Rudin states the following:
9.32 Theorem: Suppose $m,n,$ are nonnegative integers, $m\geq r,n\geq r$, $F$ is a $C^1$ mapping of an open set $E\subset R^n$ into $R^m$, and $F'(x)$ has rank $r$ for every $x\in E$.
Fix $a\in E$, put $A=F'(a)$, Let $Y_1$ be the range of A, and let $P$ be the projection in $R^m$ whose range if $Y_1$. Let $Y_2$ be the null space of $P$.
Then there are open sets $U$ and $V$ in $R^n$, with $a\in U,U\subset E$ and there is a 1-1 mapping $H$ of $V$ onto $U$(whose inverse is also of class $C^1$) such that
(66) $F(H(x))=Ax+\phi(Ax)$ $(x\in V)$ where $\phi$ is a $C^1$ mapping of open set $A(V)\subset Y_1$ into $Y_2$.
So basically, I got the intuition (from his remarks after the proof) that $F$ can basically be projected in a one-to-one manner in some neighborhood of $a$ and that this projection determines $F$. However, I don't think my intuition fully got it.
The problem however is the following, why use $H$? What's the point of introducing something this absurd? Can anyone please explain? Thanks.
Edit: I just thought that maybe this is because $P$ is arbitrary and we need to make a certain change of coordinates to accomodate for that, but I don't see why.
