No. It's not a -> f a. It's \* -> \*. There is no a -> f a here. You are simply ...

		Sniffnoy on Nov 5, 2019 \| parent \| context \| favorite \| on: Algebraic Structures: Things I wish someone had ex... 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.