Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-27
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.

The subsheaf generated by a family of sections

Statement

Let X be a topological space, let F be a sheaf of abelian groups on X (A sheaf on a topological space), and let (sα∈F(Uα))α∈A be a family of sections indexed by a set A, each Uα⊆X open. For x∈X let Gx:=⟨ (sα)x:α∈A with x∈Uα ⟩ ≤ Fx be the subgroup of the stalk generated by the germs of the generators that are defined at x (The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups, Germs of sections). A subsheaf of abelian groups means a subsheaf of the underlying sheaf of sets (Subsheaves) whose sections over every open set form a subgroup of the ambient section group, with the inherited operations. Then:

  1. the prescription ⟨sα⟩(W):={ s∈F(W):sx∈Gx for every x∈W },W⊆X open, defines a subsheaf of abelian groups ⟨sα⟩⊆F, called the subsheaf generated by the family (sα);
  2. sα∈⟨sα⟩(Uα) for every α∈A;
  3. ⟨sα⟩x=Gx for every x∈X;
  4. if H⊆F is a subsheaf of abelian groups with sα∈H(Uα) for every α, then ⟨sα⟩⊆H; conversely, if H⊆F is a subsheaf with Hx=Gx for every x∈X, then H=⟨sα⟩;
  5. if (tβ∈F(Vβ))β∈B is a further family of sections with tβ∈⟨sα⟩(Vβ) for every β∈B, then the subsheaf generated by the combined family equals ⟨sα⟩.

The construction is functorial in the family in the evident way and uses no choice principle. In particular, when A is finite and each Uα is a compact open set, the subsheaf generated by the sα is generated by finitely many sections over compact opens.

Facts & Assumptions

[F1]

A subsheaf H of a sheaf G may be identified with a choice of a subset H(U)⊆G(U) for every open U, with restriction maps inherited from G, such that these subsets define a sheaf (Subsheaves).

[F2]

A sheaf is precisely a presheaf whose compatible local sections glue uniquely (A sheaf on a topological space).

[F3]

Elements of the stalk Fx are equivalence classes of pairs (U,s) with U an open neighbourhood of x and s∈F(U), where (U,s)∼(V,t) when s and t agree on some smaller open neighbourhood of x; for a sheaf of abelian groups the stalk is the filtered colimit of the section groups over the neighbourhoods of x (The stalk of a presheaf at a point).

[F4]

The germ of a section is its class in the stalk, and if V⊆U is open and contains x then (s∣V)x=sx (Germs of sections).

[F5]

For a subset S of a group G, the subgroup ⟨S⟩ generated by S contains S and is contained in every subgroup of G that contains S; it is the smallest such subgroup and ⟨∅⟩={e} (The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups).

Proof

Given: A topological space X, a sheaf of abelian groups F on X, a family of sections sα∈F(Uα) over open sets, the subgroups Gx≤Fx generated by the germs of the generators defined at x, and the prescription H(W)={s∈F(W):sx∈Gx for all x∈W}.

1.1

Define H(W):={s∈F(W):sx∈Gx for every x∈W} for every open W⊆X, the prescription of assertion 1. Each H(W) is a subgroup: germs preserve addition and zero (compute these on representatives), and each Gx is a subgroup, so zero belongs to H(W) and s−t belongs whenever s,t do. This is a subpresheaf of abelian groups of F: for W′⊆W and s∈H(W) one has (s∣W′)x=sx for x∈W′ [F4], so s∣W′∈H(W′). [F4, given]

F4
1.2

Let H⊆F be a subsheaf and let s∈F(W) be a section over an open W such that for every x∈W the germ sx lies in the image of the germ map H(Wx)→Fx for some open neighbourhood Wx⊆W of x. Consider the family W of all pairs (V,h) with V⊆W open and h∈H(V)⊆F(V) such that h=s∣V; this is a family of sections indexed by a set and defined by a condition, and for every x∈W there is a member whose first component contains x: by hypothesis there are h∈H(Wx) and a smaller open neighbourhood V⊆Wx of x on which h and s agree [F3], and the restriction h∣V∈H(V) satisfies h∣V=s∣V, so (V,h∣V)∈W. Any two members (V,h) and (V′,h′) of W are compatible on V∩V′, because both agree there with s. By [F2] the family (h)(V,h)∈W glues to a section h0∈H(W)⊆F(W), and h0∣V=h=s∣V for every member (V,h); since the first components of the members cover W, locality in F [F2] gives h0=s, hence s∈H(W). [F2, F3]

