Surface groups and Grushko
by
John R. Stallings
University of California, Berkeley

Let the free product A*B of two groups be represented as the fundamental group of two spaces X and Y with \pi₁(X)=A and \pi₁(Y)=B, joined along an arc with midpoint p; call this union of two spaces and the arc X \/ Y. Let T be a connected compact 2-manifold with possibly empty boundary \partialT, and let G=\pi₁(T). Let f:T --> X \/ Y be a continuous map, which, on fundamental groups, is surjective (we can suppose that f behaves nicely in a neighborhood of f^-1(p) ); suppose that f maps each boundary component of T either into X or into Y, except possibly for one such component, which maps by a path of length two (that is, one arc on that component maps to X and the rest maps nicely to the complement of X ). Then f is homotopic to g such that g^-1(p) consists of a single component (a two-sided arc or simple closed curve).

The proof mimics a topological proof of Grushko's Theorem and uses a geometric idea to simplify certain paths into arcs. As corollaries it has several group-theoretic results. In addition, there is a sort of ``immersion'' result when the map on fundamental groups may not be surjective. Further generalization is possible for certain other 2-complexes than surfaces.

Date received: February 26, 1998