Pullback of locally free sheaves is locally free

Question

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.

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

Answer 1

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 $x$.

Answer 2

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.

Answer 3

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.