Each factor is a retract of a free product when all other factors are sent trivially

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).

Every canonical factor homomorphism $\iota_i:G_i\to\ast_jG_j$ is injective. (Every canonical factor map into a free product is injective).

Let $(M,\cdot,e)$ and $(M',\cdot',e')$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M'$ is a function $f : M \to M'$ such that - (H1) $f(x \cdot y) = f(x) \cdot' f(y)$ for all $x, y \in M$; - (H2) $f(e) = e'$. Let $G$ and $G'$ be groups (def-group). A group homomorphism from $G$ to $G'$ is a function $f : G \to G'$ satisfying (H1) alone: $f(xy) \;=\; f(x)\, f(y) \qquad \text{for all } x, y \in G .$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e) = e'$ 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 $(g \circ f)(e) = g(e') = e''$; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).