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Let $f \colon X \to Y$ be a generically injective, finite morphism between projective varieties over $\mathbb{C}$, with $Y$ non-singular. Does $f$ have a section i.e., a morphism $g:Y \to X$ such that $f \circ g:Y \to Y$ is the identity?

If not true in general, is there any known condition under which $f$ has a section?

EDIT. The map $f$ is assumed to be dominant.

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    $\begingroup$ Presumably you want to assume that $f$ is dominant, or else this is trivially false. If $f$ is dominant and generically injective, then $f$ is birational. A finite, birational morphism with normal target $Y$ is an isomorphism by Zariski's Main Theorem (classical formulation). $\endgroup$
    – Jason Starr
    Commented Apr 2, 2017 at 13:34
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    $\begingroup$ @JasonStarr Thank you. Yes, I assume $f$ to be dominant. Can we say something if we drop the assumption on finite morphism and replace it with projective morphism. The picture I have in mind is that when $f$ is generically injective except for a codimension $2$ locus where the fibers are rational curves. $\endgroup$
    – user45397
    Commented Apr 2, 2017 at 14:02
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    $\begingroup$ With "projective morphism" it seems to me that the answer is no in such a generality. Take the blow-up $\pi \colon X \to \mathbb{P}^2$ of the plane at at a point: of course it has no regular sections. $\endgroup$
    – Francesco Polizzi
    Commented Apr 2, 2017 at 14:25

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If $Y$ is normal, then every finite generically injective dominant morphism $f:X \to Y$ is an isomorphism. Indeed, take an open affine subset $U = Spec~ A$ of $Y$. Then $f^{-1}(Y) = Spec~B$ where $B$ is integral over $A$. Since $A$ is integrally closed and $A,B$ have the same field of fractions, $A = B$.

So the answer to your question is: ``Yes, always!''

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    $\begingroup$ "A,B have the same field of fractions" is true over complex numbers, but it requires an additional explanation why is so since over a field of positive characteristic it is simply not true. Indeed, take a Frobenius morphism. What we can prove in general is that $f$ is an universal homeomorphism, by proving it is injective, what is equivalent to being finite, radicial, and surjective. Then, in char=0, radicial means birational, and we are done. $\endgroup$
    – P. Grabowski
    Commented Apr 22, 2022 at 15:22

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