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Dear members, Given an open manifold $M$ and its completion $\bar{M}$, do we have $inj(M)=inj(\bar{M})$, Thank you Riadh

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When the completion is not a manifold, how do you define its injectivity radius? –  Bruno Martelli Apr 5 '13 at 11:15
In any case, it seems to me that the injectivity radius of $\mathbb{R}$ is infinite and the injectivity radius of $\mathbb{R}\backslash{0}$ is zero. –  Deane Yang Apr 5 '13 at 13:58
If the injectivity radius is strictly positive the metric is automatically complete. –  Mohan Ramachandran Apr 5 '13 at 21:53

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