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


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

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


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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:

\(\ds \paren {u * v} * w\) \(=\) \(\ds \paren {\map \phi x * \map \phi y} * \map \phi z\) as $\phi$ is a surjection
\(\ds \) \(=\) \(\ds \map \phi {x \circ y} * \map \phi z\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds \map \phi {\paren {x \circ y} \circ z}\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds \map \phi {x \circ \paren {y \circ z} }\) Associativity of $\circ$
\(\ds \) \(=\) \(\ds \map \phi x * \map \phi {y \circ z}\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds \map \phi x * \paren {\map \phi y * \map \phi z}\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds u * \paren {v * w}\) 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

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