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

Bryan Shih
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  • How do you define $f^* G(U)$ ? – reuns May 17 '19 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 '19 at 00:15
  • @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(?) – Bryan Shih May 18 '19 at 10:07
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    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 '19 at 17:20
  • Thanks, is what I wrote correct? – Bryan Shih May 19 '19 at 05:45

3 Answers3

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Answer fully inspired from @Kreiser's comments.

Let $x\in X, y=f(x)$. Let $V \subseteq Y$ be an open neighborhood of $y$ and $U=f^{-1}(V)$.

$G$ is a locally free $\mathcal O_X$-module, hence $G|_{V}\cong \bigoplus\limits_{i=1}^n \mathcal O_Y|_V$.

Pullback commutes with direct sum hence $(f^*G)|_{U}\cong \bigoplus\limits_{i=1}^n f^*(\mathcal O_Y|_V) \cong \bigoplus\limits_{i=1}^n \mathcal O_X|_U $ in a neighborhood of $x$.

Conjecture
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The question is local in nature and one can reduce to the case of affine schemes with free sheaves on them. In this case, it actually reduces to simple facts in commutative algebra as follows:

Given a morphism of affine schemes $f:spec(B)\rightarrow spec(A)$, and a free $\mathcal{O}_{spec(A)}$-module $\mathcal{F}$, we know that $\mathcal{F}=\widetilde{M}$ for some free $A-$module $M=\bigoplus_{i\in I} A$. In this case, we know (this is proven in Hartshorne for example) that $f^*(\mathcal{F})=f^*{\widetilde{M}}=\widetilde{M\otimes_{A}B}$, and $M\otimes_AB=B\otimes_A(\bigoplus_{i\in I} A)=\bigoplus_{i\in I}B\otimes_AA=\bigoplus_{i\in I}B$ which is a free $B$-module, hence $f^*\mathcal{F}$ is a free $\mathcal{O}_{spec(B)}$-module. We just used the facts that tensor product commutes with direct sum.

FNH
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Question: (Prove the following statement at the stack-project:) "[Lemma 17.4.3][1] 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."

Answer: Assume $E$ is a locally free sheaf on $Y$ of rank $r$. The pull back is defined using the $f^{-1}$-functor and this functor "commutes with direct sums". This fact implies the claim:

If $f^{\#}: \mathcal{O}_Y \rightarrow f_*\mathcal{O}_X$ is the map of structure sheaves, you get a corresponding map

$$ \tilde{f}: f^{-1}(\mathcal{O}_Y) \rightarrow \mathcal{O}_X$$

and you define $f^*(E):=\mathcal{O}_X \otimes_{f^{-1}(\mathcal{O}_Y)} f^{-1}(E)$. If $U \subseteq E$ is an open set where $E_U \cong \mathcal{O}_U^r$, let $F: f^{-1}(U) \rightarrow U$ be the induced map. It follows (we work with presheaves since we are taking a tensor product)

$$f^*(E)_{f^{-1}(U)} \cong \mathcal{O}_{f^{-1}(U)}\otimes_{F^{-1}(\mathcal{O}_U)}F^{-1}(E_U) \cong $$

$$\mathcal{O}_{f^{-1}(U)}\otimes_{F^{-1}(\mathcal{O}_U)} F^{-1}(\mathcal{O}_U^r) \cong \mathcal{O}_{f^{-1}(U)}\otimes_{F^{-1}(\mathcal{O}_U)} F^{-1}(\mathcal{O}_U)^r \cong $$

$$\mathcal{O}_{f^{-1}(U)}^r$$

since topological pull back commutes with direct sums. Hence the pull back $f^*(E)$ is locally trivial of the same rank $r$. If $E$ is locally free of arbitrary rank a similar argument applies.

hm2020
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