How do you show that $S^{\infty}$ is contractible?

Question

Here I mean the version with all but finitely many components zero.

Answer 1

This is the swindle, isn't it?

There's an elegant way to phrase this with lots of sines and cosines, but working it all out is too much like hard work. Here's the quick and dirty way.

Let $T: S^\infty \to S^\infty$ be the "shift everything down by 1" map.

Then for any point $x \in S^\infty$, $T(x)$ is not a multiple of $x$ and so the line between them does not go through the origin. We can therefore define a homotopy from the identity on $S^\infty$ to $T$ by taking the homotopy $t x + (1 - t)T(x)$ and renormalising so that it is always on the sphere (incidentally, although you are working in $\ell^0$, by talking about a sphere you implicitly have a norm).

Then we simply contract the image of $T$, which is a codimension 1 sphere, to a point not on it, say $(1,0,0,0,0,...)$. Again, we can use 'orrible sines and cosines, but renormalising the direct path will do.

(Incidentally, there's nothing special about which space you are taking the sphere in. So long as your space is stable in the sense that $X \oplus \mathbb{R} \cong X$ then this works)

Added a bit later: Incidentally, if you want to work in a space that doesn't support a norm (such as an infinite product of copies of $\mathbb{R}$) you can still define the sphere as the quotient of $X$ without the origin by the action of $\mathbb{R}^+$. The argument above still works in this case.

Added even later: Revisiting this in the light of the duplicate: Is $L^p(\mathbb{R})$ minus the zero function contractible?, the key property on $T$ is that it be continuous, injective, have no eigenvectors, and be not surjective. These conditions imply the following:

1. injective ⟹ the end-point of the homotopy is not the origin
2. no eigenvalues ⟹ the homotopy does not pass through the origin en route
3. not surjective ⟹ there is a point not in the image to which the image can be contracted
4. continuous ⟹ the homotopy is jointly continuous

Finally, there's no difference between the sphere and the space minus a point (indeed, without a norm the "space minus a point" is easier to deal with). Indeed, the homotopy described here actually works on the "space minus a point" and is just renormalised to work on the sphere.

Answer 2

Kind of late to the party, but the (weak) contractibility follows from $\pi_i(S^\infty) = 0$ for $i>0$.

Answer 3

General fact: Let $A_{-1} \subset \ldots A_n \subset \ldots$ be a filtration of cellular inclusions of $CW$ complexes. (More generally, let this be a filtration of cofibrations). Then $A_n$ contractible in $A_{n+1}$ $\implies$ $A:=\operatorname{colim}_n A_n$ is contractible. (Here $A$ is given the weak topology.)

Proof: Consider the composition $A_n \times I \xrightarrow{\text{contraction}} A_{n+1} \to A$. Since $A_n \to A$ is a cofibration, extend the above map to a map $A \times I \xrightarrow{\alpha_n} A$. The map $f: A\times I \to A$ defined by $f|_t=\alpha_{n+1}(2^{n+1} t-2^{n+1}+2)\circ\alpha_n(1)\circ\ldots\circ\alpha_1(1)$ for $1-\frac{1}{2^n}\leq t \leq 1-\frac{1}{2^{n+1}}$, is the required retraction. It is continuous because $f|_t$ is continuous when restricted to each $A_n$ and obviously $f|_a$ is continuous for all $a \in A$.

Now give $S^\infty$ the canonical $\mathbb{Z}/2$ equivariant cell structure (i.e. the pullback of the canonical cell structure on $RP^\infty$). The skeletal filtration satisfies the hypotheses of this general fact: $S^n \xrightarrow{i} S^{n+1}$ is null homotopic: $S^{n+1}$ can be given an $n$-skeleton that is a point. By the cellular approximation theorem, the map $i$ is homotopic to one that factors through this particular $n$-skeleton.

I guess this is more complicated than the other answers but this shows that a lot of other things are contractible too (like Milnor space).

Answer 4

Another nice solution to a similar question is at http://katlas.math.toronto.edu/drorbn/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible:

Let $H=L^2([0,1])$ and define $S^\infty = \{x \in H : \|x\|=1\}$.

