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

  1. $H^ i(K_0^\bullet ) = 0$ for $i > 0$,

  2. $H^{-j}(K_ j^\bullet ) \to H^{-j}(K_{j + 1}^\bullet )$ 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

  1. $H^ i(K_0^\bullet ) = 0$ for $i < 0$,

  2. $H^ j(K_{j + 1}^\bullet ) \to H^ j(K_ j^\bullet )$ 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

Is the condition for necessary?

I think if induces zero maps on cohomologies. Then in by considering the distinguished triangle . Hence by the distinguished triangle and the fact that is a homological functor, we know factorizes through .

Comment #7405 by on

It is not true that if is zero on cohomology, then . A simple counterexample is to consider any nonzero map in . Such nonzero maps exist as the correspond -to- to nonzero extensions of by .

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