Universal uniqueness criterion?

[bo198214]

I said that the superlogarithm should be negative infinity at the fixed points,

No, its not. The super-exponential approaches L for \( z=x+iy \), \( y\to+\infty \). Thatswhy the super-logarithm has positively unbounded imaginary part, if we approach the upper fixed point in the initial region, which one could interpret as \( +i\infty \). And it has negatively unbounded imaginary when we approach \( L^\ast \) in the initial region which we could interpret as \( -i\infty \). However strictly considered there is no \( \pm i\infty \) as value of a limit.

instead of infinity*i or = -infinity*i, because the fixed point is approached via a complex iteration, and then infinite negative iteration of exp, ie iteration of log.

If you approach the fixed point via repeated logarithm, then you dont stay inside the initial region.

I guess its about deforming one initial region into the other initial region

How do we go about doing that? It is not given whether the Abel functions are holomorphic outside of their domains.

If you go slightly to the right of \( \partial_1 D_1 \) having a curve \( \gamma \) inside \( D_1 \) then you can continue \( A_1 \) to the region between \( \partial_2 D_1 \) and \( F(\gamma) \), i.e. \( A_1(F(z)):=A_1(z)+1 \) where we use the already established values of \( A_1 \) on the region between \( \partial_1 D_1 \) and \( \gamma \).

This works if \( F \) is bijective in the considered area. For example \( F=\exp \) is bijective in the strip along the real axis \( \{z: -\pi < \Im(z) < \pi \} \), which includes the fixed points \( L \) and \( L^\ast \).