next by a line segment:

The Covering space

Each sphere is mapped homeomorphically onto the $S^2$ component of $S^2 \vee S^1$, and each segment is projected onto the circle component by identifying endpoint.

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    $\begingroup$ I think it's important to note that this is not the universal cover of the space in the question, it is only homotopy equivalent to the universal cover. Much like a point is not the universal cover of the torus, but it is homotopy equivalent to the universal cover, namely the plane. $\endgroup$ – Dan Rust Feb 4 '15 at 14:19
  • $\begingroup$ Yes, good point @Daniel. I think in this case the universal cover is the space I've drawn where you collapse every line segment to a point. $\endgroup$ – Najib Idrissi Feb 4 '15 at 14:26
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    $\begingroup$ Yes, that seems reasonable. The covering map would be given by the orbit quotient of the 'shift' map $s\colon \tilde{X}\to\tilde{X}\colon (x,n)\mapsto (x,n+1)$ where $x$ is a point in the sphere and $n$ refers to which sphere in the chain that point lives in. Or if you like, let $(x,n)\sim(y,m)$ if and only if $x=y$ (by definition we say that $(q,n)=(p,n+1)$). Then $\tilde{X}/{\sim} \cong X$. $\endgroup$ – Dan Rust Feb 4 '15 at 14:54

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