Try
The straight path γ₀(t)=1+t and bent path γ₁(t)=1+t+i·sin(πt)/20 both join 1 to 2. The formula H(s,t)=1+t+i·s·sin(πt)/20 continuously changes the first into the second as s increases from zero to one. Every intermediate path stays in the right half-plane, where the initial logarithm can continue. Both endpoints remain fixed.
Draw this deformation, then distinguish its two parameters: t travels along a path; s chooses which path. Our aim is to prove that the final germ at 2 stays unchanged, not just to sample a few endpoint values.
Predict
Could the same reasoning contract a full circuit around a missing point? Starting with a cube-root germ of z−2 at z=3, one circuit around 2 changes its value from 1 to exp(2πi/3). If a fixed-endpoint contraction stayed in the punctured domain, the theorem we are about to prove would make those terminal germs equal. Locate which geometric hypothesis must fail before looking at the calculation below.
Experiment
The diagram is the parameter square, not the image of H in the complex plane. Each horizontal line represents an entire path. A finite chain for one line also works throughout a narrow strip. Different strips can initially use different t subdivisions; drawing all their dividing lines creates one finite common refinement.
Choose a rectangle and answer: which function represents its germs, and why does the neighboring rectangle agree on the seam? Now choose a shared row between strips. Does agreement there follow from the picture or from uniqueness along the entire row starting at a?
Repair the tempting picture-only argument
The picture provides no analytic functions by itself. First take continuation along a reference row, construct its segment disks and agreement disks, then use a uniform perturbation margin to supply functions throughout its strip. Within a strip the old seam disks enforce compatibility; a newly added subdivision uses the same function on both sides. Across strips, both assignments continue the same initial germ along the shared row, so path uniqueness identifies their germs there. The displayed grid illustrates a finite proof structure; it is not a numerical validation of an arbitrary homotopy.
Understand
Let D be a domain, a an initial point and G a holomorphic germ at a. Assume G admits analytic continuation along every continuous path in D starting at a. Let H:[0,1]²→D be continuous with H(s,0)=a and H(s,1)=b for every s. Such a map is a fixed-endpoint homotopy between γ₀(t)=H(0,t) and γ₁(t)=H(1,t). We will prove their terminal germs at b are equal. The existence assumption is supplied before compactness is used; it is not deduced from a geometric disk cover.
First construct a strip around one row
Fix s₀. The existence assumption supplies continuation along γₛ₀(t)=H(s₀,t). The path-stability construction supplies a finite partition in t, holomorphic representatives on disks, seam agreement disks, and a positive uniform displacement bound ε.
Uniform continuity of H on the compact square gives δ>0 such that |s−s₀|<δ implies |H(s,t)−H(s₀,t)|<ε for every t. To justify this familiar compactness step directly, if no such uniform bound existed, there would be pairs of square points at distances tending to zero whose image distances remain at least ε. Compactness gives a convergent subsequence of the first points; the second points have the same limit, and continuity contradicts the separated images.
Every row in the relative open interval Iₛ₀=(s₀−δ,s₀+δ)∩[0,1] therefore uses the same finite chain. At each seam its path lies in the agreement disk. It starts with G and ends with the same germ at b as the reference row. Uniqueness on each path identifies these constructed germs with any continuation supplied separately by the hypothesis. This is a proved neighborhood of rows, not a presumed global analytic branch.
Construct the finite homotopy subdivision
The row intervals Iₛ₀ cover [0,1]. Extract finitely many. The Lebesgue-number lemma gives a finite partition in s whose every closed slab lies in one selected row interval. Assign that interval’s chain to the slab. Each chain has finitely many t subdivision points. Take the union of all these finite sets of points: it is a finite common refinement for the entire square.
Every refined rectangle lies inside a rectangle from its slab’s original chain, so its image lies in that chain’s representative disk. A new t seam inside an original segment uses the same function on both sides. An original seam keeps its agreement disk, valid for every row in the slab. Refinement therefore preserves the certificate rather than demanding a new theorem.
At the shared row between adjacent slabs, the two assignments are both continuations of G along exactly the same path. Path uniqueness makes their germs agree at every point of that row, including the terminal point b. Equality of germs gives agreement on a small spatial neighborhood of each shared-row point. By continuity of H, that agreement also holds on a sufficiently small parameter neighborhood when comparing the adjacent local representatives. Thus the local pieces are compatible across the row; no assumption that H is injective is used.
The terminal germ is constant throughout each slab by the strip argument, and equal across adjacent slabs by the shared-row argument. There are finitely many slabs, so induction carries equality from s=0 to s=1. This proves homotopy invariance of the terminal germ. Equivalently, the endpoint-germ assignment is locally constant on the connected s interval. The finite construction explains why the local statement is justified.
