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This question stems from the discussion in:

how to define the injectivity radius of manifolds with boundary?

Suppose $(M,g)$ is a compact Riemannian manifold with boundary. In this context, let the injectivity radius of a point $x$ be the minimum distance from $x$ at which there is a point $y$ with more than one length-minimizing geodesic connecting $x$ to $y$.

Is it true that the injectivity radius as defined this way is bounded below by some nonzero value? If so, is there a standard reference for this fact?

In the discussion linked above, the following theorem is referenced:

Corollary 2. If for a complete Riemannian manifold with boundary, M, the sectional curvatures of the interior and the outward sectional curvatures of the boundary are no greater than $K$, then $N(p,\frac{\pi}{2K})$ is open in M and the distance function from p is convex on $N(p,\frac{\pi}{2K})$.

Where $N(p,\frac{\pi}{2K})$ is the set of points connected to $p$ by a unique geodesic of lenght $\frac{\pi}{2K}$ or less.

(Reference Paper) https://www.ams.org/journals/tran/1993-339-02/S0002-9947-1993-1113693-1/S0002-9947-1993-1113693-1.pdf

This seems like it is close to the result I am looking for. Earlier in the paper, it is also stated that there are no conjugate points in $N(p,\frac{\pi}{K})$.

Is there a simple step from this result that proves that the injectivity radius is nonzero?

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Yes, they show that any compact Riemannian manifold with boundary is locally $\mathrm{CAT}(\kappa)$ for some $\kappa\in\mathbb{R}$. In particular the injectivity radius is positive.

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    $\begingroup$ This definitely works but is it necessary to go the generality of $CAT$-spaces? This should be proveable purely in Riemannian geometry language. $\endgroup$ – quarague Aug 21 at 6:37
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    $\begingroup$ @quarague I do not have an example of curvature-free estimate on injectivity radius. So if there is such a proof, then it might give something new in Riemannian geometry. $\endgroup$ – Anton Petrunin Aug 21 at 16:00
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    $\begingroup$ You definitely need curvature, I just think you don't need $CAT$-spaces. A $CAT$-space is by definition a metric space that satisfies a triangle comparison inequality. Historically this is a generalisation from Riemannian manifolds with bounded curvature. One can show that a Riemannian manifold with curvature bounded from above is a $CAT$-space but this is why $CAT$-spaces are defined the way they are. The question of OP was known before $CAT$-spaces where invented/defined. So one should be able to prove it directly with Riemannian geometry machinery. $\endgroup$ – quarague Aug 22 at 6:46

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