Given a rational function $R$ from the Riemann sphere to itself and an annulus fatou component $A_0$ we can create the chain $$A_0 \xrightarrow{R} A_1 \xrightarrow{R} A_2 \xrightarrow{R} ...$$ One can show that all $A_i$ here are also annuli. Assuming that $R:A_i \rightarrow A_{i+1}$ is a covering map with degree >1, how does one show that $R$ has a critical point in each complementary component of each $A_i$.
(For context, I am working through Sullivan's non-wandering theorem, and this is one of the stages in his set up)
Any help would be greatly appreciated.
Taking for a fact that each $A_{i}$ is an annulus we can use the Riemann-Hurwitz theorem to derive your answer. The theorem reads as follows. Given two strict subsets $X,Y$ of the sphere $\hat{\mathbb{C}}$ with respective connectivities $m,n$ and a proper mapping $f:X\rightarrow Y$ of degree $d$ between them, having $r$ critical points in $X$ counted with multiplicity $$ m-2=d(n-2)+r. $$
In your case, we can take $R:A_{i}\rightarrow A_{i+1}$ as a proper mapping of degree $d_{i}>1$ between topological annuli. As such, we have connectivities $c_{i}=c_{i+1}=2$, and by the Riemann-Hurwitz formula $r=0$. So, there are no critical points in $A_{i}$, meaning that at least one component of the complement of $A_{i}$ contains one critical point.
If all critical popints were contained in just one component of the complement of $A_{i}$ we can "fill" the annulus with the other in order to obtain a simply connected set $U$ in $\hat{\mathbb{C}}$ upon which $R$ acts as a proper mapping of some degree $\lambda$. Applying the Riemann-Hurwitz formula on this simply connected set we get immediatly $$-1=\lambda(c'-2)+0,$$ for $c'$ the connectivity of $R(U)$. We get $\lambda=1$ which whould imply that $R|_{A_{i}}$ restricted on the annulus $A_{i}$ acts as a homeomorphism, contradicting $d_{i}>1$.
I hope this is helpful!