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 .