Edit: So, my original question (stated below) was to find an error in my "proof" that immediate parabolic basins for rational maps are always simply connected.
Since I have not received any answers as of yet I would ask alternatively if someone could point out an explicit example of a rational map with a parabolic fixed point which has a non-simply connected immediate basin, so that I could hopefully check by examining this example where my argument goes wrong.
Kind regards
an idiot


Hello!
Sorry if this seems stupid. I know there must be an error in my thinking.  

Let f be a rational map on the Riemann sphere with a parabolic fixed point $f(z\_0)=z\_0$, $f'(z\_0)=e^{2\pi t}$ with $t\in \mathbb{Q}$.  

I will try to demonstrate that each parabolic immediate basin is simply connected. I know this is wrong. But I don't find the mistake in my "proof". So please help me out.  

By Leau-Fatou Flower theorem, in each immediate parabolic basin of $z\_0$ there is an attractive petal $V$ such that each point in that immediate basin tends to $z\_0$ via $V$.  

Let $V\_0$ be such a petal, and for simplicity's sake let's say that $f(\overline{V\_0})\subset V\_0\cup{z\_0}$ (i.e. no periodic jumping between different petals, can be achieved by simply taking an iterate $f^n$ instead of $f$ for suitable n).  

So we have:  

- $f(\overline{V\_0})\subset V\_0\cup{z\_0}$   

- $V\_0\subset A^\*(z\_0)$ open and simply connected  

- For every $z\in A^\*(z\_0)$ there is some $n\in\mathbb{N}$ with $f^n(z)\in V\_0$  



We may slightly shrink $V\_0$ if necessary such that $\partial V\_0$ does not contain any postcritical points and $\overline{V\_0}$ is homeomorphic to a closed disk.


Now for $k\in\mathbb{N}$ let $V\_k$ be the component of $f^{-1}(V\_0)$ that contains $V\_0$. It's easy to see that $A^\*(z\_0)=\cup\_{k=0}^{\infty}V\_k$.


If $A^\*(z\_0)$ is not simply connected then there must be a minimal $m\in\mathbb{N}$ such that $V\_m$ is not simply connected.  

In that case let $B$ be a component of $ \hat{\mathbb{C}}-\overline{V\_m}$, such that $\partial B$ does not contain $z\_0$.  

Then $\partial B\subset \partial V\_m$ and so $f(\partial B)\subset\partial V\_{m-1}$.  

Since $\partial V\_0$ contains no postcritical points, $\cup\_{k=0}^m \partial V\_m$ contains no critical points.
  
Thus $f^m$ is locally injective on $\partial B\subset\partial V\_m$ and $f^m(\partial B)$ is a full component of $\partial V\_0$ (proper covering), hence $f^m(\partial B)=\partial V\_0$, since $\partial V\_0$ has only one component.  

But then there is $z\in\partial B\subset F(f)$ with $f^n(z)=z\_0\in J(f)$. That's a contradiction.


Can someone help me see my mistake? I hope it's a simple one.



parabolic immediate basins always simply connected?
The mistake is in the statement that $\partial B\subset F(f)$.
There can be points on $\partial B$ and $\partial V\_m$ which are in $J(f)$,
namely preimages of $z\_0$ :-)


An example is $f(z)=z+1-1/z$. There is one petal for the neutral point at infinity.
Let $A$ be the dmain of attraction of $\infty$.
Critical points are $\pm i$. Everything is symmetric with respect to the real line, because the function is real.
One critical point is in $A$, so by symmetry the other one is 
also in $A$. The map $f:A\to A$ is 2-to-1 (because $f$ is of degree $2$),
so Riemann and Hurwitz tell us that $A$ is infinitely connected.


