TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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$p : R \to R / Z$ is a covering map with translated interval sheets

Statement

$p : R \to R / Z$ is a covering map. More explicitly, for every $x \in R$ , let

$J_{x} = (x - 1 / 3, x + 1 / 3), U_{x} = p [J_{x}] .$

Then $U_{x}$ is an open neighbourhood of $[x]$ ,

$p^{- 1} (U_{x}) = ∐_{n \in Z} (J_{x} + n),$

and every restriction $p ∣_{J_{x} + n} : J_{x} + n \to U_{x}$ is a homeomorphism.

Facts & Assumptions

Given: The quotient map $p : R \to R / Z$ and a real representative $x$ of an arbitrary quotient class.

[L1]

The quotient map is open, and every interval shorter than one embeds in $R / Z$ . Moreover, $p^{- 1} (p [V]) = ⋃_{n \in Z} (V + n)$ for open $V \subseteq R$ (The quotient map is open, and every interval shorter than one embeds in $R / Z$ ).

[L2]

A covering map is a continuous surjection $p : E \to B$ such that every $b \in B$ has an open neighbourhood $U$ for which $p^{- 1} (U)$ is a disjoint union of open sheets $V_{j}$ , and each restriction $p ∣_{V_{j}} : V_{j} \to U$ is a homeomorphism (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).

Proof

technique · direct

1.1

The interval $J_{x} = (x - 1 / 3, x + 1 / 3)$ has length $2 / 3 < 1$ . Its image $U_{x} = p [J_{x}]$ is open by [L1] and contains $p (x) = [x]$ .

1.2

By the saturation formula in [L1], $p^{- 1} (U_{x}) = ⋃_{n \in Z} (J_{x} + n)$ . These open intervals are pairwise disjoint: if $y$ belonged to $J_{x} + m$ and $J_{x} + n$ , then $∣ (y - m) - (y - n) ∣ = ∣ n - m ∣ < 2 / 3$ , and an integer of absolute value below one is zero, so $m = n$ .

L1algebra

2.1

Every translate $J_{x} + n$ has length $2 / 3 < 1$ , so [L1] makes $p ∣_{J_{x} + n}$ a homeomorphism onto its image; its image is $p [J_{x} + n] = p [J_{x}] = U_{x}$ . The map $p$ is already a continuous surjection because it is a quotient projection, and steps 1.1 and 1.2 give an open neighbourhood with a disjoint union of open sheets. Thus every clause of [L2] holds, and $p$ is a covering map.

step 1.1step 1.2L1L2∎

Depends on

Used by

The degree of a based circle loop Definition
A covering quotient of a simply connected space need not be simply connected Example
FALSE: every continuous self-map of the circle is nullhomotopic False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 44 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, Ch. 1, Section 1.1 (standard reference, not scraped)
J. Peter May, A Concise Course in Algebraic Topology, Ch. 1, Section 5 (standard reference, not scraped)

p:R→R/Zp:R→R/Z is a covering map with translated interval sheets