MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f : S \to X$ be a dominant morphism of smooth complex compact surfaces. Let $C \subset S$ be a smooth curve such that $df$, seen as a map from $T_S \to T_X$, is generically of rank $1$ along $C$ and that in a neighborhood of $C$, $df_x$ is an isomorphism outside $C$. Suppose that $C$ is not contracted by $f$, is it true that $f_{|C}$ is an immersion?

One might want to decompose it into two sub-questions:

  1. Can one show that the rank of $df$ is exactly $1$ along $C$?
  2. Suppose the rank of $df$ is exactly $1$ along $C$, can we show that $f_{|C}$ is an immersion?
share|cite|improve this question
your surfaces are compact? – auniket Jun 12 '13 at 10:08
Yes, they are compact. Thanks for asking the question. – Ikabruob Jun 12 '13 at 10:24
What do you mean by generically rank 1 along $C$? Do you mean that on all but a finite number of points of $C$ the differential of $f$ (seen as a map from $S\to X$ has rank one or that the restriction of $f$ to $C$ (thus seen as a map from $C$ to $X$) has injective differential at all but finitely many points of $C$? – diverietti Jun 12 '13 at 12:27
It means on all but a finite number of points of $C$, the rank of $df_x$ is 1, seen as map from $T_xS \to T_xX$. – Ikabruob Jun 12 '13 at 12:49
up vote 6 down vote accepted

The question in essentially local on $X$, so the compacteness assumption is irrelevant. The answer to both questions is no: consider $S\subset \mathbb C^3$ defined by $z^3+xz+y=0$ and the map to $X=\mathbb C^2$ given by $(x,y,z)\mapsto (x,y)$.

The ramification curve $C$ is smooth (given by $z^3+xz+y=3z^2+x=0$) while its image has a cusp at $x=y=0$. The differential of the map has rank $\ge 1$ everywhere

share|cite|improve this answer
Great! Thanks for your counter-ex Rita. – Ikabruob Jun 13 '13 at 14:25
You're welcome! – rita Jun 13 '13 at 15:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.