Lemma 15.75.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $\mathcal{P}$ be a class of $R$-modules. Assume

each $P \in \mathcal{P}$ is a projective $R$-module,

if $P_1 \in \mathcal{P}$ and $P_1 \oplus P_2 \in \mathcal{P}$, then $P_2 \in \mathcal{P}$, and

if $f : P_1 \to P_2$, $P_1, P_2 \in \mathcal{P}$ is surjective modulo $I$, then $f$ is surjective.

Then given any bounded above acyclic complex $E^\bullet $ whose terms are of the form $P/IP$ for $P \in \mathcal{P}$ there exists a bounded above acyclic complex $P^\bullet $ whose terms are in $\mathcal{P}$ lifting $E^\bullet $.

**Proof.**
Say $E^ i = 0$ for $i > b$. Assume given $n$ and a morphism of complexes

\[ \xymatrix{ & & P^ n \ar[r] \ar[d] & P^{n + 1} \ar[r] \ar[d] & \ldots \ar[r] & P^ b \ar[r] \ar[d] & 0 \ar[r] \ar[d] & \ldots \\ \ldots \ar[r] & E^{n - 1} \ar[r] & E^ n \ar[r] & E^{n + 1} \ar[r] & \ldots \ar[r] & E^ b \ar[r] & 0 \ar[r] & \ldots } \]

with $P^ i \in \mathcal{P}$, with $P^ n \to P^{n + 1} \to \ldots \to P^ b$ acyclic in degrees $\geq n + 1$, and with vertical maps inducing isomorphisms $P^ i/IP^ i \to E^ i$. In this situation one can inductively choose isomorphisms $P^ i = Z^ i \oplus Z^{i + 1}$ such that the maps $P^ i \to P^{i + 1}$ are given by $Z^ i \oplus Z^{i + 1} \to Z^{i + 1} \to Z^{i + 1} \oplus Z^{i + 2}$. By property (2) and arguing inductively we see that $Z^ i \in \mathcal{P}$. Choose $P^{n - 1} \in \mathcal{P}$ and an isomorphism $P^{n - 1}/IP^{n - 1} \to E^{n - 1}$. Since $P^{n - 1}$ is projective and since $Z^ n/IZ^ n = \mathop{\mathrm{Im}}(E^{n - 1} \to E^ n)$, we can lift the map $P^{n - 1} \to E^{n - 1} \to E^ n$ to a map $P^{n - 1} \to Z^ n$. By property (3) the map $P^{n - 1} \to Z^ n$ is surjective. Thus we obtain an extension of the diagram by adding $P^{n - 1}$ and the maps just constructed to the left of $P^ n$. Since a diagram of the desired form exists for $n > b$ we conclude by induction on $n$.
$\square$

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