4
$\begingroup$

Grushko's theorem says that given an epimorphism $\phi: F \to G_1 * G_2$ where $F$ is a finitely-generated free group, there exists. subgroups $F_1$ and $F_2$ of $F$ so that $F = F_1 * F_2$ and $\phi(F_i) = G_i$.

I am trying to piece together a proof of the result that was mentioned in the introduction of a paper by Stallings that uses ideas from 3-dimensional topology. Stallings writes: "An earlier paper by Kneser [1] shows how to join together spheres in a 3-manifold to obtain a given free product structure on the fundamental groups; this can translate into a proof of Grushko's original theorem."

Here is my guess of how that proof of Grushko's theorem is supposed to go up until where I am stuck: Take some 2-dimensional CW complexes $K_1$ and $K_2$ with $\pi_1(K_i) \cong G_i$ and take the join of these two spaces by attaching an interval to the disjoint union of $K_1$ and $K_2$ with one endpoint attached to the basepoint of $K_1$ and the other to the basepoint of $K_2$. Take the midpoint of this interval in this new space as the basepoint and denote the space by $K_1 \vee K_2$. Now, there is some $k$ with $\pi_1(\sharp^k S^1 \times S^2) \cong F$.

Is there now a map $f : \sharp^k S^1 \times S^2 \to K_1 \vee K_2$ such that $\pi_1(f) = \phi$?

If we had such a map, then by mumbling something a transversality to the basepoint (which is a point I never truly understand in my heart of hearts since the codomain is not a manifold, so if anyone knows a nice way to think of that, I'd love to hear it), then $f^{-1}(*)$ is a bunch of embedded surfaces in $\sharp^k S^1 \times S^2$. We can then homotope $f$ so that $f^{-1}(*)$ is a bunch of embedded spheres (by compressing the surfaces along disks) and then we can homotope $f$ again to make $f^{-1}(*)$ just a single sphere (by combining the spheres along arcs like Kneser). From this, Grushko's result follows.

Is this what Stallings had in mind? If so, how do I prove that $f$ exists? If not, what did he have in mind? Thanks for the help.

$\endgroup$
4
  • 1
    $\begingroup$ Hint: Instead of join, connect $K_1, K_2$ by an interval and then take the preimage of a generic point if that interval under a simplicial map of your 3-manifold. This should be in the paper of Scott and Wall. $\endgroup$ Commented Jun 21, 2021 at 23:45
  • $\begingroup$ It just occurred to me (while reading a bit further in Stallings' paper) that he probably had in mind first killing off the higher homotopy groups of $K_1$ and $K_2$ first, and then proceeding as I said. This ensures that the aforementioned map $f$ actually exists, since the wedge of Eilenberg-MacLane spaces is again an Eilenberg-MacLane space. $\endgroup$
    – user101010
    Commented Jun 22, 2021 at 16:21
  • $\begingroup$ @MoisheKohan Sorry, I think I am being really thick with this, but then don't I just get the preimage to be subsimplical complex? For what I outlined above I was always thinking of it as a submanifold. Scott and Wall indeed give a fun topological proof of Gruschko's theorem using a very similar setup, but it doesn't involve this same 3-dimensional manifold-topology flair. $\endgroup$
    – user101010
    Commented Jun 22, 2021 at 16:28
  • 2
    $\begingroup$ I would do this with # S^1x D^2 instead of S^2 since a handlebody is irreducible. Transversality is purely local, so you can do it for the midpoint of an interval connecting K(π,1)’s instead of using a strict wedge. Then do compressions and boundary compressions to get down to the preimage being a collection of embedded disks. $\endgroup$
    – Ian Agol
    Commented Jun 23, 2021 at 3:17

0

You must log in to answer this question.

Browse other questions tagged .