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Apply the rational Serre spectral sequence to the path-loop fibration of $K(\mathbb{Z},3)$. Its fiber iswhose rational cohomology ring is $\mathbb{Q}[c]$ with $|c|=2$. Since the path space is contractible, $c$ must transgress to a nonzero class $x\in H^3(K(\mathbb{Z},3);\mathbb{Q})$. Multiplicativity givesOver $\mathbb{Q}$ these differentials pair and kill every positive-degree class except $x$, while graded commutativity gives $x^2=0$. Hence

Solved by gpt-5.6-sol high.

Give the closed oriented genus-$g$ surface ${\Sigma }_g$ its usual CW structure with one vertex, $2g$ one-cells $a_1,b_1,\dots ,a_g,b_g$, and one two-cell attached along ${\prod }_i[a_i,b_i]$. The cellular boundary maps vanish after abelianization, soChoose degree-one classes ${\alpha }_i,{\beta }_i$ dual to $a_i,b_i$ and orient ${\Sigma }_g$ by $\omega \in H^2$. The intersection pairing, equivalently the cellular diagonal approximation, gives the cohomology ring of a closed oriented surface:and all ${\alpha }_i⌣{\alpha }_j$ and ${\beta }_i⌣{\beta }_j$ vanish.

For the space $X$, use the genus-two CW structure and attach an additional two-cell $e_v^2$ along $[b_1,b_2]$. Both two-cell attaching words have zero exponent sum in every one-cell, so the cellular boundary $C_2(X)\to C_1(X)$ is zero. HenceLet $u,v$ be the degree-two classes dual respectively to the original surface cell and the new cell. The original relator $[a_1,b_1][a_2,b_2]$ and the new relator $[b_1,b_2]$ givetogether with the products forced by graded commutativity. Every other product of degree-one basis classes is zero, and every product of total degree greater than two is zero. These relations completely determine the cohomology ring of $X$.

Solved by gpt-5.6-sol high.