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 $f_1 = \phi \circ f_2 \circ \psi$?

  • $\begingroup$ You can rephrase your question in terms of equations over free groups: each such surjection corresponds to a solution to the equation $[x_1, y_1]\cdots[x_g, y_g]=1$ in the free group $F_g$, and you want to know if two such solutions are "equivalent". There is a well-developed theory of equations over free groups, using gadgets called "Makanin-Razborov" diagrams. These describe all solutions using automorphisms of "limit groups", and your group is a limit group (although may not be a limit group for the equations you want...). $\endgroup$
    – user1729
    Mar 21, 2018 at 10:26
  • $\begingroup$ For a reference, see: Sela, Zlil (2001), "Diophantine geometry over groups. I. Makanin-Razborov diagrams", Publications Mathématiques de l'IHÉS, 93 (1): 31–105, doi:10.1007/s10240-001-8188-y, MR 1863735 Also, this theory comes with a health warning attached - it is all very complicated and slightly controversial. But also extremely important! $\endgroup$
    – user1729
    Mar 21, 2018 at 10:27
  • $\begingroup$ This seems to be proven in Lemma 2.2 of arxiv.org/pdf/math/0202261.pdf $\endgroup$ Nov 15, 2023 at 16:53

1 Answer 1


Reduction of problem, but not an answer:

Suppose $N=\ker f_1 = \ker f_2$. Then $f_1$ and $f_2$ both induce isomorphisms, $f_1' : S/N\to F_g$ and $f_2' : S/N\to F_g$. Then $\phi =f_1'f_2^{\prime -1}$ is an automorphism of $F_g$, and $f_1=\phi f_2$. Thus we only need to find an automorphism $\psi$ of $S$ taking $\ker f_1$ to $\ker f_2$, since then we'll have $\ker f_2\psi = \ker f_1$. Thus the desired result is true if and only if the automorphism group of $S$ acts transitively on normal subgroups $N$ of $S$ with $S/N \cong F_g$.

I'm not sure how to prove that the automorphism group either does or does not act transitively on such subgroups though.

  • $\begingroup$ I think that the desired result is not whether the automorphism group acts transitively on such subgroups. Instead it is whether there is a way (a method, or an algorithm) which, given two such subgroups, determines whether there exists an automorphism taking one of those two subgroups to the other. $\endgroup$
    – Lee Mosher
    Mar 23, 2018 at 16:24

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