Exercise 7 (Chapter 2, “Categories”):
In the category of sets, the two morphisms ( and ) in a direct product are monomorphisms and the two morphisms in a direct sum are epimorphisms. Is this true in every category?
There are two mistakes in this wording.
- The author has exchanged “monomorphisms” and “epimorphisms”. In a direct product of sets, the morphisms are generally not monomorphisms (injective), and in a direct sum of sets, they are generally not epimorphisms (surjective).
- In a direct product of sets, the morphisms are almost always epimorphisms. If one of the sets is empty, the product is empty, and the morphism associated with the other set (an empty mapping) is not an epimorphism; unless this other set is empty too.
The general theorem I will prove is this:
Let and be objects of a category , and a -direct product of and . If is not empty then is an epimorphism.
We consider the above conditions satisfied. There exists a morphism .
Let be the identity morphism and the morphism . Applying the universal property of the direct product to the triplet , we obtain that there exists a unique such that both and .
The first of these two relations says that is the identity on . If we have an object and two morphisms and from to such that , then, composing this relation on the right with and eliminating the resulting identity morphisms, we get . Hence is an epimorphism.
We obtain, of course, a similar result exchanging the roles of and .
Reversing the arrows, we get the corresponding result for direct sums:
Let and be objects of a category , and a -direct sum of and . If is not empty then is a monomorphism.
Plus, of course, the similar result obtained by exchanging and .
It so happens that in the category of sets, in the case of a direct product, both and are always monomorphisms, even if , respectively , are empty.