Constant rank map means image locally a graph Suppose that $f:U \to \mathbb{R}^n$, where $U$ is an open subset of $\mathbb{R}^m$, is such that $Df(p)$ has rank $k$ for all $p\in U$. Show that for each $p\in M$, there exists a neighborhood $V$ of $p$ in $U$ such that $f(V)$ is a smooth $k$-manifold.
(Here we may take as our definition of a $k$-manifold that $f(V)$ can be locally represented as the graph of a smooth function $\mathbb{R}^k \to \mathbb{R}^{n-k}$.)
$Df(p)$ has rank $k$, so there is a $k \times k$ submatrix of $Df(p)$ which is invertible. Suppose it is the top-left submatrix. Now there is a local diffeomorphism between the $k$ variables $x_1, \dotsc, x_k$ in the source and the $k$ variables $y_1, \dotsc, y_k$ in the target. We need to show somehow that we can parameterize the image $f(V)$ by these variables.
Any ideas?
 A: There's probably a much faster, more elegant way, but if I haven't assumed something about ODE theory that's false, this should work. So you know that you can assume that the upper left $k\times k$ block of $Df(p)$ is invertible. Write $$Df(x) = \begin{bmatrix} A(x) & B(x) \\ C(x) & D(x) \end{bmatrix}.$$ Then, by first order ODE theory, (I think) we can locally construct a solution $g: \mathbb{R}^m \rightarrow \mathbb{R}^m$ that satisfies $A^{-1}(g(z_1,\ldots, z_n))$ is the upper $k \times k$ block of $Dg(z)$ at each point. Precomposing with this function reduces $Df$ to something of this form: $$ \begin{bmatrix} I & B(x) \\ C(x) & D(x) \end{bmatrix}.$$ But the same argument allows us to pre and post-compose with matrices that will do the row reduction to arrive at a matrix of the form  $$\begin{bmatrix} I & 0\\0 & D(x) \end{bmatrix}.$$ Since these functions we are pre- and post-composing with are locally diffeomorphisms, we never change the rank of $f$ in sufficiently small neighborhoods, and so $D(x)$ must vanish in the new coordinates, and we are done, as $Df$ is constantly the above matrix in some coordinates, and so is just projection.
