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Let $f,g:X\dashrightarrow\mathbb{P}^N$ be two rational maps from a complex smooth irreducible projective variety $X$ to a projective space.

Suppose that for every general point $x\in X$ we have $\overline{f^{-1}(f(x))}=\overline{g^{-1}(g(x))}$.

Is true that $\overline{f(X)}$ and $\overline{g(X)}$ are isomorphic (resp. projectively equivalent) ?

If not, is true that $\overline{f(X)}$ is smooth if and only if $\overline{g(X)}$ is smooth?


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up vote 2 down vote accepted

The answer all your questions is no. Let $X = \mathbb{P}^1$ and $N=2$. Let $f:X\to \mathbb{P}^2$ be $f(x:y) = (x:y:0)$ and $g:X\to \mathbb{P}^2$ be $g(x:y) = (x^2y:x^3:y^3)$. Then $f$ and $g$ are both bijective, but the image of $g$ is the singular curve $x_0^3 = x_1^2 x_2$.

EDIT. My guess is that when $f$ and $g$ are regular and both images are normal then the answer that $f(X)$ and $g(X)$ are isomorphic could be yes thanks to Zariski's Main Theorem, but I don't see an argument that would show this.

EDIT. Suppose that $f$ and $g$ are both regular. Let $Y_0 = f(X)$ and $Y_1 = g(X)$. Denote the image of $X$ in $Y_0\times Y_1$ under $(f, g)$ by $Z$. Then by your assumption $Z$ projects bijectively onto both $Y_0$ and $Y_1$. If $Y_0$ and $Y_1$ are normal, then by ZMT we have $Y_0 = Z = Y_1$. In any case (even if $f$ and $g$ are rational) we get that $Y_0$ and $Y_1$ are birational.

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Read carefully my question. I am not required $\overline{f^{-1}(f(x))}\simeq \overline{g^{-1}(g(x))}$ but $\overline{f^{-1}(f(x))}= \overline{g^{-1}(g(x))}$. – gio Oct 14 '12 at 20:47
In my example we have $f^{-1}(f(x)) = g^{-1}(g(x)) = x$ since $f$ and $g$ are both bijective... – Piotr Achinger Oct 14 '12 at 21:23
@gio: Piotr Achinger answered your question. In his example the equality of the closures does hold (not just for generic points but for every point $x\in X$). Or am I missing somethig? – Qfwfq Oct 14 '12 at 21:30
Pior, I'm sorry, you're right. However, I also mean equality of schemes... – gio Oct 14 '12 at 21:59

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