I am studying the book "Riemannian Manifolds: An Introduction to Curvature" by John Lee. In the chapter on Gauss-Bonnet Theorem, there is an exercise problem that outlines the proof of the fact that every compact Riemannian 2-manifold has a smooth triangulation. In this outline, the author asks us to prove the following:
Let $M$ be a smooth, compact, Riemannian $2$-manifold, $v_1, v_2, \cdots, v_k \in M$ and $\epsilon > 0$ be such that the geodesic balls $B_{3 \epsilon} \left( v_i \right)$ are convex and uniformly normal and the geodesic balls $B_{\epsilon} \left( v_i \right)$ cover $M$. Then, for each $i$, there is a convex geodesic polygon in $B_{3 \epsilon} \left( v_i \right)$ whose interior contains $B_{\epsilon} \left( v_i \right)$.
The hint given to solve it says, "Let the vertices be sufficiently nearby points on the circle of radius $2 \epsilon$."
I understand how this might work. If we consider points on the circle of radius $2 \epsilon$ in the Euclidean plane, we can get a polygon whose sides do not intersect the ball of radius $\epsilon$. While this is geometrically clear to me, I am not able to bring in the rigour and write the proof of this.
Also, another question that I have is that if we move to arbitrary Riemannian manifolds where the metric is not flat, how can we guarantee that the geodesic between points does not enter the smaller ball of radius $\epsilon$?
Any insights into this are appreciable!
Hint:
Call the points on the circle centered at, say, $v_1$ and of radius $2\epsilon$ $$ p_1,...,p_n $$ where the cyclic order is the one coming from the circle.
Make sure that for each $i$, the distance from $p_i$ to $p_{i+1}$ ($i$ is taken mod $n$) is $<\epsilon$) and apply the triangle inequality.
This will guarantee that the polygon that you formed is disjoint from $B(v_1,\epsilon)$ and is contained in $B(v_1, 3\epsilon)$.
But do not forget to verify that the polygon is simple. Here you will be using uniform normality and the fact that your manifold is 2-dimensional.