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 $Sin{g}_{.}(X)\otimes Sin{g}_{.}(X)$. Look at the component for $Sin{g}_i(X)\otimes Sin{g}_j(X)$, 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 $\beta$.

Importance

Further information: Cohomology ring of a topological space

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.