Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 11I (Topics in Analysis)

Assume (i), and suppose a map $f$ as in (ii) existed. Thenare closed, cover $T$, and contain $I,J,K$, respectively. Butsince the three sides have no common point. This contradicts (i), so (i) implies (ii).

Conversely, suppose (i) fails. Let $A,B,C$ be a counterexample and putNo point lies in all three closed sets, so $D=d_A+d_B+d_C>0$. Identify $T$ affinely with the standard simplex so that side $I$ is the side on which the first barycentric coordinate is zero, and similarly for $J,K$. Define $f(x)$ to be the point with barycentric coordinatesBecause $A\cup B\cup C$ covers $T$, at least one distance is zero, so $f(x)\in \partial T$. If $x\in I\subset A$, then $d_A(x)=0$, so $f(x)\in I$; likewise for $J,K$. Thus $f$ is a forbidden map from (ii). This proves (ii) implies (i).

Brouwer's fixed-point theorem in the plane states that every continuous map from a closed disk to itself has a fixed point. Equivalently, the same holds for every nonempty compact convex subset of ${\mathbb{R}}^2$.

On the closed unit disk defineFor $|z|\le 1$,so $F$ maps the disk continuously into itself. Brouwer supplies $z$ with $F(z)=z$, which is equivalent toThus this polynomial has a root with $|z|\le 1$.

Write $\beta (s)=({\beta }_1(s),{\beta }_2(s))$, $\gamma (t)=({\gamma }_1(t),{\gamma }_2(t))$, andThe endpoint conditions implyLet $P(x)=\max (-1,\min (1,x))$ and define the continuous self-map of $I^2$By Brouwer, $F$ has a fixed point $(s,t)$. If $s$ is interior, its fixed-point equation gives $h_2=0$. At $s=-1$ or $s=1$, the projection equation and the corresponding displayed boundary sign again force $h_2=0$. The identical argument in the second coordinate gives $h_1=0$. HencesoThe two paths intersect.