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For geodesic metric spaces, a quasi-isometry preserves the property of being a Gromov-hyperbolic metric space. Images of metric geodesic sides are uniform quasigeodesics. The Morse lemma for quasi-geodesics keeps them uniformly close to metric geodesic sides in the hyperbolic target. Thinness there and the coarse lower distance bound pull a uniform triangle-thinness constant back to the source. Coarse surjectivity supplies a quasi-inverse for the reverse implication. Consequently a metric geodesic space quasi-isometric to a tree is hyperbolic, with no local finiteness assumption on the tree.

Choose a finite generating set $T$ of $H$. A $d_T$-geodesic between two elements of $H$ maps under the inclusion to a uniform quasigeodesic in ${\mathrm{Cay}}_S(G)$ because $H$ is quasi-isometrically embedded. Since $G$ is a hyperbolic group, the Morse lemma for quasi-geodesics gives a constant $R$ such that this quasigeodesic and the ambient geodesic with the same endpoints have Hausdorff distance at most $R$. Every vertex of the former lies in $H$, so the latter lies in the closed $R$-neighborhood of $H$. Therefore $H$ is a quasiconvex subgroup.

Let $f:X\to T$ be a $(\lambda ,\epsilon )$-quasi-isometry to a tree. The image under $f$ of every geodesic segment in $X$ is a $(\lambda ,\epsilon )$-quasigeodesic in $T$. By the Morse lemma for quasi-geodesics, it lies within a constant $R=R(\lambda ,\epsilon )$ of the tree geodesic with the same endpoints.

Consider a geodesic triangle in $X$. A point $p$ on one side maps within $R$ of the corresponding side of the comparison triangle in $T$. Every geodesic triangle in a tree is $0$-thin, so that comparison side is contained in the other two sides. Those two tree sides are in turn within $R$ of the images of the other two sides of the original triangle. Hence some point $q$ on one of those sides satisfiesThe lower quasi-isometry inequality givesThus $X$ is Gromov-hyperbolic metric space with $\delta =\lambda (2R+\epsilon )$.

Retain the $(\lambda ,\epsilon )$-quasi-isometry $f:X\to T$ and the Morse lemma for quasi-geodesics constant $R$. Given $x,y\in X$, let $m$ be the midpoint of a geodesic $[x,y]$. There is a point $z$ on the tree geodesic $[f(x),f(y)]$ with

For any continuous path $\alpha$ from $x$ to $y$, choose a partition fine enough that consecutive $\alpha (t_i)$ are at distance at most one. Consecutive images under $f$ are then at distance at mostRemoving $z$ separates $f(x)$ from $f(y)$ in the tree along their geodesic, so this finite $D$-chain must contain an element within $D$ of $z$. For the corresponding $t_i$,The lower quasi-isometry inequality yieldsThis constant depends only on the chosen quasi-isometry, so every quasi-tree has the bottleneck property.