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:

Also see

 * Epimorphism Preserves Commutativity
 * Epimorphism Preserves Identity
 * Epimorphism Preserves Inverses


 * Epimorphism Preserves Semigroups
 * Epimorphism Preserves Groups


 * Epimorphism Preserves Distributivity