Claim. $S^\infty$ is contractible.

Proof. For any $t \in [0,1]$ and any $f \in H$ define $f_t(x)= f$ for $0<x<t$ and $f_t(x)=1$ for $t<x<1$. Observe that $t \mapsto f_t/\|f_t\|$ is continuous and gives the desired retraction to the point $f=1$.

Answer 5

A very nice proof is via classifying spaces of categories. It goes as follows: Take the “walking isomorphism“ category $J$, that is the unique category with two isomorphic objects and four morphisms in total. The geometric realization of its simplicial nerve is exactly the infinity-sphere! (The non-degenerate $n$-simplices are given by the functors $[n] \to J$ starting at either $0$ or $1$ and then going back and forth using the isomorphism to produce upper and lower hemisphere of the standard cell-decomposition.) Then you conclude by observing that this category is equivalent to the terminal category and thus, as nerve and realization send natural transformations to simplicial, respectively topological homotopies, $S^\infty$ is contractible.

Answer 6

Here are my thoughts on the matter. However, this is not too much more than what is done above. I think...

$\quad$ We seek to show that a homotopy from the identity map of $S^{\infty}$ ($id_{S^{\infty}}$) to a constant map can be constructed and thus it must be null-homotopic, $i.e.$, contractible. Let $T: S^\infty \to S^\infty$ be the "shift everything 'down' by 1" map given by $(x_1, x_2, x_3,...) \mapsto (0, x_1, x_2,...)$. Then for any point, $x$ and its image $T(x)$, the line between them does not go through the origin.

$\quad$ We can therefore define a homotopy from the identity on $S^\infty$ to T by taking a homotopy and renormalizing, so that it is always on the sphere, as follows. Let $f_t: \mathbb{R}^{\infty} \setminus 0 \rightarrow \mathbb{R}^{\infty} \setminus 0$ be given by
$$f_t(x_1,x_2,...) = (1-t)(x_1, x_2,...) + tT(x_1, x_2,...).$$ (Note that the vector $f_0 = id_{\mathbb{R}^{\infty}}$ and that $f_t$ takes nonzero vectors to nonzero vectors $\forall t \in \left[ 0, 1\right]$.) Then we can renormalize it to ensure everything is still on $S^{\infty}$, $i.e.$, $$\frac{f_t}{\left| f_t\right|} = F(x,t) : S^{\infty} \times I \rightarrow S^{\infty}.$$ Thus we now have that $id_{S^{\infty}} \simeq T$. Or, in other words, $F$ gives a homotopy from the identity map of $S^{\infty}$ to the map $(x_1, x_2, x_3,...) \mapsto (0, x_1, x_2,...)$.

$\quad$ Then we simply contract the image of $T$ , which is a codimension 1 sphere, to a point not on it, say $(1,0,0,0,0,...) = N$ (north pole). So let $g_{t} : \mathbb{R}^{\infty} \setminus 0 \rightarrow \mathbb{R}^{\infty} \setminus 0$ be given by $$g_t(x_1, x_2, ...) = (1-t)(0, x_1, x_2,....) + t(1, 0, 0, ...).$$ Now observe that $g_0 = f_1$, $f_0 = id_{S^{\infty}}$, and $g_1 = N$ (a constant). Again we can renormalize to guarantee everything is still on $S^{\infty}$, $i.e.$, $$\frac{g_t}{\left| g_t\right|} = G(x,t) : S^{\infty} \times I \rightarrow S^{\infty}.$$ Furthermore, we have that $\frac{g_0}{\left| g_0\right|} = \frac{f_1}{\left| f_1\right|}$. Therefore, it follows that $T \simeq N$ ($G$ gives a homotopy from
$T$ to the north pole) and since the composition of homotopies is again a homotopy we have that $id_{S^{\infty}} \simeq N$. Hence the desired result follows and we conclude that $S^{\infty}$ is contractible.

Answer 7

3 proofs on Wikipedia - basically the same arguments as above. The Hilbert space part is superfluous.