Lemma 48.10.1. In the situation above, the map (48.10.0.1) is an isomorphism if and only if the base change map

$Lf^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ Z, K) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{O}_{Z'}, Lf^*K)$

of Cohomology, Remark 20.40.13 is an isomorphism.

Proof. The statement makes sense because $\mathcal{O}_{Z'} = Lf^*\mathcal{O}_ Z$ by the assumed tor independence. Since $i'_*$ is exact and faithful we see that it suffices to show the map (48.10.0.1) is an isomorphism after applying $Ri'_*$. Since $Ri'_* \circ Lg^* = Lf^* \circ Ri_*$ by the assumed tor indepence and Derived Categories of Schemes, Lemma 36.22.9 we obtain a map

$Lf^*Ri_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \longrightarrow Ri'_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_{Z'}, Lf^*K)$

whose source and target are as in the statement of the lemma by Lemma 48.9.3. We omit the verification that this is the same map as the one constructed in Cohomology, Remark 20.40.13. $\square$

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