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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).

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  • $\begingroup$ At least in American English usage familiar to me, in the title it'd be "deformation retract to", not "of"... Just wondering... especially since your definition does have the arrow going to the larger space, etc. $\endgroup$ Commented Oct 2, 2020 at 19:29
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    $\begingroup$ If you think of $\Bbb{S}^{n+1}$ as $\Bbb{D}^{n+1}$ with $\Bbb{S}^{n}$ identified then $\Bbb{S}^{n+1}$ with two points removed is homeomorphic to open disk $\Bbb{D}^{n+1}$ with origin removed. Then the deformation retraction is straightforward $\endgroup$
    – Ivin Babu
    Commented Oct 3, 2020 at 6:36

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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.

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