Lemma 22.24.1. The functor (22.24.0.1) defines an exact functor of triangulated categories $K(\text{Mod}_{(A, \text{d})}) \to K(\text{Mod}_{(B, \text{d})})$.

## 22.24 Derived tensor product

This section is analogous to More on Algebra, Section 15.58.

Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras over $R$. Let $N$ be a $(A, B)$-bimodule equipped with a grading and differential such that $N$ is a left differential graded $A$-module and a right differential graded $B$-module. In other words, $N$ is a differential graded $A^{opp} \otimes _ R B$-module. Consider the functor

defined in Section 22.23.

**Proof.**
The functor was constructed in Lemma 22.23.1. We have to show that $- \otimes _ A N$ transforms distinguished triangles into distinguished triangles. Suppose that $0 \to K \to L \to M \to 0$ is an admissible short exact sequence of differential graded $A$-modules. Let $s : M \to L$ be a graded $A$-module homomorphism which is left inverse to $L \to M$. Then $s$ defines a graded $B$-module homomorphism $M \otimes _ A N \to L \otimes _ A N$ which is left inverse to $L \otimes _ A N \to M \otimes _ A N$.
$\square$

At this point we can consider the diagram

The dotted arrow that we will construct below will be the *left derived functor* of the composition $F$. (*Warning*: the diagram will not commute.) Namely, in the general setting of Derived Categories, Section 13.15 we want to compute the left derived functor of $F$ with respect to the multiplicative system of quasi-isomorphisms in $K(\text{Mod}_{(A, \text{d})})$.

Lemma 22.24.2. In the situation above, the left derived functor of $F$ exists. We denote it $- \otimes _ A^\mathbf {L} N : D(A, \text{d}) \to D(B, \text{d})$.

**Proof.**
We will use Derived Categories, Lemma 13.15.15 to prove this. As our collection $\mathcal{P}$ of objects we will use the objects with property (P). Property (1) was shown in Lemma 22.13.4. Property (2) holds because if $s : P \to P'$ is a quasi-isomorphism of modules with property (P), then $s$ is a homotopy equivalence by Lemma 22.15.3.
$\square$

Lemma 22.24.3. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Let $f : N \to N'$ be a homomorphism of differential graded $A^{opp} \otimes _ R B$-modules. Then $f$ induces a morphism of functors

If $f$ is a quasi-isomorphism, then $1 \otimes f$ is an isomorphism of functors.

**Proof.**
Let $M$ be a differential graded $A$-module with property (P). Then $1 \otimes f : M \otimes _ A N \to M \otimes _ A N'$ is a map of differential graded $B$-modules. Moreover, this is functorial with respect to $M$. Since the functors $- \otimes _ A^\mathbf {L} N$ and $- \otimes _ A^\mathbf {L} N'$ are computed by tensoring on objects with property (P) (Lemma 22.24.2) we obtain a transformation of functors as indicated.

Assume that $f$ is a quasi-isomorphism. Let $F_\bullet $ be the given filtration on $M$. Observe that $M \otimes _ A N = \mathop{\mathrm{colim}}\nolimits F_ i(M) \otimes _ A N$ and $M \otimes _ A N' = \mathop{\mathrm{colim}}\nolimits F_ i(M) \otimes _ A N'$. Hence it suffices to show that $F_ n(M) \otimes _ A N \to F_ n(M) \otimes _ A N'$ is a quasi-isomorphism (filtered colimits are exact, see Algebra, Lemma 10.8.8). Since the inclusions $F_ n(M) \to F_{n + 1}(M)$ are split as maps of graded $A$-modules we see that

is a short exact sequence of differential graded $B$-modules. There is a similar sequence for $N'$ and $f$ induces a map of short exact sequences. Hence by induction on $n$ (starting with $n = -1$ when $F_{-1}(M) = 0$) we conclude that it suffices to show that the map $F_{n + 1}(M)/F_ n(M) \otimes _ A N \to F_{n + 1}(M)/F_ n(M) \otimes _ A N'$ is a quasi-isomorphism. This is true because $F_{n + 1}(M)/F_ n(M)$ is a direct sum of shifts of $A$ and the result is true for $A[k]$ as $f : N \to N'$ is a quasi-isomorphism. $\square$

Lemma 22.24.4. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Let $N$ be an $(A, B)$-bimodule which comes with a grading and a differential such that it is a differential graded module for both $A$ and $B$. Then the functors

of Lemma 22.24.2 and

of Lemma 22.21.2 are adjoint.

