Timeline for Local isometry implies covering map: nonempty boundary case
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2019 at 2:20 | comment | added | Eduardo Longa | Related question: is a local isometry a local diffeomorphism, even in the boundary case? If so, a local isometry must send boundary to boundary. | |
| Dec 10, 2019 at 2:19 | comment | added | Eric Canton | Or, less complicated: take $f: M \to N$ to be the cylinder $M = [0, 1/2] \times \mathbb{S}^1$ including into $N = [0, 1] \times \mathbb{S}^1$. | |
| Dec 10, 2019 at 2:18 | comment | added | Eric Canton | @EduardoLonga: could one not use the same idea for a riemannian manifold with two isomorphic boundary components, gluing just one? More specifically, I'm thinking of a cylinder $[0, 1] \times \mathbb{S}^1$, and gluing two of these: one along $\{0\} \times \mathbb{S}^1$ and the other along $\{1\} \times \mathbb{S}^1$. | |
| Dec 10, 2019 at 2:15 | comment | added | Eduardo Longa | But in this case $N$ does not have a boundary, which is assumed in my question. | |
| Dec 10, 2019 at 2:12 | history | answered | Robbie Lyman | CC BY-SA 4.0 |