Maybe you meant to write

    (??) :: Functor f => f (a -> b) -> a -> f b
    funs ?? a = fmap ($ a) funs

from lens: https://hackage.haskell.org/package/lens-5.3.2/docs/Control-...

This is valid definition of lift along a different, less interesting but more general axis.

    lift = (??) @[]