Let $f:\Omega \to \Omega'$ be a surjective local diffeomorphism between bounded, simply connected domains $\Omega,\Omega'\subset \mathbb{R}^2$. Given a curve $\beta\subset\Omega'$, is it possible to find a curve $\alpha\subset \Omega$ with $f(\alpha)=\beta$?
Because $f$ is surjective, for any $y\in \beta$, we can find $x\in \Omega$ with $f(x)=y$. Since $f$ is a local diffeomorphism, there is a neighbourhood $U(x)$ of $x$ such that $f_{\mid_{U(x)}}$ is a diffeomorphism with $y\in f(U(x))$. In this way, we can construct an open covering $\bigcup_{i} U_i$ of $\beta$ where $f$ has an inverse on every $U_i$. By Heine-Borel, we only need finitely many open sets $U_1,..., U_n$ to do so. However, these inverses don't have to coincide on $U_i \cap U_{i+1}$, so we can't just set $\alpha = \bigcup_{i=1}^n (f_{\mid_{U(x)}})^{-1}(U_i)$, because $\alpha$ doesn't have to be a curve.