Suppose $a \in \mathbb C^n$, $U$ is a neighbourhood of $a$, and $f: U \to \mathbb C^n$ is analytic. Let $b = f(a)$ and suppose also that $f^{-1}(b) = \{a\}$. Must the image of $f$ contain a neighbourhood of $b$?

This would be some sort of local version of the open mapping theorem, which in general is not true for several complex variables. If the Jacobian determinant of $f$ is non-zero then the inverse function applies and we are fine. But this condition is not necessary, for example if $f$ is given by $f(z) = z^2$.


1 Answer 1


Yes. Expanding a previous comment into an answer: Let $M$ and $N$ be complex manifolds of the same dimension $n >0$ and let $f: M \to N$ be a holomorphic mapping. If $a \in M$ is an isolated point of its fiber $f^{-1}(f(a))$, then $m_af:=\sup\{\#f_\Omega^{-1}(w): w \in \Delta\}$, where $\Omega$ and $\Delta$ are small enough neighborhoods of $a$ and $f(a)$, respectively, is well defined (does not depend on $\Omega$ and $\Delta$) and moreover is finite. Then one can use the Remmert Open Mapping Theorem, which was already addressed on this site: in the analytic category, finite morphisms are open maps? A good reference is Chapter V of Introduction to Complex Analytic Geometry, Stanislaw Lojasiewicz, Birkhäuser 1991, ISBN 978-3-0348-7617-9-- especially if sheaves are not your cup of tea.

  • $\begingroup$ Thank you, that's very clear and helpful. $\endgroup$ Commented Oct 6, 2019 at 10:11

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.