Lemma 86.3.2. Notation and assumptions as in Lemma 86.3.1. Let $a : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ be the right adjoint to $Rf_*$. Then $a$ maps $D^+_\mathit{QCoh}(\mathcal{O}_ Y)$ into $D^+_\mathit{QCoh}(\mathcal{O}_ X)$. In fact, there exists an integer $N$ such that $H^ i(K) = 0$ for $i \leq c$ implies $H^ i(a(K)) = 0$ for $i \leq c - N$.
Proof. By Derived Categories of Spaces, Lemma 75.6.1 the functor $Rf_*$ has finite cohomological dimension. In other words, there exist an integer $N$ such that $H^ i(Rf_*L) = 0$ for $i \geq N + c$ if $H^ i(L) = 0$ for $i \geq c$. Say $K \in D^+_\mathit{QCoh}(\mathcal{O}_ Y)$ has $H^ i(K) = 0$ for $i \leq c$. Then
\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\tau _{\leq c - N}a(K), a(K)) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*\tau _{\leq c - N}a(K), K) = 0 \]
by what we said above. Clearly, this implies that $H^ i(a(K)) = 0$ for $i \leq c - N$. $\square$
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