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CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26
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A connected plane domain that is not homologically simply connected

Statement refuted

Every complex domain is homologically simply connected.

Facts & Assumptions

Given: The annulus A={z∈C:12<∣z∣<2} and the contour C(t)=exp⁡(it) on [0,2π], taken as the chain with the single term (1,C).

[L1]

A complex domain is homologically simply connected when every cycle with trace in it is null-homologous in it (Homologically simply connected complex domains).

[L2]

A cycle Γ with trace in an open Ω is null-homologous in Ω when n(Γ,q)=0 for every q∈C∖Ω (Null-homologous cycles and homologous cycles in an open set).

[L3]

For a∈C, r>0 and k∈Z, the contour a+rexp⁡(ikt) on [0,2π] is a closed complex contour with index k for ∣z−a∣<r and 0 for ∣z−a∣>r, and with trace {∣z−a∣=r} when k≠0 (A circle traversed k times has winding number k inside and 0 outside).

[L4]

For a positively oriented circle a+rexp⁡(it) with r>0, (2πi)−1∫γdz/(z−a)=1 (The normalized integral around a positively oriented circle centred at a is 1).

[L5]

A complex domain is a nonempty, connected, open subset of C (A complex domain is a nonempty connected open subset of C).

[L6]

For c∈C and R≥0 the set {z:∣z−c∣>R} is path-connected (The exterior of a closed disc in the plane is path-connected).

[L7]

A single closed contour with coefficient 1 is a cycle whose trace is that contour's trace (Complex chains, their traces, and cycles), and its index is the winding number of that contour (Integration over a complex chain and the index of a chain).

[L8]

For n≥2 the unit sphere Sn−1⊆Rn is path-connected (For n≥2, the sphere Sn−1 is path-connected and connected); the map x↦x/∥x∥2 into Sn−1 is continuous (Radial normalisation x↦x/∥x∥2 is continuous on Rn∖{0}); and Sn−1={x:∥x∥2=1} (Euclidean spheres and closed balls as subspaces of Rn).

[L9]

A subset is path-connected when any two of its points are joined by a continuous map from [0,1] with image inside it (Paths, path-connected spaces and path components), a path-connected subset is connected (Every path-connected space is connected, and every path component lies inside a component), and a function continuous on each member of a finite closed cover is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L11]

Counterexample

technique · constructive
1.1givenconstruct

Take A={12<∣z∣<2} and C(t)=exp⁡(it) on [0,2π].

1.2givenL10L11

A is open: for z∈A the ball of radius min⁡(∣z∣−12, 2−∣z∣) about z lies in A by [L10] and [L11]; and A is nonempty, since 1∈A.

2.1step 1.1L6L8L9L10

A is path-connected, hence connected by [L9]: given z,w∈A, the ambient exterior {∣u∣>12} is path-connected by [L6], and more concretely the radial paths s↦((1−s)∣z∣+s)z/∣z∣ and s↦((1−s)∣w∣+s)w/∣w∣ are continuous by [L8] and [L9], keep the modulus between 1 and the starting modulus, hence inside (12,2), and end on the unit circle; the unit circle is path-connected by [L8], and its points have modulus 1; concatenating the three pieces on closed subintervals of [0,1] gives a continuous path in A from z to w by [L9].

2.2step 1.1L3L4L7

By [L3] with a=0, r=1, k=1 the contour C is closed with trace {∣z∣=1}⊆A and n(C,0)=1, a value [L4] confirms directly; by [L7] the chain with the single term (1,C) is a cycle with the same trace and the same index.

3.1step 1.2step 2.1step 2.2L1L2L5discharge-construct∎

Steps 1.2 and 2.1 make A a complex domain in the sense of [L5]. The point 0 lies in C∖A, and step 2.2 gives n(C,0)=1≠0, so by [L2] the cycle C is not null-homologous in A; by [L1] the domain A is therefore not homologically simply connected, which refutes the claim.

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Sources