This is the situation:
Let $A = R_* \otimes C_*$ be an $R$-module where $C_*$ is a finitely generated graded ($*\geq 0$) vector space over a field $F$ which is also bounded above, and $R$ is a graded algebra over $F$ where $R_0 = F$. Suppose that $d: A \rightarrow A$ is a map of $R$-modules of degree $1$ such that
- $d^2 = 0$
- $im(d) \otimes_R F = 0$ when $F$ is consider as an $R$-module with the augmentation map $R \rightarrow F$.
- The filtration $F_pA = R_{*\geq p} \otimes C_*$ of $A$ by $R$-degree satisfies $d(F_{p} A) \subseteq F_{p+1} A$
Then I am trying to prove that
the cohomology $H^*(A)$ is free as $R$-module if and only if $d = 0$.
If $d = 0$, it is immediate from $H^*(A) = R_* \otimes C_*$ and the assumptions over $C$.
Now I suppose that $H^*(A)$ is a free $R$-module with basis $a_1, \ldots, a_n$ and I want to show that $H^*(A) \cong R_* \otimes C_*$. If the latter does not hold, then the associated spectral sequence to the given filtration does not degenerate at $E_1$.
Consider then $q$ the lowest $q$-th row where a non-zero differential occurs, and let $d_r$ be the longest of such a map. Then the map $H^*(M) \rightarrow H^*(M) \otimes_R F \cong C_*$ is surjective up to a degree $q-1$, and let $\{x_i\}$ be the image of those $a_i \in H^*(A)$ basis elements such that $\deg(a_i)<q$.
As $d$ is a map of $R$-modules, so is $d_r$ and thus we may assume that $d_r(1 \otimes y) \neq 0$ for some $y \in C_q$. Since $d_r(1 \otimes y) \in E_r^{r,q-r+1} = R_r \otimes C_{q-r+1}$ we can write $d_r(1 \otimes y) = \sum_i r_i \otimes x_i$
I am trying to show that $\sum_i r_ia_i = 0$ so it gives a linearly dependence relation between some basis elements in $H^*(M)$ , but I do not know how relate this sum with $d_r(1 \otimes y)$
