Cup product

This uses the Alexander-Whitney map

Definition
Let $$X$$ be a topological space and $$R$$ a commutative ring. The cup product can be viewed as a bilinear map:

$$H^i(X;R) \times H^j(X;R) \to H^{i+j}(X;R)$$

or equivalently as a linear map:

$$H^i(X;R) \otimes H^j(X;R) \to H^{i+j}(X;R)$$

defined as follows. Let $$a \in H^i(X;R), b \in H^j(X;R)$$. Pick representing cocycles $$\alpha$$ for $$a$$ and $$\beta$$ for $$b$$. We will now produce an $$(i+j)$$-cocycle.

To do this, let $$s$$ be any $$(i+j)$$-simplex in $$X$$. Then via the diagonal embedding of $$X$$ in $$X \times X$$, $$s$$ becomes an $$(i+j)$$-simplex in $$X \times X$$, and the Alexander-Whitney map then sends $$s$$ to an element of $$Sing_.(X) \otimes Sing_.(X)$$. Look at the component for $$Sing_i(X) \otimes Sing_j(X)$$, and evaluate $$\alpha \otimes \beta$$ on this element. This gives a scalar (element of $$R$$). This scalar is the value on the simplex $$s$$.

It needs to be checked that the cochain defined in this manner is indeed a cocycle, and that its cohomology class is independent of the choices for representatives $$\alpha$$ and $$\beta$$.

The cup product of $$a$$ and $$b$$ is denoted by:

$$a \smile b$$

Importance
The cup product does not depend specifically on the Alexander-Whitney map, but rather on the Alexander-Whitney map upto chain homotopy, and by the theory of acyclic models, there is only one such map upto chain homotopy. Thus, it yields a natural multiplication on the direct sum of all the cohomology groups.

It turns out that this multiplication is associative on the nose for the usual choice of Alexander-Whitney map (for other choices, it is associative only upto homotopy). Also, multiplication is graded-commutative (sometimes called supercommutative) if the ground ring is commutative.