A necessary condition for a proper map

Question

In Steinmetz's Rational Iteration, a holomorphic map $f$ from domain $D$ into some domain $G$ is said to be proper if there is a $k \in \mathbb{Z}^+$ such that $f:D \overset{k:1}{\longrightarrow} G$ ie. every point in $G$ has exactly $k$ preimages counted with multiplicity.
Then the author claims that boundary points are mapped to boundary points under a proper map in the following sense: given any sequence $(z_n)$ converging in $\partial D$, the image sequence $(f(z_n))$ has all its limit points in $\partial G$. He leaves it as an exercise for the reader. My initial attempt was as follows:
Let $\lim_{n \to \infty} z_n =\zeta$. Let $w$ be a limit point $(f(z_n))$. Suppose $w \in G$, then there is a open ball $V$ contained in $G$. Let $f^{-1}(w)=\{\zeta_1,\zeta_2 \ldots \zeta_r\}$. Then there are pairwise disjoint open balls $B_1,B_2 \ldots B_r$ such that:

1. $\zeta_i$ is center of $B_i$,
2. each $B_i$ is properly contained in a open ball $B_{i}^{'}$ which is contained in $D$,
3. $\bigcup_{i=1}^rB_i \subseteq f^{-1}(V)$ and for sufficiently large $n \in \mathbb{Z}^+$,
4. $f^{-1}(f(z_n)) \cap B_i \not = \emptyset$ for all $i$.

(I realised later this was a very sketchy claim)
Then one of the $B_i$ (say wlog $B_1$) contains infinitely many members of $(z_n)$, call this subsequence $(z_{n_k})$. Note $z_{n_k} \longrightarrow \zeta$. By construction $\overline{B_1} \subseteq D$. So, $\zeta \in \overline{B_1} \subseteq D$. But since $D$ is open, $D \cap \partial D = \emptyset$. So we reached a contradiction.
Thus $w \in G^{c} = \overline{G^{c}}$ (since $G^{c}$ is closed). Note $w \in \overline{G}$ (limit point of a subset of $G$). Thus, $w \in \overline{G} \cap \overline{G^{c}}=\partial G$.
The whole argument depends on existence on pairwise disjoint open balls $B_1,B_2 \ldots B_r$. I personally tried to prove this rigorously but failed. I think I have made wrong claim since I never used the full strength of the fact that $f$ is proper. (I used that fibers of $f$ are finite). How should I modify my argument? And in case my argument was wrong I would like some hints for the correct proof (I don't want the full solution, I wish to try this problem myself).

Answer 1

Then one of the $B_i$ [...] contains infinitely many members of $(z_n)$

This is where you need that $f$ is a proper map.

Without loss of generality we can assume that $f(z_n) \to w \in G$. If $f(\zeta_j) = w$ with multiplicity $k_j$ then for sufficiently small $\epsilon > 0$, all $\tilde w \in B_\epsilon(w)$ have exactly $k_j$ preimages in a neighborhood $V_j$ of $\zeta_j$. We also know that $k_1 + \cdots + k_r = k$ is the degree of the proper map. For all sufficiently large $n$ is $f(z_n) \in B_\epsilon(w)$, so that there are $k$ preimages of $f(z_n)$ in $V_1 \cup \cdots \cup V_r$. Now use that $f$ is proper to conclude that $z_n \in V_1 \cup \cdots \cup V_r$.

The rest of your proof can also be simplified a bit: If all $V_j$ have a positive distance from $\partial D$ then $z_n \in V_1 \cup \cdots \cup V_r$ gives a contradiction to $z_n \to \zeta \in \partial D$.