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There was a similar question.

Let $X=(S^2\times S^2)/\mathbb{Z}_4$ where $\mathbb{Z}_4$ acts on $S^2\times S^2$ as $(x,y)\mapsto(-y,x)$.

My question: What are the cohomology rings of $X$ with coefficients $\mathbb{Z}_4$ and $\mathbb{Z}_2$? What is the Bockstein homomorphism $$\beta:H^*(X,\mathbb{Z}_4)\to H^{*+1}(X,\mathbb{Z}_2)?$$

Note that the $\mathbb{Z}_4$ action on $S^2\times S^2$ is orientation-reversing, so $X$ is non-orientable. Also the Euler characteristic is $\chi(X)=1$.

Thank you!

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    $\begingroup$ Nice question. Try coming up with a $\Bbb Z/4$-invariant cell decomposition of $S^2 \times S^2$, cobbled together via the following antipodal-invariant cell decomposition on $S^2$: the 0-cells are the north and south pole, the 1-cells are latitude lines going from the south-to-north pole on the "east side" and the north-to-south pole on the "west side", and then the front and back faces are the 2-cells. Once you write down the cellular chain complex upstairs with its $\Bbb Z/4$ action, quotienting will give you the chain complex downstairs, and this is what you need for these computations. $\endgroup$ – user98602 Nov 3 '18 at 18:23
  • $\begingroup$ For the cohomology rings you'll need to do a bit more work, but this is a starting point. $\endgroup$ – user98602 Nov 3 '18 at 18:24
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    $\begingroup$ An unattractive spectral sequence calculation gives me $H_0 = \Bbb Z, H_1 = \Bbb Z/4, H_3 = \Bbb Z/2$ and all other homology groups zero. This implies for degree reasons that there are no interesting Bocksteins in the direction you ask, and that the $\Bbb Z/2$ cohomology ring is uninteresting. I think there is a nonzero product in the $\Bbb Z/4$ cohomology ring. $\endgroup$ – user98602 Nov 6 '18 at 5:25
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    $\begingroup$ Would it be of any value to you for me to write down the spectral sequence calculation even if I can't get the product structure? $\endgroup$ – user98602 Nov 6 '18 at 20:51

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