<Home>Qualifying Exams><Analysis>Fall 2013 - Problem 1

Fall 2013 - Problem 1

open mapping theorem

Let $U$ and $V$ be open and connected sets in the complex plane $\C$ , and $\func{f}{U}{\C}$ be a holomorphic function with $f\p{U} \subseteq V$ . Suppose that $f$ is a proper map from $U$ to $V$ , i.e., $f^{-1}\p{K} \subseteq U$ is compact, whenever $K \subseteq V$ is compact. Show that then $f$ is surjective.

Solution.

By the open mapping theorem, $f\p{U}$ is open. If $f\p{U}$ is also closed, then $f\p{U}$ is a non-empty open and closed set in the connected set $V$ , so $f\p{U} = V$ . Otherwise, there exists some $w_0 \in \cl{f\p{U}} \setminus f\p{U}$ , so let $\set{w_n}_n \subseteq f\p{U}$ be a sequence which converges to $w_0$ . Then $K = \set{w_n}_n \cup \set{w_0}$ is compact so by assumption, $f^{-1}\p{K}$ is compact as well. For each $n$ , let $z_n \in U$ be such that $f\p{z_n} = w_n$ , which exist by assumption, and by compactness, there exists a convergent subsequence $\set{z_{n_k}}_k$ which converges to some $z_0 \in U$ . But by continuity, we get

f\p{z_0} = \lim_{k\to\infty} f\p{z_{n_k}} = \lim_{k\to\infty} w_{n_k} = w_0 \notin f\p{U},

a contradiction. Thus, no $w_0$ could have existed to begin with, so $f\p{U}$ was open and closed in $V$ to begin with.