**Proof.**
The statement means that we have

bifunctorially in $M$ and $N'$. To see this we may assume that $M$ is a differential graded $A$-module with property (P) and that $N'$ is a differential graded $B$-module with property (I). The computation of the derived functors given in the lemmas referenced in the statement combined with Lemma 22.15.3 translates the above into

where $\mathcal{B} = \text{Mod}^{dg}_{(B, \text{d})}$. Thus it is certainly sufficient to show that

as differential graded $\mathbf{Z}$-modules where $\mathcal{A} = \text{Mod}^{dg}_{(A, \text{d})}$. This follows from the fact that the isomorphism (Lemma 22.23.2)

of internal homs of graded modules respects the differentials. $\square$

Example 22.24.5. Let $R$ be a ring. Let $(A, \text{d}) \to (B, \text{d})$ be a homomorphism of differential graded $R$-algebras. Then we can view $B$ as a $(A, B)$-bimodule and we get a functor

By Lemma 22.24.4 the left adjoint of this is the functor $R\mathop{\mathrm{Hom}}\nolimits (B, -)$. For a differential graded $B$-module let us denote $N_ A$ the differential graded $A$-module obtained from $N$ by restriction via $A \to B$. Then we clearly have a canonical isomorphism

functorial in the $B$-module $N$. Thus we see that $R\mathop{\mathrm{Hom}}\nolimits (B, -)$ is the restriction functor and we obtain

bifunctorially in $M$ and $N$ exactly as in the case of commutative rings. Finally, observe that restriction is a tensor functor as well, since $N_ A = N \otimes _ B {}_ BB_ A = N \otimes _ B^\mathbf {L} {}_ BB_ A$ where ${}_ BB_ A$ is $B$ viewed as an $(B, A)$-bimodule.

Lemma 22.24.6. With notation and assumptions as in Lemma 22.24.4. Assume

$N$ defines a compact object of $D(B, \text{d})$, and

the map $H^ k(A) \to \mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(N, N[k])$ is an isomorphism for all $k \in \mathbf{Z}$.

Then the functor $-\otimes _ A^\mathbf {L} N$ is fully faithful.

**Proof.**
Because our functor has a left adjoint given by $R\mathop{\mathrm{Hom}}\nolimits (N, -)$ by Lemma 22.24.4 it suffices to show that for a differential graded $A$-module $M$ the map

is an isomorphism. We may assume that $M = P$ is a differential graded $A$-module which has property (P). Since $N$ defines a compact object, we reduce using Lemma 22.13.1 to the case where $P$ has a finite filtration whose graded pieces are direct sums of $A[k]$. Again using compactness we reduce to the case $P = A[k]$. Assumption (2) on $N$ is that the result holds for these. $\square$

Lemma 22.24.7. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras. Let $N$ be a differential graded $A^{opp} \otimes _ R B$-module with property (P). Let $M$ be a differential graded $A$-module with property (P). Then $Q = M \otimes _ A N$ is a differential graded $B$-module which represents $M \otimes _ A^\mathbf {L} N$ in $D(B)$ and which has a filtration

by differential graded submodules such that $Q = \bigcup F_ pQ$, the inclusions $F_ iQ \to F_{i + 1}Q$ are admissible monomorphisms, the quotients $F_{i + 1}Q/F_ iQ$ are isomorphic as differential graded $B$-modules to a direct sum of $(A \otimes _ R B)[k]$.

**Proof.**
Choose filtrations $F_\bullet $ on $M$ and $N$. Then consider the filtration on $Q = M \otimes _ A N$ given by

This is clearly a differential graded $B$-submodule. We see that

for example because the filtration of $M$ is split in the category of graded $A$-modules. Since by assumption the quotients on the right hand side are isomorphic to direct sums of shifts of $A$ and $A^{opp} \otimes _ R B$ and since $A \otimes _ A (A^{opp} \otimes _ R B) = A \otimes _ R B$, we conclude that the left hand side is a direct sum of shifts of $A \otimes _ R B$ as a differential graded $B$-module. (Warning: $Q$ does not have a structure of $(A, B)$-bimodule.) This proves the first statement of the lemma. The second statement is immediate from the definition of the functor in Lemma 22.24.2. $\square$

