Proof of von Dyck's theorem

Proof

1. By the definition of a presentation, $G = F_S/N$, where $N$ is the smallest normal subgroup of $F_S$ that contains $R$. Let $\pi \colon F_S \to G$ be the projection $\pi(w) = wN$.
2. The homomorphism $\varphi$ exists and is unique, by the universal property of the free group.
3. By hypothesis, $\varphi(r) = e$ for each $r \in R$. So $R \subseteq \ker \varphi$, the kernel of $\varphi$.
4. Since the kernel is a normal subgroup, $\ker \varphi$ is a normal subgroup of $F_S$ that contains $R$. Since $N$ is the smallest one, $N \subseteq \ker \varphi$.
5. By the universal property of the quotient group, there is a group homomorphism $\psi \colon G \to H$ with $\psi(wN) = \varphi(w)$ for each $w \in F_S$.
6. In particular, $\psi(sN) = \varphi(s) = f(s)$ for each $s \in S$. This proves existence.
7. For uniqueness, let $\psi' \colon G \to H$ be a homomorphism with $\psi'(sN) = f(s)$ for each $s \in S$.
8. Then $\psi' \circ \pi \colon F_S \to H$ is a homomorphism with $(\psi' \circ \pi)(s) = f(s)$ for each $s \in S$.
9. By the uniqueness part of the universal property of the free group, $\psi' \circ \pi = \varphi$.
10. So $\psi'(wN) = \varphi(w) = \psi(wN)$ for each $w \in F_S$. Every element of $G$ has the form $wN$, so $\psi' = \psi$.