Here's something that seems to be true, or at least I hope it to be true, but I'm unable to prove it:
Let $S$ be a $k$-rational surface and $B$ a curve, both projective, smooth and geometrically integral, and $f:\ S\ \longrightarrow\ B$ nonconstant. Let $C\subset S$ be a $k$-rational curve such that $f\vert_C$ is nonconstant. Then the map $S\times_BC\ \longrightarrow\ S$ is dominant.
I'm only concerned with the case that $f$ is an elliptic fibration, but this property does not seem relevant here. If there are more unnecessary conditions still I'd be glad to hear so.
My first thought was that the map $f\vert_C$ is dominant, and hence its base change $S\times_BC\ \longrightarrow\ S$ is dominant as well. But being dominant is unfortunately not preserved under base change in general. Are there conditions (that are satisfied here) under which it is preserved? Or is an entirely different approach required?
The map $S\times_BC\ \longrightarrow\ C$ has a section coming from the inclusion $C\ \hookrightarrow\ S$ and the identity on $C$. This seems like a strong property, but I do not see a relation between the one morphism having a section and the other morphism being dominant. Help is appreciated.
Update: I've found a paper asserting the following without any proof:
Let $X\ \longrightarrow\ Y$ be a dominant rational projective morphism of $k$-schemes, and let $Z\subset X$ be a subscheme which dominates $Y$. Then the base change $X\times_YZ\ \longrightarrow\ X$ is dominant.
It seems the statement might concern only integral schemes. Unfortunately I still see no relation between the dominance of the maps.