I am probably not phrasing this question well, so please bear with me as I try to explain what I mean.

I am working on learning category theory, as applied to programming. So far, I understand that:

  • Objects in a category are "unbroken"; you're not supposed to peer inside them to see their internal structure.
  • Any Set is a category
  • Programming language types can be thought of as sets. (Bool is the set True, False; Int is the set of all integers; etc.)
  • Thus, all of the types in a language form a category of sets.
  • Morphisms are arrows between objects.
  • If those objects are themselves categories, then the morphisms get called functors.

Taken together, that would imply that a function from Int to Bool is a functor, because it's a map from the the set category Int to the set category Bool.

However, I have also read elsewhere (in particular https://www.johndcook.com/blog/2014/05/10/haskell-category-theory/), that thinking of it that way is wrong and we really shouldn't be talking about language types being a base category with "just" morphisms. But I don't see how that fits with my previous logic.

I therefore must conclude that my previous logic is faulty, but I'm unclear how or why. What is the right way to conceptualize this? Are Sets just extra special exceptions? Or is it really just an arbitrary matter of preference for how to view the problem space? Or am I just flat out wrong somewhere?

  • $\begingroup$ I'm no expert on this but I don't believe individual sets are categories, rather there exists a canonical category of sets: en.wikipedia.org/wiki/Category_of_sets. $\endgroup$
    – josh314
    Dec 14, 2019 at 22:36

1 Answer 1


There are several ideas floating around here. You can look at a programming language through a categorical lens, and then types are objects, and they are indeed opaque. So Bool is not a set of True and False in this picture. It's an object that has two morphisms (injections) going into it, one corresponding to True, and one to False. In Haskell these are the two constructors of Bool. They take no arguments, but that is interpreted as being morphisms from the terminal object (the unit type). It all works out nicely.

Then there is the more traditional picture where types are just sets of values and morphisms are functions between sets. Sets and functions form a category Set. That picture works nicely as long as you don't talk about partial function (ones that loop forever).

Then there is the third idea that every individual set is a category in itself. It's a discrete category, meaning there are no morphisms (other than identity) between elements. In that case, functions between sets would indeed correspond to functors between discrete categories. But a functor between discrete categories is nothing but a function on objects (elements) which is trivial on morphisms, so it's not a very interesting point of view.

  • $\begingroup$ Thanks! So if I follow correctly, it's valid to use either of the first two perspectives depending on what it is you want to describe? So an string->int function is a morphism and a string->string function an endomorphism, but they're not useful to think of as a functor even though you could. But "int with null" or "list of int" are valid alternate categories, which you can convert to with a functor. Following? (Context: I'm reading your book and trying to translate the concepts for PHP devs. Because I'm that kind of crazy.) $\endgroup$
    – Crell
    Dec 15, 2019 at 17:17
  • $\begingroup$ List of int is just like string. In fact, in Haskell, String is defined as a list of Char. List of a, where a is not specified, is an (endo-) functor because it maps all objects (types) a to objects (types). It maps int to list of int, char to list of char, and so on. It's like f(x) with x being arbitrary vs. f(42). $\endgroup$ Dec 16, 2019 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.