Epimorphism Preserves Commutativity
From ProofWiki
Contents |
Theorem
Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.
If $\circ$ is commutative, then so is $*$.
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:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle u * v\) | \(=\) | \(\displaystyle \phi \left({x}\right) * \phi \left({y}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | as $\phi$ is a surjection | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({x \circ y}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Morphism Property | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({y \circ x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Commutativity of $\circ$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({y}\right) * \phi {\left({x}\right)}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Morphism Property | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle v * u\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by definition as above |
$\blacksquare$
Warning
Note that this result is applied to epimorphisms. For a general homomorphism which is not surjective, we can say nothing definite about the behaviour of the elements of its codomain which are not part of its image.
Also see
Sources
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 12$: Theorem $12.2$