ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

How statement and proof provenance work

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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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The square-root surface is a two-sheeted covering of the punctured plane

Example

For $n = 2$ , the square-root surface is biholomorphic to $C^{\times}$ , and under that identification the projection is

$w ⟼ w^{2} .$

So each nonzero base point has exactly the two lifts $w$ and $- w$ .

Facts & Assumptions

Given: The $n$ th-root surface theorem with $n = 2$ .

[L1]

The $n$ th-root surface is biholomorphic to $C^{\times}$ , and the projection becomes $w \mapsto w^{n}$ (The Riemann surface of an nth root is the n-sheeted covering w maps to w to the nth power).

Verification

technique · direct

1.1

Specializing [L1] to $n = 2$ makes the square-root surface biholomorphic to $C^{\times}$ with projection $w \mapsto w^{2}$ .

2.1

If $z \in C^{\times}$ and $w^{2} = z$ , then also $(- w)^{2} = z$ , and these are the only lifts because $u^{2} = z$ implies $(u / w)^{2} = 1$ , hence $u / w = \pm 1$ . So the surface has exactly two sheets over each nonzero base point.

step 1.1algebra∎

Depends on

The Riemann surface of an nth root is the n-sheeted covering w maps to w to the nth power

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 3 §4.3 (standard reference, not scraped)