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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

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    $\begingroup$ I don't know the answer to your math question, but you don't sound at all like an idiot. $\endgroup$
    – tweetie-bird
    Commented Aug 19, 2012 at 17:15
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    $\begingroup$ This time your argument and the statement seem to be correct. What bothers you? $\endgroup$
    – Alexandre Eremenko
    Commented Aug 19, 2012 at 20:08
  • $\begingroup$ Thank you. I guess my insecurity bothers me. :) Sometimes I come to wrong conclusions and they stick in my head for a long time. For example, in this case, I had thought up this argument some time ago, but not carefully enough, and had concluded that every complementary component of $U$ that intersects the Julia set contains a preimage of the whole Riemann sphere. Thinking about it now I'm still not sure if that is true. But I had drawn other certainly wrong conclusions, for example that every complementary component of a complete Fatou component must contain a preimage of the whole Riemann $\endgroup$
    – idiot_1337
    Commented Aug 20, 2012 at 13:02
  • $\begingroup$ sphere, which certainly can't be true in the face of an infinitely connected Fatou component for a map of finite degree. So I do fabricate a lot of wrong arguments which contradict simple facts quite often, to the point where I don't trust my own thinking anymore at all. But I am very grateful for your clear and straightforward responses. I should have asked questions more often during the course of my student career. Then my thinking would probably be more thorough and dependable. At the moment I have to weed out a lot of confusion in order to finish my Diploma thesis. :) $\endgroup$
    – idiot_1337
    Commented Aug 20, 2012 at 13:08
  • $\begingroup$ I would try reformulating your hypothesis with $\infty$ replaced by an arbitrary point $\alpha$ and see where it goes. My guess is you will not get that every complementary component of $U$ that interests $J(f)$ contains a preimage of the whole sphere, but that if $f(U)$ is bound away from $\alpha$, then every complementary component of $U$ that intersects $J(f)$ contains the preimage of $\alpha$. $\endgroup$
    – Aaron Golden
    Commented Aug 20, 2012 at 18:51

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