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:
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:
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.
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.
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.
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.
Proof. Clearly the following three conditions are satisfied for
A/L:
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] |