Lemma 13.28.1. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset. Assume that every object of $\mathcal{A}$ is a quotient of an element of $\mathcal{P}$. Let $K^\bullet$ be a complex. There exists a commutative diagram

$\xymatrix{ P_1^\bullet \ar[d] \ar[r] & P_2^\bullet \ar[d] \ar[r] & \ldots \\ \tau _{\leq 1}K^\bullet \ar[r] & \tau _{\leq 2}K^\bullet \ar[r] & \ldots }$

in the category of complexes such that

1. the vertical arrows are quasi-isomorphisms,

2. $P_ n^\bullet$ is a bounded above complex with terms in $\mathcal{P}$,

3. the arrows $P_ n^\bullet \to P_{n + 1}^\bullet$ are termwise split injections and each cokernel $P^ i_{n + 1}/P^ i_ n$ is an element of $\mathcal{P}$.

Proof. By Lemma 13.16.5 any bounded above complex has a resolution by a bounded above complex whose terms are in $\mathcal{P}$. Thus we obtain the first complex $P_1^\bullet$. 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 $P_{n + 1}^\bullet \to \tau _{\leq n + 1}K^\bullet$. Consider the cone $C_1^\bullet$ of the composition $P_ n^\bullet \to \tau _{\leq n}K^\bullet \to \tau _{\leq n + 1}K^\bullet$. This fits into the distinguished triangle

$P_ n^\bullet \to \tau _{\leq n + 1}K^\bullet \to C_1^\bullet \to P_ n^\bullet [1]$

Note that $C_1^\bullet$ is bounded above, hence we can choose a quasi-isomorphism $Q^\bullet \to C_1^\bullet$ where $Q^\bullet$ is a bounded above complex whose terms are elements of $\mathcal{P}$. Take the cone $C_2^\bullet$ of the map of complexes $Q^\bullet \to P_ n^\bullet [1]$ to get the distinguished triangle

$Q^\bullet \to P_ n^\bullet [1] \to C_2^\bullet \to Q^\bullet [1]$

By the axioms of triangulated categories we obtain a map of distinguished triangles

$\xymatrix{ P_ n^\bullet \ar[r] \ar[d] & C_2^\bullet [-1] \ar[r] \ar[d] & Q^\bullet \ar[r] \ar[d] & P_ n^\bullet [1] \ar[d] \\ P_ n^\bullet \ar[r] & \tau _{\leq n + 1}K^\bullet \ar[r] & C_1^\bullet \ar[r] & P_ n^\bullet [1] }$

in the triangulated category $K(\mathcal{A})$. Set $P_{n + 1}^\bullet = C_2^\bullet [-1]$. Note that (3) holds by construction. 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 $Q^\bullet \to C_1^\bullet$ are. $\square$

Comment #2466 by anonymous on

Is there a reason item (2) does not state that all $P_n$ are bounded above with terms in $\mathcal{P}$?

Same in Tag 070F.

Comment #2475 by Tanya Kaushal Srivastava on

A small typo in the proof, we want to construct the map from $P^{\bullet}_{n+1} \rightarrow \tau_{\leq n+1}K^{\bullet}$ as we already have the map $P^{\bullet}_{n} \rightarrow \tau_{\leq n+1}K^{\bullet}$ just by composition.

Comment #2502 by on

@#2475: Sorry, I do not understand. @#2466: Thanks, fixed here.

Comment #2512 by Tanya Kaushal Srivastava on

Rephrasing: Why do we want to construct a map from $P_n^{\bullet} \rightarrow \tau_{\leq n +1}K^{\bullet}$ but not just $P_{n+1}^{\bullet} \rightarrow \tau_{\leq n +1}K^{\bullet}$? (Line 4 of proof)

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).