Image of basin of attraction is not an entire Fatou component.

Question

The function $E(z)=\lambda e^{z}$ with $0<\lambda<1/e$ has an attracting point in $p\in \mathbb{R}$. Using Koenings coordinates, if $U$ is the immediate basin of attraction for $p$, why is $E(U)$ not an entire Fatou component? This was part a of a proof, and I don't understand this step.

Answer 1

Koenig's coordinates are injective only on a neighborhood of the fixed point, but not on the whole immediate attracting basin $U$. So it cannot be used to prove that $E(U)$ is strictly contained in $U$ (which is indeed not the case).

Your question can be rephrased as : prove that $E(U)=U -\{0\}$. Since the Fatou set is invariant, it is clear that $E(U)$ is an open, connected subset of $U$. Let $z_0 \in U$, and let $y \in U - \{0\}$: to see that $y \in E(U)$, choose a path $\gamma$ in $U$ joining $f(z_0)$ and $y$ and avoiding $0$. You can cover $\gamma$ by finitely many disks $U_i \subset U$ on which a local inverse $f^{-1}$ is well-defined, and agree on intersections of the $U_i$, such that $f^{-1}(f(z_0))=f(z_0)$. Since each $\gamma(t)$ is in $U$, each $f^{-1}(\gamma(t))$ is in the Fatou set, so in $U$ (otherwise you would cross the boundary of $U$ at some point). So $f( \gamma(1) ) =y$ and $\gamma(1) \in U$, QED.

Note: more generally, the argument (analytic continuation of $f^{-1}$) proves that for any holomorphic map $f: \mathbb C \to \mathbb C$, the image of a Fatou component $U$ is $f(U) - O$, where $O$ is the set of omitted values (ie $f(\mathbb C)=\mathbb C- O$. By Picard's theorem, $O$ is finite and has cardinal at most one.