F2F3construct
2.1

Let W=⋃i∈IWi be an open cover and let si∈H(Wi) be compatible on the overlaps in the sense of [F2]. Each si is a section of F over Wi and the si are compatible sections of F, so by [F2] there is s∈F(W) with s∣Wi=si for all i. For x∈W pick i with x∈Wi; then sx=(s∣Wi)x=(si)x∈Gx by [F4] and the definition of H(Wi), hence s∈H(W). Locality of H is inherited from F, so H satisfies the sheaf condition and is a subsheaf of F by [F1]; this proves assertion 1. [F1, F2, F4, step 1.1, given]

F1F2F4step 1.1
2.2

Let α∈A and x∈Uα. The germ (sα)x is one of the generators of the subgroup Gx, so (sα)x∈Gx by [F5]; hence sα∈H(Uα) by the definition of H, which proves assertion 2.

F5step 1.1
2.3

Fix x∈X. An element of Hx is the class of a section s∈H(W) over an open neighbourhood W of x [F3], and its image under the germ map into Fx is sx, which lies in Gx by the definition of H(W); two sections of H over neighbourhoods of x with the same germ in Fx agree on a smaller neighbourhood [F3] and therefore define the same element of Hx, so this is well defined and Hx⊆Gx inside Fx. Conversely let g∈Gx. The finite integer combinations of the generating germs form a subgroup containing those germs, and every subgroup containing them contains all such combinations. Thus [F5] identifies this subgroup with Gx, and g is a finite integer combination g=∑jnj(sαj)x of germs of generators with x∈Uαj, 1≤j≤m. Put W:=⋂jUαj (with W=X for the empty combination), an open neighbourhood of x, and t:=∑jnj sαj∣W∈F(W). The stalk is the filtered colimit of the section groups over neighbourhoods of x [F3], so the germ of an integer combination of sections over a common open set is the corresponding integer combination of germs, and tx=g. For every y∈W one has y∈Uαj for all j, so ty=∑jnj(sαj)y∈Gy; hence t∈H(W) and g=tx∈Hx. Therefore Hx=Gx inside Fx, which proves assertion 3.

F3F5step 1.1
2.4

Let H⊆F be a subsheaf of abelian groups with sα∈H(Uα) for every α∈A. For x∈X and α with x∈Uα the germ (sα)x∈Fx is the image of the section sα∈H(Uα) under the germ map [F4], so Gx⊆Hx because Hx is a subgroup of Fx containing all these germs [F5]. Now let s∈H(W). For every x∈W the germ sx lies in Gx⊆Hx, so the hypothesis of [step 1.2] holds for H and s, and [step 1.2] gives s∈H(W). Hence H⊆H as subsheaves of F by [F1], which is the first half of assertion 4.

F1F4F5step 1.2step 1.1
2.5

Conversely let H⊆F be a subsheaf with Hx=Gx for every x∈X. If s∈H(W) for an open W, then sx∈Hx=Gx for every x∈W, so s∈H(W) and H⊆H by [F1]. For the reverse inclusion, if s∈H(W), then every germ sx belongs to Gx=Hx, so [step 1.2] gives s∈H(W). Hence H⊆H, and H=H, proving the second half of assertion 4 even when H is initially only a subsheaf of sets.

F1step 1.2step 1.1
3.1

Let (tβ∈F(Vβ))β∈B be a further family with tβ∈H(Vβ) for every β, and let Gx′≤Fx be the subgroup generated by the germs at x of the combined family. For x∈X one has Gx⊆Gx′, and every new generator (tβ)x with x∈Vβ lies in Gx because tβ∈H(Vβ); hence Gx′=Gx for every x. The subsheaf H:=H satisfies Hx=Gx=Gx′ for every x by [step 2.3], so [step 2.5] applied to the combined family gives that the subsheaf generated by the combined family equals H, which is assertion 5. ∎

step 2.3step 2.5step 1.1

Depends on

Used by

Dependency tree · two levels

13 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