The usual version of the "rigidity lemma" in algebraic geometry says something like this:
If $U, \, V, \, W$ are algebraic varieties, with $U$ proper, and $f: U \times V \rightarrow W$ is a morphism such that $f(\{u_0\} \times V)$ is a point for one $u_o \in U$, then $f( \{u\} \times V)$ is a point for every $u \in U$, and moreover there is a morphism $g: V \rightarrow X$ such that $f=g \circ \pi_V$.
Question: is this still true if we replace the product by an algebraic fibre space?
More precisely, say $\pi: X \rightarrow V$ is a proper surjective morphism of varieties. (If necessary the varieties can be smooth, and $\pi$ can be flat.) Suppose that $f: X \rightarrow W$ is a morphism such that $f(X_u)$ is a point for every $u \in U$. Does there exist a morphism $g: V \rightarrow W$ such that $f = g \circ \pi$?
Edit: cant_log provides an example to show this is not true in general.
I can't comment. Here are a few remarks:
I guess $V=U$ in the last paragraph.
Consider a morphism $g: W\to U$ which is bijective and flat but not an isomorphism. Let $f: X=\mathbb P^1_W\to W$ and let $\pi=g\circ f$. Then $\pi$ maps any fiber $X_s$ to one point of $W$ (namely the point $g^{-1}(s)$), but there is no $U\to W$.
A version of rigidity lemma I know is under the hypothesis $\pi$ is proper with geometrically integral fibers.