This is a follow-up of this question.
Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Does there exist a smooth map $f:D \to D$ such that $\det df >1$ everywhere and $f(\partial D) \subseteq \partial D$?
Does anything change if I require $f|_{\partial D}=\text{Id}_{\partial D}$?
If we do not impose boundary conditions, then the answer is positive.
Let $D^n$ denote the closed $n$-dimensional unit disk. Let $f: (D^n,\partial D^n)\to (D^n,\partial D^n)$ be a smooth map of pairs with $J_f(x)\ne 0$ for all $x\in D^n$. Then $f$ is a diffeomorphism.
To prove this, observe first that $f$ is a local homeomorphism. Then "double" $f$ across the boundary, to get a local homeomorphism $F: S^n\to S^n$, which has to be a covering map, hence, a homeomorphism (provided that $n\ge 2$). From this, conclude that $f$ is bijective, hence, a diffeomorphism. If $n=1$ then $f$ is strictly monotonic and, hence, is a diffeomorphism.
From this, conclude that maps $f$ with $J_f>1$ do not exist.