Lemma 20.44.3. 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 20.44.1. Set $U_ i = f^{-1}(V_ i)$. By Lemma 20.44.2 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 20.43.4. Thus we conclude. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)