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.