Lemma 47.14.1. In the situation above, the map (47.14.0.1) is an isomorphism if and only if the map

of More on Algebra, Lemma 15.69.5 is an isomorphism.

Lemma 47.14.1. In the situation above, the map (47.14.0.1) is an isomorphism if and only if the map

\[ R\mathop{\mathrm{Hom}}\nolimits _ R(A, K) \otimes _ R^\mathbf {L} R' \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ R(A, K \otimes _ R^\mathbf {L} R') \]

of More on Algebra, Lemma 15.69.5 is an isomorphism.

**Proof.**
To see that the map is an isomorphism, it suffices to prove it is an isomorphism after applying $\varphi '_*$. Applying the functor $\varphi '_*$ to (47.14.0.1) and using that $A' = A \otimes _ R^\mathbf {L} R'$ we obtain the base change map $R\mathop{\mathrm{Hom}}\nolimits _ R(A, K) \otimes _ R^\mathbf {L} R' \to R\mathop{\mathrm{Hom}}\nolimits _{R'}(A \otimes _ R^\mathbf {L} R', K \otimes _ R^\mathbf {L} R')$ for derived hom of More on Algebra, Equation (15.92.1.1). Unwinding the left and right hand side exactly as in the proof of More on Algebra, Lemma 15.92.2 and in particular using More on Algebra, Lemma 15.92.1 gives the desired result.
$\square$

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