Why do we have $S^2-\{N,S\}\simeq S^1\times\mathbb{R}$? Does anybody has an easy way to explain why a sphere minus its north and south pole is homeomorphic to the cartesian product of the sphere and $\mathbb{R}$? That is, $S^2-\{N,S\}\simeq S^2\times\mathbb{R}$?
Let say if during an exam, I would like to know what is $S^2-\{N,S\}$, what is the best way to guess, without actually writing out the homeomorphism explicitly?
My weakness is I really struggle with intuition and figuring out an easy explanation, I only understand any concept literally but it takes me a long time to understand the underlying idea.
Could anyone please give some light on this? Thanks.
 A: As written the claim can't be true -- your left-hand side is a two-dimensional manifold, whereas your right-hand side is three-dimensional.
But $S^2\setminus\{N,S\}\cong S^1\times \mathbb R$.
To see this, realize $S^1\times \mathbb R$ as an infinite cylinder that encloses the sphere, touching it along the equator. The map given by projecting each (non-pole) point on the sphere onto the cylinder along a ray through the center of the sphere is then a homeomorphism.
A: Delete the north and south poles. Consider the copy of $S^1$ as the equator and the meridian which passes through a point on the equator. That meridian is now broken into two pieces, that lying on the front face and that of the back face. Look at a point on the front face, then the meridian by our discussion is just an open interval since the end points are deleted. However, open intervals are homeomorphic to $\mathbb{R}$. Thus, every point $s$ on the equator contains a copy of $\mathbb{R}$ i.e $\{s\} \times \mathbb{R}$ and putting this together for every point, we have have $S^1 \times \mathbb{R}$. 
If you want some more practice in thinking about things like this, think about why an annulus is topologically equivalent to a cylinder.
A: Another way of viewing this:
$S^2 - \{N\}$ is homeomorphic to $\mathbb{R}^2$ via stereographic projection.  
Now, subtract say the origin from $\mathbb{R}^2$ (which corresponds to the south pole on $S^2$) so your new space is $\mathbb{R}^2 - \{(0,0)\} \simeq S^1 \times \mathbb{R}$.
