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In [[complex analysis]], the '''open mapping theorem''' states that if ''U'' is a [[connected space|connected]] open subset of the [[complex plane]] '''C''' and ''f'' : ''U'' → '''C''' is a non-constant [[holomorphic]] function, then ''f'' is an [[open map]] (i.e. it sends open subsets of ''U'' to open subsets of '''C''').
The open mapping theorem points to the sharp difference between holomorphy and real-differentiability. On the [[real line]], for example, the differentiable function ''f''(''x'') = ''x''<sup>2</sup> is not an open map, as the image of the [[open interval]] (
The theorem for example implies that a non-constant [[holomorphic function]] cannot map an open disk ''[[onto]]'' a portion of a line.
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[[Image: OpenMapping1.png |frame|right|Blue dots represent zeros of ''g''(''z''). Black spikes represent poles. The boundary of the open set ''U'' is given by the dashed line. Note that all poles are exterior to the open set. The smaller red circle is the set ''B'' constructed in the proof.]]
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>z_0</math> in <math>U</math>. Since ''U'' is open, we can find <math>d>0</math> such that the closed disk <math>B</math> around ''z''<sub>0</sub> 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''. Consider the [[Image (mathematics) | image]] point, <math>w_0 = f(z_0)</math>. Then <math>f(z_0)-w_0 = 0</math>, making <math>z_0</math> a [[Root (mathematics) | root]] of the function <math>g(z)=f(z)-w_0</math>.
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