TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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A quasi-isometric embedding with coarsely dense image has a quasi-inverse quasi-isometric embedding

Statement

Assume the Axiom of Choice (The Axiom of Choice).

A quasi-isometric embedding with coarsely dense image has a quasi-inverse quasi-isometric embedding.

Facts & Assumptions

Given: The hypotheses of the Statement, including the Axiom of Choice.

[F1]

A subset is coarsely dense when every point of the space is within a fixed distance of it, and a quasi-isometry is a coarse Lipschitz map admitting a coarse Lipschitz quasi-inverse (Coarsely dense subsets, quasi-inverses and quasi-isometries).

[L1]

A map is $(L, C)$ -coarse Lipschitz when $d (f (x), f (x^{'})) \leq L d (x, x^{'}) + C$ , and an $(L, C)$ -quasi-isometric embedding when in addition $L^{- 1} d (x, x^{'}) - C \leq d (f (x), f (x^{'}))$ (Coarse Lipschitz maps and quasi-isometric embeddings).

[L2]

Two maps into a metric space are at bounded distance when the distance between their values is bounded uniformly (Bounded distance between two maps into a metric space).

[A1]

Every family of nonempty sets has a choice function >. (The Axiom of Choice).

Proof

technique · direct

1.1

Let $f : X \to Y$ be an $(L, C)$ -quasi-isometric embedding whose image is $R$ -coarsely dense. By the definition of coarse density, for every $y \in Y$ the set ${x \in X : d_{Y} (f (x), y) \leq R}$ is nonempty, so the Axiom of Choice gives a map $g : Y \to X$ with $d_{Y} (f (g (y)), y) \leq R$ for every $y \in Y$ .

F1L1A1choose

2.1

For $y, y^{'} \in Y$ , the upper inequality for $f$ gives $L^{- 1} d_{X} (g (y), g (y^{'})) - C \leq d_{Y} (f (g (y)), f (g (y^{'}))) \leq d_{Y} (y, y^{'}) + 2 R,$ so $d_{X} (g (y), g (y^{'})) \leq L d_{Y} (y, y^{'}) + L (C + 2 R)$ . Likewise $d_{Y} (y, y^{'}) \leq d_{Y} (y, f (g (y))) + d_{Y} (f (g (y)), f (g (y^{'}))) + d_{Y} (f (g (y^{'})), y^{'})$ is at most $2 R + L d_{X} (g (y), g (y^{'})) + C$ , so $g$ is a quasi-isometric embedding.

L1step 1.1

3.1

By step 1.1 the composite $f \circ g$ is at bounded distance at most $R$ from ${id}_{Y}$ . Also $L^{- 1} d_{X} (g (f (x)), x) - C \leq d_{Y} (f (g (f (x))), f (x)) \leq R,$ so $d_{X} (g (f (x)), x) \leq L (C + R)$ for every $x \in X$ ; hence $g \circ f$ is at bounded distance from ${id}_{X}$ . Therefore $g$ is a coarse Lipschitz quasi-inverse of $f$ .

F1L2step 1.1step 2.1∎

Depends on

Used by

A map is a quasi-isometry exactly when it is a quasi-isometric embedding with coarsely dense image Corollary

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

C. Loh, Geometric Group Theory: An Introduction (2015 course version), 264 pp. (standard reference, not scraped)
C. Drutu and M. Kapovich, Geometric Group Theory (with an appendix by B. Nica), 837 pp. (standard reference, not scraped)