proof of Brouwer fixed point theorem

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 $g(x)$ bethe first intersection of that line with the sphere. This map is continuous and well defined onlybecause $f$ fixes no point. Also, it is not hard to see that it must be the identity on the boundarysphere. 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 inclusionmap, $g\circ i={\mathrm{id}}_{S^{n-1}}$. 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 itselffactoring through a trivial group, which is clearly impossible. Thus we have a contradiction,and no such map $f$ exists.