Distance between antipodal points in a geodesic ball Let $(M^n,g)$ be a complete Riemannian manifold. For fixed $p \in M$ and small $r > 0$ (smaller than the injectivity radius of $M$ at $p$), let $q$ and $q'$ be antipodal points in the sphere $S_R(p)$ of radius $R$ centered at $p$. This means that if $\gamma : \mathbb{R} \to M$ is the unique unit speed geodesic with $\gamma(0) = p$ and $\gamma(R) = q$, then $\gamma(-R) = q'$.
My question: is it true that the distance between $q$ and $q'$ is equal to $2 R$?
Since there is a curve of length $2R$ joining $q$ and $q'$, we know that $d(q, q') \leq 2R$. Can it be strictly smaller than $2R$?
 A: The answer is no.
Consider the unit sphere $S^n$ with $n\geqslant 1$.
Let $p= N$ be the north pole.
Its injectivity radius is $\pi$, but two antipodal points at distance $\pi/2<R<\pi$ lie in the same (south) hemisphere, hence are at distance strictly less than $\pi$ from each other, which is strictly less than $2R$.
This happens because, even though the ball $B(0;R)\subset T_pM$ is mapped diffeomorphically onto $B(p;R)\subset M$ by the exponential map, the ball $B(p;R)$ may not be convex in $M$.
The good notion for having such a property is not that of injectivity radius, but that of convexity radius.
For compact Riemannian manifolds, we can compare these two notions (also with the volume): see this article by Marcel Berger.
Remark that this latter article shows that in fact, a compact Riemannian manifold always has its convexity radius lesser than or equal to half of its injectivity radius, and the above counterexample is not an isolated case.
There is equality in the case of the unit sphere though.
See also this article where is is shown that a compact Riemannian manifold can have a ratio convexity radius/injectivity radius arbitrarily small.
