# $\mathbb{S}^n$ without two points

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.

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

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 $$S^{n-1}$$