Let $C$ be the cycle of points $\{z_1,..,z_m\}$ for a rational map $R$ of degree at least $2$. Show that either $R^{-1}(C)$ contains pre-periodic points or $C$ is completely invariant, has two elements, and lies in the Fatou set.
I'm stuck trying to pull a few ideas together to prove this. Here are some facts that I have so far:
$C$ is finite. I'm not sure how to show invariance. However, if I
can show $C$ is invariant, then the following is true: for some $n$ we have that $R^n(C)=C$. Suppose that $R^n$ has degree $d$. Then by Riemann-Hurwitz we have $k(d-1)\leq 2d-2$.$R^{-1}(C)=C$ in only a few circumstance. But, if it doesn't, it's because of pre-periodic points. In particular, $|R^{-1}(C)|<=n|C|$.
My attempt at a proof: Since $C$ has a finite orbit we know it's an invariant set. Since $R$ is a rational map and $C$ is completely invariant, then by the first bullet point we know that it has at most two elements. Furthermore, $C$ is a set of exceptional points, so it lives in the Fatou set.
I'm not sure if this is correct. Nor am I sure how the problem changes with the introduction of pre-periodic points.