Quasi-isometric embedding and Quasi-isometry Let $X$ and $Y$ be geodesic metric spaces. Suppose there are quasi-isometric embeddings $f:X \rightarrow Y$, $g:Y \rightarrow X$. Then, can we say there is a quasi-isometry from $X$ to $Y$? I tried to make a quasi-isometry by using the quasi-inverses, but it was not successful.. is there a counterexample?
 A: Consider the infinite binary rooted tree $T$, and consider the tree $T'$ obtained by gluing to $T$ an infinite ray to the root. Then there are quasi-isometric embeddings between $T$ and $T'$ in both directions but no quasi-isometry between them, since their boundary are not homeomorphic (Cantor versus (Cantor plus isolated point)).
A: In general, $X$ and $Y$ need not be quasi-isometric.  Here's a counterexample:
For each positive integer $n$, let $A_n$ be an interval of length $n$.  Let $A$ be the space obtained by identifying a single endpoint of all the (countably-many) $A_n$, with the induced path metric.  In essence, $A$ consists of a single basepoint with intervals of arbitrary length coming out of it.
Let $X$ consist of countably infinitely many copies of the interval $[0, \infty)$, with the point $0$ identified in each copy, given the induced path metric.  In essence, $X$ is a single basepoint with countably many geodesic rays coming out of it.
Let $Y = X \cup A$, where $X$ and $A$ have their basepoints identified.
Now, $X$ clearly has a quasi-isometric embedding into $Y$ (just map it to the copy of $X$).  $Y$ has a quasi-isometric embedding into $X$ as well: Map the $n^{\mathrm{th}}$ inifinite geodesic ray in $Y$ to the $(2n)^{\mathrm{th}}$ geodesic ray in $X$, and the $n^{\mathrm{th}}$ finite interval in $Y$ to an initial segment in the $(2n+1)^{\mathrm{th}}$ infinite geodesic ray in $X$.
However, there is no quasi-isometry between $X$ and $Y$; I'll leave the technical details of the proof to you but basically any quasi-isometry from $X$ to $Y$ would need to map points in one of the infinite intervals in $X$ arbitrarily far down a finite interval in $Y$, which will mean that points arbitrarily far apart in $X$ will have images that are boundedly far apart in $Y$ (since the ray has "nowhere to go" once it reaches the end of the finite interval).
