TITLE: Intuition behind Alexander duality
QUESTION [31 upvotes]: I was wondering if anyone could offer some intuition for why Alexander duality holds.  Of course, the proof is easy enough to check, and it is also easy to work out many examples by hand.  However, I don't have any feeling for why it is true.
To give you an example of what I am looking for, when I think of Poincare duality I think of the picture in terms of triangulations and dual triangulations.  Is there any picture like that for Alexander duality?  Is there at least maybe some kind of obvious bilinear pairing between the two sides of it or something?

REPLY [8 votes]: Of course Alexander duality is most exciting for arbitrary compact subspaces $X \subset S^n$, Cech cohomology etc. But there is a very direct combinatorial proof for subcomplexes $X \subset \partial \Delta^{n+1}$, due to Blakers and Massey "The homotopy groups of a triad II" (Ann. of Maths. 55, 192-201 (1952)), which harks back to Poincare's proof of his duality using dual cells. For a simplicial complex $K$ let $K'$ denote the barycentric subdivision, and let $\sigma^* \subset K'$ denote the dual cell of a simplex $\sigma \in K$. Define the supplement of the subcomplex $X\subset \partial \Delta^{n+1}$ to be the subcomplex $\overline{X} \subset \Sigma^n=(\partial \Delta^{n+1})'$ consisting of the dual cells $\sigma^* \subset \Sigma^n$ of the simplices $\sigma \in \partial \Delta^{n+1} \backslash X$. The relative cellular chain complex $C^{cell}(\Sigma^n,\bar{X})$ is isomorphic to the $n$-dual ${C^{simp}(X)}^{n-*}$ of the simplicial chain complex $C^{simp}(X)$, inducing the Alexander duality isomorphisms 
$H_* (\Sigma^n,\overline{X}) \cong H^{n-*}(X)$ .
I used this combinatorial approach to Alexander duality in my book "Algebraic L-theory and topological manifolds" (CUP (1992)) to construct a combinatorial model for the generalized homology theory with algebraic surgery spectrum coefficients.