A map is a quasi-isometry exactly when it is a quasi-isometric embedding with coarsely dense image

If a map is a quasi-isometric embedding with coarsely dense image, the previous theorem supplies a quasi-inverse quasi-isometric embedding, so the map is a quasi-isometry.

Conversely, let $g$ be a coarse Lipschitz quasi-inverse of the coarse Lipschitz map $f$. If $d_Y(f(g(y)),y)\le R$ for every $y\in Y$, then every target point lies within distance $R$ of $f[X]$, so the image of $f$ is coarsely dense.

Choose coarse Lipschitz bounds $d_Y(fx,f{x}^{\mathrm{\prime }})\le A_f{d}_X(x,x^{\mathrm{\prime }})+B_f$ and $d_X(gy,g{y}^{\mathrm{\prime }})\le A_g{d}_Y(y,y^{\mathrm{\prime }})+B_g$, enlarging $A_g$ to at least $1$. If $d_X(g(f(x)),x)\le D$ for every $x$, the triangle inequality gives $d_X(x,x^{\mathrm{\prime }})\le 2D+A_g{d}_Y(fx,f{x}^{\mathrm{\prime }})+B_g$. Thus $d_Y(fx,f{x}^{\mathrm{\prime }})\ge A_g^{-1}{d}_X(x,x^{\mathrm{\prime }})-(2D+B_g)\mathrm{/}{A}_g$. Enlarging the multiplicative and additive constants to dominate both this lower bound and the upper bound for $f$ makes $f$ a quasi-isometric embedding. Combined with step 1.2, this proves the converse. AC is used only for the forward implication through [L1].