Given a morphism $\varphi:Z\to Y$ over $X$, where $X,Y,Z$ are all varieties over an algebraically closed field of characteristic zero. Let $q:Z\to X$ be the structure morphism. Suppose that $q$ is smooth and $\varphi$ is constant on the fibers of $q$. Then is there a morphism $\rho:X\to Y$ such that $\varphi=\rho\circ q$?
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I think you want a properness assumption. Otherwise take $Z = Y = \mathbb{A}^1 \subset \mathbb{P}^1 = X$), with $\varphi$ the identity and $q$ and the structure morphism of $Y$ the inclusion maps. – Jake Levinson Aug 27 '22 at 22:07
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How about assuming $q$ is surjective? – schuming Aug 28 '22 at 01:30
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Can't you just fix @JakeLevinson example to be surjective by taking a disjoint union with a point? With properness hypotheses this is basically the Rigidity Lemma as in Mumford's book on abelian varieties, isn't it? – Alex Youcis Aug 28 '22 at 02:11