Alexander duality

Remarks

The proof uses a duality between the cohomology of a compact subset of a manifold and the homology of the manifold relative to its complement. This duality is proved like Poincaré duality. The long exact sequence of the pair $(S^n, S^n \setminus K)$ then gives the result. For an arbitrary compact $K$, the statement still holds with Čech cohomology in place of singular cohomology. With singular cohomology, local contractibility cannot be dropped: the Warsaw circle in $S^2$ has the singular cohomology of a point but separates $S^2$ into two components.

Example: if $K \subset S^n$ is homeomorphic to $S^{n-1}$, then $\tilde H_0(S^n \setminus K) \cong \tilde H^{\,n-1}(S^{n-1}) \cong \Z$. So the complement has exactly two path components, which is the Jordan–Brouwer separation theorem.

Example: for a knot $K \subset S^3$, that is, a subspace homeomorphic to $S^1$, $H_1(S^3 \setminus K) \cong \tilde H^1(S^1) \cong \Z$, whatever the knot is.

Going deeper

For a student

Alexander duality computes the homology of complements of embedded subsets, such as knot complements and complements of embedded spheres. It gives a quick proof of the Jordan–Brouwer separation theorem in all dimensions. Its proof is a direct application of the duality theory of manifolds, and its Čech version is a standard reason to introduce Čech cohomology.