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The long exact sequence in cohomology

Statement

Let $0 \to A^{∙} \to B^{∙} \to C^{∙} \to 0$ be a short exact sequence of cochain complexes in an abelian category. Then there is a natural exact sequence $\dots \to H^{n} (A) \to H^{n} (B) \to H^{n} (C) \overset{\partial^{n}}{\to} H^{n + 1} (A) \to H^{n + 1} (B) \to H^{n + 1} (C) \to \dots .$

Facts & Assumptions

Given: A short exact sequence $0 \to A^{∙} \to B^{∙} \to C^{∙} \to 0$ of cochain complexes.

[L1]

A cochain complex may be read as a reindexed chain complex with $(C^{♯})_{n} = C^{- n}$ (Cochain complex in an abelian category).

[L2]

The cohomology object $H^{n} (C)$ is the homology of that reindexed chain complex in degree $- n$ (Cohomology object of a cochain complex).

[L3]

Short exact sequences of chain complexes carry long exact sequences in homology (The long exact sequence in homology).

Proof

technique · direct

1.1L1L3givenconstruct

By [L1], the given short exact sequence of cochain complexes is the same data as a short exact sequence of reindexed chain complexes. Applying [L3] gives a long exact sequence in homology for those chain complexes.

2.1L1L2step 1.1algebra∎

Replace each term $H_{- n} (X^{♯})$ by $H^{n} (X)$ using [L2]. Under the same reindexing, the connecting map from degree $- n$ homology to degree $- n - 1$ homology becomes a map $\partial^{n} : H^{n} (C) \to H^{n + 1} (A) .$ This is the displayed long exact cohomology sequence.

The long exact sequence in cohomology

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources