TITLE: Bijective application
QUESTION [1 upvotes]: Let $F(2,4)$ be the set of all $4\times 2$ matrices of rank $2$, topologized as a subspace of $\mathbb{R}^{n\times k}$, and $\sim$ the equivalence relation
$$A \sim B \text{ iff there is a matrix } g \in GL(2,\mathbb{R}) \text{ such that } B = Ag .$$
For any $4\times 2$ matrix $A$, let $A_{ij}$ be the $2\times 2$ submatrix consisting of its $i$-th row and $j$-th row. Define
$$V_{ij}=\{A\in F(2,4): \det(A_{ij})\neq 0\},$$
which is an open set in $F(2,4)$. Define $U_{ij} = V_{ij}/\sim$ with the quotient topology. 
I want to show that the map $\overline{\phi}_{12}\colon V_{12} \to \mathbb{R}^{2\times 2}$ defined by $\overline{\phi}_{12}(A)=A_{34}A_{12}^{-1}$
induces a homeomorphism $\phi_{12}\colon U_{12} \to \mathbb{R}^{2\times 2}$.
MY TRY:
$\overline{\phi}_{12}$ is continuous because it consists of matrix product and matrix inversion. 
Now, for the injectivity, if $\overline{\phi}_{12}(A)=\overline{\phi}_{12}(B)$, then
$$A_{34}=B_{34}B_{12}^{-1}A_{12}.$$
However, $B_{12}^{-1},A_{12}\in GL(2,\mathbb{R})$, so its product too. Thus, $A_{34}=B_{34}g$ for some $g\in GL(2,\mathbb{R})$. It's enough?
For the surjectivity, let $C\in \mathbb{R}^{2\times 2}$. Then exist a regular matrix $A_{12}^{-1}$ so that $C=A_{34}A_{12}^{-1}$ (equivalence by columns). It's correct?
No idea for the inverse.

REPLY [1 votes]: By the universal property of the quotient map $q\colon V_{12}\to U_{12}=V_{12}/\sim$, the continuous map $\overline\phi_{12}\colon V_{12}\to\mathbb{R}^{2\times2}$ defined by $A\mapsto A_{34}A_{12}^{-1}$ induces a unique continuous map $\phi\colon U_{12}\to\mathbb{R}^{2\times2}$ satisfying $\overline\phi=\phi\circ q$.

To do so, we first show that $\phi$ is well-defined, that is,
$$
\overline\phi(A)=\overline\phi(B) \text{ for all $A,B\in V_{12}$ such that $A\sim B$} 
$$
Since $B=Ag$ for some $g\in GL(2,\mathbb{R})$, we have
$$
\overline\phi(B)=B_{34}B_{12}^{-1}=(A_{34}g)(A_{12}g)^{-1}=A_{34}A_{12}^{-1}=\overline\phi(A)
$$
OP already showed that $\phi$ is injective.
For surjectivity of $\phi$ it suffices to show $\overline\phi$ is surjective. For $C\in\mathbb{R}^{2\times2}$, choose a matrix $A\in V_{12}$ such that $A_{12}=I$, the identity $2\times2$ matrix and $A_{34}=C$. Then
$$
\overline\phi(A)=A_{34}A_{12}^{-1}=A_{34}=C
$$
Now let us define the inverse of $\phi$. We first define $\overline\psi\colon\mathbb{R}^{2\times2}\to V_{12}$ as follows:
$$
\text{For $C\in\mathbb{R}^{2\times2}$, choose $\overline\psi(C)\in V_{12}$ such that $\overline\psi(C)_{34}=C$ and $\overline\psi(C)_{12}=I$.}
$$
and then define $\psi=q\circ\overline\psi$ which is of course continuous. It remains to show that $\psi\circ\phi=\mathrm{id}_{U_{12}}$ and $\phi\circ\psi=\mathrm{id}_{\mathbb{R}^{2\times2}}$.


Claim A. $\overline\psi\circ\overline\phi(A)\sim A$ for all $A\in V_{12}$.
Proof. Let $B=\overline\psi\circ\overline\phi(A)$. Then $B_{34}=A_{34}A_{12}^{-1}$ and $B_{12}=I=A_{12}A_{12}^{-1}$, and thus $B=Ag$ for $g=A_{12}^{-1}$, that is, $A\sim B$.
Claim B. $\overline\phi\circ\overline\psi(C)=C$ for all $C\in\mathbb{R}^{2\times2}$.
Proof. Since $\overline\psi(C)_{34}=C$ and $\overline\psi(C)_{12}=I$, we have $\overline\phi\circ\overline\psi(C)=CI^{-1}=C$.


Show that $\psi\circ\phi=\mathrm{id}_{U_{12}}$. (In fact (2) follows from this.)


Proof. For any $[A]\in U_{12}$ choose $A\in V_{12}$ such that $\phi([A])=\overline\phi(A)$. Then
  $$
\psi\circ\phi([A]) = (q\circ\overline\psi)\circ\overline\phi(A) = [A]
$$
  since $\overline\psi\circ\overline\phi(A)\sim A$ by Claim A.


Show that $\phi\circ\psi=\mathrm{id}_{\mathbb{R}^{2\times2}}$. (In fact (3) follows from this.)


Proof. For any $C\in\mathbb{R}^{2\times2}$, Claim B implies
  $$
\phi\circ\psi(C) = \phi\circ(q\circ\overline\psi)(C) = \overline\phi\circ\overline\psi(C) = C
$$
{"set_name": "stack_exchange", "score": 1, "question_id": 2685596}