The Stacks project

Proof. We will prove this by showing that $RQ_ Y \circ Rf_*$ and $\Phi \circ RQ_ X$ are right adjoint to the same functor $D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathcal{O}_ X)$.

Since $f$ is quasi-compact and quasi-separated, we see that $f_*$ preserves quasi-coherence, see Schemes, Lemma 26.24.1. Recall that $\mathit{QCoh}(\mathcal{O}_ X)$ is a Grothendieck abelian category (Properties, Proposition 28.23.4). Hence any $K$ in $D(\mathit{QCoh}(\mathcal{O}_ X))$ can be represented by a K-injective complex $\mathcal{I}^\bullet $ of $\mathit{QCoh}(\mathcal{O}_ X)$, see Injectives, Theorem 19.12.6. Then we can define $\Phi (K) = f_*\mathcal{I}^\bullet $.

Since $f$ is flat, the functor $f^*$ is exact. Hence $f^*$ defines $f^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$ and also $f^* : D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathit{QCoh}(\mathcal{O}_ X))$. The functor $f^* = Lf^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$ is left adjoint to $Rf_* : D(\mathcal{O}_ X) \to D(\mathcal{O}_ Y)$, see Cohomology, Lemma 20.28.1. Similarly, the functor $f^* : D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathit{QCoh}(\mathcal{O}_ X))$ is left adjoint to $\Phi : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$ by Derived Categories, Lemma 13.30.3.

Let $A$ be an object of $D(\mathit{QCoh}(\mathcal{O}_ Y))$ and $E$ an object of $D(\mathcal{O}_ X)$. Then

This implies what we want. $\square$