# every congruence is the kernel of a homomorphism

Let $\Sigma$ be a fixed signature, and $\mathfrak{A}$ a structure for $\Sigma$. If $\sim$ is a congruence on $\mathfrak{A}$, then there is a homomorphism $f$ such that $\sim$ is the kernel of $f$.

###### Proof.

Define a homomorphism $f\colon\mathfrak{A}\to\mathfrak{A}/\!\sim\ :a\mapsto[\![a]\!]$. Observe that $a\sim b$ if and only if $f(a)=f(b)$, so $\sim$ is the kernel of $f$. To verify that $f$ is a homomorphism, observe that

1. 1.

For each constant symbol $c$ of $\Sigma$, $f(c^{\mathfrak{A}})=[\![c^{\mathfrak{A}}]\!]=c^{\mathfrak{A}/\!\sim}$.

2. 2.

For each $n\in\mathbb{N}$ and each $n$-ary function symbol $F$ of $\Sigma$,

 $\displaystyle f(F^{\mathfrak{A}}(a_{1},\ldots a_{n}))$ $\displaystyle=[\![F^{\mathfrak{A}}(a_{1},\ldots a_{n})]\!]$ $\displaystyle=F^{\mathfrak{A}/\!\sim}([\![a_{1}]\!],\ldots[\![a_{n}]\!])$ $\displaystyle=F^{\mathfrak{A}/\!\sim}(f(a_{1}),\ldots f(a_{n})).\qed$
3. 3.

For each $n\in\mathbb{N}$ and each $n$-ary relation symbol $R$ of $\Sigma$, if $R^{\mathfrak{A}}(a_{1},\ldots,a_{n})$ then $R^{\mathfrak{A}/\!\sim}([\![a_{1}]\!],\ldots,[\![a_{n}]\!])$, so $R^{\mathfrak{A}/\!\sim}(f(a_{1}),\ldots,f(a_{n}))$.

Title every congruence is the kernel of a homomorphism EveryCongruenceIsTheKernelOfAHomomorphism 2013-03-22 13:48:59 2013-03-22 13:48:59 almann (2526) almann (2526) 11 almann (2526) Theorem msc 03C07 msc 03C05