Epimorphism Preserves Associativity

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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 $\circ$ 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