prove: When a homomorphism $f: K \rightarrow K'$ of double complexes induces $H_d$-isomorphism, it also induces $H_D$-isomorphism.
This is in the context of proving Kunneth formula using the double complex. The map is the pullback of wedge products:
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 with closed forms basis $\{[\omega_{\alpha}]\}$ Then define
$$ \pi^*: H^*(F) \otimes C^*(U,\Omega^*)\rightarrow C^*(\pi^{-1}U,\Omega^*)$$
$$ \pi^*([\omega_{\alpha}]\otimes\phi)=\rho^*\omega_{\alpha}\wedge \pi^*\phi$$
This commutes with $d$ and $\delta$ and induces an isomorphism in $d$-cohomology.
*without use of spectral sequences which comes later in the book.
book bott and Tu: 108 and 107, http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf
Edit: I do not understand the answer below. The cohomolgy classes for me are given by "diagonals, see text. The answer below does not seem correct to me.