We know that there are compact manifolds with diffeomorphic interiors but their boundaries are not homeomorphic (see the question Manifolds with homeomorphic interiors).
My question is about a very special case:
Assume $D$ is a bounded open set in $R^n$ with smooth boundary.
If $D$ is diffeomorphic to the unit ball, is $\bar D$, the closure of $D$, diffeomorphic to the closed unit ball?
Best Answer
The case $n=4$ is open as far as I know.
The case $n=3$ follows since $\mathbb R^3$ is irreducible, so it contains no fake 3-disk, i.e. $\bar D$ must be the standard disk.
The case $n=5$ is equivalent to the smooth $4$-dimensional Poincare conjecture (which is still open). Here is why:
Any homotopy $4$-sphere embeds smoothly into $\mathbb R^5$ (Sketch: homology $4$-sphere bounds a contractible smooth manifold $C$ [Kervaire, "Smooth homology spheres and their fundamental groups", Theorem 3]. In our case $\partial C$ is simply-connected, so attaching a collar on the boundary one gets a contractible $5$-manifold that is simply-connected at infinity, and hence it is diffeomorphic to $\mathbb R^5$ by a result of Stallings).
Any embedding of the standard $4$-sphere into $\mathbb R^5$ bounds a standard disk, see [Smale, "Differentiable and Combinatorial Structures on Manifolds", Corollary 1.3]. What Smale actually states is that any embedded $S^{n-1}$ in $\mathbb R^n$ bounds a standard disk unless $n=4$ or $7$. This was before he proved the h-cobodorsm theorem hence he excludes $7$.
Finally, as mentioned in comments if $n>5$, then $\bar D$ is diffeomorphic to the standard disk by the h-cobordism theorem (sketch: since by assumption $D$ is simply-connected at infinity, $\partial D$ is a homotopy sphere and $\bar D$ is a contractible smooth manifold, so removing a small ball in its interior results in h-cobordism between then standard sphere and the embedded one. Proving that this is an h-cobordism involves standard excision considerations in homology).