I'm trying to understand, precisely, what a "quotient category" is. I've looked at several different definitions and they all seem to vary so I'm having a hard time nailing this concept down. A more-or-less typical definition is given by Rotman in Introduction to Algebraic Topology. Central to actually understanding the quotient category is understanding what congruence means in this context which Rotman defines as follows:

A congruence on a category $C$ is an equivalence relation $\sim$ on the set of all morphisms of $C$ such that $f \in \mathsf{Hom}(A,B)$ and $f \sim g$ then $g \in \mathsf{Hom}(A,B)$ and for $f_1 \sim f_2$ and $g_1 \sim g_2$ if the composite $g_1 \circ f_1$ exists then $$ g_1 \circ f_1 \sim g_2 \circ f_2 $$

The first thing I want to do with this is show that it is indeed an equivalence relation but, honestly, I'm not sure exactly what the equivalence classes are. It seems to me that in order for two functions to be related they must, first of all, have the same domain and the same codomain. The other observation is that, loosely speaking, the relation must respect composition in the obvious way. Other than these conclusions, I don't really see the point of the definition. Yes, the relation partitions the morphisms in $C$ but hom-sets in $C$ are already pairwise-disjoint by the definition of category.

I'm sure there's a subtlety of the definition I'm overlooking but would appreciate a clear explanation of exactly what's going on here, specifically, why introduce a relation to partition the morphisms of $C$ when this partition already exists by definition?


It’s perfectly true that setting $f\sim g$ iff there are objects $A$ and $B$ such that $f,g \in \mathsf{Hom}(A,B)$ gives you a congruence; it’s the coarsest possible congruence, but it’s certainly not the only one: you can have a finer congruence, whose equivalence classes partition each $\mathsf{Hom}(A,B)$.

It might be helpful to think of $\sim$ in terms of its restrictions to the $\mathsf{Hom}(A,B)$. Rotman’s definition is equivalent to the following one.

Suppose that for each pair of objects $A,B$ you have an equivalence relation $\overset{A,B}\sim$ on $\mathsf{Hom}(A,B)$. Suppose further that these equivalence relations respect composition: if $f_1,f_2\in\mathsf{Hom}(A,B)$, $g_1,g_2\in\mathsf{Hom}(B,C)$, $f_1\overset{A,B}\sim f_2$, and $g_1\overset{B,C}\sim g_2$, then $g_1\circ f_1\overset{A,C}\sim g_2\circ f_2$. Then the union $\sim$ of the $\overset{A,B}\sim$ is a congruence on the category.

In this form it’s perhaps clearer that the congruence can chop up the morphisms much more finely than the partition into the $\mathsf{Hom}(A,B)$.

Given a congruence $\sim$ on a category $\mathscr{C}$, we can then form the quotient category $\mathscr{C}/\sim$ whose objects are those of $\mathscr{C}$ and whose morphisms are equivalence $\sim$-equivalence classes of morphisms of $\mathscr{C}$: $$\mathsf{Hom}_{\mathscr{C}/\sim}(A,B)=\mathsf{Hom}_\mathscr{C}(A,B)/\overset{A,B}\sim\;.$$

I expect that you’ll be working with one of the standard examples of a quotient category. $\mathsf{Top}$, the category whose objects are topological spaces and whose morphisms are continuous maps, has as a quotient $\mathsf{hTop}$, whose objects are topological spaces and whose morphisms homotopy classes of continuous functions. Without going into details, if $X$ and $Y$ are topological spaces, and $f,g\in\mathsf{Hom}(X,Y)$ are continuous maps between them, $f$ and $g$ belong to the same homotopy class if each can be continuously deformed into the other.

For another example, note that a group $\mathcal{G}$ can be thought of as a category with one object, which I’ll call $G$, whose morphisms are the elements of the group: $\mathsf{Hom}(G,G)$ is the only $\mathsf{Hom}$-set. Composition of morphisms is simply the group multiplication. You can check that a category congruence on $\mathcal{G}$ is precisely the same thing as a group congruence, and hence that the quotient categories of $\mathcal{G}$ are precisely the same as its quotient groups.

  • $\begingroup$ Thanks for the explanation $\endgroup$ – ItsNotObvious May 23 '12 at 17:08

The definition of congruence that you quote from Rotman is pretty much the standard one and for many purposes it suffices. However, with this definition, not every functor has a meaningful kernel and so the situation $\rm {Im}(\psi ) \cong C/\rm{Ker}(\psi) $, for a functor $\psi :C\to D$ (familiar from, e.g., group theory) does not hold. This motivates the definition of generalized congruences for category, remedying the situation completely. Details can be found here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.