Let \(L(p)\) be a space whose integral homology groups are \(\mathbb{Z}\), \(\mathbb{Z}_p\), \(0\), \(\mathbb{Z}\) in dimensions \(0\), \(1\), \(2\), \(3\), and zero otherwise. Let \(\Sigma\) denote the suspension of a space. Compute the cohomology \(H^*(\Sigma L(p) \times \Sigma L(q); \mathbb{Z})\), where \(p\) and \(q\) are coprime.
Solution
Put \(A=\Sigma L(p)\) and \(B=\Sigma L(q)\).
1. The suspension isomorphism gives\[ H_i(A;\mathbb Z)\cong \begin{cases} \mathbb Z,&i=0,4,\\ \mathbb Z_p,&i=2,\\ 0,&\text{otherwise}, \end{cases} \]and the analogous formula for \(B\) with \(p\) replaced by \(q\).
2. Apply the homological Künneth theorem to \(A\times B\).
2.1. The only possible torsion–torsion tensor and Tor terms are\[ \mathbb Z_p\otimes\mathbb Z_q \quad\text{and}\quad \operatorname{Tor}_1^{\mathbb Z}(\mathbb Z_p,\mathbb Z_q), \]and both vanish because \(\gcd(p,q)=1\).
2.2. Therefore\[ H_i(A\times B;\mathbb Z)\cong \begin{cases} \mathbb Z,&i=0,8,\\ \mathbb Z_p\oplus\mathbb Z_q,&i=2,6,\\ \mathbb Z^2,&i=4,\\ 0,&\text{otherwise}. \end{cases} \]Since \(p\) and \(q\) are coprime, \(\mathbb Z_p\oplus\mathbb Z_q\cong\mathbb Z_{pq}\).
3. The universal coefficient theorem for cohomology gives a split short exact sequence\[ 0\longrightarrow \operatorname{Ext}(H_{k-1}(A\times B),\mathbb Z) \longrightarrow H^k(A\times B;\mathbb Z) \longrightarrow \operatorname{Hom}(H_k(A\times B),\mathbb Z) \longrightarrow0. \]
3.1. The free summands in homology contribute to cohomology in the same degree, while each \(\mathbb Z_{pq}\) in degrees \(2\) and \(6\) contributes an \(\operatorname{Ext}\) term \(\mathbb Z_{pq}\) in degrees \(3\) and \(7\).
3.2. Hence\[ H^k(\Sigma L(p)\times\Sigma L(q);\mathbb Z)\cong \begin{cases} \mathbb Z,&k=0,8,\\ \mathbb Z^2,&k=4,\\ \mathbb Z_{pq},&k=3,7,\\ 0,&\text{otherwise}. \end{cases} \]