ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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.

Higher homotopy groups of a product

Example

For based spaces X,Y and n≥1, $π_{n} (X \times Y, (x_{0}, y_{0})) ≅ π_{n} (X, x_{0}) \times π_{n} (Y, y_{0}),$ and their pointed component sets also correspond. No path-connectedness assumption is needed. For a concrete instance, take X=Y=R based at 0: the loop $a (t) = (t (1 - t), 2 t (1 - t))$ represents the pair of its two coordinate classes, and its product with $b (t) = (3 t (1 - t), 0)$ has first-half value $(2 t (1 - 2 t), 4 t (1 - 2 t))$ and second-half value $(3 (2 t - 1) (2 - 2 t), 0)$ .

Facts & Assumptions

[F1]

A product-valued map is continuous exactly when both coordinates are continuous. A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice

[F2]

Projections act by based postcomposition and preserve cubical products. Higher homotopy groups are functorial and based homotopy invariant

Verification

Given: The spaces, maps, and hypotheses in the statement above.

1.1

Send [a] to $([p_{X} a], [p_{Y} a])$ . Projecting a boundary-fixed homotopy gives boundary-fixed coordinate homotopies, so the map is well-defined. Conversely pair any representatives u,v to obtain $(u, v) : I^{n} \to X \times Y$ , continuous by F1 and constant at (x0,y0) on the boundary. Pairing the two homotopies proves independence of representatives. The composites are the identity because projections of (u,v) are u,v and pairing the projections of a recovers a pointwise.

F1F2

2.1

Projection commutes with both half-cube formulas, hence the bijection is a homomorphism by F2. Two product points are joined by a path exactly when both coordinate pairs are joined: project a path in one direction and pair the two paths in the other using F1. This proves the component statement. In the displayed instance, substitution of 2t and 2t−1 gives exactly the two polynomial formulas; both equal (0,0) at their common endpoint t=1/2. The homotopy $(1 - s) a (t)$ also contracts that particular loop, with both endpoints fixed.

F1F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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, Chapter 4 (standard reference, not scraped)