Let $X$ be a smooth projective variety in $\mathbb{P}^n$, over the field of complex numbers or an algebraically closed field of characteristic $0$.

EDIT (After Stefano's response): Assume $\dim X < n-1$, $\deg X > 1$ and $X$ is nondegenerate.

In particular, $X$ is not a hypersurface. Let $P\in X$ be a general point, fix a general hyperplane in $\mathbb{P}^n$ and consider the map $\varphi_P:X\setminus \{P\}\to \mathbb{P}^{n-1}$ obtained by "projecting from $P$". Let $Y$ be the closure of the image $\varphi_P(X)$ in $\mathbb{P}^{n-1}$.

(1) For general $P$, is the map $\varphi_P:X\setminus\{P\}\to Y$ going to be quasi-finite?,

(2) For general $P$, is the map $\varphi_P:X\setminus\{P\}\to Y$ going to be birational?

I am guessing that these are true. I could not find a reference for these two statements in standard texts like Harris, Hartshorne or Shararevich . Can someone please point out a reference for these. If someone could sketch a proof or explain a counterexample then that would be more helpful. Thanks.


I think that to have (1) or (2) you need to put extra assumptions (I am not sure whether it will work even then).

One condition that you need is that $X$ is non-degenerate, i.e. it is not contained in any hyperplane. Indeed you can consider a plane cubic $E$ as a curve in $\mathbb{P}^3$. Then, the projection will be $2:1$, since every line will meet two residual points. Thus, the map is not birational.

Similarly, you can consider a quadric surface $Q \subset \mathbb{P}^3$ as embedded in $\mathbb{P}^4$. Then, through $P$ there are exactly two rulings ($Q$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$), and the map contracts each ruling to a point. Thus, the map is not quasi-finite.

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