TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Being quasi-isometric is reflexive, symmetric and transitive

Statement

Being quasi-isometric is reflexive, symmetric and transitive.

Facts & Assumptions

Given: The hypotheses of the Statement.

[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]

Bounded distance is an equivalence relation, is preserved by pre-composition, and is preserved by post-composition with a coarse Lipschitz map (Bounded distance is an equivalence relation and is preserved by pre-composition and by post-composition with a coarse Lipschitz map).

[L2]

Composites of coarse Lipschitz maps and of quasi-isometric embeddings are again such, with explicit constants (Composites of coarse Lipschitz maps and of quasi-isometric embeddings are again such, with explicit constants).

Proof

technique · direct

1.1

For every metric space $X$ , the identity map ${id}_{X}$ is coarse Lipschitz and is its own quasi-inverse, so $X$ is quasi-isometric to itself.

2.1

If $g$ is a quasi-inverse of a quasi-isometry $f : X \to Y$ , then $f$ is a quasi-inverse of $g$ , so $Y$ is quasi-isometric to $X$ .

F1step 1.1

3.1

If $f : X \to Y$ has quasi-inverse $g$ and $h : Y \to Z$ has quasi-inverse $k$ , then $h \circ f$ is coarse Lipschitz and $g \circ k$ is a coarse Lipschitz quasi-inverse of it: the composites $(g \circ k) \circ (h \circ f) = g \circ (k \circ h) \circ f and (h \circ f) \circ (g \circ k) = h \circ (f \circ g) \circ k$ are at bounded distance from the relevant identities by compatibility of bounded distance with composition. Thus quasi-isometry is transitive.

F1L1L2step 2.1∎

Depends on

Used by

Dependency tree · two levels

7 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)