TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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$S^{n}$ is simply connected for every $n \geq 2$

Statement

For every natural number $n \geq 2$ , the unit sphere $S^{n} \subseteq R^{n + 1}$ is simply connected (Simply connected topological spaces).

Facts & Assumptions

Given: A natural number $n \geq 2$ and an arbitrary basepoint $x_{0} \in S^{n}$ .

[L1]

For either of two distinct coordinate axes, the complements of its antipodal coordinate poles are open and simply connected, cover $S^{n}$ , and have path-connected overlap (Antipodal complements cover $S^{n}$ by simply connected sets with path-connected overlap for $n \geq 2$ ).

[L2]

For a two-set open cover whose two members and overlap are path-connected and contain the basepoint, the inclusion-images of the two fundamental groups generate the fundamental group of the union (Loops over a two-set path-connected open cover factor through the covering sets).

[F1]

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

Proof

technique · direct

1.1

The point $x_{0}$ cannot be a pole on both the first and second coordinate axes. Choose the first axis unless $x_{0}$ is one of its two poles, and choose the second axis otherwise. For that axis, [L1] supplies an antipodal open cover $S^{n} = U \cup V$ with $x_{0} \in U \cap V$ , with $U$ and $V$ simply connected and $U \cap V$ path-connected. Any two points of $S^{n}$ can be joined by going inside their respective cover members to a point of the nonempty overlap and then inside the overlap, so $S^{n}$ is path-connected.

2.1

By [L2], every element of $π_{1} (S^{n}, x_{0})$ is a product of classes induced from $π_{1} (U, x_{0})$ and $π_{1} (V, x_{0})$ . Both groups have one element by [L1], so every factor, and hence every such product, is the identity. Thus $π_{1} (S^{n}, x_{0})$ has one element.

step 1.1L1L2

3.1

The sphere is nonempty, step 1.1 makes it path-connected, and step 2.1 applies to the arbitrary basepoint $x_{0}$ . Therefore every basepoint has a one-element fundamental group, so [F1] makes $S^{n}$ simply connected.

step 1.1step 2.1F1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

25 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)

SnSn is simply connected for every n≥2n≥2