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 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 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 co-universal (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:
(I₁) The image of an idempotent under any morphism is an idempotent.
(I₂) If A = UV is the product of two objects U and V then the element [0_U, 1_V] 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 ([0_U, 0_V], [0_U, 1_V]). This follows from the fact that p₁: A --> U factors through the coequalizer A --> U' of ([0_U, 0_V], [0_U, 1_V]), and p₂: A --> V factors through the coequalizer A --> V' of ([1_U, 0_V], [1_U, 1_V]), and A = UV = 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: ZZ --> Z is the coequalizer of a pair consisting of an idempotent and 0_{Z 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: ZZ --> Z is the coequalizer of (0_Z, 0₁) (and then the general case follows from (I₁)). Note first that p₁ is the coequalizer of (j, i), where i: Z Z --> ZZ is the identity morphism and j is the composite of p₁ with the unique morphism Z --> ZZ (recall that Z is an initial object). From the dual of [Categorical geometry, Prop. 1.3.9] the morphism t: Z[x] --> ZZ  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] = 0_{Z Z} and it(x) = [0, 1]. Thus [0, 1] is an idempotent of ZZ.
(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] --> ZZ  induced by the morphisms f₀,  f₁: Z[x} --> Z is an epimorphism.