[Haskell-cafe] Why Kleisli composition is not in the Monad signature?
Daniel Peebles
pumpkingod at gmail.com
Tue Oct 16 17:22:08 CEST 2012
Although I agree that Functor should be a superclass of Monad, the two
methods of the Monad typeclass _are_ sufficient to make any instance into a
Functor in a mechanical/automatic way. The language may not know it, but
return/bind is equivalent in power to fmap/return/join. Apart from bind
being easier to use for the things we typically do with monads in programs,
using bind is actually more "succinct" in that it doesn't require three
primitive operations.
I'm not saying bind is a better primitive than join/fmap, but
"mathematicians do it this way, therefore it's better" doesn't seem like a
particularly convincing argument either. And for a more philosophical
question, is something not a functor just because we don't have a Functor
instance for it? If we agree that the Monad class (with no Functor
superclass) does implicitly form a Functor with liftM, then I don't really
see what the problem is, apart from the inconvenience of not being able to
use fmap.
On Tue, Oct 16, 2012 at 10:37 AM, AUGER Cédric <sedrikov at gmail.com> wrote:
> Le Tue, 16 Oct 2012 09:51:29 -0400,
> Jake McArthur <jake.mcarthur at gmail.com> a écrit :
>
> > On Mon, Oct 15, 2012 at 11:29 PM, Dan Doel <dan.doel at gmail.com> wrote:
> > > I'd be down with putting join in the class, but that tends to not be
> > > terribly important for most cases, either.
> >
> > Join is not the most important, but I do think it's often easier to
> > define than bind. I often find myself implementing bind by explicitly
> > using join.
>
> join IS the most important from the categorical point of view.
> In a way it is natural to define 'bind' from 'join', but in Haskell, it
> is not always possible (see the Monad/Functor problem).
>
> As I said, from the mathematical point of view, join (often noted μ in
> category theory) is the (natural) transformation which with return (η
> that I may have erroneously written ε in some previous mail) defines a
> monad (and requires some additionnal law). As often some points it out,
> Haskellers are not very right in their definition of Monad, not because
> of the bind vs join (in fact in a Monad either of them can be defined
> from the other one), but because of the functor status of a Monad. A
> monad, should always be a functor (at least to fit its mathematical
> definition). And this problem with the functor has probably lead to the
> use of "bind" (which is polymorphic in two type variables) rather than
> "join" (which has only one type variable, and thus is simpler).
> The problem, is that when 'm' is a Haskell Monad which does not belong
> to the Functor class, we cannot define 'bind' in general from 'join'.
>
> That is in the context where you have:
>
> return:∀ a. a → (m a)
> join:∀ a. (m (m a)) → (m a)
> x:m a
> f:a → (m b)
>
> you cannot define some term of type 'm b', since you would need to use
> at the end, either 'f' (and you would require to produce a 'a' which
> would be impossible), or 'return' (and you would need to produce a 'b',
> which is impossible), or 'join' (and you would need to produce a 'm (m
> b)', and recursively for that you cannot use return which would make
> you go back to define a 'm b' term)
>
> For that, you need the 'fmap' operation of the functor.
>
> return:∀ a. a → (m a)
> join:∀ a. (m (m a)) → (m a)
> fmap:∀ a b. (a→b) → ((m a)→(m b))
> x:m a
> f:a → (m b)
>
> in this context defining a term of type 'm b' is feasible (join (fmap f
> x)), so that you can express "bind = \ x f -> join (fmap f x)".
>
> To sum up, mathematical monads are defined from 'return' and 'join' as
> a mathematical monad is always a functor (so 'fmap' is defined, and
> 'bind', which is more complex than 'join' can be defined from 'join'
> and 'fmap'). Haskell does not use a very good definition for their
> monads, as they may not be instance of the Functor class (although
> most of them can easily be defined as such), and without this 'fmap',
> 'join' and 'return' would be pretty useless, as you wouldn't be able
> to move from a type 'm a' to a type 'm b'.
>
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