**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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