Let $X,Y,Z$ be projective varieties, and let $f:X\rightarrow Y$, $g:X\rightarrow Z$ be dominant morphisms. Assume that all the fibers of $g$ have the same dimension and are connected.
If there exists a point $z_0\in Z$ such that $f(g^{-1}(z_0))\subseteq Y$ has dimension $k$ is it true that $f(g^{-1}(z))\subseteq Y$ has dimension $k$ for any $z\in Z$ ?
When $k=0$, that is $f(g^{-1}(z_0))$ is a point, this is true. It is known as the rigidity lemma.
EDIT: I've just realized that this holds under somewhat weaker assumptions. It is not necessary that the fibers of $g$ are connected. EDIT#2: Apparently, in my previous edit I weakened the conditions too far... properness of $g$ is back as it is needed, but connectivity of fibers is not as it is not. I think it is OK now. :)
Cool question. Actually, I think this is true!
Generalized Rigidity Lemma Let $X,Y,Z$ be (irreducible) varieties, $f:X\rightarrow Y$ a morphism and $g:X\rightarrow Z$ a surjective proper morphism whose fibers are of the same dimension $n$. Then $f(g^{-1}(z))\subseteq Y$ has the same dimension for any $z\in Z$.
Proof [This proof is inspired by the proof of the original Rigidity Lemma given in [Kollár-Mori, 1.6]] Let $z_0\in Z$ and $k=\dim f(g^{-1}(z_0))$. Further let $W=\mathrm{im}(f\times g)\subseteq Y\times Z$ with projection $p:W\to Z$. Let $h=f\times g: X\to W$. Notice that since $g$ is proper, and $g=p\circ h$, so $h$ is also proper and hence $W=\mathrm{im}\, h\subseteq Y\times Z$ is closed and so it is a variety. Then for all $z\in Z$, $p^{-1}(z)=h(g^{-1}(z))$ and hence $\dim p^{-1}(z_0)=k$. By semi-continuity of fiber dimension there exists a non-empty open set $z_0\in U\subseteq Z$ such that for all $z\in U$ $\dim p^{-1}(z_0)\leq k$.
This implies that $p^{-1}U\subseteq W$ is a non-empty open set (in fact, since $g$ is surjective, this is dense) such that for all $w\in p^{-1}U$, $\dim h^{-1}(w)\geq n- k$. However, then by semi-continuity of fiber dimension again the same is true for all $w\in W$.
Now fix a $z\in Z$ and consider the morphism $h:g^{-1}(z)\to p^{-1}(z)$. By assumption $g^{-1}(z)$ is $n$-dimensional and we have just observed that the fibers of this map have dimension at least $n-k$, so the image has dimension at most $k$.
In other words, we proved that for any $z\in Z$, $\dim h(g^{-1}(z))\leq k$. Finally, observe that if there was a $z_1\in Z$ for which $\dim h(g^{-1}(z_1))< k$, then by repeating the same proof with $z_0$ and $z_1$ exchanged we would get that for any $z\in Z$, $\dim h(g^{-1}(z))< k$, but this would contradict the assumption that $\dim h(g^{-1}(z_0))= k$. Therefore we must have equality for every $z\in Z$. $\square$
