## 13.23 Resolution functors

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

\[ K^{+}(\mathcal{I}) \longrightarrow D^{+}(\mathcal{A}) \]

is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulated categories.

**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 functor*^{1} 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

\[ \xymatrix{ K^{+}(\mathcal{A}) \ar[rd] \ar[rr]_ j & & K^{+}(\mathcal{I}) \ar[ld] \\ & D^{+}(\mathcal{A}) } \]

where $\mathcal{I}$ is the full additive subcategory of $\mathcal{A}$ consisting of injective objects.

**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

\[ \xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_{i_{K^\bullet }} & L^\bullet \ar[d]^{i_{L^\bullet }} \\ j(K^\bullet ) \ar[r]^{j(\alpha )} & j(L^\bullet ) } \]

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

\[ \xymatrix{ K^\bullet [1] \ar[d]^{i_{K^\bullet [1]}} \ar@{=}[rr] & & K^\bullet [1] \ar[d]^{i_{K^\bullet }[1]} \\ j(K^\bullet [1]) \ar@{..>}[rr]^{\xi _{K^\bullet }} & & j(K^\bullet )[1] } \]

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

\[ \xymatrix{ K^\bullet \ar[r]_ f \ar[d] & L^\bullet \ar[r]_ g \ar[d] & M^\bullet \ar[rr]_ h \ar[d] & & K^\bullet [1] \ar[d] \\ j(K^\bullet ) \ar[r]^{j(f)} & j(L^\bullet ) \ar[r]^{j(g)} & j(M^\bullet ) \ar[rr]^{\xi _{K^\bullet } \circ j(h)} & & j(K^\bullet )[1] } \]

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 13.5.6. This is true by the remarks following Definition 13.23.2.
$\square$

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