Many constructions in mathematics are characterized not by explicit formulas but by universal properties—conditions that specify the construction uniquely up to isomorphism. We’ve already seen this with quotient spaces: the quotient is characterized by the fact that any linear map that kills factors uniquely through it. Products and coproducts provide two more fundamental examples.
8.3.1 Products and Coproducts
Definition 8.3.1 (Product). Let be a category and . A product of and is an object together with morphisms and (called projections) satisfying the following universal property:
For any object and morphisms ,, there exists a unique morphism such that and .
We write and for the unique induced map.
Note. The universal property says: to give a map into the product is the same as giving a map into each factor. This is the categorical abstraction of the familiar fact that a function is determined by its components and .
The dual notion reverses all arrows.
Definition 8.3.2 (Coproduct). Let be a category and . A coproduct (or sum) of and is an object together with morphisms and (called inclusions or coprojections) satisfying:
For any object and morphisms ,, there exists a unique morphism such that and .
We write or and for the unique induced map.
Note. The universal property of the coproduct says: to give a map out of the coproduct is the same as giving a map out of each summand. This duality between products and coproducts—reversing all arrows—is a general phenomenon in category theory.
Theorem 8.3.3 (Uniqueness of products and coproducts). If a product (resp. coproduct) of and exists, it is unique up to unique isomorphism. That is, if and are both products of and , then there is a unique isomorphism with and ; and similarly for coproducts.
Proof. By the universal property of , applied to the maps and , there exists a unique with and .
By the universal property of , applied to and , there exists a unique with and .
Now satisfies and . But also satisfies these equations. By uniqueness in the universal property of (applied to ,,), we have . Similarly, . So is an isomorphism, and it is the only morphism with and .
The proof for coproducts is the same with all arrows reversed. ∎
Note. This proof pattern—using universal properties to construct inverse maps—is standard in category theory. The key is that universal properties give uniqueness, which forces the composites to be identities.
Products and coproducts need not exist in every category. When working in a new category, one of the first questions is whether these constructions exist and what form they take.
8.3.2 Products and Coproducts in Familiar Categories
Example 8.3.4 (Products and coproducts in ). In the category of sets:
The product is the Cartesian product with projections and .
The coproduct is the disjoint union with inclusions and .
For the coproduct, given and , the only possible is ,. The product case is Exercise 8.3(1).
Example 8.3.5 (Products and coproducts in ). In the category of groups:
The product is the direct product, with componentwise operations and the two projections.
The coproduct is the free product, a more complicated construction whose elements are reduced words alternating between nonidentity elements of and of . Exercise 8.3(2) treats the case , where it is the free group on two generators (Section 3.6).
In the subcategory of abelian groups, both the product and the coproduct of and are the direct sum (the same group, with different universal properties: projections for the product, inclusions for the coproduct).
Example 8.3.6 (Products and coproducts in ). In (and also in ), both the product and the coproduct of and are realized by the direct sum.
As a product: Define by and by . Given linear maps and , define by . It is linear, and it is the unique map with and , since these two conditions specify both components of .
As a coproduct: Define by and by . Given linear maps and , define by . It is linear and satisfies and . It is the only such map: any linear with these properties satisfies
Note. The coincidence of products and coproducts in (and ) reflects the additive structure of these categories: morphisms can be added, which is what the formula uses. In categories without such structure (like or ), products and coproducts are genuinely different constructions.
Exercise 8.3.7 (Products and coproducts).
Verify explicitly that the Cartesian product satisfies the universal property of products in .
Show that in , the coproduct of and is , while in , it is the free group on two generators.
Let and be finite-dimensional vector spaces over . Show that using only the universal property of as a coproduct (not by direct computation).
Show solution
(1) Let be sets and let be any set with functions and . We must show there exists a unique function with and .
Existence: Define . Then and .
Uniqueness: If satisfies and , then for any ,.
(2) In : take with and . Let be an abelian group and homomorphisms. Define by . This is a homomorphism because is abelian (the terms can be rearranged), and ,. It is unique: any homomorphism with and satisfies .
In : the coproduct must accommodate non-commuting images. A homomorphism is determined by , since , and every arises this way. Let be the free group on , with and . Given , put and . By the universal property of free groups (Section 3.6), there is a unique homomorphism with and . Then and . Conversely, any with and has and , so . Hence is a coproduct of and in . (The group is not, since and need not commute.)
(3) We use the following characterization of bases, which involves only maps out of .
Claim. Let be a vector space and a family of vectors in . Then is a basis of consisting of distinct vectors iff for every vector space and every function there is a unique linear map with for all .
Proof of the claim. If is a basis of distinct vectors, this is the familiar fact that a linear map can be prescribed arbitrarily on a basis, and is determined by these values. Conversely, assume the property. Spanning: let , and compare the quotient map (Definition 7.1.1) with the zero map ; both send every to , so by uniqueness , i.e., .Independence and distinctness: let be a vector space with a basis (for instance the space of finitely supported functions ), and let be the linear map with . If (a finite sum), applying gives , so all ; and forces , so . This proves the claim.
Now let be a basis of and a basis of , with and . Consider the family of vectors of indexed by the disjoint union : for and for . Let be any vector space and any function. By the claim for and , there are unique linear maps and with and . By the universal property of the coproduct, there is a unique linear with and . Thus and . Any other linear with these values satisfies and (by uniqueness of and ), hence . So the family has the property in the claim, and it is a basis of consisting of distinct vectors. Therefore .
Note. The argument in (3) illustrates a general principle: universal properties translate into bijections of Hom-sets. For instance, the universal property of the coproduct says that is a bijection (it is even linear): the contravariant functor of Example 8.2.6 turns the coproduct into a product.