Lemma 15.84.7. Let $I$ be an ideal of a ring $R$. Let $K$ be an object of $D(R)$ with $H^ i(K) = 0$ for $i \not\in \{ -1, 0\} $. Let $R \to R'$ be a ring map. If $K$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5 with respect to $(R, I)$, then $\tau _{\geq -1}(K \otimes _ R^\mathbf {L} R')$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5 with respect to $(R', IR')$
Proof. We may assume $K$ is represented by a two term complex $K^{-1} \to K^0$ with $K^0$ free such that for any $a \in I$ the map $a : K^{-1} \to K^{-1}$ is equal to $h_ a \circ d_ K^{-1}$ for some map $h_ a : K^0 \to K^{-1}$. By Lemma 15.84.6 we see that $\tau _{\geq -1}(K \otimes _ R^\mathbf {L} R')$ is represented by $K^\bullet \otimes _ R R'$. Then of course for every $a \in I$ we see that $a \otimes 1 : K^{-1} \otimes _ R R' \to K^{-1} \otimes _ R R'$ is equal to $(h_ a \otimes 1) \circ (d_ K^{-1} \otimes 1)$. Since the collection of maps $K^{-1} \otimes _ R R' \to K^{-1} \otimes _ R R'$ which factor through $\text{d}_ K^{-1} \otimes 1$ forms an $R'$-module we conclude. $\square$
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