Free groups - extending quasi isometry to commuting square Let's assume one has a finitely generated free group on $n$- generators, and two homomorphisms : $F \rightarrow F/k_1$, $F \rightarrow F/k_2)$ (for $k_i$ normal subgroup ) such that $F/k_1$ is quasi-isometric to $F/k_2$ and we have one fixed QI $G: F/k_1 \rightarrow F/k_2$. Does it mean that there is a quasi isometry $H:F \rightarrow F$ giving a commuting square up to a bounded distance ? Since $F$ is a tree it is easy to show that we can inductively find quasi isometric embedding $H:F \rightarrow F$ which gives a commuting square but is it possible to extend it somehow to full quasi isometry?
Edit:
We have one fixed quasi-isometry $G: F/k_1 \rightarrow F/k_2$ .
For anyone reading this post the comments are now irrelevant - I've updated it all to the post itself.
Here's how I do my lift.
We will denote the lift $H$. Let's denote projection $F \rightarrow F/k_1$ with $p$ and projection $F \rightarrow F/k_2$ with $q$. We equip $F$ with standard symmetric generating set $<a_1,..a_n>$ and we fix their projection images as generating sets for $F/k_i$. Additionaly let $c_i$ be a reduced word in $ k_2$ beggining with letter $a_i$ and of minimal norm (it always can be found if $k_2$ is not trivial). $e$ denotes the neutral element. We build our lift inductively. 
Start with an injective lift of $\lbrace{ e,a_1,..,a_n \rbrace}$ and assume that $H(e) =e$ (for technical reasons and also I am specifically interested in a QI $G$ which sends neutral element to itself).
By injective I mean the following : if$ |b| < |d| $ then $|H(b)| < |H(d)|$.
Now assume we have an injective lift of elements from the ball of radius $n$ -$B(e,n) = \lbrace{ d \in F : |d| \le n \rbrace}$. Take any vertex $v$ of norm $n$  - we will describe lifting of elements $\gamma \in F$ such that $|\gamma^{-1} \cdot v| =1 $ and $|\gamma| > |v|$ (i.e. vertices which share one edge with $v$ but are further from the neutral element).
Let's denote such vertices $\gamma_1,..,\gamma_{n-1}$. $G$ is a QI hence $d_{F/k_2}(G(p(v)),G(p(\gamma_{i})) \le K $ for some uniform constant $K$.
As a result $G(p(v))^{-1} \cdot G(p(\gamma_i)) = q(a_{i_{1}}) \cdot ... \cdot q(a_{i_{k_{i}}})$ for $i = \lbrace{1,..,n-1 \rbrace}$ and for all $i$ we have that $ i_{k_{i}} \le K$.( it is the definition of the word metric $G(p(v))^{-1} \cdot G(p(\gamma_i))$ can be written as a word shorter then $K$ but for each $i$ word's letters and its lenght may differ -but we know they are uniformly bounded!).
Since we have a lift of $v$  - denoted $H(v)$ , we can lift $\gamma_i$ to $H(\gamma_i) = H(v) \cdot a_{i_{1}} \cdot ... \cdot a_{i_{k_{i}}} \cdot c_{j_{i}}$.
So far the choice of $c_{j_{i}}$ may be arbitrary - the lift will give us commuting square and will be coarse Lipschitz. But it should additionaly be QI embedding hence we choose $c_{j_{i}}$ in a special way - forcing the lifts to "spread".
For given $i$ we choose $c_{j_{i}}$  in such a way that:
i) $j_i \neq i_{k_{i}}$ hence word $ H(v) \cdot a_{i_{1}} \cdot ... \cdot a_{i_{k_{i}}} \cdot c_{j_{i}}$ is a reduced word.
ii) we choose them in such a way that if $G(p(\gamma_m)) = G(p(\gamma_r)) $ then $c_{j_{m}} \neq c_{j_{r}}$ hence lifts $H(\gamma_m),H(\gamma_r$ separate at vertex $H(v)$
iii) and we even choose them so that $c_{j_{i}} \neq c_{j_{r}}$ if $ i \neq r$.
Such choices can obviously be made. It ends the construction.
Now let's take any two points $u,t \in F$ such that $d(u,t) = n$ it means that they are connected by a unique shortest path $d_0,..,d_n$ such that $d_0 = u,d_n =t$ and $d(d_i,d_{i+1})=1$. This path divides into two parts: 
$d_0,...,d_{s},d_{s+1},..,d_n$
For $i \le s $ $|d_{i}| <|d_{i-1}|$ and for $i \ge s $ we have that $|d_{i+1}| >|d_{i}|$ .
