Questions about Rudin's rank theorem I am trying to understand the rank theorem in Rudin's Principles of Mathematical Analysis.
The theorem states:

Theorem Suppose $m,n,r$ are nonnegative integers, $m\ge r, n\ge r$, $F$ is a $C^1$ mapping of an open set $E\subset \mathbb{R}^n$ into $\mathbb{R}^m$, 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 a projection in $\mathbb{R}^m$ whose range is $Y_1$ and let $Y_2$ be the kernel of $P$. Then there are open sets $U$ and $V$ in $\mathbb{R}^n$, with $a\in U\subset E$, and there is a 1-1 $C^1$ mapping $H$ of $V$ onto $U$ (whose inverse is also of $C^1$) such that $$F(H(x)) = Ax+\phi(Ax)\;\;\;\;(x\in V)$$where $\phi$ is a $C^1$ mapping of the open set $A(V)\subset Y_1$ into $Y_2$.

I have quite a few problems trying to understand this theorem, but perhaps mainly this:

*

*What kind of projection does Rudin have in mind? All he says about the projection is that its range is $Y_1$. Is this the projection of the entire  $\mathbb{R}^m$ and the $\text{im}(P)=Y_1$? Can you choose any projection with such range? Does the choice of such projection affect the null space of P?

*What does the equation $F(H(x)) = Ax + \phi(Ax)$ tell us intuitively? Why are we mapping with $\phi$ from the range of $P$ to its nullspace?

 A: (See comments...) Finally, for the intuition. Think of $H$ as a change of coordinate system in the domain. Let's go ahead and do a change of coordinate system in the range as well. But this one is easier: There is an invertible linear map $B: R^m \to R^m$ such that
$$
(B \circ A) (x_1, x_2, \ldots, x_n) = (x_1, x_2, \ldots, x_r).
$$
So, the claim of the theorem is (you need to replace your $\phi$) that  given any $C^1$ maps that has rank $r$ on an open set, we can do a local $C^1$ change of coordinate in domain and a linear change of coordinate in the range, such that in the new coordinates, the function depends only on the first $r$ coordinates:
$$
(x_1, x_2, \ldots, x_n) \mapsto (x_1, x_2, \ldots, x_r) + \phi(x_1, x_2, \ldots, x_r).
$$
This becomes extra beautiful if $m=r$ in which case we can take $\phi$ to be zero.
If you wan to avoid doing the change of coordinates in the range, then you can interpret the result as saying that "you know where the image of $f$ (after the change of coordinates in domain) is if you know where the image of the (simpler) map $A$ is, i.e. $Ax$ completely determines $f(x)$."
