4
$\begingroup$

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 $\mathbb{D}$.

$\endgroup$

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ I am not sure what you mean in the edit: since $U\cup V=D$, the union of the closures contains $\partial D$, and so $\partial U\cup \partial V=\partial D$. Since the latter is connected, $\partial U\cap \partial V\ne\varnothing$. Or did you mean how to drop the condition $U\cup V=D$? I would be very much interested in merely having $U\cap V=\varnothing$.. $\endgroup$
    – erz
    Mar 8, 2020 at 10:29
  • $\begingroup$ Annulus and concentric circle. $\endgroup$ Mar 8, 2020 at 10:32
  • $\begingroup$ Actually my intuition says this might be a counterexample, but I'm not sure yet. $\endgroup$ Mar 8, 2020 at 10:33
  • $\begingroup$ answer updated to full generality $\endgroup$ Mar 8, 2020 at 14:39
0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ I've made them bigger. Thank you $\endgroup$
    – erz
    Mar 8, 2020 at 3:33
  • $\begingroup$ @erz better, but you probably want to choose $\phi$ first and allow $U, V$ to be arbitrary with $U\cup V = \mathbb{D}$? Also is $\mathbb{D}$ open or closed for you? $\endgroup$ Mar 8, 2020 at 5:33
  • $\begingroup$ I am not sure I understand your first question. $\mathbb{D}$ is both open and closed in itself. The motivation for my question is mathoverflow.net/questions/354133/generating-h-inftyx I am allowed to take a lot of functions, and compose them with a function of two variables. This is a natural extension in my opinion $\endgroup$
    – erz
    Mar 8, 2020 at 5:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.