Let $\Omega \subseteq \mathbb R^n$ be a nice open, connected, bounded subset (say with smooth boundary), and let $p \ge 1$.
I want to know whether the following notion of "injective a.e. in the image" is well-defined:
Say that $f \in W^{1,p}(\Omega,\mathbb R^n)$ is injective a.e. in the image if $|f^{-1}(y)|=1$ for almost every point $y \in \mathbb R^n$.
Is this notion well-defined, or does it depend upon the representative we choose for $f$ (which formally is an equivalence class in $L^p$ satisfying some additional requirements...).
I guess that since we can freely change $f$ on a subset of measure zero, we can just take any uncountable subset of measure zero $\Omega_0 \subseteq \Omega$, and send it to whatever we want in $\mathbb R^n$. Somehow I am not sure if this really shows that this notion is not well-defined.
(Do we need the continuum hypothesis here? If we knew any uncountable subset had the cardinality of $\mathbb R$ then we could just map it bijectively onto $\mathbb R^n$ "twice" or something).
Yes, that is a bad choice of language here, and your counterexample would work. There is no need for continuum hypothesis here because we know there is a Cantor set that will work.
The definition of injective a.e. is: there exists a negligible set $S$ outside of which $f$ is injective. In other words, for any choice of representative $f$, there exists a negligible $S\subset\Omega$ such that $f\vert_{\Omega\setminus S}$ is injective. Note there is no mention of the image here, and it works as advertised.