I'm trying to learn how to perform diagram-chasing in abstract Abelian categories. Instead of an approach with some elements one have to use universal properties somehow in the proof. But I reckon the need of some lemmas...

A category is Abelian if:

  • it has a zero object
  • it has all binary products and binary coproducts
  • it has all kernels and cokernels
  • every monomorphism is a kernel to some morphism
  • every epimorphism is a cokernel to some morphism

The first "lemma" that coming into my mind (though I don't really know if it is true) deals with the connection between being monic and have a kernel equal to zero.

From the universal properties:

(1)$\quad f$ is a monomorphism if it given morphisms $m,n$ holds that $\beta m=\beta n\implies m=n$ $\require{AMScd}$ \begin{CD} X @>m>n> B@>\beta>> B' \end{CD} (2)$\quad k$ is a kernel to $\beta:B\to B'$ if $\beta k=0$ and for each $k'$ with $\beta k'=0$ there is a unique morphism $\phi$ such that $k\phi=k'$ \begin{CD} K'@>k'>>B\\ @V\exists!\phi VV\# @|\\ K @>k>> B@>\beta>> B' \end{CD}

How to prove that $\operatorname{ker}\beta=0\implies \beta$ is mono, using (1) and (2)?

My own approach, unfortunately, consists of staring on the diagrams above.

  • $\begingroup$ @QiaochuYuan Actually, I don't see how we get enrichment over commutative monoids. That's the only problem, though – once we have biproducts, there's a trick for getting negatives. $\endgroup$
    – Zhen Lin
    Feb 8, 2015 at 19:50
  • 2
    $\begingroup$ Hmmm. It turns out there's a trick for getting biproducts too. See Q6 here. $\endgroup$
    – Zhen Lin
    Feb 8, 2015 at 19:53
  • 1
    $\begingroup$ @Zhen: oh, that's curious. Well, in any case, the OP needs either a stronger definition or some lemmas to get an enrichment over abelian groups. $\endgroup$ Feb 8, 2015 at 19:59
  • 1
    $\begingroup$ Pity! This definition from Wikipedia seemed so gentle, but obviously it's a rather long way to prove the lemma from it. But Q6 might be a good exercise. $\endgroup$
    – Lehs
    Feb 8, 2015 at 20:26
  • 1
    $\begingroup$ The exercise that Zhen Lin links to uses the notion of a pseudomonomorphism which in this context probably means a morphism $m$ such that $mf=0$ implies $f=0$. $\endgroup$
    – tcamps
    Feb 8, 2015 at 23:57

1 Answer 1


I assume your abelian categories are also additive.

First prove that $f$ is a monomorphism if and only if $fg = 0$ implies $g = 0$ for all $g$.

Now assume $f : X \to Y$ has kernel $(0, 0 \to X)$ and suppose $g : W \to X$ with $fg = 0$. Then by (2), there is a morphism $h : W \to 0$ such that $g = 0h = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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