# Epimorphism Preserves Identity

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## Contents

## 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 $x \in S$.

$\blacksquare$

## Also see

- Epimorphism Preserves Associativity
- Epimorphism Preserves Commutativity
- Epimorphism Preserves Inverses

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 12$: Theorem $12.2$