# Expanding distances on a sphere

This came up in a discussion at AIM.

Let $\mathbb S^{n-1}\subset \mathbb R^n$ be the unit sphere. Suppose that $f\colon \mathbb S^{n-1}\to\mathbb R^n$ is a continuous map which does not decrease distances: that is, $|f(x)-f(y)|\ge |x-y|$ for all $x,y\in \mathbb S^{n-1}$. By the generalized Jordan theorem the complement of $f(\mathbb S^{n-1})$ has two components, precisely one of which, denoted $\Omega$, is bounded.

Prove that $\Omega$ contains an open ball of radius 1.

Proof. Consider the mapping $g=f^{-1}$, which is 1-Lipschitz. By Kirszbraun’s theorem $g$ has a 1-Lipschitz extension to $\mathbb R^n$, still denoted $g$. Since the topological degree $\mathrm{deg}\, (g,\Omega,0)$ is $\pm 1$ (why?), there exists a point $y\in \Omega$ such that $g(y)=0$. The open ball of radius 1 centered at $y$ cannot contain any points of $\partial\Omega$, and therefore is a subset of $\Omega$.

It remains to answer why $\mathrm{deg}\, (g,\Omega,0) =\pm 1$. In two dimensions one can invoke the Schoenflies theorem, which provides an extension of $g_{|\partial\Omega}$ to a homeomorphism of the plane. In higher dimensions this does not work, but the proof can be saved with the “chain rule” for the degree (found, e.g., in Deimling’s Nonlinear Functional Analysis). Extend $f$ to a continuous map of  $\mathbb R^n$ and note that $g\circ f=\mathrm{id}$ on $S^{n-1}$. The chain rule expresses the degree of $g\circ f$ wrt 0 in terms of the degrees of $f$ and $g$: namely, $1=\mathrm{deg}\,(g,\Omega,0)\cdot \mathrm{deg}\,(f,\mathbb S^{n-1},z)$ where $z$ is an arbitrary point of $\Omega$. Only $1$ and $-1$ have multiplicative inverses in $\mathbb Z$.