LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Natural singular-cohomology identities are detected on finite regular complexes

Statement

Assume AC. Fix a prime $p$ and $m, n \geq 0$ . Suppose that for every space $X$ there is a map

$R_{X} : H^{m} (X; F_{p}) ⟶ H^{n} (X; F_{p})$

natural in the sense that $f^{*} R_{X} (x) = R_{K} (f^{*} x)$ for every continuous $f : K \to X$ . If $R_{K} = 0$ for every finite regular cell complex $K$ , then $R_{X} = 0$ for every space $X$ . Neither additivity of $R$ nor a simultaneous finite model for all classes is required.

Facts & Assumptions

Given: AC, the prime $p$ , nonnegative degrees $m, n$ , and the natural family $R$ in the statement.

[F1]

Singular chain groups are made of finite formal sums (Singular simplices and singular chain groups with coefficients); their boundary squares to zero, and homology is cycles modulo boundaries (The singular chain complex and singular homology).

[F2]

Over $F_{p}$ , singular cochains are the full linear dual of the singular chains (Singular cochain complex with coefficients).

[F3]

The singular-cochain coboundary is $δ ψ = ψ \partial$ (Singular cochain complex with coefficients).

[F4]

A continuous map pulls a cohomology class back by precomposition with its induced singular chain map (Singular cohomology is contravariantly functorial).

[F5]

The mod- $p$ Kronecker pairing is well-defined and natural: $⟨ f^{*} α, z ⟩ = ⟨ α, f_{*} z ⟩$ (The kronecker pairing is independent of cocycle and cycle representatives).

[F6]

AC supplies a choice function for every family of nonempty sets (The Axiom of Choice).

Proof

technique · realize each individual singular cycle on a finite Delta complex, subdivide it to a finite regular complex, and use evaluation to detect the cohomology class

1.1F1

Realize a mod- $p$ singular cycle on a finite regular complex. Let $z \in Z_{n} (X; F_{p})$ and write its finite support as $z = \sum_{r = 1}^{N} a_{r} σ_{r}$ . Form the finite Delta complex $P_{z}$ generated by these labeled top simplices and all their iterated face restrictions: two face occurrences are attached to the same lower simplex exactly when they are the same singular simplex of $X$ , and all attaching maps are the corresponding order-preserving affine face maps. The simplicial identities make these attachments compatible in lower dimensions. Mapping the cell labeled by a singular simplex $τ$ by $τ$ itself gives a continuous map $g : P_{z} \to X$ .

Put $ξ = \sum_{r} a_{r} [σ_{r}]_{P_{z}}$ in the Delta-chain group. For every labeled $(n - 1)$ -simplex $τ$ , its coefficient in $\partial ξ$ is exactly the coefficient of the singular basis element $τ$ in $\partial z$ , hence is zero in $F_{p}$ . Thus $ξ$ is a mod- $p$ cycle and $g_{*} [ξ] = [z]$ . This construction also covers $n = 0$ : $P_{z}$ is the finite discrete set of labeled vertices in the support, carrying their coefficients $a_{r}$ .

The second barycentric subdivision $K_{z}$ of a Delta complex is a finite simplicial complex, hence a finite regular cell complex. The affine subdivision operator is a chain map and the cone calculation $\partial T + T \partial = 1 - S$ makes it chain-homotopic to the identity. Therefore the subdivided cycle $ζ$ and the composite $f : K_{z} \to P_{z} \to X$ satisfy

$f_{*} [ζ] = [z] .$

All face identifications, coefficient operations, and subdivisions here are finite prescribed operations; no choice principle is used.

1.2F2F3F6

Prove that evaluation detects a mod- $p$ cohomology class. Let a degree- $n$ cocycle $φ$ vanish on every degree- $n$ cycle. If $n = 0$ , every zero-chain is a cycle and hence $φ = 0$ . Suppose $n > 0$ . For $u \in B_{n - 1} = \partial C_{n} (X; F_{p})$ , choose any $c$ with $\partial c = u$ and define $b_{0} (u) = φ (c)$ . This is well-defined: two choices differ by a cycle, on which $φ$ vanishes. It is linear by using sums and scalar multiples of preimages. By [F6], choose a vector-space complement $M$ with $C_{n - 1} = B_{n - 1} \oplus M$ , and extend $b_{0}$ by zero on $M$ to a cochain $b$ . Then [F3] gives

$δ b (c) = b (\partial c) = b_{0} (\partial c) = φ (c)$

for every $c \in C_{n}$ . Thus $φ = δ b$ . Consequently, if a class in $H^{n} (X; F_{p})$ pairs to zero with every homology class, it is zero. The sole AC use is the complement of the possibly infinite-dimensional boundary subspace.

2.1givenF4F5step 1.1step 1.2

Apply the finite hypothesis to every evaluation cycle. Fix a space $X$ and $x \in H^{m} (X; F_{p})$ , and put $α = R_{X} (x)$ . For any mod- $p$ $n$ -cycle $z$ , choose $f : K_{z} \to X$ and $ζ$ as in step 1.1. Naturality of $R$ and the assumed finite-complex vanishing give

$f^{*} α = f^{*} R_{X} (x) = R_{K_{z}} (f^{*} x) = 0.$

By [F5] and $f_{*} [ζ] = [z]$ ,

$⟨ α, [z] ⟩ = ⟨ α, f_{*} [ζ] ⟩ = ⟨ f^{*} α, [ζ] ⟩ = 0.$

Step 1.2 now yields $α = 0$ . Since $X$ and $x$ were arbitrary, $R_{X} = 0$ for every space.

3.1F1F2F3F6step 1.1step 1.2step 2.1∎

Check empty, zero, endpoint, and degeneracy cases. If $X$ is empty, its chain, homology, and cohomology groups in the stated degrees are zero. The zero cycle may be represented by the empty finite complex and evaluates to zero. The case $n = 0$ was handled separately in the evaluation-detection step; degree $m = 0$ requires no change because naturality alone is used on the input. A one-term zero-cycle with any nonzero coefficient is represented by one weighted vertex. Degenerate singular simplices are still finite basis elements and may label cells whose map to $X$ is degenerate. Both implications in the displayed naturality equality are literal equalities, not directions of a biconditional. Finite face identification and subdivision use no choice; AC is used only for the complement in step 1.2.

Natural singular-cohomology identities are detected on finite regular complexes

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources