Group Homomorphism Preserves Inverses/Proof 1

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
Let $x \in G$.

Then:

So, by definition, $\phi \left({x^{-1}}\right)$ is the right inverse of $\phi \left({x}\right)$.

Similarly:

So, again by definition, $\phi \left({x^{-1}}\right)$ is the left inverse of $\phi \left({x}\right)$.

Finally, as $\phi \left({x^{-1}}\right)$ is both:
 * a left inverse of $\phi \left({x}\right)$

and:
 * a right inverse of $\phi \left({x}\right)$

it is by definition an inverse.

From Inverse in Group is Unique, $\phi \left({x^{-1}}\right)$ is the only such element.

Hence the result.