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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.

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  • $\begingroup$ I have a feeling the answer is 'no'. But it may take me a while to dig out a reference. $\endgroup$ – HJRW May 25 '10 at 22:07
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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.

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 $F$.

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  • $\begingroup$ If $F_1$ maps onto $F_2$ , surely there is a basis $A$ of $F_1$ with subset $B$ such that <$B$> maps isomorphically onto $F_2$ and $A-B$ to the identity. Is the problem then that the subgroup $H$ is being determined beforehand? $\endgroup$ – Steve D May 25 '10 at 22:58
  • $\begingroup$ What do you mean by 'surely'? There isn't. $\endgroup$ – HJRW May 25 '10 at 23:18
  • $\begingroup$ Steve: Here's where I think you're confused. It's true that a subset B of A maps surjectively to H but that doesn't mean it (or its image) freely generates H. $\endgroup$ – Vipul Naik May 25 '10 at 23:29
  • $\begingroup$ Vipul, as the example I have added shows, that isn't quite the point. In the example, the image of g does freely generate. $\endgroup$ – HJRW May 25 '10 at 23:32
  • $\begingroup$ @Henry: Yes, there is. See Proposition 2.12 in Lyndon-Schupp, pg. 9 , for example. But it's clear this depends on the complement $H$ you choose, as in your example (where the complement <a> would be OK). $\endgroup$ – Steve D May 25 '10 at 23:37

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