Let $(A, \text{d})$ be a differential graded algebra. Consider the functor $\text{Mod}_{(A, \text{d})} \to K(\text{Mod}_{(A, \text{d})})$. This functor is not a $\delta $-functor in general. However, it turns out that the functor $\text{Mod}_{(A, \text{d})} \to D(A, \text{d})$ is a $\delta $-functor. In order to see this we have to define the morphisms $\delta $ associated to a short exact sequence
in the abelian category $\text{Mod}_{(A, \text{d})}$. Consider the cone $C(a)$ of the morphism $a$. We have $C(a) = L \oplus K$ and we define $q : C(a) \to M$ via the projection to $L$ followed by $b$. Hence a homomorphism of differential graded $A$-modules
It is clear that $q \circ i = b$ where $i$ is as in Definition 22.6.1. Note that, as $a$ is injective, the kernel of $q$ is identified with the cone of $\text{id}_ K$ which is acyclic. Hence we see that $q$ is a quasi-isomorphism. According to Lemma 22.9.4 the triangle
is a distinguished triangle in $K(\text{Mod}_{(A, \text{d})})$. As the localization functor $K(\text{Mod}_{(A, \text{d})}) \to D(A, \text{d})$ is exact we see that $(K, L, C(a), a, i, -p)$ is a distinguished triangle in $D(A, \text{d})$. Since $q$ is a quasi-isomorphism we see that $q$ is an isomorphism in $D(A, \text{d})$. Hence we deduce that
is a distinguished triangle of $D(A, \text{d})$. This suggests the following lemma.
Lemma 22.23.1. Let $(A, \text{d})$ be a differential graded algebra. The functor $\text{Mod}_{(A, \text{d})} \to D(A, \text{d})$ defined has the natural structure of a $\delta $-functor, with
with $p$ and $q$ as explained above.
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 functoriality of this construction, see Derived Categories, Definition 13.3.6. This follows from Lemma 22.6.2 with a bit of work. Compare with Derived Categories, Lemma 13.12.1. $\square$
Lemma 22.23.2. Let $(A, \text{d})$ be a differential graded algebra. Let $M_ n$ be a system of differential graded modules. Then the derived colimit $\text{hocolim} M_ n$ in $D(A, \text{d})$ is represented by the differential graded module $\mathop{\mathrm{colim}}\nolimits M_ n$.
Proof. Set $M = \mathop{\mathrm{colim}}\nolimits M_ n$. We have an exact sequence of differential graded modules
by Derived Categories, Lemma 13.33.6 (applied the underlying complexes of abelian groups). The direct sums are direct sums in $D(\mathcal{A})$ by Lemma 22.22.4. Thus the result follows from the definition of derived colimits in Derived Categories, Definition 13.33.1 and the fact that a short exact sequence of complexes gives a distinguished triangle (Lemma 22.23.1). $\square$
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