Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 11H (Topics in Analysis)

Read the $2n+1$ colours from left to right. If every neighbouring pair had the same colour, mathematical induction would make every point the same colour as $-1$, contradicting the different colour of $1$. Therefore at least one adjacent pair has different colours.

Suppose for a contradiction that $A\cap B=\varnothing$. Each grid point then belongs to exactly one of $A$ and $B$; colour it red in $A$ and green in $B$. By part (a), there are adjacent points $a_n\in A$ and $b_n\in B$, in either order, withThe interval $I$ is a compact set, so a subsequence of $(a_n)$ converges to some $x\in I$. Along the same subsequence $b_n\to x$. Since $A$ and $B$ are closed sets, $x\in A\cap B$, a contradiction. Hence $A\cap B\ne \varnothing$.

If such a continuous function existed, the inverse imageswould be closed, would cover $I$, and would contain $-1$ and $1$ respectively. Part (b) gives a point in $A\cap B$, which is impossible.

Suppose that the continuous function $f:I\to I$ has no fixed point. Then $g(x)=f(x)-x$ is continuous and never zero, sois a continuous function from $I$ to $\{-1,1\}$. Since $f(-1)\ge -1$ and equality would be a fixed point, $h(-1)=1$. Similarly $h(1)=-1$. Multiplying $h$ by $-1$ contradicts part (c). Thus $f$ has a fixed point.

Conversely, suppose a continuous $f:I\to \{-1,1\}$ satisfies $f(-1)=-1$ and $f(1)=1$. The function $g=-f:I\to I$ is continuous, so part (d) gives $x=g(x)=-f(x)$. If $f(x)=1$, this forces $x=-1$, contradicting $f(-1)=-1$; if $f(x)=-1$, it forces $x=1$, contradicting $f(1)=1$. Hence no such $f$ exists.

Suppose $I=A\cup B$, where $A$ and $B$ are disjoint closed sets with $-1\in A$ and $1\in B$. Since each is the complement of the other, both are also open in $I$. Define $f=-1$ on $A$ and $f=1$ on $B$. This locally constant function is continuous and contradicts part (c). Therefore $A\cap B\ne \varnothing$.

The conclusions of (b) and (c) remain true. Indeed, the unit circle is a path-connected space, and the upper semicircle supplies a path from $-1$ to $1$; restricting the proposed closed cover or two-valued continuous function to that path reduces either claim to the corresponding result on $I$.

The fixed-point conclusion (d) fails. For example, any nontrivial spatial rotation $f(z)=e^{i\theta }z$ of the unit circle has no fixed point when $e^{i\theta }\ne 1$; the antipodal map $f(z)=-z$ is the simplest example.