Wedge product of basis elements of cohomology Let $M$ be a compact, connected, oriented 4-manifold without boundary. If $H^2(M)\cong \mathbb{R}^2$ and I have a basis $\{[\omega_1],[\omega_2]\}$ for $H^2(M)$, is it the case that $[\omega_1\wedge \omega_2]$ is a basis for $H^4(M)$? (Supposing $H^4(M)\cong \mathbb{R}$). Here $H^k(M)$ is the $k$th de Rham cohomology group for $M$.
I know that if $\{\omega_1,\omega_2\}$ is a basis for the space of 2 forms on $M$, then $\omega_1\wedge \omega_2$ is a basis for the space of 4 forms on $M$. But I'm not sure of the relationship between bases for the spaces of differential forms and the cohomologies.

I guess the asnwer is no. For example, let $M$ be the Hopf surface $H = \mathbb S^1 \times \mathbb S^3$ with two blow up points $p, q\in H$ . Then $H^2(M) \cong \mathbb R^2$ and is spanned by the Poincare dual $w_p$, $w_q$ of the two exceptional divisor $E_p$, $E_q$. But $w_p \wedge w_q$ corresponds to the intersection of $E_p$ and $E_q$, which is zero. Thus $w_p \wedge w_q$ is not a basis. 
