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Disk-pair cohomology over an arbitrary commutative ring

Statement

For every commutative ring $R$ and $n \geq 0$ , $H^{q} (D^{n}, S^{n - 1}; R) ≅ {\begin{cases} R & q = n, \\ 0 & q \neq n . \end{cases}$ where $S^{- 1} = \emptyset$ . The iterated connecting maps, with the ordered coordinate orientation, normalize the element corresponding to $1_{R}$ . Every linear automorphism of the disk pair acts on the top group by multiplication by a unit. The calculation is choice-free.

Facts & Assumptions

Given: A commutative ring $R$ and $n \geq 0$ .

[F1]

Singular cohomology satisfies the Eilenberg Steenrod cohomology axioms gives choice-free homotopy invariance, dimension, exactness, excision, and finite additivity for singular cohomology.

[F2]

Mayer vietoris sequence in singular cohomology gives the natural two-open Mayer–Vietoris sequence.

[F3]

Long exact sequence of a pair in singular cohomology gives the natural disk/sphere pair sequence.

Proof

technique · direct suspension recursion and the pair sequence

1.1F1

For $n = 0$ , $(D^{0}, S^{- 1}) = (*, \emptyset)$ , so [F1]'s dimension axiom gives $R$ in degree zero and zero in every other degree. Its distinguished element is $1_{R}$ .

1.2F1

The two points of $S^{0}$ have $H^{0} (S^{0}; R) = R \oplus R$ by finite additivity. The reduced group is the cokernel of the diagonal constants and is freely generated by $(1, 0)$ modulo the diagonal, hence is $R$ ; all other reduced groups vanish.

2.1F1F2step 1.2

Suppose $m \geq 1$ and the reduced cohomology of $S^{m - 1}$ is $R$ in degree $m - 1$ and zero otherwise. Cover $S^{m}$ by two slightly enlarged hemispheres. Each is contractible and their intersection deformation retracts onto the equator $S^{m - 1}$ . In the Mayer–Vietoris sequence [F2], the map on degree-zero constants is the diagonal-difference map and the higher groups of the hemispheres vanish. Exactness therefore makes the connector an isomorphism ${\tilde{H}}^{q - 1} (S^{m - 1}; R) ≅ {\tilde{H}}^{q} (S^{m}; R)$ in every $q$ . Thus the asserted sphere calculation propagates from step 1.2.

3.1F1F2F3step 1.1step 1.2step 2.1

For $n \geq 1$ , $D^{n}$ is contractible and $S^{n - 1}$ is nonempty. The pair sequence [F3], together with the sphere calculation of step 2.1, identifies $H^{q} (D^{n}, S^{n - 1}; R)$ naturally with ${\tilde{H}}^{q - 1} (S^{n - 1}; R)$ . It is therefore $R$ exactly at $q = n$ and zero otherwise. Following $1_{R}$ through the ordered hemisphere connectors fixes the stated coordinate generator; reversing one ordered coordinate reverses the first difference and hence negates that generator.

4.1F1F2F3step 1.1step 1.2step 2.1step 3.1∎

A linear automorphism is a homeomorphism of the pair, so functoriality in [F1] makes its top-degree action an $R$ -module automorphism of the rank-one module calculated in step 3.1. Such an automorphism is multiplication by a unique unit: the image of $1$ is $u$ , and the inverse image supplies $v$ with $u v = 1$ . At $n = 0$ this is the identity. Empty boundary, zero ring, ranks zero and one, both hemisphere endpoints, repeated/degenerate cover pieces, and the two possible coordinate signs are all included above. Only a fixed finite cover and its canonical maps occur, so neither arbitrary additivity nor AC is used.

Depends on

Dependency tree · 2 levels

Sources

May, A Concise Course in Algebraic Topology, Chapter 23 §5