Epimorphism Preserves Identity

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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 element $e_S$.


Then $\struct {T, *}$ has the identity element $\map \phi {e_S}$.


Proof

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

Then:

$\forall x \in S: x \circ e_S = x = e_S \circ x$

Thus:

$\forall x \in S: x \circ e_S, e_S \circ x \in \Dom \phi$

Hence:

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

and:

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

The result follows because every element $y \in T$ is of the form $\map \phi x$ with $x \in S$.

$\blacksquare$


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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