Show that f: $\mathbb{R}$/$\mathbb{Z}$ $\to$ $\mathbb{R}$/$\mathbb{Z}$ orientation reversing. Then f(x) = x has exactly 2 solutions.

Question

Im having some problem with the following question.

Show that if $f: \mathbb{R}/\mathbb{Z} \to \mathbb{R}/\mathbb{Z}$ orientation reversing, then $f(x) = x$ has exactly $2$ solutions. ($f$ has $2$ fixed points)

I was wondering if anyone can help.

Answer 1

Take a circle as $[0,1)$. (Drawing this as a picture will help!) Take $x$ such that $f(x) \neq x$. Choose this $x$ maximally so you have $[0,1]$ partitioned in the following way: $$[x,p),[f(p),f(x)),[f(x),f(q)), [q,1).$$ The reason you can write this partition in a preceding way is because that $f$ is orientation reversing. The fact this $x$ is maximally chosen means that $f(p) = p$, and $f(q) = q$.