6
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ If you know that $S^2$ minus one pole is homeomorphic to $\mathbb{R}^2$, then $S^2$ minus two poles is homeomorphic to...? $\endgroup$ – Qiaochu Yuan Mar 31 '12 at 18:04
  • 2
    $\begingroup$ $S^1 \times \mathbb{R}$ 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. $\endgroup$ – t.b. Mar 31 '12 at 18:06
  • 1
    $\begingroup$ 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. $\endgroup$ – t.b. Mar 31 '12 at 18:15
  • 1
    $\begingroup$ If $Y$ is homeomorphic to $X$ it's clearly also homotopy equivalent. Just use the homeomorphism and a constant homotopy. $\endgroup$ – user20266 Mar 31 '12 at 18:16
  • 4
    $\begingroup$ Linguistic nitpick: maps are homotopic. Spaces are homotopy equivalent. $\endgroup$ – Neal Mar 31 '12 at 18:57
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.