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.

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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:

$$\begin{align*} f(w) &= f( \varphi_{i_1}(x_{i_1})) \dots f(\varphi_{i_k}(x_{i_k})) \\ &= \varphi'_{i_1}(f_{i_1}(x_{i_1})) \dots \varphi'_{i_k}(f_{i_k}(x_{i_k})) \\ &= f_{i_1}(x_{i_1}) \dots f_{i_k}(x_{i_k}) \end{align*}$$

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.