I want to compute the cohomology ring of the space $S^1 \times \mathbb{C} P^{\infty}/(S^1 \times\{x_0\})$.

By Künneth formula, the cohomology ring of $S^1 \times \mathbb{C}P^{\infty}$ is $\mathbb{Z}[\alpha, \beta]/(\beta^2,\alpha \beta)$,where $\alpha$ is the generator of $H^2(\mathbb{C}P^{\infty})$ and $\beta$ is a generator of circle. But how to compute the quotient space really puzzles me. I also know that there is a formula $H_i(S^1 \times X)=H_i(X)\oplus H_{i-1}(X)$. However I don’t know whether this formula helps? Maybe we can use the long exact sequence for pairs, but it’s still complicated to compute. Hope someone could help. Thanks!

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    $\begingroup$ You can start by computing the cohomology groups via the long exact sequence of the pair $(S^1\times \mathbb CP^\infty, S^1\times \{x_0\})$, since $S^1\times \{x_0\} \to S^1\times \mathbb CP^\infty$ is a cofibration (as it is a sub-CW-complex). This will also allow you to get the induced morphism on cohomology by the quotient map, and this should help in the computation $\endgroup$ – Max Aug 24 at 9:26
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    $\begingroup$ Note also that the cofibration Max suggests is split. (the inclusion $S^1\hookrightarrow S^1\times \mathbb{C}P^\infty$ has a retraction) $\endgroup$ – Tyrone Aug 24 at 10:10
  • $\begingroup$ Also note that in the cohomology ring if $S^1\times\mathbb{C}P^\infty$, the product of $\alpha$ and $\beta$ is non-zero. In fact, it's a generator of $H^3$. $\endgroup$ – Jason DeVito Aug 24 at 12:14

Your space is the reduced suspension of $\mathbb{C}P^{\infty} \sqcup * $. This implies the multiplication is trivial, and Mayer-Vietoris says the additive structure of a suspension is just the cohomology shifted up.

The key geometric picture to have is that your space is the union of two cones on projective space, joined at the base and tip.

  • $\begingroup$ Interesting point of view. Thanks! $\endgroup$ – Yuyi Zhang Aug 25 at 11:31

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