Express locally the image $g(U)$ as the zero set of some functions

Given $U\subset\Bbb R^n$ open, and $g\colon U\to\Bbb R^m$ smooth and injective, with $m>n$, how can I locally express $g(U)$ as the zero-set of some smooth functions? That is, for every $w_0\in g(U)$, find $W\subset\Bbb C^m$ open and $f_1,\dots,f_d$ smooth functions on $W$ such that $$ g(U)\cap W=\{w\in W\;:\;f_1(w)=\dots=f_d(w)=0\} $$

Does this hold near every point $w_0\in g(U)$, or is there some difference between regular and singular points?

Example: consider $g\colon\Bbb R\to\Bbb R^2$ defined by $g(t)=(t,t^2)$ (which is trivially an embedding); it's clear that \begin{align*} g(\Bbb R) &=\{(x,y)\in\Bbb R^2\;:\;x=t\,,y=t^2,t\in\Bbb R\}\\ &=\{(x,y)\in\Bbb R^2\;:\;y=x^2\}\;. \end{align*} Taking $g(t)=(t^2,\sin t)$, I already have problems. I thought I can consider the auxiliary function $h(t):=(t,g(t))$ getting $$ g(\Bbb R)=\pi(h(\Bbb R)) $$ ($\pi$ is the projection on the last two coordinates) but I cannot really get a result.

I suspect it's tremendously trivial and well known, so I just request some reference.

Solution 1:

I don't have a reference, but let us take $w_0\in g(U)$ and $u_0\in U$ such that $g(u_0)= w_0$. Let us choose $d$ vectors $a_1, \ldots, a_d\in {\mathbb R}^m$ such that ${\mathbb R}^m = \text{Ran}(g'(u_0))\oplus \text{Span}(a_1,\cdots,a_d)$. Let us embed ${\mathbb R}^n$ into ${\mathbb R}^m$ and define for $x\in U\times {\mathbb R}^d$ \begin{equation} \tilde{g}(x_1, \cdots, x_m) = g(x_1,\cdots, x_n) + x_{n+1} a_1 + x_{m} a_d \end{equation} Then $\tilde g$ is locally invertible near $(u_0, 0,\ldots, 0)$. Apply the inverse function theorem to get an inverse function $h$ defined locally around $w_0$. Define $f_i = e_{n+i}^*\circ h$ for $i=1,\cdots,d$ where $e_i^*$ is the $i$-th coordinate form. These functions $f_i$ answer the question.

Edit Thinking again, this works only if we can ensure that $\text{rank}(g'(u_0)) = n$. Apparently it is not implied by the assumptions, for example we could take $g:t\to (t^2, \sin (t^3))$ which is smooth and injective near $0$, but $g'(0) = 0$.

I remember similar concerns in books about bifurcation theory.

Example In the case of $g(t) = (t^2, \sin t)$, near $t_0 = 0$, we have $g'(0) = (0, 1)$, we choose $a_1 = (1, 0)$, which leads to $\tilde g(t, s) = (t^2+s, \sin(t))$. The inverse is $h(x, y) = (\arcsin y, x - \arcsin^2 y)$, hence $f_1(x, y) = x - \arcsin^2 y$ near $(0,0)$.