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I am looking for an Open Mapping Theorem for a holomorphic function $f: U \subset \mathbb{C}^n \to \mathbb{C}^n$ where $U$ is a domain. I believe the following is true:

Let $f: U \subset \mathbb{C}^n \to \mathbb{C}^n$ be holomorphic, where $U$ is a domain. Suppose the determinant of the Jacobian of $f$ is not identically zero on $U$. Then $f(U)$ is open.

References and thoughts are welcome.

wellfedgremlin
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1 Answers1

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Not true: $(x,y)\mapsto(x, xy)$.

Daniel Fischer
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user312865
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  • The map provided is an open map. This is not a counterexample. – wellfedgremlin Feb 19 '16 at 17:46
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    Denote the given map by $f$. I show it is not open at $(0,0)$. Let $r_1<1$. I show that for any $r_2>0$, there exists $z,w\in B_{r_2}(0)$ such that $(z,w)\not\in f(B_{r_1}(0)\times B_{r_1}(0))$. Indeed, choose $z=r_2r_1, w=r_2$, then the only solution is $x=r_2r_1, y=1/r_1$. – Bananach Feb 26 '16 at 07:19