[Haskell-cafe] lawless instances of Functor

[Edward Kmett]

Given fmap id = id, fmap (f . g) = fmap f . fmap g follows from the free
theorem for fmap.

This was published as an aside in a paper a long time back, but I forget
where.
-Edward Kmett

On Mon, Jan 4, 2010 at 5:14 PM, Paul Brauner <paul.brauner at loria.fr> wrote:

> Hi,
>
> I'm trying to get a deep feeling of Functors (and then pointed Functors,
> Applicative Functors, etc.). To this end, I try to find lawless
> instances of Functor that satisfy one law but not the other.
>
> I've found one instance that satisfies fmap (f.g) = fmap f . fmap g
> but not fmap id = id:
>
> data Foo a = A | B
>
> instance Functor Foo where
>  fmap f A = B
>  fmap f B = B
>
> -- violates law 1
> fmap id A = B
>
> -- respects law 2
> fmap (f . g) A = (fmap f . fmap g) A = B
> fmap (f . g) B = (fmap f . fmap g) B = B
>
> But I can't come up with an example that satifies law 1 and not law 2.
> I'm beginning to think this isn't possible but I didn't read anything
> saying so, neither do I manage to prove it.
>
> I'm sure someone knows :)
>
> Paul
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