Let $Z = \{(L,P) | P \supset L \} \subset G(n,k+1) \times G(n,k)$ be incidence correspondence of $k$ planes contained in $k+1$ planes. Consider $p_1$, the projection onto the first factor. Is $p_1$ open (when restricted to Z)? (Of course $p_1$ is open on the domain $G(n,k+1) \times G(n,k)$, because it is flat, but this is not the question I am asking.)
If so, a hint would be nice.
Motivation:
Suppose we are trying to prove a statement $S$ about $k+1$ planes. If we can prove that for a generic $k$ plane $L$, and a generic $k+1$ plane $P$ containing this $k$ plane, $P$ has $S$, can we prove that a generic $k+1$ has $S$?
I don't think so in general, but if the failure of $S$ on $P$ above $L$ varies suitably "algerbaically" with $L$, then we get an open subset of $Z$ (the incidence correspondence defined above) of pairs $(P,L)$ where $P$ has $S$, and we would like to project this along $p_1$ to get an open subset of the Grassmanian.
(If there is a better way to get that in this condition, the set of pairs $(P,L)$ is open in $Z$, I would be happy to know...)
I think something like this is used in Lazarsfeld's Positivity in Algerbaic Geometry Book I, 3.3.1. I don't know what the precise statement is.
Everything is over $\mathbb{C}$.
Edit: I think the projection restricted to Z is flat, because the fibers are all isomorphic to projective spaces of the same dimension (the k planes in a given k+1 plane). And I think it is true that a map into an integral Noetherian scheme is flat if the fibers all have the same Euler characteristics, but I want to find a reference for the precise statement.
If I am not misunderstanding your setup, the map $p_1: Z\to \text{G}(k + 1, n)$ is a projective bundle, i.e. not only is it flat and proper, but it is actually locally a product in the Zariski topology. (On $\text{GL}(k + 1, n)$ there is a natural "tautological" vector bundle of rank $k + 1$, and your $Z$ is the projective bundle of hyperplanes in this.)
So you do not need anything fancy e.g. about general criteria for flatness. In this case it is clear. (Incidentally, I do not think flatness is implied by all the fibers having the same Euler characteristic—e.g. this is true of the blowing up of $\mathbb{P}^2$ at a point. You want them to have the same Hilbert polynomial with respect to some ample divisor, e.g.)