The canonical surjection from a free product to the direct product of its factors

Given: The objects and hypotheses in the example.

For a family $(G_i)_{i\in I}$, a free product is a group $F$ with homomorphisms $\iota_i:G_i\to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_i:G_i\to H$, there is a unique homomorphism $f:F\to H$ satisfying $f\circ\iota_i=f_i$ for all $i$. It is denoted $\ast_{i\in I}G_i$. Injectivity of the maps $\iota_i$ is not part of this definition. (The free product of an arbitrary family of groups).

Every element of $\ast_{i\in I}G_i$ has a unique reduced syllable expression. The identity is represented by the empty word, and no nonempty reduced word represents the identity. (Normal form theorem for free products).

Let $G$ and $H$ be groups. Their external direct product has underlying set $G\times H:=\{(g,h):g\in G,\ h\in H\}$ and componentwise operation $(g,h)(g',h') := (gg',hh').$ The fact that this operation makes $G\times H$ a group, with the indicated identity and inverses, is proved in thm-external-direct-product-is-a-group. Until that result is used, this definition introduces only the set and its componentwise binary operation. (The external direct product $G\times H$ with componentwise multiplication).

For groups $G$ and $H$, the componentwise operation of def-external-direct-product-of-groups makes $G\times H$ a group. Its identity is $(e_G,e_H)$, and $(g,h)^{-1}=(g^{-1},h^{-1}).$ Moreover the coordinate maps $\pi_G(g,h)=g$ and $\pi_H(g,h)=h$ are group homomorphisms. ($G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections).

Let $G$ be a group. For $g,h\in G$, their commutator is $[g,h]:=ghg^{-1}h^{-1}.$ This convention is fixed throughout; some sources use its inverse. By the inverse laws of lem-group-inverse-laws, one has $[g,h]^{-1}=hgh^{-1}g^{-1}=[h,g]$. The commutator subgroup, or derived subgroup, is the subgroup generated by all commutators: $[G,G]:=\langle\{[g,h]:g,h\in G\}\rangle.$ The generated subgroup notation is that of def-generated-subgroup. (Commutators $[g,h]=ghg^{-1}h^{-1}$ and the commutator subgroup $[G,G]$).