Show that $\mathbb{S}^d$ is homeomorphic to the the boundary of the cube $\partial I^{d+1}$.

Question

How to show that the sphere $\mathbb{S}^d$ and the boundary of the cube $\partial I^{d+1}$ are equivalent topological manifolds?

One way is to use the homeomorphism $\varphi: \mathbb S^2 \to I^3 : (x, y, z) \mapsto \frac{(x,y,z)}{\max \{ |x|,|y|,|z| \}}$ which has been defined linked to the result, and then extend it to $d$-dimensional sphere.

But the process of showing that it's a homeomorphism is not so easy.

Can some one give an another way for to show that $\mathbb{S}^d$ is homeomorphic to $\partial I^{d+1}$?

Thanks!

Answer 1

$S^d=\{x\in\Bbb R^{d+1}:\|x\|_2=1\}$ and $\partial I^{d+1}=\{x\in\Bbb R^{d+1}:\|x\|_\infty=1\}$.

There are continuous maps $x\mapsto \|x\|_\infty^{-1}x$ and $x\mapsto \|x\|_2^{-1}x$ in both directions, which are inverses of each other. Therefore, the spaces are homeomorphic.

This argument works with the unit spheres associated to any two norms on $\Bbb R^n$.