Compute all of the possible cases for \(H_*(S^a\cross S^b)\), where \(a, b\geq 0\) are not necessarily distinct.
Solution
1. For \(a,b>0\), Künneth gives\[H_k(S^a\times S^b;\mathbb Z)\cong\bigoplus_{i+j=k}H_i(S^a)\otimes H_j(S^b),\]with no Tor terms.
All homology groups of spheres are free abelian.
2. If \(a,b>0\) and \(a\ne b\), then\[\boxed{H_k\cong\begin{cases}\mathbb Z,&k=0,a,b,a+b,\\0,&\text{otherwise.}\end{cases}}\]
The four tensor products from degrees \((0,0),(a,0),(0,b),(a,b)\) occur in distinct total degrees.
3. If \(a=b>0\), then\[\boxed{H_k\cong\begin{cases}\mathbb Z,&k=0,2a,\\\mathbb Z^2,&k=a,\\0,&\text{otherwise.}\end{cases}}\]
The two middle tensor products now occur in the same degree \(a\).
4. Since \(S^0\) is two points, if exactly one of \(a,b\) is zero then \(S^a\times S^b\) is a disjoint union of two copies of the positive-dimensional sphere; if \(a=b=0\) it is four points.
Products with a finite discrete set are disjoint unions indexed by that set. Thus for, say, \(a=0<b\), \(H_0\cong H_b\cong\mathbb Z^2\) and all other groups vanish; for \(a=b=0\), \(H_0\cong\mathbb Z^4\).