Let $\mathcal{T}_{\infty}= \left\{ U \subset \mathbb{R}^{\infty}: \ U \cap \mathbb{R}^n \in \mathcal{T}_n, \text{ for } n=1,2,… \right\} $.
Of course $\mathcal{T}_{\infty}$ is topology in $\mathbb{R}^{\infty}$. How to prove that $S^{\infty} = \{ v \in \mathbb{R}^{\infty} : \ \|v\|=1 \}$ is contractible?
🙂
Can we find homeomorphism without fixed point from $D^{\infty} = \{ v \in \mathbb{R}^{\infty} : \ \|v\| \le 1 \}$ onto $D^{\infty}$? I was trying to find such homeomorphism, but I failed…
Best Answer
You'll find a proof that the infinite dimensional sphere is contractible on page 88 of Allen Hatcher's Algebraic Topology kindly hosted for free by him on his website.
The proof gives an explicit homotopy between the identity map and the constant map on the sphere $S^{\infty}$.
Let $f_t\colon\mathbb{R}^{\infty}\rightarrow \mathbb{R}^{\infty}$ be given by $f_t(x_1,x_2,\ldots)=(1-t)(x_1,x_2,\ldots)+t(0,x_1,x_2,\ldots)$. For all $t\in[0,1]$, this map sends nonzero points to nonzero points, so $f_t/|f_t|$ is a homotopy from the identity map on $S^{\infty}$ to the map $(x_1,x_2,\ldots)\mapsto (0,x_1,x_2,\ldots)$. We then define a homotopy from this map to the constant map at $(1,0,0,\ldots)$ by setting $g_t(x_1,x_2,\ldots)=(1-t)(0,x_1,x_2,\ldots)+t(1,0,0,\ldots)$. The homotopy is then given by $g_t/|g_t|$. The composition of these two homotopies then gives a homotopy from the identity map to the constant map, and so $S^{\infty}$ is contractible.