Let $f:\mathbb{R}^n \to \mathbb{R}^n$ be smooth. Can we find, for every $\epsilon>0$, a $C^1$ map $\tilde f:\mathbb{R}^n \to \mathbb{R}^n$ of full rank such that
$\|df-d\tilde f\|_{C^0}<\epsilon$?
Note that what I require is weaker than $\|f-\tilde f\|_{C^1}<\epsilon$, I only want the differentials to be uniformly close.
The point is that pointwise, for every real $n \times n$ matrix $A$, and every $\epsilon>0$, there exist an invertible matrix $\tilde A$ such that $\|A-\tilde A\|<\epsilon$. The question is whether we can always create a smooth and exact perturbation. (i.e. I want $d\tilde f$ to be everywhere invertible).
I am fine if this perturbation cannot be done on the whole space, but only on a fixed "part of it"- e.g. I am OK with restricting the domain to be the unit disk. (but I don't want to be forced to shrink the domain to a disk with a radius which depends on $\epsilon$).
This can be done locally (that is, for any point $p \in \mathbb{R}^n$ and any $\varepsilon > 0$, you can get an approximation that works on $B(p,r)$ where $r$ depends on $\varepsilon$ and $p$) but definitely cannot be done globally. Let $n = 1$ and consider $f(x) = \frac{x^2}{2}$. If $g \colon \mathbb{R} \rightarrow \mathbb{R}$ is a $C^1$ map with non-vanishing derivative then $g'(x) > 0$ for all $x \in \mathbb{R}$ or $g'(x) < 0$ for all $x \in \mathbb{R}$. In any case,
$$ \| f' - g' \|_{\infty} = \infty. $$