Zhaohua Luo (11/17/1998) Recall that an element e of a commutative ring R is an idempotent if e^{2} = e. If e is an idempotent then 1  e is also an idempotent. Since e + (1  e) = 1, any element a of R can be written uniquely as ae + a(1  e). The ideals Re and R(1e) are two rings and R = ReR(1  e) is a direct factorization of R. Conversely, if U, V are two rings then the elements [0, 1] and [1, 0] of the direct product UV are idempotents and the direct factors U, V arise as above. Since the image of an idempotent under any ring homomorphism is idempotent, it follows that the direct product of two rings is couniversal. In this note we show that for any algebraic category, direct products are couniversal iff they are defined by "idempotents", as in the case of commutative rings. Consider an algebraic category (A, U). By a difference of an object
A
we mean a notation
a  b, where a, b are elements
of A.
Remark. (a) If there is an object in A with a unit
difference then the terminal object T of A is strict (i.e.
any morphism with
T as domain is an isomorphism).
Assumption: In the following we assume Z has a unit difference 0  1. For any object A the images of 0 and 1 in A under the unique morphism Z > A are denoted by 0_{A} and 1_{A} respectively. Remark. Denote by Z[x] the free object on a singleton x. Any element a of an object A determines a morphism f_{a}: Z[x] > A sending the free element x to a. If (a, b) are two elements of A, by the coequalizer of (a, b) we mean the coequalizer of the morphisms (f_{a}, f_{b}) (it is the quotient of A by the effective congruence generated by (a, b)). Definition. An element e of A is called an idempotent of A if the coequalizer of (0_{A}, e) and (e, 1_{A}) forms a direct product of A (note that this notion depends on the choice of (0, 1)). Suppose UV is the product of two objects U and V with the projections u: U V > U and v: UV > V. A general element of U V will be denoted by [a, b], where a is an element of U and b is an element of V. Definition. A product UV is couniversal (orcostable) if for any morphism f: UV > Z, let Z > Z_{U} and Z > Z_{V} be the pushouts of u and v along f, then the induced morphism Z > Z_{U} Z_{V} is an isomorphism. Let us consider the following axioms:
Remark. Note that I_{2 }implies that the direct factor morphism p_{1}: A > U is the coequalizer of ([0_{U}, 0_{V}], [0_{U}, 1_{V}]). This follows from the fact that p_{1}: A > U factors through the coequalizer A > U' of ([0_{U}, 0_{V}], [0_{U}, 1_{V}]), and p_{2}: A > V factors through the coequalizer A > V' of ([1_{U}, 0_{V}], [1_{U}, 1_{V}]), and A = UV = U' V'. Theorem. (a) (G_{2}) is equivalent to (I_{1})
and (I_{2}).
Proof. (a) (I_{2}) and the above remark imply that the
direct factor morphism p_{1}:
ZZ
> Z is the coequalizer of a pair consisting of an idempotent and
0_{Z
Z}. Then (I_{1}) implies that the direct product of two
products is couniversal.
Remark. The proof of the above theorem shows that I_{1}
and the following I_{2}' is equivalent to G_{2}:
