map (\xs -> zip xs [1..]) list

とりあえず入力。

Prelude List> map (\xs -> zip xs [1..]) list

:1:26: Not in scope: `list'

listなんて作ってないもんな・・・

Prelude List> map (\xs -> zip xs [1..]) [1,2]

:1:27:
    No instance for (Num [a])
      arising from the literal `1' at :1:27
    Probable fix: add an instance declaration for (Num [a])
    In the list element: 1
    In the second argument of `map', namely `[1, 2]'
    In the definition of `it': it = map (\ xs -> zip xs ([1 .. ])) [1, 2]

？？？
まず、\xs -> zip xs [1..]は、引数xsをとる関数。で、zip xs [1..]を評価してその結果を返す。
zipは・・・
http://www-users.cs.york.ac.uk/~ndm/hoogle/generated/Prelude.f.zip.htm

makes a list of tuples, each tuple conteining elements of both lists occuring at the same position In: zip [1,2,3] [9,8,7]
Out: [(1,9),(2,8),(3,7)]

tupleをつくる、と。
ということはzip xs [1..]はxsと[1..]からtupleを作る。
で、mapは
http://www-users.cs.york.ac.uk/~ndm/hoogle/generated/Prelude.f.map.htm

Prelude.map :: (a -> b) -> [a] -> [b]

returns a list constructed by appling a function (the first argument) to all items in a list passed as the second argument
map f [ ] = []
map f (x:xs) = f x : map f xs
In: map abs [-1,-3,4,-12]
Out: [1,3,4,12]

型aから型bをつくる関数(a->b)が引数その１、型aのリストが引数その２で型bのリストが得られる、と・・・
この場合は、関数(a->b)がzip xs [1..]だから、リストを引数にとる。
map (\xs -> zip xs [1..]) [1,2]だと、数字の1と2が適用されるからエラーになる。

Prelude List> map (\xs -> zip xs [1..]) [ [1,2] ]
[ [(1,1),(2,2)] ]
Prelude List> map (\xs -> zip xs [1..]) [ [1,2,3],[4,5,6] ]
[ [(1,1),(2,2),(3,3)],[(4,1),(5,2),(6,3)] ]

できたー！