Proposition 10

Let M be a commutative monoid (resp. group) and f a mapping of X into M. There exists one and only one homomorphism of $\mathbf{N}^{(X)}$ (resp. $\mathbf{Z}^{(X)}$) into M such that $g(\delta_x) = f(x)$ for all $x \in X$. If M is written additively, then $g(\alpha) = \sum_{x \in X} \alpha(x) \cdot f(x)$ for all $\alpha$ in $\mathbf{N}^{(X)}$ (resp. $\mathbf{Z}^{(X)}$).

Let $g$ be a homomorphism of $\mathbf{N}^{(X)}$ (resp. $\mathbf{Z}^{(X)}$) into M such that $g(\delta_x) = f(x)$ for all $x \in X$. For all $\alpha$ in $\mathbf{N}^{(X)}$ (resp. $\mathbf{Z}^{(X)}$), it follows from (12) that

whence the uniqueness of $g$.

For all $\alpha$ in $\mathbf{N}^{(X)}$ (resp. $\mathbf{Z}^{(X)}$) we write $g(\alpha) = \sum_{x \in X} \alpha(x) \cdot f(x)$. Then obviously $g(0) = 0$; for $\alpha, \beta$ in $\mathbf{N}^{(X)}$ (resp. $\mathbf{Z}^{(X)}$),

and hence $g$ is a homomorphism of $\mathbf{N}^{(X)}$ (resp. $\mathbf{Z}^{(X)}$) into M. Also, for $y$ in $X$, now $\delta_y(x) \cdot f(x) = 0$ for $x \neq y$ and $\delta_y(y) \cdot f(y) = f(y)$, whence $g(\delta_y) = f(y)$.

Let $u : X \to Y$ be a mapping. By Proposition 10 there exists one and only one homomorphism of $\mathbf{Z}^{(X)}$ into $\mathbf{Z}^{(Y)}$ which maps $\delta_x$ to $\delta_{u(x)}$ for all $x \in X$. It is denoted by $\mathbf{Z}^{(u)}$; it is immediately seen that it maps $\alpha \in \mathbf{Z}^{(X)}$ to the element $\beta \in \mathbf{Z}^{(Y)}$ defined by

As in the case of magmas (no. 1), the formula $\mathbf{Z}^{(v \circ u)} = \mathbf{Z}^{(v)} \circ \mathbf{Z}^{(y)}$ is established for every mapping $v : Y \to Z$; it is also shown that $\mathbf{Z}^{(u)}$ is injective (resp. surjective, bijective) if $u$ is.

Let S be a subset of X; if i is the injection of S into X, the mapping $f = \mathbf{Z}^{(i)}$ is an isomorphism of $\mathbf{Z}^{(S)}$ onto the subgroup H of $\mathbf{Z}^{(X)}$ generated by the elements $\delta_s$ for $s \in S$. By (15),

and therefore $H$ is the set of elements of $\mathbf{Z}^{(X)}$ of support contained in $S$. Henceforth $\mathbf{Z}^{(S)}$ will be identified with $H$ by means of $f$.

Formula (15) shows that the restriction of $\mathbf{Z}^{(u)}$ to $\mathbf{N}^{(X)}$ induces a homomorphism $\mathbf{N}^{(u)}$ of $\mathbf{N}^{(X)}$ into $\mathbf{N}^{(Y)}$. Then $\mathbf{N}^{(v \circ u)} = \mathbf{N}^{(v)} \circ \mathbf{N}^{(u)}$ for every mapping $v : Y \to Z$; further, $\mathbf{N}^{(u)}$ is injective (resp. surjective, bijective) if $u$ is. If $S$ is a subset of $X$,