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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Circle power maps are classified by their exponent

Example

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). For $k \in Z$ let $p_{k} : S^{1} \to S^{1}$ , $p_{k} (z) = z^{k}$ (equivalently $p_{k} ([t]) = [k t]$ ), with the counterclockwise orientations. Then $\deg (p_{k}) = k$ , and by the Hopf classification for $M = S^{1}$ the maps $p_{k}$ and $p_{l}$ are homotopic if and only if $k = l$ . Hence the homotopy classes $[S^{1}, S^{1}]$ are in bijection with $Z$ and the class of a power map is determined by its exponent, consistently with the covering-space computation of circle self-maps.

Facts & Assumptions

Given: ${A C}_{ω}$ , the unit circle $S^{1}$ with its counterclockwise orientation and smooth structure, the real line as its universal cover, and the power maps $p_{k}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ , Smooth manifolds and their smooth charts, $R \to R / Z$ is a universal covering).

[F1]

For every $m \in Z$ the power map $P_{m} ([t]) = [m t]$ has $\deg (P_{m}) = m$ ; equivalently the continuous map $p_{d} (z) = z^{d}$ has degree $d$ (Degree of the power map on the circle).

[F2]

For $m \geq 1$ , degree induces a bijection from $[S^{m}, S^{m}]$ to $Z$ , so two maps are homotopic exactly when their degrees agree (Sphere self-maps are homotopic exactly when their degrees agree, Degree of a proper smooth map by compact-support cohomology).

[F3]

For the quotient covering $p : R \to R / Z$ , the explicit map ${\tilde{p}}_{k} (t) = k t$ satisfies $p \circ {\tilde{p}}_{k} = p_{k} \circ p$ and ${\tilde{p}}_{k} (t + 1) - {\tilde{p}}_{k} (t) = k$ ( $R \to R / Z$ is a universal covering). This is an explicit lift of the composite $p_{k} \circ p$ , not a lift $S^{1} \to R$ of $p_{k}$ itself.

Verification

technique · direct

1.1F1F3given

By [F1] the degree of $p_{k}$ is the integer $k$ ; the explicit lift of [F3] has period increment $k$ , so this increment agrees with the degree already computed by [F1].

1.2F1F2

Let $k, l \in Z$ . If $k = l$ then $p_{k} = p_{l}$ ; conversely if $p_{k}$ and $p_{l}$ are homotopic, [F2] with $m = 1$ gives $k = \deg (p_{k}) = \deg (p_{l}) = l$ , so the two maps are homotopic exactly when their exponents agree.

2.1F2F3step 1.1step 1.2∎

Hence the assignment $k \mapsto [p_{k}]$ is a bijection from $Z$ to $[S^{1}, S^{1}]$ , inverse to the degree, and a power map is determined up to homotopy by its exponent alone; this matches the covering-space description.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

33 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

Victor Guillemin and Alan Pollack, Differential Topology (standard reference, not scraped)
John Milnor, Topology from the Differentiable Viewpoint (standard reference, not scraped)
Daniel S. Freed, Bordism: Old and New (lecture notes, UT Austin, Fall 2012) (standard reference, not scraped)