= Quasi-inverse of a quasi-isometry
For a <quasi-isometry> $f:X\to Y$ with multiplicative constant $\lambda$ and additive/coarse-surjectivity constant $\epsilon$, choose $f'(y)$ with $d_Y(f(f'(y)),y)\le\epsilon$. The triangle inequality and the two distortion inequalities give
$$
\lambda^{-1}d_Y(y,y')-3\epsilon/\lambda\le d_X(f'(y),f'(y'))\le\lambda d_Y(y,y')+3\lambda\epsilon.
$$
Also $d_X(f'(f(x)),x)\le2\lambda\epsilon$. Hence $f'$ is a <quasi-isometry> and both composites stay uniformly close to their respective identities. No continuity is required.
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