I am studying the Escape Trichotomy for Singularly Perturbed Rational Maps of Devaney et al. (reference can be found here). I need help understanding the existence of a level set coming from a Green’s function.
In Proposition 2.1. they want to show that if some critical value $v_\lambda$ of $$F_\lambda(z) = z^{n} + \frac{\lambda}{z^{d}},\ \lambda\in\mathbb{C},\ n\geq2,\ d\geq1$$ lies in the immediate basin of attraction (denoted by $B$), then all of the critical points of $F_\lambda$ also lie in B. $F_\lambda$ is defined on $\hat{\mathbb{C}}$.
In the proof, they construct a neighbourhood $N$, on which an analytic homeomorphism $\phi_\lambda$ conjugates $F_\lambda$ to $z^n$. This neighbourhood includes all the critical values. Now they claim that there exists a level set $\gamma$ of the Green’s function associated to $\phi_\lambda$ which bounds a simply connected open set containing $N$.
The problem is that I am not familiar with Green's function, and therefore, I do not understand this step. So why there exists this level set $\gamma$? After that, the proof is clear to me.