Compatibility here lives over parameter points of the square. It does not assert that two unrelated visits to the same complex point have equal germs. That stronger conclusion requires the following path-independence argument.
Monodromy, including the missing local-holomorphy step
Suppose additionally that D is simply connected. Use the equivalent path formulation: any two paths in D with the same endpoints are homotopic with endpoints fixed. For any z in D, choose a path from a to z and continue G along it. Homotopy invariance makes the resulting germ independent of the chosen path. Let F(z) be its value at z.
Here is the topological step behind that formulation. Simple connectivity means path connectedness and trivial based-loop homotopy classes. For paths α and β from a to z, the loop α followed by reversed β contracts with base point a fixed. Concatenate that homotopy with β. Cancelling the reversed-β/β backtracking and removing constant pauses gives α homotopic to β with endpoints fixed. The endpoint-preserving reparameterizations and backtracking contraction are proved in fundamental-group loop laws; endpoint conditions explain why an unrestricted free homotopy is not the assumption used here.
Values alone do not prove that F is holomorphic. Fix z and choose a disk U about z, contained in D, supporting a representative f of its terminal germ. For any w in U, append to a path from a to z the straight segment from z to w, contained in U. The continuation along the appended segment uses f, so path independence gives F(w)=f(w). Therefore F agrees with one holomorphic function throughout a neighborhood of every z and is holomorphic on D. At a it agrees with the original germ. Identity uniqueness also makes this extension unique on the connected domain.
Changed branch behavior: one return is not all-path independence
Start with the cube root of z−2 taking value 1 at 3. Along γ(t)=2+exp(2πit), a continuation is exp(2πit/3). At the end it equals −1/2+i√3/2. After three turns it returns to 1 and to the original germ. Local roots exist on small disks avoiding 2: a local logarithm of z−2 gives its root by exponentiating one third of that logarithm. Along any continuous path avoiding 2, the compact path can be covered successively by such disks with the branch chosen to match the preceding germ. A fixed-endpoint contraction of one turn inside that domain would contradict homotopy invariance. The puncture cannot be crossed during a contraction.
To justify that branch matching, pull the local disks back to open parameter intervals and choose a finite subordinate partition. At a seam, the two nonzero cube-root values differ by a cube root of unity. Multiply the next representative by that constant to match the value. On a connected small overlap disk the ratio of the two representatives is holomorphic and takes values in the finite set of cube roots of unity, so it is constant; the matched value makes the germs agree. This constructs a coherent continuation, rather than assuming a disk cover already supplies one. For square roots the same argument uses the constants +1 and −1.
Now start the square root of z²−1 at z=2 with value √3. On the large circle γ(t)=2exp(2πit), each factor z−1 and z+1 winds once around zero. Their continued logarithms each gain 2πi. Half their sum gains 2πi, so the root returns to its original germ. On the other based loop η(t)=1+exp(2πit), the factor z−1 winds once, but z+1 traces a radius-one circle centered at 2 and winds zero times. The root therefore changes sign and ends at −√3. Both paths avoid ±1 and start at 2.
The winding counts can be checked by the circles’ centers and radii, or by the existing argument-principle work. On any small disk avoiding ±1, continue local logarithms of both factors and exponentiate half their sum. This also establishes pathwise continuation; the product may have global branch ambiguity despite local existence. Return after one chosen loop does not establish a global branch on the twice-punctured plane.
Apply
Write a proof-repair dossier: first state the mistaken claim “compactness alone produces all analytic extensions,” then replace it with the precise existence hypothesis, segment/seam construction, uniform strip, finite refinement, shared-row uniqueness and local-holomorphy argument. Include one changed loop calculation and preserve your original attempt before revision. Use the mixed continuation audit without opening its solution first.
As a separate hypothesis check, the germ 1/(1−z) at zero cannot extend through 1 even though the whole plane is simply connected. A global extension would agree with that rational function away from 1 and contradict continuity near its pole. Simple connectivity does not supply the missing continuation existence.
Continue
Return to analytic elements and path stability to repair a missing margin or seam. The complex-analysis unit retains contour, mapping, series and assessment obligations. This finite monodromy proof does not prove the Riemann mapping theorem, Picard theory or classify Riemann surfaces; public written practice still requires independent mathematical review.
Primary comparison: Ko Honda, Math 520 notes, Lemma 31.1, Theorem 31.2 and Corollary 31.3, undated PDF, retrieved 28 October 2026. The explicit margins, strip refinement and local patching here make the compactness steps inspectable. Berkeley Lecture 18, 3 April 2026 supplies additional course context for continuation and branch examples; it is not evidence that every program has the same required proof depth.
