Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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Hyperbolicity is transported by a quasi isometry

Statement

Assume AC for the Morse supplier. If f:X→Y is a (λ,ε)-quasi-isometric embedding of geodesic spaces and Y is δ-slim, then X is λ(2M(λ,ε,δ)+δ+ε)-slim. Hence quasi-isometric geodesic spaces share hyperbolicity.

Facts & Assumptions

Given: The displayed quasi-isometric embedding, with λ≥1, ε,δ≥0; write M=92λ2(ε+3δ).

[F1]

Under AC, Morse stability with explicit parameter dependence gives both Hausdorff inclusions with this M, for every specified endpoint geodesic, without continuity or properness.

[F2]

Under AC, attained coarse density supplies a controlled quasi-isometric inverse by A quasi isometry of geodesic spaces has a controlled coarse inverse.

[F3]

A quasi-isometry admits a coarse Lipschitz quasi-inverse with both composites at bounded distance from the identities (Coarsely dense subsets, quasi-inverses and quasi-isometries).

[A1]

AC has the meaning in The Axiom of Choice and is used only through F1 and, for the final converse, F2.

Proof

1.1givenF1A1

Fix any geodesic triangle in X with vertices a,b,c, and any point p on its specified side [a,b]. Compose the isometric parametrization of each of the three sides with f. The given inequalities make each composition a (λ,ε)-quasi-geodesic on a nonempty compact interval, even if f is discontinuous. Choose three target endpoint geodesics Gab,Gbc,Gca. F1 applies separately to every one of them.

2.1step 1.1givenF1algebra

Fix h>0. The first Hausdorff inclusion gives p′∈Gab with dY(f(p),p′)<M+h. Target slimness gives q′∈Gbc∪Gca with dY(p′,q′)<δ+h. The reverse Hausdorff inclusion on whichever side contains q′ gives q∈[b,c]∪[c,a] with dY(q′,f(q))<M+h. Hence dY(f(p),f(q))<2M+δ+3h, and the lower embedding inequality gives dX(p,q)<λ(2M+δ+ε+3h). Only finitely many approximate witnesses are used; there is no assumption that the image of f or of a side has closest points.

3.1step 2.1givenalgebra

Taking the infimum over the other two source sides and then letting h decrease to zero proves their distance from p is at most λ(2M+δ+ε). The chosen source triangle, side and point were arbitrary, proving the exact stated slimness constant. Repeated vertices and zero parameters are included: a side can have a one-point interval, and the argument divides only by the positive λ.

4.1step 3.1F2F3A1algebra∎

Finally let f be a quasi-isometry in F3's convention, with supplied coarse inverse h. If h has coarse Lipschitz constants A,B and dX(hf(x),x)≤D, then dX(x,x′)≤2D+A dY(fx,fx′)+B. Replace A by max⁡{1,A}, combine this lower bound for f with its coarse Lipschitz upper bound, and enlarge the two constants to obtain some (λ,ε) embedding inequalities. The other composite bound gives attained coarse density. F2 therefore supplies a controlled inverse g which is a quasi-isometric embedding. If Y is hyperbolic, step 3.1 applied to f proves X hyperbolic; if X is hyperbolic, apply the same result to g to prove Y hyperbolic. If the spaces are empty, F3 forces both empty and there are no triangles to check. This proves invariance with the stated AC assumption.

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