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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?
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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
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1 Answer

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

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