TITLE: Approximation of smooth diffeomorphisms by polynomial diffeomorphisms?
QUESTION [17 upvotes]: Is it possible to (locally) approximate an arbitrary smooth diffeomorphism by a polynomial diffeomorphism?
More precisely: Let $f:\mathbb{R}^d\rightarrow\mathbb{R}^d$ be a smooth diffeomorphism for $d>1$. For $U\subset\mathbb{R}^d$ bounded and open and $\varepsilon>0$, is there a diffeomorphism $p=(p_1, \cdots, p_d) : U\rightarrow\mathbb{R}^d$ (with inverse $q:=p^{-1} : p(U)\rightarrow U$) such that both

$\|f - p\|_{\infty;\,U}:=\sup_{x\in U}|f(x) - p(x)| < \varepsilon$, $\ \textbf{and}$
each component of $p$ and of $q=(q_1,\cdots,q_d)$is a polynomial, i.e. $p_i, q_i\in\mathbb{R}[x_1, \ldots, x_d]$ for each $i=1, \ldots, d$?

Clearly, by Stone-Weierstrass there is a polynomial map $p : \mathbb{R}^d\rightarrow\mathbb{R}^d$ with $\|f - p\|_{\infty;\,U} < \varepsilon$ and such that $q:=(\left.p\right|_U)^{-1}$ exists; in general, however, this $q$ will not be a polynomial map.
Do you have any ideas/references under which conditions on $f$ an approximation of the above kind can be guaranteed nonetheless?
$\textbf{Note:}$ This is a crosspost from https://math.stackexchange.com/questions/3689873/approximation-of-smooth-diffeomorphisms-by-polynomial-diffeomorphisms

REPLY [3 votes]: An illustration for one of the examples in the answer by Robert Bryant.
It is supposed to convey the feeling of something extremely rigid, unyielding and inflexible.
Image of the square $[-1,1]\times[-1,1]$ under the map $(x,y)\mapsto(x-y^2-2x^2y-x^4,y+x^2)$ (composite of $(x,y)\mapsto(x-y^2,y)$ with $(x,y)\mapsto(x,y+x^2)$).