Retract of $F_2$ onto $[F_2,F_2]$

Question

I'm trying to show that the commutator subgroup $[F_2,F_2]$ is not a retract of $F_2$. I was trying to do the proof by contradiction so I assume there is a retract $f: F_2 \to [F_2,F_2]$, then I know this is surjective and then I tried to apply abelianization to get the map $f^{ab}: \mathbb{Z}^2 \to {1}$ but this doesn't give me any contradictions since this map is surjective. Then I was thinking to apply the universal property of quotient groups somehow because I know $F_2/ [F_2,F_2] = \mathbb{Z}^2$ and that is a surjective map but I can't find anything that gives me a contradiction. I'm not sure where to go from here.

Answer 1

Abelianization does give you a contradiction; $[F_2, F_2]$ is a free group on countably many generators, so its abelianization is the free abelian group on countably many generators, and $f^{ab}$ can't be surjective.

Answer 2

We can make the previous argument more precise. The group $[F_2,F_2]$ is a free group. It is freely generated by elements of the form $a^n b^m [a,b]b^{-m}a^{-n}$, for $n, m\in \bf Z$, as it is the fundmantal group of the Cayley graph of $\bf Z^2$ viewed as a quotient $F_2/[F_2,F_2]$, so it is not finitely generated.