Proof verification on the openness of the space of matrices of full rank Let be $m<n$ and $M_m(m\times n, \mathbb{R})$ be the set of the matrices $m \times n$ of full rank $m$. 
I want to show that $M_m(m\times n,\mathbb{R})$ is an open subset of $M(m\times n,\mathbb{R})$ 
I reasoned as follows. Let $A$ be an element in $M_m(m\times n,\mathbb{R})$. Then $A$ has some $m\times m$ submatrix whose determinant is non zero. Let be $j_1,\dots,j_m$ the columns of one such submatrix of $A$, and let $\pi\colon M(m\times n,\mathbb{R})\to M(m,\mathbb{R})$  be the map who takes a matrix $B$ and associates the $m\times m$ submatrix obtained by taking the columns $j_1,\dots, j_m$. Then $\pi$ is smooth, and then continuous. I already know that $GL(m,\mathbb{R})$ is open in $M(m,\mathbb{R})$. So I conclude that $\pi^{-1}(GL(m,\mathbb{R}))$ is an open subset of $M(m\times n,\mathbb{R})$ contained in $M_m(m\times n,\mathbb{R})$ and containing $A$.
So $M_m(m\times n,\mathbb{R})$ is a neighbor of all its points, then it is open. 

Is my proof correct? 

 A: Obviously you are right.
I give here something different. I show $S:=$ set of all matrices of rank strictly less than $m$ , is a closed subset of $M(m\times n, \mathbb{R})$ . So let $\{A_p\}_{p\in \Bbb N}$ be a sequence in $S$ converging to some $A\in M(m\times n, \mathbb{R})$. If possible let $A\in M_m(m\times n, \mathbb{R})$. Then we have a matrix $B\in M_m(n\times m, \mathbb{R})$ such that $AB=I_m$. Now $\{A_pB\}_{p\in \Bbb N}$ converges to $AB$. But notice that $det(A_pB)=0$ for each $p$ , since $rank(A_pB)≤rank(A_p)<m$ , also $rank(AB)=rank(I_m)=1$.
Now $det$ is a continuous function i.e. $0=det(A_pB)\rightarrow det(AB)=1$ as $p\rightarrow \infty$, contradiction. Therefore $S$ is closed in  $M(m\times n, \mathbb{R})$ i.e. $ M_m(m\times n, \mathbb{R})$ is open in $ M(m\times n, \mathbb{R})$.
Existence of $B\in M_m(n\times m, \mathbb{R})$ for $A\in  M_m(m\times n, \mathbb{R})$ with $AB=I_m$ can be given as : $ dim(column\ space(A))=dim(row\ space (A))=m$ , so that j-th column $e_j$ of $I_m$ can be written as $Ab_j=e_j$ where $b_j$ is column matrix, for each $j=1,....,m$. Therefore define $B=[b_1,....,b_m]$.
