The Stacks project

Lemma 24.29.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $(\mathcal{A}, \text{d})$ be a sheaf of differential graded algebras on $(\mathcal{C}, \mathcal{O})$. Then any exact functor

\[ T : K(\textit{Mod}(\mathcal{A}, \text{d})) \longrightarrow \mathcal{D} \]

of triangulated categories has a right derived extension $RT : D(\mathcal{A}, \text{d}) \to \mathcal{D}$ whose value on a graded injective and K-injective differential graded $\mathcal{A}$-module $\mathcal{I}$ is $T(\mathcal{I})$.

Proof. By Theorem 24.25.13 for any (right) 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. Hence by Derived Categories, Lemma 13.14.15 it suffices to show that given a quasi-isomorphism $\mathcal{I} \to \mathcal{I}'$ of differential graded $\mathcal{A}$-modules which are both graded injective and K-injective then $T(\mathcal{I}) \to T(\mathcal{I}')$ is an isomorphism. This is true because the map $\mathcal{I} \to \mathcal{I}'$ is an isomorphism in $K(\textit{Mod}(\mathcal{A}, \text{d}))$ as follows for example from Lemma 24.26.7 (or one can deduce it from Lemma 24.25.10). $\square$

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Your email address will not be published.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 0FTN. Beware of the difference between the letter 'O' and the digit '0'.