Alexander duality formulation + Jordan-Brouwer separation

Question

In Davis & Kirk LNAT p.71 there is written:

(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?

Answer 1

You need this version of Poincaré-Lefschetz duality using Čech cohomology group.

Then you can prove the Alexander Duality

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

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