LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Sphere endomorphisms act by the same integer in every ordinary theory

Statement

Let $n \geq 0$ and $u : S^{n} \to S^{n}$ be continuous. If $u_{*}$ on ${\tilde{H}}_{n} (S^{n}; Z)$ is multiplication by $d$ , then $u_{*}$ on ${\tilde{h}}_{n} (S^{n}) ≅ G$ for every ordinary theory $h$ is $d {i d}_{G}$ . The identifications use the same oriented sphere generator; for $n = 0$ use the difference of the two point classes.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

For ordinary homology theories $h, k$ and a specified isomorphism $u : h_{0} (*) \to k_{0} (*)$ , there is a unique natural equivalence on finite CW pairs normalized by $u$ and commuting with connecting homomorphisms. (Coefficient comparison on finite cw pairs)

[F2]

Let $h$ be an ordinary theory with coefficient group $G = h_{0} (*)$ . For every $n \geq 0$ and $k \in Z$ , $h_{k} (D^{n}, S^{n - 1}) ≅ {\begin{cases} G & k = n, \\ 0 & k \neq n . \end{cases}$ At $n = 0$ the pair means $(*, \emptyset)$ . Also ${\tilde{h}}_{k} (S^{n}) ≅ G$ for $k = n$ and zero otherwise, including $n = 0$ . Choose the disk identifications by ordered orientations and iterated cone boundaries, so they commute with these boundaries. (Any ordinary homology theory computes relative cell groups from its coefficient group)

[F3]

For $n \geq 1$ , ${\tilde{H}}_{k} (S^{n}; G)$ is $G$ for $k = n$ and $0$ otherwise. For $S^{0}$ , ${\tilde{H}}_{0} (S^{0}; G) ≅ G$ and all other reduced groups vanish. Thus $H_{0} (S^{n}; G) ≅ G$ for $n \geq 1$ , whereas $H_{0} (S^{0}; G) ≅ G \oplus G$ . (Homology of spheres)

Proof

1.1

The sphere is a finite CW complex. F1, applied with the identity of $G$ , identifies $h$ naturally with singular homology with $G$ on it; the point splitting identifies their reduced groups as well. F2 and F3 identify those groups with $G$ in the stated dimension.

F1F2F3

2.1

For a triangulated oriented sphere the top integral cycle is the sum of its consistently oriented top simplices. The equation for a top cycle forces the coefficients of adjacent simplices to agree, so with any abelian $G$ every reduced top cycle is this same fundamental cycle with a common coefficient $g$ . There are no chains one degree higher in this triangulation. After simplicial approximation and subdivision, the integer matrix of the sphere map therefore sends that coefficient to $d g$ , because its action on the integral fundamental cycle is $d$ . This is the coefficient-chain comparison used in F1, and does not assert that tensor product preserves arbitrary exact sequences.

F1step 1.1

3.1

When $n = 0$ , the reduced generator is $[+] - [-]$ . The identity, transposition and two constant maps act on it by $1, - 1, 0, 0$ , respectively; the same computation on $(- g, g)$ gives $g, - g, 0, 0$ . This proves the assertion also for $S^{0}$ and completes all cases, including $G = 0$ and $d = 0$ .

F2F3algebra∎

Depends on

Used by

Axiomatic cellular boundaries are integral incidence matrices with coefficients Lemma

Dependency tree · two levels

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Sources

May, A Concise Course in Algebraic Topology, 15§2, first theorem and coefficient paragraph p.119 (standard reference, not scraped)
Hatcher, Algebraic Topology, Theorem 4.59 homology argument, printed pp.399–401 (in chapter 4); alternative simplicial route specified in notes (standard reference, not scraped)