does Alexander duality commute with cup product?
How could the dimensions add up? Unless I'm interpreting this the wrong way, taking the cup product of the images would be in a different cohomology group than the image of of the cup product.
On second thought, I think I can make the dimensions add up if you're also invoking Poincare duality. Can you tell me the statement of Alexander duality that you're using?
I am really thinking about a special case of Alexander duality, the case of a circle embedded in S3,
H^*(S3-C) iso Hn-*(C)
If there are two embedded circles then cup product maps H^1(S3-C1)xH^1(S3-C2) -> H^2(S3-C1UC2).
The Alexander maps take these two groups into H1(C1)xH1(C2) and H0(C1UC2).
These are ZxZ and Z.
I was really wondering if there is a natural homomorphism from H1(C1)xH1(C2) into Z that completes the square. The conjecture is that it is the degree of the linking map of the torus C1xC2 into the 2 sphere.
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