Free product of algebras

Remarks

The free product exists for every family in a variety, and it is unique up to a unique isomorphism compatible with the maps $\iota_i$. It is the coproduct of the family in the category of algebras of $\mathcal{V}$. Some authors require in addition that every $\iota_i$ be injective.

Example: in the variety of abelian groups, $A_1 \times A_2$ with $\iota_1(a) = (a, 0)$ and $\iota_2(b) = (0, b)$ is a free product. Given $f_1, f_2$ into an abelian group $B$, the map $f(a, b) = f_1(a) + f_2(b)$ is a homomorphism because $B$ is abelian, it satisfies $f \circ \iota_k = f_k$, and it is the only such map since $(a, b) = \iota_1(a) + \iota_2(b)$.

Non-example: in the variety of all groups, $\Z \times \Z$ with the two coordinate inclusions is not a free product of $\Z$ and $\Z$. Take $f_1, f_2 \colon \Z \to S_3$ sending $1$ to the transpositions $(1\,2)$ and $(2\,3)$. A homomorphism $f$ with $f \circ \iota_k = f_k$ would send the commuting elements $(1, 0)$ and $(0, 1)$ to these two transpositions, which do not commute.

In the variety of groups, the free product of a family is the group of reduced words in the factors.