Epimorphism Preserves Commutativity
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 a commutative operation.
Then $*$ is also a commutative operation.
Proof
![]() | This article needs proofreading. In particular: Check validity of proof for the case of $S$ being the empty set If you believe all issues are dealt with, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
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 commutative on $S$, when $S$ is empty, as required.
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Suppose $S$ is non-empty.
As an epimorphism is surjective, it follows that:
- $\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$
So:
\(\ds u * v\) | \(=\) | \(\ds \map \phi x * \map \phi y\) | Definition of Surjection | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x \circ y}\) | Definition of Morphism Property | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {y \circ x}\) | Definition of Commutative Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi y * \map \phi x\) | Definition of Morphism Property | |||||||||||
\(\ds \) | \(=\) | \(\ds v * u\) | 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): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$