Reduced syllable words form the free product of a family of groups

Given: The objects and hypotheses in the statement.

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).

For groups as in def-group, a syllable is a tagged pair $(i,g)$ with $i\in I$ and $g\in G_i\setminus \{e_i\}$. A reduced syllable word is a finite list of syllables, indexed by a natural length as in def-natural-numbers, in which adjacent tags differ. The empty list is allowed. At a concatenation seam, adjacent syllables from the same factor are multiplied and an identity result is deleted; this elementary reduction is repeated until the seam is reduced. (Reduced syllable words in a family of groups).

For each $i\in I$ and $g\in G_i$, left multiplication at the first syllable defines a permutation $P_{i,g}$ of the set of reduced words. One has $P_{i,g}^{-1}=P_{i,g^{-1}}$ and $P_{i,gh}=P_{i,g}\circ P_{i,h}$, so $g\mapsto P_{i,g}$ is a group homomorphism. (Factor elements act by mutually inverse permutations on reduced syllable words).

Let $(M,\cdot ,e)$ and $(M^{\mathrm{\prime }},{\cdot }^{\mathrm{\prime }},e^{\mathrm{\prime }})$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M^{\mathrm{\prime }}$ is a function $f:M\to M^{\mathrm{\prime }}$ such that - (H1) $f(x\cdot y)=f(x){\cdot }^{\mathrm{\prime }}f(y)$ for all $x,y\in M$; - (H2) $f(e)=e^{\mathrm{\prime }}$. Let $G$ and $G^{\mathrm{\prime }}$ be groups (def-group). A group homomorphism from $G$ to $G^{\mathrm{\prime }}$ is a function $f:G\to G^{\mathrm{\prime }}$ satisfying (H1) alone: $f(xy)=f(x)f(y)\text{for all }x,y\in G\mathrm{.}$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e)=e^{\mathrm{\prime }}$ and $f(x^{-1})=f(x{)}^{-1}$ (lem-group-homomorphism-basic-properties). For monoids it does not follow and must be assumed, which is why the two definitions differ. A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g\circ f)(xy)=g(f(x)f(y))=g(f(x))g(f(y))$ and $(g\circ f)(e)=g(e^{\mathrm{\prime }})=e^{\mathrm{\prime }\mathrm{\prime }}$; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).