Idempotents Zhaohua Luo (11/17/1998) 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 = Re R(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 U V 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 A we mean a notation a - b, where a, b are elements of A.  (a) A difference a - b is a zero if a = b. (b) A difference a - b is a unit if for any morphism f: A --> B, f(a) = f(b) implies that B is a terminal object.  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). (b) If t: A --> B is morphism and a - b is a unit (resp. zero) difference of A then so is the difference t(a) - t(b) of B. (c) It follows from (b) that if the initial object Z of A has a unit difference, then any object has a unit difference.  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 U V is the product of two objects U and V with the projections u: U V --> U and v: U V --> 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 U V is co-universal (orcostable) if for any morphism f: U V --> 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: (I1) The image of an idempotent under any morphism is an idempotent. (I2) If A = U V is the product of two objects U and V then the element [0U, 1V] of A is an idempotent of A. (G2) The product of any two objects is co-universal. 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 = U V = U' V'. Theorem. (a) (G2) is equivalent to (I1) and (I2). (b) If the conditions of (a) hold then the set of idempotents of any object is a Boolean algebra. Proof. (a) (I2) and the above remark imply that the direct factor morphism p1: Z Z --> Z is the coequalizer of a pair consisting of an idempotent and 0Z Z. Then (I1) implies that the direct product of two products is co-universal. Conversely, assume (G2). Then clearly (I1) holds. To see that (I2) holds it suffices to show that the direct factor morphism p1: Z Z --> Z is the coequalizer of (0Z, 01) (and then the general case follows from (I1)). Note first that p1 is the coequalizer of (j, i), where i: Z Z --> Z Z is the identity morphism and j is the composite of p1 with the unique morphism Z --> Z Z (recall that Z is an initial object). From the dual of [Categorical geometry, Prop. 1.3.9] the morphism t: Z[x] --> Z Z  induced by the morphisms f0,  f1: Z[x} --> Z is an epimorphism. Thus p1 is also the coequalizer of (jt, it). But jt(x) = [0, 0] = 0Z Z and it(x) = [0, 1]. Thus [0, 1] is an idempotent of Z Z. (b) follows from [Pierce Topologies]  and the fact that there is a one-to-one correspondence between the set of direct factors of an object A and the set of  idempotents of A. Remark. The proof of the above theorem shows that I1 and the following I2' is equivalent to G2: (I2') The morphism t: Z[x] --> Z Z  induced by the morphisms f0,  f1: Z[x} --> Z is an epimorphism. 