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Let $f: X\to Y$ be a holomorphic map between two complex manifolds, and $\mathcal{E}$ a finite rank complex holomorphic vector bundle (or even a general sheaf) over $Y$. Let $E$ be the corresponding locally trivial $\mathcal{O}_Y$-module (the objects $\mathcal{E}$ and $E$ determine each other by the GAGA theorems of Serre from the 1950s). Here $\mathcal{O}_Y$ is the sheaf of complex holomorphic functions on $Y$.

Ques:Is $f^*E|_{f^{-1}(y)}$ a trivial sheaf (i.e. $=\mathcal{O}_{f^{-1}(y)}$)? How can we give a precise proof?

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Question: "Is $(f^∗E)|_{f^{−1}(y)}$ a trivial sheaf (i.e. $=\mathcal{O}_{f^{−1}(y)}$)? How can we give a precise proof?"

Note: The user speaks of "(finite rank) holomorphic vector bundle" and "(locally) trivial (finite rank) sheaf". There is an "equivalence of categories" between the category of finite rank complex holomorphic vector bundles on $X$ and the category of finite rank locally trivial $\mathcal{O}_X$-modules, hence the vector bundle $\mathcal{E}$ and the locally trivial sheaf $E$ determine each other.

Answer: No, not in general. Assume for simplicity that $X,Y$ are projective (hence algebraic varieties) and let $i: X_y \rightarrow X$ be the inclusion map. Let $g:=f \circ i$ be the composed map. If $E$ is finite rank locally trivial (or "locally free") sheaf of rank $r$ on $Y$ we may assume $E$ to be algebraic and then $g^*(E)\cong (f^*E)_{X_y}$ is a locally trivial sheaf on $X_y$ of the same rank $r$. Hence $g^*(E)$ is not trivial in general.

If $f^{\#}: \mathcal{O}_Y \rightarrow f_*\mathcal{O}_X$ is the structure map, 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 comutes with direct sums. Hence the pull back $f^*(E)$ is locally trivial of the same rank $r$.

Note: There is an "equivalence of categories" between the category of finite rank complex holomorphic vector bundles on $X$ and the category of finite rank locally trivial $\mathcal{O}_X$-modules, hence the above argument may be used for arbitrary holomorphic maps of complex manifolds. Note also that $f^{-1}(y):=X_y$ is not a point in general - moreover the fiber $f^{-1}(y) \subseteq X$ is not a complex manifold in general.

Comment: "@hm2020 I mean there should be an explicit description of X, Y, f and E, such that the above statement is wrong."

Answer: Let $E:=\mathcal{O}_Y^2$. It follows $f^*(E)_{X_y} \cong \mathcal{O}_{X_y}^2 \neq \mathcal{O}_{X_y}$.

Comment: Thank you. What do you mean by "locally trivial sheaf "? Do you mean "locally free sheaf "? If so, Can you say a little more on proving the claim that g∗(E) is a locally trivial sheaf on Xy of the same rankr? Do you use the isomorphism formula like g−1(E)x≅Eg(x) or the projection formula? – Lelong Wang Oct 17 at 16:03

Answer: The claim follows from the fact that topological inverse image preserves direct sums of sheaves - see the above proof.

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  • $\begingroup$ Thank you. What do you mean by "locally trivial sheaf "? Do you mean "locally free sheaf "? If so, Can you say a little more on proving the claim that $g^{*}(E) $ is a locally trivial sheaf on $X_{y}$ of the same $\operatorname{rank} r$? Do you use the isomorphism formula like $g^{-1}(E)_x \cong E_{g(x)}$ or the projection formula? $\endgroup$ Oct 17, 2021 at 16:03
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    $\begingroup$ I think there needs to be a concrete counterexample to disprove the claim $\endgroup$
    – Notone
    Oct 17, 2021 at 17:04
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    $\begingroup$ I agree with @Notone : Locally trivial hence not trivial in general is not really convincing, and I am pretty sure that in the situation of the OP, then the claim holds : Let $i:\{y\}\to Y$ and $i':f^{-1}(y)\to X$ be the inclusions and $g=f|_{f^{-1}}(y):X_y\to\{y\}$ the canonical map. The question is whether $i'^*f^*E*$ is trivial, but $i'^*f^*E=(fi')^*E=(ig)^*E=g^*i^*E$. Since $i^*E$ is trivial (as any vector bundle over a point) and since the pullback of a trivial bundle is trivial, the pullback is indeed trivial. $\endgroup$
    – Roland
    Oct 17, 2021 at 20:38
  • $\begingroup$ @Roland Thank you. I think your argument is strictly correct. $\left(f i^{\prime}\right)^{*} E=(i g)^{*} E$ is very important. $\endgroup$ Oct 18, 2021 at 6:37
  • $\begingroup$ @Notone: "I think there needs to be a concrete counterexample to disprove the claim" Take any locally trivial sheaf $E$ of rank $r\geq 2$. $\endgroup$
    – hm2020
    Oct 18, 2021 at 8:25

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