Lemma 15.84.6. Let $R$ be a ring. Let $K$ be an object of $D(R)$ with $H^ i(K) = 0$ for $i \not\in \{ -1, 0\} $. Let $K^{-1} \to K^0$ be a two term complex of $R$-modules representing $K$ such that $K^0$ is a flat $R$-module (for example projective or free). Let $R \to R'$ be a ring map. Then the complex $K^\bullet \otimes _ R R'$ represents $\tau _{\geq -1}(K \otimes _ R^\mathbf {L} R')$.
Proof. We have a distinguished triangle
\[ K^0 \to K^\bullet \to K^{-1}[1] \to K^0[1] \]
in $D(R)$. This determines a map of distinguished triangles
\[ \xymatrix{ K^0 \otimes _ R^\mathbf {L} R' \ar[d] \ar[r] & K^\bullet \otimes _ R^\mathbf {L} R' \ar[r] \ar[d] & K^{-1} \otimes _ R^\mathbf {L} R'[1] \ar[r] \ar[d] & K^0 \otimes _ R^\mathbf {L} R'[1] \ar[d] \\ K^0 \otimes _ R R' \ar[r] & K^\bullet \otimes _ R R' \ar[r] & K^{-1} \otimes _ R R'[1] \ar[r] & K^0 \otimes _ R R'[1] } \]
The left and right vertical arrows are isomorphisms as $K^0$ is flat. Since $K^{-1} \otimes _ R^\mathbf {L} R' \to K^{-1} \otimes _ R R'$ is an isomorphism on cohomology in degree $0$ we conclude. $\square$
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