Suppose $\Omega\subset \mathbb{S}^n$ is an open set with $\Sigma=\partial \Omega$ a smooth hypersurface.  If $\Omega$ is a homology ball, in the sense that $H_i(\Omega)=H_i(\mathbb{D}^n)$ for $0\leq i \leq n$, then is it possible for $\Omega$ to be non-contractible?

Some (possibly incorrect) observations:

 - When $n=1,2$ it is clear any such $\Omega$ is contractible.
 - For all $n>1$, I believe $\Sigma$ is a homology sphere (via Poincare duality) so when $n=3$ there is no such non-contractible $\Omega$ (using the classification of surfaces and Alexander's theorem -- is there a simpler argument?)
 - By removing a small ball from a homology sphere one can produce a manifold with boundary, $X$, that is a homology ball that is not contractible.  However, by the generalized Schoenflies conjecture of Brown and Mazur, such an  $X$ can't embed in $\mathbb{S}^n$.

EDIT:
For what it is worth, this was essentially asked (and answered) previously in this [question][1]


  [1]: https://mathoverflow.net/questions/209774/can-a-homology-n-1-sphere-divide-mathbbsn-into-non-contractible-compon