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Suppose we have a bounded open set $U \subset \mathbb{C}^n$ and a surjective holomorhpic map $f: U \rightarrow V$, where $V$ is open in $\mathbb{C}^n$ and not necessarily bounded. Also suppose we have a sequence $z_n \in U$ such that $z_n \rightarrow w \in \partial U$. Then is it true that $\lim_{n \rightarrow \infty} f(z_n) \in \partial V \cup \{\infty\}$?

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    $\begingroup$ Surely you don't mean exactly what you've written. First of all, you could take $V = \mathbb{C}^n$ and a bounded (or even constant) map $f$. Are you assuming $f$ is biholomorphic and surjective? Second, if $f$ is only assumed to be holomorphic on $U$, what do you mean by $f(\partial U)$? ($f$ is not defined on $\partial U$). $\endgroup$ – mrf Sep 25 '15 at 20:25
  • $\begingroup$ I edited the question. It really wasn't worded very well. $\endgroup$ – Mcjosher Sep 28 '15 at 16:50
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No, this is not even true for $n=1$.

Take $U = \{ z = x+iy \in \mathbb{C} : 0 < x < 1, 0 < y < 10 \}$ and $f(z) = e^z$.

Then $f$ maps $U$ surjectively onto the annulus $V = \{ w: 1 < |w| < e \}$, but $f(1/2) \notin \partial V$.

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