Let $G$ and $H$ are groups and $f:G\to H$ is group homomorphism. The kernel $f$ is defined by \begin{align*} \ker f =\{ g\in G \ : \ f(g)=1_{H}\}. \end{align*} And let $S$ and $T$ are Semigroups and $\varphi:S\to T$ is semigroup homomorphism. The kernel $\varphi$ is defined by \begin{align*} \ker \varphi=\{(s, s') \in S\times S \ : \ \varphi(s)=\varphi(s')\}. \end{align*} I know category of groups has a zero morphism, then the equalizer of zero morphism $0_{GH}:G\to H$ and $f:G\to H$ is kernel of $f$. But category of Semigroups doesn't has zero morphism.

I have two questions:

  1. What is the relation between these kernels.

  2. What is the definition of kernels in the category has no zero object or zero morphism.

Thank you

  • 2
    $\begingroup$ Well, if $\varphi$ is a group homomorphism, and writing $\ker_G(\varphi)$ for the group definition of kernel and $\ker_S(\varphi)$ for the semigroup definition, we have that $(s,s')\in\ker_S(\varphi)\Leftrightarrow s^{-1}s'\in\ker_G(\varphi)$. (In semigroups, $s^{-1}$ does not necessarily make sense.) Does that help with your question (1)? $\endgroup$
    – user1729
    Sep 24, 2019 at 10:07

1 Answer 1


In general algebraic structures, the kernel of a homomorphism $f:X\to Y$ is defined just as for semigroups: $$\ker f:=\{(x, x') : f(x) =f(x')\}$$ which is always a congruence relation on $X$, i.e. an equivalence relation closed under the operations (meaning that it's a subalgebra of $X\times X$).

For groups [or rings or vector spaces, Boolean algebras, etc.], there is a one-to-one correspondence between congruence relations and normal subgroups [ideals, subspaces, Boolean ideals, etc], namely the equivalence class of the identity element [or, of $0$] already determines the whole relation.

In category theory, we can reflect this general notion of kernel by a kernel pair: a pair of arrows $k_1,k_2:K\to X$ (where $K$ plays the role of the congruence relation by the induced arrow $K\to X\times X$), which satisfy $f\circ k_1=f\circ k_2$, and whenever $f\circ u=f\circ v$ with $u,v:A\to X$, there's a unique $s:A\to K$ satisfying $u=k_1\circ s$ and $v=k_2\circ s$.

Working in a category of general algebras, $s$ is simply $A\ni\ a\mapsto (u(a),\, v(a))\ \in\ker f$.


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