Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 11F (Topics in Analysis)

The planar Brouwer fixed-point theorem states that every continuous map from a closed disc to itself has a fixed point.

Suppose first that a retraction $R:D\to C$ exists. The continuous map $p\mapsto -R(p)$ takes $D$ to itself, so the Brouwer fixed-point theorem gives $p=-R(p)$. Such $p$ lies on $C$, where $R(p)=p$, contradicting $p=-p$.

Conversely, suppose a continuous map $T:D\to D$ has no fixed point. For each $p$, follow the ray from $T(p)$ through $p$ to its last intersection with $C$. This defines a retraction. More explicitly, put $a=p-T(p)\ne 0$ and setThe positive-root formula is continuous, $t(p)\ge 1$, and $R(p)$ is on the unit circle. If $p\in C$, the last intersection is $p$ itself, so $R(p)=p$. Thus the nonexistence of a retraction rules out a fixed-point-free $T$, proving the equivalence.

Fix $z$ with $|z|<1-\epsilon$, and let $\Pi (w)=w/\max (1,|w|)$ be radial projection to the closed disc. The continuous map $p\mapsto \Pi (p-G(p)+z)$ has a fixed point $p$ by the Brouwer fixed-point theorem. On the boundary, $|p-G(p)+z|<\epsilon +|z|<1$, so its image under $\Pi$ is interior and cannot equal the boundary point $p$. Therefore $p\in B$. An interior fixed point of this projection must satisfy $p-G(p)+z=p$, because points outside the disc project to its boundary. Thus $G(p)=z$. Every $z\in B_{1-\epsilon }(0)$ belongs to $G(B)$.

No such map exists. If $H$ fixed the boundary, it would satisfy the hypothesis of part (iii) for every $\epsilon >0$. Choose $0<\epsilon <1-|q|$. Then part (iii) would give $q\in H(B)$, contradicting the stipulated omission of $q$. Equivalently, radial projection from the omitted point would turn $H$ into a forbidden retraction.