A concrete object is a pair (X, T_X) consisting of a set X and a notation T_X ; X is called the carrier and T_X the structure of (X, T_X) respectively. In the following we often write X for a concrete object (X, T_X). 

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, T_X) and (Y, T_Y) in A, a set of functions from Y to X, each is called a morphism from Y to X. These morphisms satisfy the following conditions: 
(a) Suppose (X, T_X), (Y, T_Y) and (Z, T_Z) are three concrete objects in A. If f: Y --> X and g: Z --> Y are two morphisms, then fg: Z --> X is a morphism. 
(b) If (X, T_X) is a concrete object and t: X --> S is a bijection from the carrier X to a set S, there is a unique structure T_S on S such that t and its inverse t^-1 are morphisms between (X, T_X) and (S, T_S). 

Let A be a concrete category. A concrete object in A is simply called an object. A morphism f: (Y, T_Y) --> (X, T_X) 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, T_X) is an object in A, the identity function 1_X: X --> X is a morphism from (X, T_X) to itself. 
(b) If (X, T_X) and (X, S_X) are two objects in A and the identity function 1_X: X --> X is a morphism from (X, T_X) to (X, S_X) and from (X, S_X) to (X, T_X), then T_X = S_X. 

Proof. (a) Let f: X --> Y be a bijection. Then by (1.1.b) there is a unique A -structure T_Y on Y such that f and f^-1 are morphisms between (Y, T_Y) and (X, T_X). By (1.1.a) the composition 1_X = f^-1f: X --> X is a morphism from (X, T_X) to itself. 
(b) Under the assumption 1_X: X --> X is a morphisms among (X, T_X) and (X, S_X). But it is also a morphism among (X, T_X) and (X, T_X) by (a). By the uniqueness of T_X in (1.1.b) we have T_X = S_X. 

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 F_A 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 X --> Y whose restriction on S is t. 
(b) If S is a free generating set of X then we say that X is a free object on S. 
(c) An object is called free if it has a free generating set. 
(d) We say A has free objects if the free objects on any set exists. 
(e) A concrete category is called free if any object is free. 

Proposition 1.5. Suppose X and Y are two free objects with S and T as free generating sets respectively. If t: S --> T is a bijection of sets, then t induces an isomorphism from X to Y. 

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 1_X: X --> X also sends S bijectively to itself. By the definition of a generating set this implies that vu = 1_X. Similarly we have uv = 1_Y. 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 . 

Example 1.5.1. Suppose X is an object and the free object Z[X] on the set X exists, then the identity X --> X determines a morphism d: Z[X] --> X , called the canonical morphism. 

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). 
(b) An object X is the free object on a singleton iff the structure functor F_A on A is represented by X (i.e. F_A is equivalent to hom (X, ~)). 

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: 
(a) Any pair of functions from a set to an object has a surjective coequalizer. 
(b) Any pair of parallel morphisms of objects has a surjective coequalizer. 
(c) Any surjective function s: Y --> X from an object to a set has a surjective generic extension g: X --> G (and the composition gs is a coequalizer). 

Proof Clearly (a) implies (b). 
Assume (b) is satisfied. Consider a surjective function s: Y --> X from an object Y to a set X. Let u, v: C = Y × _X Y --> Y be the projections of the equivalence relation on the set Y determined by the surjective function s. Let u', v': Z[C] --> Y be the morphisms determined by u and v. Then by (b) the pair u', v' has a surjective coequalizer q: Y --> Z. Then we have qu' = qv', which implies that qu = qv. Since s: Y --> X is the coequalizer of u, v in the category of sets, we have a function g: X --> Z such that q = gs. Since q is surjective, g is surjective. If g': X --> Z' is a function to an object Z' such that g's is a morphism, then g'su' = g'sv', so there is a function t: Z --> Z' such that g's = tq. Thus g's = tgs. Since s is surjective we have g' =tg. Thus g' factors through g. This shows that g is a surjective generic extension of f, and q = gs is a coequalizer. This proves that (b) implies (c). 
Finally we show that (c) implies (a). Suppose u, v: Y --> X is a pair of functions from a set Y to an object X. Let s: X --> Z be the coequalizer of u, v in the category of sets. Then s is a surjective function. Let g: Z --> W be the surjective generic extension of s. Then gs is a morphism and we have gsu =gsv. Suppose t: X --> W' is another morphism of objects such that tu = tv. Then t factors through s by a function r: Z --> W' (i.e. t = rs ). Now by the definition of a generic extension we have a morphism r': W --> W' such that r = r'g. Thus t = r'gs, and r' is unique as gs is surjective. This shows that gs is the coequalizer of u, v. 

Proposition 1.10. Consider the following three conditions for A: 
(a) Any surjective morphism is a coequalizer. 
(b) A function f: Y --> X of objects is a morphism if there is a surjective morphism g: Z --> Y such that fg is a morphism. 
(c) Any bijective morphism is an isomorphism. 
Then (a) implies (b) and (b) implies (c). If A has free objects and any pair of parallel morphisms has a surjective coequalizer, then (c) implies (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 f^-1f is the identity morphism of Y, f^-1 is a morphism by (b). Thus f is an isomorphism. This shows that (b) implies (c). 
Finally we show that (c) implies (a) under the assumptions that A has free object and any pair of morphisms has a surjective coequalizer. Consider a surjective morphism f: Y --> X. According to (1.9) the surjective function f has a surjective generic extension g: X --> G such that gf is a coequalizer. But f is already a morphism, thus by the definition of a generic extension there is a morphism r: G --> X such that rg = 1_X. Thus g is injective, therefore a bijection. By (c) this means that g: X --> G is an isomorphism. Thus f = g^-1gf is also a coequalizer.