5. Algebraic Subcategories Suppose A is an algebraic category. Consider a class B of objects of A.  Definition 5.1. We say B is a full subcategory of A if any object which isomorphic to an object of B is contained in B.  Any full subcategory B of A with morphism induced from A is naturally a concrete category. Definition 5.2. Suppose X is an object of A. An object J(X) of B together with a morphism r: X --> J(X) is called a (right) associated object of X in B if it has the following universal property:  For any morphism f: X --> Z with Z in B there exists a unique morphism g: J(X) --> Z such that f = gr.  An associated object of X is uniquely determined up to isomorphism. The morphism r is called the unit of X.  Definition 5.3. B is called areflective subcategory of A if any object of A has a right associated object in B.  Proposition 5.4. Any non-trivial reflective subcategory B of an algebraic category A. has free objects. Proof. Suppose Z[S] is a free object of A on a non-empty set S. First we show that the restriction of the unit r: Z[S] --> J(Z[X]) on S is  injective, and then we show that the associated object J(Z[S]) of Z[S] in B is a free object on the set r(S). By assumption B has a non-terminal object T, which has at least a pair a, b of distinct elements. Since Z[S] is free on S, for any two distinct elements u, v of S we can find a map t: S --> T such that t(u) and t(v) are distinct. Let f: Z[S] --> T be the extension of t to Z[S]. Now f = gr, where g: J(Z[S]) --> T is a morphism (as J(Z[S]) is the associated object of Z[S]). It follows that a = gr(u) = f(u) = t(u) and b = gr(v) = f(u) = t(v) are distinct. Thus r(a) and r(t) are distinct. Therefore the restriction of r on S is an injection. If w: r(S) --> W is any map with W in B, there is a unique morphism z: Z[S] --> W extending the composite S --> r(S) --> W of the bijection S --> r(S) with w.Since J(Z[S]) is the associated object of Z[S] in B, z factors through r uniquely in a morphism J[Z[S]) --> W, which extends w: r(S) --> W. Thus J(Z[S]) is a free object on r(S).  Definition 5.5. A full subcategory B of A is called an algebraic subcategory if the following conditions are satisfied: (a) Any subobject of an object in B is in B.  (b) Any product of objects in B is in B. Proposition 5.6. Suppose B is a subcategory of A. Suppose any subobject of an object in B is in B. The following conditions are satisfied. (a) B contains any product of objects in B (i.e. B is an algebraic subcategory of A). (b) B is a reflective subcategory of A with surjective units. (c) B is an algebraic category. If A is a variety then B is a variety if it satisfies (a) and the following (d). (d) B contains any quotient of its object. Proof. (a) => (b) We show that any object X of A has an associated object in B. Consider the set {pi: X --> Zi} of quotients of X with codomains in B. Let P(X) be the product of {Zi} with the projections qi: P(X) --> Zi, and let p: X --> P(X) be the morphism induced by {pi}. By (a) P(X) is an object in B. Consider the factorization st of p, where t: X --> p(X) is the projection and s: p(X) --> P(X) is the injection. We have pi = qip =  qist. By hypothesis the subobject p(X) of P(X) is in B. We show that p(X) is the associated object of X with the unit t. Any morphism f: X --> T with T in B factors as the composite of a surjection u: X --> f(X) and an injection v(X) --> T. By hypothesis the subobject f(X) of T is in B. Thus u is isomorphic to a quotient pi of X for some i. Thus f = vu  = vpi = vqist factors through t uniquely as t: X --> p(X) is a surjection. Hence B is a reflective subcategory of A.  (b) => (c) We already know that B has free objects by (5.4).  Clearly any bijection is an isomorphism in B because this is true in A. Suppose f, g: Y --> X is a pair of morphisms in B and k: X --> K is the surjective  coequalizer of f, g, then the composite of k with the surjective unit K --> J(K) is a surjective coequalizer of f, g in B. Thus B is an algebraic category by definition. (c) => (a) is obvious. Finally any congruence of an object X in B is also a congruence of X as an object in A. If (d) holds then A is a variety implies that B is also a variety. Definition 5.7. (a) A law (resp. finitary law) in A is a pair (a, b) of elements in a free object Z[S] over a set (resp. finite set) S. We also say that (a, b) is a law overS. (b) We say an object X of A satisfies a law  (a, b) over a set S if for any morphism f: Z[S] --> X we have f(a) = f(b). Let L be a  set of laws in A. Denote by A/L the class of objects in A satisfying the laws in L. Proposition 5.8. (a) A/L is an algebraic subcategory of A. (b) If A is a variety then A/L is a variety. Proof. Clearly the following three conditions are satisfied for A/L: (i) Any closed subobject of an object in A/L is in A/L.  (ii) A/L contains any product of objects in A/L. (iii) Any quotient of an object in A/L is in A/L. The assertions then follow from (5.6). Remark 5.9. Suppose X is any object in A. Let I be the congruence generated by all the images of the laws in L under any morphism to X. Then one can show that the quotient X/I is the associated object of X in A/L.     [References][Notations][Home] 