ProofOfBrouwerFixedPointTheorem 2002-12-04 00:11:40 2003-09-05 01:01:10 Proof Brouwer fixed point theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Proof of the Brouwer fixed point theorem: Assume that there does exist a map from $f:B^n\to B^n$ with no fixed point. 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 intersection of that line with the sphere. This map is continuous and well defined only because $f$ fixes no point. Also, it is not hard to see that it must be the identity on the boundary sphere. Thus we have a map $g:B^n\to S^{n-1}$, which is the identity on $S^{n-1}=\partial B^n$, that is, a retraction. Now, if $i:S^{n-1}\to B^n$ is the inclusion map, $g\circ i=\mathrm{id}_{S^{n-1}}$. Applying the reduced homology functor, we find that $g_*\circ i_*=\mathrm{id}_{\tilde{H}_{n-1}(S^{n-1})}$, where $_*$ indicates the induced map on homology. But, it is a well-known fact that $\tilde{H}_{n-1}(B^n)=0$ (since $B^n$ is contractible), and that $\tilde{H}_{n-1}(S^{n-1})=\mathbb{Z}$. Thus we have an isomorphism of a non-zero group onto itself factoring through a trivial group, which is clearly impossible. Thus we have a contradiction, and no such map $f$ exists.