Math 2.0

Algebraic topology Definition

Cup product

Builds on Singular cohomologyCommutative ring

Definition

Let XX be a topological space and let RR be a commutative ring with 11. For a singular simplex σ ⁣:Δn→X\sigma \colon \Delta^{n} \to X with vertices v0,…,vnv_0, \ldots, v_n, write σ∣[vi,…,vj]\sigma|_{[v_i, \ldots, v_j]} for its restriction to the face spanned by vi,…,vjv_i, \ldots, v_j. For cochains φ∈Ck(X;R)\varphi \in C^k(X; R) and ψ∈Cl(X;R)\psi \in C^l(X; R), the cup product φ⌣ψ∈Ck+l(X;R)\varphi \smile \psi \in C^{k+l}(X; R) is (φ⌣ψ)(σ)=φ(σ∣[v0,…,vk])⋅ψ(σ∣[vk,…,vk+l])(\varphi \smile \psi)(\sigma) = \varphi\big(\sigma|_{[v_0, \ldots, v_k]}\big) \cdot \psi\big(\sigma|_{[v_k, \ldots, v_{k+l}]}\big) for every singular simplex σ ⁣:Δk+l→X\sigma \colon \Delta^{k+l} \to X. It satisfies δ(φ⌣ψ)=δφ⌣ψ+(−1)kφ⌣δψ\delta(\varphi \smile \psi) = \delta\varphi \smile \psi + (-1)^k \varphi \smile \delta\psi. Hence it induces a bilinear map ⌣ ⁣:Hk(X;R)×Hl(X;R)→Hk+l(X;R)\smile \colon H^k(X; R) \times H^l(X; R) \to H^{k+l}(X; R) on singular cohomology, also called the cup product.

Remarks

The cup product on cohomology is associative, the class of the constant 00-cochain 11 is a unit, and a⌣b=(−1)kl b⌣aa \smile b = (-1)^{kl}\, b \smile a for a∈Hka \in H^k and b∈Hlb \in H^l. Graded commutativity holds on cohomology but not on cochains. It is natural: f∗(a⌣b)=f∗(a)⌣f∗(b)f^*(a \smile b) = f^*(a) \smile f^*(b) for a continuous map ff. The same formula gives relative products such as Hk(X;R)×Hl(X,A;R)→Hk+l(X,A;R)H^k(X; R) \times H^l(X, A; R) \to H^{k+l}(X, A; R).

Example: for the torus T2=S1×S1T^2 = S^1 \times S^1, H1(T2;ℤ)≅ℤ2H^1(T^2; \Z) \cong \Z^2 with generators α\alpha and β\beta dual to the two circles. Then α⌣β\alpha \smile \beta generates H2(T2;ℤ)≅ℤH^2(T^2; \Z) \cong \Z, and α⌣α=0\alpha \smile \alpha = 0.

Non-example: the pointwise product (φψ)(σ)=φ(σ)ψ(σ)(\varphi\psi)(\sigma) = \varphi(\sigma)\psi(\sigma) only makes sense when φ\varphi, ψ\psi and σ\sigma all have the same dimension. So it does not define a product Ck×Cl→Ck+lC^k \times C^l \to C^{k+l}. The cup product avoids this by splitting each simplex into a front face and a back face.

Reading path

To understand this page from the ground up, read these 3 pages in order. The foundations come first.

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  1. Commutative ring concept
  2. Singular cohomology concept
  3. Cup product definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Cup_product
About this page
TagW10ERK1ADE
URIm2://W10ERK1ADE
Kinddefinition
WritingW0 stub
MSC55N45
DefinesCup product
UsesSingular cohomology, Commutative ring
Changed2 November 2026

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