TITLE: When does a cohomology theory have a ring structure?
QUESTION [11 upvotes]: I've looked around and I can't quite seem to find an answer to this question. When does a cohomology theory admit a non trivial product structure? I was trying to compute a cohomology ring from a CW structure and I simply couldn't find any reasonable definition for the cup product on CW spaces. Does such a thing exist? If not is there some conditions on a cohomology theory that allows for a natural definition of a cup product?
Obviously you could define some contrived cup product structure via singular or simplicial homology, but what about Cech Cohomology or any of the extraordinary cohomology theories?

REPLY [3 votes]: The ring structure in cohomology arrises in the following way. We have a natural map
$$
\triangle:X\rightarrow X\times X
$$
which sends $x\mapsto (x,x)$ the diagonal. Thus in cohomology, we get a nice map
$$
\triangle^*:H^k(X\times X)\rightarrow H^k(X),
$$
since cohomology is contravariant. Now, for nice cohomology theories one can make sense of a map
$$
P=H^i(X)\times H^j(Y)\rightarrow H^{i+j}(X\times Y)
$$
Taking $X=Y$ one can compose these maps to get a ring structure
$$
\triangle P:H^i(X)\times H^j(X)\rightarrow H^{i+j}(X).
$$
This is explained a bit in Hatchers book http://www.math.cornell.edu/~hatcher/AT/AT.pdf on page 185. So for any cohomology theory you have to be able to make sense of the map $P$.