Remark 86.3.5. In the situation of Lemma 86.3.3 we have
by Derived Categories of Spaces, Lemma 75.19.2. Thus if $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \in D_\mathit{QCoh}(\mathcal{O}_ X)$, then we can “erase” the $DQ_ Y$ on the left hand side of the arrow. On the other hand, if we know that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K) \in D_\mathit{QCoh}(\mathcal{O}_ Y)$, then we can “erase” the $DQ_ Y$ from the right hand side of the arrow. If both are true then we see that (86.3.2.1) is an isomorphism. Combining this with Derived Categories of Spaces, Lemma 75.13.10 we see that $Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)$ is an isomorphism if
$L$ and $Rf_*L$ are perfect, or
$K$ is bounded below and $L$ and $Rf_*L$ are pseudo-coherent.
For (2) we use that $a(K)$ is bounded below if $K$ is bounded below, see Lemma 86.3.2.
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