1. Concrete Categories A X and a notation T ; _{X}X is called
the carrier
and T the _{X}structure
of (X, T) respectively. In the following we often write
_{X}X
for a concrete object (X, T).
_{X}
Y, T) in _{Y}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)
and (_{Y}Z, T) are three concrete objects in _{Z}A. If
f:
Y --> X and g: Z --> Y are two morphisms, then fg:
Z --> X is a morphism.
(b) If ( X, T) is a concrete object and _{X}t:
X --> S is a bijection from the carrier X to a set S,
there is a unique structure T on _{S}S such that
t
and its inverse t are morphisms between (^{-1}X, T)
and (_{X}S, T).
_{S}Let --> (X,
T) of objects is called _{X}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 are both morphisms. We
say two objects are ^{-1}isomorphic
if there is an isomorphism between them.
A, the identity function 1:_{X}
X --> X is a morphism from (X, T) to itself.
_{X}(b) If ( X, T) and (_{X}X, S)
are two objects in _{X}A and the identity function 1:_{X}
X --> X is a morphism from (X, T) to (_{X}X,
S) and from (_{X}X, S) to (_{X}X,
T), then _{X}T.
_{X} = S_{X}
Y such that f and f are morphisms
between (^{-1}Y, T) and (_{Y}X, T).
By (1.1.a) the composition _{X}1:_{X} = f^{-1}f
X --> X is a morphism from (X, T) to itself.
_{X}(b) Under the assumption 1:_{X} X --> X is a
morphisms among (X, T) and (_{X}X, S).
But it is also a morphism among (_{X}X, T) and (_{X}X,
T) by (a). By the uniqueness of _{X}T in
(1.1.b) we have _{X}T.
_{X} = S_{X}
A to the category
Set
of sets sending each object to its carrier, called the structural
functor on A.
Suppose
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. Similarly
we have _{X}uv = 1. Thus _{Y}u is a bijective morphism,
therefore an isomorphism by (1.2.b).
It follows from (1.5) that the free object on a set
A is represented by X
(i.e. F is equivalent to _{A}hom (X,
~)).
The coequalizer of a pair of functions, if exists, is uniquely determined up to an isomorphism.
Clearly a generic extension of
Proof Clearly (a) implies (b).
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.
Y, f is a morphism
by (b). Thus ^{-1}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. Thus _{X}g is injective,
therefore a bijection. By (c) this means that g: X --> G
is an isomorphism. Thus f = g is also a coequalizer.
^{-1}gf |