Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-14
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Vector bundles are glued from transition cocycles

Statement

Let (Ui)iI be an open cover of X, and let gji:UiUjGLn(F) be continuous maps such that

gii=I,gki=gkjgji.

The quotient of iUi×Fn by (x,v,i)(x,gji(x)v,j) is a rank-n F-vector bundle. Replacing gji by gji=hjgjihi1 for continuous hi:UiGLn(F) gives an isomorphic bundle. Every rank-n bundle is recovered from the cocycle of any linear atlas.

Facts & Assumptions

Given: The cover and cocycle in the statement.

[F1]

A vector bundle is locally a product by fiberwise-linear charts, and its transition order is gki=gkjgji (Real and complex topological vector bundles).

Proof

technique · direct
1.1

The cocycle with k=i gives gijgji=I. Hence the displayed relation is reflexive, symmetric, and transitive: the transitive calculation is gkj(gjiv)=gkiv. It therefore defines a quotient q:iUi×FnE and a map p:EX by p[x,v,i]=x. Since pq(x,v,i)=x is continuous on every summand, the quotient property [F2] makes p continuous.

F2givenalgebra
2.1

The quotient map q is open. Indeed, if O is open in the coproduct, then the part of its saturation in the jth summand is the union over k of the images of O((UjUk)×Fn×{k}) under the homeomorphism (x,v,k)(x,gjk(x)v,j); its inverse uses gkj=gjk1. Hence every such part is open. The restriction of q over the saturated open set p1(Ui) is therefore again a quotient map. Define Φi[x,v,k]=(x,gik(x)v). The cocycle makes this independent of the representative, and its composite with the restricted quotient map is continuous on every summand, so [F2] makes Φi continuous. Its inverse is (x,w)[x,w,i] and is continuous as the ith-summand inclusion followed by q. Thus Φi:p1(Ui)Ui×Fn is a fiberwise-linear chart. Its overlap from i to j is gji, so [F1] proves that E is the claimed bundle.

F1F2step 1.1
3.1

For the primed cocycle, the maps on summands (x,v,i)[x,hi(x)v,i] respect the relation because gjihi=hjgji. By [F2] they descend to a continuous fiberwise-linear map EE. Replacing hi by hi1 gives its continuous inverse, so it is a bundle isomorphism.

F2step 2.1algebra
4.1

Finally, a linear atlas of a rank-n bundle supplies the functions gji and their cocycle law by [F1]. Sending the quotient class [x,v,i] to ϕi1(x,v) is well-defined, continuous by [F2], and in each chart is the identity map on Ui×Fn. It is therefore a bundle isomorphism from the reconstructed quotient to the original bundle.

F1F2step 2.1

Depends on

Used by

Dependency tree · two levels

10 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