CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

De Rham vector-space comparison with continuous singular cohomology

Statement

Assume ${A C}_{ω}$ . For a finite-dimensional Hausdorff second-countable smooth manifold $M$ , possibly with boundary, let $r_{M} : H_{s i n g}^{k} (M; R) \to H_{\infty}^{k} (M; R)$ be restriction to smooth simplices. The map $J_{M} = r_{M}^{- 1} I_{M} : H_{d R}^{k} (M) ⟶ H_{s i n g}^{k} (M; R)$ is a linear isomorphism in every integer degree, natural for smooth maps. Here $I_{M}$ is integration on smooth singular simplices. No integration over an arbitrary continuous simplex is asserted.

Facts & Assumptions

[F1]

De Rham theorem for smooth singular cohomology gives the natural linear isomorphism $I_{M}$ under countable choice, including boundary manifolds.

[F2]

Smooth and continuous real singular cohomology agree gives the natural linear isomorphism $r_{M}$ , in the displayed direction, under the same assumption.

[F3]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is the choice principle assumed here.

Proof

Given: $M$ as stated, an integer $k$ , and ${A C}_{ω}$ .

1.1

By [F1] and [F2], each class $a \in H_{d R}^{k} (M)$ has a unique class $b \in H_{s i n g}^{k} (M; R)$ with $r_{M} b = I_{M} a$ . Define $J_{M} a = b$ . Uniqueness defines this function without choosing representatives or selecting preimages from non-singleton fibres. Since $r_{M}$ and $I_{M}$ are linear, applying $r_{M}$ to $J_{M} (s a + t c)$ and to $s J_{M} a + t J_{M} c$ gives the same class; injectivity of $r_{M}$ proves linearity. The inverse is $I_{M}^{- 1} r_{M}$ , as both composites reduce to identity.

F1F2F3given

2.1

For a smooth map $f : M \to N$ , write $f_{s i n g}^{*}, f_{\infty}^{*}, f_{d R}^{*}$ for the three pullbacks. The two naturality equalities give $r_{M} f_{s i n g}^{*} J_{N} = f_{\infty}^{*} r_{N} J_{N} = f_{\infty}^{*} I_{N} = I_{M} f_{d R}^{*} = r_{M} J_{M} f_{d R}^{*} .$ Cancel the injective $r_{M}$ to obtain $f_{s i n g}^{*} J_{N} = J_{M} f_{d R}^{*}$ . This proves the contravariant naturality claimed, also for maps whose image lies in a target boundary.

F1F2step 1.1

3.1

Empty manifolds and negative degrees give the unique maps between zero spaces. Degree zero, degree one, dimension zero and top degree are already included in both isomorphisms, so their composite and the cancellation proof apply without new endpoint assumptions. Smooth and continuous complexes retain degenerate simplices; the construction uses their actual restriction map. Countable choice is inherited exactly from the two suppliers' partition, countable-product and globalization arguments. Inverting their bijections requires no further choice, and no full AC is used.

F1F2F3step 1.1step 2.1∎

Depends on

Used by

De Rham cohomology depends only on the underlying homotopy type Corollary
The de Rham theorem Theorem

Dependency tree · two levels

21 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

Peter S. Park, Proof of de Rham's Theorem (standard reference, not scraped)