Let $\pi \colon X\to Y$ be a morphism, where $X$ is projective. Suppose the existence of an open subset $U\subset X$ such that $\pi^{-1}(\pi(U))=U$, that $\pi$ is injective on $U$, and that its derivative is injective.


I would like to say that $\pi$ restricts to an isomorphism between $U$ and its image. And I suspect that this should be classical. Does someone knows a proof or a reference for this? (Or a counterexample if it is false..)


Remark: Note that if we omit the assertion $\pi^{-1}(\pi(U))=U$ or $X$ projective, we can easily get counterexamples. We take $\mathbb{P}^1\to C$ the normalisation of a nodal cubic curve $C\subset\mathbb{P}^2$, and let $U\subset \mathbb{P}^1$ be an open subset which is the complement of one point going to the singular point of $C$, then $\pi$ does not restrict to an isomorphism from $U$ to its image (this latter being $C$, which is singular). Taking $X=\mathbb{P}^1$ or $X=U$ shows that both assertions are necessary.



Birational maps and open subsets inducing isomorphisms
The useful theorem here is "a proper, universally injective, unramified morphism is a closed immersion." see: <http://math.columbia.edu/~dejong/wordpress/?p=1358>


$X$ being projective and $\pi^{-1}(\pi(U))=u$ ensures that $U$ is proper over $\pi(U)$, by the base change property of properness.


You need to assume that $U \to \pi(U)$ is universally injective, not just injective, because of field extensions. The natural map $\operatorname{Spec} \mathbb C \to \operatorname{Spec} \mathbb R$ is injective on the point set but not an isomorphism onto its image.


"Unramified" should be what you mean by "the derivative is injective". 


Thus $U\to \pi(U)$ is a closed immersion. A closed immmersion is indeed an isomorphism onto its image.


However, note that this image need not be open. One could take $X$ to be a point - a point is certainly a projective variety - the map $X \to Y$ the inclusion of a point, and $U=X$. 


