Relation between range and kernel of vector valued function Consider a continously differentiable function $f:\mathbb{R}^n\mapsto\mathbb{R}^m$ with $n<m$ and ${\cal R}_f$ denote its range. Is there always a nontrivial function $g:\mathbb{R}^m\mapsto\mathbb{R}^n$ such that 
$$\{x\in\mathbb{R}^m|g(x)=0\}={\cal R}_f?$$
If $f$ is linear there of course we can find $g$ by taking the transpose of the corresponding matrix. It is still true for any function? Any counter example?
 A: If you don't have any require on $g$, you can construct it in the following way.
Let $v \in \mathbb{R}^n \setminus \{ 0\}$. 
Then you can define:
$$
g = \chi_{\mathcal{R}_f} v.
$$
Id est $g$ is the characteristic function of the image of $f$ multiplied by  the non-zero vector $v$. 
Since you required $g$ to be somehow "nontrivial", I guess you wanted to preserve some feature of $f$. By hypothesis we have that $f$ is continuously differentiable. I guess you would like $g$ to be at least continuous. But in some cases you can't construct such a $g$. This is because $\mathcal{R}_f$ could be not closed but if you want $g$ to vanish on  $\mathcal{R}_f$ and to be continuous, then $g$ must vanish also on the boundary of $\mathcal{R}_f$.
EDIT: Think about this example. Let $f \colon \mathbb{R} \to \mathbb{R}^2$ defined as $$f(x) = (\arctan(x), 1).$$
Then $\mathcal{R}_f = \{(x, y) \in \mathbb{R}^2 \colon -\frac{\pi}{2} < x < \frac{\pi}{2} \text{ and } y = 1\}$ is not closed.
EDIT: Here another example. Let $f \colon \mathbb{R} \to \mathbb{R}^2$ defined as $$f(x) = (\cos(2\arctan(x)), \sin(2\arctan(x))).$$
It's easy to see that $f$ is injective and smooth and its image is 
$$
\mathcal{R_f}= S^1 \setminus \{(-1, 0)\}. 
$$
Therefore $\mathcal{R_f}$ is not closed and if we want define a continuous function $g$ which vanishes on $\mathcal{R_f}$, then it must vanish also on $\{(-1, 0)\}$.
