TITLE: Why is it that the congruence relations usually correspond to some type of subobject?
QUESTION [13 upvotes]: From the perspective of universal algebra, quotient structures of algebraic structures are built using congruence relations. If $A$ is an algebraic structure (a set with a bunch of operations on the set) und $R$ congruence relation on a set, then the quotient $A/R$ is well-defined and it will be an algebraic structure of the same type.
Now, as it turns out, in particular algebraic categories, these congruence relations on $A$ correspond exactly to some type of subobject of $A$. For instance, the congruence relations on a ring correspond precisely to the ideals of that ring; the congruence relations on a group correspond precisely to the normal subgroups of that group; the congruence relations on a module correspond precisely to the submodules of that module.
Why is it that the congruence relations usually correspond to some type of subobject? Is this a general phenomenon that can be generalized to all algebraic structures (as studied in this generality by universal algebra)?

REPLY [8 votes]: This is generally false. The fact that it holds in groups and rings is due to the fact that in each case we have an operation with identity and inverses, namely the group operation or the ring addition respectively.
As long as we have such an operation, every congruence is indeed determined by a single class. To see this, suppose $S$ is a structure with an invertible operation $*$, $a\in S$, and $\sim,\approx$ are congruences on $S$ with $[a]_\sim=[a]_\approx$. Fix $b\in S$; we want to show $[b]_\sim\subseteq[b]_\approx$ (from which by symmetry and universal generalization we'll get $\sim=\approx$).
Suppose $b\sim c$. Then $b b'a\sim cb'a$, where $x'$ denotes the $*$-inverse of $x$. This means $a\sim cb'a$, and so since $[a]_\sim=[a]_\approx$ we get $a\approx cb'a$. Now we undo the previous step: we multiply on the right by $a'b$ to get $b\approx c$ as desired.