Geodesically convexity and convexity in normal coordinates

Question

This question regards an assertion in the proof of Theorem 1.3.1 of Douglas Moore's book "Introduction to Global Analysis, Minimal Surfaces in Riemannian Manifolds" (AMS, 2017).

Assumptions/Definitions:

$M$ is a complete, compact Riemannian manifold $M$. Consider $y_1, y_2 \in M$ two points, $\varepsilon_i>0$, $i=1,2.$

Set $D_i := \{v \in T_{y_i}M | \ |v|< \varepsilon_i\} $, and $B_i := \{z\in M \ | d_M(z,y_i) < \varepsilon_i\}$.

We suppose that:

1. the exponential map gives diffeomorphisms $$\exp_{y_i}|_{D_i}:D_i\to B_i$$

2. $B_1, B_2, $ and $B_1\cap B_2$ are geodesically convex (i.e. any two points can be joined by a unique minimizing geodesic)

Problem

At page 20 of the book there is a claim about convexity which I don't know how to prove, let me state it precisely.

We would like to show that the domain $U$ where the composition $ (\exp_{y_2}|_{D_2})^{-1}\circ \exp_{y_1}|_{D_1}$ is defined is convex in $T_{y_1}M$. Notice that $$U := \exp_{y_1}^{-1} (B_2)\cap D_1$$

Question: why is $U$ convex?

My initial idea was to show a property which turned out to be false, as showed by @Kajelad on Exponential map and convexity (Riemannian geometry) that is why I decided to add the geodetically convex hypothesis.

We know that $B_1\cap B_2$ is geodetically convex, is not clear how this implies that the preimage of it under the exponential map at $y_1$ is convex.

Answer 1

Even with the added requirement of geodesic convexity, this is false. A similar counterexample is to let $M$ be the unit sphere and choose $\varepsilon_1=\varepsilon_2=\pi/2$. This will always satisfy the convexity requirements, and if $y_1$ and $y_2$ are chosen sufficiently close to antipodal, the resulting $U$ will be a nonconvex "half-crescent" shape.

Glancing at the cited book, the discussion of convexity don't seem to make sense. Unless I'm mistaken, it can be disregarded entirely with a slight change in argument: Fiberwise convexity is only needed in order for the Taylor remainder integral to be well-defined, but even without it, the integral will be well-defined locally (due to compactness), which suffices for the local computations in the proof.