It's not entirely clear what you want to happen here with `f` and `g`. It's trivially easy to apply `g` to the `b` in the result of f. You just use `fmap`.

It's not entirely clear what you want to happen here with `f` and `g`. It's trivially easy to apply `g` to the `b` in the result of f. You just use `fmap`. -- suppose we have some x :: a fmap g (f x) :: IO [c] Of course, you wouldn't want the [c] to be able to escape the IO container, because that would break referential transparency - you could no longer guarantee that a function returns the same result given the same arguments. Of course, that's no problem for manipulating the [c] value by passing it to pure code. It's true that "monad's don't compose" in the sense that you need to manually write monad transformers to correctly combine monadic effects, whereas functors compose automatically. However, that's not the same thing as saying it's difficult to use functions which happen to return different monadic results together, if the types match.