kernel of a homomorphism is a congruence

Let $\Sigma$ be a fixed signature, and $\mathfrak{A}$ and $\mathfrak{B}$ two structures for $\Sigma$ . If $f\colon\mathfrak{A}\to\mathfrak{B}$ is a homomorphism, then $\ker(f)$ is a congruence on $\mathfrak{A}$ .

Proof.

If $F$ is an $n$ -ary function symbol of $\Sigma$ , and $f(a_{i})=f(a_{i}^{\prime})$ , then

	$\displaystyle f(F^{\mathfrak{A}}(a_{1},\ldots,a_{n}))$	$\displaystyle=F^{\mathfrak{B}}(f(a_{1}),\ldots,f(a_{n}))$
		$\displaystyle=F^{\mathfrak{B}}(f(a_{1}^{\prime}),\ldots,f(a_{n}^{\prime}))$
		$\displaystyle=f(F^{\mathfrak{A}}(a_{1}^{\prime},\ldots,a_{n}^{\prime})).\qed$

Title	kernel of a homomorphism is a congruence
Canonical name	KernelOfAHomomorphismIsACongruence
Date of creation	2013-03-22 13:48:03
Last modified on	2013-03-22 13:48:03
Owner	almann (2526)
Last modified by	almann (2526)
Numerical id	8
Author	almann (2526)
Entry type	Theorem
Classification	msc 03C05
Classification	msc 03C07
Related topic	KernelOfAHomomorphismBetweenAlgebraicSystems