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Characterizing Lipschitz images of injective metric spaces

A metric space \(X\) is {\em injective} if every non-expanding map \(f:B\to X\) defined on a subspace \(B\) of a metric space \(A\) can be extended to a non-expanding map \(\bar f:A\to X\). We prove that a metric space \(X\) is a Lipschitz image of an injective metric space if and only if \(X\) is L...

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Bibliographic Details
Published in:arXiv.org 2024-05
Main Authors: Bąk, Judyta, Banakh, Taras, Garbulińska-Węgrzyn, Joanna, Nowak, Magdalena, Popławski, Michał
Format: Article
Language:English
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Summary:A metric space \(X\) is {\em injective} if every non-expanding map \(f:B\to X\) defined on a subspace \(B\) of a metric space \(A\) can be extended to a non-expanding map \(\bar f:A\to X\). We prove that a metric space \(X\) is a Lipschitz image of an injective metric space if and only if \(X\) is Lipschitz connected in the sense that for every points \(x,y\in X\), there exists a Lipschitz map \(f:[0,1]\to X\) such that \(f(0)=x\) and \(f(1)=y\). In this case the metric space \(X\) carries a well-defined intrinsic metric. A metric space \(X\) is a Lipschitz image of a compact injective metric space if and only if \(X\) is compact, Lipschitz connected and its intrinsic metric is totally bounded. A metric space \(X\) is a Lipschitz image of a separable injective metric space if and only if \(X\) is a Lipschitz image of the Urysohn universal metric space if and only if \(X\) is analytic, Lipschitz connected and its intrinsic metric is separable.
ISSN:2331-8422