# Understanding the proof that $S^\infty$ is contractible.

Proposition $$:$$ $$S^\infty$$ is contractible.

Proof $$:$$ First define $$f_t : \Bbb R^\infty \longrightarrow \Bbb R^\infty$$ by $$f_t(x_1,x_2,\cdots) = (1-t) (x_1,x_2,\cdots) + t(0,x_1,x_2,\cdots).$$ This takes non-zero vectors to non-zero vectors for all $$t \in [0,1],$$ so $$\frac {f_t} {|f_t|}$$ gives a homotopy from the identity map of $$S^\infty$$ to the map $$(x_1,x_2,\cdots) \mapsto (0,x_1,x_2,\cdots).$$ Then the homotopy from this map to a constant map is given by $$\frac {g_t} {|g_t|}$$ where $$g_t(x_1,x_2,\cdots) = (1-t) (0,x_1,x_2,\cdots) + t(1,0,0,\cdots).$$

This proof is given in my lecture note and it is copied from Hatcher's book. I don't understand the construction of the homotopy. I know that a topological space $$X$$ is contractible if there exists $$x_0 \in X$$ such that $$Id_X \simeq C_{x_0},$$ where $$C_{x_0}$$ is the constant function taking every element of $$X$$ to the specific point $$x_0.$$ Here $$X = S^{\infty}.$$ So the homotopy $$H$$ should be defined on $$S^{\infty} \times I$$ and should take values in $$S^{\infty}.$$ But instead it is defined on $$\Bbb R^{\infty} \times I.$$ I don't understand what's going here? Why are we taking $$\Bbb R^{\infty}$$ here and exploiting convexity of the space? Can anybody please shed some light on it? Any suggestion regarding this will be warmly appreciated.

• Note that $S^\infty \subset \mathbb R^\infty$; and if you restrict $H(t,x) = f_t(x)$ to $I \times S^\infty$, the image lies in $S^\infty$. May 11, 2021 at 15:21
• @guidoar Yeah now I can see. I think $f_t$ restricted to $S^{\infty}$ will give a homotopy between the identity on $S^{\infty}$ and the right shift where the intermediate functions may take values in $\Bbb R^{\infty}.$ So when we divide $f_t$ by it's norm then we are only allowing the intermediate functions to take values in $S^{\infty}.$ Similar thing happens for $g_t.$ May 11, 2021 at 15:33

Noting that $$S^\infty$$ is a subspace of $$\mathbb R^\infty$$, this proof describes several different homotopies of the identity map on $$S^\infty$$:

1. $$f_t$$ which is a "straight line" homotopy in the space $$\mathbb R^{\infty}$$ from the identity map to a certain unnamed map which I shall name $$h:(x_1,x_2,\cdots) \mapsto (0,x_1,x_2,\cdots)$$; and next
2. $$\frac{f_t}{|f_t|}$$ which is a homotopy in $$S^\infty$$ from the identity map to $$h$$; and next
3. $$g_t$$ which is a homotopy in $$\mathbb R^\infty$$ from $$h$$ to a constant map; and next
4. $$\frac{g_t}{|g_t|}$$ which is a homotopy in $$S^\infty$$ from $$h$$ to the constant map.

Unsaid here is the final step, namely that you must concatenate homotopy 2 and homotopy 4 to get the final homotopy in $$S^\infty$$ from the constant map to the identity map.

• Don't $f_t$ and $\frac {f_t} {|f_t|}$ give homotopy between $\text {id}_{\Bbb R^{\infty}}$ and the right shift $h$ in $\Bbb R^{\infty}$ and in $S^{\infty}$ respectively? May 11, 2021 at 15:41
• The homotopies with denominators (2 and 4) are not defined in $\mathbb R^\infty$. They could be defined in $\mathbb R^\infty - \{0\}$, but we only care about them in $S^\infty$. May 11, 2021 at 16:04
• Yeah I get your point. Now another question came to my mind in this context. We know that $\Bbb RP^n = S^n / x \sim -x.$ Can we do the same for $\Bbb RP^{\infty},$ when regarded as a CW complex? What I mean is that can it be said that $\Bbb RP^{\infty} = S^{\infty} / x \sim -x\$? May 11, 2021 at 16:06
• Yes, $\Bbb RP^{\infty} = S^{\infty} / x \sim -x$. But in case you were wondering, that doesn't mean $\Bbb RP^{\infty}$ is contractible (it's not). If you have some followup question about $\Bbb RP^{\infty}$, it is best to post it as another question so that others can see it besides you and me. May 11, 2021 at 16:20
• Given any $x \in S^{\infty}$ there exists $n \geq 0$ such that $x \in S^n.$ Then if we take the map $x \mapsto \{x,-x\}$ then that will give us a continuous surjective map from $S^{\infty}$ to $\Bbb RP^{\infty}.$ So by universal property of quotient topology it induces a bijective continuous map from $S^{\infty} / x \sim -x$ onto $\Bbb RP^{\infty}.$ But how to show that it's a homeomorphism? May 11, 2021 at 16:20