Inverse of Group Isomorphism is Isomorphism/Proof 2

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

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

Then $\phi$ is an isomorphism iff $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is also an isomorphism.

Necessary Condition
Let $\phi: G \to H$ be an isomorphism.

Then by definition $\phi$ is a bijection.

From Bijection iff Inverse is Bijection it follows that:
 * $\exists \phi^{-1}: \left({H, *}\right) \to \left({G, \circ}\right)$

such that $\phi^{-1}$ is also a bijection.

Thus:

So $\phi^{-1}: \left({H, *}\right) \to \left({G, \circ}\right)$ is a homomorphism.

$\phi^{-1}$ is also (from above) a bijection.

Thus, by definition, $\phi^{-1}$ is an isomorphism.

Sufficient Condition
Let $\phi^{-1}: \left({H, *}\right) \to \left({G, \circ}\right)$ be an isomorphism.

Applying the same result as above in reverse, we have that $\left({\phi^{-1}}\right)^{-1}: \left({G, \circ}\right) \to \left({H, *}\right)$ is also an isomorphism.

But by Inverse of Inverse of Bijection:
 * $\left({\phi^{-1}}\right)^{-1} = \phi$

and hence the result.