The Stacks project

Lemma 22.34.1. Let $R$ be a ring. Let $(A, \text{d})$, $(B, \text{d})$, and $(C, \text{d})$ be differential graded $R$-algebras. Let $N$ be a differential graded $(A, B)$-bimodule. Let $N'$ be a differential graded $(B, C)$-module. Assume ( is an isomorphism. Then the composition

\[ \xymatrix{ D(A, \text{d}) \ar[rr]^{- \otimes _ A^\mathbf {L} N} & & D(B, \text{d}) \ar[rr]^{- \otimes _ B^\mathbf {L} N'} & & D(C, \text{d}) } \]

is isomorphic to $- \otimes _ A^\mathbf {L} N''$ with $N'' = N \otimes _ B N'$ viewed as $(A, C)$-bimodule.

Proof. Let us define a transformation of functors

\[ (- \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} N' \longrightarrow - \otimes _ A^\mathbf {L} N'' \]

To do this, let $M$ be a differential graded $A$-module with property (P). According to the construction of the functor $- \otimes _ A^\mathbf {L} N''$ of the proof of Lemma 22.33.2 the plain tensor product $M \otimes _ A N''$ represents $M \otimes _ A^\mathbf {L} N''$ in $D(C, \text{d})$. Then we write

\[ M \otimes _ A N'' = M \otimes _ A (N \otimes _ B N') = (M \otimes _ A N) \otimes _ B N' \]

The module $M \otimes _ A N$ represents $M \otimes _ A^\mathbf {L} N$ in $D(B, \text{d})$. Choose a quasi-isomorphism $Q \to M \otimes _ A N$ where $Q$ is a differential graded $B$-module with property (P). Then $Q \otimes _ B N'$ represents $(M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} N'$ in $D(C, \text{d})$. Thus we can define our map via

\[ (M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} N' = Q \otimes _ B N' \to M \otimes _ A N \otimes _ B N' = M \otimes _ A^\mathbf {L} N'' \]

The construction of this map is functorial in $M$ and compatible with distinguished triangles and direct sums; we omit the details. Consider the property $T$ of objects $M$ of $D(A, \text{d})$ expressing that this map is an isomorphism. Then

  1. if $T$ holds for $M_ i$ then $T$ holds for $\bigoplus M_ i$,

  2. if $T$ holds for $2$-out-of-$3$ in a distinguished triangle, then it holds for the third, and

  3. $T$ holds for $A[k]$ because here we obtain a shift of the map ( which we have assumed is an isomorphism.

Thus by Remark 22.22.5 property $T$ always holds and the proof is complete. $\square$

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