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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 $f(z) = z^2$.

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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.

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

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