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May 6, 2021 at 16:39 comment added Anton Petrunin A simpler way: it is well known that any noncontracting map from compact metric space to itself is an isometry --- apply it to the partial inverse of $f\colon X \to X$.
May 6, 2021 at 16:26 comment added Anton Petrunin Oh, it seems that you expect me reading your post to the end :)
May 6, 2021 at 11:07 comment added Moishe Kohan @AntonPetrunin: As I said, it works for $X=Y$. As for your example, I do not understand it.
May 6, 2021 at 5:02 comment added Anton Petrunin Take the standard sphere $X$. Shrink its equator by factor 2; denote the obtained space by $Y$. The induced map $X\to Y$ is measure-preserving and short. So, something wrong with your argument. So you need to assume more about spaces. For Alexandrov spaces it was done by Nan Li arxiv.org/abs/1110.5498
May 5, 2021 at 13:30 comment added Moishe Kohan @JialongDeng: No, you need also the assumption that the map is measure-preserving and that it is a map from a manifold to itself.
May 5, 2021 at 12:22 comment added Jialong Deng @ Moishe: Does it mean that a 1-Lipschitz map between Riemannian $n$-manifolds is an isometric map by the argument?
May 4, 2021 at 13:10 history edited Moishe Kohan CC BY-SA 4.0
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May 4, 2021 at 13:08 comment added Moishe Kohan @JohannesHahn: You are right: The argument was originally written in the case of self-maps, $X=Y$ and I did not think through the general case. I will correct...
May 4, 2021 at 11:49 comment added Johannes Hahn $f^2(\Delta_r)$ need not be a proper subset for non-isometries. Extreme counterexample: Constant maps. You need to use the volume-condition or manifold-ness somewhere to infer that.
May 3, 2021 at 16:11 history edited Moishe Kohan CC BY-SA 4.0
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May 3, 2021 at 15:07 history answered Moishe Kohan CC BY-SA 4.0