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

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.
