Exercise 0.16

Proof following Prop.  $0.16$. Let $r_n:S^n\times \left[\frac{1}{2^{n+1}},\frac{1}{2^n}\right]\to S^n$ be the homotopy pushing the equator $S^{n-1}$ to a point $p\in e_2^n$; note that this also contracts $e_2^n$ to the point $p$. Then, combining the homotopies $r_n$ gives a homotopy $r:S^{\infty }\times I\to S^{\infty }$ between the identity map and a constant map. There is no continuity issues at $t=0$ since $r$ is continuous on $S^n\times I$, since it is stationary on $\left[0,\frac{1}{2^{n+1}}\right]$, and since CW complexes have the weak topology with respect to skeleta. □

Proof following Ex. 1B.3. Let $I=[0,1]$. Define

Note $f$ is continuous by [?, Thm. 19.6] since it is continuous in each coordinate. $f_t$ takes nonzero vectors to nonzero vectors for all $t\in I$, hence $f_t∕|f_t|$ gives a homotopy from the identity map of $S^{\infty }$ to the map $(x_1,x_2,\dots )\mapsto (0,x_1,x_2,\dots )$. Next, define

$g$ is continuous for the same reason as for $f$, and $g_t$ takes nonzero vectors to nonzero vectors for all $t\in I$. Thus, $g_t∕|g_t|$ gives a homotopy from the map $(x_1,x_2,\dots )\mapsto (0,x_1,x_2,\dots )$ to a constant map. Composing these two homotopies together gives a homotopy between the identity map and a constant map, so $S^{\infty }$ is contractible. □