# holomorphic non constant function, and the surjectivity of some boundaries.

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? – 23rd Nov 6 '12 at 16:10

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, $\partial f(U) = f(\bar U)\setminus f(U)\subset f(\bar U\setminus U) = f(\partial U)$.
I have a question with the same. Please help me )=!. Let's suppose the particular case, that $U=D$ is the unit disk. And let's suppose that $f (\partial D) \subset \partial \triangle$ , for some triangle subset of $\Bbb C$. I want to know if this triangle is related with the boundary of $f(D)$ for example if $\partial \triangle$ \subset \partial f(D) ?$If I have this , then in this case I'll have the equality that I want. – Daniel Nov 7 '12 at 1:04 Unfortunately, the other inclusion$f(\partial U) \subset \partial f(U)$is not always true, as also pointed out in another comment. – WimC Nov 8 '12 at 17:56 Please )=!! I need an example of an holomorphic function, defined on a domain D, and continuous on$\overline{D}$such that , does not hold$ f(\partial(U)) \subset \partial (f(U))\$ – Daniel Nov 10 '12 at 18:06