Proof of Kunneth's formula in Bott & Tu Let $M, F$ be smooth manifolds and let's assume all (henceforth, de Rham) cohomologies of $F$ are finite-dimensional. Let $\pi : M \times F \rightarrow M$ and $\rho : M \times F \rightarrow F$ be the projections. There is a pre-sheaf $\mathcal{H}^q$ on $M$, namely $\mathcal{H}^q(U) = H^q(\pi^{-1} U)$. Let's take a good cover $\mathfrak{U}$ of $M$. Then $\mathcal{H}^q$ is locally constant on $\mathfrak{U}$. In fact it is constant; moreso, there are global forms on $M \times F$ which, when restricted to each fiber $F$, freely generate the cohomology. Indeed, we need only take forms $\sigma_1, \ldots, \sigma_n$ in $F$ freely generating $H^q(F)$ and set $\widetilde{\sigma_i} = \rho^{*} \sigma_i$.
There is a spectral sequence converging to the cohomology of $M \times F$, and given that the pre-sheaf $\mathcal{H}^q$ is constant (for every $q$) we see that the $E_{2}^{p, q}$ term is $H^{p}(M) \otimes H^{q}(F)$. So far, so good. Now Bott & Tu wish to argue that the $d_2$ on $E_2$ is $0$; they claims that forms on $E_2$ are already global. I don't understand this statement. I'm aware that for any element $\omega \in E_{2}^{p, q} = H^p(\mathfrak{U}, \mathcal{H}^q)$, we will have $\omega(U_{\alpha_0 \ldots \alpha_p}) = \sum_{i = 1}^{n} a^{\alpha_0 \ldots \alpha_p}_{i} [\widetilde{\sigma}_i]$, so that each $\omega(U_{\alpha_0 \ldots \alpha_p})$ is a global form on $M \times F$. But what guarantees that $\omega$ itself is a global form?
 A: I believe I've found the answer myself. Let me now speak of $\omega$ on the level of forms, recalling that $E_{2} = H_{\delta} H_{d}(C^{*}(\pi^{-1} \mathfrak{U}, \Omega^{*}))$. By this I mean $\omega_{\alpha_0 \ldots \alpha_p} = \sum_{i = 1}^{n} a^{\alpha_0 \ldots \alpha_p}_{i} \sigma_i$ represents something in $H_{\delta} H_d$ cohomology, so $d\omega = 0$ and $[\delta \omega]_{d} = 0$.
Calculating $\delta \omega$ we get $(\delta \omega)_{\alpha_0 \ldots \alpha_{p+1}} = \sum_{i = 1}^{n} (\sum_{j = 0}^{p + 1} (-1)^{j} a^{\alpha_0 \ldots \widehat{\alpha_j} \ldots \alpha_{p + 1}}_{i}) \sigma_i$. So $0 = [(\delta \omega)_{\alpha_0 \ldots \alpha_{p + 1}}]_d = \sum_{i = 1}^{n} (\sum_{j = 0}^{p + 1} (-1)^{j} a^{\alpha_0 \ldots \widehat{\alpha_j} \ldots \alpha_{p + 1}}_{i}) [\sigma_i]$. But since the $[\sigma_i]$ are freely generating the cohomology, we see $\sum_{j = 0}^{p + 1} (-1)^{j} a^{\alpha_0 \ldots \widehat{\alpha_j} \ldots \alpha_{p + 1}}_{i} = 0$. Thus $\delta \omega = 0$ already on the level of forms. Then it's easy to see $d_2 = d_3 = \ldots = 0$.
