# Pullback of locally free sheaves is locally free

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, $$\bigoplus O_{f^{-1}(U)} \simeq (f^*G)|_{f^{-1}(U)}$$

• How do you define $f^* G(U)$ ? – reuns May 17 at 18:03
• 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? – KReiser May 18 at 0:15
• @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(?) – CL. May 18 at 10:07
• 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 $f^{-1}(U)$? – KReiser May 18 at 17:20
• Thanks, is what I wrote correct? – CL. May 19 at 5:45