proof of Brouwer fixed point theorem

Assume that there does exist a map from f:BnBn with no fixed pointPlanetmathPlanetmath. Then let g(x) be the following map: Start at f(x), draw the ray going through x and then let g(x) be the first intersectionMathworldPlanetmathPlanetmath of that line with the sphere. This map is continuousMathworldPlanetmathPlanetmath and well defined only because f fixes no point. Also, it is not hard to see that it must be the identityPlanetmathPlanetmath on the boundary sphere. Thus we have a map g:BnSn-1, which is the identity on Sn-1=Bn, that is, a retractionMathworldPlanetmathPlanetmathPlanetmath. Now, if i:Sn-1Bn is the inclusion mapMathworldPlanetmath, gi=idSn-1. Applying the reduced homology functorMathworldPlanetmath, we find that g*i*=idH~n-1(Sn-1), where * indicates the induced map on homologyMathworldPlanetmathPlanetmath.

But, it is a well-known fact that H~n-1(Bn)=0 (since Bn is contractibleMathworldPlanetmath), and that H~n-1(Sn-1)=. Thus we have an isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of a non-zero group onto itself factoring through a trivial group, which is clearly impossible. Thus we have a contradictionMathworldPlanetmathPlanetmath, and no such map f exists.

Title proof of Brouwer fixed point theoremPlanetmathPlanetmath
Canonical name ProofOfBrouwerFixedPointTheorem
Date of creation 2013-03-22 13:11:24
Last modified on 2013-03-22 13:11:24
Owner bwebste (988)
Last modified by bwebste (988)
Numerical id 6
Author bwebste (988)
Entry type Proof
Classification msc 47H10
Classification msc 54H25
Classification msc 55M20