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.

Could you please clarify what exactly you're assuming about $f$? (It's a homeomorphism, it's smooth, ....) — Andrew D. Hwang, Apr 23 '15 at 14:06
Assume $f$ is continuous and for the sake of an example, define the orientation reversing map $f = -x$. Then $f(0) = -0 = 0$ and $f(0.5) = -0.5~\mathrm{mod}~1 = 0.5$. What intuition does this give you for other orientation-reversing maps? — bashfuloctopus, Apr 23 '15 at 14:11
So I get that we can write $f = \alpha -x$ and this gives us fixed points $0.5\alpha$ and $-0.5 + 0.5\alpha$ is this enough? — user149134, Apr 23 '15 at 14:25

score 1 · Accepted Answer · edited Oct 13 '18 at 19:13

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 $f(q) = q$.

edited Oct 13 '18 at 19:13

answered Apr 23 '15 at 18:29

Usman Qureshi

26

Please learn how to properly format equations on this site by reading this : http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference . No one likes to read poorly formatted maths. – Alexandre Halm Apr 23 '15 at 18:32

score 0 · Answer 2 · answered Oct 13 '18 at 20:50

Translating the assumptions into simpler terms it means that $f$ can be lifted into a continuous function $g:\Bbb R\to\Bbb R$, which

is monotonically falling (from orientation reversing),
can be assumed to have $g(0)\in [0,1)$ and
$g(x+1)=g(x)-n$ for some $n\in \Bbb N_{>0}$.

Then by the intermediate value theorem this $g$ will have intersections over $[0,1]$ with any of the functions $h_k(x)=x-k$, $k=0,1,...,n$. Any of these intersection points represents one fixed points of $f$.

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

2 Answers2