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General fact: Let $A_{-1} \subset ... A_n \subset ..$ be a filtration of cellular inclusions of $CW$ complexes.(More generally let this be a filtration of cofibrations). Then $A_n$ contractible in $A_{n+1}$ $\implies$ $A:=colim_n A_n$ 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 $f|_t=\alpha_{n+1}(2^{n+1} t-2^{n+1}+2)\circ\alpha_n(1)...\circ\alpha_1(1)$ 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 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 a lot of other things are contractible too(like milnor space).

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).

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

Proof: Consider the composition $A_n \times I \to A_{n+1} \to A$. Since this is a cofibration, extend it to a map $A \times I \xrightarrow{\alpha_n} A$. The map $f: A\times I \to A$ defined by $f|_t=\alpha_{n+1}(2^{n+1} t-2^{n+1}+2)\circ\alpha_n(1)...\circ\alpha_1(1)$ 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 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 a lot of other things are contractible too(like milnor space).

General fact: Let $A_{-1} \subset ... A_n \subset ..$ be a filtration of cellular inclusions of $CW$ complexes.(More generally let this be a filtration of cofibrations). Then $A_n$ contractible in $A_{n+1}$ $\implies$ $A:=colim_n A_n$ 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 $f|_t=\alpha_{n+1}(2^{n+1} t-2^{n+1}+2)\circ\alpha_n(1)...\circ\alpha_1(1)$ 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 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 a lot of other things are contractible too(like milnor space).

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

Proof: Consider the composition $A_n \times I \to A_{n+1} \to A$. Since this is a cofibration, extend it to a map $A \times I \xrightarrow{\alpha_n} A$. The map $f: A\times I \to A$ defined by $f|_t=\alpha_{n+1}(2^{n+1} t-2^{n+1}+2)\circ\alpha_n(1)...\circ\alpha_1(1)$ 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 $a \in A$.

$S^\infty$ satisfies the hypotheses of this general fact: $S^n \to S^{n+1}$ is contractible - $S^{n+1}$ can be given an $n-skeleton$ that is a point, and cellular approximation implies that this map is this inclusion is homotopic to the null map.

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

Proof: Consider the composition $A_n \times I \to A_{n+1} \to A$. Since this is a cofibration, extend it to a map $A \times I \xrightarrow{\alpha_n} A$. The map $f: A\times I \to A$ defined by $f|_t=\alpha_{n+1}(2^{n+1} t-2^{n+1}+2)\circ\alpha_n(1)...\circ\alpha_1(1)$ 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 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 a lot of other things are contractible too(like milnor space).