Linear version of the submersion theorem

Question

I'm trying to prove the constant rank theorem for smooth functions in euclidean spaces and I've stumbled into a problem:

Given a direct sum decomposition $\mathbb{R}^{n+m} = E \oplus F$ such that $\dim E = n$ and the canonical projection $\pi \colon \mathbb{R}^{n + m} \to E$, is there a way to construct linear isomorphisms $\phi \colon \mathbb{R}^{n + m} \to \mathbb{R}^{n + m}$ and $\psi \colon E \to \mathbb{R}^n$ such that $\psi \circ \pi \circ \phi \colon \mathbb{R}^{n + m} \to \mathbb{R}^n$ is given by $(x,y) \mapsto x$?

I've tried using a choice of basis to explicitly construct the isomorphisms but I'm getting nowhere.

Answer 1

So, a friend helped me with it and the answer is actually quite simple:

Take a basis $B = \{b_1, \dots, b_n\}$ for $E$ and complete it for a basis $B' = {b_1, \dots, b_n, f_1, \dots, f_m}$ of $\mathbb{R}^{n + m}$. If $e_1, \dots, e_{n + m}$ is the canonical basis, define \begin{equation} \psi(b_i) = e_i \quad \text{and} \quad \phi(e_i) = \begin{cases} b_i, &\text{if } i \leq n, \\ f_{i - n}, &\text{if }i > n. \end{cases} \end{equation} We have \begin{align} (\psi \circ \pi \circ \phi)(x^1, \dots, x^{n + m}) &= (\psi \circ \pi)(x^1b_1 + \cdots + x^nb_n + x^{n + 1}f_1 + \cdots x^{n + m}f_m) \\ &= \psi(x^1b_1 + \cdots + x^nb_n) = (x^1, \dots, x^n), \end{align} as desired.