TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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$Deg : π_{1} (R / Z, [0]) \to (Z, +)$ is an isomorphism

Statement

$Deg : π_{1} (R / Z, [0]) \to (Z, +)$ is an isomorphism. Its inverse is

$n ⟼ [ω_{n}] .$

Facts & Assumptions

Given: The degree function on based loop classes of the quotient circle.

[L1]

$Deg : π_{1} (S^{1}, [0]) \to (Z, +)$ is a group homomorphism ( $Deg : π_{1} (S^{1}, [0]) \to (Z, +)$ is a group homomorphism).

[L2]

Two based circle loops are path-homotopic if and only if they have equal degree (Two based circle loops are path-homotopic if and only if they have equal degree).

[L3]

$\deg (ω_{n}) = n$ for every integer $n$ ( $\deg (ω_{n}) = n$ for every integer $n$ ).

[L4]

An isomorphism $f : G \to H$ is a bijective group homomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[L5]

The integers form a commutative ring, and hence $(Z, +)$ is a group (The integers form a commutative ring).

[L6]

$π_{1} (X, x_{0})$ consists of endpoint-fixed path-homotopy classes $[α]$ of based loops (Based loops and the fundamental group).

Proof

technique · direct

1.1L1L5

By [L1], $Deg$ is a group homomorphism into the additive group of integers supplied by [L5].

1.2L2L6

If $Deg ([α]) = Deg ([β])$ , then $\deg (α) = \deg (β)$ , so [L2] makes $α$ and $β$ path-homotopic. Their classes are equal by [L6], and $Deg$ is injective.

1.3L3

Let $n \in Z$ . By [L3], $Deg ([ω_{n}]) = \deg (ω_{n}) = n$ , so every integer is attained and $Deg$ is surjective. This includes $n = 0$ , $n = 1$ , and negative integers.

2.1step 1.1step 1.2step 1.3L3L4∎

Steps 1.1, 1.2, and 1.3 show that $Deg$ is a bijective group homomorphism, hence an isomorphism by [L4]. Step 1.3 also shows that $n \mapsto [ω_{n}]$ is its inverse, since injectivity makes this preimage unique.

Depends on

Used by

Connected coverings of the circle are classified by the subgroups nℤ for n≥0 Corollary
Every connected covering of the circle is regular Corollary
The trigonometric loops give π₁({(x,y):x²+y²=1},(1,0))≅ℤ Corollary
π₁(T²)≅ℤ×ℤ Corollary
A complete manifold with zero global injectivity radius Counterexample
A homology equivalence need not be a homotopy equivalence without simple connectivity Counterexample
Ignoring monodromy gives the wrong Serre E2 page Counterexample
Deck groups of connected circle coverings: ℤ/nℤ for n≥1 and ℤ for the universal cover Example
Mapping cone of a degree d circle map Example
Oriented two-plane bundles over the two-sphere by winding number Example
Sign local system on real projective space Example
π₁(ℝ²∖{0})≅ℤ Example
FALSE: the two-set van Kampen conclusion needs no path-connectedness hypothesis on the overlap False statement
Circle and path-loop models for Eilenberg–Mac Lane induction Lemma
The boundary label of a van Kampen diagram is trivial in the presented group Lemma
The fundamental group of a finite wedge of circles is free of that rank Theorem

Dependency tree · two levels

29 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, Ch. 1, Section 1.1, Theorem 1.7 (standard reference, not scraped)
J. Peter May, A Concise Course in Algebraic Topology, Ch. 1, Section 5 (standard reference, not scraped)

Deg⁡:π1(R/Z,[0])→(Z,+)Deg:π1​(R/Z,[0])→(Z,+) is an isomorphism