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  "body": "\n## Proof\n\nWe first build $\\varphi$ on all words, then show that reduction does not change its value.\n\n1. By the definition of the [[free group|020Z5SEN27]], the elements of $F_S$ are the [[reduced words|0FPXK5CHQ3]] on $S$. To get the product $u \\cdot v$, form the concatenation $uv$. Then delete adjacent pairs $s s^{-1}$ or $s^{-1} s$, one pair at a time, until the word is reduced. The identity is the empty word.\n2. Extend $f$ to the letters by $\\hat f(s) = f(s)$ and $\\hat f(s^{-1}) = f(s)^{-1}$ for $s \\in S$.\n3. For any word $w = t_1 \\cdots t_n$, reduced or not, let $\\Phi(w) = \\hat f(t_1) \\cdots \\hat f(t_n)$ in $G$. For the empty word, $\\Phi$ gives $e$.\n4. For words $w$ and $w'$, the concatenation satisfies $\\Phi(ww') = \\Phi(w)\\Phi(w')$, directly from step 3.\n5. Let $w'$ arise from $w$ by deleting one adjacent pair. Write $w = a\\,t\\,t'\\,b$ and $w' = ab$, where $tt'$ is $ss^{-1}$ or $s^{-1}s$. Then $\\hat f(t)\\hat f(t') = e$, since it is $f(s)f(s)^{-1}$ or $f(s)^{-1}f(s)$.\n6. By steps 4 and 5, $\\Phi(w) = \\Phi(a)\\,\\hat f(t)\\hat f(t')\\,\\Phi(b) = \\Phi(a)\\Phi(b) = \\Phi(w')$.\n7. By step 6 and induction on the number of deletions, $\\Phi$ takes the same value on $uv$ and on $u \\cdot v$.\n8. Let $\\varphi \\colon F_S \\to G$ be the restriction of $\\Phi$ to reduced words. For $u, v \\in F_S$, steps 4 and 7 give $\\varphi(u \\cdot v) = \\Phi(uv) = \\varphi(u)\\varphi(v)$. So $\\varphi$ is a [[group homomorphism|0HV9Y7WG5R]].\n9. For $s \\in S$, the word $s$ of length $1$ is reduced, and $\\varphi(s) = \\hat f(s) = f(s)$. This proves existence.\n10. For uniqueness, let $\\psi \\colon F_S \\to G$ be a homomorphism with $\\psi(s) = f(s)$ for each $s \\in S$.\n11. For $s \\in S$, the product $s \\cdot s^{-1}$ in $F_S$ is the empty word, the identity. So $s^{-1}$ is the inverse of $s$ in $F_S$, and $\\psi(s^{-1}) = \\psi(s)^{-1} = f(s)^{-1}$. Hence $\\psi(t) = \\hat f(t)$ for each letter $t$.\n12. Let $w = t_1 \\cdots t_n$ be a reduced word with $n \\geq 1$. The concatenation of $t_1$ and $t_2 \\cdots t_n$ is $w$ itself, which is reduced. So $w = t_1 \\cdot (t_2 \\cdots t_n)$ in $F_S$.\n13. By induction on $n$, using step 12, $w = t_1 \\cdot t_2 \\cdots t_n$ is the product of its letters in $F_S$.\n14. Since $\\psi$ is a homomorphism, steps 11 and 13 give $\\psi(w) = \\hat f(t_1) \\cdots \\hat f(t_n) = \\varphi(w)$. For the empty word, both give $e$.\n15. So $\\psi = \\varphi$, and $\\varphi$ is the only such homomorphism.\n\n## Idea\n\nThe value of a word is forced: send each letter to $f(s)$ or its inverse and multiply. The only thing to check is that cancelling a pair $ss^{-1}$ does not change this value. It does not, because $f(s)f(s)^{-1} = e$ in $G$.\n",
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