Lemma 22.35.1. In the situation above there is a functor
of differential graded categories. This functor sends $E$ to $K^\bullet $ and commutes with direct sums.
Lemma 22.35.1. In the situation above there is a functor
of differential graded categories. This functor sends $E$ to $K^\bullet $ and commutes with direct sums.
Proof. Let $M$ be a differential graded $E$-module. For every object $U$ of $\mathcal{C}$ the complex $K^\bullet (U)$ is a left differential graded $E$-module as well as a right $\mathcal{O}(U)$-module. The actions commute, so we have a bimodule. Thus, by the constructions in Sections 22.12 and 22.28 we can form the tensor product
which is a differential graded $\mathcal{O}(U)$-module, i.e., a complex of $\mathcal{O}(U)$-modules. This construction is functorial with respect to $U$, hence we can sheafify to get a complex of $\mathcal{O}$-modules which we denote
Moreover, for each $U$ the construction determines a functor $\text{Mod}^{dg}_{(E, \text{d})} \to \text{Comp}^{dg}(\mathcal{O}(U))$ of differential graded categories by Lemma 22.29.1. It is therefore clear that we obtain a functor as stated in the lemma. $\square$
Your email address will not be published.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)