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 $C^1$ be bijective maps with $\det(df_n)>0$, and suppose that $f_n$ converges to a continuous function $f: \bar\Omega_1 \to \bar\Omega_2$ strongly in $W^{1,2}$.
Question: Must $f$ be surjective?
Note that $f$ is surjective if and only if $|f^{-1}(y)| \le 1$ a.e. on $\Omega_2$:
By the area formula $$ \int_{\Omega_1} \det df_n = \int_{\Omega_2} |f_n^{-1}(y)|=\text{Vol}(\Omega_2), $$ so $$ \int_{\Omega_2} |f^{-1}(y)|= \int_{\Omega_1} \det df =\lim_{n \to \infty} \int_{\Omega_1} \det df_n=\text{Vol}(\Omega_2). $$ This implies the claim.
Suppose for contradiction that $f$ misses a point $q\in\Omega_2.$ It must miss some neighborhood $U\subset \Omega_2$ of $q.$ Choose a bump function $\alpha$ with $\int \alpha=1$ whose support lies in $U.$
The Jacobians $\det f_n$ converge in $L^1$ to $\det df.$ The functions $\alpha\circ f_n$ are uniformly bounded and converge in measure to $\alpha\circ f.$ By the result at Convergence in measure and $L_p$ implies product converges in $L_p$,
$$\int \alpha(f_n(x))\det df_n\to \int \alpha(f(x))\det df$$
The right hand side is zero because $\alpha\circ f\equiv 0.$ But by the change of variables formula along the orientation-preserving diffeomorphisms $f_n,$ the left hand side is $\int_{\Omega_2}\alpha = 1.$ This is a contradiction.