Let's fix some terminology first. A category $\mathcal{C}$ is preabelian if:

1) $Hom_{\mathcal{C}}(A,B)$ is an abelian group for every $A,B$ such that composition is biadditive,

2) $\mathcal{C}$ has a zero object,

3) $\mathcal{C}$ has binary products,

4) $\mathcal{C}$ has kernels and cokernels.

A category $\mathcal{C}$ is abelian if it is preabelian and satisfies:

5) every monomorphism is a kernel and every epimorphism is a cokernel.

Define the coimage of a map to be the cokernel of its kernel, and the image to be the kernel of its cokernel. We have the following commutative diagram:

enter image description here

where $\overline{f}$ is the only existing map (because of universality of kernel and cokernel).

I'm having trouble proving the following:

A preabelian category $\mathcal{C}$ is abelian iff $\overline{f}$ is an isomorphism.

The converse is easily shown, I'm having trouble proving $\Rightarrow$...

  • $\begingroup$ Can you prove that $\overline{f}$ is both a monomorphism and an epimorphism if $\mathcal{C}$ is abelian? $\endgroup$
    – t.b.
    Jun 12, 2011 at 21:01
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    $\begingroup$ Sorry for not following up earlier. It's not trivial at all. You can extract a proof from chapter 2 of Freyd's book where it basically occupies the entire section. It's getting late here and there's the danger that I get myself hopelessly confused when trying to find some short cuts. $\endgroup$
    – t.b.
    Jun 12, 2011 at 22:32
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    $\begingroup$ That the map $\operatorname{Coim}f \to B$ is monic is Freyd's Theorem 2.18$^\ast$ on page 70 of the pdf (page 44 of the book). This implies that $\bar{f}$ is monic. Dually, $\bar{f}$ is epic (use Thm 2.18 on page 69 (43)). Then Theorem 2.12 on page 66 (37) shows that $\bar{f}$ is an isomorphism because it is monic and epic. $\endgroup$
    – t.b.
    Jun 12, 2011 at 22:48
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    $\begingroup$ a) Consider $0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \xrightarrow{\cdot 2} \mathbb{Z}/2 \to 0$ for a counterexample to the last question (abelian groups are an abelian category, of course). b) If $g: A \to B$ is mono-epi then its kernel is zero $0 \to A$ and epis are cokernels of their kernels in an abelian category. Now the composition $0 \to A \xrightarrow{1_A} A$ is zero, so $1_A = hg$ for some $h: B \to A$. On the other hand $ghg = 1_Bg$, so $gh = 1_B$ and hence $h$ is inverse to $g$. $\endgroup$
    – t.b.
    Jun 14, 2011 at 1:07
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    $\begingroup$ Yes, I know you don't like Freyd's way of writing :), yes that's what I'm using (that's not hard). No In the "on the other hand..." I'm just using that $g$ is epi: If $kg = lg$ then $k = l$ applied to $(gh)g = g(hg) = g1_A = g = 1_Bg$ thus $gh = 1_B$. $\endgroup$
    – t.b.
    Jun 14, 2011 at 1:54

1 Answer 1


Here is an argument for $\Rightarrow$. There is not much more to it than chasing diagrams (as it should be). Also, I didn't really bother to check which (parts of the) axioms are actually needed:

  1. In presence of 1),2),3) we have that $\mathcal{C}$ has biproducts as well: every binary coproduct is also a binary product. (This is not used below but I added it for the sake of completeness)

  2. Assuming 1)-5), an epi $e:B \to C$ is the cokernel of its kernel. cokernel of kernel diagram
    Indeed, let $f$ be a morphism such that $e = \operatorname{coker}\,{f}$ and let $k = \operatorname{ker}{e}$. Since $ef = 0$, we see that $f = kf'$. If $y$ is such that $yk =0$ then $ykf' = yf = 0$ and hence $y = y'e$, and thus $e$ is a cokernel of $k$.

