Let $f\colon X\to Y$ be a surjective algebraic map between two projective $k$-varieties, where $k$ is algebraically closed. Let $n=\dim(X),\,m=\dim(Y)$. Suppose furthermore that X,Y are irreducible. From surjectivity, we have - $n\geq m$. Is it always true that we can find a closed subset $X_0\subset X$ s.t. $\dim(X_0)=m$ and $f|_{X_0}\colon X_0\to Y$ is surjective? If not, under what conditions is it true and can we forget about the irreducibility hypothesis?
Thanks!
If $\dim X>\dim Y$, taking $\mathcal{O}(1)$ be a very ample line bundle on $X$, then in the linear system $|\mathcal{O}(1)|$ there is a divisor dominates $Y$.
This is because $|\mathcal{O}(1)|$ intersects the fiber of $f$ non-trivially.
By this, we can cut down the dimension of $X$ and hence get an affirmative answer to your problem.
On the other hand, irreducibility hypothesis is not necessary because we can just take irreducible component, then the problem is reduced to the irreducible case.