The Stacks project

Proof. Write $\mathcal{B} = \text{Mod}^{dg}_{(B, \text{d})}$. The right $A$-module structure on $S$ comes from the map $A \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, T)$ and the composition $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, B) \otimes \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, T) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, B)$ defined in Example 22.26.8. Using this multiplication a second time there is a map

functorial in $N$. Given $N$ we can choose quasi-isomorphisms $P \to N \to I$ where $P$, resp. $I$ is a differential graded $B$-module with property (P), resp. (I). Then using $c_ N$ we obtain a map $P \otimes _ B S \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, I)$ between the objects representing $S \otimes _ B^\mathbf {L} N$ and $R\mathop{\mathrm{Hom}}\nolimits (T, N)$. Clearly this defines a transformation of functors $c$ as in the lemma.

To prove that $c$ is an isomorphism of functors, we may assume $N$ is a differential graded $B$-module which has property (P). Since $T$ defines a compact object in $D(B, \text{d})$ and since both sides of the arrow define exact functors of triangulated categories, we reduce using Lemma 22.20.1 to the case where $N$ has a finite filtration whose graded pieces are direct sums of $B[k]$. Using again that both sides of the arrow are exact functors of triangulated categories and compactness of $T$ we reduce to the case $N = B[k]$. Assumption (2) is exactly the assumption that $c$ is an isomorphism in this case. $\square$