Here is the problem:
Let $S$ be a complete oriented regular surface and $\gamma : [a, b] \rightarrow S$ a regular curve in $S$. Let $a = t_0 < t_1 <... < t_n = b$ be a regular partition of $[a, b]$, which means that $t_i = a + i \Delta t$, where $\Delta = \frac{b−a}{n}$. Explain why for sufficiently large $n$, there exists a unique geodesic from $\gamma(t_i)$ to $\gamma(t_{i+1})$ for each $i$, which together form a piecewise geodesic curve $\beta$ from $\gamma(a)$ to $\gamma(b)$. As $n\to \infty$, prove that the sum of the signed angles of $\beta$ converges to the angle displacement along $\gamma$.
First of all, when he says "unique geodesic", I think he means "uniquely minimizing geodesic". Also, I think the assumption that $S$ is complete is unessecary, because the existence of these uniquely minimizing geodesics can be guaranteed by covering the trace of $\gamma$ with convex neighborhoods.
Here is my progress: I can show that the angle displacements of $\gamma$ and $\beta$ are close, given that they intersect at finitely many points. Under this hypothesis, we can apply Gauss Bonnet to each small loop created by the two curves between consecutive points of intersection. Then, by summing, we can show that the angle displacements are close.
I am not sure how to proceed if the points of intersection are infinite. Untuitively this should make things easier, because the curves will be "closer". Observe that the issue persists even if $S$ is the plane, so $\beta$ is just the inscribed polygon of $\gamma$. In this case though, the result can be shown with elementary means, without resorting to Gauss Bonnet.
Any help is appreciated!
