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Home/ Questions/Q 4605478
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T00:26:15+00:00 2026-05-22T00:26:15+00:00

I can define a natural transformation in Haskell as: h :: [a] -> Maybe

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I can define a natural transformation in Haskell as:

h :: [a] -> Maybe a
h []    = Nothing
h (x:_) = Just x

and with a function k:

k :: Char -> Int
k = ord

the naturality condition is met due to the fact that:

h . fmap k == fmap k . h

Can the naturality condition of the List monad’s join function be demonstrated in a similar way? I’m having some trouble understanding how join, say concat in particular, is a natural transformation.

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  1. Editorial Team
    Editorial Team
    2026-05-22T00:26:15+00:00Added an answer on May 22, 2026 at 12:26 am

    Okay, let’s look at concat.

    First, here’s the implementation:

    concat :: [[a]] -> [a]
    concat = foldr (++) []
    

    This parallels the structure of your h where Maybe is replaced by [] and, more significantly, [] is replaced by–to abuse syntax for a moment–[[]].

    [[]] is a functor as well, of course, but it’s not a Functor instance in the way that the naturality condition uses it. Translating your example directly won’t work:

    concat . fmap k =/= fmap k . concat

    …because both fmaps are working on only the outermost [].

    And although [[]] is hypothetically a valid instance of Functor you can’t make it one directly, for practical reasons that are probably obvious.

    However, you can reconstruct the correct lifting as so:

    concat . (fmap . fmap) k == fmap k . concat

    …where fmap . fmap is equivalent to the implementation of fmap for a hypothetical Functor instance for [[]].

    As a related addendum, return is awkward for the opposite reason: a -> f a is a natural transformation from an elided identity functor. Using : [] the identity would be written as so:

    (:[]) . ($) k == fmap k . (:[])

    …where the completely superfluous ($) is standing in for what would be fmap over the elided identity functor.

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