Quasi isometries extend to boundary homeomorphisms

First put $f$ in quantitative form. If $h$ is a coarse Lipschitz inverse with constants $A,B$ and $d_X(hf(x),x)\le D$, then $d_X(x,x^{\mathrm{\prime }})\le 2D+A{d}_Y(fx,f{x}^{\mathrm{\prime }})+B\mathrm{.}$ Enlarging $A$ to at least one and combining with the upper coarse Lipschitz bound for $f$ gives $(\lambda ,\epsilon )$ embedding inequalities for some $\lambda \ge 1$, $\epsilon \ge 0$. The bound on $fh$ gives attained coarse density with a finite radius $R$. If either space is empty, F4 forces both empty; there are no Gromov sequences and the boundary assertion is the empty homeomorphism. Henceforth take $o\in X$, use $fo$ as target basepoint, choose target slimness ${\delta }_Y$, and put ${\kappa }_Y=3{\delta }_Y$ and $C_f=\epsilon +M(\lambda ,\epsilon ,{\delta }_Y)+2{\kappa }_Y$.

For any $x,y\in X$, let $P=(x\mathrm{\mid }y{)}_o$ and choose a source segment $[x,y]$. For each $z$ on it, the two triangle inequalities through $z$ give $P\le d_X(o,z)$. Put $a=fx$, $b=fy$, $v=fo$ and choose a target segment $G=[a,b]$. Let $m\in G$ have distance $(v\mathrm{\mid }b{)}_a$ from $a$. This parameter belongs to $[0,d_Y(a,b)]$. Expanding products yields $d_Y(a,v)+d_Y(b,m)=d_Y(b,v)+d_Y(a,m)=d_Y(a,b)+(a\mathrm{\mid }b{)}_v\mathrm{.}$ The four-point condition in F5 gives $d_Y(v,m)\le (a\mathrm{\mid }b{)}_v+2{\kappa }_Y$. F2, applied to the image of the source segment, gives for each $t>0$ a $z\in [x,y]$ with $d_Y(m,fz)<M+t$. Therefore ${\lambda }^{-1}P-\epsilon \le d_Y(fo,fz)<(fx\mathrm{\mid }fy{)}_{fo}+2{\kappa }_Y+M+t\mathrm{.}$ Letting $t$ decrease to zero proves the finite-point threshold estimate $(fx\mathrm{\mid }fy{)}_{fo}\ge {\lambda }^{-1}(x\mathrm{\mid }y{)}_o-C_f\mathrm{.}$ This uses only finite segments; neither rays, compact balls nor properness enter.

If $(x_n)$ is Gromov, the estimate in step 2.1 sends every joint cutoff for $(x_n\mathrm{\mid }{x}_m{)}_o$ at threshold $\lambda (T+C_f)$ to a joint cutoff at threshold $T$ in $Y$. Thus $(f{x}_n)$ is Gromov. The same calculation with two different sequences proves that equivalent representatives have equivalent images. Define $\mathrm{\partial }f([x_n])=[f{x}_n]$; the equivalence just proved makes this a function without selecting representatives simultaneously.

For each fixed representative pair $x\in \xi$, $y\in \eta$, take tail infima and then their supremum in step 2.1. For a finite representative product this gives $P_{fo}(fx,fy)\ge {\lambda }^{-1}{P}_o(x,y)-C_f$; if it is infinite, the same conclusion means that every finite threshold holds. The image pair is among the pairs defining $B_{fo}(\mathrm{\partial }f\xi ,\mathrm{\partial }f\eta )$. Taking the supremum over source representative pairs therefore yields $B_{fo}(\mathrm{\partial }f\xi ,\mathrm{\partial }f\eta )\ge {\lambda }^{-1}{B}_o(\xi ,\eta )-C_f,$ with the infinite case understood through finite thresholds. Let $O\subseteq \mathrm{\partial }Y$ be open and $\mathrm{\partial }f\xi \in O$. F1 gives a $T$ with $U_T(\mathrm{\partial }f\xi )\subseteq O$. The displayed estimate gives $U_{\lambda (T+C_f)}(\xi )\subseteq (\mathrm{\partial }f{)}^{-1}(O)$. Applying F1's open-set criterion at every such $\xi$ proves continuity. Basepoint independence in F1 removes the special choice $fo$. This proof does not assert that threshold sets are open.

Suppose $f^{\mathrm{\prime }}$ is at distance at most $E$ from $f$ at every point. At the fixed target basepoint $fo$, changing one argument of a finite product by at most $E$ changes that product by at most $E$, by its formula and the reverse triangle inequality. Changing both arguments costs at most $2E$. Thus images under $f^{\mathrm{\prime }}$ of a Gromov sequence are Gromov, and the mixed products $(f{x}_n\mathrm{\mid }{f}^{\mathrm{\prime }}{x}_m{)}_{fo}$ differ from $(f{x}_n\mathrm{\mid }f{x}_m{)}_{fo}$ by at most $E$. They jointly diverge, so $[f{x}_n]=[f^{\mathrm{\prime }}{x}_n]$. In particular bounded-distance quasi-isometries induce the same boundary map. The identity map induces the identity on classes.

If $k:Y\to Z$ is another quasi-isometry, the embedding inequalities compose: with parameters $(\mu ,\eta )$ for $k$, the composite has parameters $(\mu \lambda ,\mu \epsilon +\eta )$. Its attained density follows by choosing an image point within the density radius for $k$, then one within the density radius for $f$, and using the upper bound for $k$. Alternatively its supplied coarse inverse is the composition of the two supplied inverses, since the coarse Lipschitz inequalities preserve bounded errors. Thus the preceding construction applies to $k\circ f$, and on every representing sequence it gives $[(k\circ f)(x_n)]=[k(f(x_n))]$. Consequently $\mathrm{\partial }(k\circ f)=\mathrm{\partial }k\circ \mathrm{\partial }f$.

Apply F3 with the attained radius $R$ from step 1.1. Under A1 it supplies $g:Y\to X$ with $d_Y(fg(y),y)\le R$, $d_X(gf(x),x)\le \lambda (R+\epsilon )$, and embedding constants $(\lambda ,\lambda (2R+\epsilon ))$. The second bound makes its image coarsely dense, and these two bounds make it a quasi-inverse in F4. Thus steps 2.1–4.1 apply with the roles of $X,Y$ reversed and prove that $\mathrm{\partial }g$ is continuous. Steps 4.2–4.3 show $\mathrm{\partial }g\circ \mathrm{\partial }f={\mathrm{id}}_{\mathrm{\partial }X}$ and $\mathrm{\partial }f\circ \mathrm{\partial }g={\mathrm{id}}_{\mathrm{\partial }Y}$. Therefore $\mathrm{\partial }f$ is a homeomorphism. Empty or singleton boundaries and zero slimness/additive errors obey the same threshold and inverse arguments. AC has been used precisely through the Morse projection families and the coarse-inverse selection; no properness assumption was added.