Open mapping theorem: Difference between revisions

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Suppose <math>U</math> is an open subset of <math>\mathbb{C}</math>, and <math>f: U \to \mathbb{C}</math> is a [[holomorphic function]]. Then, <math>f</math> is either a constant map (i.e. maps all elements of <math>U</math> to the same complex number) or an [[open map]]: the image of any open subset of <math>U</math> is open.
Suppose <math>U</math> is an open subset of <math>\mathbb{C}</math>, and <math>f: U \to \mathbb{C}</math> is a [[holomorphic function]]. Then, <math>f</math> is either a constant map (i.e. maps all elements of <math>U</math> to the same complex number) or an [[open map]]: the image of any open subset of <math>U</math> is open.
===For Riemann surfaces===
Suppose <math>M,N</math> are [[Riemann surface]]s and <math>f:M \to N</math> is an [[analytic mapping]]. Then <math>f</math> is either constant, or an open map.

Revision as of 01:05, 1 May 2008

This article gives the statement, and possibly proof, of a basic fact in complex analysis.
View a complete list of basic facts in complex analysis

Statement

For an open subset in the complex numbers

Suppose is an open subset of , and is a holomorphic function. Then, is either a constant map (i.e. maps all elements of to the same complex number) or an open map: the image of any open subset of is open.

For Riemann surfaces

Suppose are Riemann surfaces and is an analytic mapping. Then is either constant, or an open map.