# Epimorphism Preserves Identity

## 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:

 $\displaystyle \map \phi x$ $=$ $\displaystyle \map \phi {x \circ e_S}$ $e_S$ is an identity for $\circ$ $\displaystyle$ $=$ $\displaystyle \map \phi x * \map \phi {e_S}$ Morphism property of $\circ$ under $\phi$

and:

 $\displaystyle \map \phi x$ $=$ $\displaystyle \map \phi {e_S \circ x}$ $e_S$ is an identity for $\circ$ $\displaystyle$ $=$ $\displaystyle \map \phi {e_S} * \map \phi x$ Morphism property of $\circ$ under $\phi$

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

$\blacksquare$