This question is from my course of smooth manifolds.
Let $G(k,n)$ denote $\{k\text{-dimensional vector space in } \mathbb{R}^n\}$, which is equal to $\{n \times k \text{ matrix of rank }= k\} \big/ \sim$
where $A \sim B \iff \exists g \in GL_k(\mathbb{R}) ~\text{such that} ~B = Ag$.
I want to show that the equivalent relation $\sim$ is open, which means that the canonical (projection) map is open.
The below is my trial :
Let $F(k,n)$ denote $\{n \times k \text{ matrix of rank }= k\}$, and $\pi$ denote the canonical (projection) map derived from $\sim$.
Assume that $U$ is open in $F(k,n)$.
To show that $\sim$ is open, it is enough to show that $\pi^{-1}(\pi(U))$ is open in $F(k,n)$.
$$\pi^{-1}(\pi(U)) = \{Ag : A \in U, ~ g \in GL_k(\mathbb{R})\} = \bigcup_{g \in GL_k(\mathbb{R})} Ug .$$
Thus if $Ug$ is open for all $g \in GL_k(\mathbb{R})$, then $\pi^{-1}(\pi(U))$ is open, and so I obtain the fact that $\sim$ is open.
I tried to prove it directly.
For any $A \in Ug$, we know $Ag^{-1} \in U$. Because $U$ is open, there exists $\epsilon >0$ such that $B_\epsilon(Ag^{-1}) \subset U$.
Hence $B_\epsilon(Ag^{-1})g \subset Ug$, and we know $A \in B_\epsilon(Ag^{-1})g$.
Thus if we can find $\delta > 0$ satisfying $B_\delta(A) \subset B_\epsilon(Ag^{-1})g$, then we can conclude that $Ug$ is open.
However, I'm stuck to proceed.
I would really appreciate to concern and advice.
The map $g: M_{n, k}(\mathbb R) \to M_{n, k}(\mathbb R)$ given by $A \mapsto Ag$ is continuous; you know the rules of matrix multiplication, the entries of $Ag$ end up being polynomials in the entries of $A$, polynomials are continuous.
Moreover, it has a continuous inverse $A \mapsto A g^{-1}$, so it is a homeomorphism. Homeomorphisms are open maps, so $Ug$ is open.