I think I found a proof of Brouwer's fixed point theorem which is much simpler than any of the proofs in my books.

One part is standard: Suppose there is an $f:D^n \rightarrow D^n$ with no fixed points. Then we can draw the ray from $f(x)$ through $x$ to get a retraction $r:D^n \rightarrow S^{n-1}$.

Now, suppose such an $r$ existed. Then we obviously have $S^{n-1} \stackrel{i}{\rightarrow} D^n \stackrel{r}\rightarrow S^{n-1}$ where $ri=\text{id}_{S^{n-1}}$.

Taking de Rahm cohomology and writing $f^*\equiv H^p(f)$ for maps, we get that $i^*r^*=\text{id}^*_{S^{n-1}}$ is an isomorphism, such that $i^*:H^p(D^n)\rightarrow H^p(S^{n-1})$ is an epimorphism for all $p$, but this is impossible for $p=n-1$.

Thus there cannot exist such an $r$ and we have proved Brouwer's fixed point theorem.

So this proof uses nothing except that $H^p$ is a contravariant functor. If we were to do this with homology, we would need to use the notion of degree of maps, but my book on de Rahm cohomology does this by using contractibility and homotopy invariance. Is there some heavy stuff hidden under the surface here that I'm not seeing?

-
This is a good example of the extent to which heavy is partly a matter of taste and background: from my point of view this argument hides an enormous amount of unfamiliar background, and the combinatorial proof via Sperner’s lemma is simple! – Brian M. Scott Nov 30 '12 at 9:48

There is no need of the use of degree. Just apply the covariant functor $H_{n-1}(-)$ and you will get your desired contradiction. More precisely on homology we would have

$$\begin{array}{ccccc} &H_{n-1}(S^{n-1}) &\stackrel{ i_\ast}{\longrightarrow} &H_{n-1}(D^n)& \stackrel{r_\ast}{\longrightarrow} & H_{n-1}(S^{n-1}) \\ \end{array}$$

which are respectively (left to right) $\Bbb{Z}$, 0, $\Bbb{Z}$ contradicting the fact that the identity map factors through zero. I should add that this is one of the standard proofs of the Brouwer fixed point theorem. For example Rotman does it this way in his book on Algebraic Topology, while Hatcher proves it in the case $n=2$ by applying the functor $\pi_1(-)$ instead of homology.

If you want a proof that uses degree, it is not so hard. Suppose you have a map $f:D^n \to D^n$ that has no fixed points. Then we can treat $f$ as a map from the northern hemisphere $D^n_+$ of $S^n$ to itself. Now we can extend $f$ to a map on $S^n$ as follows. We define

$$g(x) = \begin{cases} f(x), & \text{if}\hspace{2mm} x \in D^n_+,\\ f\circ r(x), & \text{if}\hspace{2mm} x \in D^n_{-} \end{cases}$$

where $r(x)$ is reflection about the plane through the equator and $D^n_{-}$ is the southern hemisphere. It is clear that $g(x)$ is a continuous function; furthermore $g(x)$ has no fixed points. Hence we can homotope $i\circ g$ to the antipodal map on $S^n$ that has degree $(-1)^{n+1}$, where $i : D^n_{+} \hookrightarrow S^n$ is inclusion.

However $i \circ g$ is not surjective because for example no point in the southern hemisphere is in the image. It follows that $\deg i \circ g = 0$ contradicting the fact that we found it to be $(-1)^{n+1}$.

-
Ah, of course, the identity map on $H_{n-1}(S^{n-1})=\mathbb{Z}$ cannot factor through $H_{n-1}(D^n)=0$. – Espen Nielsen Nov 30 '12 at 9:37
@espen180 Yes basically you can even take singular (co)homology and you will get the same contradiction. – user38268 Nov 30 '12 at 10:39

de Rham cohomology is only functorial with respect to smooth maps.

-
I see your point. It should not be a problem if I take singular cohomology instead. – Espen Nielsen Nov 30 '12 at 13:06