Quasi-inverse of a quasi-isometry

For a quasi-isometry $f:X\to Y$ with multiplicative constant $\lambda$ and additive/coarse-surjectivity constant $ϵ$, choose $f^{\prime }(y)$ with $d_Y(f(f^{\prime }(y)),y)\le ϵ$. The triangle inequality and the two distortion inequalities giveAlso $d_X(f^{\prime }(f(x)),x)\le 2\lambda ϵ$. Hence $f^{\prime }$ is a quasi-isometry and both composites stay uniformly close to their respective identities. No continuity is required.