On pg. 134 of Joe Harris' Algebraic Geometry: A First Course, the author writes the following:
Let $X\subseteq P^n$ be an irreducible variety. The natural place to look for a finite map to a projective space is among the projections. Now, when we project $X$ from a point $p\in P^n$ not on $X$, the fibers are necessarily finite---since $p\notin X$, no line through $p$ can meet $X$ in more than a finite number of points.
Can somebody explain as to what is the author trying to say here? What is meant by projecting $X$ from a point $p$ not on $X$?
Thanks.
Pick $p \notin X$, for simplicity $p = [1:0: \dots : 0]$. The projection is the map $\pi : \Bbb P^n \to \Bbb P^{n-1}, [x_0 : \dots : x_n] \mapsto [x_1: \dots : x_n]$. Then take $Y := p(X)$.
The general procedure is as follows : pick $p \in \Bbb P^n$ and an hyperplane $H$ with $p \notin H$. Then, for each $x$, $\pi(x)$ is the intersection of $H$ and the line between $p$ and $x$.