TheoremProved

Brouwer fixed-point theorem

Statement

Every continuous function $f: D^n \to D^n$ from the closed unit ball in $\mathbb{R}^n$ to itself has at least one fixed point, that is, a point $x$ with $f(x) = x$ .

Why is it true?

Stir a cup of coffee however you like, as long as the motion is continuous and the coffee stays inside the cup: some point of the liquid always ends up exactly where it started. Crumple a sheet of paper and lay it flat again inside its original outline, without tearing it: some point of the paper lands exactly on top of itself.

Proof sketch

Suppose, for contradiction, that $f$ has no fixed point. For each $x \in D^n$ , draw the ray starting at $f(x)$ through $x$ ; let $r(x)$ be the point where this ray exits through the boundary sphere $S^{n-1}$ . This $r$ is a continuous map $D^n \to S^{n-1}$ that is the identity on the boundary - a retraction of the ball onto its boundary sphere. Algebraic topology (the fact that the boundary sphere is not a retract of the ball, detectable via homology or, for $n=1$ , directly via the intermediate value theorem) shows no such retraction exists, giving a contradiction.

Topics that use this theorem

Homology and cohomology

Related theorems

Borsuk–Ulam theorem Intermediate value theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

L. E. J. Brouwer (1911). Über Abbildung von Mannigfaltigkeiten · DOI:10.1007/bf01456931
John Milnor (1978). Analytic proofs of the 'hairy ball theorem' and the Brouwer fixed point theorem