Lemma 22.24.8. Let $R \to R'$ be a ring map. Let $(A, \text{d})$ be a differential graded $R$-algebra. Let $(A', \text{d})$ be the base change, i.e., $A' = A \otimes _ R R'$. If $A$ is K-flat as a complex of $R$-modules, then

$- \otimes _ A^\mathbf {L} A' : D(A, \text{d}) \to D(A', \text{d})$ is equal the right derived functor of

\[ K(A, \text{d}) \longrightarrow K(A', \text{d}),\quad M \longmapsto M \otimes _ R R' \]the diagram

\[ \xymatrix{ D(A, \text{d}) \ar[r]_{- \otimes _ A^\mathbf {L} A'} \ar[d]_{restriction} & D(A', \text{d}) \ar[d]^{restriction} \\ D(R) \ar[r]^{- \otimes _ R^\mathbf {L} R'} & D(R') } \]commutes, and

if $M$ is K-flat as a complex of $R$-modules, then the differential graded $A'$-module $M \otimes _ R R'$ represents $M \otimes _ A^\mathbf {L} A'$.

**Proof.**
For any differential graded $A$-module $M$ there is a canonical map

Let $P$ be a differential graded $A$-module with property (P). We claim that $c_ P$ is an isomorphism and that $P$ is K-flat as a complex of $R$-modules. This will prove all the results stated in the lemma by formal arguments using the definition of derived tensor product in Lemma 22.24.2 and More on Algebra, Section 15.57.

Let $F_\bullet $ be the filtration on $P$ showing that $P$ has property (P). Note that $c_ A$ is an isomorphism and $A$ is K-flat as a complex of $R$-modules by assumption. Hence the same is true for direct sums of shifts of $A$ (you can use More on Algebra, Lemma 15.57.10 to deal with direct sums if you like). Hence this holds for the complexes $F_{p + 1}P/F_ pP$. Since the short exact sequences

are split exact as sequences of graded modules, we can argue by induction that $c_{F_ pP}$ is an isomorphism for all $p$ and that $F_ pP$ is K-flat as a complex of $R$-modules (use More on Algebra, Lemma 15.57.7). Finally, using that $P = \mathop{\mathrm{colim}}\nolimits F_ pP$ we conclude that $c_ P$ is an isomorphism and that $P$ is K-flat as a complex of $R$-modules (use More on Algebra, Lemma 15.57.10). $\square$

Lemma 22.24.9. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Let $T$ be an $(A, B)$-bimodule which comes with a grading and a differential such that it is a differential graded module for both $A$ and $B$. Assume

$T$ defines a compact object of $D(B, \text{d})$, and

$S = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(T, B)$ represents $R\mathop{\mathrm{Hom}}\nolimits (T, B)$ in $D(A, \text{d})$.

Then $S$ has a structure of a $(B, A)$-bimodule which comes with a grading and a differential such that it is a differential graded module for both $A$ and $B$ and there is an isomorphism

functorial in $N$ in $D(B, \text{d})$.

**Proof.**
Write $\mathcal{B} = \text{Mod}^{dg}_{(B, \text{d})}$. The right $A$-module structure on $S$ comes from the map $A \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, T)$ and the composition $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, B) \otimes \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, T) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, B)$ defined in Example 22.19.8. Using this multiplication a second time there is a map

functorial in $N$. Given $N$ we can choose quasi-isomorphisms $P \to N \to I$ where $P$, resp. $I$ is a differential graded $B$-module with property (P), resp. (I). Then using $c_ N$ we obtain a map $P \otimes _ B S \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, I)$ between the objects representing $S \otimes _ B^\mathbf {L} N$ and $R\mathop{\mathrm{Hom}}\nolimits (T, N)$. Clearly this defines a transformation of functors $c$ as in the lemma.

To prove that $c$ is an isomorphism of functors, we may assume $N$ is a differential graded $B$-module which has property (P). Since $T$ defines a compact object in $D(B, \text{d})$ and since both sides of the arrow define exact functors of triangulated categories, we reduce using Lemma 22.13.1 to the case where $N$ has a finite filtration whose graded pieces are direct sums of $B[k]$. Using again that both sides of the arrow are exact functors of triangulated categories and compactness of $T$ we reduce to the case $N = B[k]$. Assumption (2) is exactly the assumption that $c$ is an isomorphism in this case. $\square$

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