Homeomorphism between quotient spaces Let $ A = \{\left(0,0,1\right), \left(0,0,-1\right) \} \subset S^2 $
And let $ B \subset T^2$ be the image of $ \Bbb R \times 0 \subset \Bbb R^2 $, where we view $T^2 = \Bbb R^2/\Bbb Z^2$
Now I am asked to show that $S^2/A \simeq T^2/B$ but I'm not too sure how to go about this, or what these quotient spaces look like. 
I think $S^2/A$ is the set of pairs of directly opposite points on $S^2$ and that $T^2/B$ is the set of all horizontal lines in $T^2$ viewed as $\Bbb R^2/\Bbb Z^2$ but I'm not sure if that's right or how to show that they are homeomorphic if it is. 
Any help or advice you might be able to offer would be greatly appreciated, thank you! 
 A: Writing down the homeomorphism explicitly gets a little messy, but getting a helpful visualization isn’t too bad; I’ll work through one for you.
I assume that when you write that $T^2=\Bbb R^2/\Bbb Z^2$, you’re talking about an algebraic rather than a topological quotient; $S^2/A$ and $T^2/B$ are clearly topological quotients.
$A$ is the set consisting of the north and south poles of the sphere $S^2$, so $S^2/A$ is what you get when you identify these poles. The obvious way to carry out the identification is to imagine taking a spherical balloon and pushing the top and bottom points together until they coincide. What you get can be viewed as a torus whose inner radius is $0$; this is known as a horn torus. (The pictures here may be more helpful.) However, another visualization makes this problem much easier. Imagine pulling the poles apart, so that you get a sort of hollow cigar shape with a point on each end. Now bend the cigar around in a circle until the points touch. The result is the same as if you took the torus shown below (attribution: By Krishnavedala (Own work) [CC0], via Wikimedia Commons) and squeezed the red circle to a point.

Now consider $T^2$. It can more transparently be obtained by starting with the closed unit square $[0,1]\times[0,1]$, identifying the left and right edges to form a horizontal cylinder, and then identifying the circular ends of the cylinder (which originally were the top and bottom edges of the square) to make the torus. (This PDF may be helpful.) This is essentially what taking $\Bbb R^2/\Bbb Z^2$ does, though it does it in a more roundabout fashion. It first identifies all of the squares of the form $[n,n+1]\times[m,m+1]$ for $n,m\in\Bbb Z$ with $[0,1]\times[0,1]$ in the natural way and then performs the edge identifications on that square. 
Now think about $T^2/B$ in terms of what happens to the unit square. First identify the left and right edges as before to get the vertical cylinder. Then identify the bottom edge to a single point. Of course the bottom and top edges are going to be identified to make the torus, so we must also identify the top edge to a single point. After these identifications we have the same hollow cigar shape with pointed ends that I described above, and we do exactly the same thing to it: we identify the top and bottom points.
