Epimorphism Preserves Commutativity

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Theorem

Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.

Let $\circ$ be a commutative operation.


Then $*$ is also a commutative operation.


Proof

Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ is commutative.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.

As an epimorphism is surjective, it follows that:

$\forall u, v \in T: \exists x, y \in S: \phi \left({x}\right) = u, \phi \left({y}\right) = v$

So:

\(\displaystyle u * v\) \(=\) \(\displaystyle \phi \left({x}\right) * \phi \left({y}\right)\) $\phi$ is a surjection
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({x \circ y}\right)\) Morphism Property
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({y \circ x}\right)\) Commutativity of $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({y}\right) * \phi {\left({x}\right)}\) Morphism Property
\(\displaystyle \) \(=\) \(\displaystyle v * u\) 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