Applicative is to monad what X is to comonad

Question

Can we solve this equation for X ?

Applicative is to monad what X is to comonad

Answer 1

To me it seems that Apply class should not be a part of the picture at all.

For example the definition of apply in @Zeta's answer does not seem to be well-behaved. In particular, it always discards the context of the first argument and only uses the context of the second argument.

Intuitively, it seems that comonad is about "splitting" the context instead of combining, and so "co-applicative" should be the same.

This question seems to have better answers: Is there a concept of something like co-applicative functors sitting between comonads and functors?.