Lemma 15.60.2. The comparison map (15.60.0.1) is an isomorphism if $A' = A \otimes _ R R'$ and $A$ and $R'$ are Tor independent over $R$.

Proof. To prove this we choose a free resolution $F^\bullet \to R'$ of $R'$ as an $R$-module. Because $A$ and $R'$ are Tor independent over $R$ we see that $F^\bullet \otimes _ R A$ is a free $A$-module resolution of $A'$ over $A$. By our general construction of the derived tensor product above we see that

$K^\bullet \otimes _ A A' \cong \text{Tot}(K^\bullet \otimes _ A (F^\bullet \otimes _ R A)) = \text{Tot}(K^\bullet \otimes _ R F^\bullet ) \cong \text{Tot}(E^\bullet \otimes _ R F^\bullet ) \cong E^\bullet \otimes _ R R'$

as desired. $\square$

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