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.



Does every retraction of free groups arise from projection to a subset of a freely generating set?
No. This is explicitly stated in the paragraph above Theorem 1 of:


[Turner, Edward C, Test words for automorphisms of free groups.
Bull. London Math. Soc. 28 (1996), no. 3, 255--263.](http://blms.oxfordjournals.org/cgi/reprint/28/3/255.pdf)


The author refers to Proposition 1.


**EDIT:**


Let's give an explicit example. Let $F=\langle a,b\rangle$ and let $g=a[b,a]=ab^{-1}a^{-1}ba$. Clearly $\langle g\rangle$ is a retract of $F$. But the Whitehead graph of $g$ is two triangles glued along an edge. This is Whitehead-reduced, so $g$ is not part of a free basis for $F$.


