Epimorphism Preserves Inverses

Theorem
Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.

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

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 $e_S$.

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

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

Then:

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

Warning
Note that this result is applied to epimorphisms. For a general homomorphism which is not surjective, we can say nothing definite about the behaviour of the elements of its codomain which are not part of its image.

However, also see: for when the algebraic structure is actually a group.
 * Group Homomorphism Preserves Inverses

Also see

 * Epimorphism Preserves Associativity
 * Epimorphism Preserves Commutativity
 * Epimorphism Preserves Identity


 * Epimorphism Preserves Semigroups
 * Epimorphism Preserves Groups


 * Epimorphism Preserves Distributivity