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LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Homotopy invariance of singular cohomology with field coefficients

Statement

Let k be a field, let X,Y be topological spaces, and let f,g:X→Y be continuous maps joined by a continuous homotopy H:X×[0,1]→Y. Then f∗=g∗:Hn(Y;k)→Hn(X;k) for every integer n. Consequently every homotopy equivalence induces an isomorphism on singular cohomology with coefficients in k.

Facts & Assumptions

Given: A field k, spaces X,Y, continuous maps f,g:X→Y, and the homotopy H from f to g.

[F1]

Singular cochains are homomorphisms on singular chains, their coboundary is positive precomposition with the chain boundary, and cohomology is cocycles modulo coboundaries. Singular cochain complex with coefficients Singular cohomology with coefficients

[F2]

The prism operator PH:Cn−1(X;k)→Cn(Y;k) satisfies g#−f#=∂PH+PH∂, with PH=0 in degree −1. The singular chain homotopy formula

Proof

Given: The data in the statement.

1.1F1F2algebra

For n≥1, define Kn:Cn(Y;k)→Cn−1(X;k) by Knφ=φ∘PH, and set K0=0. Evaluating the chain identity [F2] on an n-chain and using the positive coboundary convention in [F1] gives

(g∗−f∗)φ=δ(Knφ)+Kn+1(δφ).

For n=0, the first summand is zero because C−1=0, and the degree-zero case of [F2] gives the same identity.

2.1F1step 1.1algebra∎

If φ is a cocycle, then [F1] gives δφ=0, so step 1.1 says g∗φ−f∗φ=δ(Knφ), a coboundary; hence the induced cohomology classes agree. Negative cohomology groups are zero by [F1]. If f has homotopy inverse h, apply the equality to hf≃idX and fh≃idY; functoriality of pullback then makes h∗ a two-sided inverse for f∗.

Depends on

Dependency tree · 2 levels

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