How to prove the existence of this homomorphism $G_1*G_2\to H$?

Question

Say there is a homomorphism from a group $G_1$ to another group $H$, and there is a homomorphism from $G_2$ to $H$, then there is a homomorphism from $G_1*G_2$ to $H$ induced by the former to homomorphisms? Thanks. (Where $G_1*G_2$ is the free product.)

Answer 1

Assuming you mean "group homomorphisms" as opposed to "homotopies"

Let our homomorphisms be $\phi_1:G_1 \rightarrow H$ and $\phi_2:G_2 \rightarrow H$. Elements of the free product $G_1 * G_2$ can be represented as alternating products of elements of $G_1$ and $G_2$. The element $a_1a_2b_1b_2$ where $a_1,b_1 \in G_1$ and $a_2, b_2 \in G_2$ is one such example.

We can construct a homomorphism $\phi: G_1 * G_2 \rightarrow H$ just by mapping an element $g$ to $\phi_1(g)$ if $g \in G_1$ and $\phi_2(g)$ if $g \in G_2$. For example, $\phi(a_1a_2b_1b_2) = \phi_1(a_1)\phi_2(a_2)\phi_1(b_1)\phi_2(b_2) = \phi(a_1a_2)\phi(b_1b_2)$.

It is clear from the above example that $\phi$ is a homomorphism.

Answer 2

The elements of $G_1*G_2$ look like words whose letters are elements of $G_1$ and $G_2$. By fiat, the identity elements of these two groups are equal in $G_1*G_2$, while any other two elements are distinct. The multiplication tables of $G_1$ and $G_2$ are the same inside $G_1*G_2$ but otherwise there are no relations satisfied between the elements of $G_1$ and $G_2$.

Suppose $\rho: G_1*G_2\to H$ is any group homomorphism. Prove that $\rho$ is determined by its values on $G_1$ and $G_2$. That should inspire you to go in reverse and show that any two maps $G_1,G_2\to H$ can be put together to get a map $G_1*G_2\to H$. How is this map defined?

(The free product satisfies the universal property of coproducts in the category of groups.)