# kernel of cokernel is cokernel of kernel [duplicate]

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 YuanJul 13 '11 at 21:31

• 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... – t.b. Jul 13 '11 at 21:29