Epimorphism 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 an epimorphism.

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

Let $x^{-1}$ be an inverse element of $x$ for $\circ$.

Then $\phi \left({x^{-1}}\right)$ is an inverse element 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 element $e_S$.

From Epimorphism Preserves Identity, it follows that $\left({T, *}\right)$ also has an identity element, which is $\phi \left({e_S}\right)$.

Let $y$ be an inverse of $x$ in $\left({S, \circ}\right)$.

Then:

 $\displaystyle \phi \left({x}\right) * \phi \left({y}\right)$ $=$ $\displaystyle \phi \left({x \circ y}\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({e_S}\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({y \circ x}\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({y}\right) * \phi \left({x}\right)$

So $\phi \left({y}\right)$ is an inverse of $\phi \left({x}\right)$ in $\left({T, *}\right)$.

$\blacksquare$