A discrete category is one whose only morphisms are identity morphisms.

The category of discrete categories would have as objects the discrete categories and as morphisms the functors between those discrete categories.

A functor from one discrete category A to another one B can map any object from A to any object from B, since there are no arrows in either category to restrict a mapping.

But that is exactly the same when mapping objects between sets.


1 Answer 1


Almost. The category of small discrete categories (i.e. discrete categories whose class of objects is a set) would be equivalent (actually isomorphic) to the class of sets. The functor in question would be the forgetful one (send each category to its underlying set of objects, "forgetting" that the morphisms are there).

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    $\begingroup$ When using the "objects and arrows" definition of a category, the functor you mention is not an isomorphism; to each underlying set of objects, there are many different choices of one-element sets to use for the various $\hom(x,x)$. I suppose your functor would give an isomorphism for the "arrows only" definition of a category. (assuming composition is defined in terms of a specific product functor on hom-sets) $\endgroup$
    – user14972
    Oct 14, 2017 at 21:06
  • $\begingroup$ Why is the distinction between one-sorted ("arrows only definition") and two-sorted ("objects and arrows" definition) definitions of category supposed to be relevant here? The fact that there are many different choices of one-element sets would seem to merely imply that there are many different choices of isomorphism. That seems irrelevant to the answer though, which only asserts existence of an isomorphism, not existence of a unique isomorphism. It seems I am misunderstanding the point being made. Is this something about anafunctors/not assuming axiom of choice? $\endgroup$ Apr 6, 2022 at 16:36

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