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Let $H_g$ be the standard $3$-dimensional handle-body, whose boundary is denoted $S_g$, the oriented closed surface of genus $g\geq 1$. Call $F_g$ be the free group of rank $g$.

Denote by $i:S_g \to H_g$ the inclusion map. This map induces an epimorphism $i_*: \pi_1(S_g) \to \pi_1(H_g) \simeq F_g$.

Is any epimorphism from $\pi_1(S_g)$ to $F_g$ conjugate to $i_*$?

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  • $\begingroup$ Up to twisting by a mapping class of S_g, yes. This is an old result of Zieschang, Stallings etc. $\endgroup$
    – HJRW
    Commented Feb 5, 2015 at 15:09

2 Answers 2

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$\newcommand{\from}{\colon}\newcommand{\Aut}{\rm{Aut}}\newcommand{\bdy}{\partial}$Here is a fairly hands-on proof. I'll use the following notation: $H = H_g$, $S = S_g = \bdy H$, and $F = F_g$. I'll write $g(S) = g(H) = g$. Also, $i \from S \to H$ is the inclusion map.

Suppose that $f_* \from \pi_1(S) \to F = \pi_1(H)$ is an epimorphism. Then there is a homeomorphism $k \from S \to S$ so that $i_* \circ k_* = f_*$.

Fact number one: since $S$ and $H$ are Eilenberg-MacLane spaces, there is a map $f \from S \to H$ inducing $f_*$. Fix a cut system $D \subset H$: that is, a collection of $g(H)$ disks so that $H - n(D)$ is a three-ball. Homotope $f$ to make it transverse to $D$. Thus $C = f^{-1}(D)$ is a collection of simple closed curves in $S$. For any $\alpha \subset C$, let $D(\alpha)$ be the disk containing $f(\alpha)$.

We now construct a graph $G$. We have a vertex $v(X)$ for every component $X$ of $S - C$. We have an edge $e(\alpha)$ for every component $\alpha$ of $C$. We connect one (both) end(s) of $e(\alpha)$ to $v(X)$ if one (both) side(s) of $\alpha$ are contained in $X$. Note that $G$ is a finite connected graph.

Recall that the genus $g(S)$ is the number of curves we must cut $S$ along to obtain a connected planar surface. In similar fashion we define the genus $g(G)$ to be the number of edges we must remove from $G$ to obtain a tree. Define a map $r \from S \to G$ as follows. All points of $X - n(C)$ are mapped to $v(X)$. The open annulus $n(\alpha)$ is sent to the edge $e(\alpha)$ by crushing the circle coordinate. Deduce that $g(G) \leq g(S)$. We also observe that $r$ induces an epimorphism $r_* \from \pi_1(S) \to \pi_1(G)$, a free group of rank $g(G)$. [This much of the proof implies that $\pi_1(G)$ can only surject free groups of at most half the rank of $\pi_1(S)$.]

Fix a point $x \in H - D$. We now define a map $h \from G \to H$ as follows. The map $h$ sends all vertices of $G$ to $x$. The edge $e(\alpha)$ is sent to any loop based at $x$ and meeting $D$ exactly once, transversely, at a point of $D(\alpha)$. (I am ignoring a small issue about orientations here.) We deduce that $f_* = h_* \circ r_*$. Thus $h_* \from \pi_1(G) \to F$ is an epimorphism. Fact number two: since free groups are Hopfian, $g(G) = g(S)$ and $h_*$ is an isomorphism.

We now construct a handlebody $H'$, homotopy equivalent to $G$. Take $S$, thicken to get $S \times [0,1]$, and attach two-handles along the curves $C \times \{0\}$. We next attach three-handles to the sphere components of the lower boundary. We identify the groups $\pi_1(H')$ and $\pi_1(G)$. Let $i' \from S \to H'$ be the inclusion map.

Fact number three: using Nielsen's theorem (giving generators of $\Aut(F)$), there is a homeomorphism $h' \from H' \to H$ inducing $h_*$. Consider the homeomorphism $k = (\bdy i)^{-1} \circ \bdy h' \circ \bdy i'$, where all domains and ranges are restricted to be $S$. Chasing the diagram of spaces shows that $i_* \circ k_* = f_*$, and we are done.

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  • $\begingroup$ Thank you very much for your detailed and constructive answer Sam! In my opinion, the construction of the graph $G$ is definitely the crucial point. Thank you also for the very elegant argument about the Hopfian property. $\endgroup$
    – Moraga
    Commented Feb 8, 2015 at 18:29
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In Example 1.5 of their Notes on Sela's work, Bestvina and Feighn write:

When $G$ is the fundamental group of a closed genus $g$ orientable surface, let $\mu:G\to F_g$ denote the homomorphism to a free group of rank $g$ induced by the (obvious) retraction of the surface to the rank $g$ graph. It is a folk theorem that for every homomorphism $f:G\to F$ there is an automorphism $\alpha: G\to G$ (induced by a homeomorphism of the surface) so that $f\circ \alpha$ factors through $\mu$.

They refer to:

John Stallings. How not to prove the Poincar\'e conjecture. In Topology Seminar, Wisconsin, 1965, Edited by R. H. Bing and R. J. Bean. Annals of Mathematics Studies, No. 60, pages ix+246. Princeton University Press, 1966.

and

Heiner Zieschang. Alternierende Produkte in freien Gruppen. Abh. Math. Sem. Univ. Hamburg, 27:13–31, 1964.

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    $\begingroup$ It is a bit more than a folk theorem! The arguments seem similar to those of Stallings' papers on Grushko's theorem and fibration over $S^1$ that I read more than forty years ago. Recently a colleague asked me about the equivalence between Poincaré's conjecture and an algebraic conjecture (conjecture D in "How not to prove..." by Stallings, 1965). There Stallings proved that Conjecture D implies P. IMO, the converse requires Waldhausen's paper on Heegaard splitting of the 3-sphere (1967) and of course, the geometrisation of epimorphisms $\pi_1(S_g)\to F_g$ (not proved in the mentioned paper). $\endgroup$
    – user66870
    Commented Feb 8, 2015 at 15:49
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    $\begingroup$ @FrancoisLaudenbach: Welcome to MO, Francois! $\endgroup$ Commented Feb 8, 2015 at 19:55
  • $\begingroup$ @FrancoisLaudenbach, I would say that a theorem referred to several times in the literature but with no proof written down is the definition of a folk theorem! Clearly the proof was known to Stallings, anyway. $\endgroup$
    – HJRW
    Commented Feb 9, 2015 at 22:16

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