15
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • $\begingroup$ A naive question: is it clear such a surjection exists? $\endgroup$
    – PseudoNeo
    Commented Oct 15, 2018 at 2:00
  • 3
    $\begingroup$ @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. $\endgroup$
    – mme
    Commented Oct 15, 2018 at 2:02
  • $\begingroup$ Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_{2g}$) and was very confused. $\endgroup$
    – PseudoNeo
    Commented Oct 15, 2018 at 2:03
  • 2
    $\begingroup$ 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? $\endgroup$
    – YCor
    Commented Oct 15, 2018 at 8:15

1 Answer 1

Reset to default
9
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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 $\pi_1(S)$. $\endgroup$
    – YCor
    Commented Oct 15, 2018 at 8:17
  • $\begingroup$ I would say no, because it seems unlikely to me that any automorphism of the free group induces an homeomorphism of the handlebody, and barring that I have no idea about your lifting problem and I see no indication about it in Leininger--Reid's arguments (but I'm not a specialist so I might be missing something obvious). $\endgroup$
    – Jean Raimbault
    Commented Oct 15, 2018 at 16:45
  • $\begingroup$ @JeanRaimbault Thanks this is perfect. I am trying to do this algorithmically also - is it clear how lemma 2.2 can be made effective? I.e. given the surface with the labeled generators and a surjection to a free group, how do we find the set of curves to attach to the surface to realize the handlebody that realizes the surjection? $\endgroup$
    – user101010
    Commented Oct 18, 2018 at 2:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .