We write $cl$ for the commutator length, i.e. the least number of commutators which multiply to a given element of a group. 


Given an element $g$ in the commutator subgroup of the free group $G=F\_2$ on two generators, is it true that 
$$cl\_G(g) = \displaystyle \max\_{\mbox{H < G finite index normal}} cl\_{G/H} (g \mod H)$$ ?



Commutator length modulo finite index subgroups
I think the answer is no and that actually the "finite quotient" commutator length (defined by your formula) is bounded on $[F\_2,F\_2]$.


Indeed by Nikolov-Segal, in the profinite completion $P$ of $F\_2$, the derived subgroup $[P,P]$ is closed; since the set $C$ of commutators is compact, it follows by a Baire argument that $[P,P]$ is boundedly generated by $C$ (first get by Baire that some open neighborhood of the identity of $[P,P]$ has bounded commutator length in $P$ and then use compactness). Observe on the other hand that $[F\_2,F\_2]=[P,P]\cap F\_2$, which is essentially trivial since the abelian group $F\_2/[F\_2,F\_2]$ is residually finite. So in any finite quotient of $F\_2$, the commutator length of $w\in [F\_2,F\_2]$ is bounded by the universal number that arises as upper bound of the commutator length of $[P,P]$ in $P$.


On the other hand, the commutator length of $F\_2=\langle x,y\rangle$ is unbounded on $[F\_2,F\_2]$, as the commutator length of $[x,y]^n$ grows linearly (I think it's due to Bavard).


