Let (X,d) and (Y,d') be metric spaces. Let f be an (L,C) quasi-isometry from X to Y. That is,
(1/L)d(x₁,x₂)-C ≤ d'(f(x₁),f(x₂)) ≤ Ld(x₁,x₂)+C for all x₁, x₂ in X. Also,
for every y in Y, there is an x in X such that d'(f(x),y) ≤ C.
We define a quasi-inverse g from Y to X by setting g(y)=x where x satisfies d'(f(x),y) ≤ C. Such an x exists because of condition 2 above; pick any such x. How do we show that this is a quasi-isometry? I know that its constants, call them L' and C', depend only on L and C. In a previous problem, I showed that the composition of two quasi-isometries is a quasi-isometry. I thought the idea would be similar for this problem, but I couldn't work it out. Any suggestions?