Lemma 22.30.1. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras over $R$. Let $N$ be a differential graded $(A, B)$-bimodule. The construction above defines a functor

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, -) : \text{Mod}^{dg}_{(B, \text{d})} \longrightarrow \text{Mod}^{dg}_{(A, \text{d})} \]

of differential graded categories. This functor induces functors

\[ \text{Mod}_{(B, \text{d})} \to \text{Mod}_{(A, \text{d})} \quad \text{and}\quad K(\text{Mod}_{(B, \text{d})}) \to K(\text{Mod}_{(A, \text{d})}) \]

by an application of Lemma 22.26.5.

**Proof.**
Above we have seen how the construction defines a functor of underlying graded categories. Thus it suffices to show that the construction is compatible with differentials. Let $N_1$ and $N_2$ be differential graded $B$-modules. Write

\[ H_{12} = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N_1, N_2),\quad H_1 = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, N_1),\quad H_2 = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, N_2) \]

Consider the composition

\[ c : H_{12} \otimes _ R H_1 \longrightarrow H_2 \]

in the differential graded category $\text{Mod}^{dg}_{(B, \text{d})}$. Let $f : N_1 \to N_2$ be a $B$-module homomorphism which is homogeneous of degree $n$, in other words, $f \in H_{12}^ n$. The functor in the lemma sends $f$ to $c_ f : H_1 \to H_2$, $g \mapsto c(f, g)$. Similarly for $\text{d}(f)$. On the other hand, the differential on

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(A, \text{d})}}(H_1, H_2) \]

sends $c_ f$ to $\text{d}_{H_2} \circ c_ f - (-1)^ n c_ f \circ \text{d}_{H_1}$. As $c$ is a morphism of complexes of $R$-modules we have $\text{d} c(f, g) = c(\text{d}f, g) + (-1)^ n c(f, \text{d}g)$. Hence we see that

\begin{align*} (\text{d}c_ f)(g) & = \text{d}c(f,g) - (-1)^ n c(f, \text{d}g) \\ & = c(\text{d}f, g) + (-1)^ n c(f, \text{d}g) - (-1)^ n c(f, \text{d}g) \\ & = c(\text{d}f, g) = c_{\text{d}f}(g) \end{align*}

and the proof is complete.
$\square$

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