(In following we are working in "classical" complex setting: i.e. all involved schemes are considered to be varieties over $k=\mathbb{C}$)  

Let $X \subset \mathbb{P}^n$ be irreducible *surface* and $L $ some general $(n-3)$-plane disjoint from $X$.
We consider now the projection map $f = \pi\_L: X \to \mathbb{P}^2$ from $L$.


Let $Z \subset X$ be the locus of *critical points*, i.e. points $p \in X$ that are singular or such that the tangent map $df\_p: T\_pX \to T\_{f(p)}\mathbb{P}^2$ has a non trivial kernel
$\text{Ker}(df\_p) \neq 0 $. It's a fact that this is a proper closed subset of $X$.  

Let $B = f(Z)$ be it's image in $\mathbb{P}^2$ which by properness of $f$ has to be closed too. $B$ is from naive point of
view given as union of a plane curve and a finite collection of some points.


There is an interesting remark in Harris' book *[Algebraic Geometry](http://www.springer.com/it/book/9780387977164)*, on p 290 in proof of Prop. 18.10
that one knows moreover that $B$ has *no* $0$-dimensional components. Unfortunately this fact is not needed for the rest
of the proof, so the author not gave a justification for this.


*Question:* How to check this last statement? Ot looks rather counterintuitive, since thinking of contractions of curves one might expect $B$ might have $0$-dimensional components.


Thoughts: It seems not to be true for all generically finite projective morphisms $f: X \to \mathbb{P}^2$ of surfaces,
since for example the blowup at center $0 \in \mathbb{P}^2$ gives $B=0$. So if the statement is true then it must be based on
special structure of projections from linear subspaces, but I not find an argument. I also not know how "deep" this result is, ie which "tools" are required to prove it. Even if the quoted book is for undergrades, Harris often quotes there some facts going far beyond the scope of the book, and so I'm not sure if this property of $B$ is a result of advanced research or can it be seen with rather "elmentary" tools.



Generically finite projection $\pi_L: X \to \mathbb{P}^2$ from plane $L$ and critical points
The morphism $f$ is not just generically finite but finite. I think this resolves all the difficulties.


To see this, since $f$ is proper it just suffices to check it is quasi-finite, and thus to check the fiber over a point can't contain a curve. But the fiber over a point is a $(n-2)$-plane containing $L$, and any curve in an $(n-2)$-plane intersects every $(n-3)$-plane, so if such a curve were contained in the fiber it would contradict the assumption that $X$ is disjoint from $L$.


Alternately, we can argue that $\mathbb P^n \setminus L \to \mathbb P^2$ is affine and thus $f$ is affine, and, since affine and proper, is finite.


