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For odd $r\ge 1$ and $W_g={\#}_{i=1}^g(S^r\times S^r)$, the integral cohomology has one copy of $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^{2g}$ in degree $r$, and zero elsewhere. Choose the top orientation class $\omega$ and degree-$r$ classes ${\alpha }_i,{\beta }_i$ from the two factors of summand $i$. Their nonzero positive-degree basis products areAll ${\alpha }_i{\alpha }_j$ and ${\beta }_i{\beta }_j$ vanish, and $\omega$ annihilates positive-degree classes. The Künneth theorem computes each sphere product; the Mayer–Vietoris sequence computes the connected sum groups, and its degree-one pinch maps determine the products. Thus its middle-degree Poincare duality pairing is a symplectic vector space after changing coefficients to a field of characteristic different from two. The case $r=1$ is the usual cohomology ring of a closed oriented surface.

Here closed oriented surfaces are connected, as in the usual single-genus convention. Let $f:X\to Y$ have nonzero mapping degree $d$. Work over $\mathbb{Q}$, so multiplication by $d$ is invertible and cannot vanish.

The Poincare duality pairing on $H^1(Y;\mathbb{Q})$ is nondegenerate. Thus for every $0\ne a\in H^1(Y;\mathbb{Q})$ there is $b\in H^1(Y;\mathbb{Q})$ with $⟨a⌣b,[Y]⟩\ne 0$. Naturality of the cup product and evaluation on the fundamental class giveIn particular $f^{\ast }a\ne 0$, so $f^{\ast }:H^1(Y;\mathbb{Q})\to H^1(X;\mathbb{Q})$ is injective. The cohomology ring of a closed oriented surface has $\dim H^1({\Sigma }_g;\mathbb{Q})=2g$. Comparing dimensions now givesThis proves genus monotonicity under nonzero-degree surface maps by rational cohomology injectivity of a nonzero-degree map, without requiring $f$ to be a covering map or a smooth map.

For singular cochains $a\in C^p(X;R)$ and $b\in C^q(X;R)$, define the cup product on a singular $(p+q)$-simplex byThe identity $\delta (a⌣b)=\delta a⌣b+(-1{)}^{p}a⌣\delta b$ makes this descend to cohomology. The induced product is associative, unital and graded-commutative: $ab=(-1{)}^{pq}ba$ in degrees $p,q$.

For finite CW complexes, the integral cohomological Künneth theorem for finite cell complexes is the split short exact sequenceThe splitting need not be natural. If the factor cohomology is torsion-free, the cohomological cross product is an isomorphism, with multiplication $(a\times b)(c\times d)=(-1{)}^{|b||c|}ac\times bd$.

A genus-$g$ closed oriented surface has cellular groups $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^{2g}$, $H^2=\mathbb{Z}$, and no higher cohomology. Choose the positive orientation class $\omega$ and a symplectic basis $a_i,b_i$ of $H^1$. The cohomology ring of a closed oriented surface is specified bywith unit $1\in H^0$. These products follow by representing their Poincare duality classes by oriented handle curves: the $i$th meridian meets the $i$th longitude once positively, curves on distinct handles have zero intersection, and swapping two oriented curves reverses the intersection sign. In general, for closed oriented $n$-manifolds and transverse oriented submanifolds of codimensions $p,q$, the cup product of their Poincare-dual classes is the dual class of their oriented intersection. Complementary-dimensional intersections give the signed point count $⟨\alpha \beta ,[M]⟩$.

For ${\Sigma }_2$, the four curves $A_1,B_1,A_2,B_2$ give $a_1{b}_1=a_2{b}_2=\omega$ and all other pairings zero apart from their negatives in reversed order. Thus the degree-one intersection pairing on an oriented surface is two symplectic blocks, and the point class is dual to $\omega$.

For ${\Sigma }_2\times {\Sigma }_2$, let $a_i,b_i,{\omega }_1$ come from the first factor and $c_j,d_j,{\omega }_2$ from the second. The Künneth theorem gives their graded tensor-product ring, whose additive ranks in degrees $0,1,2,3,4$ areThe top orientation class is $\Omega ={\omega }_1{\omega }_2$. The two fiber surfaces $\{p\}\times {\Sigma }_2$ and ${\Sigma }_2\times \{q\}$ have dual classes ${\omega }_1,{\omega }_2$, intersect once positively and have zero self-intersection. The other sixteen degree-two classes are products of degree-one classes and are dual, up to the chosen orientation convention, to products of handle curves. Their intersections are encoded byThe minus sign in the first mixed formula comes from commuting $c_j$ past $b_k$. These classes together with the fiber classes give the intersection form of a product of two closed oriented surfaces, here nine hyperbolic blocks. Complementary degrees one and three are similarly paired, for example $a_i(b_k{\omega }_2)={\delta }_{ik}\Omega$; geometrically this pairs a three-dimensional product submanifold with a transverse loop. This exhibits the full duality, not just the additive groups.

The Lefschetz fixed-point theorem says that a self-map of a compact polyhedron with nonzero Lefschetz number has a fixed point, where $L(f)={\sum }_q(-1{)}^q\mathrm{tr}(f_{\ast }:H_q(-;\mathbb{Q})\to H_q(-;\mathbb{Q}))$. A map homotopic to the identity has $L(f)=\chi$, and Euler characteristic of a product givesConsequentlyThe product $g_1{g}_2$ itself is not determined by the printed assumptions. For every $h\ge 0$, translation by a nonzero torus point, times the identity on ${\Sigma }_h$, is a fixed-point-free map of ${\Sigma }_1\times {\Sigma }_h$ homotopic to the identity, and its genus product is $h$. Thus the exact conclusion is the criterion for fixed-point-free self-maps homotopic to the identity on products of surfaces, rather than a single value of $g_1{g}_2$.

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. HenceHere we use the cellular homology theorem, identifying cellular and singular homology, and the universal coefficient theorem for cohomology: its exact sequence has terms $\mathrm{Ext}(H_{q-1},\mathbb{Z})$ and $\mathrm{Hom}(H_q,\mathbb{Z})$. 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 $a_1,b_1,\dots ,a_g,b_g$ so that, for the positive orientation class $\omega$,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.

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 $X$.