Cohomology ring and cup product can be defined on simplicial complex (ie a triangulation of a manifold). Can we define cohomology ring and cup product on a more general complex? In particulate, I am interested in defining cohomology ring and cup product on a complex and its dual complex. The dual of a simplicial complex, in general, is not a simplicial complex, which causes a problem.



Cohomology ring on non-simplicial complex
If $H^\ast(-,R)$ is cohomology with coefficients in a ring, the cup product may be defined purely via the functoriality of $H^\ast(-,-)$ and certain compatibilities of tensor products. There is a map
$$H^\ast(X,A) \otimes H^\ast(Y,B) \to H^\ast(X \times Y, A \otimes B)$$
coming from taking the tensor product of cochain complexes. Then, when $R$ is a ring, we have the multiplication $m: R \otimes R \to R$ and the diagonal map $\Delta: X \to X \times X$ defined by $\Delta(x) = (x,x)$. Combining these with the above gives a map
$$ H^\ast(X,R) \otimes H^\ast(X,R) \to H^\ast(X\times X, R \otimes R) \to H^\ast(X\times X, R) \to H^\ast (X,R)$$
where the middle map is $m\_\ast$ and the final map is $\Delta^\ast$.
For more details, see May, *A Concise Course in Algebraic Topology*, p. 139.


There is also a construction of cup products on singular chains. See Hatcher, *Algebraic Topology*, §3.2.


