I need help fixing a broken example I've come up with. In particular, I wanted to use the Arf invariant to distinguish two non-homeomorphic surfaces. That's the first part that's broken since there aren't any surfaces that have the same homology groups but aren't homeomorphic. So I gain nothing because the homology groups already let me distinguish two surfaces.
Concretely, what I was trying to do was this: observe that the intersection number defines a bilinear pairing (an intersection form in this case) and then compute the Arf invariant of $q(x)=\langle x,x \rangle$. Stupidly, for orientable surfaces, this always gives me $q(x)=\langle x,x \rangle = 0$ and hence $\mathrm{Arf}(q) = 0$.
On the other hand, for $\mathbb R P^2$, there is only one basis vector hence I can't even define the Arf invariant (since I can't define an intersection form since that would need even dimension). So this is the second part that is broken.
Now I'm stuck. How can I fix this? What is the simplest example of two manifolds (or topological spaces if you like) that I can distinguish using the Arf invariant? Thanks for help.