4. Congruences Let A be an algebraic category.  Let X be an object. Denote by X2 the product X X. If U is a subset of X2 we denote by p, q: U --> X the maps induced by the projections X2 --> X.  Remark 4.1. Recall that an equavlence relation on a set X is a subset U of X2 satisfying the following conditions: (1) (a, a) U for any a in X.  (2) If (a, b) U then (b, a) U.  (3) If (a, b), (b, c) U, then (a, c) U.  (a) If U is an equavelence relation on X, we denote by X/U the quotient set of X under U, and q: X --> X/U the canonical surjective map. Then U = X X/U X.  (b) A subset U of X2 is an equivalence relation iff U = X T X for a map t: X --> T (we may assume t is surjective).  (c) Any intersection of equivalence relations is an equivalence.  Definition 4.2. (a) An equavelence relation on X is called a congruence if it is a closed subset of X2.  (b) A subset U of X2 is called an effective congruence on X if there is a (surjective) morphism t: X --> T such that U = X T X.  Proposition 4.3. An effective congruence is a congruence.  Proof. If t: X --> T is a morphism then X T X is a closed subset of X2 by (3.6), which is also an equavelence relation on X by (4.1.c).  Example 4.3.1. If X is a set in Set, a subset U of X2 is a congruence iff it is an equivalence relation on X .  Proposition 4.4. An equivalence relation u, v: U --> X is an effective congruence on X iff there is a structure on the quotient X/U such that q: X --> X/U is a morphism.  Proof. Since U = X X/U X by (4.1.a), if q is a morphism then U is an effective congruence. Conversely, assume U = X T X for a morphism t: X --> T, then X/U = t(X) by (4.1) and X --> t(X) is a morphism by (2.5).  Proposition 4.5. Any intersection of congruences is a congruence.  Proof. This follows from the fact that the classes of equavelnce relations and closed subsets are closed under intersections.  Proposition 4.6. Any intersection of effective congruences is an effective congruence.  Proof. For any equivalence relation u, v: U -->X let q: X --> X/U be the quotient map. Let g: X/U --> G be the generic extensiion of q. Then gq: X --> G is a morphism. Let C(U) = X G X. Then C(U) is an effective congruence. We show that C(U) is the smallest effective congruence containing U. Suppose V is another effective congruence on X containing U. The quotient map p: X --> X/V factors through q: X --> X/U by a map t: X/U --> X/V. Then by the property of the generic extention g there is a unique morphism s: G --> X/V such that t = sg. Thus p = tq =sgq, which shows that V contains C(U).  Next consider the intersection U of a set {Ui} of effective congruences on X. We have C(U) C(Ui) = Ui for any Ui. It follows that C(U) is contained in the intersection of {Ui}, which is U. This shows that U is an effective congruence.  Definition 4.7. A variety is an algebraic category such that any congruence is effective.      [References][Notations][Home] 