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 $\lbrace K(b)\mid b\in B\rbrace$ 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 $\ker(f)=\lbrace (x,y)\in A\times A\mid f(x)=f(y)\rbrace$ . Since it is a subset of $A\times A$ , it is relation on $A$ . Furthermore, it is an equivalence relation on $A$ : ¹

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 $a_1 \equiv a_2 \pmod {\ker(f)}$ 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 $c_1,\ldots,c_n$ and $d_1,\ldots,d_n$ are two sets of elements in $A$ with $c_i\equiv d_i \mod \ker(f)$ . Then $$f(\omega_A(c_1,\ldots,c_n) = \omega_B(f(c_1),\ldots,f(c_n))=\omega_B(f(d_1),\ldots,f(d_n)) = f(\omega_A(d_1,\ldots,d_n)),$$ so $\omega_A(c_1,\ldots,c_n)\equiv \omega_A(d_1,\ldots,d_n) \pmod {\ker(f)}$ . 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 $$N=\lbrace x\mid f(x)=e\rbrace,$$ 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 in model theory.

Footnotes

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In general, $\lbrace N_i\rbrace$ is a partition of a set $A$ iff $\bigcup N_i^2$ is an equivalence relation on $A$ .