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).



$(n-1)$-dimensional sphere in $S^n$ such that the closure of a component of complement is not contractible
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.


