Lemma 22.31.5. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Let $N$ be a differential graded $(A, B)$-bimodule. If $\mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(N, N') = \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(B, \text{d})})}(N, N')$ for all $N' \in K(B, \text{d})$, for example if $N$ has property (P) as a differential graded $B$-module, then

$R\mathop{\mathrm{Hom}}\nolimits (N, M) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, M)$

functorially in $M$ in $D(B, \text{d})$.

Proof. By construction (Lemma 22.31.2) to find $R\mathop{\mathrm{Hom}}\nolimits (N, M)$ we choose a quasi-isomorphism $M \to I$ where $I$ is a differential graded $B$-module with property (I) and we set $R\mathop{\mathrm{Hom}}\nolimits (N, M) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, I)$. By assumption the map

$\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, M) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, I)$

induced by $M \to I$ is a quasi-isomorphism, see discussion in Example 22.26.8. This proves the lemma. If $N$ has property (P) as a $B$-module, then we see that the assumption is satisfied by Lemma 22.22.3. $\square$

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