The Stacks project

Proof. This is true in any Grothendieck abelian category, see Injectives, Lemma 19.11.6. By Lemma 24.11.1 the category $\textit{Mod}(\mathcal{A})$ is a Grothendieck abelian category. $\square$

Proof. This is true because products of injectives are injectives, see Homology, Lemma 12.27.3, and because products in $\textit{Mod}(\mathcal{A}, \text{d})$ are compatible with products in $\textit{Mod}(\mathcal{A})$ via the forgetful functor. $\square$

Proof. Let $T$ and $\mathcal{N}_ t \to \mathcal{N}'_ t$ be as in Lemma 24.25.1. Denote $F : \textit{Mod}(\mathcal{A}, \text{d}) \to \textit{Mod}(\mathcal{A})$ the forgetful functor. Let $G$ be the left adjoint functor to $F$ as in Lemma 24.24.1. Set

This is an injective map of acyclic differential graded $\mathcal{A}$-modules by Lemma 24.24.2. Since $G$ is the left adjoint to $F$ we see that there exists a dotted arrow in the diagram

if and only if there exists a dotted arrow in the diagram

Hence the result follows from the choice of our collection of arrows $\mathcal{N}_ t \to \mathcal{N}_ t'$. $\square$

Proof. Before we start recall that our conventions guarantee the site $\mathcal{C}$ has a set of objects and morphisms and a set $\text{Cov}(\mathcal{C})$ of coverings. If $\mathcal{F}$ is a differential graded $\mathcal{A}$-module, let us define $|\mathcal{F}|$ to be the sum of the cardinality of

as $U$ ranges over the objects of $\mathcal{C}$ and $n \in \mathbf{Z}$. Choose an infinite cardinal $\kappa $ bigger than the cardinals $|\mathop{\mathrm{Ob}}\nolimits (\mathcal{C})|$, $|\text{Arrows}(\mathcal{C})|$, $|\text{Cov}(\mathcal{C})|$, $\sup |I|$ for $\{ U_ i \to U\} _{i \in I} \in \text{Cov}(\mathcal{C})$, and $|\mathcal{A}|$.

Let $\mathcal{F} \subset \mathcal{M}$ be an inclusion of differential graded $\mathcal{A}$-modules. Suppose given a set $K$ and for each $k \in K$ a triple $(U_ k, n_ k, x_ k)$ consisting of an object $U_ k$ of $\mathcal{C}$, integer $n_ k$, and a section $x_ k \in \mathcal{M}^{n_ k}(U_ k)$. Then we can consider the smallest differential graded $\mathcal{A}$-submodule $\mathcal{F}' \subset \mathcal{M}$ containing $\mathcal{F}$ and the sections $x_ k$ for $k \in K$. We can describe

as the set of elements $x \in \mathcal{M}^ n(U)$ such that there exists $\{ f_ i : U_ i \to U\} _{i \in I} \in \text{Cov}(\mathcal{C})$ such that for each $i \in I$ there is a finite set $T_ i$ and morphisms $g_ t : U_ i \to U_{k_ t}$

for some section $y_ i \in \mathcal{F}^ n(U)$ and sections $a_{it} \in \mathcal{A}^{n - n_{k_ t}}(U_ i)$ and $b_{it} \in \mathcal{A}^{n - n_{k_ t} - 1}(U_ i)$. (Details omitted; hints: these sections are certainly in $\mathcal{F}'$ and you show conversely that this rule defines a differential graded $\mathcal{A}$-submodule.) It follows from this description that $|\mathcal{F}'| \leq \max (|\mathcal{F}|, |K|, \kappa )$.

Let $\mathcal{M}$ be a nonzero acyclic differential graded $\mathcal{A}$-module. Then we can find an integer $n$ and a nonzero section $x$ of $\mathcal{M}^ n$ over some object $U$ of $\mathcal{C}$. Let

be the smallest differential graded $\mathcal{A}$-submodule containing $x$. By the previous paragraph we have $|\mathcal{F}_0| \leq \kappa $. By induction, given $\mathcal{F}_0, \ldots , \mathcal{F}_ n$ define $\mathcal{F}_{n + 1}$ as follows. Consider the set

of triples where $U$ is an object of $\mathcal{C}$, $n \in \mathbf{Z}$, and $x \in \mathcal{F}_ n(U)$ with $\text{d}(x) = 0$. Since $\mathcal{M}$ is acyclic for each triple $l = (U_ l, n_ l, x_ l) \in L$ we can choose $\{ (U_{l, i} \to U_ l\} _{i \in I_ l} \in \text{Cov}(\mathcal{C})$ and $x_{l, i} \in \mathcal{M}^{n_ l - 1}(U_{l, i})$ such that $\text{d}(x_{l, i}) = x|_{U_{l, i}}$. Then we set

and we let $\mathcal{F}_{n + 1}$ be the smallest differential graded $\mathcal{A}$-submodule of $\mathcal{M}$ containing $\mathcal{F}_ n$ and the sections $x_{l, i}$. Since $|K| \leq \max (\kappa , |\mathcal{F}_ n|)$ we conclude that $|\mathcal{F}_{n + 1}| \leq \kappa $ by induction.

