Category Theory, Product and Coproduct

Product

The product in a category is inspired by the direct product of groups, rings, and modules. Projections map the product onto its components, and a function into the product defines, and is defined by, a function into each of the components. Let's rephrase this in terms of objects and morphisms.

If A_i is an indexed set of objects in a category, its product is an object C, along with projection morphisms p_i taking C to each A_i. Furthermore, for every object B, and collection of morphisms f_i from B into A_i, there is a unique morphism f from B to C such that p_i(f) = f_i. In other words, there is some f so that each triangle commutes, taking B to A_i directly via f_i or indirectly via f followed by p_i.

If the category is concrete, and the index is finite, the product is simply the cartesian product, with the usual component projections. You need to verify that f is a morphism in the category, but this is usually the case. For instance, if each f_i is a group homomorphism, then the direct product is a group, and f is also a group homomorphism.

If we accept the axiom of choice, the indexing set may have any cardinality. This admits infinite direct products, e.g. the product of infinitely many copies of the group Z.

The product of topological spaces is also valid, assuming we use the weak product topology. In fact, that is the default topology, so that the product of spaces becomes a product in the categorical sense.

Products over the Same Components are Equivalent

Suppose C and D are both products, with projections p_i and q_i onto the same components A_i. There is a map p_i from C onto A_i, for each i, so put this all together and there is a morphism f from C to D such that p_i = fq_i. We can run the projection p_i from C, or we can go over to D via f and then run its projection q_i. There is likewise a morphism g from D to C such that q_i = gp_i. Compose these to get p_i = fgp_i for each i. If I is the identity morphism on C, we have p_i = Ip_i. The definition of product says there is a unique morphism from C into C that makes all the triangles commute. That function has to be I, hence fg = I. Run this argument in both directions and C and D are equivalent. The product of a collection of objects is unique up to equivalence.

Permuting Components

If the component objects are permuted, does that change the product? Let C be the product of objects A_i, and let D be the product over B_i, where i ranges over the same indexing set, and the components are the same, but presented in a different order. Let S be the permutation map that describes the rearrangement. Now the projection p_i from C onto A_i is equal to q_S(i). Let r_i = q_S(i), and two products exhibit projections onto the same components A_i, and are thus equivalent. Components can be rearranged as you please, provided the rearrangement is described by a permutation function, and the new projections r_i are valid morphisms in the category.

Coproduct

The coproduct is the obverse of the product. The arrows in the commutative triangles are reversed. Let's see how this works.

Injections p_i carry each A_i into C, and for every B and family of morphisms f_i from A_i to B, we have a unique morphism f from C to B such that f_i = p_if. We can go straight from A_i over to B, via f_i, or we can inject A_i into C via p_i, and then apply f. The triangle commutes.

A proof, similar to the one given above, shows that the coproduct is unique up to equivalence.

If injections remain morphisms through a permutation function S, then the order of the component objects is not significant.

In the category of sets, the coproduct is the disjoint union. The projection p_i maps A_i into a copy of itself in C. Let f map this copy of A_i into B just as f_i maps A_i into B. This definition holds for each i, hence f is defined on all of C. And if f is defined any other way, on any point, one of the triangles will not commute; hence the morphism is unique. This makes C the coproduct.

In most concrete categories, the image of the identity element, under any morphism, is forced, hence there is no contradiction if p_i maps the identity element of each A_i to the identity element of C. In other words, the injections of the components are not strictly disjoint; they share a common element, namely the identity element of C. If you are familiar with groups and modules, then you know that these structures are called disjoint if they have nothing in common except for the identity element. This is somewhat confusing, and a new word would have been preferable, but that's not how math evolved. At any rate, C is the smallest structure that contains disjoint components A_i.

In the category of abelian groups, rings, or R modules, the coproduct of finitely many components is the same as the product. Let the components be disjoint (other than the common identity element), and C is the span of these components, which is the direct product. Each component maps into C by setting all the other components to the identity element. Verify that this injection is a morphism. Given functions f_i from the components into another object B, build f so that the triangles commute. Although f has only been defined on the components of C in isolation, The morphism rules define f across the rest of C. That is why f is unique.

The coproduct of infinitely many rings or R modules is the direct sum. When f is defined on all the components, it is implied on the sums of finitely many components, and when C is the direct sum, that is all of C.

In the category of groups, the product may not be the coproduct. consider Z and Z, with a product of Z². Can Z² be the coproduct? Let B be a group containing x and y, such that x and y do not commute. Map 1 (in the first component) to x and map 1 (in the second component) to y. This means f(1,0) = x and f(0,1) = y. However, f(1,1) forces xy = yx, which is a contradiction. Thus Z² is not the coproduct. There is a coproduct however, namely the free group on two generators. Map 1, in A₁ and A₂ respectively, to these two generators.

Internal Direct Product

Assuming a concrete category, an object is an internal direct product if it is the direct product of subobjects. Let's apply this in the category of groups.

Let a group G be generated by a set of subgroups. This is an internal direct product iff subgroups commute with each other, and are independent. By independent we mean a subgroup must not intersect the group generated by the remaining subgroups, except for the identity element. This is clear when G is known to be a direct product. Conversely, if the subgroups commute, and are independent, we can show that G, i.e. the span of these subgroups, is isomorphic to their direct product. Independence assures us that different combinations of component values will not lead to the same element in G.

If subgroups are normal and disjoint, they commute. If x is in H₁ and y is in H₂, xy/x/y is in both H₁ and H₂ by normality, and H₁ and H₂ intersect in the identity element, so x and y commute.

Showing that subgroups or submodules are independent can be a dificult task. In the category of vector spaces, or modules over a pid, we can sometimes use a dimensionality argument.

In the case of finite groups, it is enough to show that the components commute, and their orders are pairwise coprime. Any two subgroups have no intersection, else the order of the common element would divide the order of both subgroups. Combine subgroups in a direct product, and its order remains coprime to any subgroup outside this product. Thus every subgroup is disjoint from the product of the others.