Can I say that Monad makes it possible to see some types as isomorphic?

user_18070831 0
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Monad can pass Just [1,2], which is a different type from what the original length function takes, to >>= return . length.

Just [1,2] >>= return . length

Can I say that Monad makes it possible to see Maybe [a] as isomorphic with [a] on length using (>>=, return)? (Of course they are not really isomorphic.)

Can I choose the term "isomorphic" this situation?

isomorphic monads haskell
2 Answers

What your example ultimately illustrates is that Maybe is a functor: if you have some f :: a -> b, you can use fmap to turn it into fmap f :: Maybe a -> Maybe b in a way that preserves identities and composition. Monads are functors, with \f m -> m >>= return . f being the same as fmap f m. In your case, we have the length function being transformed by the Maybe functor.

can I choose term "isomorphic" this situation?

Not really. fmap for Maybe is not an isomorphism. An isomorphism requires there being a two-sided inverse that undoes it, which in this case would be something like:

unFmapMaybe :: (Maybe a -> Maybe b) -> (a -> b)

-- For it to be a two-sided inverse to `fmap`, we should have:
unFmapMaybe . fmap = id
fmap . unFmapMaybe = id

However, there are no (Maybe a -> Maybe b) -> (a -> b) functions, as there is no way to obtain a b result if the input Maybe a -> Maybe b function gives out a Nothing. While there are specific functors whose fmap is an isomorphism (Identity is one example), that is not the case in general.

user_27518510

[a] is isomorphic to the quotient type of Maybe [a] with Nothing and Just [] considered equivalent. Alternatively it is isomorphic to Maybe (NonEmpty a), which simply eliminates the Just [] case.
In other words, [a] can be factorized as a composition of the Maybe and NonEmpty functors.

A result of this is that you can lift any function on NonEmpty a to a function on [a]:

liftEmptyable :: (NonEmpty a -> r) -> [a] -> Maybe r
liftEmptyable _ [] = Nothing
liftEmptyable f (x:xs) = Just $ f (x:|xs)

Not sure that actually has much to do with your question though. As duplode answered, you don't really do anything but a simple functor mapping. We could at most elaborate that the monad laws ensure that the fmap really behaves as if length acted directly on the contained list:

Just [1,2] >>= return . length
  ≡ return [1,2] >>= return . length  -- def. of `Monad Maybe`
  ≡ return (length [1,2])             -- left-identity monad law
user_7459030
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