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

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

1 Answer 1

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 $X$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.