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Let $f: U\to f(U)\subset \Bbb C$ be a holomorphic and nonconstant ( thus in particular an open map), and such that it can be extended continuously on $\overline{U}$.

Where $U$ is a bounded domain of $\Bbb C$ i.e an open and connectedness bounded set. Well I want to know if it's true that the boundary of $U$ is mapped onto ( surjective) the boundary $ f(U)$.

Well at least we know that the image of the boundary of $U$, is contained in the boundary of $f(U)$, i.e $f(\partial(U)) \subset \partial (f(U))$. And that is clear from the fact that $f$ is an open map. (I only used that). I want to know the other containment.

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    $f(\partial U)\subset \partial f(U)$ is incorrect. Can you find some counter-example?2012-11-06

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Since $U$ is bounded, $\bar U$ is compact. It follows that $f(\bar U)$ is compact, and in particular it is a closed subset of $\overline{f(U)}$. But it contains $f(U)$, which is dense in $\overline{f(U)}$, and so we have $f(\bar U) = \overline{f(U)}$. In particular, $\partial f(U) = f(\bar U)\setminus f(U)\subset f(\bar U\setminus U) = f(\partial U)$.

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    Please )=!! I need an example of an holomorphic function, defined on a domain D, and continuous on $\overline{D}$ such that , does not hold $ f(\partial(U)) \subset \partial (f(U))$2012-11-10
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The claim is wrong. E.g., if $U = \{z = x+iy: |z|<1, \, y>0 \}$ is the upper part of the unit disk and $f(z) = z^3$, then $f(U) = \mathbb{D} \setminus \{0 \}$ is the punctured unit disk, and part of $\partial U$ is mapped into the interior of $f(U)$. It is generally true, though, that $\partial f(U) \subseteq f(\partial U)$.