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The Stacks project

13.24 Functorial injective embeddings and resolution functors

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

s : \text{InjRes}(\mathcal{A}) \longrightarrow \text{Comp}^{+}(\mathcal{A})

defined by (\alpha : K^\bullet \to I^\bullet ) \mapsto K^\bullet . There is also a functor

t : \text{InjRes}(\mathcal{A}) \longrightarrow K^{+}(\mathcal{I})

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.

  1. There exists a functor inj : \text{Comp}^{+}(\mathcal{A}) \to \text{InjRes}(\mathcal{A}) such that s \circ inj = \text{id}.

  2. 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

\alpha : K^\bullet \longrightarrow \text{Tot}(I^{\bullet , \bullet })

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

inj(K^\bullet ) = (i_{K^\bullet } : K^\bullet \to j(K^\bullet ))

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

\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 }

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

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 ))

Hence we see that j matches t \circ inj in this case, i.e., the diagram

\xymatrix{ \text{Comp}^{+}(\mathcal{A}) \ar[rr]_{t \circ inj} \ar[rd] & & K^{+}(\mathcal{I}) \\ & K^{+}(\mathcal{A}) \ar[ru]_ j }

is commutative.

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

j : K^{+}(\textit{Mod}(\mathcal{O}_ X)) \longrightarrow K^{+}(\mathcal{I})

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:

  1. The category \text{Mod}_ R of R-modules over a ring R.

  2. The category \textit{PMod}(\mathcal{O}) of presheaves of \mathcal{O}-modules on a site endowed with a presheaf of rings.

  3. The category \textit{Mod}(\mathcal{O}) of sheaves of \mathcal{O}-modules on a ringed site.

  4. Add more here as needed.


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