False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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De rham cohomology is a covariant functor

Statement

False claim: the pullback construction makes de Rham cohomology covariant.

Facts & Assumptions

Given: On the discrete three-point manifold $X = {1, 2, 3}$ , let $F$ swap $1, 2$ and $G$ swap $2, 3$ . Let $u$ be the indicator of ${1}$ .

[F1]

De rham cohomology is a contravariant functor: De Rham cohomology is contravariant: for smooth $F : M \to N$ and $G : N \to P$ , $(G \circ F)^{*} = F^{*} \circ G^{*}$ , and ${id}_{M}^{*} = {id}_{H^{k} (M)}$ .

Refutation

technique · direct

1.1

Every function on $X$ is smooth and closed, with no nonzero degree-zero boundaries, so $u$ represents itself in $H^{0}$ . Pullback is composition. Hence $(F^{*} G^{*} u) (2) = u (G (F (2))) = u (1) = 1$ , while $(G^{*} F^{*} u) (2) = u (F (G (2))) = u (3) = 0$ .

F1given

2.1

Thus these pullback operators do not commute, and $(G \circ F)^{*} = F^{*} G^{*}$ cannot be replaced by $G^{*} F^{*}$ . In general $F : M \to N$ induces $F^{*} : H^{∙} (N) \to H^{∙} (M)$ , with the reversed source and target, exactly as the contravariant functor theorem states.

F1step 1.1∎

Source locator

Lee, Proposition 17.2(a) and Corollary 17.3, p.442; explicit noncommuting permutation pullbacks supply the witness.

Depends on

De rham cohomology is a contravariant functor

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)