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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.

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What do you mean that the derivative is injective? –  name Aug 11 '12 at 18:24
What do you mean by "image" ? Scheme-theoretic image ? - but then it would be closed... My guess is that you are probably looking for something like Lemma 2.4 in the article "Abelian varieties" by Milne in the volume edited by Cornell-Silverman. –  Damian Rössler Aug 11 '12 at 19:38
derivative: linear application from the Tangent space to the tangent space of the image. derivative is injective corresponds to say that there is no ramification at the point. For Lemma 2.4 of "Abelian varieties" of Milne, I found something on Hermitian forms on vector spaces.. jmilne.org/math/CourseNotes/AV.pdf –  Jérémy Blanc Aug 12 '12 at 8:48
This is not the text published in the book edited by Cornell-Silverman... –  Damian Rössler Aug 13 '12 at 7:43
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1 Answer

up vote 4 down vote accepted

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 $U=X$.

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Thanks for the answer, but in your theorem, you need that the source is projective. In my question, I want an isomorphism of $U$ (open) with its image, which will be an OPEN immersion. It cannot be closed of course. –  Jérémy Blanc Aug 12 '12 at 8:07
(I think) he's saying that $\pi_U : U \to \pi(U)$ is a closed immersion. –  Tom Bachmann Aug 12 '12 at 10:41
I don't need that the source is projective, I need the map to be proper. The map $X \to Y$ is proper because $X$ is projective, and the map $U \to \pi(U)$ is proper because it's the base change of that map by $\pi(U) \to Y$. There is no reason to suspect $U \to Y$ will be open, unless you assume $X\to Y$ surjective or something like that. $U\to \pi(U)$ is open, of course, because we show it's an isomorphism. –  Will Sawin Aug 12 '12 at 13:05
Ok. thanks for the correction. –  Jérémy Blanc Aug 12 '12 at 13:26
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