Two versions of constant rank theorem Let $f:\mathbb{R}^n\to\mathbb{R}^d$ be a $C^1$ function with the property that there is an $x_0\in\mathbb{R}^n$ such that on an open neighborhood of $x_0$, then rank of $Df(x)$ is a constant, say $j$. Then the constant rank theorem says that there are $C^1$ diffeomoprhisms $\alpha,\beta$ such that $f=\beta^{-1}\circ\pi\circ\alpha$, where $\pi(x_1,...,x_n)=(x_1,...,x_j,0,...,0)$.
My question is, I've seen a similar statement for this theorem but with $\pi$ with replaced by $Df(x)$, so $f$ locally looks like $Df(x)$. How are these statements equivalent? Is it because if $T:\mathbb{R}^n\to\mathbb{R}^d$ has rank $j$, then there is a $C^1$ change of coordinates so that in these new coordinates $T$ is $\pi$? If this is the case, how would you show this?
 A: Your guess is correct, and in fact the change of coordinates can be chosen to be linear.  This is just pure linear algebra: if $T:\mathbb{R}^n\to\mathbb{R}^d$ has rank $j$, then there are invertible linear maps $S:\mathbb{R}^n\to\mathbb{R}^n$ and $U:\mathbb{R}^d\to\mathbb{R}^d$ such that $UTS=\pi$.  To prove this, pick a basis $f_1,\dots,f_j$ for the image of $T$, and choose $e_1,\dots,e_j\in\mathbb{R}^n$ such that $T(e_i)=f_i$ for each $i$.  Also, choose $e_{j+1},\dots,e_n$ to be a basis of the kernel of $T$.  Then $e_1,\dots,e_n$ will be a basis for $\mathbb{R}^n$.  Finally, extend the $f_i$ to a basis $f_1,\dots,f_d$ of $\mathbb{R}^d$.
Now observe that we have $T(e_i)=f_i$ for $i\leq j$ and $T(e_i)=0$ for $i>j$.  That is, $T$ behaves just like $\pi$, if we use the $e_i$ and $f_i$ as our bases for $\mathbb{R}^n$ and $\mathbb{R}^d$.  So if $S$ is the change-of-basis map sending the standard basis of $\mathbb{R}^n$ to the $e_i$ and $U$ is the change-of-basis map sending the $f_i$ to the standard basis of $\mathbb{R}^d$, then $UTS=\pi$.
