Epimorphism Preserves Inverses

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Theorem

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

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

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

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


Then $\map \phi {x^{-1} }$ is an inverse element of $\map \phi x$ for $*$.

That is:

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


Proof

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

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

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

By definition of inverse element:

$x \circ y = e_S = y \circ x$

Then:

\(\ds \map \phi x * \map \phi y\) \(=\) \(\ds \map \phi {x \circ y}\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds \map \phi {e_S}\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds \map \phi {y \circ x}\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds \map \phi y * \map \phi x\) Definition of Morphism Property


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

$\blacksquare$


Warning

Note that this result is applied to epimorphisms.

For a general homomorphism which is not surjective, nothing definite can be said about the behaviour of the elements of its codomain which are not part of its image.


Also see


Sources

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