Lemma 15.75.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $\mathcal{P}$ be a class of $R$-modules. Let $K \in D(R)$ and let $E^\bullet $ be a complex of $R/I$-modules representing $K \otimes _ R^\mathbf {L} R/I$. Assume
each $P \in \mathcal{P}$ is a projective $R$-module,
$P_1 \in \mathcal{P}$ and $P_1 \oplus P_2 \in \mathcal{P}$ if and only if $P_1, P_2 \in \mathcal{P}$,
if $f : P_1 \to P_2$, $P_1, P_2 \in \mathcal{P}$ is surjective modulo $I$, then $f$ is surjective,
$E^\bullet $ is bounded above and $E^ i$ is of the form $P/IP$ for $P \in \mathcal{P}$, and
$K$ can be represented by a bounded above complex whose terms are in $\mathcal{P}$.
Then there exists a bounded above complex $P^\bullet $ whose terms are in $\mathcal{P}$ with $P^\bullet /IP^\bullet $ isomorphic to $E^\bullet $ and representing $K$ in $D(R)$.
Proof.
By assumption (5) we can represent $K$ by a bounded above complex $K^\bullet $ whose terms are in $\mathcal{P}$. Then $K \otimes _ R^\mathbf {L} R/I$ is represented by $K^\bullet /IK^\bullet $. Since $E^\bullet $ is a bounded above complex of projective $R/I$-modules by (4), we can choose a quasi-isomorphism $\delta : E^\bullet \to K^\bullet /IK^\bullet $ (Derived Categories, Lemma 13.19.8). Let $C^\bullet $ be cone on $\delta $ (Derived Categories, Definition 13.9.1). The module $C^ i$ is the direct sum $K^ i/IK^ i \oplus E^{i + 1}$ hence is of the form $P/IP$ for some $P \in \mathcal{P}$ as (2) says in particular that $\mathcal{P}$ is preserved under taking sums. Since $C^\bullet $ is acyclic, we can apply Lemma 15.75.1 and find a acyclic lift $A^\bullet $ of $C^\bullet $. The complex $A^\bullet $ is bounded above and has terms in $\mathcal{P}$. In
\[ \xymatrix{ K^\bullet \ar@{..>}[r] \ar[d] & A^\bullet \ar[d] \\ K^\bullet /IK^\bullet \ar[r] & C^\bullet \ar[r] & E^\bullet [1] } \]
we can find the dotted arrow making the diagram commute by Derived Categories, Lemma 13.19.6. We will show below that it follows from (1), (2), (3) that $K^ i \to A^ i$ is the inclusion of a direct summand for every $i$. By property (2) we see that $P^ i = \mathop{\mathrm{Coker}}(K^ i \to A^ i)$ is in $\mathcal{P}$. Thus we can take $P^\bullet = \mathop{\mathrm{Coker}}(K^\bullet \to A^\bullet )[-1]$ to conclude.
To finish the proof we have to show the following: Let $f : P_1 \to P_2$, $P_1, P_2 \in \mathcal{P}$ and $P_1/IP_1 \to P_2/IP_2$ is split injective with cokernel of the form $P_3/IP_3$ for some $P_3 \in \mathcal{P}$, then $f$ is split injective. Write $E_ i = P_ i/IP_ i$. Then $E_2 = E_1 \oplus E_3$. Since $P_2$ is projective we can choose a map $g : P_2 \to P_3$ lifting the map $E_2 \to E_3$. By condition (3) the map $g$ is surjective, hence split as $P_3$ is projective. Set $P_1' = \mathop{\mathrm{Ker}}(g)$ and choose a splitting $P_2 = P'_1 \oplus P_3$. Then $P'_1 \in \mathcal{P}$ by (2). We do not know that $g \circ f = 0$, but we can consider the map
\[ P_1 \xrightarrow {f} P_2 \xrightarrow {projection} P'_1 \]
The composition modulo $I$ is an isomorphism. Since $P'_1$ is projective we can split $P_1 = T \oplus P'_1$. If $T = 0$, then we are done, because then $P_2 \to P'_1$ is a splitting of $f$. We see that $T \in \mathcal{P}$ by (2). Calculating modulo $I$ we see that $T/IT = 0$. Since $0 \in \mathcal{P}$ (as the summand of any $P$ in $\mathcal{P}$) we see the map $0 \to T$ is surjective and we conclude that $T = 0$ as desired.
$\square$
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