PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Singular chain cross products are natural

Statement

If $f : X \to X^{'}$ and $g : Y \to Y^{'}$ are continuous maps, then for singular chains $a \in C_{p} (X; Z)$ and $b \in C_{q} (Y; Z)$ , $(f \times g)_{#} (a \times b) = f_{#} (a) \times g_{#} (b) .$

Facts & Assumptions

Given: Continuous maps $f : X \to X^{'}$ and $g : Y \to Y^{'}$ , and singular chains $a \in C_{p} (X; Z)$ and $b \in C_{q} (Y; Z)$ .

[L1]

The chain cross product is the alternating shuffle sum on generators (The singular chain cross product on generators).

[L2]

The induced singular chain map is postcomposition on each singular simplex (The induced singular chain map of a continuous map).

Proof

technique · direct

1.1

By bilinearity from [L1], it is enough to prove the identity for generators $a = σ$ and $b = τ$ . For each shuffle simplex $λ_{θ}$ , [L2] gives $(f \times g)_{#} ((σ \times τ) \circ λ_{θ}) = ((f \circ σ) \times (g \circ τ)) \circ λ_{θ} .$

L1L2given

2.1

Summing the equality of step 1.1 over all shuffles with the same signs as in [L1] yields $(f \times g)_{#} (σ \times τ) = (f_{#} σ) \times (g_{#} τ) .$ Extending bilinearly gives the formula for arbitrary integral chains.

L1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

Haynes Miller, Algebraic Topology I, Lecture 7 (standard reference, not scraped)