While analyzing a variational problem, I came to the following question:

Let $\mathbb D^n \subseteq \mathbb{R}^n$ be the closed $n$-dimensional unit ball, and let $f: \mathbb D^n \to \mathbb{R}^n$ be a smooth orientation-preserving immersion. Denote by $\omega_f :\mathbb D^n \to \mathbb{R}^n$ the unique harmonic map satisfying $\omega_f|_{\partial \mathbb D^n}=f|_{\partial \mathbb D^n}$.

$d\omega_f$ must be invertible outside a set of measure zero in $\mathbb D^n$. Indeed, $\omega_f$ is real-analytic, and so is $\det d\omega_f$, which is not identically zero, since $$ \int_{\mathbb D^n} \det d\omega_f = \int_{\mathbb D^n} \det df>0.$$

Now, the zero-set of a real-analytic function which is not identically zero has measure zero.

Question: Do there exist $f_k \in C^{\infty}(\mathbb D^n, \mathbb{R}^n)$ such that $d\omega_{f_k}$ are everywhere invertible and $f_k \to f$ in $W^{1,2}$?

($\omega_{f_k}$ is the harmonic map corresponding to the Dirichlet problem imposed by $f_k$.)

Are there any tools which might help decide this problem one way or another?

Edit:

Note that even though $d\omega_f$ must be invertible almost everywhere, this can be "pretty far" from being everywhere invertible. Indeed, an everywhere invertible matrix field must lie either in $\text{GL}^+$ or in $\text{GL}^-$. However, the original $d\omega_f$ can spend time in both.

Here is an example on the unit disc. Let $f : \mathbb D^2 \to \mathbb R^2$ be defined by $ f(x,y) = (x-2y^2,y). $ We have $$df=\left(\begin{matrix}1 & -4y \\ 0 & 1\end{matrix}\right)$$

and thus $f$ is an orientation-preserving immersion.

The solution to the Dirichlet problem in this case is $\omega_f(x,y) = (x^2 - y^2 + x - 1,y)$, so $$d\omega_f=\left(\begin{matrix}1+2x & -2y \\ 0 & 1\end{matrix}\right)$$ and $\det(d\omega_f)=1+2x>0 \iff x>-\frac{1}{2}$.