$(n-1)$-dimensional sphere in $S^n$ such that the closure of a component of complement is not contractible 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).
 A: I want to thank Ian Agol for his comments which helped point me in the right direction.  In particular, he told me that if $f:S^{n-1} \rightarrow S^n$ is a topological embedding, then the closures of the components of $S^n \setminus f(S^{n-1})$ are known as crumpled $n$-cubes.
It turns out that Bing originally proved that crumpled $n$-cubes are contractible.  See Theorem 4 of his paper
Bing, R. H.
Retractions onto spheres.
Amer. Math. Monthly 71 1964 481–484.
which proves that a crumpled $n$-cube in $\mathbb{R}^n$ is a retract of $\mathbb{R}^n$.
It also turns out that the technique I described in the question for proving that crumpled $n$-cubes are deformation retracts of standard $n$-discs works in general: if $C$ is a crumpled $n$-cube and $M$ is obtained by gluing a standard $n$-disc to the "boundary" of $C$, then $M$ is homeomorphic to $S^n$.  This was proved for $n=3$ in 
Lininger, Lloyd L. Some results on crumpled cubes. Trans. Amer. Math. Soc. 118 1965 534–549.
for $n \geq 5$ in
Daverman, Robert J. Every crumpled n-cube is a closed n-cell-complement. Michigan Math. J. 24 (1977), no. 2, 225–241
and finally for $n=4$ in
Daverman, Robert J. Each crumpled 4-cube is a closed 4-cell-complement. Topology Appl. 26 (1987), no. 2, 107–113.
