# Epimorphism Preserves Associativity

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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 $\circ$ be an associative operation.

Then $*$ is also an associative operation.

## Proof

Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.

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

As an epimorphism is surjective, it follows that:

- $\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \, \map \phi y = v, \, \map \phi z = w$

So:

\(\displaystyle \paren {u * v} * w\) | \(=\) | \(\displaystyle \paren {\map \phi x * \map \phi y} * \map \phi z\) | as $\phi$ is a surjection | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {x \circ y} * \map \phi z\) | Definition of Morphism Property | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {\paren {x \circ y} \circ z}\) | Definition of Morphism Property | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {x \circ \paren {y \circ z} }\) | Associativity of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi x * \map \phi {y \circ z}\) | Definition of Morphism Property | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi x * \paren {\map \phi y * \map \phi z}\) | Definition of Morphism Property | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle u * \paren {v * w}\) | by definition as above |

$\blacksquare$

## 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

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