Lemma 48.10.2. In the situation above, assume $f$ is flat and $i$ pseudo-coherent. Then (48.10.0.1) is an isomorphism for $K$ in $D^+_\mathit{QCoh}(\mathcal{O}_ X)$.

Proof. First proof. To prove this map is an isomorphism, we may work locally. Hence we may assume $X$, $X'$, $Z$, $Z'$ are affine, say corresponding to the rings $A$, $A'$, $B$, $B'$. Then $B$ and $A'$ are tor independent over $A$. By Lemma 48.10.1 it suffices to check that

$R\mathop{\mathrm{Hom}}\nolimits _ A(B, K) \otimes _ A^\mathbf {L} A' = R\mathop{\mathrm{Hom}}\nolimits _{A'}(B', K \otimes _ A^\mathbf {L} A')$

in $D(A')$ for all $K \in D^+(A)$. Here we use Derived Categories of Schemes, Lemma 36.10.8 and the fact that $B$, resp. $B'$ is pseudo-coherent as an $A$-module, resp. $A'$-module to compare derived hom on the level of rings and schemes. The displayed equality follows from More on Algebra, Lemma 15.98.3 part (3). See also the discussion in Dualizing Complexes, Section 47.14.

Second proof1. Let $z' \in Z'$ with image $z \in Z$. First show that (48.10.0.1) on stalks at $z'$ induces the map

$R\mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_{Z, z}, K_ z) \otimes _{\mathcal{O}_{Z, x}}^\mathbf {L} \mathcal{O}_{Z', z'} \longrightarrow R\mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_{Z', z'}, K_ z \otimes _{\mathcal{O}_{X, z}}^\mathbf {L} \mathcal{O}_{X', z'})$

from Dualizing Complexes, Equation (47.14.0.1). Namely, the constructions of these maps are identical. Then apply Dualizing Complexes, Lemma 47.14.2. $\square$

 This proof shows it suffices to assume $K$ is in $D^+(\mathcal{O}_ X)$.

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