Let $f:S^{n-1} \rightarrow S^n$ be a topological embedding and let $A_f$ and $B_f$ be the components of $S^n \setminus f(S^{n-1})$. If $\overline{A}_f$ and $\overline{B}_f$ are manifolds with boundary $f(S^{n-1})$, then the locally flat Schoenflies conjecture (proved by Mazur and Brown) says that $\overline{A}_f$ and $\overline{B}_f$ are both topological discs. The classical example of the Alexander horned sphere shows that the assumption that $\overline{A}_f$ and $\overline{B}_f$ are manifolds is necessary. Indeed, letting $\alpha:S^2 \rightarrow S^3$ be the Alexander horned sphere, it turns out that one of the components (say, $A_{\alpha}$) is not even simply connected. However, $\overline{B}_{\alpha}$ is a manifold. Using the Schoenflies conjecture, this implies that if we attach a collar neighborhood $\alpha(S^2) \times [0,1]$ to $\overline{A}_{\alpha}$, then we get a $3$-ball which deformation retracts to $\overline{A}_{\alpha}$. In particular, $\overline{A}_{\alpha}$ is contractible.

Question: Does there exist a topological embedding $f^{n-1} \rightarrow S^n$ (preferably with $n=3$) such that neither $\overline{A}_f$ nor $\overline{B}_f$ is contractible?

There definitely exist such embeddings where neither $\overline{A}_f$ nor $\overline{B}_f$ are manifolds. For example, Bing proved that the double of the Alexander horned ball $\overline{A}_{\alpha}$ along its "boundary" $\alpha(S^2)$ equals $S^3$. This means that the above trick will not work in general (but of course we already know that $\overline{A}_{\alpha}$ is contractible, so Bing's theorem does not answer our question).