CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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.

Based connected coverings are isomorphic exactly when their induced subgroups are equal

Statement

Under the hypotheses of A based morphism between connected coverings exists exactly when the induced subgroups are included, the based connected coverings $(E_{1}, e_{1})$ and $(E_{2}, e_{2})$ are isomorphic over $B$ if and only if

$(p_{1})_{*} π_{1} (E_{1}, e_{1}) = (p_{2})_{*} π_{1} (E_{2}, e_{2}) .$

The based isomorphism, when it exists, is unique.

Facts & Assumptions

Given: Two based connected coverings of the same path-connected locally path-connected base.

[L1]

A unique based covering morphism exists exactly when the source induced subgroup is contained in the target induced subgroup (A based morphism between connected coverings exists exactly when the induced subgroups are included).

[F1]

Two lifts from a connected space that agree at one point are equal (Two lifts from a connected space that agree at one point agree everywhere).

Proof

technique · direct

1.1

For the direction from subgroup equality to isomorphism, [L1] gives unique based morphisms $f : E_{1} \to E_{2}$ and $g : E_{2} \to E_{1}$ .

2.1

The composite $g \circ f$ and ${id}_{E_{1}}$ are lifts of $p_{1}$ through $p_{1}$ and agree at $e_{1}$ , so [F1] makes them equal. Likewise $f \circ g = {id}_{E_{2}}$ . Hence $f$ and $g$ are inverse based covering isomorphisms, and uniqueness follows from [L1].

step 1.1F1L1

3.1

For the converse direction, a based isomorphism and its inverse are covering morphisms, so [L1] gives both subgroup inclusions and therefore equality.

L1∎

Depends on

Used by

Dependency tree · two levels

17 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, Proposition 1.37 (standard reference, not scraped)