Lemma 22.14.4. Let $(A, \text{d})$ be a differential graded algebra. Let $M$ be a differential graded $A$-module. There exists a homomorphism $M \to I$ of differential graded $A$-modules such that

1. $M \to I$ is a quasi-isomorphism, and

2. $I$ has property (I).

Proof. Set $M = M_0$. We inductively choose short exact sequences

$0 \to M_ i \to I_ i \to M_{i + 1} \to 0$

where the maps $M_ i \to I_ i$ are chosen as in Lemma 22.14.3. This gives a “resolution”

$0 \to M \to I_0 \xrightarrow {f_0} I_1 \xrightarrow {f_1} I_1 \to \ldots$

Then we set

$I = \prod \nolimits _{i \geq 0} I_ i$

where we take the product in the category of graded $A$-modules and differential defined by

$\text{d}_ I(x) = f_ a(x) + (-1)^ a \text{d}_{I_ a}(x)$

for $x \in I_ a^ b$. With these conventions $I$ is indeed a differential graded $A$-module. Recalling that each $I_ i$ has a two step filtration $0 \to I_ i' \to I_ i \to I_ i'' \to 0$ we set

$F_{2i}P = \prod \nolimits _{j \geq i} I_ j \subset \prod \nolimits _{i \geq 0} I_ i = I$

and we add a factor $I'_{i + 1}$ to $F_{2i}I$ to get $F_{2i + 1}I$. These are differential graded submodules and the successive quotients are products of shifts of $A^\vee$. By Lemma 22.12.1 we see that the inclusions $F_{i + 1}I \to F_ iI$ are admissible monomorphisms. Finally, we have to show that the map $M \to I$ (given by the augmentation $M \to I_0$) is a quasi-isomorphism. This follows from Homology, Lemma 12.23.3. $\square$

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