Is the set of images of an open subset of full-rank matrices an open subset of the Grassmannian?

Question

$\DeclareMathOperator\Gr{Gr}$ Let $\Gr(k,n)$ be the set of $k$-dimensional subspaces in affine space $\mathbb{A}^n$ over an algebraically closed field. If $U\subseteq (\mathbb{A}^n)^{\times k}$ is an open subset of $n \times k$ matrices of rank $k$, is $\{\text{Im}(u): u \in U\} \subseteq \Gr(k,n)$ open?

Here is my convoluted attempt at a solution. Via the Plücker embedding, one can view $\Gr(k,n)$ as the set of totally decomposable elements of $\bigwedge^k_n$, which can be shown to form a projective variety.

$\Gr(k,n)$ can also be viewed as the set of $n \times k$ matrices of rank $k$, quotiented by the (equivalence relation induced by) the right action of $\text{GL}(k)$. Since $\text{GL}(k)$ is a reductive algebraic group, I believe this quotient is again an affine variety with induced topology given by the quotient topology. Is this correct? If so, then I don't think it's hard to prove that these two varieties corresponding to $\Gr(k,n)$ are isomorphic, which would seem to imply my desired result.

Related question.

Answer 1

The map $U \to \mathbb{G}(k,n)$ that you are interested in is flat, and a flat finitely presented map is well known to be open.

Flatness can be checked in several ways: for example, the map is generically flat, by the generic flatness theorem. But it is also $\mathrm{GL}_n$-equivariant, and the action of $\mathrm{GL}_n$ on $U \to \mathbb{G}(k,n)$ is transitive. Alternatively, it follows from the fact that the fibers are equidimensional, $\mathbb{G}(k,n)$ is regular and $U$ is also regular, hence Cohen--Macaulay.