every congruence is the kernel of a homomorphism

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

Proof.

Define a homomorphism $f:𝔄\to 𝔄/\sim :a\mapsto [[a]]$. 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. For each constant symbol $c$ of $\mathrm{\Sigma }$, $f(c^{𝔄})=[[c^{𝔄}]]=c^{𝔄/\sim }$.

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

   $f(F^{𝔄}(a_1,\mathrm{\dots }{a}_n))$$=[[F^{𝔄}(a_1,\mathrm{\dots }{a}_n)]]$

   $=F^{𝔄/\sim }([[a_1]],\mathrm{\dots }[[a_n]])$

   $=F^{𝔄/\sim }(f(a_1),\mathrm{\dots }f(a_n)).\mathit{∎}$

3. For each $n\in \mathbb{N}$ and each $n$-ary relation symbol $R$ of $\mathrm{\Sigma }$,if $R^{𝔄}(a_1,\mathrm{\dots },a_n)$ then $R^{𝔄/\sim }([[a_1]],\mathrm{\dots },[[a_n]])$, so$R^{𝔄/\sim }(f(a_1),\mathrm{\dots },f(a_n))$.