# Open mapping theorem for Riemann surfaces

What restriction must one impose on a Riemann surface M in order for all biholomorphic $f:M\to\mathbb{C}$ to be open mappings, aka mappings of $M$ onto open subsets $f(M)\subset\mathbb{C}$?

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The open mapping theorem from complex analysis carries over to Riemann surfaces basically immediately. –  Jack Huizenga Jul 16 '11 at 0:03

## 1 Answer

Every non-constant holomorphic map between Riemann surfaces is an open map.

See Corollary 2.4 and Theorem 2.1 of [Forster -- Lectures on Riemann surfaces (GTM81, 1981)].

By choosing charts it is immediate that the local behaviour of holomorphic maps between Riemann surfaces is just the same as the local behaviour of the usual holomorphic functions we study in elementary complex analysis.

Clarification: If you don't require Riemann surfaces to be connected then the correct statement would be: A holomorphic map $f : X \rightarrow Y$ between Riemann surfaces is an open map provided that it is not constant on any connected component of $X$.

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connected Riemann surfaces –  Ben McKay Jul 28 '11 at 7:17
Thanks. Some people include connectedness in the definition of a Riemann surface... –  Beren Sanders Jul 28 '11 at 23:17