This is a question that has been bothering me for quite a while. Let me put between quotation marks the terms that are used informally.

"Quotient objects" are always the same. Take groups, abelian groups, rings, topological vector spaces for example. Inside every object there are certain "subobjects" that we can divide by: normal subgroups, subgroups, ideals and subspaces with the subspace topology, respectively.

I (think I) know from an universal algebra viewpoint (of which I know nothing) that in the case of "algebraic" objects, the "subobjects" we take the quotient by are determined by congruences: equivalence relations on the cartesian product that respect the operation(s).

The first problem is:

1) How to define categorically the "subobjects" by which we take quotients?

Now, the resulting "quotient object" $G/N$ always satisfies the same universal property: it comes with a morphism $\pi$ such that $\pi:G\to G/N$ and $N\subset ker \,\pi$, which is universal with respect to this: any other such morphism factors through $\pi$.

The second (intimately linked) problem is:

2) How to define categorically these "quotient objects" and express in fancy terms the universal property they satisfy?

Now, in "algebraic" categories there are well-known isomorphism theorems. In the Wikipedia link, it is explained that they are all a special case of the universal algebra statement.

3) How can we express them in categorical terms?

Any insightful answer, even if partial, will be very welcome.

  • 3
    $\begingroup$ Just a technical quibble: topological quotients are by equivalence relations, not subspaces. Even if your equivalence relation is defined to identify all points in a given subspace, the subspace topology isn't really involved. This may or may not have implications for whether topological quotients fit into the general scheme you're asking for. $\endgroup$
    – Noah Stein
    Commented Sep 1, 2011 at 0:14
  • 2
    $\begingroup$ @Bruno: There is already a notion of quotient object (it is the dual of a subobject), but the ones you are referring to seem to be coequalisers of some kind. Is this the kind of answer you want? $\endgroup$
    – Zhen Lin
    Commented Sep 1, 2011 at 15:04
  • 1
    $\begingroup$ @Zhen: The problem is mainly that in non-"algebraic" categories there are at least 5 notions of epimorphisms see e.g. Adámek-Herrlich-Strecker page 121 that deserve to be called such (the cancellation property of epimorphisms is far too weak for many purposes). Somewhere in SGA 4/1 there are even more such notions. Calling the codomain of an epimorphism a quotient is most of the time not what you want (as witnessed by this question). As you suspect, Bruno seems to be after regular epimorphisms (those arising as coequalizers). $\endgroup$
    – t.b.
    Commented Sep 1, 2011 at 15:39
  • 1
    $\begingroup$ I should have thought of it earlier, but since I never really read that book, I forgot about it. Unification of isomorphism theorems and similar algebraic lemmas is one of the main aims of Borceux-Bourn, Mal'cev, protomodular, homological and semi-abelian categories, maybe it contains some of the results you're looking for. See also the n-lab entry on that book. $\endgroup$
    – t.b.
    Commented Sep 1, 2011 at 15:51
  • 2
    $\begingroup$ Let me also point to this answer, which will give you some of the flavor of congruences, and recommend, yet again, George Bergman's excellent Invitation to General Algebra and Universal Constructions which will give you the universal algebra in a categorical context. $\endgroup$ Commented Sep 1, 2011 at 16:15

1 Answer 1


First of all, let me say that there already well-defined notions of subobjects and quotient objects in category theory. I'll state them here for reference:

Definition. A subobject of an object $A$ in a category $\mathbf{C}$ is an equivalence class of monomorphisms with codomain $A$, where we identify two monomorphisms $m : B \to A$, $m' : B' \to A$ as equivalent just if there is an isomorphism $f : B \to B'$ such that $m = m' \circ f$.

Definition. A quotient object of an object $A$ in a category $\mathbf{C}$ is a subobject of $A$ in the opposite category $\mathbf{C}^{\textrm{op}}$. Explicitly, it is an equivalence class of epimorphisms with domain $A$, where we identify two epimorphisms $e : A \to B$ and $e' : A \to B'$ as equivalent just if there is an isomorphism $f : B' \to B$ such that $e = f \circ e'$.

But we shall see by means of examples that these aren't necessarily what we want them to be.

Example. In the category of topological spaces, subspaces are subobjects. But so are other things. Indeed, if $B$ has a finer topology than $A$, then $B$ is also a subobject of $A$.

