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A quotient object of an object $A$ is usually denoted $A/B$ (we're talking about equivalence classes of epis). It seems that in categories like $\mathsf {Grp}$ and $\mathsf {Ab}$ one can associate with the notation $A/B$ an actual subobject $B$ of $A$ and view quotient objects as constructs from subobjects instead of duals to them.

I'm confused and can't manage to clearly formulate this cleanly, so I don't really have a precise question I want to ask - I'd just like someone to shed light on what I mentioned above. Also, if it makes any sense to ask this - in what categories can we view quotient objects as constructs from subobjects?

In particular, I'm trying to think of quotient complexes as quotient objects in $\mathsf {Ch}_\bullet (\mathsf {A})$ for an abelian category $\mathsf A$ but I'm not sure how to view it next to the definition in Rotman's Introduction to Algebraic Topology as a grading-wise quotient by a subcomplex.

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In an abelian category (or in a different direction a pointed category with coequalizers so that I can talk about $\mathsf{Grp}$), you can define something called the cokernel: the cokernel of a morphism $f : X \to Y$ is the coequalizer of $f$ and the zero morphism $0 : X \to Y$. It's is generally denoted $Y / X$ when $X \to Y$ is a monomorphism, ie. when $X$ is a subobject of $Y$. This is the usual quotient you know in $\mathsf{Ab}$ or $\mathsf{Grp}$.

Then it is a general fact that the morphism $Y \to \operatorname{coker} f$ is an epimorphism. You can thus consider it as a representative of a quotient object, that you can still call $Y/X$.

If your category has equalizers, you can also construct the kernel of a morphism $f : A \to B$; it's the equalizer of $f$ and $0 : A \to B$. Again, in $\mathsf{Grp}$ and $\mathsf{Ab}$ it's the usual kernel you know. Then if $f$ is an epimorphism, ie. a representative of a quotient object of $A$, you can consider its kernel $K \to A$; it's a monomorphism, aka a representative of a subobject.

If your category is actually abelian, you ask that every monomorphism and epimorphism is normal. Then things behave like you expect: if $f: X \hookrightarrow Y$ is a monomorphism, then $\ker(Y \to Y/X) \cong X$. Similarly if $g: A \twoheadrightarrow B$ is an epimorphism, then $\operatorname{coker}(\ker g \to A) \cong B$.(*)

(*) Thanks to Jakob Werner for the correction. You actually need an abelian category at this point, and the claim isn't true in $\mathsf{Grp}$ for example: a non-normal subgroup isn't the kernel of any morphism...

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    $\begingroup$ "So if your category is pointed and has both equalizers and coequalizers, all quotient objects come from subobjects" I think this is only true, if our category is conormal - as $\mathbf {Grp} $ is. But the dual statement fails in $\mathbf {Grp} $: Not every subgroup is the kernel of a quotient. Every subgroup is an equalizer, but even this is not totally trivial. $\endgroup$ – user158047 Jan 7 '15 at 14:53
  • $\begingroup$ @JakobWerner Ah, you're right. I was about to write down the condition that holds in an abelian category, but then (wrongly) thought at the last second it wasn't needed... My bad... $\endgroup$ – Najib Idrissi Jan 7 '15 at 14:54

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