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The article "Quasimöbius maps" by Jussi Väisälä states that one always has the implication QM $\implies$ QC. But a proof is only given in for maps of the form $f:\dot{A} \to \dot{Y}$ where $A \subset \dot{X}$ and where $\dot{X} = X \cup \{\infty\}$ denotes the one-point extension of the metric space $X$.

Does the statement hold for all QM maps between arbitrary metric spaces? Does anyone know a reference to a paper where the relationship between QM and QC for arbitrary metric spaces is shown more clear?

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2 Answers 2

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It holds in general. I can't see exactly what Vaisala writes, since it's behind a paywall, but since quasiconformality is an infinitesimal property and quasi-Mobius is a global one, it is easy to prove the former from the latter. (The general idea to prove quasiconformality at $x$ is to use the cross-ratio condition on four points: $x$, 2 points on (or near) a small sphere around $x$, and a fourth point which is extremely far away from all of this action.)

Igor Rivin's answer above is a bit backwards; the Loewner condition is used for the harder task of improving the infinitesimal property of quasiconformality to the global quasisymmetry or quasi-Mobius condition.

(Since my answer is not very precise, this should be a comment, but I hope it was mildly helpful.)

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  • $\begingroup$ Right, I did not read carefully, and was assuming the OP wanted equivalence. $\endgroup$
    – Igor Rivin
    Commented Nov 10, 2015 at 0:36
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This is true in Loewner spaces. For the definitions and exciting properties, see Heinonen's 2001 book Lectures on Analysis in Metric Spaces.

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