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A based circle loop is nullhomotopic exactly when its degree is zero

Statement

A based circle loop is nullhomotopic exactly when its degree is zero. Here nullhomotopic means path-homotopic relative to the endpoints to the constant loop at $[0]$ .

Facts & Assumptions

Given: A based loop $α$ at $[0]$ .

[L1]

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).

[L2]

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

[L3]

For every integer $n$ , define ${\tilde{ω}}_{n} (t) = n t$ and $ω_{n} = p \circ {\tilde{ω}}_{n}$ ; in particular, $ω_{0}$ is the constant loop at $[0]$ (The standard circle loops $ω_{n} (t) = [n t]$ for $n \in Z$ ).

Proof

technique · direct

1.1

If $α$ is nullhomotopic, then it is path-homotopic to the constant loop $ω_{0}$ by [L3]. The forward implication of [L1] and [L2] give $\deg (α) = \deg (ω_{0}) = 0$ .

L1L2L3

1.2

Conversely, if $\deg (α) = 0$ , then [L2] gives $\deg (α) = \deg (ω_{0})$ . The reverse implication of [L1] makes $α$ path-homotopic to $ω_{0}$ , which is the required based nullhomotopy by [L3].

L1L2L3

2.1

Steps 1.1 and 1.2 establish both directions of the degree-zero criterion.

step 1.1step 1.2∎

Depends on

Used by

ℝ/ℤ is not simply connected Corollary
Based circle loops with the same endpoints need not be path-homotopic Counterexample
A surjective circle loop can have degree zero and be nullhomotopic Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 42 results over 12 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)