area-preserving iff $|\det |=+1$ Why is a (not necessarily linear) mapping $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ area- and orientation preserving iff the determinant of its jacobian is $\pm 1$ ?
(I understand by an area-preserving mapping $f$ a mapping $f$ such that the measure $m(f^{-1}(A))=m(A)$, where $m(\cdot)$ denotes the measure of a measurable set $A$.)
I have no idea how to prove this... but I'd also be happy with a reference.
 (I'd also be happy for a sketch of the proof for a less general definition of "area-preserving", where $A$ is not just any measurable set, but a polytope - this definition would work easier with the concept of determinant since the volumen of a polytope is just the absolute value of the determinant of the vector that represent it's edges.)
Googling didn't get me anything.
 A: In dynamical systems or ergodic theory it is preferable to call a map $f:\>X\to Y$ measure preserving (or  area preserving when $X$ and $Y$ are surfaces) if  $$\mu\bigl(f^{-1}(B)\bigr)=\mu(B)\qquad\forall B\subset Y\ .\tag{1}$$ This allows for functions that are many-to-one to be measure preseving nevertheless. But using this definition the Jacobian determinant need not have absolute value $1$. For instance, the map $$f:\quad S^1\to S^1,\qquad e^{it}\mapsto e^{2it}$$
is measure preserving, but  its Jacobian determinant is $=2$.
When $f$ is essentially injective then $(1)$ can be replaced by
$$\mu\bigl(f(A)\bigr)=\mu(A)\qquad\forall A\subset X\ .$$
Now it is proven in calculus that when $f$ is essentially injective and  $f(A)=B$ then for any reasonable function $g:B\to{\mathbb R}$ one has
$$\int_B g(x)\ {\rm d}(x)=\int_Ag\bigl(f(u)\bigr)\>|J_f(u)|\>{\rm d}(u)\ .$$
Putting $g(x):\equiv 1$ here gives
$$\mu\bigl(f(A)\bigr)=\mu(B)=\int_B 1\ {\rm d}(x)=\int_A |J_f(u)|\>{\rm d}(u)\ .$$
Here the right hand side can only be $=\mu(A)$ for every $A\subset X$  if $|J_f(u)|\equiv1$.
