Possible Duplicate:
Equivalent conditions for a preabelian category to be abelian

Let $\mathcal{C}$ be an abelian category, and consider an arrow $f:A\rightarrow B$. In a number of sources (Vakil's notes, the appendix of Weibel's book, Wikipedia, personal conversation), I've seen it noted that $\operatorname{Ker}{(\operatorname{coker}{(f)})}\cong \operatorname{Coker}{(\operatorname{ker}{(f))}}$ (as objects), so that we may safely call either object $\operatorname{Im}(f)$. But none of these sources worked out the details. It seems like it should be a very easy exercise, but I'm getting stuck.

Here is a plan for the proof:

  1. Construct a map $\hat{\!f}:\operatorname{Coim}{f}\rightarrow \operatorname{Im}{f}$ using the universal properties.

  2. Prove that the arrow $A\rightarrow \operatorname{Im}{f}$ is epic and the arrow $\operatorname{Coim}{f}\rightarrow B$ is monic.

  3. Since $\operatorname{coim}{f}$ is epic and $\operatorname{im}{f}$ is monic, this implies that $\hat{\!f}$ is both epic and monic.

  4. Make sure you've proven the lemma that an arrow which is both epic and monic is iso.

I've done all but step 2.

The question: Why are the arrows $A\rightarrow \operatorname{Im}{f}$ and $\operatorname{Coim}{f}\rightarrow B$ epic and monic, respectively?

Note 1: This is trivial for modules, so we could just apply the Freyd-Mitchell Embedding Theorem. I'm really looking for an elementary proof from the axioms.

Note 2: To avoid ambiguity, I am taking as my axioms for an abelian category:

  • Additive structure on Hom sets, with distributivity
  • $0$ object
  • Finite products
  • Every arrow has a kernel and cokernel
  • A monic is the kernel of its cokernel
  • An epic is the cokernel of its kernel

Edit: (TB) The question proper is answered in point 4. of the answer in the thread linked to above.


marked as duplicate by t.b., Qiaochu Yuan Jul 13 '11 at 21:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ I wrote a rather elaborate answer to your question in the link above. Don't worry, it's not trivial at all! $\endgroup$ – t.b. Jul 13 '11 at 15:34
  • 1
    $\begingroup$ Also, appealing to Freyd-Mitchell does not help, as its proof relies heavily on the category being abelian and the localization theory involved in its proof uses several applications of the fact you want to prove. You could resort to proving that your axioms imply that your category is exact (in the sense of Quillen) and apply a refinement of the Gabriel-Quillen embedding theorem, but that's considerable overkill (also proving the exact category axioms from the axioms you gave is about as difficult as the problem you try to solve). $\endgroup$ – t.b. Jul 13 '11 at 15:40
  • $\begingroup$ Thanks, the question is more or less the same... the answer to my more specific question is just part 4 of your lengthy answer to Bruno's question. Not sure what the etiquette here is for duplicates - I would accept your answer, but I can't accept a comment! $\endgroup$ – Alex Kruckman Jul 13 '11 at 15:41
  • $\begingroup$ I voted to close your question as a duplicate, and it will be closed at some point in the near future (it doesn't make much sense that I repost that part of my answer and you accept it). There is no harm in having the question closed, it's mere information for the next visitors. I'll add a remark to that extent to your question as soon as the question is closed. $\endgroup$ – t.b. Jul 13 '11 at 15:43
  • 1
    $\begingroup$ I edited your post to make it look a bit nicer. One further remark: I find it rather nice to denote the kernel object by $\operatorname{Ker}{f}$ (uppercase because it is an object) and the kernel morphism by $\operatorname{ker}{f}$ (lowercase because it is a morphism). Similarly for coimage, image and cokernel. Because $\hat{\!f}$ is an isomorphism people usually don't introduce the coimage, but they should for reasons of symmetry. For instance, homology can... $\endgroup$ – t.b. Jul 13 '11 at 21:29