Let $B^n$ denote the unit ball in $\mathbb{R}^n$ (wrt the standard euclidean metric) and $\bar{B}^n$ denote the unit closed ball. Suppose that $\Sigma$ is a a smooth embedded hypersurface with boundary in $\bar{B}^4$ which is topologically $\bar{B}^3$ and so that $\Sigma$ meets $\partial \bar{B}^4$ transversely and $\partial \Sigma\subset \partial \bar{B}^4$. It should follow from known results that $B^4\backslash \Sigma=U\_+\cup U\_-$ where $U\_\pm$ are open domains homeomorphic to $B^4$. Is it also known whether they are diffeomorphic, or is this equivalent to the smooth 4D Schoenflies problem (it is clearly weaker)?


I'm not a topologist so I apologize if this is a silly question.



Can a 3-ball divide a standard 4-ball into two exotic 4-balls?
I think this is equivalent to the smooth Schoenflies conjecture; the executive summary is that this is true because smooth balls (in any dimension) are isotopic to standard ones. 


Here are some details, starting with some preliminary remarks. Your $\bar{B}^3$ is diffeomorphic to a closed 3-ball, so I'll take that as given. Its boundary is a 2-sphere in $\partial \bar{B}^4$, and so is isotopic to the standard 2-sphere in $S^3$ by the 3D Schoenflies theorem (Alexander's theorem). Also, the standard result about balls in a manifold being standard applies to a codimension $0$ ball $\bar{B}^n$ embedded in an $n$-manifold in a non-proper (or neat) way. In other words the boundary of $\bar{B}^n$ is divided into $\bar{B}^{n-1}\_\pm$ with $\bar{B}^n \cap \partial M = \bar{B}^{n-1}\_-$ lying in the boundary of $M$, and $\bar{B}^{n-1}\_+$ properly embedded. You really should worry a bit about corners here, but I'm going to ignore this (and in what follows).


Suppose first that we have $\bar{B}^4$ as described and that the Schoenflies conjecture holds. By the remarks above, do a preliminary isotopy so the boundary is standard. Now add a 4-ball $B\_0^4$, divided into balls $U^0\_\pm$, where $\bar{U}^0\_+ \cap \bar{U}^0\_-$ is a standard $\bar{B}\_0^3$. Then $\bar{B}\_0^3 \cup \bar{B}^3$ is an embedded sphere, and assuming the Schoenflies theorem, bounds balls on either side. Each of these balls is of the form $U^0\_\pm \cup U\_\pm$. But since we know that $U^0\_\pm$ is a ball, its embedding is standard, so it follows that $U\_\pm$ is a ball as well. 


The converse is also straightforward. Given $S^3 \subset S^4$, dividing $S^4$ into regions $A\_+ \cup A\_-$, remove a small ball $B^4\_0$ centered at a point of $S^3$. The complement of $B^4\_0$ is a ball divided into regions $U\_\pm$ as in your description, and by hypothesis each of $U\_\pm$ is a ball. It follows that $A\_\pm$ are balls as well. 


