I'm currently going through a proof of the Künneth formula via the Cech-de Rham complex. More precisely, the statement is the following:
If $M$ and $F$ are two manifolds and $F$ has finite-dimensional cohomology, then $H^{\bullet}(M\times F)=H^{\bullet}(M)\otimes H^{\bullet}(F)$.
Let $\mathfrak{U}=\{U_{\alpha}\}_{\alpha\in A}$ be a countable good open cover of $M$ (which we suppose to be ordered). Then we know that there is an isomorphism $H^{\bullet}(M)=H^{\bullet}(C^{\bullet}(\mathfrak{U},\Omega^{\bullet}))$.
Choose closed forms $\omega_{1},...,\omega_{r}\in \Omega^{\bullet}(F)$ such that $\{[\omega_{1}],...,[\omega_{r}]\}$ is a basis of $H^{\bullet}(F)$. The authors then define a map $\pi_{\mathfrak{U}}^{*}\colon H^{\bullet}(F)\otimes C^{\bullet}(\mathfrak{U},\Omega^{\bullet})\to C^{\bullet}(\mathfrak{\pi^{-1}\mathfrak{U}},\Omega^{\bullet})$ given by $[\omega_{i}]\otimes \eta \to \rho^{*}\omega_{i}\wedge\pi^{*}\eta$, where $\rho\colon M\times F\to F$ and $\pi\colon M\times F\to M$ are the projections.
Here is the confusing part to me: the authors claim that $H^{\bullet}(F)\otimes C^{\bullet}(\mathfrak{U},\Omega^{\bullet})$ can be seen as a (double, I think) complex with cohomology $H^{\bullet}(F)\otimes H^{\bullet}(C^{\bullet}(\mathfrak{U},\Omega^{\bullet}))$ by using the differential of the second factor. That is, if we denote by $d$ and $\delta$ the differentials of $C^{\bullet}(\mathfrak{U},\Omega^{\bullet})$, then $1\otimes d$ and $1\otimes \delta$ are the differentials of $H^{\bullet}(F)\otimes C^{\bullet}(\mathfrak{U},\Omega^{\bullet})$. At first, this made sense to me, but later it is stated that $\pi^{*}_{\mathfrak{U}}$ induces an isomorphism in cohomology. But if we take some $\eta\in C^{p}(\mathfrak{U},\Omega^{q})$, then $\rho^{*}\omega_{i}\wedge \pi^{*}\eta\in C^{p}(\pi^{-1}\mathfrak{U},\Omega^{q+\operatorname{deg}\omega_{i}})$, so $\pi^{*}_{\mathfrak{U}}$ is not a homomorphism of double complexes.
What am I missing here? Is it that the structure of complex in $H^{\bullet}(F)\otimes C^{\bullet}(\mathfrak{U},\Omega^{\bullet})$ is different than the one I'm thinking of?
Thank you in advance!
EDIT: I believe the problem here lies on the structure given to $H^{\bullet}(F)\otimes C^{\bullet}(\mathfrak{U},\Omega^{\bullet})$. I thought that maybe we can consider it as a product complex by using the fact that $H^{\bullet}(F)$ is finite-dimensional, but I'm not sure on how to proceed.
This is to long for a comment. We can always look at a chain complex $C^{*}$ as a double complex by defining the double complex $\tilde{C}^{*,*}$ to be $$\tilde{C}^{i,0}=C^i$$ with the vertical boundary map $d$. For $j>0$ we define $\tilde{C}^{i,j}=0_{Ab}$ and the horizontal maps to be the zero homomorphism. In general The tensor product of two chain complexes $X^*, Y^*$ is
$$(X \otimes Y)^{n}:=\oplus_{i+j=n} X^{i} \otimes Y^{j}.$$ The double complex case is the same $$(X\otimes Y)^{i,j}=\oplus_{a+b=i} \oplus_{c+d=j}X^{a,c}\otimes Y^{b,d} $$ The vertical maps are given by $$id_X\otimes d_Y+d_X\otimes id_Y,$$ and the horizontal maps are given by $$id_X\otimes \delta_Y+\delta_X\otimes id_Y.$$
In your example we have $d_{H^*(F)}=0$ and $\delta_{H^*(F)}=0$. Thus we have $$(H^{\bullet,\bullet}(F)\otimes C^{\bullet}(\mathfrak{U},\Omega^{\bullet}))^{i,j}=\oplus_{a+b=i}\oplus_{c+d=j}H^{a,c}(F)\otimes C^{d}(\mathfrak{U},\Omega^{b})$$ This is given by $$(H^{\bullet,\bullet}(F)\otimes C^{\bullet}(\mathfrak{U},\Omega^{\bullet}))^{i,j}=\oplus_{a+b=i}\oplus_{0+d=j}H^{a,0}(F)\otimes C^{d}(\mathfrak{U},\Omega^{b})=\oplus_{a+b=i}H^{a,0}(F)\otimes C^{j}(\mathfrak{U},\Omega^{b}).$$ If we now compute the differentials we get that they are $id_{H(F)}\otimes d_{C^{\bullet}(\mathfrak{U},\Omega^{\bullet})}$ and $id_{H(F)}\otimes \delta_{C^{\bullet}(\mathfrak{U},\Omega^{\bullet})}$.