Let $\Omega \subseteq \mathbb{R}^d$ be an open bounded domain. Fix an integer $1<k<d$.
Let $f \in W^{1,k}(\Omega;\mathbb{R}^d)$ be a continuous map with $\det df > 0$ a.e.
Consider the map $\Omega \to \text{End}(\bigwedge^k \mathbb{R}^d)$ defined by $x \to \bigwedge^k df_x $. Since $\det df > 0$ a.e. $\bigwedge^k df_x \in \text{GL}(\bigwedge^k \mathbb{R}^d)$ a.e.
Think of $\text{GL}(\bigwedge^k \mathbb{R}^d)$ as a subset of $\text{End}(\bigwedge^k \mathbb{R}^d)\cong \mathbb{R}^{\binom{d}{k}^2} $. With this identification in mind, suppose that $$\bigwedge^k df \in C\big(\Omega,\mathbb{R}^{\binom{d}{k}^2} \big) \cap W^{1,k}\big(\Omega,\mathbb{R}^{\binom{d}{k}^2} \big).$$
Do there exist $u_n \in C^{\infty}\big(\Omega,\text{GL}(\bigwedge^k \mathbb{R}^d)\big)$, such that $u_n \to \bigwedge^k df$ in $W^{1,k}_{loc}\big(\Omega,\text{GL}(\bigwedge^k \mathbb{R}^d)\big) $?
I am also interested in a weaker result: Are there $u_n \in C^{\infty}\big(\Omega,\text{End}(\bigwedge^k \mathbb{R}^d)\big)$, such that $u_n \in \text{GL}(\bigwedge^k \mathbb{R}^d)$ a.e. and $u_n \to \bigwedge^k df$ in $W^{1,k}_{loc}\big(\Omega,\text{GL}(\bigwedge^k \mathbb{R}^d)\big) $?
I don't really need the $u_n$ to be defined on all $\Omega$. It is enough that for every arbitrarily small compact ball in $\Omega$, there would be a neighbourhood where such a sequence $u_n$ would be defined.
Of course, there are always $u_n \in C^{\infty}(\Omega,\mathbb{R}^{\binom{d}{k}^2})$ such that $u_n \to \bigwedge^k df$ in $W^{1,k}\big(\Omega,\mathbb{R}^{\binom{d}{k}^2} \big)$. (just approximate each component of $\bigwedge^k df$ separately). The question is whether or not we can choose the approximations of the different components in such a way that their "combined" map will stay invertible almost everywhere.
The problem is that it is not always true that $\bigwedge^k df_x \in \text{GL}(\bigwedge^k \mathbb{R}^d)$ for every $x \in \Omega$. The rank can fall on a subset of measure zero.
If we knew $\bigwedge^k df \in \text{GL}(\bigwedge^k \mathbb{R}^d)$ everywhere then the answer would positive. This follows from the facts that "being invertible" is an open condition, and that continuous Sobolev maps can be approximated uniformly by smooth maps in the strong Sobolev topology.
In more detail, let $K \subseteq \Omega$ be compact. Since we assumed $\bigwedge^k df \in C\big(\Omega,\mathbb{R}^{\binom{d}{k}^2} \big) \cap \text{GL}(\bigwedge^k \mathbb{R}^d)$, the map $\psi:x \to \bigwedge^k df_x$, considered as a map $K \to \text{GL}(\bigwedge^k \mathbb{R}^d)$, is continuous. Thus $\psi(K) $ is compact. This implies that $\text{dist}\big(\psi(K),\partial \text{GL}(\bigwedge^k \mathbb{R}^d)\big)>0$, where $\partial \text{GL}(\bigwedge^k \mathbb{R}^d)$ is the boundary of the set $ \text{GL}(\bigwedge^k \mathbb{R}^d)$, considered as a subset of $ \text{End}(\bigwedge^k \mathbb{R}^d)$.
Now, consider each component of $\psi(x)=\bigwedge^k df_x \in \text{End}(\bigwedge^k \mathbb{R}^d) \cong \mathbb{R}^{\binom{d}{k}^2}$. We can approximate each component, using mollification (on an open subset of $\Omega$ containing $K$). Since each component is a continuous function, the mollifications converge uniformly to the components on compact subsets, and in particular on our $K$. This implies that from a certain point in the mollified sequence, $\text{dist}(u_n,\bigwedge^k df)<\text{dist}\big(\psi(K),\partial \text{GL}(\bigwedge^k \mathbb{R}^d)\big)$, so the $u_n$ are invertible, as required.