# 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 KernelOfAHomomorphismIsACongruence 2013-03-22 13:48:03 2013-03-22 13:48:03 almann (2526) almann (2526) 8 almann (2526) Theorem msc 03C05 msc 03C07 KernelOfAHomomorphismBetweenAlgebraicSystems