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Let $f:X \to Y$ be a surjective morphism of irreducible varieties (over an algebraically closed field) such that, for each $y \in Y$, $f^{-1}(y)$ is a vector space of dimension $r$. Is these informations enough to say that this triple is a vector bundle?

I am tempted to say that is not because of the compatibility conditions, but, I don't know any counterexamples. On the other hand the condition of both varieties to be irreducible and the fibers being of the same dimension give some hope that this could be true.

Any help will be appreciated.

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One nice counterexample is $$ (\mathbb{P}^1 \times \mathbb{P}^1) \setminus \Delta $$ (where $\Delta$ is the diagonal). It is fibered over $\mathbb{P}^1$, each fiber isomorphic to a 1-dimensional vector space, but the projection has no sections, so it is not a vector bundle.

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