# Possible condition for a many variable holomorphic map to be locally surjective

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 $$f(z) = z^2$$.

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.