Inverse of Algebraic Structure Isomorphism is Isomorphism/General Result

Theorem
Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ and $\struct {T, *_1, *_2, \ldots, *_n}$ be algebraic structures.

Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a mapping.

Then:
 * $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ is an isomorphism


 * $\phi^{-1}: \struct {T, *_1, *_2, \ldots, *_n} \to \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ is also an isomorphism.
 * $\phi^{-1}: \struct {T, *_1, *_2, \ldots, *_n} \to \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ is also an isomorphism.

Proof
As $\paren {\phi^{-1} }^{-1} = \phi$, it suffices to show the sufficient condition.

Suppose that $\phi$ is an isomorphism.

We shall show that $\phi^{-1}$ is also an isomorphism.

We have that $\phi$ is a bijection.

Hence from Inverse of Bijection is Bijection we have that $\phi^{-1}$ is also a bijection.

Then for all $x, y \in T$ and $i \in \set {1, 2, \ldots, n}$:

Therefore $\phi^{-1}$ is a homomorphism.