Some "Theorem" on the generalized superfunction

[MphLee]

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The Law 5 is not so a derivation from the "Cancel Law", it contains more generalization and symmetry. It suggests the f's symmetry should work for every nonzeroth iteration of f, it's more like \( f*h=f\to f^t*h=f^t \).

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The laws of both have something in common, like the anticommunity in multiplication and composition. I suppose these laws are not completely equivalent to those general laws, though there's much similarity, maybe someday we can really combine the multivalued iteration(which can be so sophiscated) with the general laws.

Regards, Leo

Sure, I guess because the laws you describe are just informal observations. I claim that, if you try to formalize them, you'll come necessarily to the general laws I'm talking about.

Now I don't have enough time to touch all the points atm. I'll limit myself to the remark on Law 5.
I'm not sure to understand what you mean by symmetry in that context. I won't be too technical, but I'd like to note that some of those sets could be empty so the proof of the cancel law as you state it is tricky. I'm sorry, but the way to go for that law is categorical.


Define. the sets \( [f,g] \) as the set of all the bijective functions s.t. \( \chi f=g\chi \)

Cancel Law: \( [g,h][f,g]=[f,h] \)

This is not correct in general as stated. It needs more fine tuning. Let's ignore the fine details.

General Law 5: \( [f,g][f,f]=[f,g] \)
and \( [g,g][f,g]=[f,g] \)

Read that as: if \( \alpha f=f\alpha \) and \( \beta g=g\beta \) then \( \beta [f,g] \) and \( [f,g]\alpha \) are subsets of \( [f,g] \).
To prove the other side of the inclusion there are more technical details to check. So even in this case the way to go is the categorical setting. If we do it then the proof follows trivially from cancel Law.

Now the key part. Notice that the set \( [f,f] \) is a group. It is called the centralizer of \( f \) and contains all the functions that commute with f. Among them there is the identity \( f^0 \) and all the integer iterations \( f^{n} \). But the set can be richer than that: it must contain all the fractional iterations, if they exist, and could be even richer and contain functions that are not iterations (think about commutative groups)!
The key point to derive Law 5 from the previous is to notice that iterations of \( f \) are in \( [f,f] \) and iterations of \( g \) are in \( [g,g] \). We conclude that

Corollary: for every n,m \( g^m [f,g] \) and \( [f,g]f^n \) are subsets of \( [f,g] \).


Note: n need not to be non-zero in this case. But in the Invariant Law it is crucial. Notice in the end that also the cancel law we are subject to even subtler technical details if we want a formal proof (probably we could only aim at an inclusion.).