Lemma 22.33.3. 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 $(A, B)$-bimodules. Then $f$ induces a morphism of functors

\[ 1\otimes f : - \otimes _ A^\mathbf {L} N \longrightarrow - \otimes _ A^\mathbf {L} N' \]

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 22.33.2) 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

\[ 0 \to F_ n(M) \otimes _ A N \to F_{n + 1}(M) \otimes _ A N \to F_{n + 1}(M)/F_ n(M) \otimes _ A N \to 0 \]

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.
$\square$

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