Let $\alpha, \beta$ two forms continuous with compact supports and maximum degree on surfaces oriented $M, N $ respectively. consider $\pi_M:M\times N \rightarrow M$ and $\pi_N:M\times N \rightarrow N$. Prove that
$$\int_{M\times N}\pi^*_M\alpha \wedge\pi^*_N\beta=(\int_M \alpha )\ .(\int_N \beta ) $$
any suggestions to address the problem, is welcome
Here's a lazy answer, that needs fact-checking (or a better answer):
I wouldn't think there is a "brutal" proof because because the reason this is true when $M = \mathbb{R}^n$ and $N = \mathbb{R}^m$ is a property of the Lebesgue measure.
So I would probably say: this is true when $M$ and $N$ are open sets in Euclidean spaces, so for the general case use a partition of unity, etc etc