# Alexander duality for Homology sphere which is the Geometric realization of a finite simplicial complex

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

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

## 1 Answer

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).