Does the Bergman kernel always arise as the Jacobian of a biholomorphism?

Question

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 $F: D \to \mathbb{C}^n$?

Answer 1

Consider the case $n=1$. In this case the question actually is:

"For a fixed $w\in D$, is the antiderivative of $K(z,w)$ a well defined univalent function on $D$?"

For a simply connected domain the answer is yes, because there exists a biholomorphic map to a disk. But for a general domain the answer is no. To see this, let $D$ be an annulus. It is not difficult to see that the antiderivative of $K(z,w)$ is not even well defined, because
$\int K(z,w)dz\neq 0$ for any path which is not homotopic to zero. (This follows directly from the Laurent series of the kernel).

I am convinced that for $n\gt 1$ your conjecture is not true even for a domain homeomorphic to a ball, but I do not know how to construct a counterexample.