Understanding why removing two points from the 3-sphere ($S^{3}$) is homeomorphic to $S^{2} \times I$ I understand why removing one point from $S^{3}$ gives a surface that is homeomorphic to a three-dimensional space (i.e. $S^{2}$). But can someone please explain (the intuition of) why removing a second point from the 3-sphere is homeomorphic to a thickened 2-sphere?
 A: With an open interval, remove the north and south poles. Then use the sideways projection of Archimedes to the cylinder.
i.e. from $x^2 + y^2 + z^2 + w^2 = 1,$ with $w \neq \pm 1,$ project $(x,y,z,w)$ to $$  \left( \frac{x}{\sqrt {x^2 + y^2 + z^2}}, \; \frac{y}{\sqrt {x^2 + y^2 + z^2}}, \; \frac{z}{\sqrt {x^2 + y^2 + z^2}}, \; w \; \right) $$
A: Wlog. (why?) the points removed from $S^n$ are $(0,\ldots,0,\pm1)$.
Then $$(x_0,\ldots, x_n)\mapsto \left(\frac{(x_0,\ldots, x_{n-1})}{\sqrt{1-x_n^2}},x_n\right)\in S^{n-1}\times (-1,1)$$
is the desired homeomorphism.
A: The answers given by Hagen and Will are straight to the point. A slightly more long-winded explanation is that removal of a point from $S^3$ produces a space homeomorphic to $\mathbb{R}^3$ (by stereographic projection), and that removal of a second point produces a space that can be identified with $\mathbb{R}^3 - \{0\}$. The map $\mathbb{R}^3 - \{0\} \to S^2 \times (0, \infty)$ that sends $v$ to $(\frac{v}{\|v\|}, \|v\|)$ is a homeomorphism with inverse $(u, t) \mapsto tu$. Finally, $(0, \infty)$ is homeomorphic to $(0, 1)$. 
