[Haskell-beginners] which Applicative instance?

[Graham Gill]

I'm following the CIS194 lectures and exercises. In particular, in one 
of the Applicative lectures [0] we're asked to implement mapA and 
sequenceA, from the name and given type signature alone. I renamed them 
mapA' and sequenceA' to avoid name clashes, and defined:

import Control.Applicative

mapA' :: Applicative f => (a -> f b) -> ([a] -> f [b])
mapA' f = sequenceA' . map f

sequenceA' :: Applicative f => [f a] -> f [a]
sequenceA' = foldr (\fa fla -> (:) <$> fa <*> fla) (pure [])

These implementations seem to be correct, though undoubtedly not the nicest.

Now, I ask for a type in ghci:

 > :t mapA' pure [1..5]
mapA' pure [1..5] :: (Enum b, Num b, Applicative f) => f [b]

And so I try the following:
 > mapA' pure [1..5] :: [[Int]]
[[1,2,3,4,5]]
 > take 5 . getZipList $ (mapA' pure [1..5] :: ZipList [Int])
[[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5]]
 > mapA' pure [1..5] :: Maybe [Int]
Just [1,2,3,4,5]

That makes sense. But I don't understand the following:

 > mapA' pure [1..5]
[1,2,3,4,5]

Which Applicative instance is being used here? The type test above says 
my expression should have type f [b], where f is an Applicative and b is 
a Num, but all I see is a result of type [b] when I compute mapA' pure 
[1..5]. What's going on?

More simply
 > :t pure
pure :: Applicative f => a -> f a
 > pure [1..5]
[1,2,3,4,5]

a must match a list of Nums, but then where did f go? Is ghci using some 
default instance?

Actually, after I add the following lines to my source file:
ggg = pure [1..5]
hhh = mapA' pure [1..5]

I get the errors:
No instance for (Applicative f0) arising from a use of ‘pure’
No instance for (Applicative f1) arising from a use of ‘mapA'’

respectively. That makes sense to me.

Graham

[0] 
http://www.seas.upenn.edu/~cis194/fall14/spring13/lectures/11-applicative2.html

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