The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 12.24.5.

  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 12.24.5. 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 12.22.7 to get that the map

\[ \alpha : K^\bullet \to sI^\bullet \]

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.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0140. Beware of the difference between the letter 'O' and the digit '0'.