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.

Thanks in advance.

  • $\begingroup$ 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 $S^\infty$. $\endgroup$
    – qualcuno
    Commented May 11, 2021 at 15:21
  • $\begingroup$ @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 $g_t.$ $\endgroup$ Commented May 11, 2021 at 15:33

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Commented May 11, 2021 at 15:41
  • $\begingroup$ 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 $S^\infty$. $\endgroup$
    – Lee Mosher
    Commented May 11, 2021 at 16:04
  • $\begingroup$ 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\ $? $\endgroup$ Commented May 11, 2021 at 16:06
  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Commented May 11, 2021 at 16:20
  • $\begingroup$ 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? $\endgroup$ Commented May 11, 2021 at 16:20

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