Define the projection maps, $$\pi_M:M\times N \rightarrow M,\pi_N:M\times N \rightarrow N$$ Let $M^{2m}$ and $N^{2n}$ be smooth compact orientable manifolds with no boundary of dimension $2m$ and $2n$. Prove that the map below is well defined: $$H^2_{dR}(M)\times H^2_{dR}(N)\rightarrow \mathbb{R}$$ $$([\gamma],[\eta])\rightarrow \int_{M\times N}(\pi^*_{M}\gamma+\pi^*_{N}\eta)^{m+n}$$
Can anyone help me solve this problem? Or at least give me some hints to how to begin? I know all the definitions and the idea but I just don't seem to get what they mean by a well-defined map?
My thoughts towards the solution: Let us take the cohomology classes and let $[\eta]=[\eta^{'}]$ and $[\gamma]=[\gamma^{'}]$ which are respectively in $H^2(M)$ and $H^2(N)$. As the hint suggests we wish to show that:
$$∫_{M×N}(π^*_Mγ+π^*_N\eta)^{m+n}=∫_{M×N}(π^*_Mγ^{'}+π^*_N\eta^{'})^{m+n}$$
We make a small obersvation, and throguh binomical expansion that we can easily obtain:
$$(π^*_Mγ+π^*_N\eta)^{m+n}=C^{m+n}_m(π^*_Mγ)^m\wedge (π^*_M\eta)^n$$ $$∫_{M×N}(π^*_Mγ)^m\wedge (π^*_M\eta)^n=∫_{M×N}(π^*_Mγ^{'})^m\wedge (π^*_M\eta^{'})^n$$
So we start from the RHS: $$∫_{M×N}(π^*_Mγ)^m\wedge (π^*_M\eta)^n=∫_{M×N}\big(π^*_M(\gamma^{'}+d\tau))^m\wedge (π^*_M\eta)^n\big)$$ $$∫_{M×N}\big(π^*_M(\gamma^{'}+d\tau))^m\wedge (π^*_M\eta)^n\big)=∫_{M×N}π^*_M((\gamma^{'})^m+C^m_1(\gamma^{'})^{m-1}\wedge d\tau+...+(d\tau)^m)\wedge (π^*_M\eta)^n=∫_{M×N}(π^*_M\gamma^{'})^m\wedge (π^*_M\eta)^n+∫_{M×N}π^*_M(C^m_1(\gamma^{'})^{m-1}\wedge d\tau+...+(d\tau)^m)\wedge (π^*_M\eta)^n$$
It is not hard to see that the second term $C^m_1(\gamma^{'})^{m-1}\wedge d\tau+...+(d\tau)^m=d\big(C^m_1(\gamma^{'})^{m-1}\wedge \tau+...+\tau \wedge d\tau \wedge ... \wedge d\tau\big)$. This holds since $\gamma^{'}$ since it belongs to the cohomology class. Therfore our above integral becomes:
$$∫_{M×N}π^*_M(C^m_1(\gamma^{'})^{m-1}\wedge d\tau+...+(d\tau)^m)\wedge (π^*_M\eta)^n=∫_{M×N}π^*_M(d\Lambda) \wedge (π^*_M\eta)^n=∫_{M×N}d(π^*_M\Lambda) \wedge (π^*_M\eta)^n=∫_{M×N}d\big((π^*_M\Lambda) \wedge (π^*_M\eta)^n\big)=0$$
It is $0$ via the generalised stokes theorem for manifolds with no boundary. And therefore we have conclusively proved that: $$∫_{M×N}(π^*_Mγ)^m\wedge (π^*_M\eta)^n=∫_{M×N}(π^*_M\gamma^{'})^m\wedge (π^*_M\eta)^n+∫_{M×N}π^*_M(C^m_1(\gamma^{'})^{m-1}\wedge d\tau+...+(d\tau)^m)\wedge (π^*_M\eta)^n$$ $$∫_{M×N}(π^*_Mγ)^m\wedge (π^*_M\eta)^n=∫_{M×N}(π^*_M\gamma^{'})^m\wedge (π^*_M\eta)^n$$
From the last line, if we continue and let $\eta = \eta^{'} + d\psi$, then we can prove that: $$∫_{M×N}(π^*_M\gamma^{'})^m\wedge (π^*_M\eta)^n=∫_{M×N}(π^*_M\gamma^{'})^m\wedge (π^*_M\eta^{'})^n$$ $$\implies ∫_{M×N}(π^*_Mγ)^m\wedge (π^*_M\eta)^n= ∫_{M×N}(π^*_M\gamma^{'})^m\wedge (π^*_M\eta^{'})^n$$
Hence we are done.