= Solution
Give the closed oriented genus-$g$ surface $\Sigma_g$ its usual CW structure with one vertex, $2g$ one-cells $a_1,b_1,\ldots,a_g,b_g$, and one two-cell attached along $\prod_i[a_i,b_i]$. The cellular boundary maps vanish after abelianization, so
$$
H^q(\Sigma_g;\mathbb Z)\cong
\begin{cases}
\mathbb Z,&q=0,2,\\
\mathbb Z^{2g},&q=1,\\
0,&\text{otherwise}.
\end{cases}
$$
Choose 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>:
$$
\alpha_i\smile\beta_j=\delta_{ij}\omega,
\qquad
\beta_j\smile\alpha_i=-\delta_{ij}\omega,
$$
and all $\alpha_i\smile\alpha_j$ and $\beta_i\smile\beta_j$ 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. Hence
$$
H^0(X;\mathbb Z)=\mathbb Z,qquad
H^1(X;\mathbb Z)=\mathbb Z^4,qquad
H^2(X;\mathbb Z)=\mathbb Z^2.
$$
Let $u,v$ 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 $[b_1,b_2]$ give
$$
\alpha_1\smile\beta_1=alpha_2\smile\beta_2=u,
\qquad
\beta_1\smile\beta_2=v,
$$
together 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.
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