A concrete object is a pair (X, TX) consisting of a set X and a notation TX ; X is called the carrier and TX the structure of (X, TX) respectively. In the following we often write X for a concrete object (X, TX).
Definition 1.1. A concrete
category A is a system consisting of a class of concrete
objects, and for each pair of concrete objects (X, TX)
and (Y, TY) in A, a set of functions from
to X, each is called a morphism
from Y to X. These morphisms satisfy the following conditions:
Let A be a concrete category. A concrete object in A is simply called an object. A morphism f: (Y, TY) --> (X, TX) of objects is called surjective (resp, injective, resp. bijective) if the function f: Y --> X is so. A bijective function f: Y --> X is an isomorphism if f and its inverse f-1 are both morphisms. We say two objects are isomorphic if there is an isomorphism between them.
Proposition 1.2. (a) If (X,
TX) is an object in A, the identity function 1X:
X --> X is a morphism from (X, TX) to itself.
Proof. (a) Let f: X --> Y be a bijection. Then
by (1.1.b) there is a unique A -structure TY
on Y such that f and f-1 are morphisms
between (Y, TY) and (X, TX).
By (1.1.a) the composition 1X = f-1f:
X --> X is a morphism from (X, TX) to itself.
Remark 1.3. It follows from (1.1) and (1.2) that any concrete category A together with the morphisms forms a category in the usual sense. There is a natural faithful functor FA from A to the category Set of sets sending each object to its carrier, called the structural functor on A.
Example 1.3.1. The category Set is naturally a concrete category with the identity functor Set --> Set as the structural functor. Here for each set X there is a unique structure on it, and any function of sets is a morphism.
Example 1.3.2. A class D of objects in a concrete category A is called isomorphically closed if any object which is isomorphic to an object D is in D. Any such class with the morphisms between them is a concrete category, denoted also by D , called a concrete subcategory of D.
Suppose X is an object in A.
Definition 1.4. (a) A subset S
of X is called a free
generating set of X if for any function t:
S --> Y from S to an object Y there is a unique morphism
--> Y whose restriction on S is t.
Proof. Since by definition the generating set T is a subset of Y, t: S --> T may be viewed as a function to Y, which then induces a morphism u: X --> Y as S is a generating set of X. Similarly f-1: T --> S induces a morphism v: Y --> X. The composition vu: X --> X then sends S bijectively into itself. But the identity morphism 1X: X --> X also sends S bijectively to itself. By the definition of a generating set this implies that vu = 1X. Similarly we have uv = 1Y. Thus u is a bijective morphism, therefore an isomorphism by (1.2.b).
It follows from (1.5) that the free object on a set S is uniquely determined up to an isomorphism. We shall write Z[S] for the free object on a set S .
Remark 1.6. (a) An object 0 is
the free object on the empty set iff for any object X there is a
unique morphism from 0 to X (i.e. 0 is the initial
of the category A).
Definition 1.7. Suppose f, g: Y --> X is a pair of functions from a set Y to an object X. A coequalizer of f, g is a morphism k: X --> K of objects such that kf = kg , which has the universal property that, if m: X --> M is another morphism of objects with mf = mg, then there is a unique morphism n: K --> M such that m = nk.
The coequalizer of a pair of functions, if exists, is uniquely determined up to an isomorphism.
Definition 1.8. Suppose s: Y --> X is a function from an object Y to a set X. A function g: X --> G from X to an object G is called a generic extension of s if gs is a morphism of objects, and any function h: X --> H from the set X to an object H such that hs is a morphism factors through g uniquely.
Clearly a generic extension of s: Y --> X , if exists, is uniquely determined up to an isomorphism.
Proposition 1.9. Suppose A has
free objects. Then the following conditions are equivalent:
Proof Clearly (a) implies (b).
Proposition 1.10. Consider the following
three conditions for A:
Proof. First assume any surjective morphism in A is a
coequalizer. Consider a function f: Y --> X of objects. Suppose
there is a surjective morphism g: Z --> Y such that fg
is a morphism. Then by assumption the surjective morphism g is the
coequalizer of a pair of parallel morphisms (u, v), for which gu
= gv, and therefore (fg)u = (fg)v. By the
definition of a coequalizer there is a morphism d: Y --> X
such that fg = dg. Since g is surjective we have f = d,
thus d is a morphism. This shows that (a) implies (b). Next assume
(b) holds. Consider a bijective morphism f: Y --> X. Since
the identity morphism of Y, f-1 is a morphism
by (b). Thus f is an isomorphism. This shows that (b) implies (c).