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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