How are monoid and applicative connected?

Question

I am reading in the haskellbook about applicative and trying to understand it.

In the book, the author mentioned:

So, with Applicative, we have

Answer 1

Although this question got a great answer long ago, I would like to add a bit.

Take a look at the following class:

class Functor f => Monoidal f where
  unit :: f ()
  (**) :: f a -> f b -> f (a, b)

Before explaining why we need some Monoidal class for a question about Applicatives, let us first take a look at its laws, abiding by which gives us a monoid:

• f a (x) is isomorphic to f ((), a) (unit ** x), which gives us the left identity. (** unit) :: f a -> f ((), a), fmap snd :: f ((), a) -> f a.
• f a (x) is also isomorphic f (a, ()) (x ** unit), which gives us the right identity. (unit **) :: f a -> f (a, ()), fmap fst :: f (a, ()) -> f a.
• f ((a, b), c) ((x ** y) ** z) is isomorphic to f (a, (b, c)) (x ** (y ** z)), which gives us the associativity. fmap assoc :: f ((a, b), c) -> f (a, (b, c)), fmap assoc' :: f (a, (b, c)) -> f ((a, b), c).

As you might have guessed, one can write down Applicative's methods with Monoidal's and the other way around:

unit   = pure ()
f ** g = (,) <$> f <*> g = liftA2 (,) f g

pure x  = const x <$> unit
f <*> g = uncurry id <$> (f ** g)
liftA2 f x y = uncurry f <$> (x ** y)

Moreover, one can prove that Monoidal and Applicative laws are telling us the same thing. I asked a question about this a while ago.