Definition 2.1. An algebraic
category is a concrete category satisfied the following conditions:
Combining (1.9) and (1.10) we have the following variants of (2.1): Proposition 2.2. A concrete category is
an algebraic category iff the following conditions are satisfied:
Proposition 2.3. A concrete category is
an algebraic category iff the following conditions are satisfied:
Proposition 2.4. (a) Any algebraic category
A has a terminate object.
Proof. (a) First we note that A has an object whose carrier
is a singleton. Let Z[Q] be the free object on a singleton
Q. Let Z[Q] > Q be the surjective function.
By (1.9) Z[Q] > Q
has a surjective generic extension Q > P. Then P is an
object whose carrier is a singleton.
Suppose A is an algebraic category. Proposition 2.5. Any morphism f: Y > X of objects factors uniquely as a surjective morphism followed by an injective morphism. Proof. Let f(Y) be the image of the set Y under the function f. Let q: Y > f(Y) be the surjective function and e: f(Y) > X the inclusion. Let t: f(Y) > B be the surjective generic extension of q (see (2.3.c)). As f = eq is a morphism, by the definition of a generic extension there is a unique morphism h: B > X such that e = ht. This implies that the function t is injective, therefore it is bijective. Since B is only defined up to isomorphism, and f(Y) carries a A structure such that t is an isomorphism of objects by (1.2), we may assume B = f(Y) . This shows that f factors as a surjective morphism q: Y > f(Y) followed by an injective morphism f(Y) > X of objects. Definition 2.6. (a) By a subobject
of X we mean an object (V, T_{V}) such that V
is a subset of X, and the inclusion V > X is a morphism
of objects, called the inclusion of a subobject;
T_{V} is a subobjectic structure
on V.
Proposition 2.7. A subset of X is closed iff it is the image of a morphism to X . Proof. This follows from (2.5). Proposition 2.8. (a) Suppose (U, T_{U})
and (V, T_{V}) are two subobjects of an object X
and V U. Then the inclusion
V > U is a morphism.
Proof. (a) Consider the morphism d: Z[V]
> X, s: Z[V] > (V, T_{V})
and t: Z[V] > (U, T_{U}) determined
by the inclusions V > X, the identity V > V, and the
inclusion V > U of sets respectively. Then d = ee's = e't
where e: U > X and e': V > U are the inclusions
of sets. Since e is injective, we have e's = t. Since s
is surjective and t is a morphism, by (2.3.b)
the inclusion e' is a morphism.
Remark 2.9. It follows from (2.8.b) that a subobject is uniquely determined by its carrier. This justifies the common practice in algebra that the carrier of a subobject is simply called a subobject (i.e. subalgebra). If U is a subset of an object X, let g: Z[U] > X be the morphism determined by the inclusion U > X. We shall write G(U) for the image g(Z[U]), which is a subset of X. Proposition 2.10. (a) A subset U
of an object X is closed iff U = G(U).
Proof. (a) If U = G(U) then U is the image
of the morphism g: Z[U] > X, so it is closed
by (2.7). Conversely, assume U = f(Y) where f:
Y > X is a morphism of objects. Let d: Z[f(Y)]
> X be the morphism determined by the inclusion d': f(Y)
> X. Let s: f(Y) > Y be a section of
f': Y > f(Y). Let s': Z[f(Y)]
> Y be the morophism determined by s. Then d = fs'.
Thus d(Z[f(Y)]) = f(s'(Z[f(Y)]))
f(Y). Thus G(f(Y)) = f(Y).
Definition 2.11. (a) G(U)
is called the closed subset generated by
U.
Example 2.11.1. If Z[S] is a free object on a set S, then S is a set of generators of Z[S]. Proposition 2.12. (a) Suppose f:
Y > X and g: Y > Z are two surjective morphisms.
If there is a bijection h: X > Z such that g = hf,
then h is an isomorphism.
Proof (a) Note that g and G^{1} are morphisms
by (2.3.b).
Example 2.11.1. We verify that the
concrete category Set of sets is a free algebraic category.
