# Equivalence of surjections from a surface group to a free group

Let $$g \geq 2$$. Let $$S = \langle a_1,b_2,...,a_g,b_g | [a_1,b_1] \cdots [a_g,b_g] \rangle$$ be the fundamental group of a genus $$g$$ surface and let $$F_g$$ be a free group with $$g$$ generators. Given two surjections $$f_1,f_2 : S \to F_g$$ is there a way to determine if there are automophisms $$\phi: S \to S$$ and $$\psi: F_g \to F_g$$ so that $$f_1 = \phi \circ f_2 \circ \psi$$?

Is there an example of two surjections $$f_1,f_2$$ that are not equivalent in the above way?

I asked the question on MSE before but didn't get much.

• A naive question: is it clear such a surjection exists? – PseudoNeo Oct 15 '18 at 2:00
• @PseudoNeo Algebraically, yes: kill all of the $b_i$. Geometrically, yes: the surface is the boundary of a handlebody, equivalent to a wedge of $g$ circles. – Mike Miller Oct 15 '18 at 2:02
• Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_{2g}$) and was very confused. – PseudoNeo Oct 15 '18 at 2:03
• For context, there's no homomorphism onto $F_{g+1}$. Out of curiosity, what can be said of the set of surjective homomorphisms $\pi_1(S)\to F_k$ modulo $Aut(\pi_1(S))\times Aut(F_k)$, when $1\le k<g$? is it infinite? – YCor Oct 15 '18 at 8:15

This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly different language, that is they prove that any such epimorphism is induced by choosing a genus $$g$$ handlebody.
• Do you know if the set of surjective homomorphisms onto $F_g$ is a singleton modulo $Aut(\pi_1(S))$ (instead of modding out by $Aut(\pi_1(S))\times Aut(F_g)$)? this is related to the question whether automorphisms of $F_k$ can be lifted to automorphisms of $\pi_1(S)$. – YCor Oct 15 '18 at 8:17