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Let $X$ be a topological space and $F$ a field. Suppose the singular cohomology of $X$, $H^n(X;F)$, is finitely generated in every dimension. In this case, by Künneth's formula: $$H^*(X\times X;F)\cong H^*(X;F)\otimes H^*(X;F),$$ we know that the cup product: $$\smile\colon H^*(X;F)\otimes H^*(X;F)\to H^*(X;F)$$ is induced by the diagonal map $\Delta\colon X\to X\times X$. I want to show that the kernel of the cup product is exactly the image of $(\pi_1^* - \pi_2^*)$, where $\pi_1,\pi_2\colon X\times X\to X$ are the projections.

The inclusion $\mathrm{Im}(\pi_1^* - \pi_2^*)\subseteq\ker\Delta^*$ is straightforward. However, I am having trouble showing the other inclusion. My attempt so far is the following: let $\gamma = \sum_{i,j} c_{ij} \alpha_i \otimes \beta_j \in \ker\Delta^*$. This means $\sum_{i,j} c_{ij} \alpha_i \smile \beta_j = 0$. Let $(e_k)$ be a graded base of $H^*(X;R)$. We can write: $$\alpha_i = \sum_k a_{ik} e_k, \quad \beta_j = \sum_l b_{jl} e_l.$$ By substituting in $\gamma$: $$\gamma = \sum_{i,j,k,l} c_{ij} a_{ik} b_{jl} \, e_k \otimes e_l$$ the condition $\Delta^*(\gamma) = 0$ means that: $$\sum_{k,l} \left( \sum_{i,j} c_{ij} a_{ik} b_{jl} \right) e_k \smile e_l = 0.$$ Can we deduce from here that $$\sum_{i,j} c_{ij} a_{ik} b_{jl} = 0, \quad \text{ for every } \, k,l\geq 1?$$ Then maybe we can use the fact that $e_k\otimes e_l-e_l\otimes e_k=(\pi_1^*-\pi_2^*)(e_k)\cdot e_l$? I don't really know how to justify this properly and continue from here. Any hint/reference is appreciated!

YM78
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    I don't think this is true. The kernel you are looking for is an ideal in the cohomology ring, but the image of $\pi_1^{\ast}-\pi_2^{\ast}$ is typically just a subgroup. E.g. if $X=\mathbb{RP}^{\infty}$ and $F=\mathbb{F}_2$, then you're looking at the map $\mathbb{F}_2[x,y]\rightarrow\mathbb{F}_2[x]$ given by $x,y\mapsto x$ and the kernel contains $x^2-xy=x(x-y)$, which is not in the image of $\mathbb{F}_2[x]\rightarrow\mathbb{F}_2[x,y],,p(x)\mapsto p(x)-p(y)$. So perhaps you should ask about the ideal generated by $\mathrm{Im}(\pi_1^{\ast}-\pi_2^{\ast})$. – Thorgott Apr 22 '25 at 10:55
  • It looks like we could identify the kernel as the image of $H^\ast(X\times X,X;F)\to H^\ast(X\times X;F)$ and then some kind of Mayer-Vietoris thing could produce the $\pi_2-\pi_1$, but I can't tie it up – FShrike Apr 22 '25 at 11:02
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    This already fails for $X=S^1$ (for reasons of degree). – Tyrone Apr 22 '25 at 12:31

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As @Thorgott and @Tyrone suggested in the comments, the claim is false. However, it is true if we consider the ideal generated by $\mathrm{Im}(\pi_1^*-\pi_2^*)$.

If $z\in\mathrm{Im}(\pi_1^* - \pi_2^*)$, then $z=x\otimes 1-1\otimes x$ for some $x\in H^*(X;R)$. Hence: $$\Delta^*(z)=\Delta^*(x\otimes 1)-\Delta^*(1\otimes x)=x\smile 1-1\smile x=x-x=0.$$

On the other hand, if $z=\sum_ia_i\otimes b_i\in\ker\Delta^*$, then $\sum_ia_i\smile b_i=0$. We can rewrite each term as: $$a_i\otimes b_i=a_i(1\otimes b_i-b_i\otimes 1)+(a_i\smile b_i)\otimes 1,$$ so if we add them all we obtain: \begin{equation*}\begin{split} z = \sum_ia_i\otimes b_i &=\sum_i(a_i(1\otimes b_i-b_i\otimes 1))+(a_i\smile b_i)\otimes 1 \\ &= \sum_i(a_i(1\otimes b_i-b_i\otimes 1))+\sum_i ((a_i\smile b_i)\otimes 1) \\ &= \sum_ia_i(1\otimes b_i-b_i\otimes 1)\in\left\langle\mathrm{Im}(\pi_1^*-\pi_2^*)\right\rangle. \end{split}\end{equation*}

YM78
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