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

\[ - \otimes _ E K^\bullet : \text{Mod}^{dg}_{(E, \text{d})} \longrightarrow \text{Comp}^{dg}(\mathcal{O}) \]

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

\[ M \otimes _ E K^\bullet (U) \]

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

\[ M \otimes _ E K^\bullet \]

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. $\square$

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