# Homomorphism with Identity Preserves Inverses

## Theorem

Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

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

Let $\left({S, \circ}\right)$ have an identity $e_S$.

Let $\left({T, *}\right)$ also have an identity $e_T = \phi \left({e_S}\right)$.

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

That is:

$\phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$

## Proof

Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ has an identity $e_S$.

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

Let $\left({T, *}\right)$ be an algebraic structure in which $*$ has an identity $e_T = \phi \left({e_S}\right)$.

 $\displaystyle \phi \left({x}\right) * \phi \left({x^{-1} }\right)$ $=$ $\displaystyle \phi \left({x \circ x^{-1} }\right)$ Morphism property of $\phi$ $\displaystyle$ $=$ $\displaystyle \phi \left({e_S}\right) = e_T$ $\displaystyle$ $=$ $\displaystyle \phi \left({x^{-1} \circ x}\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({x^{-1} }\right) * \phi \left({x}\right)$ Morphism property of $\phi$

$\blacksquare$