Local Submersion Theorem - Differential Topology of Guillemin and Pollack

Local Submersion Theorem: Suppose that $f:X \to Y$ is a submersion at $x$, and $y=f(x)$. Then there exist local coordinates around $x$ and $y$ such that $f(x_1,\dotsc,x_k)=(x_1,\dotsc,x_{\ell})$. That is, $f$ is locally equivalent to the canonical submersion near $x$.

Definitions:

1. The canonical submersion is the standard projection of $\mathbb{R}^k$ onto $\mathbb{R}^{\ell}$ for $k\geq \ell$, in which $(a_1,\dotsc,a_k) \to (a_1,\dotsc,a_{\ell})$.

2. We shall say that two maps $f: X \to Y$ and $f' : X' \to Y'$ are equivalent (or locally equivalent) if there exist diffeomorphisms $\alpha$ and $\beta$ such that $f=\beta\circ f'\circ\alpha$ (commutative)

I understand roughly that $f$ is locally equivalent to the canonical submersion near $x$. However, it is not clear why $f(x_1,\dotsc,x_k)=(x_1,\dotsc,x_{\ell})$. The vectors $(x_1,\dotsc,x_k)$ and $(x_1,\dotsc,x_{\ell})$ are not vectors of $X$ and $Y$. These $X$ and $Y$ are manifolds of dimension $k$ and dimension $\ell$. This does not necessarily imply that $X \subseteq \mathbb{R}^k$ and $Y \subseteq \mathbb{R}^{\ell}$. Otherwise, local coordinates (system) around $x$ and $y$ implies that there exist the coordinate systems $\phi^{-1}=(x_1,\dotsc,x_k)$ and $\psi^{-1}=(x_1,\dotsc,x_{\ell})$ such that $f= \psi \circ \mathrm{canonical\ submersion} \circ \phi^{-1}$, but $f\circ \phi^{-1} = f(x_1,\dotsc,x_k)=(x_1,\dotsc,x_{\ell}) =\psi^{-1}$ is not true. Could someone explain the nature of $f (x_1,\dotsc,x_k) = (x_1,\dotsc,x_{\ell})$? It's not clear to me what it means.

• I don't understand what's your quibble. It means that there are charts $\phi : U \subset X \rightarrow \widetilde{U} \subset \mathbb{R}^{k}$ and $\psi : V \subset Y \rightarrow \widetilde{V} \subset \mathbb{R}^{l}$ such that $f(U) \subset V$ and $\psi \circ f \circ \phi^{-1} : \widetilde{U} \rightarrow \widetilde{V}$ is the canonical submersion as you call it. – Pedro Jun 11 '15 at 16:21
• This is clearly abuse of notation, someday you will get used to it. – Gustavo Apr 26 '16 at 20:41

Please allow me to introduce another notation. Let $\ m,n \in \mathbb{N}^*$, $M$ and $N$ be smooth manifolds with dimensions $\, m \,$ and $\, n \,$ respectively, $p \in M \$ and $\ f : M \to N \$ be a smooth function such that $\, f \,$ is a submersion at $\, p$, ie, the differential $\ df|_p : TM|_p \to TN|_{f(p)} \$ is surjective. Clearly one has $\ m \geqslant n$. The local submersion theorem says that there exists open sets $\ U \subset M$, $V \subset N$, $Z \subset \mathbb{R}^{m-n} \$ and $\ W \subset \mathbb{R}^n \$ and there exists charts $\ x: U \to \mathbb{R}^m \$ and $\ y : V \to \mathbb{R}^n \$ such that $\ p \in U$, $f(p) \in f[U] \subset V$, $im(y) = W$, $im(x) = W \times Z \subset \mathbb{R}^{m-n} \times \mathbb{R}^n = \mathbb{R}^m \$ and the local representation $\ f_{xy} = y \circ f \circ x^{-1} : W \times Z \to W \$ is of the form $$f_{xy} (a^1 , ... , a^m) = (a^1 , ... , a^n) \, ,$$ for all $\ (a^1 , ... , a^m) \in W \times Z \subset \mathbb{R}^m$. We can write this last equation as $$(y \circ f)(q) = f_{xy} \big( x(q) \big) = f_{xy} \big( x^1(q) , ... , x^m(q) \big) = \big( x^1(q) , ... , x^n(q) \big) = (x^1 , ... , x^n)(q) \, ,$$ for all $\ q \in U \subset M$. Where we have the coordinate functions $\ x = (x^1,...,x^m)$. So, in functional terms, we have $$y \circ f = (x^1 , ... , x^n) \, .$$
For each $\ \mu \in \{ 1,...,m \}$, let $\ \pi^{\mu} : \mathbb{R}^m \to \mathbb{R} \$ be the projection onto the $\mu$-th coordinate, ie, $\pi^{\mu} (a^1,...,a^m) = a^{\mu}$, $\forall (a^1,...,a^m) \in \mathbb{R}^m$, and a restriction $\ i^{\mu} = \pi^{\mu}|_{W \times Z} : W \times Z \to \mathbb{R}$, ie, $i^{\mu} (a^1,...,a^m) = a^{\mu}$, $\forall (a^1,...,a^m) \in W \times Z$. Then, the function $\ i = (i^1,...,i^m) : W \times Z \hookrightarrow \mathbb{R}^m \$ is the inclusion, ie, $i(a)=a$, $\forall a \in W \times Z$, and we have that $\ im(i)= W \times Z$. So, we can write the above equality as $$f_{xy} = (i^1 , ... , i^n) \, .$$