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It's an easy result that if we have two quasi isometric hyperbolic spaces, then their Gromov boundaries at infinity are homeomorphic.

I found online these notes where at page 8, prop 2.20 they seem to drop the hypothesis on hyperbolicity. They give two references (french articles) for the proof but I didn't found anything.

Hence, can someone give me a reference/proof/counterexample to the following statement?

Let X,Y proper geodesic spaces, and let $f\colon X \to Y$ be a quasi isometry between them, then $\partial_pX \cong \partial_{f(p)}Y$

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  • $\begingroup$ This is not exactly the answer you are looking for but it looks like boundary in those notes is only defined for $\delta$-hyperbolic so I am guessing that statement implicitly has the spaces hyperbolic. $\endgroup$ – Paul Plummer Mar 13 '15 at 1:34
  • $\begingroup$ that was I thinking actually. Because in the two references I just spot the word hyperbolic in the statements of the proposition :( Anyway I hope someone can provide a counterexample, I'm curious now :) Thanks for the comment! $\endgroup$ – Riccardo Mar 13 '15 at 7:42
  • $\begingroup$ Which definition of boundary do you use? I think they are not all equivalent for non-hyperbolic spaces. $\endgroup$ – Seirios Mar 13 '15 at 8:26
  • $\begingroup$ @Seirios I'm used to "working" (actually studying) with the boundary as the set of geodesic rays passing through a point p in which we identify two geodesics iff their distance is bounded by a costant $\endgroup$ – Riccardo Mar 13 '15 at 9:12
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Counter-examples are due to Buyalo in his paper "Geodesics in Hadamard spaces" and to Croke and Kleiner in an unpublished preprint. See also the later paper of Croke and Kleiner entitled "Spaces with nonpositive curvature and their ideal boundaries" which cites their preprint and gives further information.

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  • $\begingroup$ Thank you for the references! So it turns out that they are not easy counterexamples $\endgroup$ – Riccardo Mar 13 '15 at 17:36

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