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Relative degree-four cup products are symmetric

Statement

Let $(X, A)$ be a topological pair and let $R$ be a commutative unital ring. Then the relative cup product of Relative cup product for an excisive triad restricts to a symmetric pairing in degree four: $H^{4} (X, A; R) \times H^{4} (X, A; R) \to H^{8} (X, A; R), a ⌣ b = b ⌣ a .$

Facts & Assumptions

Given: A topological pair $(X, A)$ and a commutative unital ring $R$ .

[L1]

Relative singular cochains $C^{k} (X, A; R)$ are the $R$ -linear functions on $C_{k} (X, A; R) = C_{k} (X; R) / C_{k} (A; R)$ , identified with the cochains on $X$ vanishing on simplices in $A$ ; relative cohomology is their cohomology (Relative singular cochain complex).

[L2]

For $A, B$ open in $U = A \cup B$ the relative cup product is built from the front/back cochain product, which vanishes on $N = C_{*} (A; R) + C_{*} (B; R)$ , followed by the inverse of the comparison isomorphism $q^{*} : H^{*} (X, U; R) \to H^{*} ({Hom}_{R} (C_{*} (X; R) / N, R))$ ; when $A = B$ this comparison is the identity because $N = C_{*} (A; R) = C_{*} (U; R)$ (Relative cup product for an excisive triad).

[L3]

The singular cup product on cochains is the front/back formula $(φ ⌣ ψ) (σ) = φ (σ [0, \dots, p]) ψ (σ [p, \dots, p + q])$ , extended $R$ -linearly, and it is $R$ -bilinear (Singular cup product on cochains).

[L4]

There is a natural chain homotopy $H = K Δ_{#}$ with $d H + H \partial = W D_{X} - D_{X}$ , where $D_{X} = AW Δ_{#}$ and $W (x \otimes y) = (- 1)^{∣ x ∣ ∣ y ∣} y \otimes x$ ; in particular $H$ is natural for continuous maps $X \to Y$ (Factor reversal gives the commutativity chain homotopy).

[L5]

In the absolute case the same primitive proves $a ⌣ b = (- 1)^{p q} b ⌣ a$ for $a \in H^{p} (X; R)$ , $b \in H^{q} (X; R)$ (Singular cohomology is graded commutative).

Proof

technique · direct

1.1L1L2L5given

Taking $A = B$ in [L2], the two factors are open in $U = A$ and $N = C_{*} (A; R) = C_{*} (U; R)$ , so the comparison $q$ is the identity and the relative product $H^{4} (X, A; R) \times H^{4} (X, A; R) \to H^{8} (X, A; R)$ is represented by the front/back product of relative cocycle representatives; this is the relative form of the absolute computation of [L5].

2.1step 1.1L1L3

For relative cocycles $φ \in Z^{4} (X, A; R)$ and $ψ \in Z^{4} (X, A; R)$ , the cochain $φ ⌣ ψ$ vanishes on $C_{*} (A; R)$ : on a simplex $σ$ with image in $A$ the front face $σ [0, \dots, 4]$ also lies in $A$ , so $φ (σ [0, \dots, 4]) = 0$ ; hence $φ ⌣ ψ$ is a well-defined relative $8$ -cochain, and likewise $ψ ⌣ φ$ .

3.1step 2.1L4

Let $i : A \to X$ be the inclusion. By naturality in [L4], $H i_{#} = (i_{#} \otimes i_{#}) H$ on $C_{*} (A; R)$ ; therefore for every chain $c$ in $C_{*} (A; R)$ the chain $H (c)$ lies in $C_{*} (A; R) \otimes_{R} C_{*} (A; R)$ .

4.1step 3.1L1L4

Define the tensor functional $J$ on bidegree $(4, 4)$ tensors by $J (x \otimes y) = φ (x) ψ (y)$ . Then $J d = 0$ : in bidegree $(5, 4)$ it is $δ φ (x) ψ (y) = 0$ and in bidegree $(4, 5)$ it is $(- 1)^{4} φ (x) δ ψ (y) = 0$ . Evaluating the homotopy identity of [L4] gives $J W D_{X} - J D_{X} = J H \partial = δ (J H)$ , and $J H$ vanishes on $C_{*} (A; R)$ by step 3.1 because $φ$ and $ψ$ do.

5.1step 4.1L1L4

The functionals $J D_{X}$ and $J W D_{X}$ also vanish on $C_{*} (A; R)$ : for a simplex $σ$ with image in $A$ , the chains $D_{X} (σ)$ and $W D_{X} (σ)$ are combinations of tensors whose two factors are chains in $A$ , so every evaluation factor $φ (-)$ or $ψ (-)$ vanishes. Hence $J D_{X}$ , $J W D_{X}$ and $δ (J H)$ are relative cochains and the identity of step 4.1 holds in $C^{8} (X, A; R)$ .

6.1step 5.1L3

By [L3], $J D_{X}$ is the cochain $φ ⌣ ψ$ ; on a simplex $σ$ the functional $J W D_{X}$ takes the value $(- 1)^{4 \cdot 4} ψ (σ [0, \dots, 4]) φ (σ [4, \dots, 8]) = ψ (σ [0, \dots, 4]) φ (σ [4, \dots, 8])$ , which is $(ψ ⌣ φ) (σ)$ because $R$ is commutative, so $J W D_{X} = ψ ⌣ φ$ .

7.1step 6.1L1

Therefore $ψ ⌣ φ - φ ⌣ ψ = δ (J H)$ is a coboundary in the relative complex, so the two products agree in $H^{8} (X, A; R)$ ; the same computation with one vanishing condition dropped gives the corresponding relative/absolute symmetry.

8.1step 7.1L1L5∎

Hence the degree-four relative cup product is symmetric on $H^{4} (X, A; R)$ , as asserted; for $A = \emptyset$ this recovers the absolute statement of [L5], for $A = X$ one source group is zero, and for the zero ring all products are zero.

Depends on

Used by

Boundary middle form and boundary signature Definition
The boundary middle form is well defined and glues Lemma

Dependency tree · two levels

22 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

Allen Hatcher, Algebraic Topology, Cambridge University Press 2002 (complete book) (standard reference, not scraped)
Glen E. Bredon, Topology and Geometry, Graduate Texts in Mathematics 139, Chapter VI (cohomology and products) (standard reference, not scraped)