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.

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