Re: Open sets and proper mapping
- From: José Carlos Santos <jcsantos@xxxxxxxx>
- Date: Tue, 27 Jan 2009 17:09:17 +0000
On 27-01-2009 15:17, jane wrote:
I have the following question:
Let f be a holomorphic map on C and U,V are open subsets of C such that closure(U) in V.
Is it true that then closure(f(U)) in f(V).
Here is my argument: since f is holomorphic it is proper and therefore is closed, i.e. maps closed sets to closed sets.
Who told you that holomorphic maps are proper? The exponential is a non-proper entire function (and in fact it gives a counterexample to
your original question if you choose the right U and V. Hint:
0 is not in exp(V).
Ok. Thanks, i see your point. Well, actually i need this fact nor for general holomorphic functions, but for rational functions. If i am not mistaken a rational function has a finite degree and it is a finite degree branched covering, therefore it must be proper and the argument should work in this case.
Am i right now in the case f is a rational function? (Or even finite-degree branched covering).
No. Take f(z) = 1/z and take D = { z in C | |z| >= 1 }. Then D is
closed, but f(D) is not.
Best regards,
Jose Carlos Santos
.
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- Re: Open sets and proper mapping
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- Re: Open sets and proper mapping
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