Inverse of Algebraic Structure Isomorphism is Isomorphism

Theorem
Let $$\phi: \left({S, \circ}\right) \to \left({T, *}\right)$$ be a bijection.

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

Proof
As $$\phi$$ is a bijection, then $$\exists \phi^{-1}$$ such that $$\phi^{-1}$$ is also a bijection from Bijection iff Inverse is Bijection. That is:


 * $$\exists \phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$$

It follows that:

$$ $$ $$

So $$\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$$ is a homomorphism, and (from above) bijective, and thus an isomorphism.

Applying the same result in reverse, we have that if $$\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$$ is an isomorphism, then $$\left({\phi^{-1}}\right)^{-1}: \left({S, \circ}\right) \to \left({T, *}\right)$$ is an isomorphism.

But by Inverse of Inverse of Bijection, $$\left({\phi^{-1}}\right)^{-1} = \phi$$ and hence the result.