The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 13.12.1. Let $\mathcal{A}$ be an abelian category. The functor $\text{Comp}(\mathcal{A}) \to D(\mathcal{A})$ defined has the natural structure of a $\delta $-functor, with

\[ \delta _{A^\bullet \to B^\bullet \to C^\bullet } = - p \circ q^{-1} \]

with $p$ and $q$ as explained above. The same construction turns the functors $\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$, $\text{Comp}^{-}(\mathcal{A}) \to D^{-}(\mathcal{A})$, and $\text{Comp}^ b(\mathcal{A}) \to D^ b(\mathcal{A})$ into $\delta $-functors.

Proof. We have already seen that this choice leads to a distinguished triangle whenever given a short exact sequence of complexes. We have to show that given a commutative diagram

\[ \xymatrix{ 0 \ar[r] & A^\bullet \ar[r]_ a \ar[d]_ f & B^\bullet \ar[r]_ b \ar[d]_ g & C^\bullet \ar[r] \ar[d]_ h & 0 \\ 0 \ar[r] & (A')^\bullet \ar[r]^{a'} & (B')^\bullet \ar[r]^{b'} & (C')^\bullet \ar[r] & 0 } \]

we get the desired commutative diagram of Definition 13.3.6 (2). By Lemma 13.9.2 the pair $(f, g)$ induces a canonical morphism $c : C(a)^\bullet \to C(a')^\bullet $. It is a simple computation to show that $q' \circ c = h \circ q$ and $f[1] \circ p = p' \circ c$. From this the result follows directly. $\square$


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