
## 22.25 Composition of derived tensor products

We encourage the reader to skip this section.

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 $A^{opp} \otimes _ R B$-module. Let $N'$ be a differential graded $B^{opp} \otimes _ R C$-module. We denote $N_ B$ the bimodule $N$ viewed as a differential graded $B$-module (forgetting about the $A$-structure). There is a canonical map

22.25.0.1
$$\label{dga-equation-plain-versus-derived} N_ B \otimes _ B^\mathbf {L} N' \longrightarrow (N \otimes _ B N')_ C$$

in $D(C, \text{d})$. Here $(N \otimes _ B N')_ C$ denotes the $(A, C)$-bimodule $N \otimes _ B N'$ viewed as a differential graded $C$-module. Namely, this map comes from the fact that the derived tensor product always maps to the plain tensor product (as it is a left derived functor).

Lemma 22.25.1. 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 $A^{opp} \otimes _ R B$-module. Let $N'$ be a differential graded $B^{opp} \otimes _ R C$-module. Assume (22.25.0.1) 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 $M$ 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 22.24.2 the plain tensor product $M \otimes _ A N''$ represents $M \otimes _ A^\mathbf {L} N''$. Then we write

$M \otimes _ A N'' = M \otimes _ A (N \otimes _ B N') = (M \otimes _ A N) \otimes _ B N'$

The module $Q = M \otimes _ A N$ represents $M \otimes _ A^\mathbf {L} N$. Hence it suffices to show that

$Q \otimes _ B^\mathbf {L} N' \longrightarrow Q \otimes _ B N'$

is a quasi-isomorphism. The filtration $F_\bullet$ on $M$ induces a filtration $F_\bullet$ on $Q$ whose transition maps are admissible monomorphisms and whose graded quotients are direct sums of shifts of $N_ B$. Exactly as in the proof of Lemma 22.13.1 this implies there is an admissible short exact sequence

$0 \to \bigoplus \nolimits F_ iQ \to \bigoplus \nolimits F_ iQ \to Q \to 0$

of differential graded $B$-modules. Using the fact that $- \otimes _ B^\mathbf {L} N'$ is an exact functor and commutes with direct sums and using that $- \otimes _ B N'$ transforms admissible exact sequences into admissible exact sequences and commutes with direct sums we reduce to proving that

$F_ pQ \otimes _ B^\mathbf {L} N' \longrightarrow F_ pQ \otimes _ B N'$

is a quasi-isomorphism for all $p$. Repeating the argument with the admissible short exact sequences

$0 \to F_ pQ \to F_{p + 1}Q \to F_{p + 1}Q/F_ pQ \to 0$

we reduce to showing the same statement for $F_{p + 1}Q/F_ pQ$. Using the structure of these modules (see above) we reduce to showing that

$N_ B \otimes _ B^\mathbf {L} N' \longrightarrow N_ B \otimes _ B N'$

is a quasi-isomorphism, which is what we assumed. $\square$

Let $R$ be a ring. Let $(A, \text{d})$, $(B, \text{d})$, and $(C, \text{d})$ be differential graded $R$-algebras. We temporarily denote $(A \otimes _ R B)_ B$ the differential graded algebra $A \otimes _ R B$ viewed as a (right) differential graded $B$-module, and ${}_ B(B \otimes _ R C)_ C$ the differential graded algebra $B \otimes _ R C$ viewed as a $(B, C)$-bimodule. Then there is a canonical map

22.25.1.1
$$\label{dga-equation-plain-versus-derived-algebras} (A \otimes _ R B)_ B \otimes _ B^\mathbf {L} {}_ B(B \otimes _ R C)_ C \longrightarrow (A \otimes _ R B \otimes _ R C)_ C$$

where $(A \otimes _ R B \otimes _ R C)_ C$ denotes the differential graded $R$-algebra $A \otimes _ R B \otimes _ R C$ viewed as a (right) differential graded $C$-module. Namely, this map comes from the identification

$(A \otimes _ R B)_ B \otimes _ B {}_ B(B \otimes _ R C)_ C = (A \otimes _ R B \otimes _ R C)_ C$

and the fact that the derived tensor product always maps to the plain tensor product (as it is a left derived functor).

Lemma 22.25.2. Let $R$ be a ring. Let $(A, \text{d})$, $(B, \text{d})$, and $(C, \text{d})$ be differential graded $R$-algebras. Assume that (22.25.1.1) is an isomorphism. Let $N$ be a differential graded $A^{opp} \otimes _ R B$-module. Let $N'$ be a differential graded $B^{opp} \otimes _ R C$-module. 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 $- \otimes _ A^\mathbf {L} N''$ for a differential graded $A^{opp} \otimes _ R C$-module $N''$ described in the proof.

