Let $W,X$ be nice schemes over algebraically closed $k = \overline{k}$ (separated, geometrically reduced, so that $k$-points are dense), $S$ be an arbitrary $k$-scheme, and $\phi, \psi : W \times_k S \to X \times_k S$ be two $S$-morphisms that agree on the vertical fibers of $W \times_k S$ corresponding to $k$-points of $W$ (a dense subset of the domain). Are these morphisms necessarily equal?
If $S$ is not reduced, then $W \times_k S$ is not either, so the reduced-to-separated theorem (as in Vakil, for example) doesn't apply. But my intuition thinks that since all the nonreducedness is ``in the $S$ direction," the morphisms should agree. Reducedness is not a relative notion, but here it feels like it should be!
For an example, consider $W = X = \mathbb{A}^1 = \operatorname{Spec} k[t]$ and $S = \operatorname{Spec} k[\epsilon]/\epsilon^2$, and $\phi$ the identity. Any map of $k[\epsilon]$-algebras $\psi^\# : k[\epsilon, t] \to k[\epsilon, t]$ is determined by $t \mapsto p(t) + \epsilon q(t)$, and by checking infinitely many fibers $t = c$, we can see that it must be equal to $t$. Hence any $\psi$ as above is equal to $\phi$. We can see that nothing interesting could happen in the $S$-direction because the morphism was over $S$.