I'll be delighted to get some help in understanding the proof of the first theorem here:
<http://www.math.utah.edu/~malone/QI/notes.pdf>
"If G acts geometrically on X and Y (proper geodesic metric spaces)
then X and Y are quasi-isometric."


In his proof, he fixed $a,b \in X$ and took an arbitrary $q \in Y$ . He then proved that 
$ d\_Y (g\_0 q , g\_n q) \leq R' (d(a,b)+1) $ , which (he claims) finishes the proof.


Can you please explain me how does it implies that X,Y are quasi-isometric?


The definition of Q.I I know is that if we have a map $f:X \to Y$ such that there exist some constants $L,A$ , such that for every $x\_1 , x\_2 \in X$ , $y \in Y$ :
$ \frac{1}{L} d(x\_1,x\_2) -A \leq d(f(x\_1) , f(x\_2) ) \leq Ld(x\_1,x\_2) +A$ and $d(y, f(X) \leq A$ . 


I'll be glad to receive an explanation .


Thanks in advance !



Fundamental Lemma Of Geometric Group Theory
In the context of that proof, $p$ is a base point for $X$, $q$ is a base point for $Y$, and the function $f : X \to Y$ is defined on each $x \in X$ by first choosing $g \in G$ so that $d(x,gp) \le R$ and then defining $f(x) = gq$. The proof is showing that the function $f$ is a quasi-isometry.


