I am using McMullen-Complex Dynamics and Renormalization (p. 99). There you can find the definitions of renormalization and polynomial-like mapping.
Assume $f(z) := z^2 + c$, where $c\in\mathbb{C}$ is fixed. By hypothesis I know that $f^n:U\to V$ is a renormalization. Defining $V(i) := f^i(U)$, it can be written \begin{equation} U\overset{f}{\longrightarrow}V(1) \overset{f}{\longrightarrow} V(2) \overset{f}{\longrightarrow} \dots \overset{f}{\longrightarrow} V(n) = V, \end{equation}
where the first map $U\to V(1)$ has degree 2 (because the critical point $0$ is in $U$) and the remaining maps are univalent (because $f^n$ and the first map of the composition have degree $2$).
- Here is the first doubt: it is also stated that the first map $U\to V(1)$ has to be proper (this means that the preimage of a compact set is compact). Is it trivial?
- Moreover, on the next property stated in the book (p. 99), it refers to the connected component of $f^{i-n}(U)$ contained in $V(i)$. My question: Why do we know that $f^{i-n}(U)\cap V(i)\neq \emptyset$? And, once proved that $f^{i-n}(U)\cap V(i)\neq \emptyset$, why is this intersection a connected component of $f^{i-n}(U)$?
I have tried to imagine the sets $V(i)$ or at least to relate them. If I am not wrong, although $U\subset V$, $U\cap V(1)\neq U$ because if the critical point $0\in U$ is in some $V(i)\;(1\leq i < n)$ then the corresponding iteration should have degree $2$ and this is not possible as I argued above.
I hope someone can help me. Thank you.