The (0,1,∞) Theorem
If L = (α,β,γ) is a list of distinct points of E^{+},
then there is a unique t in I^{+}(2) which maps L to (0,1,∞).


Proof
Suppose first that none of α, β and γ is ∞.
By Theorem I1, any element t of I^{+}(2) is a Mobius transformation.
Then t maps α to 0 if and only if the numerator has a factor (zα),
and t maps γ to ∞ if and only if the numerator has a factor (zγ).
To achieve both, t must have the form t(z) = κ(zα)/(zγ), with κ ≠ 0.
Such a t maps β to 1 if and only if 
1 = t(β) = κ(βα)/(βγ) 
i.e. 
κ = (βγ)/(βα). 
Thus, there is a unique t in this case.
Now suppose that one of the points is ∞.
We can choose δ not equal to α, β or γ (so, in particular, not ∞).
Let r be the Mobius transformation r(z) = 1/(zδ).
As r(δ) = ∞, r maps α, β and γ to noninfinte points
α', β' and γ'.
By the first part, there is a transformation s mapping (α',β',γ') to (0,1,∞).
Then t = sor maps (α,β,γ) to (0,1,∞), as required.
If u has the same effect on (α,β,γ), the uot^{1} maps
(0,1,∞) to itself.
By the first part, there is a unique such map, obviously the identity.
Thus u = t, so t is unique.

