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Theorem. Let $\phi:X\rightarrow Y$ be a quasi-isometry between two (Gromov) hyperbolic spaces $X$ and $Y$. If $X$ and $Y$ are proper, then ϕ induces a homeomorphism between their boundaries.

The proof of the above statement is well-written in Bridson and Haefliger's book.

My question is that `can we drop the condition that $X$ and $Y$ are proper?'. In some papers about boundaries of hyperbolic spaces, the authors usually say that the above theorem is true without mentioning that $X$ and $Y$ are proper. If you know the answer or any references, then let me know.

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  • $\begingroup$ Gromov takes care of addressing non-proper spaces in his original paper. It would be useful to indeed have a reference with detailed proofs in the general setting. $\endgroup$
    – YCor
    Commented Mar 29, 2022 at 10:13
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    $\begingroup$ Sorry for this late comment. But I think Theorem 5.35 from this reference treating intrinsic hyperbolic spaces might help. $\endgroup$
    – Muduri
    Commented Nov 3, 2023 at 13:27

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Properness is already needed to have a well-defined boundary at infinity, i.e., with a topology not depending on the chosen base point. This is Proposition III.3.7 in Bridson-Haefliger, which builds on some previous lemmata that are applications of the Arzela-Ascoli theorem. To apply the Arzela-Ascoli theorem one needs bounded sets to be compact.

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  • $\begingroup$ Thank you for your answer. So you mean that as long as we consider the visual boundary of a hyperbolic geodesic space, we cannot drop the properness condition in order to say that the induced map is a homeomorphism, right? $\endgroup$
    – Sangrok Oh
    Commented Mar 29, 2022 at 11:55
  • $\begingroup$ I‘m saying that the proofs do not work because they rely on compactness of bounded sets. $\endgroup$
    – ThiKu
    Commented Mar 29, 2022 at 12:37
  • $\begingroup$ Gromov gives an example of a nonproper hyperbolic space with empty boundary: the 1-point-union of Intervalls [0,n] over all n. I guess it is the easiest such space. I‘m not sure what to do with this example. $\endgroup$
    – ThiKu
    Commented Mar 29, 2022 at 12:42
  • $\begingroup$ If a Gromov-hyperbolic space has an empty boundary, then its isometry group has bounded orbits (because there is no loxodromic as these fix two points, and no global fixed point at infinity). (I'm assuming that the "classification" of isometric group actions on geodesic hyperbolic spaces is valid in a non-proper setting.) $\endgroup$
    – YCor
    Commented Mar 29, 2022 at 14:47
  • $\begingroup$ In the first comment, I was comparing the visual boundary to the boundary of a (geodesic) hyperbolic space which is defined as sequences of points (using Gromov product). My question moves to whether the following one is true: If $\phi:X\rightarrow Y$ is a quasi-isometry between (geodesic) hyperbolic spaces (without properness), then $\phi$ induces a homeomorphism between their boundaries obtained by considering sequences. $\endgroup$
    – Sangrok Oh
    Commented Mar 30, 2022 at 2:33

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