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

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.

Extending over one cell is equivalent to nullhomotoping the attaching sphere

Statement

Let $n \geq 0$ , and let $B = A \cup_{α} D^{n + 1}$ be obtained by attaching one cell along $α : S^{n} \to A$ , and let $f : A \to Y$ be continuous. Then $f$ extends to $B$ if and only if $f α : S^{n} \to Y$ is nullhomotopic.

Facts & Assumptions

[F1]

The attached space is the pushout of $S^{n} ↪ D^{n + 1}$ and $α : S^{n} \to A$ .

[F2]

$D^{n + 1}$ is the cone on $S^{n}$ , and a nullhomotopy of a sphere map descends to a map on that cone.

Proof

Given: The attachment and map in the statement.

1.1F1

Suppose $\overset{ˉ}{f} : B \to Y$ extends $f$ . Its restriction to the characteristic disk, composed with a radial contraction of $D^{n + 1}$ to its center, is a nullhomotopy of $\overset{ˉ}{f} ∣_{S^{n}} = f α$ .

1.2F2

Conversely, let $H : S^{n} \times I \to Y$ satisfy $H (-, 0) = f α$ and have constant terminal map. Collapsing $S^{n} \times {1}$ turns the cylinder into $C S^{n} ≅ D^{n + 1}$ , and [F2] makes $H$ descend to a map $F : D^{n + 1} \to Y$ with $F ∣_{S^{n}} = f α$ .

2.1F1step 1.1step 1.2

The maps $f$ on $A$ and $F$ on $D^{n + 1}$ agree on the attaching boundary. By [F1]'s pushout universal property they glue uniquely to a continuous map $\overset{ˉ}{f} : B \to Y$ extending $f$ . The two constructions are inverse existence implications and require no choice. $□$

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

James Davis and Paul Kirk, Lecture Notes in Algebraic Topology