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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:
Construct a map $\hat{\!f}:\operatorname{Coim}{f}\rightarrow \operatorname{Im}{f}$ using the universal properties.
Prove that the arrow $A\rightarrow \operatorname{Im}{f}$ is epic and the arrow $\operatorname{Coim}{f}\rightarrow B$ is monic.
Since $\operatorname{coim}{f}$ is epic and $\operatorname{im}{f}$ is monic, this implies that $\hat{\!f}$ is both epic and monic.
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.