Proof. By Lemma 22.24.3 we may replace $N$ and $N'$ by quasi-isomorphic bimodules. Thus we may assume $N$, resp. $N'$ has property (P) as differential graded $A^{opp} \otimes _ R B$, resp. $B^{opp} \otimes _ R C$-module, see Lemma 22.13.4. We claim the lemma holds with the $(A, C)$-bimodule $N'' = N \otimes _ B N'$. To prove this, it suffices to show that

$N_ B \otimes _ B^\mathbf {L} N' \longrightarrow (N \otimes _ B N')_ C$

is an isomorphism, see Lemma 22.25.1.

By Lemma 22.13.1 there is an admissible short exact sequence

$0 \to \bigoplus \nolimits F_ iN \to \bigoplus \nolimits F_ iN \to N \to 0$

of differential graded $A^{opp} \otimes _ R B$-modules. Using the fact that $- \otimes _ B^\mathbf {L} N'$ is an exact functor and commutes with direct sums and using that $- \otimes _ B N'$ transforms admissible exact sequences into admissible exact sequences and commutes with direct sums we reduce to proving that

$(F_ pN)_ B \otimes _ B^\mathbf {L} N' \longrightarrow (F_ pN)_ B \otimes _ B N'$

is a quasi-isomorphism for all $p$. Repeating the argument with the admissible short exact sequences

$0 \to F_ pN \to F_{p + 1}N \to F_{p + 1}N/F_ pN \to 0$

we reduce to showing the same statement for $F_{p + 1}N/F_ pN$. Since these modules are direct sums of shifts of $(A \otimes _ R B)_ B$ we reduce to showing that

$(A \otimes _ R B)_ B \otimes _ B^\mathbf {L} N' \longrightarrow (A \otimes _ R B)_ B \otimes _ B N'$

is a quasi-isomorphism. Now we choose a quasi-isomorphism $P \to (A \otimes _ R B)_ B$ of differential graded $B$-modules where $P$ has property (P). Then we have to show that $P \otimes _ B N' \to (A \otimes _ R B)_ B \otimes _ B N'$ is a quasi-isomorphism because $P \otimes _ B N'$ represents $(A \otimes _ R B)_ B \otimes _ B^\mathbf {L} N'$ by the construction in Lemma 22.24.2. As $N' = \mathop{\mathrm{colim}}\nolimits F_ pN'$ we find that it suffices to show that $P \otimes _ B F_ pN' \to (A \otimes _ R B)_ B \otimes _ B F_ pN'$. Using the admissible short exact sequences $0 \to F_ pN' \to F_{p + 1}N' \to F_{p + 1}N'/F_ pN' \to 0$ we reduce to showing $P \otimes _ B F_{p + 1}N'/F_ pN' \to (A \otimes _ R B)_ B \otimes _ B F_{p + 1}N'/F_ pN'$ is a quasi-isomorphism for all $p$. Then finally using that $F_{p + 1}N'/F_ pN'$ is a direct sum of shifts of ${}_ B(B \otimes _ R C)_ C$ we conclude that it suffices to show that

$P \otimes _ B {}_ B(B \otimes _ R C)_ C \to (A \otimes _ R B)_ B \otimes _ B {}_ B(B \otimes _ R C)_ C$

is an isomorphism. Since $P \to (A \otimes _ R B)_ B$ is a resolution by a module satisfying property (P) this map of differential graded $C$-modules represents the morphism (22.25.1.1) and the proof is complete. $\square$

Lemma 22.25.3. Let $R$ be a ring. Let $(A, \text{d})$, $(B, \text{d})$, and $(C, \text{d})$ be differential graded $R$-algebras. If $C$ is K-flat as a complex of $R$-modules, then (22.25.1.1) is an isomorphism and the conclusion of Lemma 22.25.2 is valid.

Proof. Choose a quasi-isomorphism $P \to (A \otimes _ R B)_ B$ of differential graded $B$-modules, where $P$ has property (P). Then we have to show that

$P \otimes _ B (B \otimes _ R C) \longrightarrow (A \otimes _ R B) \otimes _ B (B \otimes _ R C)$

is a quasi-isomorphism. Equivalently we are looking at

$P \otimes _ R C \longrightarrow A \otimes _ R B \otimes _ R C$

This is a quasi-isomorphism if $C$ is K-flat as a complex of $R$-modules by More on Algebra, Lemma 15.57.4. $\square$

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