This is a [cross-post](https://math.stackexchange.com/questions/3873192/is-a-locally-invertible-weak-limit-of-injective-maps-injective-almost-everywhere).


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 $K \subset \subset \Omega\_1$?




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The condition $\det(df)>0$ rules out degenerate counterexamples such as $f\_n(x)=\frac{x}{n}$ which converge to a constant.



Is a locally invertible weak limit of injective maps injective almost everywhere?
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.


