Lemma 13.29.1. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset. Assume $\mathcal{P}$ contains $0$, is closed under (finite) direct sums, and 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 and termwise surjective,

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. 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 $ 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 $ into a distinguished triangle $P_ n^\bullet \to P_{n + 1}^\bullet \to Q^\bullet \to P_ n^\bullet $ 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

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

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

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

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

$g : P^\bullet = (\ldots \to 0 \to P \xrightarrow {1} P \to 0 \to \ldots ) \longrightarrow \tau _{\leq n + 1}K^\bullet$

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. $\square$

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)

Comment #4827 by awllower on

Why do the terms of $P_{n+1}^\bullet$ belong to $\mathcal P$?

It is the cone of a map between elements of $\mathcal P$ (shifted by $-1$), so each term is a direct sum of elements of $\mathcal P$, but is it true that direct sums of elements of $\mathcal P$ also belong to $\mathcal P$?

Comment #4828 by on

OMG! Thanks very much. Fixed here. I will update the website in 10 minutes...

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