In a category with zero, a simple object is one that has only two quotients - itself and zero.

Firstly - a point of confusion. The definition above says that quotient object requires a congruence, what is the one we should use here? I don't see a natural one, but I'm probably missing something obvious.

b. When the category is abelian, quotients can be replaced by subobjects. Is there a more precise condition?


The notion of congruence being used here is linked to from the pages you linked to; it's this one.

The reason that we can replace quotients with subobjects for abelian categories is that we can quotient an object by a nontrivial subobject and expect a nontrivial quotient in return. This isn't true in general; for example, the quotient of a group $G$ by a subgroup $H$ (by which I mean the cokernel of the inclusion $H \to G$) may be trivial even if $H$ is not all of $G$ (since its normal closure may be all of $G$). I am not aware of a useful criterion substantially more general than abelian here; $\text{Grp}$ is one of the nicest not-quite-abelian categories, so the situation doesn't look hopeful.

  • $\begingroup$ Ok, I understood that a congruence should be used but which specific one is actually used? Thats the obvious bit I was missing... $\endgroup$ – Mozibur Ullah Jun 26 '13 at 3:08
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    $\begingroup$ @Mozibur: it's not clear to me what "here" means, but for quotienting by a subobject, take the kernel pair (ncatlab.org/nlab/show/kernel+pair) of the cokernel of the inclusion of the subobject. For example, if $A$ is a submodule of a module $B$, the corresponding congruence is the submodule of pairs $(b_1, b_2) \in B^2$ such that $b_1 \equiv b_2 \bmod A$. (In abelian categories it isn't necessary to work with congruences because kernel pairs can be replaced by kernels and quotients by congruences reduce to quotients by subobjects; the congruence machinery is for the general case.) $\endgroup$ – Qiaochu Yuan Jun 26 '13 at 6:01
  • $\begingroup$ Thanks, for your explanation. I just realised the obvious point I was missing. $\endgroup$ – Mozibur Ullah Jun 26 '13 at 6:18

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