Homomorphism of Generated Group

Theorem
Let $\left({G, \circ}\right)$ and $\left({H, \circ}\right)$ be groups.

Let $\phi: G \to H$ and $\psi: G \to H$ be homomorphisms.

Let $\left \langle {S} \right \rangle = G$ be the group generated by $S$.

Let $\forall x \in S: \phi \left({x}\right) = \psi \left({x}\right)$

Then $\phi = \psi$.

Proof
Let $H = \left\{{x \in G: \phi \left({x}\right) = \psi \left({x}\right)}\right\}$.

From Set of Equal Image Elements under Homomorphisms on Same Group form Subgroup, $H$ is a subgroup of $G$.

But from the definition of the group generated by $S$, the smallest subgroup that contains $S$ is $G$ itself.

Thus $G = \left\{{x \in G: \phi \left({x}\right) = \psi \left({x}\right)}\right\}$ and so $\phi = \psi$.