Can we perturb the Dirichlet boundary conditions to make harmonic maps locally invertible? 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 $d\omega_{f_k} \in \text{GL}^+$ are everywhere and $f_k \to f$ in $W^{1,2}$?

($\omega_{f_k}$ is the harmonic map corresponding to the Dirichlet problem imposed by $f_k$.) 
Note that even though $d\omega_f \in \text{GL}$ a.e., it can "spend time" in both $\text{GL}^+$ and $\text{GL}^-$. Here is an example:
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 $\det(d\omega_f)=1+2x>0 \iff x>-\frac{1}{2}$.
 A: It seems that the answer is negative for dimension $n=2$ . I am not sure if higher dimensions can be reduced to the $2D$ case.
Here is the argument for $n=2$:
Suppose that there exist $f_k \in C^{\infty}(\mathbb D^2, \mathbb{R}^2)$ such that $d\omega_{f_k} \in \text{GL}^+$ everywhere and $f_k \to f$ in $W^{1,2}$.
The convergence $f_k \to f$ in $W^{1,2}$ implies that
$d\omega_{f_k} \to d\omega_f$ in $L^2$. (This follows from the fact that the trace operator is a continuous surjection onto the fractional Sobolev space $W^{1/2,2}(\partial \Omega)$, see here for details).

This implies that $\det(d\omega_{f_k}) \to \det(d\omega_f)$ in $L^1$ (here use the fact that the dimension is $2$), so up to passing to a subsequence, $\det(d\omega_{f_k})$ converges pointwise a.e. to $\det(d\omega_f)$.

By our assumption, $\det(d\omega_{f_k}) > 0$ everywhere, so this implies $\det(d\omega_f) \ge 0$  a.e.
Now, taking any $f$ whose $\omega_f$ does not satisfy this, we get a contradiction. (e.g. the example described in the question).
