fmap . fmap :: (c->d) -> foo(bar c) -> foo(bar d) http://stackoverflow.com/questions/15029843/how-can-i-understand/15030970#15030970

gistfile1.txt

fmap :: (a->b) -> foo a -> foo b             (a->b) -> (p->a) -> (p->b)
fmap :: (c->d) -> bar c -> bar d             (c->d) -> (q->c) -> (q->d)
(.)  :: (t->u) -> (s->t) -> (s->u)

(.) fmap :: -- t ~ (a->b)  ;  u ~ (foo a->foo b)
            (s->a->b) -> (s->foo a->foo b)             (s->a->b) -> (s->(p->a)->(p->b))

(.) fmap fmap :: s ~ (c->d)  ;  a ~ (bar c)  b ~ (bar d)
             ( (c->d) -> foo(bar c) -> foo(bar d) ) 

LET foo :: ((->) p)  ;  bar :: ((->) q)  THEN

(fmap . fmap) :: (c->d) -> (p->q->c) -> (p->q->d)
(fmap . fmap) f g p q = f (g p q)