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Deck groups of connected circle coverings: $Z / n Z$ for $n \geq 1$ and $Z$ for the universal cover

Example

Let $E_{n} \to R / Z$ be the connected circle covering classified by $n Z$ .

For $n \geq 1$ , it has $n$ sheets and $Deck (E_{n} / (R / Z)) ≅ (Z / n, +) .$
For $n = 0$ , it is the real-line universal cover and its deck group is $(Z, +)$ .

The case $n = 1$ has the trivial deck group.

Facts & Assumptions

Given: The connected circle coverings $E_{n}$ from Connected coverings of the circle are classified by the subgroups $n Z$ for $n \geq 0$ .

[L1]

A regular connected covering with base group $G$ and induced subgroup $H$ has deck group $G / H$ (A regular connected covering has deck group $π_{1} (B, b_{0}) / p_{*} π_{1} (E, e_{0})$ ).

[F1]

For every natural $n$ , $(Z, +) / n Z$ is the same group as $(Z / n, +)$ , including $n = 0, 1$ (For every $n \in N$ , the congruence-class group $(Z / n, +)$ is the quotient group $(Z, +) / n Z$ ).

[L2]

Every connected covering of the quotient circle is regular (Every connected covering of the circle is regular).

[L3]

The $n = 0$ classified cover is the real-line universal cover (Connected coverings of the circle are classified by the subgroups $n Z$ for $n \geq 0$ , $R \to R / Z$ is a universal covering).

[F2]

Degree gives an isomorphism from the circle fundamental group to $(Z, +)$ ( $Deg : π_{1} (R / Z, [0]) \to (Z, +)$ is an isomorphism).

[F3]

The quotient circle is path-connected, and its open quotient arcs are homeomorphic to convex intervals and give arbitrarily small path-connected neighbourhoods, so it is locally path-connected ( $R / Z$ is compact and path-connected, The quotient map is open, and every interval shorter than one embeds in $R / Z$ , Every nonempty convex subset of $R^{n}$ is simply connected, Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point).

Verification

technique · direct

1.1

Let $n \geq 1$ . The base hypotheses for [L1] hold by [F3], and [L2] makes $E_{n}$ regular, so [L1] gives the quotient of the circle group by $n Z$ . The degree isomorphism [F2] identifies this with $(Z, +) / n Z$ , and [F1] identifies that quotient with $(Z / n, +)$ . At $n = 1$ this group has one element.

L1L2F1F2F3

2.1

For $n = 0$ , [L3] identifies $E_{0}$ with the real-line universal cover. It is regular by [L2], and [L1], [F1], and [F2] give its deck group as $(Z, +) / 0 Z = (Z, +)$ .

L1L2L3F1F2F3∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Allen Hatcher, Algebraic Topology, Section 1.3 and Proposition 1.39 (standard reference, not scraped)

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