## Vitalii Guzeev

# Platonic Hask overview: construction of the category

*Aug 10, 2022*

Let’s start with a basic definition:

##### Definition

An entity \(\mathrm{C}\) is a category if all of the following holds:

- There exists class \(\operatorname{Ob}(\mathrm{C})\) of objects of \(\mathrm{C}\).
- \(\forall A, B \in \operatorname{Ob}(\mathrm{C})\) there exists class \(\operatorname{Hom}_{\mathrm{C}}(A,B)\) of morphisms between \(A\) and \(B\) in \(\mathrm{C}\).
- There exists a binary operation \(\circ\) called composition such that:
- \(\forall A, B, C \in \operatorname{Ob}(\mathrm{C}),\; f \in \operatorname{Hom}_{\mathrm{C}}(A,B),\; g \in \operatorname{Hom}_{\mathrm{C}}(B,C)\) there exists \(g \circ f \in \operatorname{Hom}_{\mathrm{C}}(A,C)\).
- \(\forall A \in \operatorname{Ob}(\mathrm{C})\) there exists morphism \(Id_A \in \operatorname{Hom}_{\mathrm{C}}(A,A)\) such that \(\forall f \in \operatorname{Hom}(A,B)\) holds \(f = f \circ Id_A = Id_A \circ f\).
- \(\circ\) is associative.

Category with class \(\operatorname{Ob}\) and all \(\operatorname{Hom}\)-classes being sets is called small.

##### Examples

A basic example of a category is \(Set\) — category with objects — sets and morphisms — functions between sets.

We want to construct a reasonable category out of Haskell types. We might want to construct a reasonable category out of abstract types. For this purpose Haskell is a good yardstick and support — it was designed with respect to category theory and we can go far by judging the existence of some constructions in our category by the existence of their GHC implementations.

**First attempt**

Let’s consider \(Hask'\) with \(\operatorname{Ob}(Hask')\) — types of Haskell and \(\operatorname{Hom}_{Hask'}(A,B)\) — all functions (closed expressions) of type $A \to B$.

It is not a category — `seq undefined () = _|_`

and `seq (undefined . id) () = ()`

, hence `undefined`

\(\neq\) `undefined . id`

. See post by Andrej Bauer and discussion. Haskell wiki knows several more examples where the bottom breaks abstractions.

Failure of this attempt makes whole categorical reasoning about Haskell limited but still useful. Unfortunately, the real world is contradictory.

**Second attempt**

Consider \(\operatorname{Ob}(Hask)\) — Haskell types without \(\bot\) with natural \(\operatorname{Hom}\)-sets — all functions between these platonic types excluding partial and nonterminating functions. To shorten notation here we overload terms of Haskell Wiki. `Wiki.Hask = Hask'; Wiki.Platonic Hask = Hask`

.

Note: I’m not using the term “Maximal total subset of Haskell” in the sense of Wikipedia article since I’m not talking about provability here. The set of functions we take seems not to have a constructive definition and is broader.

##### Claim

\(Hask\) is a category.

##### Proof

With the assumption of totality and termination of all functions equational reasoning is legal (Church-Rosser property holds — it was the exact property broken by \(\bot\)).

We only have to check the properties of the composition:

- Composition exists.
- Composition is associative:
`f . (g . h) =(1) \y -> f (\x -> g (h x) $ y) =(2) \y -> f (g (h y))`

`(f . g) . h =(1) \y -> (\x -> f (g x) $ h y) =(2) \y -> f (g (h y))`

- Identity exists.
`seq`

in a previous attempt violated identity law. Proof of the identity law by equational reasoning is trivial.

(1) in equations represents taking definition, (2) refers to \(\beta\)-reduction, in both cases, we follow the applicative order of evaluation. Equality is the \(\alpha\)-congruence relation.

Following posts will operate \(Hask\).

The category \(Hask\) is not equivalent to \(Set\): consider the type `data Foo = Foo (Foo -> Bool)`

. The map `Foo :: (Foo -> Bool) -> Foo`

is an injective map from \(2^{Foo}\) to \(Foo\). This situation is impossible in \(Set\). Here is an older Reference.

##### Remarks

It is also common to denote \(Hom_{\mathrm{C}}(A,A)\) as \(End_{\mathrm{C}}(A)\) — set of endomorphisms of \(A\).

I will frequently use the following notation:

`f : A`

for element \(f\) of \(A \in \operatorname{Ob}(Hask)\)`f : A -> B`

for \(f \in \operatorname{Hom}_{Hask}(A,B)\)