This is a follow-up to my earlier question.
Let $\Sigma\subset \mathbb{S}^{n}$ be a hypersurface -- here $\mathbb{S}^{n}$ is a smooth sphere (possibly exotic ... if this makes a difference). If $\Sigma$ is a homology sphere, then is it possible for the components of $\mathbb{S}^{n+1}\backslash \Sigma$ to be non-contractible?
Yes; this is not hard to arrange. For instance, take any knot K in the 3-sphere with an n-fold branched cover, say M that is a homology sphere. A good example would be to take M to be the $k$-fold cover of the $(p,q)$-torus knot, where p, q, and $k$ are pairwise relatively prime. Then the amazing twist-spinning construction of Zeeman (Twisting Spun Knots, Trans. AMS 115 (1965), pp. 471-495) constructs a knot in the 4-sphere with fiber $M_0 = M - B^3$. (Zeeman also describes the monodromy as the generator of the covering group of the branched cover.)
Let $N = M_0 \cup -M_0$, embedded in $S^4$ as the boundary of a product neighborhood $M_0 \times I$. Then both components of $S^4 - N$ are diffeomorphic to $M_0 \times I$, so neither is simply-connected, and hence not contractible.
A similar construction works in any dimension; you just need knots in $S^n$ with homology spheres for branched covers. You can get these by spinning. Alternately, you can do the pairwise spin of the homology sphere $M$ above. By definition, for a manifold $M$, the spin of $M$, say $S(M)$, is gotten from $S^1 \times M$ by surgery on $S^1 \times x$ for some point $x \in M$. Two salient points are that $S(M)$ has the same fundamental group as $M$ and that $S(S^n) = S^{n+1}$. Moreover (I'm expecting the Spanish inquisition next) if $M$ is a homology sphere then so is $S(M)$.
If $M \subset S^n$, then you can spin both simultaneously to get $S(M) \subset S^{n+1}$ and the complementary regions have the same fundamental groups as those of $M$. This gives examples in all (ambient) dimensions $4$ and up.
