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I have some confusion that i would like to clarify. The product in cohomology is not the cup product $\smile$ but it is another product $*$ that is constructed from cup product. Indeed wrinte $H^\bullet(X;R) = \bigoplus_{k\in\mathbb{N}} H^k(X; R)$ then an element $x$ in $H^\bullet(X;R)$ is of the form $x=x_0+x_1+x_2+\cdots $ where each $x_i\in H^i(X;R)$ hence the product of two elements $x$ and $y$ in $H^*(X;R)$ is a mutliplication $*$ defined via cup product as follows (we take an example $x=x_1+x_2$ and $y=y_2+y_3$)

$x*y=(x_1+x_2)*(y_2+y_3)=x_1\smile y_2 + x_1\smile y_3+ x_2\smile y_2+x_2\smile y_3$ is this correct?

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Yes. (filler..) – Gunnar Þór Magnússon Dec 13 '12 at 14:11

The multiplication on the left-hand side is the multiplication in the ring $H^\ast(X)$. Elements in this ring consist of linear combinations of cocycles of certain degrees. So we expect the multiplication to take in two such elements and return another, just as in your example.

The cup product $\cup$ is not a ring multiplication. Given $p,q$, it's a bilinear pairing on abelian groups: $H^p(X)\times H^q(X)\to H^{p+q}(X)$. It takes in a p-cocycle and a q-cocycle and spits out a (p+q)-degree cocycle.

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