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Given two metric spaces $(M_1, d_1)$ and $(M_2, d_2)$, a map $\phi \colon (M_1, d_1) \to (M_2, d_2)$ is a large-scale Lipschitz essentially surjective map if there exist constants $A \geq 1, B \geq 0$, and $C \geq 0$ such that

  1. For all $x, y \in M_1$, we have $\frac{1}{A} d_1(x, y) - B \leq d_2(f(x), f(y)) \leq A \: d_1(x, y) + B$.
  2. For all $z \in M_2$, there exists $x \in M_1$ such that $d_2(z, f(x)) \leq C$.

We define the relation $\mathcal{R}$ on metric spaces by $M_1 \mathcal{R} M_2$ iff there exists a large-scale Lipschitz essentially surjective map $\phi \colon (M_1, d_1) \to (M_2, d_2)$. In ZFC, $\mathcal{R}$ is a symmetric relation (even an equivalence relation) and we normally say that $M_1$ and $M_2$ are quasi-isometric spaces.

However, it is known (see arXiv:1609.01353 for the below cited results) that while $\mathcal{R}$ is always reflexive and transitive, it turns out that symmetry of $\mathcal{R}$ implies the axiom of choice, and hence does not follow from ZF alone (Corollary 2).

It is therefore natural to ask whether if one restricts $\mathcal{R}$ to certain classes of metric spaces one may still have symmetry follow from ZF alone. Even in this case, things are messy: for example, the symmetry of $\mathcal{R}$ on hyperbolic metric spaces fails in ZF (Theorem 1). However, if one restricts one's attention to the class of $\mathbb{R}$-trees, one can do away with choice, and can show in ZF that $\mathcal{R}$ on $\mathbb{R}$-trees is an equivalence relation (Theorem 3). Thus, my question:

For what (interesting) classes of metric spaces, other than $\mathbb{R}$-trees, does symmetry of $\mathcal{R}$ follow from ZF?

Any references and/or thoughts would be appreciated (especially from those set theorists on this site who like to do away with the axiom of choice...).

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    $\begingroup$ Clearly, "be quasi-isometric" is not an appropriate term if it fails to be an equivalence relation. You should better refer to it as the relation $X\,\mathcal{R}\,Y$ "there exists a large-scale Lipschitz essentially surjective map $X\to Y$". Corson says "the QI-relation", which is OK. $\endgroup$
    – YCor
    Commented Apr 30, 2020 at 13:05
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    $\begingroup$ Also your definition (even in ZFC) is not the usual one: the identity map $(\mathbf{R},d)\to(\mathbf{R},\sqrt{d})$ is coarsely Lipschitz (even large-scale Lipschitz), coarsely surjective (= essentially surjective), but not a quasi-isometry. Either you want to stick to coarse equivalence, or restrict to spaces for which these notions are equivalent (that is to say, large-scale geodesic spaces). I note that the definition used by Corson in the paper you link is the usual one. $\endgroup$
    – YCor
    Commented Apr 30, 2020 at 13:07
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    $\begingroup$ Well, interesting is subjective. So I'd daresay "none of them"? :P $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 30, 2020 at 13:22
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    $\begingroup$ @YCor Thank you, you are right of course. I have edited the question to reflect this. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Apr 30, 2020 at 13:25
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    $\begingroup$ I have no idea how much you can get in ZF, but you should be able to go quite far with weak forms of choice. I would guess that symmetry of $\mathcal{R}$ over separable metric spaces follows from a weaker form of choice, possibly countable choice (CC). CC doesn't imply the pathologies of full choice, in particular it is consistent with the axiom of determinacy (AD rules out all choice pathologies). And ZF + CC is equivalent to ZF + "countable products of compact pseudometric spaces are Baire", so it looks like CC is necessary to develop metric geometry over separable spaces. $\endgroup$
    – Erik Walsberg
    Commented Oct 10, 2020 at 18:42

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