Understanding the Construction of the Quotient Category

Solution 1:

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 $\sim$-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.