Proposition 8

Let G be a group and f a mapping of $X$ into G. There exists one and only one homomorphism $\bar{f}$ of $F(X)$ into G which extends f.

The uniqueness of $\bar{f}$ follows from the fact that the group $F(X)$ is generated by $X$. For all $x$ in $X$, let $f_x$ be the homomorphism $n \mapsto f(x)^n$ of $\mathbf{Z}$ into $G$. By no. 3, Proposition 4, there exists a homomorphism $\bar{f}$ of $F(X)$ into $G$ such that $\bar{f}(x^n) = f_x(n)$ for $x \in X$ and $n \in \mathbf{Z}$; in particular, $\bar{f}(x) = f_x(1) = f(x)$ for all $x \in X$ and hence $\bar{f}$ extends $f$.

Let $u : X \to Y$ be a mapping. By Proposition 8 there exists one and only one homomorphism of $F(X)$ into $F(Y)$ which coincides with $u$ on $X$; it is denoted by $F(u)$. As in the case of magmas (no. 1) the formula

is established for every mapping $v : Y \to Z$ and it is shown that $F(u)$ is injective (resp. surjective, bijective) if $u$ is. For every subset $S$ of $X$, $F(S)$ will be identified with the subgroup of $F(X)$ generated by $S$.

Let I be a set. In certain cases it is of interest not to identify $i$ in I with its canonical image $\phi_i(1)$ in the free group $F(I)$; the latter will be denoted by $T_i$ (or $T'_i, X_i, \ldots$ as the case may be) and called the indeterminate of index $i$. The free group $F(I)$ is then denoted by $F((T_i)_{i \in I})$ or $F(T_1, \ldots, T_n)$ if $I = \{1, 2, \ldots, n\}$.

Let $G$ be a group and $t = (t_i)_{i \in I}$ a family of elements of $G$. By Proposition 8 there exists a homomorphism $f_t$ of $F((T_i)_{i \in I})$ into $G$ characterized by $f_t(T_i) = t_i$ for all $i \in I$. The image of an element $w$ of $F((T_i)_{i \in I})$ under $f_t$ will be denoted by $w(t)$ or $w(t_1, \ldots, t_n)$ if $I = \{1, 2, \ldots, n\}$; $w(t)$ is said to result from the substitution $T_i \mapsto t_i$ in $w$. In particular, if we take $G = F((T_i)_{i \in I})$ and $(t_i) = (T_i) = T, f_T$ is the identity homomorphism of $G$, whence $w(T) = w$; for $I = \{1, 2, \ldots, n\}$, then $w(T_1, \ldots, T_n) = w$.

Let $G$ and $G'$ be two groups, $u$ a homomorphism of $G$ into $G'$ and $t = (t_1, \ldots, t_n)$ a finite sequence of elements in $G$. Let $t' = (u(t_1), \ldots, u(t_n))$; the homomorphism $u \circ f_t$ of $F(T_1, \ldots, T_n)$ into $G'$ maps $T_i$ to $u(t_i)$ for $1 \leq i \leq n$ and hence is equal to $f_{t'}$; for $w$ in $F(T_1, \ldots, T_n)$, then

Let $w$ be given in $F(T_1, \ldots, T_n)$ and elements $v_1, \ldots, v_n$ in the free group $F(T'_1, \ldots, T'_m)$. The substitution $T_i \mapsto v_i$ defines an element $w' = w(v_1, \ldots, v_n)$ of $F(T'_1, \ldots, T'_m)$. Let $G$ be a group, $t_1, \ldots, t_m$ elements of $G$ and $u$ the homomorphism of $F(T'_1, \ldots, T'_m)$ into $G$ characterized by $u(T'_j) = t_j$ for $1 \leq j \leq m$. Then $u(v_i) = v_i(t_1, \ldots, t_m)$ and $u(w') = w(t_1, \ldots, t_m)$; formula (8) thus implies

This justifies the "functional notation" $w(t_1, \ldots, t_n)$. The reader is left to extend formulae (8) and (9) to the case of arbitrary indexing sets.