-4
$\begingroup$

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 $f(M)\subset\mathbb{C}$?

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ The open mapping theorem from complex analysis carries over to Riemann surfaces basically immediately. $\endgroup$
    – Jack Huizenga
    Commented Jul 16, 2011 at 0:03

1 Answer 1

Reset to default
2
$\begingroup$

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 $X$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ connected Riemann surfaces $\endgroup$
    – Ben McKay
    Commented Jul 28, 2011 at 7:17
  • 2
    $\begingroup$ Thanks. Some people include connectedness in the definition of a Riemann surface... $\endgroup$
    – Beren Sanders
    Commented Jul 28, 2011 at 23:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .