kernel of a homomorphism between algebraic systems

Let $f:(A,O)\to (B,O)$ be a homomorphism between two algebraic systems $A$ and $B$ (with $O$ as the operator set). Each element $b\in B$ corresponds to a subset $K(b):=f^{-1}(b)$ in $A$. Then $\{K(b)\mid b\in B\}$ forms a partition of $A$. The kernel $\ker (f)$ of $f$ is defined to be

$$\ker (f):=\bigcup _{b\in B}K(b)\times K(b).$$

It is easy to see that $\ker (f)=\{(x,y)\in A\times A\mid f(x)=f(y)\}$. Since it is a subset of $A\times A$, it is relation on $A$. Furthermore, it is an equivalence relation on $A$:

1. $\ker (f)$ is reflexive: for any $a\in A$, $a\in K(f(a))$, so that $(a,a)\in K{(f(a))}^2\subseteq \ker (f)$

2. $\ker (f)$ is symmetric: if $(a_1,a_2)\in \ker (f)$, then $f(a_1)=f(a_2)$, so that $(a_2,a_1)\in \ker (f)$

3. $\ker (f)$ is transitive: if $(a_1,a_2),(a_2,a_3)\in \ker (f)$, then $f(a_1)=f(a_2)=f(a_3)$, so $(a_1,a_3)\in \ker (f)$.

We write $a_1\equiv a_2(mod\ker (f))$ to denote $(a_1,a_2)\in \ker (f)$.

In fact, $\ker (f)$ is a congruence relation: for any $n$-ary operator symbol $\omega \in O$, suppose $c_1,\mathrm{\dots },c_n$ and $d_1,\mathrm{\dots },d_n$ are two sets of elements in $A$ with $c_i\equiv d_{i}mod\ker (f)$. Then

$$f({\omega }_A(c_1,\mathrm{\dots },c_n)={\omega }_B(f(c_1),\mathrm{\dots },f(c_n))={\omega }_B(f(d_1),\mathrm{\dots },f(d_n))=f({\omega }_A(d_1,\mathrm{\dots },d_n)),$$

so ${\omega }_A(c_1,\mathrm{\dots },c_n)\equiv {\omega }_A(d_1,\mathrm{\dots },d_n)(mod\ker (f))$. For this reason, $\ker (f)$ is also called the congruence induced by $f$.

Example. If $A,B$ are groups and $f:A\to B$ is a group homomorphism. Then the kernel of $f$, using the definition above is just the union of the square of the cosets of

$$N=\{x\mid f(x)=e\},$$

the traditional definition of the kernel of a group homomorphism (where $e$ is the identity of $B$).

Remark. The above can be generalized. See the analog (http://planetmath.org/KernelOfAHomomorphismIsACongruence) in model theory.