proof of 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.

Title proof of Brouwer fixed point theorem  ProofOfBrouwerFixedPointTheorem 2013-03-22 13:11:24 2013-03-22 13:11:24 bwebste (988) bwebste (988) 6 bwebste (988) Proof msc 47H10 msc 54H25 msc 55M20