parabolic immediate basins always simply connected?

Question

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.

Answer 1

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.