Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 2F (Topics in Analysis)

The planar Brouwer fixed-point theorem says that every continuous map from the closed disk $D$ to itself has a fixed point. If a retraction $r:D\to \partial D$ existed, $x\mapsto -r(x)$ would be a continuous self-map of $D$. At a fixed point $x=-r(x)$, $x$ lies on the boundary, where $r(x)=x$; this would give $x=-x$, impossible on the unit circle.

Conversely, suppose a continuous self-map $f:D\to D$ has no fixed point. For each $x$, take the ray starting at $f(x)$ through $x$, and let $r(x)$ be its exit point on $\partial D$. The denominator $|x-f(x)|$ never vanishes, and solving the quadratic intersection with the circle shows that the exit point depends continuously on $x$. More explicitly, with $v=x-f(x)$,The ray exits beyond or at $x$, so $\lambda \ge 1$, and when $x\in \partial D$ its exit point is $x$. Hence $r$ is a retraction. Nonexistence of a boundary retraction and the planar fixed-point theorem are equivalent.

Define the continuous map $T(z)=-(z^6-z^5+2z^2+1)/6$. For $|z|\le 1$, the triangle inequality gives $|T(z)|\le (1+1+2+1)/6=5/6$. Thus $T$ maps the closed unit disk into itself. By the Brouwer fixed-point theorem, some $z$ has $T(z)=z$, which rearranges toThis supplies the requested existence and even places a root strictly inside the unit disk.