Group Homomorphism Preserves Inverses/Proof 3

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

Let $\phi: \left({G, \circ}\right) \to \left({H, *}\right)$ be a group homomorphism.

Let:
 * $e_G$ be the identity of $G$
 * $e_H$ be the identity of $H$.

Then:
 * $\forall x \in G: \phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$

Proof
From Group Homomorphism of Product with Inverse, we have:
 * $\forall x, y \in G: \phi \left({x \circ y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$

Putting $x = e_G$ and $y = x$ we have: