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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