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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-05
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Global sections are left exact but need not preserve epimorphisms

Statement

Let X be a topological space. If

0→F′→F→F′′

is exact in sheaves of abelian groups on X, then

0→Γ(X,F′)→Γ(X,F)→Γ(X,F′′)

is exact. However, there exists an epimorphism of sheaves of abelian groups on some space X whose induced map on global sections is not surjective.

Facts & Assumptions

Given: A topological space X.

[F1]

Global sections are sections over the whole space: Γ(X,F)=F(X) (Sections, restrictions, and global sections of a presheaf).

[L2]

Exactness of sheaves means image equals kernel at each interior term (Exact sequences of sheaves).

Proof

technique · direct
1.1F1L1L2given

Suppose 0→F′→F→F′′ is exact. By [L1], the kernel of the map Γ(X,F)→Γ(X,F′′) is the section group of the kernel sheaf at X. By [L2], that kernel sheaf is the image of F′→F, so its sections over X are exactly the image of Γ(X,F′)→Γ(X,F). Together with [F1], this is the displayed left exactness.

1.2F1construct

For the failure of surjectivity, let X=S1, let C be the sheaf of continuous real-valued functions on S1, and let Z‾ be the subsheaf of locally constant integer-valued functions. Let Q be the quotient sheaf obtained by sheafifying the presheaf V↦C(V)/Z‾(V). Every germ of a section of Q is represented locally by a continuous real-valued function, so the canonical map C→Q is locally surjective and hence an epimorphism of sheaves.

1.3constructcontradiction

Cover the circle by the arcs U0=S1∖{(−1,0)} and U1=S1∖{(1,0)}. Choose continuous angle functions θ0:U0→(−1/2,1/2) and θ1:U1→(0,1) with e2πiθi(z)=z. On one connected component of U0∩U1 the difference θ1−θ0 is 0, and on the other it is 1, so the classes of θ0 and θ1 agree in the quotient presheaf and glue to a global section q∈Γ(X,Q). If q had a lift g∈Γ(X,C), then g−θi would be an integer-valued continuous function on the connected arc Ui, hence a constant ni. On the two components of U0∩U1 this would force n0−n1 to be both 0 and 1, a contradiction. Thus Γ(X,C)→Γ(X,Q) is not surjective.

2.1step 1.1step 1.3∎

Step 1.1 gives left exactness, and step 1.3 gives an epimorphism whose global sections map is not surjective.

Depends on

Used by

Dependency tree · two levels

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Sources