TITLE: Diffeomorphism imply nonzero Jacobian?
QUESTION [5 upvotes]: Why if $\varphi:U\subseteq\mathbb{R}^N\to V\subseteq\mathbb{R}^N$ is bijective and $\varphi, \varphi^{-1}$ are of class $C^1$ then the jacobian:
$\det(D\varphi (x))\neq 0,\ \forall\ x\in U$???

REPLY [6 votes]: Because both $\varphi$ and $\varphi^{-1}$ are $C^1$ you can apply the chain rule and you get $$D\varphi^{-1}(\varphi(x))\circ D\varphi(x)=D(\varphi^{-1}\circ \varphi)(x)=D(id_U)(x)=Id.$$
Hence $$\det(D\varphi^{-1}(\varphi(x)))\det\left ( D\varphi(x)\right)=\det(Id)=1$$ so $\det\left ( D\varphi(x)\right)\neq 0$.