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If there exists a homomorphism $f: A \rightarrow B$ between two groups $A$ and $B$, and a homomorphism $g: C \rightarrow D$ between two groups $C$ and $D$, then will there exist a homomorphism $h: A*C \rightarrow B*D$ between the free products $A*C$ and $B*D$?

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What are the conditions on $h$, with regards to $f$ and $g$? Maybe take a look at, since the free product is the coproduct in the category of groups you might find your answer there. – shamovic Mar 21 '11 at 9:23
@shamovic: you should post this as an answer, so it can be accepted and upvoted :D – Alexander Thumm Mar 21 '11 at 9:26
I would, but I'm not sure it is the one h988h looks for. So I want some clarification. – shamovic Mar 21 '11 at 9:48

I assume your definition of the free product $A \ast C$ of $A$ and $C$ is the following: It is a group $G$ together with two homomorphisms $i_{A}: A \to G$ and $i_{C}: C \to G$ such that for every pair of homomorphisms $k: A \to H$ and $l: C \to H$ there is a unique homomorphisms $m: G \to H$ such that $mi_{A} = k$ and $mi_{C} = l$.

By definition of the free product we have homomorphisms $i_B : B \to B \ast D$ and $i_{D}: D \to B \ast D$. Now apply the defining property of $A \ast C$ and the inclusions $i_{A} : A \to A \ast C$ and $i_{C}: C \to A \ast C$ to the homomorphisms $i_B f$ and $i_D g$ to get a unique homomorphism $h: A \ast C \to B \ast D$ such that $hi_{A} = i_B f$ and $hi_{C} = i_D g$. The last two properties are crucial in ensuring that $h$ is unique and has anything to do with $f$ and $g$.

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+1: Thanks for that, I was too lazy to spell that out. ;-) – joriki Mar 21 '11 at 11:09
@joriki: :) $\phantom{x}$ – t.b. Mar 21 '11 at 11:14

Theo's answer using the universal property of the free product is the more elegant one, but if you want to work with the construction of the free group as the group of reduced words in the group elements, a concrete homomorphism is given by the function $h$ that maps a reduced word by mapping each letter using $f$ or $g$ according as the letter is in $A$ or $B$.

The letters meeting at the ends of a product of reduced words can either be from the same group or from different groups. If they're from different groups, the homomorphism property of $h$ is trivial, and we don't actually need the homomorphism properties of $f$ and $g$ to prove it:

$$\begin{eqnarray} h((\ldots a_1c_1)(a_2c_2\ldots))&=&h(\ldots a_1c_1a_2c_2\ldots)\\\ &=&\ldots f(a_1)g(c_1)f(a_2)g(c_2)\ldots\\\ &=&(\ldots f(a_1)g(c_1))(f(a_2)g(c_2)\ldots)\\\ &=&h(\ldots a_1c_1)h(a_2c_2\ldots)\;. \end{eqnarray}$$

If the letters meeting at the ends are from the same group, say, $C$, and their product isn't the identity, we only need one invocation of the homomorphism property of $g$ to prove the homomorphism property of $h$:

$$\begin{eqnarray} h((\ldots a_1c_1)(c_2a_2\ldots))&=&h(\ldots a_1(c_1c_2)a_2\ldots)\\\ &=&\ldots f(a_1)g(c_1c_2)f(a_2)\ldots\\\ &=&\ldots f(a_1)g(c_1)g(c_2)f(a_2)\ldots\\\ &=&(\ldots f(a_1)g(c_1))(g(c_2)f(a_2)\ldots)\\\ &=&h(\ldots a_1c_1)h(c_2a_2\ldots)\;. \end{eqnarray}$$

Now we can use this as a basis for induction, assuming that the homomorphism property holds if $n$ products at the meeting point resolve to the identity, and concluding that it holds if one more product, say $c_1c_2$, resolves to the identity:

$$\begin{eqnarray} h((\ldots a_1c_1)(c_2a_2\ldots))&=&h((\ldots a_1)(a_2\ldots))\\\ &=&h(\ldots a_1)h(a_2\ldots)\\\ &=&h(\ldots a_1)g(c_1c_2)h(a_2\ldots)\\\ &=&h(\ldots a_1)g(c_1)g(c_2)h(a_2\ldots)\\\ &=&h(\ldots a_1c_1)h(c_2a_2\ldots)\;. \end{eqnarray}$$

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+1: Thanks for that, I was too lazy to spell that out. – t.b. Mar 21 '11 at 10:57

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