Let $A,B$ denote domains in $\mathbb{C}$, and $f:A\rightarrow B$ is a holomorphic mapping. Suppose that $f$ is proper ($f^{-1}$ takes compact sets to compact sets). Prove that $f(A)=B.$
Proper Holomorphic Mappings
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complex-analysis
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0$f$ must be non-constant... – 2011-08-23
1 Answers
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A proper map is easily seen to be a closed map. Moreover, a non-constant holomorphic map on a domain in $\mathbb C$ is always an open map (see Wikipedia, Open Mapping Theorem for example).
Therefore, $f(A)$ is both open and closed (and non-empty of course), so by connectedness of $B$, $f(A)=B$.
Edit: Closedness of $f$. If $f(x_n)\to y$, then $\{f(x_n), n \in \mathbb N\}\cup \{y\}$ is compact thus $\{x_n, n \in \mathbb N\}\subset f^{-1}(\{f(x_n), n \in \mathbb N\}\cup \{y\})$ lives in a compact set, therefore a subsequence of $(x_n)$ is convergent to some $x\in A$, satisfying of course $f(x)=y$, so $f$ is closed.