 Solution

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.