Lemma 20.43.3. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $E$ be an object of $D(\mathcal{O}_ Y)$. If $E$ is $m$-pseudo-coherent, then $Lf^*E$ is $m$-pseudo-coherent.

**Proof.**
Represent $E$ by a complex $\mathcal{E}^\bullet $ of $\mathcal{O}_ Y$-modules and choose an open covering $Y = \bigcup V_ i$ and $\alpha _ i : \mathcal{E}_ i^\bullet \to \mathcal{E}^\bullet |_{V_ i}$ as in Definition 20.43.1. Set $U_ i = f^{-1}(V_ i)$. By Lemma 20.43.2 it suffices to show that $Lf^*\mathcal{E}^\bullet |_{U_ i}$ is $m$-pseudo-coherent. Choose a distinguished triangle

The assumption on $\alpha _ i$ means exactly that the cohomology sheaves $H^ j(C)$ are zero for all $j \geq m$. Denote $f_ i : U_ i \to V_ i$ the restriction of $f$. Note that $Lf^*\mathcal{E}^\bullet |_{U_ i} = Lf_ i^*(\mathcal{E}|_{V_ i})$. Applying $Lf_ i^*$ we obtain the distinguished triangle

By the construction of $Lf_ i^*$ as a left derived functor we see that $H^ j(Lf_ i^*C) = 0$ for $j \geq m$ (by the dual of Derived Categories, Lemma 13.16.1). Hence $H^ j(Lf_ i^*\alpha _ i)$ is an isomorphism for $j > m$ and $H^ m(Lf^*\alpha _ i)$ is surjective. On the other hand, $Lf_ i^*\mathcal{E}_ i^\bullet = f_ i^*\mathcal{E}_ i^\bullet $. is strictly perfect by Lemma 20.42.4. Thus we conclude. $\square$

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