# Is $S^{\infty}$ contractible?

Recently I was reading this post:

Unit sphere in $\mathbb{R}^\infty$ is contractible?

Then a doubt came across to me: why I can't consider the linear homotopy $H:I\times S^{\infty}\to S^{\infty}$ given by the restriction of $$H_t= \dfrac{F_t}{|F_t|}$$ to the sphere $S^{\infty},$ where $F:I\times \mathbb{R}^{\infty}\to \mathbb{R}^{\infty}$ given by $F_t(x)=(1-t)(x_1, x_2, \ldots)+t(1,0,0,\ldots)$??

What is the need of the two steps? That is first get a homotopy between $Id$ and the shift $\sigma$ and then the homotopy between $\sigma$ and the constant map equals $(1,0,0,0, \ldots)$ ?.

• Try applying the same argument to $S^n$, which is not contractible. Commented Jan 28, 2016 at 4:09

The current answer already explains why the proposed homotopy cannot work. Let me give a geometric interpretation of the two-step homotopy on the linked answer.

Trying to contract $S^\infty$ to $(1, 0, 0,\cdots)$ directly using straightline homotopy cannot possibly work: The situation is the same as that of trying to contract $S^2$ in $\Bbb R^2$. Straightline homotopy at some point of time will run through the origin, in which case normalizing gives you undefined things.

So the point of the shift map $\sigma : S^\infty \to S^\infty$, $(x_0, x_1, x_2, \cdots) \mapsto (0, x_1, x_2, \cdots)$ is to pull $S^\infty$ up one dimension. Now you can contract the image of $\sigma$ to $(1, 0, 0, \cdots)$, because it lives in codimension one and $(1, 0, 0, \cdots)$ is just some other point outside it. The situation is the same as that of contracting $S^1 \subset \Bbb R^2$ inside $\Bbb R^3$ to a point outside the hyperplane it lives. This can easily be done using straightline homotopy.

Irrelevant to the question, but here's a different way to do it. $S^\infty$ is the same a the colimit $\bigcup_n S^n$ with $S^i \subset S^{i-1}$ being inclusion as an equator. Note that each $S^n$ bounds a disk (i.e., hemisphere) on each side in $S^\infty$. Consider the homotopy which contracts $S^n$ through those. To make this work, one needs a $[1/2^{n+1}, 1/2^n]$ trick so that the composition is continuous.

• Do I understand it right that this proof is to show that $S^\infty$ is contractible as a subset of $\mathbb{R}^\infty$? Because if you need to prove that $S^\infty$ is a contractible space, you should find a homotopy within the sphere, right? Commented Apr 24, 2017 at 13:55
• @Igor Yes, according to the question, $S^\infty$ is the unit sphere in the normed space $\Bbb R^\infty$. The homotopy is withing the sphere, it's just built out of a homotopy on $\Bbb R^\infty$. On the other hand if $S^\infty$ to you is a CW complex (obtained by attaching two n-disks to $S^{n-1}$, for each $n$) then the second proof-sketch is what you want. Commented Apr 24, 2017 at 15:19
• I somehow missed the second sketch, thanks! Commented Apr 24, 2017 at 17:19
• @BalarkaSen Couldn't we just translate the sphere so that none of its points are negative, and thus $\frac{F_t}{|F_t|}$ would never have the problem as stated above? Say consider $S^{\infty}=\{x\in R^{\infty} | d((1,0,0,...),x)=1\}$. Commented Jul 22, 2021 at 20:47
• @Larry: What is $F_t$ in this case? You cannot take a point $p \in S^\infty$, and define $F_t$ to be $t \mathbf{x} + (1 - t)p$ normalized, no, no matter however much you translate the sphere. This is because if $p \in S^\infty$, the antipodal $-p \in S^\infty$ as well, and causes the same issue at $t = 1/2$. Just try it for a finite-dimensional sphere, and you'll see the issue. Commented Jul 22, 2021 at 20:53

The problem arises when applying $H_{t}$ to the point $(-1,0,0,...)$. We get $$H_t(-1,0,0,...)=\frac{(2t-1,0,0,...)}{\Vert (2t-1,0,0,...)\Vert}=\begin{cases} (-1,0,0,...) & t<1/2\\ (1,0,0,...) & t>1/2\\ \text{undefined} & t=1/2.\end{cases}$$ This is not continuous. We do not need a two step description for a homotopy, but the one given in this example is a simple one which includes no singularities associated with dividing by zero.