Day could be a comonad

[safareli]

Not sure how useful it could be, but I think Day f g is a Comonad given f and g are Comonads.

extract :: (Comonad f, Comonad g) => Day f g a -> a
extract (Day fx gy xya) = xya (extract fx) (extract gy)

duplicate :: (Comonad f, Comonad g) => Day f g a -> Day f g (Day f g a)
duplicate (Day fx gy xya) = (Day fx gy \_ _ -> (Day fx gy xya))

This seems to be lawful.

[safareli]

http://blog.functorial.com/posts/2016-08-08-Comonad-And-Day-Convolution.html

[paf31]

I think you want _co_variant Day.

[ekmett]

Day fx gy \_ _ -> (Day fx gy xya) <- the _'s there make me look askance at this definition.

You need to duplicate f and g separately, then you can fix this definition though, at least, as Phil notes, for covariant Day convolution.

[sjoerdvisscher]

There is a comonad instance already: https://hackage.haskell.org/package/kan-extensions-5.2/docs/src/Data.Functor.Day.html#line-103