# Exercise 5.2

`map map [x] = [map x]`

Therefore:

`x :: a -> b`

because x is the first argument to map. Consider the type of map:

`map :: (a -> b) -> [a] -> [b]`

This implies:

```map x :: [a] -> [b]
[map x] :: [[a] -> [b]]```

And hence:

`map map :: [a -> b] -> [[a] -> [b]]`

â€“

`foldl foldl x [y] = ?`

This is a type error. See technical error 10.

â€“

`map foldl [x] = [foldl x]`

And as before:

```foldl :: (a -> b -> a) -> a -> [b] -> a
x :: a -> b -> a
foldl x :: a -> [b] -> a
[foldl x] :: [a -> [b] -> a]
map foldl :: [a -> b -> a] -> [a -> [b] -> a]```

### 2 responses to “Exercise 5.2”

1. Mike says:

I do not understand the first one, map map. The first argument to map is a function (a->b). I don’t see how “map” can serve as the first argument. Hope you can help. I don’t follow the leap from the second last line to the conclusion.

2. admin says:

Using different type variables to make it clearer, the type (a -> b) can be unified with the type (c -> d) -> [c] -> [d] by considering a = (c -> d) and b = [c] -> [d]. So it is perfectly OK for map to be the first argument to map.