[Haskell-beginners] fmap composition

[Ut Primum]

Hi,
as you probably know, fmap ha type
fmap :: Functor f => (a -> b) -> f a -> f b
So, you can see fmap as a unary funcrion that takes an argument of type
(a->b) and returns a result of type (f a -> f b).

For example, if you define
identity x = x
then
fmap identity :: Functor f => f b -> f b
(here a = b because identity :: p -> p )

Now, the type of (.) is;
(.) :: (b -> c) -> (a -> b) -> (a -> c)

if you write  fmap . fmap, in general (a priori) the type of the fmaps
could be
FIRSTFMAP                 fmap :: Functor f1 => (d -> e) -> f1 d -> f1 e
SECONDFMAP            fmap :: Functor f2 => (g -> h) -> f2 g -> f2 h

let's see if there must be some bonds between types d,e,g,h, due to the
type of (.)

(b -> c) = FIRSTFMAP (the first argument of (.) )
(a -> b) = SECONDFMAP  (the second argument of (.) )

so, we have (I'll not write "Functor f1" or "Functor f2", we know f1 and f2
are functors)

b = (d -> e)
c = (f1 d -> f1 e)
a = (g -> h)
b = (f2 g -> f2 h)

now, it must be b=b, so the type (d -> e) must be the same type of (f2 g ->
f2 h)
So
d = f2 g
e = f2 h

The result of  fmap . fmap, will be of type (a -> c) that is
(g -> h) -> (f1 d -> f1 e) = (g -> h) -> f1 d -> f1 e
that, remebering the conditions on d and e, is
(g -> h) -> f1 (f2 g) -> f1 (f2 h)

which is the type you found on the book
(you only need to add that f1, f2 are Functors, and the names of f and g
can of course be replaced with a and b)

Cheers,
Ut

2018-06-12 18:12 GMT+02:00 Marco Turchetto <marco.turchetto.mt at gmail.com>:

> I'm reading the "Haskell programming" book and in chapter 16 about
> functors they combine two "fmap" with the "(.)" function.
> I really don't understand how the type of "fmap . fmap" is "(Functor f2,
> Functor f1) => (a -> b) -> f1 (f2 a) -> f1 (f2 b)".
>
> The thing that really freaks me out is that I thought that "(.)" arguments
> are ONLY two unary function.
> Maybe there is some curring magic underline, there is someone that can
> explain to me how the type of "fmap . fmap" is derived from the type of
> "fmap" and "(.)"?
>
>
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