Lemma 20.49.3. Let $f : X \to Y$ be a morphism of ringed spaces. Let $E \in D(\mathcal{O}_ X)$ and $K \in D(\mathcal{O}_ Y)$. If $K$ is perfect, then

in $D(\mathcal{O}_ Y)$.

Lemma 20.49.3. Let $f : X \to Y$ be a morphism of ringed spaces. Let $E \in D(\mathcal{O}_ X)$ and $K \in D(\mathcal{O}_ Y)$. If $K$ is perfect, then

\[ Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_ Y} K = Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K) \]

in $D(\mathcal{O}_ Y)$.

**Proof.**
To check (20.49.2.1) is an isomorphism we may work locally on $Y$, i.e., we have to find a covering $\{ V_ j \to Y\} $ such that the map restricts to an isomorphism on $V_ j$. By definition of perfect objects, this means we may assume $K$ is represented by a strictly perfect complex of $\mathcal{O}_ Y$-modules. Note that, completely generally, the statement is true for $K = K_1 \oplus K_2$, if and only if the statement is true for $K_1$ and $K_2$. Hence we may assume $K$ is a finite complex of finite free $\mathcal{O}_ Y$-modules. In this case a simple argument involving stupid truncations reduces the statement to the case where $K$ is represented by a finite free $\mathcal{O}_ Y$-module. Since the statement is invariant under finite direct summands in the $K$ variable, we conclude it suffices to prove it for $K = \mathcal{O}_ Y[n]$ in which case it is trivial.
$\square$

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