If $\dim(X)=\dim(Y)>0$, and $X\to Y$ is onto, does affineness of $X$ imply affineness of $Y$?

Question

Suppose you have a pair of integral varieties $X$ and $Y$ such that $\dim(X)=\dim(Y)>0$, and there exists a surjective morphism $X\to Y$ between them. I was wondering, if $X$ is affine, does this force $Y$ to also be affine?

Answer 1

This looks plausible, but the answer is no.

Everything is over an algebraically closed field. Let $Y=\mathbf P^1$, which is not affine. Take a smooth conic curve $C \subset \mathbf P^2$ and a point $p$ on $C$. Projection away from $p$ gives a surjective map $X :=C \setminus \{p\} \rightarrow Y$. But any smooth conic is isomorphic to $\mathbf P^1$, so $X$ is just $\mathbf A^1$!

(I learned this nice example from an answer of Georges Elencwajg which I am too lazy to track down right now.)