# Arnolds geometric formulation of the implicit function theorem

I have trouble understanding why Arnold's geometric formulation of the implicit function theorem is equivalent to the usual one. In his book "Ordinary Differential Equations", Vladimir Arnold gives the following variant of the implicit function theorem (p. 92 of the third edition):

Implicit function theorem: In some neighborhood of a nondegenerate point any two smooth mappings (of spaces of fixed dimension $m$ and $n$) are equivalent.

He calls a smooth mapping $f:\mathbb{R}^m\to\mathbb{R}^n$ nondegenerate in the origin if its derivative has maximal rank there (i.e. the rank equals the minimum of $m$ and $n$). He supposes that $f(0)=0$ and calls two such mappings $f$ and $g$ equivalent in the origin if there exist diffeomorphisms $h:\mathbb{R}^m\to\mathbb{R}^m$ and $k:\mathbb{R}^n\to\mathbb{R}^n$ in the domain and target space that leave the origin fixed and such that $f\circ h = k\circ g$. Local equivalence is thus given if the two functions can be written by the same formulas if suitable local coordinates systems are used in the domain an target space. Arnold writes

The reader accustomed to more complicated statements of the implicit function theorem will easily verify that these more complicated statements are equivalent to the simple geometric statement above.

In spite of being advertized as an easy exercise, I have not succeeded in proving either that the usual formulation implies Arnold's formulation nor the other way around.

If $f:\mathbb{R}^M\to \mathbb{R}^N$ is a smooth map with constant rank $k$ in a neighbourhood of $p\in\mathbb{R}^M$ then there exist diffeomorphisms \begin{align} h&:U\ni p\overset{\sim}{\to} h(U)\subset\mathbb{R}^M\\ k&:V\ni f(p)\overset{\sim}{\to} k(V)\subset\mathbb{R}^N \end{align} such that $h(p)=0,\,k(f(p))=0$ and $$k\circ f\circ h^{-1}(x_1,\dots,x_M)=(x_1,\dots,x_k,0,\dots,0),$$ in a neighbourhood where all the maps are defined.
Then if $f,g$ have maximal rank at $p$ ($N>M$) $$k\circ f\circ h^{-1}(x_1,\dots,x_M)=(x_1,\dots,x_M,0,\dots,0)=k'\circ g\circ h'^{-1}(x_1,\dots,x_M)$$ define $K=k^{-1}\circ k'$ and $H=h'^{-1}\circ h$, then $$f\circ H^{-1}=K\circ g.$$