# cohomology of product

I shall be thankful to you for helping me understand what I have highlighted in yellow. I see that $\gamma, \alpha$ and $\beta$ are not the same as the generators of homologies but rather the cohomologies given that cohomology modules are duals to homology modules because the homology modules are all finitely generated with no torsion. that gives $\gamma^{*}(q_{*} (\alpha))=1$ and $\gamma^{*}(q_{*} (\beta))=1$ thus $\gamma^{*}(q_{*} (\alpha + \beta))=2$ thus $q^{*}(\gamma^{*}(\alpha + \beta))=2$ so it seems like my argument contains a mistake somewhere.

Consider $\alpha$ and $\beta$ to be homology classes, then the dual classes $\alpha^*, \beta^*$ are a $\mathbb Z$-basis of $H^2(S^2 \times S^2)$ and $q^*$ is the dual map of $q_*$ (by the universal coefficient theorem).
Now expand $q^*(\gamma^*)$ in the basis $\{\alpha^*, \beta^*\}$. The first coefficient will be ${\alpha^*}^*(q^*(\gamma^*)) = q^*(\gamma^*)(\alpha) = \gamma^*(q_*(\alpha)) = \gamma^*(\gamma) = 1$. Analogously the second one will be $q^*(\gamma^*)(\beta) = \gamma^*(q_*(\beta)) = 1$