In this section we redo the construction of a resolution functor $K^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})$ in case the category $\mathcal{A}$ has functorial injective embeddings. There are two reasons for this: (1) the proof is easier and (2) the construction also works if $\mathcal{A}$ is a “big” abelian category. See Remark 13.24.3 below.
Let $\mathcal{A}$ be an abelian category. As before denote $\mathcal{I}$ the additive full subcategory of $\mathcal{A}$ consisting of injective objects. Consider the category $\text{InjRes}(\mathcal{A})$ of arrows $\alpha : K^\bullet \to I^\bullet $ where $K^\bullet $ is a bounded below complex of $\mathcal{A}$, $I^\bullet $ is a bounded below complex of injectives of $\mathcal{A}$ and $\alpha $ is a quasi-isomorphism. In other words, $\alpha $ is an injective resolution and $K^\bullet $ is bounded below. There is an obvious functor
defined by $(\alpha : K^\bullet \to I^\bullet ) \mapsto K^\bullet $. There is also a functor
defined by $(\alpha : K^\bullet \to I^\bullet ) \mapsto I^\bullet $.
Lemma 13.24.1. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has functorial injective embeddings, see Homology, Definition 12.27.5.
There exists a functor $inj : \text{Comp}^{+}(\mathcal{A}) \to \text{InjRes}(\mathcal{A})$ such that $s \circ inj = \text{id}$.
For any functor $inj : \text{Comp}^{+}(\mathcal{A}) \to \text{InjRes}(\mathcal{A})$ such that $s \circ inj = \text{id}$ we obtain a resolution functor, see Definition 13.23.2.
Proof. Let $A \mapsto (A \to J(A))$ be a functorial injective embedding, see Homology, Definition 12.27.5. We first note that we may assume $J(0) = 0$. Namely, if not then for any object $A$ we have $0 \to A \to 0$ which gives a direct sum decomposition $J(A) = J(0) \oplus \mathop{\mathrm{Ker}}(J(A) \to J(0))$. Note that the functorial morphism $A \to J(A)$ has to map into the second summand. Hence we can replace our functor by $J'(A) = \mathop{\mathrm{Ker}}(J(A) \to J(0))$ if needed.
Let $K^\bullet $ be a bounded below complex of $\mathcal{A}$. Say $K^ p = 0$ if $p < B$. We are going to construct a double complex $I^{\bullet , \bullet }$ of injectives, together with a map $\alpha : K^\bullet \to I^{\bullet , 0}$ such that $\alpha $ induces a quasi-isomorphism of $K^\bullet $ with the associated total complex of $I^{\bullet , \bullet }$. First we set $I^{p, q} = 0$ whenever $q < 0$. Next, we set $I^{p, 0} = J(K^ p)$ and $\alpha ^ p : K^ p \to I^{p, 0}$ the functorial embedding. Since $J$ is a functor we see that $I^{\bullet , 0}$ is a complex and that $\alpha $ is a morphism of complexes. Each $\alpha ^ p$ is injective. And $I^{p, 0} = 0$ for $p < B$ because $J(0) = 0$. Next, we set $I^{p, 1} = J(\mathop{\mathrm{Coker}}(K^ p \to I^{p, 0}))$. Again by functoriality we see that $I^{\bullet , 1}$ is a complex. And again we get that $I^{p, 1} = 0$ for $p < B$. It is also clear that $K^ p$ maps isomorphically onto $\mathop{\mathrm{Ker}}(I^{p, 0} \to I^{p, 1})$. As our third step we take $I^{p, 2} = J(\mathop{\mathrm{Coker}}(I^{p, 0} \to I^{p, 1}))$. And so on and so forth.
At this point we can apply Homology, Lemma 12.25.4 to get that the map
is a quasi-isomorphism. To prove we get a functor $inj$ it rests to show that the construction above is functorial. This verification is omitted.
Suppose we have a functor $inj$ such that $s \circ inj = \text{id}$. For every object $K^\bullet $ of $\text{Comp}^{+}(\mathcal{A})$ we can write
This provides us with a resolution functor as in Definition 13.23.2. $\square$
Remark 13.24.2. Suppose $inj$ is a functor such that $s \circ inj = \text{id}$ as in part (2) of Lemma 13.24.1. Write $inj(K^\bullet ) = (i_{K^\bullet } : K^\bullet \to j(K^\bullet ))$ as in the proof of that lemma. Suppose $\alpha : K^\bullet \to L^\bullet $ is a map of bounded below complexes. Consider the map $inj(\alpha )$ in the category $\text{InjRes}(\mathcal{A})$. It induces a commutative diagram of morphisms of complexes. Hence, looking at the proof of Lemma 13.23.3 we see that the functor $j : K^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})$ is given by the rule Hence we see that $j$ matches $t \circ inj$ in this case, i.e., the diagram is commutative.
\[ \xymatrix{ K^\bullet \ar[rr]^-{\alpha } \ar[d]_{i_ K} & & L^\bullet \ar[d]^{i_ L} \\ j(K)^\bullet \ar[rr]^-{inj(\alpha )} & & j(L)^\bullet } \]
\[ j(\alpha \text{ up to homotopy}) = inj(\alpha )\text{ up to homotopy}\in \mathop{\mathrm{Hom}}\nolimits _{K^{+}(\mathcal{I})}(j(K^\bullet ), j(L^\bullet )) \]
\[ \xymatrix{ \text{Comp}^{+}(\mathcal{A}) \ar[rr]_{t \circ inj} \ar[rd] & & K^{+}(\mathcal{I}) \\ & K^{+}(\mathcal{A}) \ar[ru]_ j } \]
Remark 13.24.3. Let $\textit{Mod}(\mathcal{O}_ X)$ be the category of $\mathcal{O}_ X$-modules on a ringed space $(X, \mathcal{O}_ X)$ (or more generally on a ringed site). We will see later that $\textit{Mod}(\mathcal{O}_ X)$ has enough injectives and in fact functorial injective embeddings, see Injectives, Theorem 19.8.4. Note that the proof of Lemma 13.23.4 does not apply to $\textit{Mod}(\mathcal{O}_ X)$. But the proof of Lemma 13.24.1 does apply to $\textit{Mod}(\mathcal{O}_ X)$. Thus we obtain which is a resolution functor where $\mathcal{I}$ is the additive category of injective $\mathcal{O}_ X$-modules. This argument also works in the following cases:
The category $\text{Mod}_ R$ of $R$-modules over a ring $R$.
The category $\textit{PMod}(\mathcal{O})$ of presheaves of $\mathcal{O}$-modules on a site endowed with a presheaf of rings.
The category $\textit{Mod}(\mathcal{O})$ of sheaves of $\mathcal{O}$-modules on a ringed site.
Add more here as needed.
\[ j : K^{+}(\textit{Mod}(\mathcal{O}_ X)) \longrightarrow K^{+}(\mathcal{I}) \]
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