Stallings' proof of Grushko Theorem follows from the following lemma.
Lemma Let
F be a finitely generated free group, with
n generators. Let
G1 and
G2 be two finitely presented groups. Suppose there exists a surjective homomorphism \phi:F\rightarrow G_1\ast G_2. Then there exists two subgroups
F1 and
F2 of
F with \phi(F_1)=G_1 and \phi(F_2)=G_2, such that F=F_1\ast F_2.
Proof: We give the proof assuming that
F has no generator which is mapped to the identity of G_1\ast G_2, for if there are such generators, they may be added to any of F_1 or F_2. The following general results are used in the proof. 1. There is a one or two dimensional
CW complex,
Z with
fundamental group F. By
Van Kampen theorem, the
wedge of n circles is one such space. 2. There exists a two complex X=X_1\cup X_2 where \{p\}=X_1\cap X_2 is a point on a one cell of
X such that
X1 and
X2 are two complexes with fundamental groups
G1 and
G2 respectively. Note that by the Van Kampen theorem, this implies that the fundamental group of
X is G_1\ast G_2. 3. There exists a map f:Z\rightarrow X such that the induced map f_\ast on the fundamental groups is same as \phi For the sake of convenience, let us denote f^{-1}(X_1)=:Z_1 and f^{-1}(X_2)=:Z_2. Since no generator of
F maps to identity, the set Z_1\cap Z_2 has no loops, for if it does, these will correspond to circles of
Z which map to p\in X, which in turn correspond to generators of
F which go to the identity. So, the components of Z_1\cap Z_2 are contractible. In the case where Z_1\cap Z_2 has only one component, by Van Kampen's theorem, we are done, as in that case, :F=\Pi_1(Z_1)\ast\Pi_1(Z_2). The general proof follows by reducing
Z to a space homotopically equivalent to it, but with fewer components in Z_1\cap Z_2, and thus by induction on the components of Z_1\cap Z_2. Such a reduction of
Z is done by attaching discs along binding ties. We call a map \gamma :[0,1]\rightarrow Z a
binding tie if it satisfies the following properties 1. It is
monochromatic i.e. \gamma([0,1])\subseteq Z_1 or \gamma([0,1])\subseteq Z_2 2. It is a
tie i.e. \gamma(0) and \gamma(1) lie in different components of Z_1\cap Z_2. 3. It is
null i.e. f \circ \gamma([0,1]) is null homotopic in
X. Let us assume that such a binding tie exists. Let \gamma be the binding tie. Consider the map g:[0,1]\rightarrow D^2 given by g(t)= e^{it}. This map is a
homeomorphism onto its image. Define the space Z' as :Z'= Z \coprod\! D^2/\! \sim where :x\!\!\sim y \text{ iff} \begin{cases} x=y, \mbox{ or }\\ x=\gamma (t) \text{ and } y= g(t) \text{ for some } t\in [0,1]\mbox{ or }\\ x=g (t) \text{ and } y= \gamma (t) \text{ for some } t\in [0,1] \end{cases} Note that the space ''Z'
deformation retracts to Z'' We first extend
f to a function
f:Z\coprod \partial D^2/\!\sim as : f''(x) = \begin{cases}f(x),\ x\in Z\\ p \text{ otherwise.}\end{cases} Since the f(\gamma) is null homotopic, f'' further extends to the interior of the disc, and therefore, to Z'. Let Z_i' = f'^{-1}(X_i)
i = 1,2. As \gamma(0) and \gamma(1) lay in different components of Z_1\cap Z_2, Z_1' \cap Z_2' has one less component than Z_1\cap Z_2.
Construction of binding tie The binding tie is constructed in two steps.
Step 1: Constructing a
null tie: Consider a map \gamma' :[0,1]\rightarrow Z with \gamma' (0) and \gamma' (1) in different components of Z_1\cap Z_2. Since f_\ast is surjective, there exits a loop \!\lambda based at γ'(1) such that \! f(\gamma') and \! f(\lambda) are homotopically equivalent in
X. If we define a curve \gamma :[0,1]\rightarrow Z as \gamma(t)= \gamma'\ast\lambda(t) for all t\in [0,1], then \!\gamma is a null tie.
Step 2: Making the null tie
monochromatic: The tie \!\gamma may be written as \gamma_1\ast \gamma_2\ast \cdots \ast\gamma_m where each \gamma_i is a curve in Z_1 or Z_2 such that if \gamma_i is in Z_1, then \gamma_{i+1} is in Z_2 and vice versa. This also implies that f(\gamma_i) is a loop based at
p in
X. So, : [e]=[f(\gamma)]=[f(\gamma_1)]\ast\cdots\ast [f(\gamma_m)] Hence, [f(\gamma_j)]=[e] for some
j. If this \!\gamma_j is a tie, then we have a monochromatic, null tie. If \!\gamma_j is not a tie, then the end points of \!\gamma_j are in the same component of Z_1\cap Z_2. In this case, we replace \!\gamma_j by a path in Z_1\cap Z_2, say \!\gamma_j'. This path may be appended to \!\gamma_{j-1} and we get a new null tie \gamma '' = \gamma_1\ast \cdots \ast \gamma_{j-1}'\ast\gamma_{j+1} \cdots \gamma_m, where \!\gamma_{j-1}' = \gamma_{j-1}\ast\gamma_j'. Thus, by induction on
m, we prove the existence of a binding tie.
Proof of Grushko theorem Suppose that G = A*B is generated by \{g_1, g_2,\ldots, g_n\}. Let F be the free group with n-generators, viz. \{f_1, f_2,\ldots, f_n\}. Consider the homomorphism h:F\rightarrow G given by h(f_i) = g_i, where i=1,\ldots, n. By the lemma, there exists free groups F_1 and F_2 with F=F_1\ast F_2 such that h(F_1)=A and h(F_2)=B. Therefore, \text{Rank }(A) \leq \text{Rank }(F_1) and \text{Rank }(B) \leq \text{Rank }(F_2). Therefore, \text{Rank }(A) + \text{Rank }(B)\leq\text{Rank }(F_1) + \text{Rank }(F_2) = \text{Rank }(F) = \text{Rank } (A\ast B). ==See also==