# [Haskell-cafe] determine a type by universal property

Joachim Breitner mail at joachim-breitner.de
Thu Jan 17 14:46:01 UTC 2019

```Hi,

Am Donnerstag, den 17.01.2019, 14:33 +0100 schrieb Olaf Klinke:
> I have two type signatures where I conjecture the only functors satisfying these
> are the Identity functor and functors of the form (,)s. Can anyone give hints as
> how to tackle a proof?
>
> Signature 1: What functors t admit a function
> forall f. Functor f => t (f a) -> f (t a)

data Void1 a

It seems I can write

g :: forall f. Functor f => t (f a) -> f (t a)
g x = case x of {}

but Void1 is not the identity. (I guess it is `(,) Void`, if you want…)

So if you allow the latter, let’s try a proof. Assume we have t, and

g :: forall f. Functor f => t (f a) -> f (t a)

We want to prove that there is an isomorphism from t to ((,) s) for
some type s. Define

s = t ()

(because what else could it be.) Now we need functions

f1 :: forall a. t a -> (t (), a)
f2 :: forall a. (t (), a) -> t a

that are isomorphisms. Here is one implementation:

f1 :: forall a. t a -> (t (), a)
f1 = swap  . g . fmap (λx → (x,()))

{- note:
x1 :: t a
x2 :: t ((,) a ())
x2 = fmap (λx → (x,())) x1
x3 :: (,) a (t ())
x3 = g x2
x4 :: (t (), a)
x4 = swap x3
-}

f2 :: forall a. (t (), a) -> t a
f2 (s, x) = x <\$ s

Now, are these isomorphisms? At this point I am not sure how to
proceed… probably invoking the free theorem of g? But even that will
not work nicely. Consider

type T = (,) Integer
g :: forall f. Functor => T (f a) -> f (T a)
g (c, fx) = (,) (c + 1) <\$> fx

here we count the number of invocations of g. Surely with this g,
f2 . f1 cannot be the identity.

Cheers,
Joachim
--
Joachim Breitner
mail at joachim-breitner.de
http://www.joachim-breitner.de/

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