ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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The trivial principal bundle has a nullhomotopic classifying map

Claim

Assume AC. Let $G$ be a well-pointed topological group of CW type and let $X$ be a CGWH space. Under the classification of numerable principal $G$ -bundles over $X$ , the product bundle $X \times G \to X$ corresponds to the constant homotopy class in $[X, B G]$ . Consequently a numerable principal $G$ -bundle over $X$ is trivial if and only if its classifying map is nullhomotopic.

Facts & Assumptions

[F1]

Assuming AC, for $G$ well-pointed of CW type and $X$ CGWH, pullback along homotopic maps gives isomorphic numerable principal bundles, and pullback gives a bijection from $[X, B G]$ to their isomorphism classes (Numerable principal bundles are classified by maps to BG).

[F2]

The fiber of $E G \to B G$ over $b_{0}$ is the right $G$ -torsor ${e_{0} g : g \in G}$ (Milnor's infinite-join model of EG).

Verification

Given: AC, $G$ and $X$ as in the Claim, and the constant map $c : X \to B G$ with value $b_{0}$ .

1.1

The constant pullback has total space

$c^{*} E G = {(x, e) : p (e) = b_{0}} ⟶ X$

is equivariantly isomorphic to $X \times G$ by $(x, g) \mapsto (x, e_{0} g)$ . Thus the constant homotopy class maps to the trivial bundle. [F2]

2.1

If $f$ is nullhomotopic, [F1] gives $f^{*} E G ≅ c^{*} E G$ , so its pullback is trivial. Conversely, if $f^{*} E G$ is trivial, then it has the same bundle class as $c^{*} E G$ ; injectivity of the classification bijection gives $[f] = [c]$ . This proves both directions, including disconnected $X$ because the constant map uses the same based orbit on every component. $□$

F1step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Dale Husemoller, Fibre Bundles, Third Edition (standard reference, not scraped)