Suppose $R:\widetilde{\mathbb{C}} \rightarrow \widetilde{\mathbb{C}}$ where $R(A) = B$ is a rational mapping from one annulus to another. Assume that one of the components of the complement of $A$ has arbitrarily small spherical diameter and it maps to a component of $B^c$ which also has arbitrarily small spherical diameter.
The paper I am reading states that by the openness of $R$, apparently the only other possibility is that this complementary component maps onto the entire $\widetilde{\mathbb{C}}$. I am struggling to see why this is.
Any help would be greatly appreciated. Thank you.
Note: For further context, I am looking at the proof of Sullivan's No Wandering Theorem. The issue in question is on page 408.
The case phrased in Sullivan's paper is stated different to your question:
We have $R \colon A_n \to A_{n+1}$ where $R$ has bounded derivative (in the spherical metric) and one component of $A_n^c$, which we'll denote by $C_n$ has diameter going to $0$ (it's important to have a family of annuli depending on $n$ in order to "arbitrarily small component" make sense).
Now we know that $\partial C_n$ gets mapped to one of the closed curves in $\partial A_{n+1}$, let $S$ be that closed curve and $S^c = D_1\cup D_2$ be the connected components. Suppose that $R(C_n) \cap D_1 \neq \emptyset$, then we claim that $D_1 \subset R(C_n)$. Indeed, $R(\bar{C_n})$ is compact, therefore, $R(\bar{C_n}) \cap D_1 = (R(C_n) \cup S) \cap D_1 = R(C_n) \cap D_1$ is closed in $D_1$. Now by openness of $R$, $R(C_n) \cap D_1$ is open in $D_1$. Since $D_1$ is connected and $R(C_n) \cap D_1$ is nonempty, then $R(C_n) \cap D_1 = D_1$, which proves our claim.
This is basically said that the image of $C_n$ has to cover whichever component of $S^c$ that it touches. Since the derivative of $R$ is bounded and $C_n$ is shrinking, then the image of $C_n$ also has to be small, therefore, it can only cover the small component of $S^c$ (which is to say the image cannot "cover the entire sphere").