Free groups generated by the same free set are isomorphic

Source: Geometric Group Theory: An Introduction, Clara Löh, but is a standard exercise.

Problem: Let $S$ be a set. Then, up to isomorphism, there is at most one group freely generated by $S$ .

Proof: Let $F$ and $G$ be groups with free generating set $S$ . Let $\phi_F$ and $\phi_G$ be the inclusion maps from $S$ into $F$ and $G$ respectively. By the universal property of free groups, there are unique homomorphisms $\varphi_F : F \to G$ and $\varphi_G : G \to F$ such that for all $s \in S$ , $s\varphi_F = s\varphi_G = s$ . Hence $\varphi_G\varphi_F$ and $\varphi_F\varphi_G$ are homomorphisms which fix $S$ . Using the universal property again, there is a unique homomorphism from $F$ to itself which fixes $S$ – the identity map. Hence, because of uniqueness, $\varphi_F\varphi_G$ is the identity map. An analogous argument shows that $\varphi_G\varphi_F$ is the identity map, and therefore $\varphi_F\varphi_G$ is an isomorphism.

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Commutative diagram showing the above, with curly arrows
denoting inclusion in the usual places in universal property diagrams