Proving that grassmannians are smooth manifolds I'm trying to show that real grassmannians $G(k, n)$ are smooth manifolds of dimension $k(n-k)$.
The problem is set in this way:
Identify the set of all real matrices with $n$ rows and $k$ columns with $\mathbb{R}^{kn}$. Then consider the following equivalence relation on $\mathbb{R}^{kn}$:
$M_1 \sim M_2$ if and only if there exists $L \in GL(k,\mathbb{R})$ such that $M_1=M_2L$. Then define $G(k, n)=\mathbb{R}^{kn}/ \sim$. $G(k, n)$ becomes a topological space whose topology is the quotient topology. 
Now, for each subset of $J$ of $\{1, \ldots, n\}$ consisting of $k$ indices, consider the matrix $M_J$ obtained from $M \in \mathbb{R}^{kn}$ taking only the the $k$ rows whose indices are in $J$ and consider the set
$$U_J=\{[M]\in G(k, n): detM_J \neq 0\}. $$
Finally, for each $J$, define the map from $U_J$ to $\mathbb{R}^{k(n-k)}$ given by
$$\phi_J([M])= (MM^{-1}_J)_{J^c}$$
where $J^c$ denotes the complement of $J$ in $\{1, \ldots, n\}$.
I think I've managed to show that this map is a homeomorphism, but I'm in trouble when I hace to prove that transitions map between the charts are smooth, i.e. for each $J_1,J_2$ the composition $\phi_{J_1} \circ \phi_{J_2}^{-1}$ is smooth. 
Any help would be greatly appreciated. 
 A: Let $X\in\mathbb{R}^{(n-k)k}$. I will first denote $[M]=\phi_{J_2}^{-1}(X)$ and rewrite $\phi_{J_1}([M])$ as follow:
\begin{align}
\label{comp:1}
\phi_{J_1}([M])&=(MM_{J_1}^{-1})_{J_1^c} \\
&=M_{J_1^c}M_{J_1}^{-1} \\
&=M_{J_1^c}\cdot\frac{1}{\det M_{J_1}}\text{adj}(M_{j_1})
\end{align}
In case it is not clear, the second step is justified by noting that $M_J$ can be obtained from $M$ by a pre-multiplication of the $k\times n$ matrix $E_J$, whose $j$-th column is equal to the standard basis column vector $e_r\in\mathbb{R}^k$ if $j=i_r\in J$ and is equal to $0$ otherwise. For example, if $n=7$, $k=3$ and $J=(1,4,6)$, then
\begin{align}
E_j=\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 
\end{pmatrix}
\end{align}
Then $M_J=E_JM$ as you can check easily. Thus
\begin{align}
(MM_{J_1}^{-1})_{J_1^c}&=E_{J_1^c}(MM_{J_1}^{-1}) \\
&=(E_{J_1^c}M)M_{J_1}^{-1} \\
&=M_{J_1^c}M_{J_1}^{-1}
\end{align}
which justified the second step.
By rewriting $\phi_{J_1}([M])$ in this way, it is clear that each entry of $\phi_{J_1}([M])$ is a rational function of entries of $M$. In particular, the denominator of this rational function is equal to $\det M_{J_1}$, which is nonzero since $[M]\in U_{J_1}$. Therefore, each entry of $\phi_{J_1}([M])$ depends smoothly on the entries of $M$.
Note that the whole computation does not depend on the choice of the representative of the equivalence class $[M]$. So we may choose the representative $M$ to be a convenient one. When you were proving the surjectivity of $\phi_{J_2}$, I believe you have already seen how to obtain such $M$ from $X$: Simply inserting back the row vectors $(0 \cdots 1 \cdots 0)$ to the position indexed by $J_2$. For example, when $n=4$, $k=2$ and $J_2=\{1,3\}$, if
\begin{align}
X=\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\end{align}
then
\begin{align}
M=\begin{pmatrix}
1 & 0 \\
a & b \\
0 & 1 \\
c & d
\end{pmatrix}
\end{align}
Thus entries of $M$ are either $0$, $1$ or entries of $X$. Hence by choosing such $M$, it follows that
\begin{align}
\phi_{J_1}([M])=\phi_{J_1}\circ\phi_{J_2}^{-1}(X)
\end{align}
is a matrix whose entries depend smoothly on the entries of $X$. This proves the smoothness of $\phi_{J_1}\circ\phi_{J_2}^{-1}$.
