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Mar 29, 2022 at 12:51 vote accept Carlos_Petterson
Mar 29, 2022 at 12:48 answer added Vitali Kapovitch timeline score: 1
Mar 29, 2022 at 6:58 comment added Carlos_Petterson @YCor and VitaliKpovitch I have reduced the earlier incarnation of my question down to the current formulation: when are uniform embeddings quasi-symmetries (if their moduli are well-behaved).
Mar 29, 2022 at 6:57 history edited Carlos_Petterson CC BY-SA 4.0
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Mar 28, 2022 at 18:39 history edited Carlos_Petterson CC BY-SA 4.0
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Mar 28, 2022 at 18:01 comment added Vitali Kapovitch this question has Lipschitz in the title but not in the question itself. It is also strangely stated. are you assuming that $f$ is a bijection? Please restate the question clearly.
Mar 28, 2022 at 16:53 comment added YCor But the map need not be injective, and injectivity is not a necessary condition. And even for bijective maps between doubling spaces, uniform continuity can fail...
Mar 28, 2022 at 15:54 comment added YCor For a Lipschitz map the inverse simply doesn't exist.
Mar 28, 2022 at 15:46 history undeleted Carlos_Petterson
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Mar 28, 2022 at 15:20 history edited Carlos_Petterson CC BY-SA 4.0
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Mar 28, 2022 at 15:19 history undeleted Carlos_Petterson
Mar 28, 2022 at 9:20 history deleted Carlos_Petterson via Vote
Mar 28, 2022 at 9:15 history edited Carlos_Petterson CC BY-SA 4.0
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Mar 28, 2022 at 9:12 comment added YCor If it were clear, the proof would also most likely provide a clear estimate. But it's just false. Just playing zigzag, you have a surjective 1-Lipschitz map from the reals (or half-reals) onto the 1-skeleton of a regular trivalent tree, which is not doubling.
Mar 28, 2022 at 8:07 history asked Carlos_Petterson CC BY-SA 4.0