Does weak continuity of Jacobians hold for non nondegenerate maps? $\newcommand{\M}{\mathcal{M}}$
$\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be two-dimensional smooth, compact, connected, oriented Riemannian manifolds. (with or without boundaries).

Let $f_n \rightharpoonup f$ in $W^{1,2}(\M,\N) $ with $Jf_n > 0$ a.e., and suppose that the volume $V(\{x \in \M \, | \, Jf_n \le  r\}) \to 0$ when $n \to \infty$, for some $0<r<1$. Does $ Jf_n \rightharpoonup Jf $ in $L^1(K)$ for every $K \subset \subset \operatorname{Int}(\M)$.?

I am fine with assuming that $f_n$ are Lipschits and injective, and that $V(f_n(\M)) \to V(\N) $.

The "higher integrability property of determinants" implies that if $\M,\N$ are open Euclidean domains, then $ Jf_n \rightharpoonup Jf $ in $L^1(K)$ for any compact $K \subset \subset \M$.

Without the assumption $V(Jf_n \le  r) \to 0$, the answer can be negative even when $f_n$ are diffeomorphisms:
Take $\M=\N=\mathbb{S}^2$. Let $s: \mathbb{S}^2 \to \mathbb{R}^2 \cup \{\infty\}$ be the stereographic projection, and let $g_k(x) = k x$ for $x \in R^2$ (and $g_n(\infty) = \infty$.).
Set $ f_n = s^{-1} \circ g_n \circ s$. $f_k$ are conformal, orientation preserving, smooth diffeomorphisms
and thus $ \int_{\mathbb{S}^2 }Jf_n=V(\mathbb{S}^2 )$. By conformality $\int_{\mathbb{S}^2 } |Df_n|^2 =2\int_{\mathbb{S}^2 }Jf_n$ is uniformly bounded, so $f_n$ is bounded in $W^{1,2}$, and converges to a constant function. (asymptotically we squeeze bigger and bigger parts of the sphere to a small region around the pole).
So, we do not have weak convergence of $Jf_n$ to $Jf=0$. (the
Jacobians converge as measures to a Dirac mass at the pole.) The question is if by adding the non-degeneracy constraint $V(Jf_n \le  r) \to 0$ we recover this 'Jacobian continuity' under weak convergence.

*(In my case of application $r=\frac{1}{4}$ but I don't think it matters).
 A: There is a counterexample, however there might be ways to avoid it.
Take $\mathcal{M} = \mathcal{N} =\mathbb{S}^2$, but now consider sequence of maps that cover the sphere twice, where you shrink the preimage of one of them to a point. Specifically consider using the stereographic projection as you did, consider $g_n: \mathbb{C} \cup \{\infty\} \to\mathbb{C} \cup \{\infty\}$, that map $0$ to $0$, $\partial B_{1/n}(0)$ to $\infty$ and $\infty$ to $0$ again, with positive Jacobian in between, e.g. something like
$$g_n(x) := \begin{cases} \frac{x}{|x|}\tan(\pi nx/2)& \text{ for } |x| < \frac{1}{n} \\ \frac{\overline{x}}{|x|} \frac{1}{|x|-\frac{1}{n}} & \text{ for } |x| \geq \frac{1}{n} \end{cases}.$$
This fulfills all your conditions (however with $Jf_n =0$ on a circle, which is a set of measure $0$), and after a slightly tedious calculation you should get $g_n \rightharpoonup \frac{1}{x}$ in $W^{1,2}$, $Jf_n \rightharpoonup 1 +  4\pi\delta_0$ in the sense of distributions and $Jf_n \geq \frac{1}{2}$ anywhere except on the small shrinking ball that corresponds to $B_{1/n}(0)$. In fact $Jf_n \to 1$ almost everywhere and uniformly on any set excluding a neighborhood of $0$.
For manifolds without boundaries, a good way to avoid this might be to require $\int_{\mathcal{M}} Jf_n dx = V(\mathcal{N})$. There are results in the theory of bubbling that roughly tell you that the above is the only thing that happens, i.e. if $f_n \rightharpoonup f$ and $Jf_n$ converges in the sense of measures, then $Jf_n dx \rightharpoonup Jf dx + V(\mathcal{N})\sum_{i\in I} a_i  \delta_{x_i}$, where $I$ is a finite set, $a_i \in \mathbb{Z} \setminus \{0\}$ and $x_i \in \mathcal{M}$, i.e. the only thing that can happen is whole copies of the target manifold "bubbling off" in single points. So if you only have one copy available, and your condition on $Jf_n > r$ requires you to keep one, there are none to do that.
Specifically, since $Jf_n$ is non-negative, the same is true for the limit as a measure, so $Jf \geq 0$ and $a_i > 0$, but by testing with the constant function $1$ you get
$$V(\mathcal{N}) = \int_{\mathcal{M}} Jf_n dx \to \int_{\mathcal{M}} Jf dx + V(\mathcal{N})\sum_{i\in I} a_i.$$
So either $I = 0$, which gives you the $L^1$-convergence, or $Jf=0$, which implies convergence a.e. and thus contradicts the assumption.
