Assuming as known the cup product structure on the torus \(S^1 \times S^1\), compute the cup product structure in \(H^*(M_g)\) for \(M_g\) the closed orientable surface of genus \(g\) by using the quotient map from \(M_g\) to a wedge sum of \(g\) tori.
Solution
Choose the standard symplectic basis \[ a_1,b_1,\dots,a_g,b_g\in H^1(M_g;\mathbb Z) \] and let \(u\in H^2(M_g;\mathbb Z)\cong\mathbb Z\) be the orientation class.
Let \[ q:M_g\longrightarrow T_1\vee\cdots\vee T_g \] be the quotient map obtained by collapsing the separating circles between the \(g\) handles. On \(H^1\), \(q^*\) identifies the two degree-one generators of \(H^1(T_i)\) with \(a_i,b_i\). If \(u_i\in H^2(T_i)\) is the orientation class of the \(i\)th torus, then \(q^*(u_i)=u\): restricting \(q\) to the \(i\)th handle and collapsing its boundary has degree \(1\) onto \(T_i\).
On each torus, with the usual orientation, \[ x_i\smile y_i=u_i,\qquad y_i\smile x_i=-u_i, \qquad x_i^2=y_i^2=0. \] Products of classes supported on distinct wedge summands vanish. Naturality of the cup product therefore gives \[ \boxed{a_i\smile b_j=\delta_{ij}u,\qquad b_i\smile a_j=-\delta_{ij}u,} \] and \[ \boxed{a_i\smile a_j=0,\qquad b_i\smile b_j=0} \] for all \(i,j\). Since \(H^k(M_g;\mathbb Z)=0\) for \(k>2\), these relations, together with the unit in \(H^0\), determine the whole ring.