I am trying to understand the proof of the no wandering domain theorem from Beardon's iterations of rational functions and thought a good start would be to omit the quasiconformal structures part and prove the statement for circles only. So I assume there is a rational map $R: \mathbb{C}_\infty \rightarrow \mathbb{C}_\infty$ and a disjoint sequence of balls $B_j:= B_{r_j}(a_j)$ of radius $r_j$ around $a_j$ with $R|_{B_j} = B_{j+1}$ and want to show that this gives a contradiction. There is a Möbius-map $$\varphi_j: B_1(0) \rightarrow B_j \\ z \mapsto r_j z +a_j$$ which maps the j-th circle to the unit-circle which should replace the conformal map but I don't know how to proceed from here and translate the condition that Beltrami Coefficients extend to the whole Riemann sphere as Möbius-maps are already biholomorphic on the sphere.
So do you guys think this approach is fruitful? Or is there already a contradiction I can use?
Edit: My broad understanding of the actual proof from Beardon's Iteration of rational functions is as follows.
If there is a wandering domain there is also a domain $W$ where for any $n$, $R$ maps $R^{\circ n}(W)$ to $R^{\circ n+1}(W)$ homeomorphically and the domains are all simply connected. Also their diameter tends to 0.
Let's fix $W$. There is a conformal equivalence $g$ from $B_1(0)$ to $W$. Any Beltramicoefficient $\mu$ on $B_1(0)$ can be translated to a Beltramicoefficient $\nu$ on $W$ using $$ \nu (g(z) = \left(\frac{g'(z)}{\bar{g'(z)}}\right) \mu(z) $$
Let's also fix $R$. Any Beltramicoefficient $\nu$ on $W$ extends on $\mathbb{C}_\infty$ in a sense that if $\varphi$ is $\nu$-conformal, then $\varphi \circ R \circ \varphi^{-1}$ is rational and of degree $\deg R$. This means $\mu$ on $B_1(0)$ corresponds to a rational map with $\deg R$ on $\mathbb{C}_\infty$
For each $t \in [0,1]$ there are Beltramicoefficients $\mu_t$ on $B_1(0)$ such that $||\mu_t||_\infty < \epsilon$ and for $\varphi_t$ the in 3) constructed quasiconformal map on $\mathbb{C}_\infty$ $$\varphi_{_t} \circ R \circ \varphi_{_t}^{-1} = S$$ is always the same rational map and even for all $z$ the map $ t \mapsto \varphi_t(z)$ is continuous.
If we define $\Phi_t := \varphi_0^{-1} \circ \varphi_t$ we see that $\Phi_0$ is the Identity and for each $t$, the map $ z \mapsto \Phi_t(z)$ is a homeomorphism on $\mathbb{C}_\infty$. Also for each $z$ the map $t \mapsto \Phi_t(z)$ is continuous on [0,1]. Thus for each $t \in [0,1]$ $\Phi_t$ is the Identity on $J(R)$ the Julia set and maps every component of $F(R)$ the Fatou set onto itself.
Let's define $W_0 := \phi_0(W)$. Each $\phi_t$ must map $W$ to $W_0$. Furthermore there is a conformal equivalence $h: W_0 \longrightarrow B_1(0)$ and thus for $g: B_1(0) \longrightarrow W$ the map that translates a Beltramicoefficent from $(B_1(0), \mu_t)$ to $(W,\nu_t)$ we can map $$ (B_1(0), \mu_t) \overset{g}{\longrightarrow} (W,\nu_t) \overset{\varphi_t}{\longrightarrow} W_0 \overset{h}{\longrightarrow} B_1(0) $$ where $(A, f)$ is the riemann surface defined by $A \subset \mathbb{C}_\infty$ and $f: \mathbb{C}_\infty \longrightarrow \mathbb{C}_\infty$ . This mapping is analytic and thus $\psi_t := h \circ \varphi_t \circ g$ is $\mu_t$-conformal from $B_1(0)$ to itself and extends to a homeomorphism on the closed disc. Moreover, since $\psi_0^{-1} \psi_t = g^{-1}\Phi_t g$ we have $$ \psi_0^{-1}\psi_t = Id \Rightarrow \psi_t = \psi_0$$
For any $\mu_t$-conformal map $\Psi_t: B_1(0) \longrightarrow B_1(0)$ there is a Möbius-map $M_t$ such that $\psi_t = M_t \Psi_t$. This means that $$ M_t \Psi_t = \psi_t = \psi_0 = M_0 \Psi_0. $$ But we can construct enough pairwise different $\Psi_t$ that fullfill $\Psi_t = \Psi_0$ on some arc of the unit sphere. This means $M_0 = M_t$, so $\Psi_0 = \Psi_t$ which is the contradiction we were looking for.
As of the length of this post I omitted a lot. The problem of following this proof is, that the contradiction lies in the number Beltramicoefficients. And I have no idea of how to find a different contradiction when using Möbius maps.