Massey Exercise 3.4.3


Theorem: Let and be two different families of groups indexed by the same set. Let and be their free products. Suppose that for each there is a homomorphism . Then there is a unique homomorphism such that the diagram commutes (where the ’s are the canonical injections):

Furthermore, if each is injective, is injective, and if each is surjective, is surjective.


Proof:

Notice that is a homomorphism from to . That is, we have:

And since is a free group, the definition tells us that there is a unique homomorphism that makes this diagram commute. Since this really is the same diagram, this also makes the diagram in the original statement commute.

Let for some . Then

is an arbitrary element of . Since is a homomorphism and the diagram commutes, we can write:

Suppose that the are injective. Observe that the only way could be the identity is if each is the identity (otherwise, it would be some definite irreducible word in ). But, this is true only if each is the identity. But then is the identity. So, is trivial and is injective.

Suppose the are surjective. Let . Then there is an such that for all . Then, . So there is an element in that maps to , and is surjective.