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Let $f\colon X\to Y$ be a morphism of complex algebraic varieties. I am interested in those $f$ that have the following property: For any smooth point $x\in X$, the image $y=f(x)$ is a smooth point of $Y$. Certainly, open immersions have this property, closed immersions certainly do not.

Which well-known, larger classes of morphisms have this property?

Jesko Hüttenhain
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  • This is true for etale morphisms over any base field--this is obvious in the complex world though since complex analytically $f$ being etale implies it's locally an open embedding. It's also true for smooth morphisms, by using that it's obviously true from projection from the trivial bundle combined with the etale result above. These are both iff statements, not shockingly--smooth if and only if image point is smooth. PS, both of these statements are only obviously true to me if we're working over a perfect field because I'm thinking about when the image of a regular local ring is regular. – Alex Youcis Nov 16 '16 at 11:53
  • @AlexYoucis I'm probably misunderstanding your statement. If $X$ is smooth over $\mathbb C$, $Y$ is smooth over $\mathbb C$, and $X\to Y$ is a closed immerison, the image of each (smooth) point is smooth. However, $X\to Y$ is not smooth.... – Ariyan Javanpeykar Nov 16 '16 at 12:57
  • @AriyanJavanpeykar Hey Ariyan! Did I claim that somewhere? Oh, I see. The 'smooth if and only if image point is smooth' meant 'domain point smooth if and only if codomain point smooth'. Is that what you meant? – Alex Youcis Nov 16 '16 at 12:59
  • @AlexYoucis Hi Alex, yes, that's the statement I meant. You are saying that, if $f:X\to Y$ is smooth, then domain point is smooth if and only if target point smooth? (If that's the case, then I agree.) – Ariyan Javanpeykar Nov 16 '16 at 13:03
  • @AriyanJavanpeykar Yep, that's what I meant. I agree it's not clear from my first comment--I ran out my allowed character limit so I cut some words. :P – Alex Youcis Nov 16 '16 at 13:04
  • Regular immersions also (I only mean $x$ smooth implies $y$ smooth, not the reverse), and so, composing them with smooth morphisms, you obtain more or less the class locally complete intersection morphisms. – A.G Nov 16 '16 at 15:08

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The largest class of morphisms which takes smooth points to smooth points is local complete intersection morphisms. Indeed the property of being smooth is an open condition, so if source and target point are smooth then the locally the map of varieties is a map of smooth varieties—and any map of smooth varieties is local complete intersection.

(You can prove this factoring the morphism $X \to Y$ through its graph in $X \times Y$—a closed embedding of smooth varieties is always a regular embedding so your map $X \to Y$ is factoring a regular embedding with the smooth projection $X \times Y \to Y$.)

As noted a local complete intersection morphism takes smooth points to smooth points, so that is the complete characterization.

Simone Weil
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