Homomorphism with Identity Preserves Inverses

Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.

Let $\struct {S, \circ}$ have an identity $e_S$.

Let $\struct {T, *}$ also have an identity $e_T = \map \phi {e_S}$.

If $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.

That is:
 * $\map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$

Proof
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.

Let $\struct {T, *}$ be an algebraic structure in which $*$ has an identity $e_T = \map \phi {e_S}$.

If $x^{-1}$ is an inverse of $x$ for $\circ$, then by existence of identity $e_S$,


 * $\forall x \in S: x \circ x^{-1}, x^{-1} \circ x \in \Dom \phi$.

Hence: