Q:

A map of Jacobian $1$ everywhere

Let $D\subset \Bbb{R}^n$ be a bounded connected open set. Assume that $f:\overline{D}\to\overline{D}$ is a smooth map such that its Jacobian $\det\frac{\partial f}{\partial x}=1$ at any point. Is $f$ surjective? As $f:D\to D$ is an open map, we need to show that it is a closed map.

A:

You have assumed that $f$ is area preserving, but not that it is injective. Therefore the image set could cover parts of $D$ several times and leave out other parts of $D$. Consider the following example:
Let $D$ be the unit disc, and describe $f$ in polar coordinates as follows:
$$f:\quad(r,\phi)\mapsto(\rho,\psi):=\left({r\over\sqrt{2}}, \>2\phi\right)\qquad{\rm (polar\ coordinates)}\ .$$
In terms of cartesian coordinates this $f$ is given by
$$f:\quad(x,y)\mapsto(u,v):=\left({x^2-y^2\over\sqrt{2(x^2+y^2)}},\>{2xy\over\sqrt{2(x^2+y^2)}}\right)\ ,$$
and has Jacobian  $\equiv1$ (as expected!). On the other hand $f$ maps $D$ onto the (twice covered) disc $D'\subset D$ of radius ${1\over\sqrt{2}}$.

