Is a quasi-finite endomorphism of affine $n$-space finite? Let $\mathbb A^n$ be the affine space over $\mathbb C$ and let $f:\mathbb A^n\rightarrow \mathbb A^n$ be a quasi-finite morphism. In particular it is dominant. 

Is $f$ necessarily a finite morphism? 

 A: If it is finite, then it is surjective. (Indeed, finite morphisms are proper.)
Thus, it suffices to exhibit a quasi-finite non-surjective endomorphism. Let $f:\mathbb{A}^2\to \mathbb{A}^2$ be given by $(x,y)\mapsto (g(x,y),h(x,y))$. Composing with an automorphism if necessary, $f$ is non-surjective if and only if $Z(g)\cap Z(h) =\emptyset$. (This means that the origin is not in the image.)
So, how about $g= x+y$ and $h=x+y+1$? Then $f$ is given by 
$f(x,y) = (x+y, x+y+1)$. 
That's not surjective, hence not finite. And it is even injective. 
A: $\newcommand{\bA}{\mathbb{A}}\newcommand{\bC}{\mathbb{C}}$Take $n=2, f:\bA^2_{\bC}\to\bA^2_{\bC}$ defined by $$f(x, y)=(x(1+xy),y)$$
For a point $(a, b)\in \bA^2(\bC)$ the preimage $f^{-1}(a, b)$ consists of one point if $b=0$ or $1+4ab=0$ and it has two points in all other cases. So $f$ is quasi-finite, but it is not proper: the limit of $f(1/t,-t)$ as $t$ tends to $0$ exists. Hence $f$ isn't finite.
This example appears e.g. as Example 23 in The set of points at which the polynomial mapping is not proper, by Z. Jelonek. I've learned about it through a startling excercise (Problem 3 here http://achinger.impan.pl/rigid/ps10a.pdf) from a course by Piotr Achinger.
