# Homomorphism with Identity Preserves Inverses

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## 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 $e_T = \map \phi {e_S}$.

Let $x^{-1}$ be 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:

 $\ds \map \phi x * \map \phi {x^{-1} }$ $=$ $\ds \map \phi {x \circ x^{-1} }$ Definition of Morphism Property $\ds$ $=$ $\ds \map \phi {e_S} = e_T$ $\ds$ $=$ $\ds \map \phi {x^{-1} \circ x}$ $\ds$ $=$ $\ds \map \phi {x^{-1} } * \map \phi x$ Definition of Morphism Property

$\blacksquare$