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

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De rham cohomology of the circle from mayer vietoris

Example

Under countable choice, $H^{0} (S^{1}) = H^{1} (S^{1}) = R$ and all other de Rham groups vanish.

Facts & Assumptions

Given: Assume countable choice. The unit circle covered by the complements $U, V$ of two opposite points.

[F1]

Mayer vietoris sequence in de rham cohomology: Under countable choice the de Rham Mayer–Vietoris sequence is exact: $\dots \to H^{k} (M) \overset{r^{*}}{\to} H^{k} (U) \oplus H^{k} (V) \overset{s^{*}}{\to} H^{k} (U \cap V) \overset{δ}{\to} H^{k + 1} (M) \to \dots$ , beginning with $0 \to H^{0} (M)$ .

[F2]

Zero th de rham cohomology is locally constant functions: $H_{d R}^{0} (M)$ is the algebra of locally constant real functions. For nonempty connected $M$ it is canonically $R$ .

[F3]

De rham cohomology of a contractible smooth manifold: Under countable choice, a nonempty contractible smooth manifold has $H_{d R}^{0} ≅ R$ and vanishing positive-degree de Rham cohomology.

Verification

technique · direct

1.1

Stereographic projection from the omitted point makes each of $U, V$ diffeomorphic to $R$ . Their overlap has two open-arc components $W_{0}, W_{1}$ , each diffeomorphic to an open interval and hence contractible by linear contraction in that coordinate. F3 makes their positive-degree groups vanish, while F2 identifies their degree-zero groups with constants on components. Thus the initial Mayer–Vietoris segment is $0 \to H^{0} (S^{1}) \to R^{2} \overset{s}{\to} R^{2} \to H^{1} (S^{1}) \to 0$ , where $s (a, b) = (b - a, b - a)$ .

F1F2F3given

2.1

The kernel is ${(a, a)}$ and the image is the diagonal. The functional $(c, d) \mapsto d - c$ has exactly that kernel and is onto, so the cokernel is $R$ . Exactness computes both groups. Degrees above one and negative degrees have zero form spaces.

step 1.1algebra∎

Source locator

Lee, Theorems 17.20–17.21, pp.449–451; the two-component overlap map is calculated explicitly.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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)