Lemma 22.30.3. Let $R$ be a ring. Let $A$ and $B$ be $R$-algebras. Let $M$ be a right $A$-module, $N$ an $(A, B)$-bimodule, and $N'$ a right $B$-module. Then we have a canonical isomorphism

$\mathop{\mathrm{Hom}}\nolimits _ B(M \otimes _ A N, N') = \mathop{\mathrm{Hom}}\nolimits _ A(M, \mathop{\mathrm{Hom}}\nolimits _ B(N, N'))$

of $R$-modules. If $A$, $B$, $M$, $N$, $N'$ are compatibly graded, then we have a canonical isomorphism

$\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ B^{gr}}(M \otimes _ A N, N') = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ A^{gr}}(M, \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ B^{gr}}(N, N'))$

of graded $R$-modules If $A$, $B$, $M$, $N$, $N'$ are compatibly differential graded, then we have a canonical isomorphism

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

of complexes of $R$-modules.

Proof. Omitted. Hint: in the ungraded case interpret both sides as $A$-bilinear maps $\psi : M \times N \to N'$ which are $B$-linear on the right. In the (differential) graded case, use the isomorphism of More on Algebra, Lemma 15.71.1 and check it is compatible with the module structures. Alternatively, use the isomorphism of Lemma 22.13.2 and show that it is compatible with the $B$-module structures. $\square$

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