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Suppose $G\subset\mathbb{C}$ is open and connected,let $\left\{ f_{n}:n=1,2\ldots \right\}$ be a uniformly bounded sequence of holomorphic functions on $G$ that convergences uniformly on compact subsets to the function $f$. Assume that each $f_{n}$ is one-to-one on $G$ and satisfies $f_{n}(G)\subset G$. Show that if $f$ is not constant,then

$a)$ $f(G)\subset G$ ,and $b)$ $f$ is one-to-one on $G$.

My thought was to use the properties that sequence of holomorphic functions have to prove that $\lim_{n}f_{n}\left(x\right)\neq\partial\Omega$ and $f_{n}\left(x\right)\neq f_{n}\left(y\right)$. I don't know how to use the theorems we all know in complex analysis to approach the result.

Arthur
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Jack
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  • Part (b) is for example answered here: http://math.stackexchange.com/questions/1737017/show-that-f-is-either-injective-or-a-constant-function. – Martin R Jul 25 '16 at 11:20
  • Part (a) can be concluded from the https://en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis). – Martin R Jul 25 '16 at 11:21

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