Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks in advance

enter image description here

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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 $q^*(\gamma^*)(\beta) = \gamma^*(q_*(\beta)) = 1$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.