Dear fellows, I have come to another conclusion which must be wrong. Let $f$ be a rational map and let $U$ be a connected but not simply connected open subset of the Fatou set such that $f(U)$ is simply connected and bounded away from infinity as well. I have come to understand thus: Every component of $\hat{\mathbb{C}}-U$ that intersects the Julia set must contain a pole. A proof of this, as far as I understand, would be the following: If there was a component $B$ in the complement of $U$ which does not contain a pole but intersects the Julia set, then take a small simply connected open neighbourhood $V$ of $B$, the boundary of which lies in $U$. Since $V$ contains no poles, $f$ is holomorphic on $V$, so $f|_{V}$ takes its maximum on the boundary of $V$ which lies in the simply connected bounded $f(V)$. So the whole of $f(V)$ must lie in $f(U)$ and thus in the Fatou set.

Here is what's wrong with it: If this is true we could conjugate $f$ by a Möbius transformation and exchange $\infty$ by any point which is bounded away from $f(U)$, i.e. every point in $\hat{\mathbb{C}}-\overline{f(U)}$.

Is that correct? Sorry, I am confused. Now I think it might be true, but I don't trust my own reasoning so much. And if this is true there's probably a much less confused proof for it which I would be glad to see.

Any answer much appreciated. Kind regards, an idiot