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If $\mathcal{C}$ is an abelian category, we can consider the quotient $B/A$ when $A$ is a subobject of $B$ (i.e. there is a mono from $A$ to $B$.) It satisfies following universal property:

For every object $C$ and morphism $f:B\to C$ with $fi=0$ there is the unique morphism $f':A/B\to C$ such that $f=f'p$. (Where $p : B\to B/A$ be the natural projection.)

I have tried to dualize of "quotient of $A$ by $B$". Since the dual object of a quotient object is a subobject, the dualization of it should be a dual object.

If there is a epi $\pi$ from $B$ to $A$, we can find "the dual of quotient" of $B$ "modulo"(?) $A$; I call it $B; A$ temporarily. $B; A$ shall satisfy following universal property:

For every object $C$ and morphism $f:C\to B$ with $\pi f=0$ there is the unique morphism $f':C\to B; A$ such that $\iota f' = f$. (Where $\iota : B;A\to B$ should be the natural inclusion.)

In fact, $B;A=\ker \pi$ for some concrete categories like the category of abelian groups or $R$-modules. (In contrast with $B/A = \operatorname{coker} i$.)

It seems like that the annihilator is quite related to the above one. But I can't find references which says how it is called. Thanks for any help!

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The hypothesis that the map involved is a mono or an epi is unnecessary. The quotient is always the cokernel, and the categorical dual of the cokernel is the kernel.

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  • $\begingroup$ Good answer. Also, it is not necessary to restrict to abelian categories. We only need zero morphisms. $\endgroup$ May 14, 2015 at 10:59
  • $\begingroup$ Thanks for your answer, but it is not an answer I want. We do not call $G/H$ a cokernel ordinarily (I don't certain it since I don't know well about category theory and related fields.) My question I intended is, there is known notations of $A;B$ in my question used widely. (I am sorry if you misunderstand my question because of the ambiguity of my asking.) $\endgroup$
    – Hanul Jeon
    May 16, 2015 at 3:40
  • $\begingroup$ @tetori: $G/H$ is absolutely the cokernel. The known notation that is used widely is that the construction you describe is called the kernel and denoted $\text{ker}(\pi)$, regardless of whether $\pi$ is an epimorphism. $\endgroup$ May 16, 2015 at 3:44

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