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In Haskell, there is a type class called Extend.

The class is defined as the following

class Functor w => Extend w where
    extended :: (w a -> b) -> w a -> w b

Every instance of the Extend class should have the following properties:

extended f . extended g = extended (f . extended g)

I can see its similarities to Functor. Particularly, Functor's property fmap f . fmap g == fmap (f . g) looks similar to Extend.

How would you interpret Extend? What is the significance of it? Does it make any computations easier? What abstractions are made when using Extend?

melpomene
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mtber75
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    That looks like half of a [comonad](https://hackage.haskell.org/package/comonad-5.0.2/docs/Control-Comonad.html). – melpomene Jan 18 '18 at 02:38

1 Answers1

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Extend is a Comonad without the ability to extract. It's an "almost comonad", if you want to think of it like that. It's probably more helpful to ask the question "what is the meaning of comonads". Then, when you find something that's almost a comonad, you know you can use Extend to represent it. I recommend Neighborhood of Infinity for an introduction to comonads by example.

We have a similar thing for Monad and Applicative, by the way. Bind is Monad but without return, and Apply is Applicative but without pure. You can find both of these classes in the same semigroupoids package you linked.

As an example, nonempty lists form a comonad, with duplicate = tails and extract = head. Then extend f = fmap f . duplicate. This is fine if we have NonEmpty, but if the list might be empty, extract = head is no longer a total function. We still have duplicate and extend, so [] can be Extend but it can't be Comonad. (Thanks @phadej for this example!)

Silvio Mayolo
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    You should mention `[]` as an obvious example of something which is `Extend` and not `Comonad`. In fact, the `duplicate` for list is very easy to understand! – phadej Jan 18 '18 at 22:05