Let $X$ be a Riemann surface and $\pi:\mathbb{H}\rightarrow X$ a universal covering. Fix $x\in X$ and $\widetilde x\in \mathbb{H}$ such that $\pi(\widetilde x)=x$. For every loop $\gamma$ in $X$ with $\gamma(0)=\gamma(1)=x$ one can consider the lifting $\widetilde \gamma$ of $\gamma$ to $\mathbb{H}$ such that $\widetilde \gamma(0)=\widetilde x$ and one deduces that if two loops $\gamma_1$ and $\gamma_2$ based in $x$ are homotopic then $\widetilde{\gamma_1}(1)=\widetilde{\gamma_2}(1)$. In this way one defines the monodromy action $$\rho:\pi_1(X)\rightarrow Aut(F)$$ where $F:=\pi^{-1}(x)$, $\rho([\gamma]):=\widetilde\gamma(1)$.
Question: is this action faithful? In other words: given two loops $\gamma_1$ and $\gamma_2$ based in $x$, if $\widetilde{\gamma_1}(1)=\widetilde{\gamma_2}(1)$ can one deduce that $\gamma_1$ and $\gamma_2$ are homotopic? If the answer is "no" then what can one deduce from $\widetilde{\gamma_1}(1)=\widetilde{\gamma_2}(1)$?
The answer is yes. If $\widetilde{\gamma_1}(1) = \widetilde{\gamma_2}(1)$ then $\widetilde {\gamma_1}$ and $\widetilde {\gamma_2}$ are homotopic (because $\mathbb H$ is simply connected). And you can "push" the homotopy with $\pi$.
The action is free