Recall that an element e of a commutative ring R is an idempotent if e2 = 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(1-e) 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 co-universal. In this note we show that for any algebraic category, direct products are co-universal 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
we mean a notation
a - b, where a, b are elements
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 0A and 1A respectively.
Remark. Denote by Z[x] the free object on a singleton x. Any element a of an object A determines a morphism fa: 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 (fa, fb) (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 (0A, e) and (e, 1A) 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 co-universal (orcostable) if for any morphism f: UV --> Z, let Z --> ZU and Z --> ZV be the pushouts of u and v along f, then the induced morphism Z --> ZU ZV is an isomorphism.
Let us consider the following axioms:
Remark. Note that I2 implies that the direct factor morphism p1: A --> U is the coequalizer of ([0U, 0V], [0U, 1V]). This follows from the fact that p1: A --> U factors through the coequalizer A --> U' of ([0U, 0V], [0U, 1V]), and p2: A --> V factors through the coequalizer A --> V' of ([1U, 0V], [1U, 1V]), and A = UV = U' V'.
Theorem. (a) (G2) is equivalent to (I1)
Proof. (a) (I2) and the above remark imply that the
direct factor morphism p1:
--> Z is the coequalizer of a pair consisting of an idempotent and
Z. Then (I1) implies that the direct product of two
products is co-universal.
Remark. The proof of the above theorem shows that I1
and the following I2' is equivalent to G2: