TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-29

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The sphere, the plane, and the disc are pairwise non-biholomorphic

Statement

The Riemann sphere $\hat{C}$ , the complex plane $C$ , and the unit disc $D$ are pairwise non-biholomorphic.

Facts & Assumptions

Given: The three domains $\hat{C}$ , $C$ , and $D$ .

[F1]

A conformal equivalence is a biholomorphism between domains (Conformal equivalence and the automorphism group of a domain).

[F2]

The continuous image of a compact space is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[F3]

Every bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).

Proof

technique · direct

1.1

If there were a biholomorphism from $\hat{C}$ onto $C$ or onto $D$ , then [F2] would make the target compact because $\hat{C}$ is compact, but neither $C$ nor $D$ is compact. Hence the sphere is biholomorphic to neither the plane nor the disc.

F1F2given

1.2

If there were a biholomorphism $f : C \to D$ , then $f$ would be a bounded entire function and [F3] would make it constant, contradicting bijectivity. Hence $C$ and $D$ are not biholomorphic.

F1F3algebra

2.1

Steps 1.1 and 1.2 cover all three pairs, so $\hat{C}$ , $C$ , and $D$ are pairwise non-biholomorphic.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 8 §1 (standard reference, not scraped)