CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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Cohomology over a field is dual to homology for finite-dimensional complexes

Statement

For a chain complex of finite-dimensional vector spaces over $k$ , evaluation gives $H^{n} {Hom}_{k} (C, k) ≅ {Hom}_{k} (H_{n} C, k)$ .

Proof

Given: A chain complex $C$ of finite-dimensional vector spaces over $k$ , with $Z_{n} = \ker d_{n}$ , $B_{n} = im d_{n + 1}$ , and $H_{n} C = Z_{n} / B_{n}$ .

[L1] Evaluation sends the cohomology class of a cocycle $f : C_{n} \to k$ to the functional $[z] \mapsto f (z)$ on $H_{n} C$ (The evaluation map from cohomology to Hom of homology).

[L2] In a finite-dimensional vector space, every basis of a subspace extends to a basis of the whole space by finitely many exchanges; hence a linear functional on a subspace extends to the whole space by assigning zero on the added basis vectors (The Steinitz exchange lemma: if $L \subseteq V$ is linearly independent and $S \subseteq V$ spans $V$ with $S$ finite of size $n$ , then $L$ is finite with $∣ L ∣ = m \leq n$ , and there is $T \subseteq S$ of size $n - m$ such that $L \cup T$ spans $V$ ).

1.1givenL1algebra

Fix $n$ . A cochain $f : C_{n} \to k$ is a cocycle exactly when $f (B_{n}) = 0$ , because its coboundary is $f \circ d_{n + 1}$ . Such an $f$ restricts to $Z_{n}$ and factors through $Z_{n} / B_{n} = H_{n} C$ , giving the evaluation map [L1]. A coboundary $h \circ d_{n}$ vanishes on $Z_{n}$ , so evaluation is well defined on cohomology classes.

2.1L2step 1.1construct

Given $λ \in {Hom}_{k} (H_{n} C, k)$ , compose it with $Z_{n} ↠ H_{n} C$ to get a functional on $Z_{n}$ that vanishes on $B_{n}$ . Extend it to $f : C_{n} \to k$ by [L2]. Then $f (B_{n}) = 0$ , so $f$ is a cocycle, and its evaluation is $λ$ . Thus evaluation is surjective.

2.2L2step 1.1construct

Suppose a cocycle $f$ evaluates to zero. Then $f ∣_{Z_{n}} = 0$ . Define $h : B_{n - 1} = im d_{n} \to k$ by $h (d_{n} x) = f (x)$ ; if $d_{n} x = d_{n} x^{'}$ , then $x - x^{'} \in Z_{n}$ , so this is well defined. Extend $h$ to $\tilde{h} : C_{n - 1} \to k$ by [L2]. Now $f = \tilde{h} \circ d_{n}$ , so $f$ is a coboundary and its cohomology class is zero. Thus evaluation is injective.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 show that the evaluation map is a bijective linear map, hence the claimed isomorphism. Each basis extension involved only the two fixed finite-dimensional spaces $C_{n}$ and $C_{n - 1}$ ; no choice principle or general UCT is used.

Depends on

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Sources

Weibel, An Introduction to Homological Algebra (standard reference, not scraped)