Proposition 11

The canonical homomorphism $\nu$ of Mo(X) into $N^{(X)}$ is surjective. Let $w = x_1 \ldots x_n$ and $w' = x'_1 \ldots x'_m$ be two elements of Mo(X); in order that $\nu(w) = \nu(w')$, it is necessary and sufficient that $m = n$ and that there exist a permutation $\sigma \in S_n$ with $x'_i = x_{\sigma(i)}$ for $1 \leq i \leq n$.

The image of $\nu$ is a submonoid I of $N^{(X)}$ containing the elements $\delta_x$ (for $x \in X$). Formula (12) (no. 7) shows that $N^{(X)}$ is generated by the family $(\delta_x)_{x \in X}$, where $I = N^{(X)}$. Therefore $\nu$ is surjective.

If $m = n$ and $x'_i = x_{\sigma(i)}$ for $1 \leq i \leq n$, then

by formula (21) and the commutativity theorem (§ 1, no. 5, Theorem 2).

Conversely, suppose that $v(w)$ and $v(w')$ are equal to the same element $\alpha$ of $\mathbf{N}^{(X)}$; by formula (22), $n = |\alpha| = m$. For all $x \in X$, let $I_x$ (resp. $I'_x$) be the set of integers $i$ such that $1 \leq i \leq n$ and $x_i = x$ (resp. $x'_i = x$). Hence $(I_x)_{x \in X}$ and $(I'_x)_{x \in X}$ are partitions of the interval $[1, n]$ of $\mathbf{N}$; further, the formula

shows that $\alpha(x)$ is the cardinal of $I_x$; similarly the formula shows that $\alpha(x)$ is the cardinal of $I'_x$. There thus exists a permutation $\sigma$ of $[1, n]$ such that $\sigma(I'_x) = I_x$ for all $x \in X$, that is $x'_i = x_{\sigma(i)}$ for $i = 1, \ldots, n$.