Q:

Free group with 3 generators and free group with 2 generators

Let $F=<a,b>$ the free group with 2 generators and $G=<u,v,w>$ the free group with 3 generators.  How can I show that the map $f:G\rightarrow F$ that sends $u\rightarrow ab,v\rightarrow a^2b^2, w\rightarrow a^3b^3$ and extends to $G$ with the obvious way  is a monomorphism? 

I can show that $f$ is a homomorphism but I can't show that it is $1-1$.
  Is there a way to maybe show it has an inverse? 

A:

Note that the subgroup $H$ generated by $ab$, $a^2b^2$, $a^3b^3$ is also generated by $x = ab$, $y = a^2b^2(ab)^{-1} = a^2ba^{-1}$, and $z = a^3(ab)^{-2}= a^3ba^{-2}$. It is sufficient to prove that $H$ is freely generated by $x,y,z$ because if it is then it is also freely generated by $ab$, $a^2b^2$, and $a^3b^3$.
The point about doing that replacement is that the set $x,y,z$ is Nielsen reduced, but you do not need to know what that means to prove the result. 
Note that in any non-cancelling product of two of $x,y,z,x^{-1},y^{-1},z^{-1}$, the terms $b$ or $b^{-1}$ in the generators do not cancel.  For example $yx = a^2b^2$, or $yz^{-1}= a^2bab^{-1}a^{-3}$.
So in any reduced word of length $n$ in the generators $x,y,z$, there will be exactly $n$ letters equal to $b$ or $b^{-1}$. Hence no such word reduces to the identity and it follows that $H$ is freely generated by $x,y,z$.

