Use cellular chain complexes to compute the homology of \(S^m \vee S^n\) and \(S^m \times S^n\).
Solution
1. Give each sphere its CW structure with one \(0\)-cell and one top-dimensional cell. Then \(S^m\vee S^n\) has one \(0\)-cell, one \(m\)-cell and one \(n\)-cell, with all cellular differentials zero.
Each positive-dimensional cell is attached to the common basepoint, so its cellular boundary is zero. If \(m=n\), there are two cells in that common dimension.
2. Hence, for \(m,n>0\),\[ \boxed{\widetilde H_k(S^m\vee S^n;\mathbb Z)\cong \begin{cases} \mathbb Z,&k=m\ne n\text{ or }k=n\ne m,\\ \mathbb Z^2,&k=m=n,\\ 0,&\text{otherwise.} \end{cases}} \]
This is the homology of the cellular chain complex in 1.
3. The product CW structure on \(S^m\times S^n\) has one cell in dimensions \(0,m,n,m+n\), with two cells in dimension \(m=n\) when the middle dimensions coincide, and all cellular differentials zero.
The cells are products of the \(0\)- and top cells of the two sphere factors. Cellular boundaries obey the product boundary formula, and the differentials in each sphere factor are zero.
4. Therefore, for \(m,n>0\),\[ \boxed{H_k(S^m\times S^n;\mathbb Z)\cong \begin{cases} \mathbb Z,&k=0,m,n,m+n\text{ when these are distinct},\\ \mathbb Z^2,&k=m=n,\\ 0,&\text{otherwise,} \end{cases}} \]with \(H_{2m}\cong\mathbb Z\) in the case \(m=n\).
Read the homology from the zero-differential cellular chain complex of 3.