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Let $f : X \to Y$ be a morphism of noetherian $k$-schemes. Under what condition is the functor $f^* : \textbf{Coh}\ Y \to \textbf{Coh} \ X$ exact on locally free finite rank sheaves?

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  • $\begingroup$ A short exact sequence $0 \to E_{1} \to E_{2} \to E_{3} \to 0$ of finite locally free sheaves on $Y$ is locally split, so its inverse image $0 \to f^{\ast}E_{1} \to f^{\ast}E_{2} \to f^{\ast}E_{3} \to 0$ is also locally split (in particular exact). $\endgroup$ Jul 12, 2019 at 18:09
  • $\begingroup$ It's enough that the sequence is "locally split" (as opposed to (globally) split). By this I mean that there is a Zariski covering $Y = \bigcup_{i \in I} Y_{i}$ for which the restriction $0 \to E_{1}|_{Y_{i}} \to E_{2}|_{Y_{i}} \to E_{3}|_{Y_{i}} \to 0$ is split. The Euler sequence is not split but it is locally split. $\endgroup$ Jul 12, 2019 at 18:17
  • $\begingroup$ You are right. I deleted my comments right after I realized this. Thank you for your comments! $\endgroup$
    – Ruben
    Jul 12, 2019 at 18:18

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