= Solution
Give the <closed orientable surface> its standard <CW complex> structure: one zero-cell, $2g$ one-cells, and one two-cell attached by the product of $g$ commutators. The cellular boundary of the two-cell is zero, since every edge occurs once with each orientation in that word; the one-cell boundaries are also zero. Hence
$$
H^q(\Sigma_g;\mathbb Z)\cong\begin{cases}\mathbb Z,&q=0,2,\\\mathbb Z^{2g},&q=1,\\0,&\text{otherwise}.\end{cases}
$$
Here we use the <cellular homology theorem>, identifying cellular and singular homology, and the <universal coefficient theorem for cohomology>: its exact sequence has terms $\operatorname{Ext}(H_{q-1},\mathbb Z)$ and $\operatorname{Hom}(H_q,\mathbb Z)$. All the homology groups here are free, so the Ext terms vanish.
The ring structure comes from <Poincare duality> and <algebraic intersection number of curves on an oriented surface>. For a closed oriented surface, cap product with its <fundamental class> identifies degree-one <cohomology> with degree-one homology; evaluating the <cup product> of two such classes equals the signed intersection number of their dual one-cycles. Choose the usual $g$ pairs of handle curves, each pair meeting positively once, and distinct pairs disjoint. Their dual classes can accordingly be named $a_1,b_1,\ldots,a_g,b_g$ so that, for the positive orientation class $\omega$,
$$
\boxed{a_i\smile b_j=\delta_{ij}\omega,\qquad b_j\smile a_i=-\delta_{ij}\omega,\qquad a_i\smile a_j=b_i\smile b_j=0.}
$$
These formulas include squares. More generally, <graded commutativity of the cup product> kills every degree-one square here because $H^2$ is torsion-free. The unit generates $H^0$, and products involving $\omega$ and any positive-degree class vanish for dimensional reasons. These additive groups and multiplication rules completely describe the <cohomology ring of a closed oriented surface>, including $g=0$, when there are no degree-one generators. The intersection pairing is integral and unimodular, rather than merely nondegenerate over a field.
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