Dominant morphisms between projective varieties

Question

Suppose $f : X\to Y$ is a finite surjective morphism of projective integral complex varieties, and let $g : X'\to X$ and $h : Y'\to Y$ be surjective birational morphisms from smooth projective irreducible varieties $X'$ and $Y'$.

Suppose there is a morphism $f' : X'\to Y'$ making the diagram commute, i.e. such that $h \circ f' = f\circ g$ as maps $X'\to Y$.

Is $f'$ dominant? (hence surjective)

For more context, I have a diagram as in the question where $g,h$ are resolutions of singularities, but I have no control on how the regular locus of $X$ and $Y$ are related, so it is unclear if one can produce open subsets in $X'$ and $Y'$ such that $f'$ restricts to a surjection between the two, although I'd think irreducibility of all the four varieties should be key. I must be missing something simple.

Answer 1

There is an open subset $U\subset X$ such that $g\colon g^{-1}(U)\to U$ is an isomorphism, and there is an open subset $V\subset Y$ such that $h\colon h^{-1}V\to V$ is an isomorphism. Now there is a commutative diagram

$$\require{AMScd} \begin{CD} g^{-1}(U\cap f^{-1}(V))=g^{-1}(U)\cap f'^{-1}(h^{-1}(V)) @>{f'}>> h^{-1}(V)\\ @V{g}VV @V{h}VV \\ U\cap f^{-1}(V) @>{f}>> V, \end{CD}$$ where the vertical maps are isomorphisms. Since $f$ is dominant, $f(U\cap f^{-1}(V))$ is dense in $X$. Thus the image $g^{-1}(U)\cap f'^{-1}(h^{-1}(V))$ under $f'$ is dense, and so $f'$ is dominant.

Answer 2

Here is a proof of a slightly different flavour: As all varieties are projective, all the maps are closed. Now suppose $f'$ is not dominant, then its image $Z = f'(X')$ is a closed subvariety of $Y'$, hence $\dim Z < \dim Y'$. As $Y' \to Y$ is birational, both have the same dimension, so also $\dim h(Z) < \dim Y$, which means for the total image $(h \circ f')(X') = (f \circ g)(X') \neq Y$. But that contradicts the dominance and closedness of the maps $f$ and $g$, which requires that $(f \circ g)(X') = f(X) = Y$.