`bind`等同于加入（fmap fm）吗？

Question

This excellent answer in this question demonstrates how bind can be written in terms of join and fmap :

(>>=) :: mv -> (v -> mw) -> mw

says "if you have a strategy to produce av, and for each va follow-on strategy to produce aw, then you have a strategy to produce aw". How can we capture that in terms of join?

mv >>= v2mw = join (fmap v2mw mv)

But, I don't understand how v2mw , which has a type of a -> mb type checks to the first argument of fmap .

fmap :: Functor f => (a -> b) -> fa -> fb

Answer 1

Let's say v2mw :: c -> md , just so things aren't ambiguous, and

fmap :: Functor f => (a -> b) -> f a -> f b

Then fmap v2mw works out so that f ~ m , a ~ c and b ~ md , so

fmap v2mw :: m c -> m (m d)

and join :: m (me) -> me , so join (fmap v2mw mv) has type md as expected.