This is a follow-up question of this one.
Let $M,N$ be smooth connected, compact two-dimensional Riemannian manifolds, such that $M$ has a non-empty Lipschitz boundary. Suppose that $\operatorname{Vol}(M)=\operatorname{Vol}(N)$.
Question: Let $f:M \to N$ be a smooth isometric immersion (i.e. $df_p$ is an isometry for every $p \in M$). Suppose also that there exist injective $C^1$ maps $f_n:M \to N$ such that $\det df_n>0$ and $f_n \to f$ in $W^{1,2}(M,N)$. Must $f$ be surjective?
This is equivalent to $f$ being injective "a.e. in the image"- i.e. $|f^{-1}(q)| \le 1$ for a.e. $q \in N$. (see below).
Some amount of non-injectivity is definitely possible: Take for example $M=[-1,1]^2$, and let $N=M/\sim$ be the flat $2$-torus with $\sim$ the standard equivalence relation. Then $f_n:M\to N$ given by $f_n(x)=(1-1/n)x$ are injective and converge to the quotient map $\pi:M\to N$ which is not everywhere injective.
Proof of that $f$ is surjective if and only if $|f^{-1}(q)| \le 1$ a.e. on $N$:
By the area formula $$ \text{Vol}(M) = \int_M 1=\int_M \det df = \int_N |f^{-1}(y)|=\int_{f(M)} |f^{-1}(y)|. $$ So, if $|f^{-1}(y)| \le 1$ a.e. on $N$, then $ \text{Vol}(N)=\text{Vol}(M) = \text{Vol}(f(M))$. On the other hand, if $\text{Vol}(f(M))=\text{Vol}(M)$, then $$\text{Vol}(f(M))=\text{Vol}(M)= \int_{f(M)} |f^{-1}(y)| \ge \int_{f(M)} 1= \text{Vol}(f(M)), $$ so $|f^{-1}(y)| \le 1$ a.e. on $f(M)$, hence also a.e. on $N$.
We proved that $|f^{-1}(y)| \le 1$ a.e. on $N$ if and only if $\text{Vol}(f(M))=\text{Vol}(N)$. Since $f(M) $ is compact, being of full measure in $N$ is equivalent to being equal to $N$.
