I am considering rational functions $R:\overline{\mathbb{C}}\rightarrow\overline{\mathbb{C}}$ of degree $d\geq 2$.
A completely invariant set $U$ is a set for which it and its complement are invariant under $R$. Equivalently $R^{-1}(U)=U$. By a domain I mean an open connected set (not necessarily simply connected).
Now, I found the following argument that there can be at most $2$ simply connected completely invariant domains: First, note that $R$ has at most $2d-2$ critical points. Since $U$ is completely invariant, $R$ is $d$-to-$1$ on $U$. Then $U$ must contain $d-1$ critical points and thus there are at most $2$ such domains. My first question is, why exactly does it follow that $U$ contains $d-1$ critical points of $R$? (I found a relation to the Riemann-Hurwitz formula somewhere, but is there a more elementary reason?)
My second questions is, what can we say if we drop the simply connected assumption? As an example $R(z)=z^2$ has the four completely invariant domains $\Delta$, $\overline{\mathbb{C}}-\overline{\Delta}$, $\Delta-\{0\}$, $\overline{\mathbb{C}}-\overline{\Delta}-\{\infty\}$ where $\Delta$ denotes the unit disk and the last two of these domains are not simply connected. Can we still find a bound on the number of completely invariant domains?
Edit:
Here is my try for a proof of the existence of $d-1$ critical points in $U$. I would appreciate if anyone could comment on that.
Let $\phi: U \rightarrow \Delta$ be the Riemann mapping, which exists since $U$ is simply connected. Now, for $F=\phi\circ R\circ \phi^{-1}$ we have $F(z)=0$ if and only if $R(\phi^{-1}(z))=\phi^{-1}(0)$. Since $R$ is $d$-to-$1$ on U, there are $d$ such points and $F$ is a self map of the unit disk with $d$ zeroes in $\Delta$. Thus $F$ must be a Blaschke product of degree $d$. As such it has $d-1$ critical points in $\Delta$. Then by the chain rule, there are $d-1$ critical points of $R$ in $U$, since $\phi^\prime\neq 0$.
It is well-known that there are at most two completely invariant Fatou components. You are correct about how to prove this for simply-connected domains (to me using Riemann-Hurwitz seems easier than your approach using Blaschke products). Now, it is easy to see that, if there are two completely invariant Fatou components, then they must be simply connected. (This follows because they both have the whole Julia set as their boundary e.g. by Montel's theorem. Obviously, if there is a curve in one of the components that separates the boundary, then this curve would also separate the other Fatou component.)
So your question comes down to studying completely invariant subsets of one of these Fatou components. There can be many. Indeed, consider your example $z\mapsto z^2$. Take a point in the unit disc, and remove it, its grand orbit, and $0$ from the unit disc. Observe that these points accumulate only at $0$ and at the boundary, so the result is indeed a completely invariant domain. You can remove any closed completely invariant subset, as long as you make sure that it does not separate the disc. If you use an attracting rather than super-attracting example, you can even remove sets with interior (this doesn't work in the superattracting case, as there eventually the forward image of an open set will contain an annulus separating the Fatou component).
But the natural question is about Fatou components, in any case.