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

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Global sphere degree is the sum of local degrees

Statement

Let $f : S^{n} \to S^{n}$ be continuous, $n \geq 1$ , with source and target orientations fixed. If $f^{- 1} (y)$ is finite, then $\deg (f) = \sum_{x \in f^{- 1} (y)} \deg_{x} f .$ The sum over an empty fibre is $0$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

For $n \geq 1$ , $H_{n} (S^{n}, S^{n} ∖ {x}; Z) ≅ Z$ , with generator the restriction of the global sphere orientation. The local degree in def-local-degree-at-an-isolated-preimage is independent of shrinking its neighborhood. For every finite nonempty $F \subset S^{n}$ , $H_{n} (S^{n}, S^{n} ∖ F; Z) ≅ ⨁_{x \in F} H_{n} (S^{n}, S^{n} ∖ {x}; Z),$ and the global orientation maps to the tuple of local orientation generators. (Local sphere orientations and finite puncture excision)

[F2]

A continuous map between oriented $n$ -spheres, $n \geq 1$ , whose degree is nonzero must be surjective. (A map of nonzero degree between spheres is surjective)

Proof

1.1

If the fibre is empty, $f$ omits a point, so its degree is zero by F2. This equals the empty sum.

1.2

For a nonempty finite fibre $F$ , use the global-to-relative maps for $(S^{n}, S^{n} ∖ F)$ and $(S^{n}, S^{n} ∖ {y})$ . Functoriality makes the square with $f$ commute. By F1, the source global generator maps to $(1)_{x \in F}$ and the target global generator maps to $1$ .

2.1

On the summand indexed by $x$ , the lower map in this square is multiplication by $\deg_{x} f$ , by the local definition and excision. Its value on the diagonal is the sum of these integers. The other route through the square gives $\deg (f)$ , proving the formula, also for a singleton fibre.

F1step 1.2algebra∎

Depends on

Used by

Degree zero does not imply a sphere map is constant Counterexample
Degree of the circle power map Example
Local degrees of a polynomial map on the riemann sphere Example
Every integer occurs as the degree of a sphere map Proposition

Dependency tree · two levels

8 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, Proposition 2.30, p.136 (standard reference, not scraped)