LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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A quasi isometry of geodesic spaces has a controlled coarse inverse

Statement

Let $λ \geq 1$ , $ε, R \geq 0$ , and let $f : X \to Y$ satisfy $λ^{- 1} d_{X} (x, x^{'}) - ε \leq d_{Y} (f x, f x^{'}) \leq λ d_{X} (x, x^{'}) + ε .$ Assume the attained coarse-density condition: for every $y \in Y$ some $x \in X$ satisfies $d_{Y} (f x, y) \leq R$ . Assuming AC, there is $g : Y \to X$ such that $d_{Y} (f g (y), y) \leq R, d_{X} (g f (x), x) \leq λ (R + ε) .$ It is a $(λ, λ (2 R + ε))$ -quasi-isometric embedding. If such a selector $g$ is already supplied, all the estimates are choice-free. Geodesicity is unnecessary for this lemma.

Facts & Assumptions

Given: The displayed inequalities and attained density condition.

[F1]

The definition and qualitative inverse construction are given in A quasi-isometric embedding with coarsely dense image has a quasi-inverse quasi-isometric embedding; we calculate its constants explicitly below.

[A1]

AC says every family of nonempty sets has a choice function (The Axiom of Choice).

Proof

1.1givenA1F1choose

For $y \in Y$ put $E_{y} = {x \in X : d_{Y} (f x, y) \leq R}$ . Attained density says $E_{y} \neq \emptyset$ . Apply A1 to the family of these sets and set $g (y)$ equal to the selected member of $E_{y}$ . This is the sole choice use. If $Y$ is empty, existence of $f$ forces $X$ empty, and take the empty function. If a selector is supplied, start with it. In each case $d_{Y} (f g (y), y) \leq R$ .

2.1step 1.1givenalgebra

For $y, y^{'} \in Y$ , the triangle inequality and step 1.1 give $d_{Y} (y, y^{'}) - 2 R \leq d_{Y} (f g (y), f g (y^{'})) \leq d_{Y} (y, y^{'}) + 2 R .$ Combining the right inequality with the lower bound for $f$ yields $d_{X} (g (y), g (y^{'})) \leq λ d_{Y} (y, y^{'}) + λ (2 R + ε)$ . Combining the left inequality with the upper bound for $f$ yields $d_{X} (g (y), g (y^{'})) \geq λ^{- 1} d_{Y} (y, y^{'}) - (2 R + ε) / λ$ . Since $λ \geq 1$ , this implies the asserted lower bound with additive constant $λ (2 R + ε)$ .

2.2step 1.1givenalgebra

Apply the lower inequality for $f$ to $g f (x), x$ . Its image distance is at most $R$ by step 1.1 with $y = f (x)$ , so $d_{X} (g f (x), x) \leq λ (R + ε)$ . The estimates include $R = 0$ , $ε = 0$ and $λ = 1$ without division by zero or loss of attainment.

3.1step 1.1step 2.1step 2.2∎

Steps 1.1–2.2 prove all assertions. Apart from the family selection in step 1.1, only inequalities for specified points were used, proving the supplied-selector qualification.

Depends on

Used by

Dependency tree · two levels

6 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

Druţu–Kapovich Corollary 9.39; local published coarse-inverse theorem (standard reference, not scraped)