Turning existence into a contest
Guide 3 stated the Riemann mapping theorem and was upfront that the proof would be non-constructive: it shows a conformal map onto the disk exists without ever handing you a formula for it. The natural reaction is bafflement — how can you prove something exists without building it? The answer is one of the great moves in analysis. Instead of constructing the map directly, you set up a contest among all the candidate maps and prove the winner has to be exactly the map you want. You never write the winner down; you only prove the contest has one.
Fix the setup. Let D be a simply connected domain that is not the whole plane, and pin down a basepoint z_0 inside it. We want a holomorphic bijection from D onto the unit disk. Consider the family F of all holomorphic maps f from D into the disk that are injective (one-to-one, so they are univalent functions) and normalized by f(z_0) = 0 and f'(z_0) > 0 (a real positive number). The question is no longer 'does a map onto the disk exist?' It is now: among all the maps in F, which one stretches the most at z_0 — which one maximizes the derivative |f'(z_0)|?
Does the contest even have entrants?
Before chasing a winner, you must check the contest isn't empty — F has to contain at least one map, or the supremum of |f'(z_0)| is over nothing. This is the single place the hypotheses 'D is not the whole plane' and 'D is simply connected' do real work, and the construction is clever. Because D misses some point a, the function z - a is non-vanishing on D, and because D is simply connected you may extract a single-valued holomorphic square root g(z) = sqrt(z - a) — this is exactly the branch-of-the-logarithm machinery from the elementary-functions rung, and it is why simple connectivity matters.
That square root g is injective (if g(z_1) = g(z_2) then squaring gives z_1 = z_2), and a neat geometric fact falls out: g and -g have disjoint images, so g's image avoids a whole disk around some point -g(z_0). Composing g with an inversion and a translation that send the avoided disk's complement into the unit disk, then scaling, you manufacture an honest injective holomorphic map of D into the disk. Adjust by a disk automorphism to force f(z_0) = 0 and f'(z_0) > 0. The contest has at least one entrant — F is non-empty.
Catching the winner with a normal family
Now run the contest. Let M be the supremum of |f'(z_0)| over all f in F. Two worries hover. First, is M finite? Yes — every f in F maps into the unit disk, so it is bounded by 1, and the holomorphic Cauchy estimates turn a bound on |f| into a bound on |f'(z_0)|. The family is uniformly bounded, hence M < infinity. Second, and trickier: a supremum need not be attained. You can have a sequence of maps with |f_n'(z_0)| climbing toward M without any single map hitting it. This is exactly where the previous two guides pay off.
- Pick a sequence f_n in F with |f_n'(z_0)| -> M (a maximizing sequence — one always exists by the definition of a supremum).
- The f_n all map into the disk, so they form a uniformly bounded — hence locally bounded — family.
- By Montel's theorem, the family is a normal family: some subsequence converges locally uniformly on D to a limit f.
- By guide 1 of this rung, the locally uniform limit f is again holomorphic, and the limits of derivatives converge: f'(z_0) = lim f_n'(z_0), so |f'(z_0)| = M.
Read what just happened. The supremum is attained — the limit map f actually achieves the maximum derivative M. Montel's theorem was the net that caught a winner out of a sequence that only approached one, and guide 1's theorem guaranteed the catch was still holomorphic with the right derivative. This is the whole reason the rung built normal families before touching the mapping theorem: they are the existence engine. Without them, the contest would have a target M nobody reaches.
Two gaps Schwarz and Hurwitz close
We have a winner f with |f'(z_0)| = M and f(z_0) = 0, but we are not done — two things could still go wrong, and both must be ruled out for f to be the Riemann map. Gap one: is f even injective? A locally uniform limit of injective maps could in principle collapse two points together or degenerate to a constant. Gap two: does f map onto the whole disk, or might it miss some of it? Each gap is closed by one of the rung's earlier tools, and the arguments are gems.
Injectivity is secured by Hurwitz's theorem. The trick: pick any value w and look at the maps f_n(z) - w. None of the injective f_n takes the value w twice, so f_n(z) - w has at most one zero in any region avoiding the relevant points. Hurwitz says that as f_n -> f locally uniformly, zeros can only persist or disappear, never spontaneously appear in pairs — so the limit f(z) - w also has at most one zero. That is precisely injectivity. The one escape hatch Hurwitz leaves open is that f could be constant, but |f'(z_0)| = M > 0 (we already produced an entrant with positive derivative) rules that out. So f is injective.
Surjectivity is the slickest step of all, and it is pure Schwarz lemma reasoning — a proof by contradiction that uses the extremality. Suppose f missed some point alpha in the disk. Then, using the same square-root construction as before (now legitimate because the disk minus a point is no obstacle), you can build a new map g in F whose derivative at z_0 is strictly larger than M — you squeeze out extra stretching precisely from the hole f left behind. But M was the maximum, so no map can beat it. Contradiction. Therefore f misses nothing: it maps onto the entire disk.
Why the map is unique, and what the proof doesn't give you
The extremal map is also the only normalized one — that is the uniqueness of the Riemann map, and it is a one-line Schwarz argument. Suppose f and g both map D onto the disk with f(z_0) = g(z_0) = 0 and positive derivative at z_0. Then h = g composed with f-inverse is a holomorphic bijection of the disk to itself fixing 0, so it is a disk automorphism fixing the origin — by Schwarz, a pure rotation z -> e^(i theta) z. The positive-derivative normalization forces e^(i theta) = 1, so h is the identity and g = f. The three normalizations (where z_0 goes, and the sign of the derivative) nail down the map completely.
THE PROOF AT A GLANCE
F = { f : D -> disk holomorphic, injective, f(z_0)=0, f'(z_0)>0 }
M = sup over F of |f'(z_0)|
square-root trick ........ F is non-empty (needs: D != C, simply connected)
Cauchy estimate .......... M < infinity
Montel + guide 1 ......... M is attained by some f (existence)
Hurwitz .................. f is injective (no gluing)
Schwarz contradiction .... f is onto the whole disk (no hole)
Schwarz again ............ f is the unique such map (uniqueness)
===> f : D -> disk is a conformal bijection. QED.Be honest about what this argument is and isn't. It is an existence-and-uniqueness proof, not a recipe: nowhere did we write a formula for f, and for a generic domain D no elementary formula exists. The proof tells you the magic map is there and is unique, but computing it usually means numerics or special tricks (the next guide's Schwarz-Christoffel formula is one rare case where you can write it down, for polygons). The theorem also rigidly excludes the whole plane C — and rightly so, since by Liouville-type reasoning no bounded non-constant map can come from C, so the contest F would be empty there. Two simply connected regions of the plane (neither being all of C) can always be mapped to each other, by composing each one's Riemann map; that corollary is what makes the theorem the master classification tool it is.