$T_3$ is quasi isometric to $T_4$

Question

I have a question which asks me to show that $T_3$ is quasi isometric to $T_4$, that is the three and 4 valence trees.

I know that this means that I have to define a map $f:T_3\rightarrow T_4$ such that we have $\lambda,C,K$ such that for all $x,y\in T_3$ we have:

$\frac{1}{\lambda}d(f(x),f(y))\leq d(x,y)\leq \lambda(f(x),f(y)) +C$ and $\forall x\in T_4$ we have a $y\in T_3$ such that $d(x,f(y))\leq K$

But I am really not sure how I want to be defining my map, never mind showing that it is in fact a quasi isometry

Answer 1

Here is a hint: What is the effect of taking a quotient of $T_3$ by collapsing some edge to a point?