Characterization of quasi-isometric subsets.

Question

Suppose $(X,d)$ is a metric space and $Y \subset X$ is quasi-isometric to $X$. That is, suppose \begin{equation} X \sim_{QI} Y \subset X. \end{equation} I was wondering, is the inclusion $i: Y \to X$ necessarily a quasi-isometry? Equivalently, does there exists $M>0$ such that \begin{equation} \forall x \in X \hspace{3pt} \exists y_x \in Y \hspace{3pt} d(y_x,x)< M ? \end{equation} Perhaps I should mention that the metric on $Y$ is the restriction of the metric on $X$, that is, $d|_{Y\times Y}$. I believe I'm having difficulties answering this because I don't know many examples. I would also be very grateful if someone could share what kind of examples do they think about when solving this kind of problems. By the way, I just started studying geometric group theory (I'm currently reading section 5.4 on Clara Loh's book), so an elementary argument would be greatly appreciated.

Answer 1

For a counterexample, consider the rank 2 free group $F\langle a,b \rangle$ with Cayley graph $\Gamma\langle a,b \rangle$. Inside $F\langle a,b \rangle$ we have a rank 2 free group $F \langle a^2,b^2 \rangle$ of infinite index. One could take our counterexample to be simply inclusion $F \langle a^2,b^2 \rangle \hookrightarrow F\langle a,b \rangle$, using word metrics on each.

But to make this more geometric, let's consider the corresponding inclusion of Cayley graphs $i : \Gamma\langle a^2,b^2 \rangle \hookrightarrow \Gamma\langle a,b\rangle$. Clearly $i$ is not a quasi-isometry, because for example points on the ray $ababababababab....$ in $\Gamma\langle a,b\rangle$ get farther and farther from $\Gamma\langle a^2,b^2 \rangle$.

But there does exist a quasi-isometry $h : \Gamma\langle a,b \rangle \to \Gamma\langle a^2,b^2 \rangle$, in fact we can take $h$ to be a homeomorphism that multiplies the metric by a factor of $2$, taking each $a$ edge of the domain to an $a^2$ edge of the range, and similarly for $b$.

Answer 2

A simple counterexample:

$X$: the upper half-plane $\{re^{i\theta}:r\ge 0,0\le\theta\le\pi\}$, with the Euclidean distance

$Y$: the upper right corner $\{re^{i\theta}:r\ge 0,0\le\theta\le\pi/2\}$.

Clearly the isometric inclusion $Y\subset X$ is not a QI. But $Y$ is QI to $X$ via the mapping $f:re^{i\theta}\mapsto re^{2i\theta}$ (that is, $f(z)=z^2/|z|$): indeed $f$ is a $(1,2)$-bilipschitz homeomorphism.