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Let $U=\{U_{\alpha}\}$ be a good cover for $M$ and $\pi: M\times F\rightarrow M$ projection into $M$ and $\rho$ likewise into $F$. Then $\{\pi^{-1}U_{\alpha}\}$ is some cover for $ M\times F$. Assume cohomology $H^*(F)$ is finite dimensional. The $p^{th}$ column $C^p(\pi^{-1}U,\Omega^*)$ consists of forms on the $(p+1)$ fold intersections $\bigsqcup \pi^{-1} U_{\alpha_0...\alpha_p}$ and $C^p(U,\Omega^*)$ consists of forms on $\bigsqcup U_{\alpha_0...\alpha_p}$. The $d$-cohomology of $C^p(\pi^{-1}U,\Omega^*)$ is

$$ \prod H^*( \pi^{-1} U_{\alpha_0...\alpha_p}) \simeq H^*(F)\otimes \prod H^*( U_{\alpha_0...\alpha_p})$$

The isomorphism being given by the wedge product of pullbacks.

But this itself require the Kunneth formula, does it not? Can someone please give detail proof of this isomorphism?

book bott and Tu: 108 and 107, http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf

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    It does not since $U = U_{\alpha_0 \cdots \alpha_p}$ is contractible, $H^*(U)$ is trivial and $\pi^{-1} (U)\cong U\times F$ is homotopic to $F$ : So both sides of your equation is isomorphic to $H^*(F)$, and the isomorphism is given by pullback, that can be checked easily (using that the inclusion $i : F\to \pi^{-1}(U)$ induces isomorphisms on cohomology). – 2017-01-29
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    oh I see. thanks. – 2017-01-29
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    @John Ma please see http://math.stackexchange.com/questions/2123526/kunneth-formula-using-the-double-complex I am stuck in the other part of the theorem as well :( – 2017-02-01

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