ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-06

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.

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Cover-small chains for the two-arc cover of a circle

Example

Fix an abelian group $G$ and an element $g \in G$ . Let $U, V$ be proper overlapping open arcs covering $S^{1}$ , so $U \cap V$ has two components. Choose one point $a_{i}$ in each overlap component. Parameterize the closed subarc from $a_{0}$ to $a_{1}$ carried by $U$ as a singular path $u$ , and the complementary closed subarc from $a_{1}$ to $a_{0}$ carried by $V$ as $v$ . Thus $z_{Z} = u + v$ traverses the circle once with the indicated orientation. With coefficients in $G$ , put $z_{g} = u \otimes g + v \otimes g \in C_{1}^{{U, V}} (S^{1}; G) .$ This is the oriented two-edge circle cycle with coefficient $g$ ; no distinguished element of $G$ is assumed.

Facts & Assumptions

Given: The two-arc cover, overlap points, paths, abelian group $G$ , and supplied coefficient $g$ above.

Verification

technique · direct

1.1givenconstruct

Each proper open arc is an interval in the cyclic order. The interval inside $U$ joining the overlap points is one of the two closed subarcs between them, and the complementary subarc lies in $V$ because the omitted parts of $U$ must be covered by $V$ . Both endpoints are in both open sets. Thus the entire images of $u$ and $v$ , including their endpoints, lie in their respective cover members. Also $\partial (u \otimes g) = ([a_{1}] - [a_{0}]) \otimes g$ and $\partial (v \otimes g) = ([a_{0}] - [a_{1}]) \otimes g$ , so $z_{g}$ is cover-small and is a cycle.

2.1step 1.1algebra∎

By Mayer–Vietoris connecting class, $δ [z_{g}] = [([a_{1}] - [a_{0}]) \otimes g] \in H_{0} (U \cap V; G)$ . Identifying the two components in the order containing $a_{0}, a_{1}$ gives the coefficient pair $(- g, g)$ . Indeed, paths within each arc component identify its point classes, and no path crosses the components. Replacing the oriented integral cycle by its negative negates this class. If $g = 0$ (including $G = 0$ ), both chain and connecting class are zero.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Allen Hatcher, Algebraic Topology, §2.2 (standard reference, not scraped)