By construction the inclusion $\mathcal{F}_ n \to \mathcal{F}_{n + 1}$ induces the zero map on cohomology sheaves. Hence we see that $\mathcal{F} = \bigcup \mathcal{F}_ n$ is a nonzero acyclic submodule with $|\mathcal{F}| \leq \kappa $. Since there is only a set of isomorphism classes of differential graded $\mathcal{A}$-modules $\mathcal{F}$ with $|\mathcal{F}|$ bounded, we conclude. $\square$

Proof. Let $\mathcal{K}$ be an acyclic differential graded $\mathcal{A}$-module. Then we have

because taking products in $\textit{Mod}(\mathcal{A}, \text{d})$ commutes with the forgetful functor to graded $\mathcal{A}$-modules. Since taking products is an exact functor on the category of abelian groups we conclude. $\square$

Proof. Since $\mathcal{M}$ is acyclic and $\mathcal{I}$ is K-injective, there exists a graded $\mathcal{A}$-module map $h : \mathcal{M} \to \mathcal{I}$ of degree $-1$ such that $a = \text{d}(h)$. Since $\mathcal{I}$ is graded injective and $b$ is injective, there exists a graded $\mathcal{A}$-module map $h' : \mathcal{M}' \to \mathcal{I}$ of degree $-1$ such that $h = h' \circ b$. Then we can take $a' = \text{d}(h')$ as the dotted arrow. $\square$

Proof. After replacing $\mathcal{M}'$ by the direct sum of $\mathcal{M}'$ and the cone on the identity on $\mathcal{M}$ (which is acyclic) we may assume $b$ is also injective. Then the cokernel $\mathcal{Q}$ of $b$ is acyclic. Thus we see that

as $\mathcal{I}$ is K-injective. As $\mathcal{I}$ is graded injective by Remark 24.25.3 we see that

is bijective and the proof is complete. $\square$

Proof. Let $T$ and $\mathcal{M}_ t \to \mathcal{M}'_ t$ be as in Lemma 24.25.5. Let $S$ and $\mathcal{M}_ s$ be as in Lemma 24.25.6. Choose an injective map $\mathcal{M}_ s \to \mathcal{M}'_ s$ of acyclic differential graded $\mathcal{A}$-modules which is homotopic to zero. This is possible because we may take $\mathcal{M}'_ s$ to be the cone on the identity; in that case it is even true that the identity on $\mathcal{M}'_ s$ is homotopic to zero, see Differential Graded Algebra, Lemma 22.27.4 which applies by the discussion in Section 24.22. We claim that $R = T \coprod S$ with the given maps works.

The implication (1) $\Rightarrow $ (2) holds by Lemma 24.25.9.

Assume (2). First, by Lemma 24.25.5 we see that $\mathcal{I}$ is graded injective. Next, let $\mathcal{M}$ be an acyclic differential graded $\mathcal{A}$-module. We have to show that

The proof will be exactly the same as the proof of Injectives, Lemma 19.12.3.

We are going to construct by induction on the ordinal $\alpha $ an acyclic differential graded submodule $\mathcal{K}_\alpha \subset \mathcal{M}$ as follows. For $\alpha = 0$ we set $\mathcal{K}_0 = 0$. For $\alpha > 0$ we proceed as follows:

1. If $\alpha = \beta + 1$ and $\mathcal{K}_\beta = \mathcal{M}$ then we choose $\mathcal{K}_\alpha = \mathcal{K}_\beta $.

2. If $\alpha = \beta + 1$ and $\mathcal{K}_\beta \not= \mathcal{M}$ then $\mathcal{M}/\mathcal{K}_\beta $ is a nonzero acyclic differential graded $\mathcal{A}$-module. We choose a differential graded $\mathcal{A}$ submodule $\mathcal{N}_\alpha \subset \mathcal{M}/\mathcal{K}_\beta $ isomorphic to $\mathcal{M}_ s$ for some $s \in S$, see Lemma 24.25.6. Finally, we let $\mathcal{K}_\alpha \subset \mathcal{M}$ be the inverse image of $\mathcal{N}_\alpha $.

