Let $\mathcal{M}$ be a Riemannian manifold, and let $\operatorname{inj} \mathrel\colon \mathcal{M} \to (0, \infty]$ be its injectivity radius function.


It is known that if $\mathcal M$ is connected and complete, then $\operatorname{inj}$ is a continuous function: see for example [Lee, [Introduction to Riemannian Manifolds](https://link.springer.com/book/10.1007/978-3-319-91755-9), 2018, Prop. 10.37].


What is known in the case where $\mathcal M$ is not complete? Is $\operatorname{inj}$ also continuous? If not, is there a known counter-example? Would $\operatorname{inj}$ still be semi-continuous?


This question is similar to the question "[The continuity of Injectivity radius](https://mathoverflow.net/questions/53381/the-continuity-of-injectivity-radius)", but the discussion there focuses on compact or complete manifolds.



Is the injectivity radius (semi) continuous on a non-complete Riemannian manifold?
I believe the following proof shows that it is lower semi-continuous, but I do not know a whether it's continuous (or a counterexample), and like OP, would be interested.


The idea is to take a ball just slightly less than the injectivity radius at a point, push everything to the unit ball, then use a bump function to change the metric near the boundary and embed it smoothly in the sphere, at which point the theorem for complete manifolds will give the result.


Let $x \in M$ and let $i(x)$ be its injectivity radius. Let $0 < \varepsilon < i(x)$. We want to show that for some $\delta > 0$, every point $y$ in $B\_{\delta}(x)$ has $i(y) > i(x) - \varepsilon$, i.e., that the exponential map $\exp\_y: T\_y(M) \to M$ is a diffeomorphism on the ball of radius $i(x) - \varepsilon$.


The exponential map $\exp\_x$ is a diffeomorphism from the closed ball of radius $r = i(x) - \varepsilon/4$ to the closed ball in $\mathbb{R}^n$ of radius $r$, say $B\_r$. Let $\hat{x}$ be the origin of the latter, corresponding to $x$. Let $\hat{g} = (\exp\_x)\_\*g$ be the pushforward of the metric to this ball. Let $\chi$ be a bump function on $B\_r$ that is $1$ on $B\_{r - \varepsilon/2}$ and supported on $B\_{r - 3\varepsilon/4}$. Let $h = \chi\hat{g} + (1 - \chi)g\_0$, where $g\_0$ is the round metric on the closed upper hemisphere $S^n\_+$. Then we can clearly double $B\_r$ to $S^n$, extending $h$ to be the usual round metric on the lower hemisphere. So $h$ is a smooth metric on $S^n$ that coincides with $\hat{g}$ on $B\_{r - \varepsilon/2}$.


Thus, by the usual result for complete manifolds (e.g. in Lee), the injectivity radius of $h$ is a continuous function -- let's call it $\hat{i}$. So there's some $\delta > 0$ so that


* $\delta < \varepsilon/4$, and
* for $y$ in the $h$-ball of radius $\delta$ about $\hat{x}$, $\hat{i}(y) > \hat{i}(x) - \varepsilon/4 > i(x) - \varepsilon/2$.


Observe now that for such $y$, the ball $A$ of radius $r - 3\varepsilon/4 = i(x) - \varepsilon$ about $y$ is contained in $B\_{r - \varepsilon/2}$, and the exponential map at $y$ is a diffeomorphism onto $A$. Since $(B\_{r - \varepsilon/2},h|\_{B\_{r - \varepsilon/2}})$ is isometric to $B\_{r - \varepsilon/2}(x)$ in $M$, this yields the claim.


