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, $Asubset X$ is a deformation retract if there exists a function $r:Xto A$ such that $rcirc i = id_A$ and $icirc r simeq id_X$ ($i$ is the inclusion and "$simeq$" is the homotopy between two maps).

Tyrone · Accepted Answer

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 $j:S^nhookrightarrow S^{n+1}setminus{N,S}$ is the inclusion, then $rcirc j=id_{S^n}$. On the other hand we have $H:(S^{n+1}setminus{N,S})times Irightarrow S^{n+1}setminus{N,S}$ given by
$$H_t(x_0,x_1,dots,x_n,x_{n+1})=frac{1}{sqrt{1-|tcdot x_{n+1}|^2}}left(x_0,x_1,dots,x_n,sqrt{1-t^2}cdot x_{n+1}right).$$
We check that $H$ is a homotopy $idsimeq jcirc r$.
I suppose the intuition for this is that you can obtain $S^{n+1}$ by taking the cylinder $S^ntimes[-1,1]$ and identifying $S^ntimes{-1}$  and $S^ntimes {+1}$ to separate points. It we cut out the resulting points, then what is left is $S^{n+1}setminus{N,S}cong S^ntimes(-1,1)$. Clearly the inclusion $S^nhookrightarrow S^ntimes(-1,1)$, $zmapsto (z,0)$, is a deformation retract. The maps above just spell out the details of this.

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

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