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PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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Star-shaped plane domains are homologically simply connected

Statement

Let U⊆C be nonempty, open, and star-shaped with respect to some a∈U (Star-shaped open subsets of Euclidean space). Then U is a complex domain and is homologically simply connected (Homologically simply connected complex domains).

In particular every nonempty convex open subset of C is homologically simply connected, since it is star-shaped with respect to each of its points; this covers every open disc and C itself.

Facts & Assumptions

Given: A nonempty open U⊆C star-shaped with respect to a∈U; the plane identification and its segments are those of Complex star-shaped and convex domains are the published Euclidean notions under the identification C=R2.

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

[L2]

If Γ is a cycle whose trace lies in an open V and F is a primitive on V of a continuous f with F′=f continuous, then ∫Γf dz=0 (The integral of a continuous derivative over a cycle is zero).

[L3]

If U⊆C is open and star-shaped with respect to a∈U, every holomorphic f:U→C has the primitive F(z)=∫ℓazf(ζ) dζ (Every holomorphic function on a star-shaped domain has a primitive).

[L4]

A nonempty open U⊆Rn is star-shaped with respect to a∈U when a+t(x−a)∈U for every x∈U and 0≤t≤1; every convex open set is star-shaped with respect to each of its points (Star-shaped open subsets of Euclidean space, A convex subset of Rm contains every line segment between two of its points).

[L5]

Constants and the identity are complex differentiable, and linear combinations, products and nonvanishing quotients of functions complex differentiable at a point are complex differentiable there (Linearity, product, reciprocal, and quotient rules for complex derivatives); a complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L6]

n(Γ,p)=(2πi)−1∫Γdz/(z−p) for a chain Γ and p∉Γ∗ (Integration over a complex chain and the index of a chain), a chain being a finite list of integer-weighted contours (Complex chains, their traces, and cycles).

[L7]

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

[L8]

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), and a path-connected subset is connected (Every path-connected space is connected, and every path component lies inside a component); a composite of continuous maps is continuous 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).

[L9]

A primitive of f on V is a holomorphic F with F′=f on V (A primitive of a complex function on an open set).

Proof

technique · direct
1.1givenL4L7L8

For x,y∈U the maps t↦x+2t(a−x) on [0,12] and t↦a+(2t−1)(y−a) on [12,1] are continuous, take values in U by [L4], and agree at t=12 with the value a; the first begins at x and the second ends at y. Thus [L8] joins x to y inside U, making U path-connected, hence connected. With U nonempty and open, [L7] makes it a complex domain.

1.2givenL5L6

Let Γ be a cycle with trace in U and let p∈C∖U. Then z↦1/(z−p) is holomorphic on U by [L5], since z−p≠0 there.

2.1step 1.2L3L5L9

By [L3] the function F(z)=∫ℓazdζζ−p is a primitive on U of z↦1/(z−p), and F′ equals that function, which is continuous by [L5].

3.1step 1.2step 2.1L2L6

The trace of Γ lies in the open set U, so [L2] applied with V=U, f(z)=1/(z−p) and F of step 2.1 gives ∫Γdzz−p=0, whence n(Γ,p)=0 by [L6].

4.1step 1.1step 3.1L1L4∎

Since Γ and p∉U were arbitrary, step 3.1 makes every cycle in U null-homologous in U, so the complex domain of step 1.1 is homologically simply connected by [L1]. A nonempty convex open set is star-shaped with respect to each of its points by [L4], so the same conclusion applies to it.

Depends on

Used by

Dependency tree · two levels

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Sources