Let $\mathbb{D}$ be the unit disk on the plane and let $U,V\subset \mathbb{D}$ be open and such that $U\cup V=\mathbb{D}$.
Is there a holomorphic map $\varphi:\mathbb{D}\times \mathbb{D}\to \mathbb{D}$ such that $\varphi(U\times V)=\mathbb{D}$?
Of course one can ask the same question for "ambient domains" other than $\mathbb{D}$.
Edit2: everything works, updating the answer.
Yes. Consider two cases.
Case 1. There is a point on the boundary contained both in the closure of $U$ and $V$. Do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.
Now take the map $(x, y) \mapsto x+y$.
It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.
Case 2. One of the subsets (say $U$ without loss of generality) contains $\partial\mathbb{D}$ in it's closure. Then, it contains a small annulus $1-\varepsilon<|r|<1$. There is a map from disk to itself which is surjective from this annulus: one can do such an automorphism of a disk that $0$ is in the image of the annulus and then use $z \mapsto z^2$. Composing this map with the projection on the first component we obtain the desired result.
Certainly not. If you take U, V to be small neighborhoods of distinct points this will fail. Are you sure this is what you need?
