Let $\mathbb{F}$ be an algebraically closed field of characteristic zero, let $k\leq n$ be positive integers, and let $V=\mathbb{F}^n$. One can view the Grassmannian $G(k,V)$ of $k$-planes in $V$ as an algebraic variety via the Plücker embedding, which sends a $k$-dimensional subspace $W \in G(k,V)$ to $[w_1 \wedge w_2 \wedge \dots \wedge w_k] \subseteq \mathbb{P}(\bigwedge ^k (V))$, where $w_1,\dots, w_k \in V$ are any vectors such that $W=\text{span}\{w_1,\dots, w_k\}$. Indeed, it is easily seen that this map is injective, and its image is the set of decomposable elements of $\mathbb{P}(\bigwedge ^k (V))$, which forms a projective variety. In particular, this gives a Zariski topology on $G(k,V)$.
Another natural way put a topology on $G(k,V)$ is via a quotient topology. Note that $G(k,V)$ is naturally bijective as a set with $S/\equiv$, where $S \subseteq \mathbb{F}^{n k}$ is the set of full-rank $n \times k$ matrices, and $\equiv$ is the equivalence relation on $S$ given by: $A \equiv B$ if and only if there exists $X \in GL_k(\mathbb{F})$ such that $A=BX$. The bijection is given by $W \mapsto (w_1,\dots, w_k)$. The set $S$ forms an algebraic variety, and hence carries a (Zariski) topology.
Does the topology on $G(k,V)$ induced by the quotient topology on $S/\equiv$ agree with the topology on $G(k,V)$ induced by the Plücker embedding?