Recall that a Cartan–Hadamard manifold is a complete, simply-connected Riemannian manifold $(M,g)$ which has non-positive sectional curvature everywhere. One property of a Cartan–Hadamard manifold is that, given any two points $p,q \in M$, (up to reparameterisation) there is a unique geodesic segment which connects $p$ and $q$. This allows us to define a geodesic triangle: given $p,q,r \in M$ which are not colinear (i.e. not contained in the image of a single geodesic), we define the geodesic triangle $\Delta pqr$ to be the union of the images of the unique geodesic segments connecting the three points.
I am interested in geodesic triangles in complete Riemanian manifolds. However, this definition of a geodesic triangle isn't well-defined for a general complete Riemannian manifold, because there may be distinct distance-minimising geodesics connecting two points. For example, on $\mathbb S^2$, there are infinitely many distinct distance-minimising geodesics connecting antipodal points.
This motivates my question:
Question. Let $(M,g)$ be a complete Riemannian manifold, and let $p \in M$. Does there exist $R > 0$ such that for any two points in the geodesic ball $B_R(p)$, there is a unique geodesic segment connecting the two points, whose image is contained in $B_R(p)$?
I suspect that this is true: if we consider a geodesic ball of radius $R \leq \pi/2$ on $\mathbb S^2$, then any two points in the geodesic ball are connected by a unique distance-minimising geodesic segment (whose image is contained in the ball). However, I don't know where to start with the general case.