Morphism of finite type schemes not surjective. Is there a closed point in the complement of the image?

Question

I have the following proplem: Let $k$ be an algebraically closed field and let $X,Y$ be schemes of finte type over $k$.

Now let $f:X\to Y$ be a morphism of schemes that is not surjective.

Question: Is there a closed point in $Y\setminus f(X)$?

What I tried so far: I assumed $X,Y$ are affine and tried to play a little bit on the ring level using that the coordinate rings are jacobian rings. But I haven't got very far yet. Any help would be appreciated.

Answer 1

By Chevalley's theorem, $f$ is surjective if and only if $f$ is surjective on closed points (see there).

Assume all closed points of $Y$ are in $f(X)$. Let $y \in Y$ be a closed point. The scheme-theoretic fiber $X_y $ of $y$ is nonempty and is a closed subscheme of $X$. Thus $X_y$ contains a closed point $x$ and by definition we have $f(x)=y$. But $x$ is also a closed point of $X$. Consequently, $f$ is surjective on closed points.