Is a locally invertible weak limit of injective maps injective almost everywhere? This is a cross-post.
Let $\Omega_1,\Omega_2 \subseteq \mathbb R^2$ be open, connected, bounded, with non-empty $C^1$ boundaries.
Let $f_n:\bar\Omega_1 \to \bar\Omega_2$ be Lipschitz injective maps with $\det(df_n)>0$, and suppose that $f_n$ converges to a $C^1$ function $f: \bar\Omega_1 \to \bar\Omega_2$ weakly in $W^{1,2}$, and that $\det(df)>0$ everywhere on $\bar\Omega_1$.

Is it true that $|f^{-1}(y)| \le 1$ a.e. on $\Omega_2$?

Does the answer change if we assume in addition that $f_n|_{K} \to f|_{K}$ strongly in $W^{1,2}$ for every $K \subset \subset \Omega_1$?

The condition $\det(df)>0$ rules out degenerate counterexamples such as $f_n(x)=\frac{x}{n}$ which converge to a constant.
 A: Okay, let me try a writeup of the comment chain. For any reasonable subset $A\subset \Omega_2$ and $B := f^{-1}(A)$ you get
$$\int_A |f^{-1}(y)| dy = \int_B \det df dx \leq \liminf_{n\to\infty} \int_B \det df_n dx = \liminf_{n\to\infty} \mathcal{H}^2(f_n(B)). $$
Then if we know that $\mathcal{H}^2(f_n(B)) \to \mathcal{H}^2(f(B)) \leq \mathcal{H}^2(A)$ ($A$ can have points with no preimage), we get that $|f^{-1}(y)| \leq 1$ a.e. as $A$ was arbitrary.
Now using the existence of a pointwise a.e. converging subsequence (never relabeled) and Egorov's theorem, for any $\epsilon > 0$ there is $B_\epsilon$ such that $\mathcal{H}^2(B \setminus B_\epsilon) < \epsilon$ and $f_n$ converges uniformly on $B_\epsilon$. But then a quick argument shows that $f_n(B_\epsilon)$ converges in the Hausdorff-sense and thus $\mathcal{H}^2(f_n(B_\epsilon)) \to \mathcal{H}^2(f(B_\epsilon))$. Now the leftover set is small and Müller's famous result gives us that $\det df_n$ converges weakly in $L^1$ (see ¹). So in particular as $\chi_{B\setminus B_\epsilon} \in (L^1)^*$
$$\mathcal{H}^2(f_n(B\setminus B_\epsilon)) = \int_{B\setminus B_\epsilon} \det d f_n dy \to \int_{B\setminus B_\epsilon} \det df dy $$
which is small for small enough $\epsilon$ as $f \in C^1$. Similarly
$$\mathcal{H}^2(f(B\setminus B_\epsilon)) \leq \int_{B\setminus B_\epsilon} \det df dy.$$
¹As remarked by Asaf in the comments, the result gives convergence on compact subsets. However as $\Omega_1$ is $C^1$, there exists an extension operator to a larger domain $\Omega \supset \Omega_1$ and thus $\tilde{f}_n,\tilde{f} \in W^{1,2}(\Omega)$ such that $\tilde{f}_n \to \tilde{f}$ in the same sense and $\tilde{f}_n|_{\Omega_1} = f_n, \tilde{f}|_{\Omega_1} = f$. Now $\overline{\Omega_1} \subset \Omega$ is the required compact set.
