Let $R:\mathbb{C}\to\mathbb{C}$ be a rational function. Consider the pre-image of its real values $R^{-1}(\mathbb{R})$. This set should be a finite union of open analytic arcs and finitely many critical points of $R$ (quating some notes I am trying to go through). For example, for $R(z)=z+1/z$, the set $R^{-1}(\mathbb{R})$ is the unit circle and $\mathbb{R}\setminus\{0\}$. The "exceptional points" are the intersection points $\pm 1$ which are the critical points of $R$.
Now, it should be true that if $R$ is restricted to any open connected arc of $R^{-1}(\mathbb{R})$ which contains no critical point of $R$, then $R$ restricted to this arc is a real-valued injective function. I can't see a clear argument for the injectivity claim. Can anyone help me to understand? Thanks.
$R$ is locally injective on the arc. If you parametrize the arc by an interval of real numbers, you have a real-valued function on an interval that is locally injective, and thus in a neighbourhood of each point either locally increasing or locally decreasing. But then by connectedness of the interval it is either globally increasing or globally decreasing, and thus globally one-to-one.