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Lemma 17.4.3 states that if $f:X \rightarrow Y$ is a morphism of ringed space, $G$ is a locally free $O_Y$-module, then $f^*G$ is a locally free $O_X$ module.


Suppose that $G$ is a locally free $O_Y$ module, hence it is a free $O_Y|_U=O_U$ module. We have an induced map $$f^{-1}(U) \rightarrow U$$ given by restriction of morphism of ringed spaces.

Hence, as $\bigoplus O_U \simeq G|_U$, $f^*:Mod(O_U) \rightarrow Mod(O_{f^{-1}(U)})$ is functorial, $f^*(G|_U) = (f^*G)|_{f^{-1}(U)}$, we have isomoprhism, $$ \bigoplus O_{f^{-1}(U)} \simeq (f^*G)|_{f^{-1}(U)} $$

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  • $\begingroup$ How do you define $f^* G(U)$ ? $\endgroup$ – reuns May 17 at 18:03
  • $\begingroup$ Hint: for any morphism $f:X\to Y$, the pullback of the structure sheaf $f^*\mathcal{O}_Y$ is the structure sheaf $\mathcal{O}_X$. Do you see where to go from here? $\endgroup$ – KReiser May 18 at 0:15
  • $\begingroup$ @KReiser that's exactly what I used in the sequence. I took a presentation of $O_X$ as $$ 0 \rightarrow \bigoplus O_Y \rightarrow G \rightarrow 0 $$ But puilling back does not guarantee exactness(?) $\endgroup$ – CL. May 18 at 10:07
  • $\begingroup$ You're thinking too big here. What does locally free mean? Every point $y\in Y$ has an open neighborhood $U\subset Y$ on which the sheaf $\mathcal{F}$ is isomorphic to $\mathcal{O}^n$ - if the pullback of the structure sheaf is the structure sheaf and the preimage of an open set is open, what can you say about $f^*\mathcal{F}$ on $f^{-1}(U)$? $\endgroup$ – KReiser May 18 at 17:20
  • $\begingroup$ Thanks, is what I wrote correct? $\endgroup$ – CL. May 19 at 5:45

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