-1

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 $\lim_{n \rightarrow \infty} f(z_n) \in \partial V \cup \{\infty\}$?

Mcjosher
  • 21
  • 1
    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 $\partial U$). – mrf Sep 25 '15 at 20:25
  • I edited the question. It really wasn't worded very well. – Mcjosher Sep 28 '15 at 16:50

1 Answers1

0

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 $f(1/2) \notin \partial V$.

mrf
  • 43,639