This came up in a discussion at AIM.
Let be the unit sphere. Suppose that is a continuous map which does not decrease distances: that is, for all . By the generalized Jordan theorem the complement of has two components, precisely one of which, denoted , is bounded.
Prove that contains an open ball of radius 1.
Proof. Consider the mapping , which is 1-Lipschitz. By Kirszbraun’s theorem has a 1-Lipschitz extension to , still denoted . Since the topological degree is (why?), there exists a point such that . The open ball of radius 1 centered at cannot contain any points of , and therefore is a subset of .
It remains to answer why . In two dimensions one can invoke the Schoenflies theorem, which provides an extension of 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 to a continuous map of and note that on . The chain rule expresses the degree of wrt 0 in terms of the degrees of and : namely, where is an arbitrary point of . Only and have multiplicative inverses in .