# Hatcher 0.16

Hatcher, Algebraic Topology, Chapter 0

16. Show that $S^\infty$ is contractible.

Proof. Dr. Hatcher himself gives the proof of this in Example 1B.3 on Page 88, a proof the ideas of which are replicated here.

The goal is to show that $X=S^\infty$ is contractible, i.e. that the identity map $1_X$ is homotopic to a constant map. It will be shown that $1_X\simeq (1,0,0,\ldots)$ where $(1,0,0,\ldots)$ is the map $\mathbf{x}\mapsto(1,0,0,\ldots)\in S^\infty$ for all $\mathbf{x}\in S^\infty$. First, note that $S^\infty\subset\mathbb{R}^\infty$ is the collection of all points $\mathbf{x}$ for which $\|\mathbf{x}\|=1$: This fact will be used to build the desired homotopy, the construction of which will take place in two parts.

Let $F_t:\mathbb{R}^\infty\to\mathbb{R}^\infty$ be the homotopy

$F_t:(x_1,x_2,\ldots)\mapsto (1-t)(x_1,x_2,\ldots)+t(0,x_1,x_2,\ldots)$

for all $t\in I$. This is a straight-line homotopy between the point $\mathbf{x}=(x_1,\ldots)$ and the point $(0,x_1,\ldots)$ from which we define the restriction $f_t=F_t/|F_t|:S^\infty\to \mathbb{R}^\infty$ for all $t\in I$. Clearly, then, $f_t$ is a homotopy between points $y=(y_1,y_2,\ldots)\in S^\infty$ and $(0,y_1,\ldots)\in\mathbb{R}^\infty$. Next, let $G_t:\mathbb{R}^\infty\to S^\infty$ be given by

$G_t:(0,x_1,x_2,\ldots)\mapsto (1-t)(0,x_1,x_2,\ldots)+t(1,0,0,\ldots)$.

As with $F_t$, $G_t$ is a straight-line homotopy, this time between the points $(0,x_1,x_2,\ldots)\in\mathbb{R}^\infty$ and $(1,0,0,\ldots)\in S^\infty$. As above, defining the restriction $g_t=G_t/|G_t|$ for all $t\in I$ gives a chain of homotopies

$\left\{1_X\right\} \mapsto \left\{\mathbf{x}\mapsto (0,x_1,x_2,\ldots)\right\} \mapsto \left\{(0,x_1,x_2,\ldots)\mapsto(1,0,0,\ldots)\right\}$.

In particular, then, the composition $h_t$ is the desired homotopy between $1_X$ and the constant map $\mathbf{x}\mapsto(1,0,0,\ldots)$, where

$h_t = \left\{ \begin{array}{rl} f_{1-2t}, & 0\leq t\leq 1/2\\ g_{2t-1}, & 1/2\leq t\leq 1 \end{array} \right.$.   $\square$

Given $f_t (X)(1-t)(x_1,x_2…..) + t(1,0,0,0…..)$, wouldnt $F=f_t(X) / |f_t(X)|$ be enough?