    Dually, a mono is the kernel of its cokernel.

  3. Assuming 1)-5) a morphism which is both an epimorphism and a monomorphism is an isomorphism.

    I leave that as an easy exercise (I gave the argument in the comments above).

  4. Let $f: A \to B$. The morphism $i: \operatorname{Coim}{f} \to B$ is monic.i monic
    To this end, let $x: X \to \operatorname{Coim}{f}$ be such that $ix = 0$. Let $q = \operatorname{coker}{x}$ and let $j: \operatorname{Coker}{x} \to B$ be the unique map such that $i = jq$. Since $qp$ is epi we have a morphism $h: H \to A$ such that $qp = \operatorname{coker}{{h}}$. Now $fh = iph = jqph = 0$ so $h = kh'$. This gives that $ph = pkh' = 0$, factorization of p
    so $p$ factors as $p = p'(qp) = (p'q)p$. But $p$ is epi, so $p'q = 1_{\operatorname{Coim}{f}}$. This implies that $q$ is a monomorphism and finally $qx = 0$ implies that $x = 0$. We have shown that $ix = 0$ implies $x = 0$ and thus $i$ is a monomorphism.

    Dually $j: A \to \operatorname{Im}f$ is an epimorphism.

  5. Consider the factorization of $f$: Analysis of a morphism By step 4 we have that $A \to \operatorname{Im}{f}$ and $\operatorname{Coim}{f} \to B$ are epi and mono, respectively. Therefore $\bar{f}$ is both epi and mono and we're done by step 3.

  • 1
    $\begingroup$ @Theo: Sorry, I hadn't had time to finish checking the argument. I did now, it's perfect. Thank you once again, you've been of huge help! $\endgroup$ Jun 15, 2011 at 15:43
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    $\begingroup$ @Bruno: Thank you, that's nice to hear! By the way: If you want to test your understanding of this argument try to prove the following: Let $k: A' \to A$ be a monomorphism and let $f': A' \to B'$ be an arbitrary morphism. Consider the push-out diagram $$\begin{array}{ccc} A' & \xrightarrow{k\phantom{l}} & A \\ \downarrow{\scriptstyle{f'}} & \tiny{\text{PO}} & \downarrow{\scriptstyle{f}} \\ B' & \xrightarrow{\phantom{k}l} & B. \end{array}$$ Then $l$ is a monomorphism. "The push-out of a mono along an arbitrary morphism is a mono." Dually the pull-back of an epi is an epi. $\endgroup$
    – t.b.
    Jun 15, 2011 at 22:23
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    $\begingroup$ @Bruno: Third and last comment (self-advertisement): As a motivation for this exercise, look at exact categories on Wikipedia. I've written a survey on these some years ago - one point is that exact categories are very convenient for proving the standard diagram lemmas and you can do practically all the homological algebra in them. If you can't access it, there's a preliminary version of it on the ArXiV. $\endgroup$
    – t.b.
    Jun 15, 2011 at 22:34
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    $\begingroup$ Part 3 is proven in your comments above, and it is perfectly fine. Here is a proof which does not use that the coimage of an epimorphism is the same morphism. Let $f:A\to B$ be mono-epi. Then $f=ker g$ for some $g:B\to X$. Then $gf=g(kerg)=0=0f$: since $f$ is epi this implies $g=0$. But then $f=ker g=ker 0=Id_B$ hence $f$ is an isomorphism. @Theo $\endgroup$ Jul 6, 2011 at 0:49
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    $\begingroup$ @t.b. What I really liked about your answer was your outline. I was able to read the steps necessary, and show them myself, instead of just reading the proof. That was very valuable! Actually, I was able to do your steps 1-3, and 5 myself, but have not yet cracked step 4, which I guess is the big one. But I'll keep working on it for a while, and if I don't get it myself eventually, I'll read it here. This was a really great answer. Thanks! $\endgroup$
    – ziggurism
    May 30, 2012 at 3:46

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