Theorem 24.25.13. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $(\mathcal{A}, \text{d})$ be a sheaf of differential graded algebras on $(\mathcal{C}, \mathcal{O})$. For every differential graded $\mathcal{A}$-module $\mathcal{M}$ there exists a quasi-isomorphism $\mathcal{M} \to \mathcal{I}$ where $\mathcal{I}$ is a graded injective and K-injective differential graded $\mathcal{A}$-module. Moreover, the construction is functorial in $\mathcal{M}$.
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
$M_0 = \text{id}$,
$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
$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. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)