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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Assuming the Axiom of Choice, a plane domain is simply connected exactly when its spherical complement is connected

Statement

Assume the Axiom of Choice. Let $Ω \subseteq C$ be a complex domain. Then $Ω$ is simply connected if and only if $\hat{C} ∖ Ω$ is connected.

Facts & Assumptions

Given: The Axiom of Choice and a complex domain $Ω$ .

[L1]

Under the grand equivalence theorem, connected spherical complement, homological simple connectivity, and trivial fundamental group are equivalent conditions on $Ω$ (For a plane domain, the complement, homology, primitive, logarithm, conjugate, conformal, homotopy, and contractibility conditions are equivalent).

[L2]

Connected spherical complement implies null homology, and null homology implies connected spherical complement (A connected spherical complement forces every cycle in the domain to be null-homologous, A homologically simply connected plane domain has connected spherical complement).

Proof

technique · direct

1.1

If $\hat{C} ∖ Ω$ is connected, then [L2] gives homological simple connectivity, and [L1] identifies that with simple connectivity under the current Axiom-of-Choice hypothesis.

L1L2

1.2

If $Ω$ is simply connected, then [L1] places it under the grand-equivalent conditions, so it is homologically simply connected; [L2] then gives connected spherical complement.

L1L2

2.1

Steps 1.1 and 1.2 prove both directions.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

21 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

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4, §4.2 (standard reference, not scraped)
J. Lebl, Guide to Cultivating Complex Analysis, Proposition 4.3.7 (standard reference, not scraped)