Isomorphism Preserves Inverses

Theorem

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

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

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

Then $x^{-1}$ is an inverse of $x$ for $\circ$ if and only if $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.

$\map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$

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

From Epimorphism Preserves Identity, it follows that $\struct {T, *}$ also has an identity, which is $\map \phi {e_S}$.

Let $y$ be an inverse of $x$ in $\struct {S, \circ}$.

Then:

$\ds \map \phi x * \map \phi y$

$=$

$\ds \map \phi {x \circ y}$

$\ds $

$=$

$\ds \map \phi {e_S}$

$\ds $

$=$

$\ds \map \phi {y \circ x}$

$\ds $

$=$

$\ds \map \phi y * \map \phi x$

So $\map \phi y$ is an inverse of $\map \phi x$ in $\struct {T, *}$.

Thus the result for $\map \phi x$ can be applied to $\map {\phi^{-1} } {\map \phi x}$.

$\blacksquare$

$\blacksquare$