Locally free sheaves on affine bundles

Question

Let $f:Y \to X$ be an affine bundle (for example, the total space of a vector bundle) and $s: X \to Y$ be a global section. Given any locally free sheaf $E$ on $Y$, is $E$ isomorphic to $f^*s^*E$? Roughly, I would think this is true as there should be a morphism from $E$ to $f^*s^*E$ (locally given by the canonical map from $M$ to $M \otimes_{A[X_1,....,X_n]} A \otimes_A A[X_1,...,X_n]$ given by $m \mapsto m \otimes 1$, where the first morphism $A[X_1,...,X_n] \to A$ sends $X_i$ to 0 and the second morphism $A \to A[X_1,...,X_n]$ is the canonical inclusion). Furthermore, applying $- \otimes k(y)$ to this morphism will keep it injective, which will imply that the cokernel of $E \to f^*s^*E$ is flat. By rank argument this would imply the two locally free sheaves are isomorphic. Is this correct? Probably, there could be some gluing issue.

NB. If necessary, assume that $X$ is non-singular and the underlying field is $\mathbb{C}$.

Answer 1

In general this is not true. Consider $X=\mathbb{P}^1$ and $Y=X\times \mathbb{A}^1$, the trivial bundle. Since $\operatorname{Ext}^1(O_X(1),O_X(-1))=k$ (say an algebraically closed field), we can identify elements of $k$ with closed points of the affine line. This gives a vector bundle $E$ on $Y$ which fits into an exact sequence $0\to O_X(-1) \to E\to O_X(1)\to 0$. Restricted to $X\times \{0\}$, $E=O_X(-1)\oplus O_X(1)$, but for $t\neq 0$, restricted to $X\times \{t\}$, it is just $O_X\oplus O_X$. Thus, it cannot be of the form you suggest.

Answer 2

Question: "Furthermore, applying $−⊗k(y)$ to this morphism will keep it injective, which will imply that the cokernel of $E→f^∗s^∗E$ is flat. By rank argument this would imply the two locally free sheaves are isomorphic. Is this correct?"

Answer: Assume $\pi: Y:=Spec(B) \rightarrow X:=Spec(A)$ has a section $s:X \rightarrow Y$ with $\pi \circ s =Id_X$. It follows we get maps of rings

$$A \rightarrow^f B \rightarrow^g A$$

with $g \circ f =Id_A$ hence the map $g:B \rightarrow A$ is surjective with $I:=ker(g)$ and $A \cong B/I$. The map $\phi:= f \circ g$ satisfies

$$ \phi^2 =\phi.$$

Hence you get an idempotent endomorphism of rings $\phi: B \rightarrow B$. The pull back $\phi^*E$ satisfies

$$\phi^*(E):=B\otimes_{B/I} E/IE$$

and since $I \subseteq ann(\phi^*(E))$ it follows $B/I\otimes_B \phi^*(E)=\phi^*(E)$. We get

$$\phi^*(E) \cong B/I\otimes_B \phi^*(E) \cong B/I\otimes_B B\otimes_{B/I} E/IE \cong E/IE$$

hence if $(0)\neq I$ and $I \subsetneq ann(E)$ it follows $\phi^*(E) \neq E$.

Question: "Thanks for the nice example. Can we at least say that there is a non-trivial morphism from $E$ to $f^∗s^∗E$?"

There is a canonical map

$$q:E \rightarrow \phi^*(E)\cong E/IE$$

and $q$ is an isomorphism iff $I \subseteq ann(E)=(0)$ - If $E$ is projective $ann(E)=(0)$.