I want to show that Grassmannian is a manifold in a specific case the $2$-planes in $\mathbb{R}^4$.
I'm in the following context:
$G(2,4)$ are the $2$-planes in $\mathbb{R}^4$ that we can identify with an matrix of two vectors that generate the plane (they are not unique). Considering $L(2,4)=\{A \in M_{4 \times 2 } : \text{rank} A =2\}$ and equivalence relation $A \sim B $ iff exist $g \in Gl_2(\mathbb{R})$ such that $B=Ag$.
$$G(2,4) \cong L(2,4)/\sim$$
Given $A \in M_{4 \times2}$ define $A_{ij}$ submatrix of $A$ eliminating rows $i$ and $j$ of $A$ and $V_{ij} = \{ A \in L(2,4)| A_{ij} \text{ is invertible}\}$. We can see that $V_{ij}$ is open in $L(2,4)$. Therefore $U_{ij}=\pi(V_{ij})$ is open in $G(2,4)$, where $\pi$ is a natural projection in quotient.
In this context we define the possible charts $(U_{ij}, \phi_{ij})$, $\phi_{ij}: U_{ij} \to \mathbb{R}^4$ $$\phi_{ij}([A])=A_{kl}A_{ij}^{-1}$$ where $\{1,2,3,4\}=\{i,j,k,l\}$. I'm having a hard time showing that it's a homeomorphism