Lemma 22.13.4. Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded $R$-algebra. Let $M$ be a right differential graded $A$-module and let $M'$ be a left differential graded $A$-module. Let $N^\bullet$ be a complex of $R$-modules. Then we have

$\mathop{\mathrm{Hom}}\nolimits _{\text{left diff graded }A\text{-modules}}(M', \mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}(R)}(M \otimes _ A M', N^\bullet )$

where $M \otimes _ A M'$ is viewed as a complex of $R$-modules as in Section 22.12.

Proof. Let us show that both sides correspond to graded $A$-bilinear maps

$M \times M' \longrightarrow N^\bullet$

compatible with differentials. We have seen this is true for the right hand side in Section 22.12. Given an element $g$ of the left hand side, the equality of More on Algebra, Lemma 15.71.1 determines a map of complexes $g' : \text{Tot}(M' \otimes _ R M) \to N^\bullet$. We precompose with the commutativity constraint to get

$\text{Tot}(M \otimes _ R M') \xrightarrow {\psi } \text{Tot}(M' \otimes _ R M) \xrightarrow {g'} N^\bullet$

which corresponds to a graded $R$-bilinear map $g'' : M \times M' \to N^\bullet$ compatible with differentials. The $A$-linearity of $g$ translates immediately into $A$-bilinarity of $g''$. Namely, say $x \in M^ e$ and $x' \in (M')^{e'}$ and $a \in A^ n$. Then on the one hand we have

\begin{align*} g''(x, ax') & = (-1)^{e(n + e')} g'(ax' \otimes x) \\ & = (-1)^{e(n + e')} g(ax')(x) \\ & = (-1)^{e(n + e')} (a \cdot g(x'))(x) \\ & = (-1)^{e(n + e') + n(n + e + e') + n} g(x')(xa) \end{align*}

and on the other hand we have

$g''(xa, x') = (-1)^{(e + n)e'} g'(x' \otimes xa) = (-1)^{(e + n)e'} g(x')(xa)$

which is the same thing by a trivial mod $2$ calculation of the exponents. $\square$

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