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Oct 20, 2010 at 23:31 comment added Sean Tilson So suppose you have your space $X$ and call $G=\pi_1(X)$. Then you can do the group cohomology computation and if you have cocycles you can compute the yoneda product which wont be too bad (theoretically). The map that charles mentions that induces an iso on $H^1$ is a ring map since it comes from a map of spaces. I think that will work.
Oct 20, 2010 at 21:46 comment added Charles Rezk Maybe. My point is that your calculation only depends on the group cohomology calculation, which suggests that (however you carry it out), you should only need to worry about things related to the fundamental group. Since it is aspherical, this is certainly the case.
Oct 20, 2010 at 21:39 comment added DanT Is computing cup products in group cohomology any easier than for spaces? Actually, the 2-complex I'm working with is already aspherical, so that would do it.
Oct 20, 2010 at 21:34 history answered Charles Rezk CC BY-SA 2.5