Do non-continuous Sobolev maps pull back closed forms to weakly closed forms? $\newcommand{\R}{\mathbb R}$
$\newcommand{\N}{\mathbb N}$
$\newcommand{\de}{\delta}$
$\newcommand{\sig}{\sigma}$
$\newcommand{\Average}[1]{\left\langle#1\right\rangle}  $
$\newcommand{\IP}[2]{\Average{#1,#2}}$
Let $n,d \in \mathbb{N}$, and let $\Omega \subseteq \R^n$ be  open.
Let $f \in W^{1,k}(\Omega,\R^d)$. Let $\omega \in \Omega^k(\R^d)$ be a smooth closed $k$-form, such that $\omega$ and its derivative $T\omega$ are both uniformly bounded globally; Is it true that $f^*\omega$ is weakly closed? i.e. does
$$\int_{\Omega} \IP{f^*  \omega}{\de \sig}=0$$
hold for every compactly-supported $k+1$-form $\sigma \in \Omega^{k+1}(\Omega)$?
I have read that this should be true, but I am only able to show this in two special cases:


*

*The form $\omega$ is constant.

*$f$ is continuous. 



Does this hold for non-continuous Sobolev maps in general?

Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n \in C^{\infty}(\Omega,\R^d)$ satisfy $f_n \to f$ in $W^{1,k}(\Omega,\R^d)$. Then
$$
\int_{\Omega} \IP{f^*  \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{ f_n^*  \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{d f_n^*  \omega}{ \sig}=0.
$$
Now, we need to justify the passage to the limit $\int_{\Omega} \IP{f^*  \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{ f_n^*  \omega}{\de \sig}$.
If $\omega$ is constant, that is $\omega_q= \alpha$ independently of $q \in \R^d$, where $\alpha$ is a fixed element in $\bigwedge^k (\R^d)^* $, then we have
$$
|f^*  \omega-f_n^*  \omega| \le  |\alpha| \, \left|\bigwedge^{k} df-\bigwedge^{k} df_n\right|_{op},
$$
thus
$$
\left|\int_{\Omega} \IP{f^*  \omega}{\de \sig}- \IP{ f_n^*  \omega}{\de \sig}\right| \le  |\alpha| \|\de \sig\|_{\sup} \int_{\Omega} \left|\bigwedge^{k} df-\bigwedge^{k} df_n\right|.
$$
and The RHS tends to zero since Sobolev approximation lifts to exterior powers.
When $\omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
$$
\begin{split}
& |f^*  \omega-f_n^*  \omega|(p)= \\
&\left|\omega_{f(p)} \circ \bigwedge^{k} df_p -\omega_{f_n(p)} \circ \bigwedge^{k} (df_n)_p \right| \le \\
&\left|\big(\omega_{f(p)}-\omega_{f_n(p)}) \circ \bigwedge^{k} df_p +\omega_{f_n(p)} \circ \big( \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p) \right| \le \\
&|\omega_{f(p)}-\omega_{f_n(p)}| \, \, \cdot \, \, \left| \bigwedge^{k} df_p\right| +|\omega_{f_n(p)} | \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| 
\end{split}
$$ 
So, we we need to estimate $\omega_{f(p)}-\omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:
$$
\begin{split}
&|\omega_{f(p)}-\omega_{f_n(p)}| \, \, \cdot \, \, \left| \bigwedge^{k} df_p\right| +|\omega_{f_n(p)} | \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \le \\
&|T\omega|_{sup} \,  \cdot  \, |f(p)-f_n(p)| \,  \cdot  \, \left| \bigwedge^{k} df_p\right| +|\omega |_{sup} \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \le \\
&|T\omega|_{sup} \,  \cdot  \, |f-f_n|_{sup} \,  \cdot  \, \left| \bigwedge^{k} df_p\right| +|\omega |_{sup} \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \stackrel{L^1}{\to} 0,
%&|\alpha \circ  \brk{\bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p} | \le |\alpha| \cdot |\bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p|_{op}.
\end{split}
$$ 
So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?
 A: This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf which I state below. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal.

Theorem. Let $\mathcal{M}$ be a smooth, $k$-dimensional oriented manifold with or without boundary.
  
  
*
  
*If $f\in W^{1,1}_{\rm loc}(\mathcal{M},\mathbb{R}^m)$, then  $ f^\ast (\omega \wedge \eta) =  f^\ast\omega \wedge f^\ast\eta $ holds
  pointwise a.e.
  
*If $f\in W^{1,p}_{\rm loc}(\mathcal{M},\mathbb{R}^m)$, $p\geq\ell+1$, $0\leq\ell\leq k-1$, and $\eta\in
 C^\infty(\bigwedge\nolimits^\ell\mathbb{R}^m)\cap W^{1,\infty}$ (i.e.
  $\eta$ and $|\nabla \eta|$ are bounded), then $d (f^\ast\eta) = f^\ast
 (d\eta)$ holds in the weak sense, i.e. $$ \int_{\mathcal{M}}
 f^*\eta\wedge d\varphi = (-1)^{\ell+1}\int_{\mathcal M}
 f^*(d\eta)\wedge \varphi $$ for all $\varphi\in
 C_0^\infty(\bigwedge\nolimits^{k-\ell-1}\mathcal M)$.
  
*If $\eta\in W^{1,p}_{\rm loc}(\bigwedge\nolimits^{\ell_1}\mathcal M)$, $\omega\in W^{1,p}_{\rm loc}(\bigwedge\nolimits^{\ell_2}\mathcal
 M)$, $\ell_1+\ell_2\leq k-2$, $p\geq 2$, then $d(\eta\wedge
 d\omega)=d\eta\wedge d\omega$ weakly in the sense that $$
 \int_{\mathcal{M}} \eta\wedge d\omega\wedge d\varphi =
 (-1)^{\ell_1+\ell_2}\int_{\mathcal{M}} d\eta\wedge
 d\omega\wedge\varphi $$ for all $\varphi\in
 C_0^\infty(\bigwedge\nolimits^{k-\ell_1-\ell_2-2}\mathcal M)$.
  

The proof is very easy. We simply approximate $f$ by smooth mappings $f_\epsilon$ (convolution type approximation). Then the corresponding formulas are clearly true with $f$ replaced by $f_\epsilon$ and we pass to the limit as $\epsilon\to 0$.