Example. In the category of monoids, $\mathbb{Z}$ is a quotient of $\mathbb{N}$, because the natural inclusion map $\mathbb{N} \hookrightarrow \mathbb{Z}$ is epic.

The problem is that the notions of monomorphism and epimorphism are a bit too general to capture the properties we want. So we should begin by defining some ‘stronger’ notions of mono/epi.

Definition. A regular monomorphism is an monomorphism which is an equaliser of some pair of parallel morphisms. A regular epimorphism is a coequaliser of some pair of parallel morphisms.

Example. The kernel of a group/ring/module/etc. homomorphism is a regular monomorphism: indeed, $\ker f$ is the equaliser of $f$ and the zero morphism.

Example. Conversely, in an abelian category, every regular monomorphism is a kernel: indeed, the equaliser of $f, g : A \to B$ is precisely the same thing as $\ker (f-g)$, by the above observation.

Example. In the category of topological spaces, the inclusion of a subspace $B \hookrightarrow A$ is a regular monomorphism: it is the equaliser of its characteristic map $A \to 2$ (where $2$ is given the indiscrete topology) and a constant map. Conversely, every regular monomorphism is (isomorphic to) the inclusion of a subspace.

This suggests we're on the right track. Let's look now at the epimorphisms.

Example. The coimage of a group/ring/module/etc. homomorphism is a regular epimorphism: indeed, $\operatorname{coim} f$ is the coequaliser of $\ker f$ and the zero morphism. In particular, the projection maps onto quotient groups/rings/modules/etc. are regular epimorphisms. The converse is true in abelian categories by an argument similar to the one before.

Example. Let $A / R$ be the quotient of a topological space $A$ by an equivalence relation $R$. Then the projection map $\pi : A \to A / R$ is a regular epimorphism. Recall that $R$ is a subset of the cartesian product $A \times A$; so let us take $A \times A$ with the induced product topology and topologise $R$ with subspace topology. Then we have reified $R$ as an object in $\textbf{Top}$. Let $p_1, p_2 : R \to A$ be the projections. It is not hard to verify that $\pi : A \to A / R$ is the coequaliser of $p_1$ and $p_2$, so it is indeed a regular epimorphism.

Unfortunately, it turns out that in the category of groups, monomorphisms and regular monomorphisms are the same thing. (For a proof, see here.) So perhaps regular monomorphisms aren't a perfect fit for what we're looking for. Nonetheless, in sufficiently nice categories, we do have the following analogue of the first isomorphism theorem:

Theorem. In any regular category, (the domain of) the image of a morphism is isomorphic (the codomain of) its regular coimage (the coequaliser of its kernel pair).

Examples. Any abelian category is regular, as is any semiabelian category and any topos.

So I think this is probably the right level of generality, even though it doesn't cover the category of groups or the category of topological spaces. (In fact, the conclusion of the above theorem is blatantly false in the category of topological spaces: intuitively, the regular coimage is topologised by the domain, while the image is topologised by the codomain; so, for example, the map from a discrete space to an indiscrete space will have non-isomorphic image and regular coimage.)

  • 1
    $\begingroup$ As you say, this notion of quotients is suitable in essentially algebraic contexts. In most of the categories I encounter the notion of quotient objects is more or less of no interest. As I pointed out in a comment above there has been considerable effort of trying to unify all the similar lemmas. I basically agree with you that what you describe is the right generality because everything else tends to become unreadable. Anyway, a very nice answer! $\endgroup$
    – t.b.
    Commented Sep 1, 2011 at 15:56
  • $\begingroup$ Thank you for your answer. What about considering a category with zero morphisms and normal monomorphisms? This would cover at least the category of groups. At any rate I'm still not fully satisfied, as per your comment on your last paragraph... maybe category theory is not the right language to express what is at hand? $\endgroup$ Commented Sep 3, 2011 at 13:58
  • 1
    $\begingroup$ @Bruno: Ah, but it wouldn't say anything about the category of sets or the category of topological spaces, which don't have zero morphisms. I had thought about saying something about normal monomorphisms but in the end it seemed like cheating. $\endgroup$
    – Zhen Lin
    Commented Sep 3, 2011 at 15:41
  • 1
    $\begingroup$ The final theorem you state: can it be stated that the coequalizer of the kernel pair is isomorphic to the equalizer of the cokernel pair? $\endgroup$
    – Eric Auld
    Commented Nov 17, 2014 at 2:50

You must log in to answer this question.

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