CounterexampleConstruction: Literature-sourcedVerification: 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.

Incompatible choices do not define one global obstruction problem

Claim

There are a finite CW pair $(X, A)$ , a fixed map $A \to S^{1}$ , and two $2$ -cells such that each cell separately can be filled after a suitable choice of the map on one common $1$ -cell, but no single choice fills both. This does not contradict cellular obstruction theory: for one fixed map on the entire $1$ -skeleton, zero obstruction on every cell does glue to a global extension.

Facts & Assumptions

[F1]

Degrees of based circle loops add under concatenation and satisfy $\deg (γ^{m}) = m \deg (γ)$ (Degree sends concatenation to addition, reversal to negation, and the constant loop to zero).

[F2]

A based circle loop is nullhomotopic exactly when its degree is zero (A based circle loop is nullhomotopic exactly when its degree is zero), and standard loops realize every integer degree ( $\deg (ω_{n}) = n$ for every integer $n$ ).

[F3]

A map over one attached $2$ -cell extends exactly when the image of its attaching loop is nullhomotopic (Extending over one cell is equivalent to nullhomotoping the attaching sphere).

Verification

Given: Let $X^{1} = S_{a}^{1} \lor S_{b}^{1}$ , let $A = S_{b}^{1}$ , and attach two $2$ -cells along the based words $a b$ and $a b^{2}$ . Fix on $A$ a based map $f_{b} : S_{b}^{1} \to S^{1}$ of degree $1$ .

1.1

A based map $f_{a} : S_{a}^{1} \to S^{1}$ has an integer degree $k$ , and every $k$ occurs by [F2]. Under the combined map on $X^{1}$ , [F1] gives

F1F2

$\deg f (a b) = k + 1, \deg f (a b^{2}) = k + 2.$

[F1, F2]

2.1

For the first $2$ -cell alone choose $k = - 1$ ; its image attaching loop has degree zero and extends by [F2, F3]. For the second alone choose $k = - 2$ and obtain the same conclusion. A simultaneous extension for one map on $X^{1}$ would force both $k + 1 = 0$ and $k + 2 = 0$ , which is impossible. The two separately vanishing values therefore came from different choices of $f_{a}$ , not from one obstruction cochain.

F2F3step 1.1

3.1

For contrast, fix one map $h : X^{1} \to Y$ and suppose its attaching loop on every $2$ -cell is nullhomotopic. By [F3], choose a disk filling for each cell and use the CW pushout to glue the fillings to $h$ . This gives a map on all of $X^{2}$ . Only two choices occur in the displayed example; an arbitrary cell family would require the separately declared choice principle. Thus the original fixed-map counterexample is false, and the example establishes only the corrected quantifier claim. $□$

F3step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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

James Davis and Paul Kirk, Lecture Notes in Algebraic Topology (standard reference, not scraped)