The lift of the first part of the path path gives us path $H(d_0),..,H(d_s)$ such that $1 \le d(H(d_i),H(d_{i+1})) \le K'$ ($K'$ is a little bit bigger then $K$ because we have added $c_j$'s but nonetheless it is uniform. Now the very construction of the lift assures us that $ H(d_{i})$ lies in the complement of $F - H(d_{i+1})$ which does not contain lift of neutral element - $H(e)$ hence If you take the shortest path connecting $H(d_0),H(d_s)$ - $f_0,..,f_k$ ($d(f_j,f_{j+1}) =1$) then $H(d_i) $ lay on this path and if $ f_{j}$ is a vertex equal to $H(d_i)$ and $ f_l$ is a vertex equal to $H(d_{i+1})$ then $ l > j$ which means that $ d(H(d_0),H(d_s)) = k \ge \frac{1}{M+1} \cdot s = \frac{1}{M+1} \cdot d(d_0,d_s)$ ( informally speaking path $H(d_0,...,H(d_s)$ always flows forward from $H(e)$, never backwards
)
We can conduct the same reasoning with the second part - $d_s,...,d_n$ - hence $ d(H(d_s),H(d_n)) = k \ge \frac{1}{M+1} \cdot (n-s) = \frac{1}{M+1} \cdot d(d_s,d_n)$
All that is left is to analyse the critical three points - $d_{s-1},d_{s},d_{s+1}$.
We know that $|H(d_{s-1})| < |H(d_{s})| $ and that $ |H(d_{s+1}) < |H(d_{s})| $
But we chose $c_j$ in such a way that points $ H(d_{s-1}),H(d_{s}),H(d_{s+1})$ do not belong to one geodesic (. Hence if $r_0,..,r_m$ is the shortest path connecting $d_{s-1}$ with $d_{s+1}$  then $1 \le m \le K' $ hence path $H(d_0),..,H(d_s),r_0,..,r_m,H(d_{s+1}),..,H(d_n)$ is path joining $H(d_0)$ with $H(d_n)$ and it only "flows" in one direction hence $d(H(d_0),H(d_n)) \ge \frac{1}{M+1} \cdot (s + n-s) $ 
 A: What you're asking for is not possible in general.  You can't extend the quasi-isometry of the quotient groups to a quasi-isometry of the free groups.  Here is an example when it cannot be done:
$$
\newcommand{\lra}{\longrightarrow}
\newcommand{\da}{\left\downarrow\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{ccccccccc}
1 & \lra & 1 & \lra & \langle x,y,z\rangle & \lra & \langle x,y,z\rangle & \lra & 1\\
\da & & \da & & \da & & \,\,\,\,\da\phi & & \da & \\
1 & \lra & \langle a\rangle & \lra & \langle a,b,c\rangle & \lra & \langle b,c\rangle & \lra & 1
\end{array}
$$
The map $\phi$ is defined not just as a quasi-isometry, but a group homomorphism: $\phi(x)=b$, $\phi(y)=cbc^{-1}$, and $\phi(z)=c^2$.  This is an injective group homomorphism which embeds $F_3$ as a finite index subgroup of $F_2$.  Since $\phi$ is an injective homomorphism, it's a quasi-isometric embedding of $\langle x,y,z\rangle$ into $F_2$.  Since the image is finite index, $\phi$ is coarsely surjective and is therefore a quasi-isometry between $\langle x,y,z\rangle$ and $\langle b,c\rangle$.
Assume $\Phi$ is a quasi-isometry making the diagram commute. Let let $q_1$ and $q_2$ denote the quotient maps so that $q_2\circ\Phi$ is a bounded distance from $\phi\circ q_1$. Well $\phi\circ q_1$ is a QI, since $q1$ is the identity and $\phi$ is a QI.  If you post-compose a quasi-isometry with function which is bounded distance from the identity, you get another QI. Thus, $q_2\circ \Phi$ must be a QI.  But that's impossible since $\Phi$ was assumed to be a QI, and $q_2$ is most definitely NOT a QI.  So $\phi$ cannot be lifted to a quasi-isometry.
As a note, if you dislike that one of the kernels is trivial, then you can stick an extra generator in each of the $F_3$'s (replace $\langle x,y,z\rangle$ with $\langle x,y,z,w\rangle$ and similarly for the other $F_3$).  Then just kill that generator so that one of the quotients is $F_2$ and the other is $\mathbb{Z}$.  It's not as clear that the quasi-isometry can't be extended, but I believe it's still true.
