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

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