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 $(1,0,0,0, \ldots)$ ?.
 A: 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. 
A: 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.
