Is there a self-map of the disk with Jacobian everywhere greater than $1$?

Question

It might be silly, but I am not sure how to approach this problem.

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?

I don't assume $f$ maps boundary to boundary.

The area formula implies that such a map cannot be injective.

I guess this is somewhat related to the question of existence of maps of non-zero degree between disks.

Answer 1

In complex coordinates, the map $f: z \mapsto \frac{z+2}{3}$ sends the closed unit disk $D = \bar{B}(0,1)$ to $D' = \bar{B}(\frac23,\frac13)$, a disk of radius $\frac13$. The Jacobian of $f$ is constant and equals to $\frac1{3^2} = \frac19$.

In polar coordinates, the map $g: (r,\theta) \mapsto (r,10\theta)$ sends $D'$ to a subset of the annulus $A = \bar{B}(0,1) \setminus B(0,\frac13)$. Since $0 \not\in D'$, $g$ is a smooth map. The Jacobian of $g$ is another constant and equals to $10$.

If one compose $f$ with $g$, one obtain a smooth map $g\circ f : D \to A$ with constant Jabocian $\frac{10}{9} > 1$.

Answer 2

Choose an $N\gg1$, and map $D$ to an ellipse $E\subset R:=[-2N,2N]\times\bigl[-{1\over N},{1\over N}\bigr]$ by putting $$f_1:\quad (x,y)\mapsto \bigl( 2Nx,{1\over N}y\bigr)\ .$$ We have ${\rm det}(df_1)=2$. Now use a map $f_2$ with ${\rm det}(df_2)\approx1$ to wrap the long rectangle $R$ essentially area preserving around the annulus $$A_\rho:=\bigl\{(x,y)\in D\bigm| \rho\leq\sqrt{x^2+y^2}\leq1\bigr\}$$ with $\rho$ slighty smaller than $1-{2\over N}$.