ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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.

Degree of z to the m on the circle from a regular value

Example

Let $m \in Z ∖ {0}$ and $P_{m} : S^{1} \to S^{1}$ , $P_{m} (z) = z^{m}$ , with both circles counterclockwise oriented. Every target point is regular, has exactly $∣ m ∣$ preimages, and every preimage has sign $sgn (m)$ ; hence the regular-value sum is $m$ .

Facts & Assumptions

Given: The nonzero integer $m$ and a target point $y = [a] \in R / Z$ .

[F1]

Regular-value formula for compact-support degree gives degree as the finite sum of derivative signs at any supplied regular value.

[F2]

Degree of the power map on the circle verifies that $P_{m} ([t]) = [m t]$ is smooth and has degree $m$ .

[F3]

$R / Z$ is compact and path-connected makes $S^{1} = R / Z$ compact. By In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones and A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, the inverse image under the continuous $P_{m}$ of every compact target subset is compact; hence $P_{m}$ is proper.

Verification

1.1F2givenalgebra

Put $r = ∣ m ∣$ . The $r$ classes $x_{k} = [\frac{a + k}{m}], 0 \leq k < r,$ all map to $[a]$ . They are distinct: if $x_{j} = x_{k}$ , then $(j - k) / m \in Z$ , so $r$ divides $j - k$ , which is possible in the displayed range only when $j = k$ . Conversely, if $P_{m} ([t]) = [a]$ , then $m t - a \in Z$ ; reducing that integer modulo $r$ puts $[t]$ equal to exactly one $x_{k}$ . Thus this is the complete fibre.

2.1F1F2F3step 1.1

In increasing angular lift coordinates at every $x_{k}$ and at $y$ , the map is $u \mapsto m u$ plus a constant, so its derivative is the nonzero scalar $m$ . Every $x_{k}$ is therefore regular and has local sign $sgn (m)$ . Properness from [F3] lets [F1] apply, giving $\deg (P_{m}) = \sum_{k = 0}^{r - 1} sgn (m) = r sgn (m) = m,$ agreeing with [F2].

3.1F1F2step 1.1step 2.1∎

For $m = 1$ the fibre is a singleton of sign $+ 1$ ; for $m = - 1$ it is a singleton of sign $- 1$ . The excluded $m = 0$ map instead has empty fibres away from its constant value and degree zero, as [F2] records. There are no quotient-seam endpoints: all calculations use local lifts. The finite list is explicit and no choice principle is used.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 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

Robbin–Salamon, Introduction to Differential Topology, Theorem 5.4.1, pp.191–192 (standard reference, not scraped)