2 ms·
No. It's not a -> f a. It's * -> *. There is no a -> f a here. You are simply mistaken in thinking that any such thing is part of the definition of a functo
by Sniffnoy 7y ago
No. It's not a -> f a. It's * -> *. There is no a -> f a here. You are simply mistaken in thinking that any such thing is part of the definition of a functor.