Try
Follow a logarithm from 1 to 2 along the straight path γ(t)=1+t, for 0≤t≤1, starting with value zero. A nearby bent path with the same endpoints should reach the same local logarithm. What does “nearby” need to guarantee? Remaining in the right half-plane is sufficient in this example, but a general proof needs something we can construct from a finite chain of local functions.
Take disks U₁ centered at 1 and U₂ centered at 2, both of radius 3/4. On the first half of the path use the logarithm in U₁; on the second half use the compatible logarithm in U₂. Their values and derivatives agree near the shared point 3/2. Predict how far the path may move before this particular certificate stops applying.
Predict
Each closed half-segment lies at least 1/4 from its disk’s complement. At the seam, choose an agreement disk V centered at 3/2 with radius 1/5: it lies inside both disks, since 1/2+1/5 < 3/4. A uniform displacement less than 1/10 therefore leaves slack for both segment containment and seam agreement. This is a sufficient bound for this certificate, not the largest possible displacement preserving logarithm continuation.
Experiment
Draw a perturbed path η(t)=1+t+i·sin(πt)/20. Its endpoints are still 1 and 2, and its displacement from γ is at most 1/20, below 1/10. Verify the segment and seam inequalities separately before opening the explanation.
The dashed curve illustrates a small bend; it is not a numerical plot of the specified sine path. The disk radii and center distances use a common scale. Trace the first segment, seam disk and second segment with your finger or cursor. Identify the two different kinds of margin before reading the proof.
Check the worked path and certificate
The straight first half has distance at most 1/2 from center 1. Perturbing by at most 1/20 gives at most 11/20, less than 3/4. The second half has the same bound about center 2. At t=1/2 the new seam is 3/2+i/20, within distance 1/20 of 3/2 and hence inside V. Both local logarithms agree there and on a neighborhood. The final point remains 2, so the last representative gives exactly the same terminal germ, not merely the value log 2.
Understand
A germ at z records a holomorphic function on some neighborhood of z; two representatives give the same germ when they agree on a smaller neighborhood. Equal values are insufficient: z and 2z have the same value at zero but different derivatives and different germs.
Let D be an open connected subset of the complex plane and γ:[0,1]→D a continuous path. A continuation along γ assigns a germ Gₜ at γ(t), with this local coherence condition: near each parameter time t₀ there is a disk U containing the nearby path points and one holomorphic function f on U representing all the assigned germs at those times. The initial germ is prescribed. A self-intersection of γ need not have the same germ on its two visits.
Uniqueness on one path
Consider two such continuations with the same initial germ. Near any time t₀, shrink their representative domains to a common disk around γ(t₀), and shrink the parameter neighborhood so the path stays in that disk. If the two germs agree at any time in this neighborhood, the representatives agree on an open neighborhood of that path point. The identity theorem on the connected common disk makes them agree throughout it. Thus either they agree at all nearby times or at none.
The set of times where the germs agree and its complement are both open in [0,1]. The agreement set contains zero. Connectedness of [0,1] forces it to be the whole interval. This proves uniqueness when continuation exists. It constructs no extension through a singularity.
A small compactness lemma, with its reason
For an open cover of a compact metric set, there is a positive number λ such that every subset of diameter less than λ lies in one member of the cover. To see this, for each point choose a ball of radius 2r contained in a cover member; the balls of radius r cover the compact set. Extract finitely many, with radii r₁,…,rₖ, and take λ to be their minimum. For a nonempty subset of diameter less than λ, choose a point x in it and one selected radius-r ball containing x. Every other point of the subset lies within less than 2r of that ball’s center, hence in the selected cover member. The empty subset causes no difficulty. This is the Lebesgue-number lemma in the form we need; balls and openness are relative to the compact set.
Construct a finite adapted chain
Local coherence provides an open parameter interval and a disk representative around every time. These intervals cover [0,1]. Choose a finite subcover and a Lebesgue number λ. Subdivide 0=t₀<t₁<…<tₙ=1 with mesh less than λ. Each closed parameter segment lies in one coherence interval, so its entire compact path image Kⱼ=γ([tⱼ₋₁,tⱼ]) lies in a disk Uⱼ with representative fⱼ.
At every seam tⱼ, both neighboring functions represent the assigned germ G at γ(tⱼ). By the definition of a germ there is a disk Vⱼ around that point, contained in both domains, on which the functions agree. Choose it small enough to be contained in their intersection. Covering the geometric path image alone would not select compatible germs when the path revisits a point; the construction deliberately covers parameter times.
Each compact Kⱼ lies inside the open Uⱼ, so its distance dⱼ from the closed complement of Uⱼ is positive. For completeness, distance to a closed set is continuous; it is positive at every point of Kⱼ and attains a positive minimum there. A finite chain gives finitely many such distances and finitely many positive seam radii rⱼ. Choose ε smaller than all of them. If there are no internal seams, only segment margins are needed.
Stability under a uniform perturbation
Let η be another continuous path with the same initial and final points, satisfying |η(t)−γ(t)|<ε for every t. Its jth segment stays in Uⱼ; its seam point stays in Vⱼ. Use fⱼ to define its germs on that segment. Near a seam, continuity keeps η in Vⱼ for a small time interval, and the two representatives agree there. Thus the piecewise assignment is locally coherent and really is a continuation.
The first point is unchanged, so the initial germ is the prescribed one. The last point is unchanged and the last representative is the same, so the terminal germ is the same. Path uniqueness identifies this construction with any continuation already supplied along η. This proves local existence and terminal-germ stability for nearby paths; it does not establish continuation along every path in D.
Apply
Repair this argument: “All perturbed segments stay inside their disks, so compatibility follows.” State where agreement at the perturbed seam comes from and why local coherence needs a neighborhood of that seam, not only one shared value. Then explain why a local logarithm can cross a chosen cut at −1 but cannot cross zero: if exp L(z)=z on a neighborhood of zero, evaluating at zero would contradict the fact that the exponential never vanishes.
Continue
Use this finite certificate in compact homotopies and monodromy. Return to the original continuation examples for branch intuition. No earlier completion is required to open either page.
Primary comparison: Ko Honda, Math 520 notes, sections 30.2–31.2, PDF undated, retrieved 28 October 2026. Our argument expands the finite-chain and uniform-neighborhood steps directly, without requiring the sheaf topology used in that source.
