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 $$\left({T, *}\right)$$.

Comment
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 range which are not part of its image.