Let $0 \in D$ be a bounded domain. Is it true that we can always find a injective holomorphic map $F: D \to \mathbb{C}^n$ such that $JF=K(z,0)$? Here, $K$ denotes the Bergman kernel of $D$, and $JF$ denotes the complex Jacobian.

Let $0 \in D$ be a bounded domain. Is it true that we can always find a injective holomorphic map $F: D \to \mathbb{C}^n$ such that $JF=K(z,0)$? Here, $K$ denotes the Bergman kernel of $D$, and $JF$ denotes the complex Jacobian.

Edit: The question as stated is clearly not true. One could just take a domain with $K(z,0) = 0$ for some $z$. In the light of Alex Gavrilov answer, simple connectedness seems to be an important condition. So, is the conjecture true for a simply connected domain $D$ and $0$ being a point such that $K(z,0) \neq 0$ for all $z \in D$? More generally, does every non-vanishing square-integrable holomorphic function on $D$ arise as the Jacobain of some injective holomorphic map $F: D \to \mathbb{C}^n$?