Lemma 15.58.2. Let $R \to A$ be a ring map. Let $f : L^\bullet \to N^\bullet $ be a map of complexes of $A$-modules. Then $f$ induces a transformation of functors

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

Lemma 15.58.2. Let $R \to A$ be a ring map. Let $f : L^\bullet \to N^\bullet $ be a map of complexes of $A$-modules. Then $f$ induces a transformation of functors

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

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

**Proof.**
Since the functors are computing by evaluating on K-flat complexes $K^\bullet $ we can simply use the functoriality

\[ \text{Tot}(K^\bullet \otimes _ R L^\bullet ) \to \text{Tot}(K^\bullet \otimes _ R N^\bullet ) \]

to define the transformation. The last statement follows from Lemma 15.57.4. $\square$

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