CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Winding number identifies the fundamental group of C times with the integers

Statement

For a based loop $γ : [0, 1] \to C^{\times}$ at $1$ , let $r (z) = z / ∣ z ∣$ and define

$W ([γ]) : = \deg (r \circ γ) .$

Then $W$ is the standard isomorphism

$π_{1} (C^{\times}, 1) ≅ (Z, +) .$

If $γ$ is rectifiable, then

$W ([γ]) = n (γ, 0) .$

Thus the analytic winding number of any rectifiable representative is the integer classifying its loop class.

Facts & Assumptions

Given: A based loop $γ : [0, 1] \to C^{\times}$ at $1$ .

[L1]

For a rectifiable based loop, the winding number about $0$ equals the degree of its normalized circle loop (For loops in C times, the winding number about 0 equals the circle degree).

[L2]

Radial normalization is a deformation retraction of $R^{2} ∖ {0} = C^{\times}$ onto the unit circle (For $n \geq 1$ , radial normalisation is a deformation retraction of $R^{n} ∖ {0}$ onto $S^{n - 1}$ ).

[L3]

A deformation retract induces mutually inverse fundamental-group isomorphisms between the retract and the ambient space (A retract induces an injection on fundamental groups, and a deformation retract induces an isomorphism).

[L4]

The unit circle has fundamental group $Z$ with the standard trigonometric generator (The trigonometric loops give $π_{1} ({(x, y) : x^{2} + y^{2} = 1}, (1, 0)) ≅ Z$ ).

Proof

technique · direct

1.1

Let $C = {z \in C : ∣ z ∣ = 1} \subseteq C^{\times}$ and let $r : C^{\times} \to C$ be radial normalization, $r (z) = z / ∣ z ∣$ . Specializing [L2] to $n = 2$ and applying [L3], the induced map $r_{*} : π_{1} (C^{\times}, 1) ⟶ π_{1} (C, 1)$ is an isomorphism. For the given loop $γ$ , its image under $r_{*}$ is the class of the normalized circle loop $α (t) = γ (t) / ∣ γ (t) ∣$ .

givenL2L3

2.1

Fact [L4] identifies the class of $α$ in $π_{1} (C, 1)$ with the integer $\deg (α)$ . Hence $W ([γ]) = \deg (r \circ γ)$ is exactly the composite of the isomorphism $r_{*}$ from step 1.1 with the standard identification $π_{1} (C, 1) ≅ Z$ , and is therefore the standard isomorphism $π_{1} (C^{\times}, 1) ≅ (Z, +) .$ If $γ$ is rectifiable, [L1] gives $W ([γ]) = \deg (α) = n (γ, 0)$ .

step 1.1L1L4∎

Depends on

Used by

The punctured plane has connected complement in C but disconnected spherical complement Counterexample

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Exercise 4.1.7 (standard reference, not scraped)
A. Hatcher, Algebraic Topology, Theorem 1.7 (standard reference, not scraped)