Lemma 13.12.5 (08Q2)—The Stacks project

Lemma 13.12.5. Let $\mathcal{A}$ be an abelian category. Let

$K_0^\bullet \to K_1^\bullet \to \ldots \to K_ n^\bullet$

be maps of complexes such that

$H^ i(K_0^\bullet ) = 0$ for $i > 0$ ,
$H^{-j}(K_ j^\bullet ) \to H^{-j}(K_{j + 1}^\bullet )$ is zero.

Then the composition $K_0^\bullet \to K_ n^\bullet$ factors through $\tau _{\leq -n}K_ n^\bullet \to K_ n^\bullet$ in $D(\mathcal{A})$ . Dually, given maps of complexes

$K_ n^\bullet \to K_{n - 1}^\bullet \to \ldots \to K_0^\bullet$

such that

$H^ i(K_0^\bullet ) = 0$ for $i < 0$ ,
$H^ j(K_{j + 1}^\bullet ) \to H^ j(K_ j^\bullet )$ is zero,

then the composition $K_ n^\bullet \to K_0^\bullet$ factors through $K_ n^\bullet \to \tau _{\geq n}K_ n^\bullet$ in $D(\mathcal{A})$ .

Proof. The case $n = 1$ . Since $\tau _{\leq 0}K_0^\bullet = K_0^\bullet$ in $D(\mathcal{A})$ we can replace $K_0^\bullet$ by $\tau _{\leq 0}K_0^\bullet$ and $K_1^\bullet$ by $\tau _{\leq 0}K_1^\bullet$ . Consider the distinguished triangle

$\tau _{\leq -1}K_1^\bullet \to K_1^\bullet \to H^0(K_1^\bullet )[0] \to (\tau _{\leq -1}K_1^\bullet )[1]$

(Remark 13.12.4). The composition $K_0^\bullet \to K_1^\bullet \to H^0(K_1^\bullet )[0]$ is zero as it is equal to $K_0^\bullet \to H^0(K_0^\bullet )[0] \to H^0(K_1^\bullet )[0]$ which is zero by assumption. The fact that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K_0^\bullet , -)$ is a homological functor (Lemma 13.4.2), allows us to find the desired factorization. For $n = 2$ we get a factorization $K_0^\bullet \to \tau _{\leq -1}K_1^\bullet$ by the case $n = 1$ and we can apply the case $n = 1$ to the map of complexes $\tau _{\leq -1}K_1^\bullet \to \tau _{\leq -1}K_2^\bullet$ to get a factorization $\tau _{\leq -1}K_1^\bullet \to \tau _{\leq -2}K_2^\bullet$ . The general case is proved in exactly the same manner. $\square$

Comments (2)

Comment #7404 by WhatJiaranEatsTonight on June 02, 2022 at 02:28

Is the condition $H^i(K^\bullet_0)=0$ for $i>0$ necessary?

I think if $a:K^\bullet\to L^\bullet$ induces zero maps on cohomologies. Then $a=0$ in $D(\mathcal A)$ by considering the distinguished triangle $K^\bullet \to L^\bullet \to C(a)^\bullet$ . Hence by the distinguished triangle $\tau_{\leq -1} K^\bullet_1 \to K_1^\bullet \to\tau_{\geq 0} K^\bullet_1$ and the fact that $Hom_{D(A)}(K_0^\bullet,-)$ is a homological functor, we know $K_0^\bullet\to K_1^\bullet$ factorizes through $\tau_{\leq -1} K^\bullet_1 \to K_1^\bullet$ .

Comment #7405 by Johan on June 02, 2022 at 15:03

It is not true that if $a : K \to L$ is zero on cohomology, then $a = 0$ . A simple counterexample is to consider any nonzero map $\mathbf{Z}/2\mathbf{Z} \to \mathbf{Z}[1]$ in $D(\mathbf{Z})$ . Such nonzero maps exist as the correspond $1$ -to- $1$ to nonzero extensions of $\mathbf{Z}/2\mathbf{Z}$ by $\mathbf{Z}$ .

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