Massey Exercise 3.4.3 Theorem: Let \setbuilder{G_i}{i \in I} and \setbuilder{G'_i}{i \in I} be two different families of groups indexed by the same set. Let G and G' be their free products. Suppose that for each i \in I there is a homomorphism f_i : G_i \rightarrow G_i'. Then there is a unique homomorphism f : G \rightarrow G' such that the diagram commutes (where the \varphi’s are the canonical injections): Furthermore, if each f_i is injective, f is injective, and if each f_i is surjective, f is surjective. Proof: Notice that \varphi_i' \circ f_i is a homomorphism from G_i to G'. That is, we have: And since G is a free group, the definition tells us that there is a unique homomorphism f that makes this diagram commute. Since this really is the same diagram, this f also makes the diagram in the original statement commute. Let x_{i_j} \in G_{i_j} for some \set{i_1, \dots, i_k} \subset I. Then w = \varphi_{i_1}(x_{i_1}) \dots \varphi_{i_k}(x_{i_k}) = x_{i_1} \dots x_{i_k} \in G is an arbitrary element of G. Since f is a homomorphism and the diagram commutes, we can write: % Suppose that the f_i are injective. Observe that the only way f(w) could be the identity is if each f_{i_j}(x_{i_j}) is the identity (otherwise, it would be some definite irreducible word in G'). But, this is true only if each x_{i_j} is the identity. But then w is the identity. So, \ker f is trivial and f is injective. Suppose the f_i are surjective. Let w = g_{i_1} \dots g_{i_k} \in G'. Then there is an x_{i_j} such that f_{i_j}(x_{i_j}) = g_{i_j} for all j. Then, g_{i_1} \dots g_{i_k} = f_{i_1}(x_{i_1}) \dots f_{i_k}(x_{i_k}) = f(x_{i_1} \dots x_{i_k}). So there is an element in G that maps to w, and f is surjective.