Let $A,B$ be two varieties (integral finite type separated scheme) over a field $k$ and let $f : A \to B$ be a proper morphism.
Is it true that $f$ is surjective if and only if $dim(\overline{f(A)}) = dim(B)$.
If it is true I would be grateful for a reference and if it's not what would the correct statement be (and i'd still be interested in a reference) ?
Also is it true more generally ?
If $f$ is proper, then $f$ is universally closed, hence f is closed. So $\overline{f(A)}=f(A)$. That $f(A)=B$ does indeed follow from their dimensions being equal as well as $B$ being integral: a closed subscheme of equal dimension must contain an irreducible component.
The argument above shows that the statement is true any time $f(A)$ is a closed subscheme of $B$. The natural inclusion $\mathbb{A}^1 \hookrightarrow \mathbb{P}^1$ shows that it need not be true if $f(A)$ is not closed in $B$.