Continuous quasi-isometry between Riemannian manifolds

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.

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.