LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-21

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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

1.1F1algebra

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 $τ_{\pm} σ_{\pm} = id$ .

2.1step 1.1F2F3F4

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 $U_{j} \cap V_{j} ≅ R^{n} ∖ {0}$ .

3.1step 2.1F5F7F8

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.

4.1step 2.1F6∎

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

Depends on

Used by

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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