CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A sequence of chain maps is exact exactly when it is exact degreewise

Statement

Let $A_{∙} \overset{u}{\to} B_{∙} \overset{v}{\to} C_{∙}$ be composable chain maps in an abelian category. The sequence is exact at $B_{∙}$ in $Ch (A)$ if and only if for every $n \in Z$ the sequence $A_{n} \overset{u_{n}}{\to} B_{n} \overset{v_{n}}{\to} C_{n}$ is exact at $B_{n}$ in $A$ .

Facts & Assumptions

Given: Composable chain maps $A_{∙} \overset{u}{\to} B_{∙} \overset{v}{\to} C_{∙}$ .

[L1]

$Ch (A)$ is abelian (The category of complexes in an abelian category is abelian).

[L2]

Exactness at a node means equality of the image and kernel subobjects (Exactness at a node).

[L3]

Images and kernels of chain maps are computed degreewise (Images and coimages of chain maps are computed degreewise).

Proof

technique · direct

1.1L1L2L3

If the sequence is exact at $B_{∙}$ in $Ch (A)$ , then [L2] says $im (u) = \ker (v)$ as subobjects of $B_{∙}$ . By [L3], their $n$ th components are $im (u_{n})$ and $\ker (v_{n})$ , so [L2] gives exactness of $A_{n} \to B_{n} \to C_{n}$ in every degree.

2.1L2L3∎

Conversely, if every degreewise sequence is exact, then [L2] gives $im (u_{n}) = \ker (v_{n})$ inside $B_{n}$ for every $n$ . By [L3], that is exactly the degreewise comparison between the image and kernel complexes, so [L2] makes the sequence exact at $B_{∙}$ in $Ch (A)$ .

A sequence of chain maps is exact exactly when it is exact degreewise

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources