The Alexander duality Theorem is usually stated for a triangulable pair $(\mathbb S^n, Y)$ where $Y$ is a subset of the standard sphere $\mathbb S^n$. My question is: Does the duality also hold if we rather replace $\mathbb S^n$ by a compact orientable Homology sphere (without boundary) (<https://en.m.wikipedia.org/wiki/Homology_sphere>) ? I'm mainly interested in the cases $n=2$ and $3$. I'm willing to assume that the Homology sphere is the Geometric realization of a finite abstract simplicial complex (<https://en.m.wikipedia.org/wiki/Abstract_simplicial_complex>) 


Thanks



Alexander duality for Homology sphere which is the Geometric realization of a finite simplicial complex
You have to use Poincaré-Lefschetz duality : Let $M$ be a compact orientable $n$-manifold, $Y\subset M$ be a closed subset then we have an isomorphism 
$$\check{\mathrm{H}}^p(M,Y)\cong H\_{n-p}(M-Y)$$ 
induced by the cap product with the fundamental class of $M$ (the left hand side is Cech cohomology). 
You also have
$$\check{\mathrm{H}}^p(Y)\cong H\_{n-p}(M,M-Y).$$
In fact these isomorphisms are compatible with the long exact sequences of the pairs $(M,Y)$ and $(M,M-Y)$. 


In your case, if $M$ is a triangulated manifold and $Y$ is a subpolyhedron of $M$, cech cohomology groups are nothing but singular cohomology groups. 


You can have a look at Bredon's book "Topology and Geometry" (chapter VI, section 8).


