Informally, the objects of Hask are Haskell types, and the morphisms from objects `A` to `B` are Haskell functions of type `A -> B`. The identity morphism for object `A` is `id :: A -> A`, and the composition of morphisms `f` and `g` is `f . g = \x -> f (g x)`.

However, subtleties arise from questions such as the following.

• When are two morphisms equal? People often like to reason using some axioms or rewrite theory, such as beta reduction and eta conversion, but these are subtle with non-termination. A more basic notion of equality comes from contextual equivalence, or observational equivalence, which requires us to pick a basic notion of observation (e.g. termination versus non-termination).
• Do we consider the entire Haskell language, or just a fragment? Possible language fragment choices are: do we focus on terminating programs? do we consider seq? do we consider errors and exceptions? what about unsafe functions? and do we only consider the kind `*`, or also other kinds? The choice of fragment that is used affects the contextual equivalence, the notions of observation, and the universal properties within the category.

One can easily make a syntactic category for a typed lambda calculus, and there are many books about this. Perhaps the first place this becomes delicate in Haskell is the fact that Haskell types are not unboxed by default. This means we have an `undefined` expression at every type. Even the function type `A -> B`, which we would like to use as the type of morphisms, is boxed.

## Is Hask even a category?

Consider:

```undef1 = undefined :: a -> b
undef2 = \_ -> undefined
```

Note that these are not the same value:

```seq undef1 () = undefined
seq undef2 () = ()
```

This might be a problem, because `undef1 . id = undef2`. In order to make Hask a category, we define two functions `f` and `g` as the same morphism if `f x = g x` for all `x`. Thus `undef1` and `undef2` are different values, but the same morphism in Hask.

A way to resolve this would be to say that a morphism is not an arbitrary expression of type `a -> b`, but rather an expression of the form `\x -> e`, or an expression `e` of type `b` with a free variable `x:a`. Indeed the latter is the usual way to build a syntactic category.

## Does Hask have categorical coproducts and products?

Actual Hask does not have sums, products, or an initial object, and `()` is not a terminal object. The Monad identities fail for almost all instances of the Monad class.

Why Hask isn't as nice as you'd thought.
Initial Object Terminal Object Sum Product Product
Type `data Empty` `data () = ()` `data Either a b = Left a | Right b` `data (a,b) = (,) { fst :: a, snd :: b}` `data P a b = P {fstP :: !a, sndP :: !b}`
Requirement There is a unique function

`u :: Empty -> r`

There is a unique function

`u :: r -> ()`

For any functions

`f :: a -> r`
`g :: b -> r`

there is a unique function `u :: Either a b -> r`

such that: `u . Left = f`
`u . Right = g`

For any functions

`f :: r -> a`
`g :: r -> b`

there is a unique function `u :: r -> (a,b)`

such that: `fst . u = f`
`snd . u = g`

For any functions

`f :: r -> a`
`g :: r -> b`

there is a unique function `u :: r -> P a b`

such that: `fstP . u = f`
`sndP . u = g`

Candidate `u1 r = case r of {}` `u1 _ = ()` `u1 (Left a) = f a`

`u1 (Right b) = g b`

`u1 r = (f r,g r)` `u1 r = P (f r) (g r)`
Example failure condition `r ~ ()` `r ~ ()` `r ~ ()`

`f _ = ()`
`g _ = ()`

`r ~ ()`

`f _ = undefined`
`g _ = undefined`

`r ~ ()`

`f _ = ()`
`g _ = undefined`

Alternative u `u2 _ = ()` `u2 _ = undefined` `u2 _ = ()` `u2 _ = undefined`
Difference `u1 undefined = undefined`

`u2 undefined = ()`

`u1 _ = ()`

`u2 _ = undefined`

`u1 undefined = undefined`

`u2 undefined = ()`

`u1 _ = (undefined,undefined)`

`u2 _ = undefined`

`f _ = ()`

`(fstP . u1) _ = undefined`

Result FAIL FAIL FAIL FAIL FAIL

Using unboxed types might alleviate some of these problems. For example, revisiting the first column, but using the UnliftedDatatypes extension, we can define `data UEmpty :: UnliftedType where`, but now if we instist that `u1` and `u2` have type `UEmpty -> ()`, then `u2 undefined = undefined = u1 undefined`.