# Homeomorphism of the circle not conjugated to a rational rotation

The Poincaré-Birkhoff Theorem says that if the rotation number $\tau$ of an orientation-preserving homeomorphism of the circle $T$ is irrational, then $T$ is conjugate to (or a topological factor of) $R_\tau$, the rotation by $\tau$.

Now what if $\tau$ is rational? Can someone give me an example of an orientation-preserving homeomorphism $T$ with $\tau(T)\in\mathbb Q$ such that $T$ is not conjugate to $R_\tau$? And what about being just a topological factor?

• For example $x\mapsto x+\sin(2\pi x)/10$, $x\mapsto x+\sin(4\pi x)/20$, etc. – John B Jun 15 '18 at 3:16

Define the homeomorphism $f$ by the formula $$f(\cos(\theta),\sin(\theta)) = \begin{cases} \left(\cos\left(\frac{3}{2} \theta\right), \sin\left(\frac{3}{2}\theta\right)\right) &\text{if 0 \le \theta \le \pi} \\ \left(\cos\left(\frac{1}{2}\theta + \pi\right),\sin\left(\frac{1}{2} \theta + \pi\right)\right) &\text{if \pi \le \theta \le 2\pi} \end{cases}$$ Since $f(1,0)=(1,0)$, the rotation number is $0$. But $f$ has no periodic points other than $(1,0)$, so $f$ is certainly not conjugate to the rotation by angle $0$ (i.e. to the identity map). And, whatever a "topological factor" means, I'm sure it's not that either. For example, $f$ is not even semiconjugate to the identity map.