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

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

(Remark 13.12.4). The composition $K_0^\bullet \to K_1^\bullet \to H^0(K_1^\bullet )$ is zero as it is equal to $K_0^\bullet \to H^0(K_0^\bullet ) \to H^0(K_1^\bullet )$ 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. $\square$

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).