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In "An Introduction to Algebraic Topology" of Rotman, Exercise 1.31 asks to show that the equator $\mathbb{S}^{n-1}$ is a deformation retract of $\mathbb{S}^n\setminus\{a, b\}$.

I thought that if one thinks to the usual sphere in the space, then one just enlarge the holes to flatten $\mathbb{S}^3$ onto its equator, but I cannot understand how to prove this in general for $n$, probably it's the same idea but I cannot construct the required function.

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  • $\begingroup$ Do you know that $\mathbb{R}^n$ is homeomorphic to $S^n\setminus\{a\}$ ? $\endgroup$ – Max May 24 at 19:21
  • $\begingroup$ @Max Yes, is there a way to deduce what I asked from that? $\endgroup$ – W4cc0 May 24 at 19:28
  • $\begingroup$ Well do you know that $\mathbb{R}^n\setminus\{b\}$ strongly deformation retracts onto $S^{n-1}$ ? $\endgroup$ – Max May 24 at 19:29
  • $\begingroup$ No, unfortunately that is precisely what I do not know... and can't prove. $\endgroup$ – W4cc0 May 24 at 19:30
  • $\begingroup$ but, perhaps looking at math.stackexchange.com/questions/659682/… could solve this latter fact, and then probably what I asked should easily follow $\endgroup$ – W4cc0 May 24 at 19:33
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As per the comments, $S^n\setminus\{a\}$ is homeomorphic to $\mathbb{R}^n$.

Now it suffices to see that $\mathbb{R}^n\setminus\{0\}$ (strongly) deformation retracts onto $S^{n-1}$ (if you know that you just have to move things around a bit to get $b$ to align with $0$ and $S^{n-1}$ to align with the image of the equator)

For this, consider $r:x\mapsto \frac{x}{||x||}$, which is indeed a retraction. Moreover, define $H(x,t)= r(x)t+(1-t)x$. Clearly it is continuous, and it is easy to check that it does land in $\mathbb{R}^n\setminus\{0\}$ (and not just $\mathbb{R}^n$), and it is clearly a homotopy between $id$ and $r$ (I should really write $i\circ r$, where $i$ is the inclusion);

and moreover (although it is not necessary to get a deformation retract, that we already got) you can check that $H$ is constant equal to the identity on $S^{n-1}$

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