Compute \(H_i(S^n - X)\) when \(X\) is a subspace of \(S^n\) homeomorphic to \(S^k \vee S^\ell\) or to \(S^k \amalg S^\ell\).
Solution
We use Alexander duality in the form \[ \widetilde H_i(S^n-X;\mathbb Z)\cong \widetilde H^{\,n-i-1}(X;\mathbb Z) \] for these compact locally contractible subspaces of \(S^n\).
1. If \(X=S^k\vee S^\ell\), then\[ \widetilde H^j(X)\cong \begin{cases} \mathbb Z,&j=k\ne\ell,\\ \mathbb Z,&j=\ell\ne k,\\ \mathbb Z^2,&j=k=\ell,\\ 0,&\text{otherwise}. \end{cases} \]
Reduced cohomology takes a finite wedge to the direct sum of the reduced cohomologies of the two spheres. Each sphere has one copy of \(\mathbb Z\) in its dimension and zero reduced cohomology elsewhere.
2. Consequently, for \(X=S^k\vee S^\ell\),\[ \widetilde H_i(S^n-X)\cong \begin{cases} \mathbb Z,&i=n-k-1\ne n-\ell-1,\\ \mathbb Z,&i=n-\ell-1\ne n-k-1,\\ \mathbb Z^2,&i=n-k-1=n-\ell-1,\\ 0,&\text{otherwise}. \end{cases} \]
Apply Alexander duality to 1, replacing the cohomological degree \(j\) by \(n-i-1\).
3. If \(X=S^k\amalg S^\ell\), then in addition to the two positive-dimensional sphere classes there is\[ \widetilde H^0(X)\cong\mathbb Z. \]Thus\[ \widetilde H_i(S^n-X) \cong \widetilde H^{\,n-i-1}(S^k\amalg S^\ell). \]Explicitly this contributes one copy of \(\mathbb Z\) in degree \(i=n-1\), and copies of \(\mathbb Z\) in degrees \(n-k-1\) and \(n-\ell-1\) (combined to \(\mathbb Z^2\) when \(k=\ell\)).
A disjoint union of two connected spaces has reduced \(H^0\) equal to \(\mathbb Z\), while its positive-degree cohomology is the direct sum of that of its two components. Alexander duality gives the stated complement groups.
The unreduced group \(H_0\) is obtained, as usual, by adjoining one copy of \(\mathbb Z\) to \(\widetilde H_0\).