Ahlfors and Beurling (Ahlfors, Lars; Beurling, Arne Conformal invariants and function-theoretic null-sets. Acta Math. 83, (1950). 101–129.) provided an explicit example of a domain in $\mathbb C$ which allows non-constant bounded holomorphic functions but allows no injective bounded holomorphic functions. Therefore this domain is not biholomorphic to any bounded domain but allows a "light" mapping to a bounded domain, where "light mapping" means that the preimages of every point are totally disconnected. My question is about such examples in several variables. Do there exist a domain in $\mathbb C^{n}$, $n>1$ which is not biholomorphic to any bounded domain but allows a light mapping to a bounded domain? Clearly a Cartesian product of two copies of Ahlfors-Beurling domains will allow a "light" mapping to a bounded domain but there is some chance (I don't know) that it will eventually be biholomorphic to a bounded domain. A somehow simpler question is whether on some domain the algebra of bounded holomorphic functions does not separate points but does locally separate them. If such domain exists that would provide an example I request. (I actually doubt the existence of a domain as in the second question).
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1$\begingroup$ The very first sentence of the question is wrong: every simply connected region in the plane, other than the plane itself, can be mapped biholomorphically onto the unit disc. Liouville's theorem applies only to functions defined in the whole plane. The rest of the question makes little sense. $\endgroup$– Alexandre EremenkoCommented Mar 27, 2013 at 3:10
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$\begingroup$ The first sentence is actually not completely meaningless since OP does not restrict to simply-connected domains. I think, he is asking for examples of higher-dimensional domains $D$ which: (1) admit, say, "light" holomorphic maps to bounded domains, but (2) at the same time $D$ is not biholomorphic to a bounded domain. (A map is called "light" if its fibers are totally disconnected.) $\endgroup$– MishaCommented Mar 27, 2013 at 23:22
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$\begingroup$ On my opinion, the first sentence is completely meaningless. Most domains bounded or not can be mapped on bounded domains, and Liouville theorem has nothing to do with this. $\endgroup$– Alexandre EremenkoCommented Mar 28, 2013 at 1:01
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$\begingroup$ Sorry for not expressing myself clearly enough. The typical situation I have in mind is a domain of the type $\mathbb C\setminus E$, where $E$ is a "small-set"- a set for which removable singularity for bounded holomorphic functions works. Then every bounded holomorphic function on $\mathbb C\setminus E$ extends to an entire bounded function which is necessarily constant by Liouville's theorem. Playing with small-sets Ahlfors and Beurling ( Ahlfors, Lars; Beurling, Arne Conformal invariants and function-theoretic null-sets. Acta Math. 83, (1950). 101–129.) provided an explicit example of a dom $\endgroup$– ahmed sulejmaniCommented Mar 28, 2013 at 13:05
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$\begingroup$ domain which allows non-constant bounded holomorphic functions (therefore the Liouville argument does not work) but allows no injective bounded holomorphic functions (and therefore this domain is again not biholomorphic to any bounded domain). My question was about such examples in several variables. $\endgroup$– ahmed sulejmaniCommented Mar 28, 2013 at 13:07
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