Epimorphism Preserves Associativity

== 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.
Suppose $S$ is the empty set.
It follows from the definition of an epimorphism that:
$\phi$ is a surjective homomorphism
By Empty Mapping to Empty Set is Bijective, the empty map is bijective
By definition of bijection, the empty map is an epimorphism.
Therefore, suppose $\phi$ is the empty map, which is indeed an epimorphism.
By definition of a homomorphism, $\phi$ can be defined as:
:$\forall \O \in S: \map \phi {\O \circ \O} = \map \phi \O * \map \phi \O$
By Image of Empty Set is Empty Set, $T$ is also the empty set.
It follows from the definition of the homomorphism that the binary operations $\circ$ and $*$ are also the empty map.
Hence, it is vacuously true that $\circ$ is associative on $S$, when $S$ is empty, as required.
Suppose $S$ is non-empty.
It remains to be shown that:
$\forall u, v, w \in T$, $\exists x, y, z, x \circ y, y \circ z, \paren {x \circ y} \circ z, x \circ \paren {y \circ z } \in \Dom \phi$
And:
$\forall u, v, w \in T, \exists x, y, z \in S: \phi \paren{ x }, \phi \paren{ y }, \phi \paren{ z }, \phi \paren{ x \circ y }, \phi \paren{ y \circ z }, \phi \paren {\paren {x \circ y} \circ z }, \phi \paren{ x \circ \paren {y \circ z } } \in \Cdm \phi$
As an epimorphism is surjective, it follows that:
:$\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \map \phi y = v, \map \phi z = w$
Thus:
:$\forall u, v, w \in T: \exists x, y, z \in S: x, y, z \in \Dom \phi$
Similarly, by surjectivity:
:$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v \land \paren{ x \circ y \in S }$
Thus:
:$\forall u, v \in T: \exists x, y \in S: x \circ y \in \Dom \phi$
Again, by surjectivity:
:$\forall v, w \in T: \exists y, z \in S: \paren{ \map \phi y = v, \map \phi z = w } \land \paren{ y \circ z \in S }$
Thus:
:$\forall v, w \in T: \exists y, z \in S: y \circ z \in \Dom \phi$
As an epimorphism is surjective and $\circ$ is an associative operation:
:$\forall u, v, w \in T: \exists x, y, z \in S: \paren{ \map \phi x = u, \map \phi y = v, \map \phi z = w } \land \paren{ {\paren {x \circ y} \circ z} = x \circ \paren {y \circ z } }$
Thus:
:$\forall u, v, w \in T, \exists x, y, z \in S: x, y, z \in S: \paren {x \circ y} \circ z, x \circ \paren {y \circ z } \in \Dom \phi$
Hence:
$\forall u, v, w \in T, \exists x, y, z \in S: x, y, z, x \circ y, y \circ z, \paren {x \circ y} \circ z, x \circ \paren {y \circ z } \in \Dom \phi$
As an epimorphism is surjective:
$\forall u, v, w \in T, \exists x, y, z \in S: \phi \paren{ x }, \phi \paren{ y }, \phi \paren{ z }, \phi \paren{ x \circ y }, \phi \paren{ y \circ z }, \phi \paren {\paren {x \circ y} \circ z }, \phi \paren{ x \circ \paren {y \circ z } } \in \Cdm \phi$
Hence:
== Warning ==
== Also see ==
* Epimorphism Preserves Commutativity
* Epimorphism Preserves Identity
* Epimorphism Preserves Inverses
* Epimorphism Preserves Semigroups
* Epimorphism Preserves Groups
* Epimorphism Preserves Distributivity
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$
Category:Epimorphisms (Abstract Algebra)
Category:Associativity