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In Davis & Kirk LNAT p.71 there is written: enter image description here

(1) How does this imply the Alexander duality $\tilde{H}^k(A)\cong \tilde{H}_{n-k-1}(\mathbb{S}^n\!\setminus\!A)$?

(2) Is it assumed that the manifolds in 3.26 are all smooth?

(3) Does Alexander duality imply the Jordan-Brouwer separation theorem $\tilde{H}_0(\mathbb{R}^n\!\setminus\!\iota(\mathbb{S}^{n-1}))\cong\mathbb{Z}$? Can here $\iota\!:\mathbb{S}^{n-1}\rightarrow\mathbb{R}^n=\mathbb{S}^n\!\setminus\!\mathrm{pt}$ be a topological embedding or must it be smooth?

Leo
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1 Answers1

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You need this version of Poincaré-Lefschetz duality using Čech cohomology group. enter image description here

Then you can prove the Alexander Duality enter image description here

No, this duality is true for topological manifold. Here is the proof of Generalized Jordan Curve Theorem. Homeomorphism is enough. enter image description here

For more details, see Bredon's "Topology and Geometry" chapter 6 section 8.

WWK
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  • So as I have stated it, Alexander manifold duality does not imply Alexander sphere duality? This seems strange, I think it should. We have $H^k(A)\cong H_{n-k}(\mathbb{S}^n, \mathbb{S}^n!\setminus!A)$ and exact sequence of a pair $0\cong H_{n-k}(\mathbb{S}^n) \longrightarrow H_{n-k}(\mathbb{S}^n, \mathbb{S}^n!\setminus!A) \longrightarrow H_{n-k-1}(\mathbb{S}^n!\setminus!A) \longrightarrow H_{n-k-1}(\mathbb{S}^n)\cong0$, but the only problem is with cases $k=0,n!-!1,n$. Any suggestions? – Leo May 06 '14 at 21:51