3. If $\alpha $ is a limit ordinal we set $\mathcal{K}_\beta = \mathop{\mathrm{colim}}\nolimits \mathcal{K}_\alpha $.

It is clear that $\mathcal{M} = \mathcal{K}_\alpha $ for a suitably large ordinal $\alpha $. We will prove that

is zero by transfinite induction on $\alpha $. It holds for $\alpha = 0$ since $\mathcal{K}_0$ is zero. Suppose it holds for $\beta $ and $\alpha = \beta + 1$. In case (1) of the list above the result is clear. In case (2) there is a short exact sequence

By Remark 24.25.3 and since we've seen that $\mathcal{I}$ is graded injective, we obtain an exact sequence

By induction the term on the left is zero. By assumption (2) the term on the right is zero: any map $\mathcal{M}_ s \to \mathcal{I}$ factors through $\mathcal{M}'_ s$ and hence is homotopic to zero. Thus the middle group is zero too. Finally, suppose that $\alpha $ is a limit ordinal. Because we also have $\mathcal{K}_\alpha = \mathop{\mathrm{colim}}\nolimits \mathcal{K}_\alpha $ as graded $\mathcal{A}$-modules we see that

as complexes of abelian groups. The cohomology groups of these complexes compute morphisms in $K(\textit{Mod}(\mathcal{A}, \text{d}))$ between shifts. The transition maps in the system of complexes are surjective by Remark 24.25.3 because $\mathcal{I}$ is graded injective. Moreover, for a limit ordinal $\beta \leq \alpha $ we have equality of limit and value. Thus we may apply Homology, Lemma 12.31.8 to conclude. $\square$

Proof. We define $M(\mathcal{M})$ as the pushout in the following diagram

where the direct sum is over all pairs $(r, \varphi )$ with $r \in R$ and $\varphi \in \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}(\mathcal{A}, \text{d})}(\mathcal{M}_ r, \mathcal{M})$. Since the pushout of an injective map is injective, we see that $\mathcal{M} \to M(\mathcal{M})$ is injective. Since the cokernel of the left vertical arrow is acyclic, we see that the (isomorphic) cokernel of $\mathcal{M} \to M(\mathcal{M})$ is acyclic, hence $\mathcal{M} \to M(\mathcal{M})$ is a quasi-isomorphism. Property (2) holds by construction. We omit the verification that this procedure can be turned into a functor. $\square$

Proof. Let $R$ and $\mathcal{M}_ r \to \mathcal{M}'_ r$ be a set of morphisms of $\textit{Mod}(\mathcal{A}, \text{d})$ found in Lemma 24.25.11. Let $M$ with transformation $\text{id} \to M$ be as constructed in Lemma 24.25.12 using $R$ and $\mathcal{M}_ r \to \mathcal{M}'_ r$. Using transfinite recursion we define a sequence of functors $M_\alpha $ and natural transformations $M_\beta \to M_\alpha $ for $\alpha < \beta $ by setting

1. $M_0 = \text{id}$,

2. $M_{\alpha + 1} = M \circ M_\alpha $ with natural transformation $M_\beta \to M_{\alpha + 1}$ for $\beta < \alpha + 1$ coming from the already constructed $M_\beta \to M_\alpha $ and the maps $M_\alpha \to M \circ M_\alpha $ coming from $\text{id} \to M$, and

3. $M_\alpha = \mathop{\mathrm{colim}}\nolimits _{\beta < \alpha } M_\beta $ if $\alpha $ is a limit ordinal with the coprojections as transformations $M_\beta \to M_\alpha $ for $\alpha < \beta $.

Observe that for every differential graded $\mathcal{A}$-module the maps $\mathcal{M} \to M_\beta (\mathcal{M}) \to M_\alpha (\mathcal{M})$ are injective quasi-isomorphisms (as filtered colimits are exact).

Recall that $\textit{Mod}(\mathcal{A}, \text{d})$ is a Grothendieck abelian category. Thus by Injectives, Proposition 19.11.5 (applied to the direct sum of $\mathcal{M}_ r$ for all $r \in R$) there is a limit ordinal $\alpha $ such that $\mathcal{M}_ r$ is $\alpha $-small with respect to injections for every $r \in R$. We claim that $\mathcal{M} \to M_\alpha (\mathcal{M})$ is the desired functorial embedding of $\mathcal{M}$ into a graded injective K-injective module.

Namely, any map $\mathcal{M}_ r \to M_\alpha (\mathcal{M})$ factors through $M_\beta (\mathcal{M})$ for some $\beta < \alpha $. However, by the construction of $M$ we see that this means that $\mathcal{M}_ r \to M_{\beta + 1}(\mathcal{M}) = M(M_\beta (\mathcal{M}))$ factors through $\mathcal{M}'_ r$. Since $M_\beta (\mathcal{M}) \subset M_{\beta + 1}(\mathcal{M}) \subset M_\alpha (\mathcal{M})$ we get the desired factorizaton into $M_\alpha (\mathcal{M})$. We conclude by our choice of $R$ and $\mathcal{M}_ r \to \mathcal{M}'_ r$ in Lemma 24.25.11. $\square$