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A quasi-isometry $f:X\to Y$ has constants $\lambda \ge 1$, $ϵ\ge 0$ such thatfor all $x,x^{\prime }$, and every $y\in Y$ is within $ϵ$ of some point of $f(X)$. One may use a separate coarse-surjectivity constant and enlarge $ϵ$ to include it. No continuity is required.

Choose $f^{\prime }(y)\in X$ with $d_Y(f(f^{\prime }(y)),y)\le ϵ$. The triangle inequality givesCombining this with the distortion inequalities for $f$ yieldsAlso $d_Y(f(f^{\prime }(f(x))),f(x))\le ϵ$, so the lower distortion inequality gives $d_X(f^{\prime }(f(x)),x)\le 2\lambda ϵ$. Hence every point of $X$ is within $2\lambda ϵ$ of the image of $f^{\prime }$. Taking the larger common error constant shows that $f^{\prime }$ is a $(\lambda ,3\lambda ϵ)$ quasi-isometry. It is a quasi-inverse of a quasi-isometry; both composites are uniformly close to their respective identities.

For a finite symmetric generating alphabet $\Sigma$, the growth function of a finitely generated group isIt counts distinct elements, not words. Relations affect which words coincide, while the resulting coarse growth type is independent of the finite generating set.

Let $f:G\to H$ be a quasi-isometry between the vertex word metrics, with constants $\lambda ,ϵ$. A quasi-isometry of the metric graphs can be rounded to vertices with only bounded changes of these constants. If $f(g)=f(g^{\prime })$, its lower distortion bound gives $d_G(g,g^{\prime })\le \lambda ϵ$. Thus every nonempty fibre has at most $M={\beta }_G(\lceil \lambda ϵ\rceil )$ elements, independently of its centre. Also $f(B_G(1,n))$ lies in the ball of radius $\lambda n+ϵ$ about $f(1)$. Left translation in $H$ preserves ball cardinalities, soApply the same argument to a quasi-inverse of a quasi-isometry to get the reverse comparison. Enlarging all constants to an integer $K$ gives the two inequalities required by the paper's $\lesssim$ relation, and therefore ${\beta }_G{\sim }_e{\beta }_H$. This proves quasi-isometry invariance of the stated growth class.

For the exponential upper bound, put $q=|\Sigma |$. Every element in the radius-$n$ ball has a representative of length at most $n$. Hencewhere the last inequality follows by comparison with the binomial expansion of $(q+1{)}^n$; it also holds for the empty alphabet of the trivial group. Thus every finitely generated group has at most exponential growth, with no finite-relator assumption needed.