4
$\begingroup$

Etingof et. al. define abelian categories as additive categories in which for every morphism $\phi : X \to Y$ there exists a sequence $$K \xrightarrow{k} X \xrightarrow{i} I \xrightarrow{j} Y \xrightarrow{c} C$$ with the following properties

  1. $ji = \phi$,
  2. $(K,k) = \ker \phi, (C,c) = \operatorname{coker} \phi$,
  3. $(I,i) = \operatorname{coker} k, (I,j) = \ker c$.

Why is this equivalent to the standard definition of an abelian category: an additive category is abelian if

  1. every map has kernels and cokernels,

  2. every monomorphism is the kernel of its cokernel,

  3. every epimorphism is the cokernel of its kernel.

    I assume this is a standard result in some book/paper but I can't find it anywhere.

$\endgroup$
2

1 Answer 1

2
$\begingroup$

This sequence is basically giving us the image factorisation of $\phi$, and I'll assume that you've seen that you can always construct this factorisation in an abelian category.

Our task is therefore to go the other way around and show that a category with this sort of image factorisation satisfies the three conditions in the definition of abelian category. Here's a sketch of the proof:

  • Every morphism $\phi$ has a kernel and cokernel, because we need them to define the above sequence.
  • Suppose $\phi$ is a monomorphism. Its kernel is the zero morphism, so the cokernel of that kernel is isomorphic (in the suitable sense) to the identity. Therefore $\phi$ is isomorphic in the same sense to $j$, the kernel of the cokernel. Thus $\phi$ is also the kernel of its cokernel.
  • The situation for epimorphisms is dual to that for monomorphisms.

Thus a category satisfying this image factorisation condition is an abelian ctageory.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .