# Homeomorphism vs. Homotopy (Equivalence)

Trying to brush up on some geometric and algebraic topology, I got a little confused about the following:

Suppose we have the standard unit sphere $S^2$, but we remove its north and south poles. Is this topological space homeomorphic or homotopic to $S^1 \times \mathbb{R}$? I would think that they are not homotopic since I don't think both spaces are deformation retracts, are they? Now I do know that the stereographic projection is a map from $S^2$ to the plane, but that just involves the removal of either the north or the south pole, correct?

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If you know that $S^2$ minus one pole is homeomorphic to $\mathbb{R}^2$, then $S^2$ minus two poles is homeomorphic to...? – Qiaochu Yuan Mar 31 '12 at 18:04
$S^1 \times \mathbb{R}$ is a cylinder. Put the sphere inside the cylinder (north and south pole on the axis of the cylinder) and project from the center of the sphere minus north and south-pole onto that cylinder using straight lines. – t.b. Mar 31 '12 at 18:06
It makes it a straight line homeomorphism, if such a term exists :) A straight line homotopy is usually understood to be a homotopy $H$ between two functions $f,g: X \to \mathbb{R}^n$ of the form $H(x,t) = (1-t)f(x) + tg(x)$. See here for example. – t.b. Mar 31 '12 at 18:15
If $Y$ is homeomorphic to $X$ it's clearly also homotopy equivalent. Just use the homeomorphism and a constant homotopy. – user20266 Mar 31 '12 at 18:16
Linguistic nitpick: maps are homotopic. Spaces are homotopy equivalent. – Neal Mar 31 '12 at 18:57

The two spaces are homeomorphic. $S^2$ minus one point is identified by stereographic projection with $\mathbb{R}^2$, so $S^2$ minus two points is homeomorphic with say $\mathbb{R}^2 \backslash \{0\}$. Identifying $\mathbb{R}^2$ with $\mathbb{C}$ and $S^1$ with $\mathbb{R}/\mathbb{Z}$, the homeomorphism

$\mathbb{R}\times \mathbb{R}/\mathbb{Z}\cong \mathbb{C} \backslash \{0\}$

is given by $(r,\theta)\mapsto e^re^{2\pi i\theta}$.

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