TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-05

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A sequence of abelian sheaves is exact exactly when it is exact on every stalk

Statement

Let

\dots ⟶ F_{i - 1} \overset{d_{i - 1}}{\to} F_{i} \overset{d_{i}}{\to} F_{i + 1} ⟶ \dots

be a sequence of sheaves of abelian groups on a topological space $X$ . Then the sequence is exact if and only if, for every point $x \in X$ , the stalk sequence

\dots ⟶ (F_{i - 1})_{x} \overset{(d_{i - 1})_{x}}{\to} (F_{i})_{x} \overset{(d_{i})_{x}}{\to} (F_{i + 1})_{x} ⟶ \dots

is exact.

Facts & Assumptions

Given: A sequence of sheaves of abelian groups on $X$ .

[F1]

Exactness of a sequence of sheaves means exactness in the abelian category of sheaves, so at each middle term the image sheaf equals the kernel sheaf (Exact sequences of sheaves).

[F2]

Stalks are germs of neighbourhood sections (The stalk of a presheaf at a point).

[L1]

Kernels are objectwise, while images and cokernels are obtained by sheafifying the corresponding presheaves (Kernel sheaves are objectwise, while cokernels and images are sheafified).

[L2]

Sheafification preserves stalks (Sheafification preserves stalks).

[L3]

A morphism of sheaves is an isomorphism exactly when it is so on every stalk (A morphism of sheaves is an isomorphism exactly when it is an isomorphism on every stalk).

Proof

technique · direct

1.1F1F2L1L2given

Assume the sheaf sequence is exact. At each middle term $F_{i}$ , [F1] says that the canonical map $im (d_{i - 1}) \to \ker (d_{i})$ is an isomorphism. Using [F2] for the stalk construction and then [L1] and [L2], its stalk at $x$ is the canonical map $im ((d_{i - 1})_{x}) ⟶ \ker ((d_{i})_{x}),$ so the stalk sequence is exact at $x$ . Since $x$ was arbitrary, the whole stalk sequence is exact.

1.2F2L1L2L3given

Conversely, assume every stalk sequence is exact. For each middle term, consider the canonical morphism $im (d_{i - 1}) \to \ker (d_{i})$ . Using [F2] for the stalk construction and then [L1] and [L2], its stalk at $x$ is $im ((d_{i - 1})_{x}) ⟶ \ker ((d_{i})_{x}),$ which is an isomorphism by the stalkwise exactness hypothesis. Therefore [L3] implies that $im (d_{i - 1}) \to \ker (d_{i})$ is an isomorphism of sheaves.

2.1F1step 1.1step 1.2∎

By [F1], step 1.2 is exactly the assertion that the original sequence is exact. Together with step 1.1, this proves the equivalence.

A sequence of abelian sheaves is exact exactly when it is exact on every stalk

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources