# Compute cohomology ring of $S^1 \times \mathbb{C} P^{\infty}/(S^1 \times \{x_0\})$

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!

• 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 – Max Aug 24 at 9:26
• Note also that the cofibration Max suggests is split. (the inclusion $S^1\hookrightarrow S^1\times \mathbb{C}P^\infty$ has a retraction) – Tyrone Aug 24 at 10:10
• 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 $H^3$. – 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.