For which topological spaces $X$ can I define an intersection form $b(\cdot, \cdot)$?
I know at least one example: If $X$ is a closed orientable $2n$-manifold then one can define an intersection form by defining $b(c_1, c_2)$ to be the cup product of two representatives of two elements in the $k$-th homology group $H_k$.
But one can generalise and drop the condition of orientability and still define an intersection form: instead of taking an arbitrary ring $R$ of coefficients for the homologies $H_k$ one can consider $R = \mathbb Z / 2 \mathbb Z$ so that orientability becomes irrelevant since $\mathbb Z / 2 \mathbb Z$ has only one generator, $1$.
Reading the article about the cup product though suggests that the cup product can be defined on arbitrary topological spaces. Which would mean that I get a bilinear form for any topological space. (which would contradict this answer I got for one of my previous questions regarding this).
Question 1.1: What am I missing? Why can I not define the cup product on any topological space and therefore get a bilinear form (and therefore an intersection form) on any topological space?
Question 1.2: What property does a manifold have that a topological space does not have to get an intersection form? (surely the answer cannot be orientability since we want to choose coefficients mod two anyway)
Question 2: What I want to do with this is, I want to take my bilinear form on my space and then make it into a quadratic form $q(x) = b(x,x)$, choose a symplectic basis and then compute the Arf invariant of it. Does that give me additional requirements for my space $X$ in order to be able to do that?