the uniqueness of the Riemann map

The Riemann mapping theorem says a conformal map onto the disk exists, but it does not say there is only one — and in fact there are infinitely many. The natural question is how much freedom remains, and the clean answer is: exactly the freedom of the disk's own symmetries, no more and no less. Once you decide where one interior point goes and which way the map is turned there, the map is locked down completely. So 'the' Riemann map is unique after you make one sensible normalization.

Two precise statements capture this. (1) Up to symmetry: if f and g are both conformal maps of a simply connected domain D onto the unit disk, then g = phi ∘ f for some automorphism phi of the disk; the automorphisms form the three-real-parameter group of Mobius maps of the disk onto itself. So any two Riemann maps differ only by composing with a disk automorphism. (2) Normalized uniqueness: if you fix a basepoint z_0 and demand f(z_0) = 0 and f'(z_0) > 0 (a positive real derivative, i.e. no rotation at z_0), then f is unique. The proof of (2) is a one-line Schwarz-lemma argument: if f and g both satisfy the normalization, then g ∘ f^(-1) is an automorphism of the disk fixing 0 with positive derivative there, and the Schwarz lemma forces such a map to be the identity, so g = f.

Why it matters: this is what lets us speak of 'the' Riemann map and use it as a canonical coordinate system on any simply connected domain. It also explains the bookkeeping in the existence proof — the normalization f'(z_0) > 0 is exactly what removes the leftover rotational ambiguity and makes the extremal map unique. Honest caveat: the three real conditions (two for f(z_0) = 0, one for the phase of f'(z_0)) match the three real dimensions of the disk's automorphism group, which is why they pin the map down exactly — fewer conditions leave freedom, more would be inconsistent. The uniqueness is a statement about a fixed domain; it does not say different domains give 'the same' map.