How can I prove that $\mathbb S^n$ is deformation retract to $\mathbb S^{n+1}\setminus \{N,S\}$?

Question

My idea was to find a composition of functions that turnt $\mathbb S^{n+1}\setminus\{N,S\}$ into $D^{n}\setminus\{0\}$ (the proyecton of hyperplane of the first $n$ components) and then into $\mathbb S^n$ (with the normalization of the "points looked as vectors"), and move on with the properties of a deformation retract. But I don't know how further I can go with this idea.

It also sounds much more complicated that what it seems, but what do I know? Could anyone please help me out?

P.S.: To me, $A\subset X$ is a deformation retract if there exists a function $r:X\to A$ such that $r\circ i = id_A$ and $i\circ r \simeq id_X$ ($i$ is the inclusion and "$\simeq$" is the homotopy between two maps).

Answer 1

Define $r:S^{n+1}\setminus\{N,S\}\rightarrow S^n$ by $$r(x_0,x_1,\dots,x_n,x_{n+1})=\frac{1}{\sqrt{1-|x_{n+1}|^2}}(x_0,x_1,\dots,x_n).$$

If $j:S^n\hookrightarrow S^{n+1}\setminus\{N,S\}$ is the inclusion, then $r\circ j=id_{S^n}$. On the other hand we have $H:(S^{n+1}\setminus\{N,S\})\times I\rightarrow S^{n+1}\setminus\{N,S\}$ given by $$H_t(x_0,x_1,\dots,x_n,x_{n+1})=\frac{1}{\sqrt{1-|t\cdot x_{n+1}|^2}}\left(x_0,x_1,\dots,x_n,\sqrt{1-t^2}\cdot x_{n+1}\right).$$ We check that $H$ is a homotopy $id\simeq j\circ r$.

I suppose the intuition for this is that you can obtain $S^{n+1}$ by taking the cylinder $S^n\times[-1,1]$ and identifying $S^n\times\{-1\}$ and $S^n\times \{+1\}$ to separate points. It we cut out the resulting points, then what is left is $S^{n+1}\setminus\{N,S\}\cong S^n\times(-1,1)$. Clearly the inclusion $S^n\hookrightarrow S^n\times(-1,1)$, $z\mapsto (z,0)$, is a deformation retract. The maps above just spell out the details of this.