Every monad is an applicative functor — generalizing to other categories

Question

I can readily enough define general Functor and Monad classes in Haskell:

class (Category s, Category t) => Functor s t f where
    map :: s a b -> t (f a) (f b)

class Functor s s m => Monad s m where
    pure :: s a (m a)
    join :: s (m (m a)) (m a)
    join = bind id
    bind :: s a (m b) -> s (m a) (m b)
    bind f = join . map f

I'm reading this post which explains an applicative functor is a lax (closed or monoidal) functor. It does so in terms of a (exponential or monoidal) bifunctor. I know in the Haskell category, every Monad is Applicative; how can we generalize? How should we choose the (exponential or monoidal) functor in terms of which to define Applicative? What confuses me is our Monad class seems to have no notion whatsoever of the (closed or monoidal) structure.

Edit: A commenter says it is not generally possible, so now part of my question is where it is possible.

Answer 1

What confuses me is our Monad class seems to have no notion whatsoever of the (closed or monoidal) structure.

If I understood your question correctly, that would be provided via the tensorial strength of the monad. The Monad class doesn't have it because it is intrinsic to the Hask category. More concretely, it is assumed to be:

t :: Monad m => (a, m b) -> m (a,b)
t (x, my) = my >>= \y -> return (x,y)