LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Antipodal complements cover $S^{n}$ by simply connected sets with path-connected overlap for $n \geq 2$

Statement

Let $n \geq 2$ and write $S^{n} \subseteq R^{n + 1}$ for the unit sphere (Euclidean spheres and closed balls as subspaces of $R^{n}$ ). For any coordinate unit vector $e_{j}$ , put

$U_{j} = S^{n} ∖ {e_{j}}, V_{j} = S^{n} ∖ {- e_{j}} .$

Then $U_{j}$ and $V_{j}$ are open, simply connected subsets of $S^{n}$ , they cover $S^{n}$ , and their intersection is path-connected. More precisely, stereographic projection gives homeomorphisms $U_{j} ≅ R^{n}$ , $V_{j} ≅ R^{n}$ , and $U_{j} \cap V_{j} ≅ R^{n} ∖ {0}$ .

Facts & Assumptions

Given: A natural $n \geq 2$ , a coordinate index $j < n + 1$ , the unit vector $e_{j}$ , and the unit sphere $S^{n} \subseteq R^{n + 1}$ .

[F1]

The unit sphere $S^{n}$ is the set of $z \in R^{n + 1}$ with $∥ z ∥_{2} = 1$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[F2]

Sums and products of continuous real maps are continuous, and a quotient is continuous wherever its denominator is nonzero (Sums, products, absolute values, finite maxima and minima, and quotients of continuous real-valued maps on a topological space are continuous where defined).

[F3]

Continuity of maps into a finite-dimensional Euclidean space is equivalent to continuity of every coordinate map (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

[F4]

A map is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ ).

[F5]

Every nonempty convex subset of $R^{n}$ is simply connected (Every nonempty convex subset of $R^{n}$ is simply connected).

[F6]

For $n \geq 2$ , $R^{n} ∖ {0}$ is polygonally connected, hence path-connected (For $n \geq 2$ , the punctured space $R^{n} ∖ {0}$ is polygonally connected).

[F7]

Pointed homeomorphisms induce mutually inverse fundamental-group homomorphisms (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

[F8]

A space is simply connected when it is nonempty and path-connected and its fundamental group has one element at every basepoint (Simply connected topological spaces).

Proof

technique · direct

1.1

Delete the $j$ -th coordinate to identify $e_{j}^{⊥}$ with $R^{n}$ , and write $z = u + t e_{j}$ . Projection from $e_{j}$ is $σ_{+} (z) = u / (1 - t)$ on $U_{j}$ , with inverse $τ_{+} (y) = \frac{2 y}{∥ y ∥_{2}^{2} + 1} + \frac{∥ y ∥_{2}^{2} - 1}{∥ y ∥_{2}^{2} + 1} e_{j} .$ Projection from $- e_{j}$ is $σ_{-} (z) = u / (1 + t)$ on $V_{j}$ , with inverse $τ_{-} (y) = \frac{2 y}{∥ y ∥_{2}^{2} + 1} + \frac{1 - ∥ y ∥_{2}^{2}}{∥ y ∥_{2}^{2} + 1} e_{j} .$ Since $∥ y ∥_{2}^{2} \geq 0$ , the inverse denominator $∥ y ∥_{2}^{2} + 1$ is positive. The sphere equation $∥ u ∥_{2}^{2} + t^{2} = 1$ and deletion of the relevant pole give $1 - t \neq 0$ on $U_{j}$ and $1 + t \neq 0$ on $V_{j}$ . The same equation shows that both inverse formulas land on $S^{n}$ , and direct substitution gives $σ_{\pm} τ_{\pm} = id$ and $τ_{\pm} σ_{\pm} = id$ .

F1algebra

2.1

The coordinate $t$ is continuous, so $U_{j} = {z \in S^{n} : t < 1}$ and $V_{j} = {z \in S^{n} : t > - 1}$ are open by [F4]. They cover the sphere because no point has both $t = 1$ and $t = - 1$ . The formulas in step 1.1 are continuous by [F2] and [F3], so they are the asserted homeomorphisms. Moreover $σ_{+}$ sends the deleted point $- e_{j}$ to $0$ , hence restricts to a homeomorphism $U_{j} \cap V_{j} ≅ R^{n} ∖ {0}$ .

step 1.1F2F3F4

3.1

The space $R^{n}$ is nonempty and convex, so [F5] makes it simply connected. Each homeomorphism in step 2.1 transports paths, and [F7] applied to it and its inverse transports the one-element fundamental group at every basepoint. Thus [F8] makes both $U_{j}$ and $V_{j}$ simply connected.

step 2.1F5F7F8

4.1

By [F6] and the last homeomorphism in step 2.1, $U_{j} \cap V_{j}$ is path-connected. This proves every assertion.

step 2.1F6∎

Depends on

Used by

Dependency tree · two levels

62 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

Allen Hatcher, Algebraic Topology, Proposition 1.14 (standard reference, not scraped)

Antipodal complements cover SnSn by simply connected sets with path-connected overlap for n≥2n≥2