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