Let $\mathcal{A}$ be an abelian category with enough injectives. Denote $\mathcal{I}$ the full additive subcategory of $\mathcal{A}$ whose objects are the injective objects of $\mathcal{A}$. It turns out that $K^{+}(\mathcal{I})$ and $D^{+}(\mathcal{A})$ are equivalent in this case (see Proposition 13.23.1). For many purposes it therefore makes sense to think of $D^{+}(\mathcal{A})$ as the (easier to grok) category $K^{+}(\mathcal{I})$ in this case.
Proposition 13.23.1. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Denote $\mathcal{I} \subset \mathcal{A}$ the strictly full additive subcategory whose objects are the injective objects of $\mathcal{A}$. The functor is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulated categories.
\[ K^{+}(\mathcal{I}) \longrightarrow D^{+}(\mathcal{A}) \]
Proof. It is clear that the functor is exact. It is essentially surjective by Lemma 13.18.3. Fully faithfulness is a consequence of Lemma 13.18.8. $\square$
Proposition 13.23.1 implies that we can find resolution functors. It turns out that we can prove resolution functors exist even in some cases where the abelian category $\mathcal{A}$ is a “big” category, i.e., has a class of objects.
Definition 13.23.2. Let $\mathcal{A}$ be an abelian category with enough injectives. A resolution functor1 for $\mathcal{A}$ is given by the following data:
for all $K^\bullet \in \mathop{\mathrm{Ob}}\nolimits (K^{+}(\mathcal{A}))$ a bounded below complex of injectives $j(K^\bullet )$, and
for all $K^\bullet \in \mathop{\mathrm{Ob}}\nolimits (K^{+}(\mathcal{A}))$ a quasi-isomorphism $i_{K^\bullet } : K^\bullet \to j(K^\bullet )$.
Lemma 13.23.3. Let $\mathcal{A}$ be an abelian category with enough injectives. Given a resolution functor $(j, i)$ there is a unique way to turn $j$ into a functor and $i$ into a $2$-isomorphism producing a $2$-commutative diagram where $\mathcal{I}$ is the full additive subcategory of $\mathcal{A}$ consisting of injective objects.
\[ \xymatrix{ K^{+}(\mathcal{A}) \ar[rd] \ar[rr]_ j & & K^{+}(\mathcal{I}) \ar[ld] \\ & D^{+}(\mathcal{A}) } \]
Proof. For every morphism $\alpha : K^\bullet \to L^\bullet $ of $K^{+}(\mathcal{A})$ there is a unique morphism $j(\alpha ) : j(K^\bullet ) \to j(L^\bullet )$ in $K^{+}(\mathcal{I})$ such that
is commutative in $K^{+}(\mathcal{A})$. To see this either use Lemmas 13.18.6 and 13.18.7 or the equivalent Lemma 13.18.8. The uniqueness implies that $j$ is a functor, and the commutativity of the diagram implies that $i$ gives a $2$-morphism which witnesses the $2$-commutativity of the diagram of categories in the statement of the lemma. $\square$
Lemma 13.23.4. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Then a resolution functor $j$ exists and is unique up to unique isomorphism of functors.
Proof. Consider the set of all objects $K^\bullet $ of $K^{+}(\mathcal{A})$. (Recall that by our conventions any category has a set of objects unless mentioned otherwise.) By Lemma 13.18.3 every object has an injective resolution. By the axiom of choice we can choose for each $K^\bullet $ an injective resolution $i_{K^\bullet } : K^\bullet \to j(K^\bullet )$. $\square$
Lemma 13.23.5. Let $\mathcal{A}$ be an abelian category with enough injectives. Any resolution functor $j : K^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})$ is exact.
Proof. Denote $i_{K^\bullet } : K^\bullet \to j(K^\bullet )$ the canonical maps of Definition 13.23.2. First we discuss the existence of the functorial isomorphism $j(K^\bullet [1]) \to j(K^\bullet )[1]$. Consider the diagram
By Lemmas 13.18.6 and 13.18.7 there exists a unique dotted arrow $\xi _{K^\bullet }$ in $K^{+}(\mathcal{I})$ making the diagram commute in $K^{+}(\mathcal{A})$. We omit the verification that this gives a functorial isomorphism. (Hint: use Lemma 13.18.7 again.)
Let $(K^\bullet , L^\bullet , M^\bullet , f, g, h)$ be a distinguished triangle of $K^{+}(\mathcal{A})$. We have to show that $(j(K^\bullet ), j(L^\bullet ), j(M^\bullet ), j(f), j(g), \xi _{K^\bullet } \circ j(h))$ is a distinguished triangle of $K^{+}(\mathcal{I})$. Note that we have a commutative diagram
in $K^{+}(\mathcal{A})$ whose vertical arrows are the quasi-isomorphisms $i_ K, i_ L, i_ M$. Hence we see that the image of $(j(K^\bullet ), j(L^\bullet ), j(M^\bullet ), j(f), j(g), \xi _{K^\bullet } \circ j(h))$ in $D^{+}(\mathcal{A})$ is isomorphic to a distinguished triangle and hence a distinguished triangle by TR1. Thus we see from Lemma 13.4.18 that $(j(K^\bullet ), j(L^\bullet ), j(M^\bullet ), j(f), j(g), \xi _{K^\bullet } \circ j(h))$ is a distinguished triangle in $K^{+}(\mathcal{I})$. $\square$
Lemma 13.23.6. Let $\mathcal{A}$ be an abelian category which has enough injectives. Let $j$ be a resolution functor. Write $Q : K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ for the natural functor. Then $j = j' \circ Q$ for a unique functor $j' : D^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})$ which is quasi-inverse to the canonical functor $K^{+}(\mathcal{I}) \to D^{+}(\mathcal{A})$.
Proof. By Lemma 13.11.6 $Q$ is a localization functor. To prove the existence of $j'$ it suffices to show that any element of $\text{Qis}^{+}(\mathcal{A})$ is mapped to an isomorphism under the functor $j$, see Lemma 13.5.7. This is true by the remarks following Definition 13.23.2. $\square$
Remark 13.23.7. Suppose that $\mathcal{A}$ is a “big” abelian category with enough injectives such as the category of abelian groups. In this case we have to be slightly more careful in constructing our resolution functor since we cannot use the axiom of choice with a quantifier ranging over a class. But note that the proof of the lemma does show that any two localization functors are canonically isomorphic. Namely, given quasi-isomorphisms $i : K^\bullet \to I^\bullet $ and $i' : K^\bullet \to J^\bullet $ of a bounded below complex $K^\bullet $ into bounded below complexes of injectives there exists a unique(!) morphism $a : I^\bullet \to J^\bullet $ in $K^{+}(\mathcal{I})$ such that $i' = i \circ a$ as morphisms in $K^{+}(\mathcal{I})$. Hence the only issue is existence, and we will see how to deal with this in the next section.
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Comment #9854 by Elías Guisado on