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

\[ 0 \to M_{i + 1} \to P_ i \to M_ i \to 0 \]

where the maps $P_ i \to M_ i$ are chosen as in Lemma 22.13.3. This gives a “resolution”

\[ \ldots \to P_2 \xrightarrow {f_2} P_1 \xrightarrow {f_1} P_0 \to M \to 0 \]

Then we set

\[ P = \bigoplus \nolimits _{i \geq 0} P_ i \]

as an $A$-module with grading given by $P^ n = \bigoplus _{a + b = n} P_{-a}^ b$ and differential (as in the construction of the total complex associated to a double complex) by

\[ \text{d}_ P(x) = f_{-a}(x) + (-1)^ a \text{d}_{P_{-a}}(x) \]

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

\[ F_{2i}P = \bigoplus \nolimits _{i \geq j \geq 0} P_ j \subset \bigoplus \nolimits _{i \geq 0} P_ i = P \]

and we add $P'_{i + 1}$ to $F_{2i}P$ to get $F_{2i + 1}$. These are differential graded submodules and the successive quotients are direct sums of shifts of $A$. By Lemma 22.11.1 we see that the inclusions $F_ iP \to F_{i + 1}P$ are admissible monomorphisms. Finally, we have to show that the map $P \to M$ (given by the augmentation $P_0 \to M$) is a quasi-isomorphism. This follows from Homology, Lemma 12.23.2.
$\square$

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