$\newcommand\C{\mathbb C} \newcommand\A{\mathbb A}$My base field throughout is $\C$. Let $m<n$ and suppose I have dominant morphisms $$f:\A^m \to \A^m \qquad g:\A^n \to \A^n$$ with finite general fibers, i.e. there is a nonempty Zariski open $U\subset \A^m$ such that for all $P\in U$ we have $\#f^{-1}(P) < \infty$ (similarly for $g$) and such that given the canonical projection $$\pi : \A^n \to \A^m \qquad (x_1,\dots, x_n) \mapsto (x_1,\dots, x_m)$$ we have the commutative diagram $f\pi = \pi g$.
I believe then that this is true (but please correct me and help me with an argument if not):
There is a nonempty Zariski open set $U\subset \A^n$ such that for all $P\in U$ one has the equality $$\{\pi(Q) : Q\in g^{-1}(P)\} = f^{-1}(\pi(P)) $$
So my question is if this is correct or are there some easy counterexamples? I feel that we just have to make some degree argument on the morphisms.
I am not sure if this is true. But here is a motivational example that is too long to give in a comment...
Consider $m=1,n=2$ and suppose $f(x)=x^2$. Let us try to see if there is a $g$ that satisfies all the assumptions, but the equality does not hold. In this case we know the coordinates of $g$ satisfy $g_1(x,y)=x^2$ and $g_2(-x,y) \ne g_2(x,y)$ for all $(x,y)$ in a nonempty Zariski open set for which $x\ne 0$. One checks that this would imply that $g_2(x,y)$ is just a power of $x$ multiplied by a constant $c\in \mathbb C$. But the map $g:(x,y) \mapsto (x^2,cx^k)$ is clearly not dominant!