Prove the Brouwer fixed point theorem for maps \(f: D^n \to D^n\) by applying degree theory to the map \(S^n \to S^n\) that sends both the northern and southern hemispheres of \(S^n\) to the southern hemisphere via \(f\). [This was Brouwer’s original proof.]
Solution
Regard \(S^n\) as the double of \(D^n\), with northern and southern hemispheres each identified with \(D^n\) and glued along their common boundary. Given \[ f:D^n\to D^n, \] define \[ F:S^n\to S^n \] by applying \(f\) on either hemisphere and regarding the image as lying in the southern hemisphere.
1. The map \(F\) has degree zero.
Its image is contained in the closed southern hemisphere, hence misses every point in the interior of the northern hemisphere. Therefore \(F\) factors through a contractible subspace of \(S^n\), so \(F\) is nullhomotopic. Thus \[ \deg F=0. \]
2. If \(f\) had no fixed point, then \(F\) would have no fixed point on \(S^n\).
On the southern hemisphere this is exactly the assertion that \(f(x)\ne x\). On the northern hemisphere the image under \(F\) lies in the southern hemisphere, so a fixed point could occur only on the equator; there it would again give a boundary point \(x\in D^n\) with \(f(x)=x\).
3. Any fixed-point-free map \(G:S^n\to S^n\) is homotopic to the antipodal map.
For \(x\in S^n\), the vectors \(G(x)\) and \(x\) are never equal. Hence \[ H_t(x)=\frac{(1-t)G(x)-t x}{\|(1-t)G(x)-t x\|} \] is defined for all \(t\in[0,1]\): the numerator could vanish only if \(G(x)\) were a positive scalar multiple of \(x\), which for unit vectors would force \(G(x)=x\). At \(t=0\) this is \(G\), and at \(t=1\) it is the antipodal map \(x\mapsto -x\).
4. Therefore \(f\) must have a fixed point.
If \(f\) had no fixed point, then by <1>2–<1>3 \[ \deg F=\deg(-\operatorname{id}_{S^n})=(-1)^{n+1}, \] which is \(\pm1\). This contradicts 1, where \(\deg F=0\). Hence every continuous map \[ f:D^n\to D^n \] has a fixed point.