Idk if someone is interested but this night I had an eureka moment. Since years I'm trying to find the missing ingredient in the functorial approach to hyperoperations.

I found a new way to define a functor out of the subfunction operator, i.e. the one that takes a function and "de-iterate" it or, if you want, that decrease the rank by one. It is the inverse of the superfunction operator (I won't annoy you with the fine details).

The problem is that I always found some obstruction in this specific point. Tonight I had a nice idea and finally I was albe to give half of the definition... the missing half is not certain so I'm asking on MSE and if ignored on MO.

Maybe nobody can help but at least you can get a glimpse of how my work on hyperoperations looks.

https://math.stackexchange.com/questions...left-given

Regards.

UPDATE

I solved it! Subfucntion operator is functorial! That's great news. I wasn't able to see it caus I worked on it all the night so my brain was super tired but it was trivial.

What this means? We have a very general result that holds for all the groups but I can spell to you its corollary in the particular case of real and complex functions.

Take an invertible function \( f:{\mathbb C}\to{\mathbb C} \) in this case we fix our base function to be \( g(z)=S(z)=z+1 \).

The operation \( \Sigma[f](z):=f(1+f^{-1}(z)) \) is functorial!!!! To be precise be \( {\mathfrak B}({\mathbb C}) \) the group of bijective functions over the complex numbers then we have a functor

\( \Sigma: S\setminus{\mathfrak B}({\mathbb C})^{\mathbb N}\to S\setminus{\mathfrak B}({\mathbb C})^{\mathbb N} \)

It takes discrete dynamicals system over C and gives the dynamics of the subfunction. This means that subtracting one rank is functorial!!!!

I found a new way to define a functor out of the subfunction operator, i.e. the one that takes a function and "de-iterate" it or, if you want, that decrease the rank by one. It is the inverse of the superfunction operator (I won't annoy you with the fine details).

The problem is that I always found some obstruction in this specific point. Tonight I had a nice idea and finally I was albe to give half of the definition... the missing half is not certain so I'm asking on MSE and if ignored on MO.

Maybe nobody can help but at least you can get a glimpse of how my work on hyperoperations looks.

https://math.stackexchange.com/questions...left-given

Regards.

UPDATE

I solved it! Subfucntion operator is functorial! That's great news. I wasn't able to see it caus I worked on it all the night so my brain was super tired but it was trivial.

What this means? We have a very general result that holds for all the groups but I can spell to you its corollary in the particular case of real and complex functions.

Take an invertible function \( f:{\mathbb C}\to{\mathbb C} \) in this case we fix our base function to be \( g(z)=S(z)=z+1 \).

The operation \( \Sigma[f](z):=f(1+f^{-1}(z)) \) is functorial!!!! To be precise be \( {\mathfrak B}({\mathbb C}) \) the group of bijective functions over the complex numbers then we have a functor

\( \Sigma: S\setminus{\mathfrak B}({\mathbb C})^{\mathbb N}\to S\setminus{\mathfrak B}({\mathbb C})^{\mathbb N} \)

It takes discrete dynamicals system over C and gives the dynamics of the subfunction. This means that subtracting one rank is functorial!!!!

MSE MphLee

Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)