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Suppose $X,Y$ are varieties, $Y$ is projective and $f: X \to Y$ is a locally trivial fibration with fibre $\mathbb{P}^1$. Then there exists an open covering $\{U_i\}_i$ of $Y$ such that $f^{-1}(U_i) \cong U_i \times \mathbb{P}^1$ for each $i$.

Question: why is $X$ a projective variety?

Balerion_the_black
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  • If $Y$ is regular, then $f$ is a projective bundle by Hartshorne Ex. II.7.10. So $X$ is projective because $Y$ is. – Andrea Jun 10 '18 at 15:49

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