Lemma 20.54.3. 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 20.54.3. 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).
Proof. To check (20.54.2.1) is an isomorphism we may work locally on Y, i.e., we have to find an open 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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