PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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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)

Proof

1.1

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 $0$ .

F1F4

1.2

On $S^{1}$ choose the upper and lower semicircle singular paths $a, b$ from the left endpoint to the right endpoint, with parametrizations matched by reflection across the horizontal axis, and put $S^{0}$ equal to the two endpoints. CW excision and the pair sequences of the two arcs identify $H_{1} (S^{1}, S^{0}; Z)$ with $Z [a] \oplus Z [b]$ ; the connecting map sends each of $[a]$ and $[b]$ to the same generator $[r i g h t] - [l e f t]$ of ${\tilde{H}}_{0} (S^{0}; Z)$ . Its kernel is therefore generated by $[a] - [b]$ . Since $H_{1} (S^{0}; Z) = 0$ , exactness identifies $H_{1} (S^{1}; Z)$ with this kernel, so the absolute cycle $a - b$ is a fundamental cycle. Reflection interchanges $a$ and $b$ , hence sends that generator to its negative and has degree $- 1$ .

F1F4algebra

2.1

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.

F2F3step 1.2

3.1

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 $n = 1$ .

F2step 2.1∎

Depends on

Used by

A fixed point free sphere map has antipodal degree Corollary
A group acting freely on a positive even sphere has at most two elements Corollary
Degree zero does not imply a sphere map is constant Counterexample
Degree of a coordinate reflection on a sphere Example
Degree of the antipodal map in low dimensions Example
Degree of the circle power map Example
Every integer occurs as the degree of a sphere map Proposition
No nowhere zero tangent vector field on an even sphere Theorem

Dependency tree · two levels

12 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

Hatcher, Algebraic Topology, Degree properties (a),(e),(f) and (b), pp.134–135 (standard reference, not scraped)