The Stacks project

Proof. We are going to use that the homotopy category $K(\mathcal{A})$ is a triangulated category, see Proposition 13.10.3. By Lemma 13.15.4 we can find a termwise surjective map of complexes $P_1^\bullet \to \tau _{\leq 1}K^\bullet $ which is a quasi-isomorphism such that the terms of $P_1^\bullet $ are in $\mathcal{P}$. By induction it suffices, given $P_1^\bullet , \ldots , P_ n^\bullet $ to construct $P_{n + 1}^\bullet $ and the maps $P_ n^\bullet \to P_{n + 1}^\bullet $ and $P_{n + 1}^\bullet \to \tau _{\leq n + 1}K^\bullet $.

Choose a distinguished triangle $P_ n^\bullet \to \tau _{\leq n + 1}K^\bullet \to C^\bullet \to P_ n^\bullet [1]$ in $K(\mathcal{A})$. Applying Lemma 13.15.4 we choose a map of complexes $Q^\bullet \to C^\bullet $ which is a quasi-isomorphism such that the terms of $Q^\bullet $ are in $\mathcal{P}$. By the axioms of triangulated categories we may fit the composition $Q^\bullet \to C^\bullet \to P_ n^\bullet [1]$ into a distinguished triangle $P_ n^\bullet \to P_{n + 1}^\bullet \to Q^\bullet \to P_ n^\bullet [1]$ in $K(\mathcal{A})$. By Lemma 13.10.7 we may and do assume $0 \to P_ n^\bullet \to P_{n + 1}^\bullet \to Q^\bullet \to 0$ is a termwise split short exact sequence. This implies that the terms of $P_{n + 1}^\bullet $ are in $\mathcal{P}$ and that $P_ n^\bullet \to P_{n + 1}^\bullet $ is a termwise split injection whose cokernels are in $\mathcal{P}$. By the axioms of triangulated categories we obtain a map of distinguished triangles

in the triangulated category $K(\mathcal{A})$. Choose an actual morphism of complexes $f : P_{n + 1}^\bullet \to \tau _{\leq n + 1}K^\bullet $. The left square of the diagram above commutes up to homotopy, but as $P_ n^\bullet \to P_{n + 1}^\bullet $ is a termwise split injection we can lift the homotopy and modify our choice of $f$ to make it commute. Finally, $f$ is a quasi-isomorphism, because both $P_ n^\bullet \to P_ n^\bullet $ and $Q^\bullet \to C^\bullet $ are.

At this point we have all the properties we want, except we don't know that the map $f : P_{n + 1}^\bullet \to \tau _{\leq n + 1}K^\bullet $ is termwise surjective. Since we have the commutative diagram

of complexes, by induction hypothesis we see that $f$ is surjective on terms in all degrees except possibly $n$ and $n + 1$. Choose an object $P \in \mathcal{P}$ and a surjection $q : P \to K^ n$. Consider the map

with first copy of $P$ in degree $n$ and maps given by $q$ in degree $n$ and $d_ K \circ q$ in degree $n + 1$. This is a surjection in degree $n$ and the cokernel in degree $n + 1$ is $H^{n + 1}(\tau _{\leq n + 1}K^\bullet )$; to see this recall that $\tau _{\leq n + 1}K^\bullet $ has $\mathop{\mathrm{Ker}}(d_ K^{n + 1})$ in degree $n + 1$. However, since $f$ is a quasi-isomorphism we know that $H^{n + 1}(f)$ is surjective. Hence after replacing $f : P_{n + 1}^\bullet \to \tau _{\leq n + 1}K^\bullet $ by $f \oplus g : P_{n + 1}^\bullet \oplus P^\bullet \to \tau _{\leq n + 1}K^\bullet $ we win. $\square$

Proof. By (1) and Lemma 13.15.4 for any bounded above complex $K^\bullet $ there exists a quasi-isomorphism $P^\bullet \to K^\bullet $ with $P^\bullet $ bounded above and $P^ n \in \mathcal{P}$ for all $n$. Suppose that $s : P^\bullet \to (P')^\bullet $ is a quasi-isomorphism of bounded above complexes consisting of objects of $\mathcal{P}$. Then $F(P^\bullet ) \to F((P')^\bullet )$ is a quasi-isomorphism because $F(C(s)^\bullet )$ is acyclic by assumption (2). This already shows that $LF$ is defined on $D^{-}(\mathcal{A})$ and that a bounded above complex consisting of objects of $\mathcal{P}$ computes $LF$, see Lemma 13.14.15.

Next, let $K^\bullet $ be an arbitrary complex of $\mathcal{A}$. Choose a diagram

as in Lemma 13.29.1. Note that the map $\mathop{\mathrm{colim}}\nolimits P_ n^\bullet \to K^\bullet $ is a quasi-isomorphism because colimits over $\mathbf{N}$ in $\mathcal{A}$ are exact and $H^ i(P_ n^\bullet ) = H^ i(K^\bullet )$ for $n > i$. We claim that

(termwise colimits) is $LF(K^\bullet )$, i.e., that $\mathop{\mathrm{colim}}\nolimits P_ n^\bullet $ computes $LF$. To see this, by Lemma 13.14.15, it suffices to prove the following claim. Suppose that

is a quasi-isomorphism of complexes, such that each $P_ n^\bullet $, $Q_ n^\bullet $ is a bounded above complex whose terms are in $\mathcal{P}$ and the maps $P_ n^\bullet \to \tau _{\leq n}P^\bullet $ and $Q_ n^\bullet \to \tau _{\leq n}Q^\bullet $ are quasi-isomorphisms. Claim: $F(\alpha )$ is a quasi-isomorphism.

The problem is that we do not assume that $\alpha $ is given as a colimit of maps between the complexes $P_ n^\bullet $ and $Q_ n^\bullet $. However, for each $n$ we know that the solid arrows in the diagram

are quasi-isomorphisms. Because quasi-isomorphisms form a multiplicative system in $K(\mathcal{A})$ (see Lemma 13.11.2) we can find a quasi-isomorphism $L^\bullet \to P_ n^\bullet $ and map of complexes $L^\bullet \to Q_ n^\bullet $ such that the diagram above commutes up to homotopy. Then $\tau _{\leq n}L^\bullet \to L^\bullet $ is a quasi-isomorphism. Hence (by the first part of the proof) we can find a bounded above complex $R^\bullet $ whose terms are in $\mathcal{P}$ and a quasi-isomorphism $R^\bullet \to L^\bullet $ (as indicated in the diagram). Using the result of the first paragraph of the proof we see that $F(R^\bullet ) \to F(P_ n^\bullet )$ and $F(R^\bullet ) \to F(Q_ n^\bullet )$ are quasi-isomorphisms. Thus we obtain a isomorphisms $H^ i(F(P_ n^\bullet )) \to H^ i(F(Q_ n^\bullet ))$ fitting into the commutative diagram

The exact same argument shows that these maps are also compatible as $n$ varies. Since by (4) and (5) we have

and similarly for $Q^\bullet $ we conclude that $H^ i(\alpha ) : H^ i(F(P^\bullet ) \to H^ i(F(Q^\bullet )$ is an isomorphism and the claim follows. $\square$

Proof. This lemma is dual to Lemma 13.29.1. $\square$