Continuous quasi-isometry between Riemannian manifolds Let $(S,d_{1})$ and $(S',d_{2})$ be two proper geodesic metric spaces. If there exists a quasi-isometric embedding $f\colon S\rightarrow S'$, does there exist a $\textbf{continuous}$ quasi-isometric embedding $g\colon S\rightarrow S'$ as well?.
If the answer is affirmative, is $g$ canonically related to $f$?. Concretely, is there a functor relating quasi-isometric embeddings to continuous quasi-isometric embeddings?.
If the answer is negative (which I think is the case), is the statement true at least for complete Riemannian surfaces (assuming quasi-isometries in the geometric group theoretical setting)?.
Edit:
A $\textbf{quasi-isometric embedding}$ $f\colon S\rightarrow S'$ is a function for which there exist positive constants $L,C$ such that
$$L^{-1}d_{1}(x,y)-C\leq d_{2}\big(f(x),f(y)\big)\leq Ld_{1}(x,y)+C$$
for all $x,y\in S$.
 A: For a counterexample, take $(S,d_1)$ to be the Euclidean plane and $(S',d_2)$ to be the union of the integer coordinate lines in the Euclidean plane:
$$S' = (\mathbb R \times \mathbb Z) \cup (\mathbb Z \times \mathbb R)
$$
For a counterexample with complete Riemannian surfaces, start with the same $(S,d_1)$. For $(S',d_2)$, start with the Euclidean plane $S$, consider each integer coordinate square $[m-1,m] \times [n-1,n]$, and then for each $m,n$ replace that square by its connected sum with a torus.
A: What Lee Mosher wrote is, of course, correct. Nevertheless, there are some reasonable assumptions under which one can "approximate" quasi-isometries by continuous quasi-isometries:
Definition. Suppose that $X$ is a Riemannian manifold. I will say that $X$ is uniformly contractible if there is a function $\phi: {\mathbb R}_+\to {\mathbb R}_+$ such that the following holds:
For every $k=0,...,\dim(X)-1$, every continuous map $c: S^k\to  X$ such that $diam(c(S^k))\le D$, extends to a continuous map
$$
\tilde{c}: B^{k+1}\to X
$$
such that
$$
diam(\tilde{c}(B^{k+1}))\le \phi(D).
$$
Here $S^k$ is the unit sphere in ${\mathbb R}^{k+1}$ and $B^{k+1}$ is the closed unit ball bounded by this sphere.
For instance, if $X$ is contractible and is the universal covering space of a compact Riemannian manifold (with the pull-back Riemannian metric), then $X$ is uniformly contractible.
I will leave the following without a proof:
Theorem. Suppose that $X$ is a Riemannian manifold and $Y$ is a uniformly contractible Riemannian manifold. Then for each quasi-isometry $f: X\to Y$ there exists a continuous quasi-isometry $f': X\to Y$ such that
$$
\sup_{x\in X} d_Y(f(x), f'(x)) <\infty.
$$
In Lee Mosher's example, the target Riemannian surface is not simply-connected, hence, is not uniformly contractible.
