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

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).

Mathematics Asked by GreekCorpse on January 3, 2021

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.

Correct answer by Tyrone on January 3, 2021

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