Group Homomorphism Preserves Inverses/Proof 1

Theorem
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
The result follows directly from the morphism property of $\circ$ under $\phi$.

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 and a right inverse of $\phi \left({x}\right)$, it is by definition an inverse.

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

Hence the result.