Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Under what conditions, in what contexts, are kernels (resp. cokernels) finite limits (resp. finite colimits)?

share|cite|improve this question
...always? ${}{}$ – Qiaochu Yuan Oct 29 '12 at 21:59
Assuming you have the right definition of kernel and cokernel, of course, to say nothing of the assumptions needed so that such things even make sense... – Zhen Lin Oct 29 '12 at 22:11
up vote 5 down vote accepted

I misunderstood the question slightly (I thought you were asking about equalizers and coequalizers). The most general definition of a kernel or cokernel that I'm aware of takes place in a category with zero morphisms. This includes categories with zero objects as well as $\text{Ab}$-enriched categories such as abelian categories.

In such a category, the kernel (resp. cokernel) of a map $f : a \to b$ may be defined as the equalizer (resp. coequalizer) of $f$ and the zero morphism $0 : a \to b$. In particular, kernels are a finite limit and cokernels are a finite colimit. In an $\text{Ab}$-enriched category, the equalizer of an arbitrary pair of parallel morphisms $f, g : a \to b$ is the kernel of $f - g$, so being able to compute kernels (resp. cokernels) is equivalent to being able to compute equalizers (resp. coequalizers).

You can verify that this reproduces the usual notion of kernel and cokernel of modules, for example. It also gives a notion of kernel and cokernel for, say, pointed topological spaces.

A more general definition replaces the notion of kernel and cokernel with the notion of kernel pair and cokernel pair. This is a generalization of the equivalence relation on a set $X$ induced by a map $f : X \to Y$ and is a pullback, so in particular a finite limit. Cokernel pairs are dually given by pushouts, so in particular are finite colimits.

share|cite|improve this answer
Maybe trivial to add on, but for completion, you might note that the equalizer is always a finite limit since it is the limit of the finite index category * <--->*. – user21725 Oct 29 '12 at 23:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.