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  "body": "\n## Proof\n\n1. By the definition of a [[presentation|0QZCY9PE3Q]], $G = F_S/N$, where $N$ is the smallest normal subgroup of $F_S$ that contains $R$. Let $\\pi \\colon F_S \\to G$ be the projection $\\pi(w) = wN$.\n2. The homomorphism $\\varphi$ exists and is unique, by the [[universal property of the free group|0CDWKDYWN6]].\n3. By hypothesis, $\\varphi(r) = e$ for each $r \\in R$. So $R \\subseteq \\ker \\varphi$, the [[kernel|0Z424HK8CQ]] of $\\varphi$.\n4. Since [[the kernel is a normal subgroup|0C9ZKYSPYR]], $\\ker \\varphi$ is a normal subgroup of $F_S$ that contains $R$. Since $N$ is the smallest one, $N \\subseteq \\ker \\varphi$.\n5. By the [[universal property of the quotient group|0GJ83E78A5]], there is a [[group homomorphism|0HV9Y7WG5R]] $\\psi \\colon G \\to H$ with $\\psi(wN) = \\varphi(w)$ for each $w \\in F_S$.\n6. In particular, $\\psi(sN) = \\varphi(s) = f(s)$ for each $s \\in S$. This proves existence.\n7. For uniqueness, let $\\psi' \\colon G \\to H$ be a homomorphism with $\\psi'(sN) = f(s)$ for each $s \\in S$.\n8. Then $\\psi' \\circ \\pi \\colon F_S \\to H$ is a homomorphism with $(\\psi' \\circ \\pi)(s) = f(s)$ for each $s \\in S$.\n9. By the uniqueness part of the universal property of the free group, $\\psi' \\circ \\pi = \\varphi$.\n10. So $\\psi'(wN) = \\varphi(w) = \\psi(wN)$ for each $w \\in F_S$. Every element of $G$ has the form $wN$, so $\\psi' = \\psi$.\n\n## Idea\n\nThe homomorphism $\\varphi$ kills every relator, so it kills the normal subgroup they generate. Hence $\\varphi$ factors through the quotient $F_S/N = G$. Uniqueness follows from uniqueness for free groups. Two homomorphisms that agree on the cosets $sN$ give the same map on $F_S$, so they agree on $G$.\n",
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