Lemma 24.31.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $(\mathcal{B}, \text{d})$ be a differential graded $\mathcal{O}$-algebra. There exists a quasi-isomorphism of differential graded $\mathcal{O}$-algebras $(\mathcal{A}, \text{d}) \to (\mathcal{B}, \text{d})$ such that $\mathcal{A}$ is graded flat and K-flat as a complex of $\mathcal{O}$-modules and such that the same is true after pullback by any morphism of ringed topoi.
Proof. The proof is exactly the same as the first proof of Lemma 24.23.7 but now working with free graded algebras instead of free graded modules.
We will construct $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits \mathcal{A}_ i$ as in Lemma 24.31.1 by constructing
Let $\mathcal{S}_0$ be the sheaf of graded sets (Remark 24.23.5) whose degree $n$ part is $\mathop{\mathrm{Ker}}(\text{d}_\mathcal {B}^ n)$. Consider the homomorphism of differential graded modules
where map sends a local section $s$ of $\mathcal{S}_0$ to the corresponding local section of $\mathcal{A}^{\deg (s)}$ (which is in the kernel of the differential, so our map is a map of differential graded algebras indeed). By construction the induced maps on cohomology sheaves $H^ n(\mathcal{A}_0) \to H^ n(\mathcal{B})$ are surjective and hence the same will remain true for all $i$.
Induction step of the construction. Given $\mathcal{A}_ i \to \mathcal{B}$ denote $\mathcal{S}_{i + 1}$ the sheaf of graded sets whose degree $n$ part is
This comes equipped with a canonical map
whose image is contained in the kernel of $\text{d}_{\mathcal{A}_ i}$ by construction. Hence $\mathcal{A}_{i + 1} = \mathcal{O}\langle \mathcal{S}_0 \amalg \ldots \mathcal{S}_{i + 1}\rangle $ has a differential exteding the differential on $\mathcal{A}_ i$, see discussion at the start of this section. The map from $\mathcal{A}_{i + 1}$ to $\mathcal{B}$ is the unique map of graded algebras which restricts to the given map on $\mathcal{A}_ i$ and sends a local section $s = (a, b)$ of $\mathcal{S}_{i + 1}$ to $b$ in $\mathcal{B}$. This is compatible with differentials exactly because $\text{d}(b)$ is the image of $a$ in $\mathcal{B}$.
The map $\mathcal{A} \to \mathcal{B}$ is a quasi-isomorphism: we have $H^ n(\mathcal{A}) = \mathop{\mathrm{colim}}\nolimits H^ n(\mathcal{A}_ i)$ and for each $i$ the map $H^ n(\mathcal{A}_ i) \to H^ n(\mathcal{B})$ is surjective with kernel annihilated by the map $H^ n(\mathcal{A}_ i) \to H^ n(\mathcal{A}_{i + 1})$ by construction. Finally, the flatness condition for $\mathcal{A}$ where shown in Lemma 24.31.1. $\square$
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