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)


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    $\begingroup$ This seems to be true, cf. Massey, "A generalization of the Alexander duality theorem" Indiana Univ. Math. J. 30 (1981). $\endgroup$ – Jens Reinhold May 26 '20 at 7:09

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).


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