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.
$\square$
Comments (0)