Vtyjkyuk

Example : (1) If $f : mathbbR^2rightarrow mathbbR^2,
f(x)=Ccdot x$ is dilation, then it is bi-Lipschitz map.

(2) More generally, we consider an inversion
$f:mathbbR^2-orightarrow mathbbR^2-o,
(x)=fracxx$. By a direct computation, we have $$ 0<|
Df|^2leq 2cdot J ast $$ at any point where $|A|=sqrtsum_i,j
A_ij^2$ and $J=|rm det Df|$.

If $g: mathbbR^2times
0rightarrow mathbbS^2, g(x,y) = frac11+r^2
(x,y,frac-1+r^22)$, i.e. stereographic projection, then $g$
satisfies $ast$.

(3) If $f=id: (M^n,g) rightarrow (M^n,kg)$ where $k>0$, then $J=k^n=n^-n/2|Df|^n$.

Definition : $f:Xrightarrow Y$ between $n$-dimensional manifolds is a quasi-regular map if $ 0<| Df|^nleq Ccdot J$ for global constant $C$.

Question : In the definition, any smooth map $f: Xrightarrow Y$ with $Jneq 0 $ is a quasi-regular map ?

SHORT ANSWER. The answer is negative. The mapping $fcolon (0, infty)^2to (0, infty)^2$ defined as $f(x, y)=(x, sqrt y)$ is smooth and not quasi-regular.

LONG ANSWER (Prior to edit).

If I understand it correctly, the answer is negative. Let $fcolon mathbb R^nto mathbb R^n$ be a smooth function, let $xin mathbb R^n$ be fixed and denote
$$
A:=Df(x).$$
Now $A$ is an $ntimes n$ matrix, and the question asks whether
$$tag1|A|^nle C|det A|$$
holds for an absolute constant $C>0$, where
$$
|A|:=max |Ax|_mathbb R^n / |x|_mathbb R^n.$$
But (1) cannot hold, for the matrix
$$
A_epsilon:=beginbmatrix 1 & 0 \ 0 & epsilonendbmatrix$$
is such that $|A_epsilon|=1$ and $det A_epsilon=epsilon$, and $epsilon$ can be made as small as we wish.