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

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A regular connected covering has deck group $π_{1} (B, b_{0}) / p_{*} π_{1} (E, e_{0})$

Statement

Let $p : (E, e_{0}) \to (B, b_{0})$ be a regular connected covering of a path-connected locally path-connected base. Put $G = π_{1} (B, b_{0})$ and $H = p_{*} π_{1} (E, e_{0})$ . Then

$Deck (E / B) ≅ G / H .$

Facts & Assumptions

Given: The regular connected covering and groups $G, H$ in the Statement.

[L1]

For a connected covering, $Deck (E / B) ≅ N_{G} (H) / H$ ( $Deck (E / B) ≅ N_{G} (H) / H$ for a connected covering).

[L2]

A connected covering is regular exactly when $H$ is normal in $G$ (A connected covering is regular exactly when its induced subgroup is normal, exactly when deck transformations act transitively on a fibre).

[F1]

The normalizer is $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

Proof

technique · direct

1.1

By [L2], regularity gives $H ⊴ G$ , so every $g \in G$ preserves $H$ under conjugation and [F1] gives $N_{G} (H) = G$ .

L2F1

2.1

Substitution of $N_{G} (H) = G$ in [L1] yields $Deck (E / B) ≅ G / H$ .

step 1.1L1∎

Depends on

Used by

Deck groups of connected circle coverings: ℤ/nℤ for n≥1 and ℤ for the universal cover Example

Dependency tree · two levels

15 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 1.39 (standard reference, not scraped)

A regular connected covering has deck group π1(B,b0)/p∗π1(E,e0)π1​(B,b0​)/p∗​π1​(E,e0​)