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. $\square$