Suppose $F_1$ and $F_2$ are free groups, and suppose $\alpha:F_1 \to F_2$ is a surjective homomorphism. Then, because $F_2$ is free, the homomorphism splits, and we get a subgroup $H$ of $F_1$ isomorphic to $F_2$ and a retraction of $F_1$ onto $H$, i.e., a surjective map to $H$ that restricts to the identity on $H$ (with kernel a normal complement to $H$).
Question: Can we find a freely generating set $A$ for $F_1$ and a freely generating set $B$ for $H$ such that $B$ is a subset of $A$ and the retraction sends all elements of $B$ to themselves and sends all elements of $A \setminus B$ to the identity element?
The corresponding statement for free abelian groups is true: simply pick a (free abelian) generating set for the retraction image and the kernel and take their union to get a freely generating set for the whole group. [Note: Any subgroup of a free abelian group is free abelian.] But this technique of taking a freely generating set for the kernel fails in the non-abelian case, because the kernel is too big.