Hi.


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?


Thanks.



Equality of rational maps
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.


