Open mapping theorem (complex analysis): Difference between revisions

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Assume ''f'':''U'' → '''C''' is a non-constant holomorphic function and <math>U</math> is a connected [[open subset]] of the complex plane. We have to show that every [[Point (geometry)|point]] in <math>f(U)</math> is an [[interior point]] of <math>f(U)</math>, i.e. that every point in <math>f(U)</math> is contained in a disk which is contained in <math>f(U)</math>.

Consider an arbitrary <math>w_0</math> in <math>f(U)</math>. Then there exists a point <math>z_0</math> in ''U'' such that <math>w_0 = f(z_0)</math>. Since ''U'' is open, we can find <math>d>0</math> such that the closed disk <math>B</math> around <math>z_0</math> with radius ''d'' is fully contained in ''U''. Since ''U'' is connected and ''f'' is not constant on ''U'', we then know that ''f'' is not constant on ''B'' because of [[Analytic continuation]]. Consider the function <math>g(z)=f(z)-w_0</math>. Note that <math>z_0</math> is a [[root of a function|root]] of the function.

We know that ''g''(''z'') is not constant and holomorphic. The reciprocal of any holomorphic ''g''(''z'') is [[meromorphic]] and has isolated poles. Thus the roots of holomorphic non-constant functions are isolated. In particular, the roots of ''g'' are isolated and by further decreasing the radius of the image disk ''d'', we can assure that ''g''(''z'') has only a single root in ''B'' (although this single root may have multiplicity greater than 1).