I'm trying to understand the proof of this theorem (p.45, A comprehensive introduction to DG by Spivak):
If $n \leq m$ and $f:M^n \to N^m$ has rank $n$ at $p$, then for any coordinate system $(x,U)$ around $p$, there is a coordinate system $(y,V)$ around $f(p)$ with $y \: \circ f \: \circ x^{-1} (a^1, ... ,a^n) = (a^1, ... ,a^n,0,...,0).$
The case $k<n$ has been proved in the previous page (Theorem 9). But what confused me is the proof:
Question: Why did the points get moved to the wrong place?
Sorry if it is a basic question. I've thinking it for hours and still can't find out why.
What we are looking for is a coordinate system $\psi_0$ on the codomain $\Bbb{R}^m$ of $f : \Bbb{R}^n \to \Bbb{R}^m$ such that $\psi_0 \circ f (a^1,\dots,a^n) = (a^1,\dots,a^n,0,\dots,0)$. As above, by Theorem 9 we have coordinates $\psi$ and $\phi$ such that $$ \psi \circ f \circ \phi^{-1} (b^1,\dots,b^n) = (b^1,\dots,b^n,0,\dots,0), \quad \forall (b^i) \in \Bbb{R}^n $$ which is not quite what we want, since we don't want to change the coordinate domain of $f$. Having this information, we investigate what happen if we do not perform $\phi^{-1}$ \begin{align} \psi \circ f (a^1,\dots,a^n) &= \psi \circ f \circ \phi^{-1} \circ \phi (a^1,\dots,a^n) \\&= \psi \circ f \circ \phi^{-1} (b^1,\dots,b^n), \quad \text{with } \phi(a^1,\dots,a^n) = (b^1,\dots,b^n) \\&= (b^1,\dots,b^n,0,\dots,0). \end{align} So $\psi$ is a candidate for our desired coordinate system $\psi_0$, only with an issue that the image $(b^1,\dots,b^n,0,\dots,0)$ is not the original position of the point $(a^1,\dots,a^n)$. This is what the author meant by $\textbf{"points in $\Bbb{R}^n$ get moved to the wrong place $\cdots$}"$. That's why we correct it by composing with $\lambda$ (moved back these wrong coordinates to the original position). So the desired coordinate system is $\psi_0 = \lambda \circ \psi$, since $$ \psi_0 \circ f (a^1,\dots,a^n) = \lambda \circ \psi \circ f (a^1,\dots,a^n) =\lambda (b^1,\dots,b^n,0,\dots,0) = (a^1,\dots,a^n,0,\dots,0). $$