TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-28

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Stereographic projection identifies the Riemann sphere with the unit two-sphere

Statement

Let $S^{2} : = {(x, y, t) \in R^{3} : x^{2} + y^{2} + t^{2} = 1} .$ Define $Σ (z) = (\frac{2 Re z}{1 + ∣ z ∣^{2}}, \frac{2 Im z}{1 + ∣ z ∣^{2}}, \frac{∣ z ∣^{2} - 1}{1 + ∣ z ∣^{2}}) (z \in C)$ and $Σ (\infty) = (0, 0, 1)$ . Then $Σ : \hat{C} \to S^{2}$ is a homeomorphism, with inverse $Π (x, y, t) = {\begin{cases} \frac{x + i y}{1 - t}, & (x, y, t) \neq (0, 0, 1), \\ \infty, & (x, y, t) = (0, 0, 1) . \end{cases}$

Facts & Assumptions

Given: The Riemann sphere $\hat{C} = C \cup {\infty}$ , the unit sphere $S^{2}$ , and the displayed formulas for $Σ$ and $Π$ .

[L1]

In the one-point compactification, a neighbourhood of $\infty$ is exactly the complement of a closed compact subset of $C$ , and $\hat{C}$ is compact Hausdorff (The one-point (Alexandroff) compactification $X^{*} = X \cup {\infty}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ , $X^{*}$ is compact and contains $X$ as an open subspace; $X$ is dense in $X^{*}$ exactly when $X$ is not compact; and $X^{*}$ is Hausdorff exactly when $X$ is locally compact and Hausdorff).

Proof

technique · direct

1.1

Direct algebra gives $Σ (z) \in S^{2}$ for every finite $z$ , and the displayed formulas satisfy $Σ (Π (x, y, t)) = (x, y, t)$ for $(x, y, t) \neq (0, 0, 1)$ and $Π (Σ (z)) = z$ for finite $z$ , with $Σ (\infty) = (0, 0, 1)$ and $Π (0, 0, 1) = \infty$ .

givenalgebra

1.2

On $C$ and on $S^{2} ∖ {(0, 0, 1)}$ the formulas are rational with nonzero denominator, so both restrictions are continuous; and for the cap $U_{t} = {(x, y, s) : s > t} \cup {(0, 0, 1)}$ one has $Σ^{- 1} (U_{t}) = {z : ∣ z ∣ > \sqrt{(1 + t) / (1 - t)}} \cup {\infty}$ , which is a neighbourhood of $\infty$ by [L1].

L1givenalgebra

1.3

If $V = \hat{C} ∖ K$ is a neighbourhood of $\infty$ , compactness of $K$ gives $R > 0$ with $K \subseteq \overline{D} (0, R)$ , so the cap $U_{(R^{2} - 1) / (R^{2} + 1)}$ satisfies $Π (U_{(R^{2} - 1) / (R^{2} + 1)}) \subseteq V$ ; therefore $Π$ is continuous at the north pole.

L1choosealgebra

2.1

The maps $Σ$ and $Π$ are continuous inverse bijections by the preceding three steps, so $Σ$ is a homeomorphism.

given∎

Depends on

Used by

Complex conjugation is a homeomorphism of the Riemann sphere that is not holomorphic Counterexample
The chordal metric on the Riemann sphere Definition
Stereographic projection and its inverse are explicit in coordinates Example
The chordal distance has the standard coordinate formula on the finite plane Example
FALSE: the Riemann sphere is homeomorphic to the complex plane False statement
The chordal metric induces the standard topology of the Riemann sphere Theorem

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Sources

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 3 §§2.2-3.5 (standard reference, not scraped)
Matthias Weber, Complex Analysis, Ch. 1 §§1.3-1.4 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 8 §§1-2 (standard reference, not scraped)