PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-10 (Codex supervising session)

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Degree of identity constant reflection and antipodal sphere maps

Statement

On $S^{n} \subset R^{n + 1}$ for $n \geq 1$ , the identity, a constant map, any single coordinate reflection, and the antipodal map have degrees $1$ , $0$ , $- 1$ , and $(- 1)^{n + 1}$ respectively.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

Let $n \geq 1$ . Choose a generator $[S^{n}]$ of $H_{n} (S^{n}; Z) ≅ Z$ , using cor-homology-of-spheres. For a continuous self-map $f : S^{n} \to S^{n}$ , its degree is the unique integer satisfying $f_{*} [S^{n}] = \deg (f) [S^{n}] .$ The induced map is furnished by prop-relative-homology-is-functorial-for-maps-of-pairs with empty subspaces. Replacing the same generator in source and target by its negative does not change the integer. For a map between separately oriented copies of $S^{n}$ , use their separately specified generators; reversing just one orientation changes the sign. The unreduced definition here is restricted to $n \geq 1$ . (Degree of a self map of an oriented sphere)

[F2]

For $n \geq 1$ and continuous sphere self-maps $f, g$ , homotopic maps have the same degree and $\deg (g \circ f) = \deg (g) \deg (f)$ . Every homotopy equivalence $S^{n} \to S^{n}$ has degree $1$ or $- 1$ . (Degree is homotopy invariant and multiplicative under composition)

[F3]

For $n \geq 1$ and $f : S^{n} \to S^{n}$ , let $Σ f$ be its two-cone (unreduced) suspension. Orient $Σ S^{n} ≅ S^{n + 1}$ by the natural suspension isomorphism. Then $\deg (Σ f) = \deg (f)$ . (Suspension preserves sphere map degree)

[F4]

For any abelian group $G$ , $H_{0} (*; G) ≅ G$ and $H_{n} (*; G) = 0$ for every integer $n \neq 0$ . For every set-indexed family of pairs the canonical map $⨁_{α} H_{n} (X_{α}, A_{α}; G) ⟶ H_{n} (⨆_{α} X_{α}, ⨆_{α} A_{α}; G)$ is an isomorphism. Together with the structural axioms, singular homology is an ordinary theory with coefficient group $G$ . (Singular homology satisfies dimension and arbitrary additivity)

[F5]

In an integral cellular chain complex, an oriented edge has boundary equal to its terminal vertex minus its initial vertex. (Cellular boundary is the incidence degree matrix)

[F6]

Cellular homology is naturally isomorphic to singular homology for cellular maps. (Cellular homology computes singular homology)

Proof

1.1F1F4

The identity fixes the chosen generator. A constant map factors through a point whose $H_{n}$ is zero because $n \geq 1$ . Their degrees are therefore $1$ and $0$ .

1.2F1F5F6algebra

Give $S^{1}$ the CW structure with left and right vertices and with the upper and lower semicircles as oriented $1$ -cells $a, b$ , both directed from left to right. Its cellular chain complex has $C_{1} = Z [a] \oplus Z [b]$ , $C_{0} = Z [l e f t] \oplus Z [r i g h t]$ , no $2$ -cells, and $d_{1} [a] = d_{1} [b] = [r i g h t] - [l e f t] .$ Thus $H_{1}^{c e l l} (S^{1}; Z) = \ker d_{1} = Z ([a] - [b])$ . Reflection across the horizontal axis is cellular, fixes both vertices, and interchanges the equally oriented cells $a, b$ , so it sends $[a] - [b]$ to its negative. The natural cellular-to-singular comparison carries this generator to a generator of $H_{1} (S^{1}; Z)$ and intertwines the reflection maps. Hence the reflection has degree $- 1$ .

2.1F2F3step 1.2

Suspending this reflection $n - 1$ times gives a coordinate reflection of $S^{n}$ , still of degree $- 1$ by F3. Every other coordinate reflection is conjugate to it by a coordinate permutation. The conjugating homeomorphism has degree $\pm 1$ , and multiplicativity cancels its degree with that of its inverse.

3.1F2step 2.1∎

The antipodal map is the composite of the $n + 1$ coordinate reflections. Its degree is their product $(- 1)^{n + 1}$ . This also covers the starting case $n = 1$ .

Depends on

Dependency tree · 2 levels

Sources

Hatcher, Algebraic Topology, Degree properties (a),(e),(f) and (b), pp.134–135