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Oct 4, 2010 at 19:15 comment added Torsten Ekedahl If you mean examples such as you gave in your question, then yes your formula is OK. I have to admit that they go beyond what I expected but I still think that they are quite artificial. Even though the multiplicative structure of the rings involved can be arbitarily complicated, their interrelations are given by tensor products of rings and ring homomorphisms of the type $R\bigotimes R\to R$ given by multiplication. Also they are verified after having computed explicitly the cohomology rings involved.
Oct 4, 2010 at 16:59 comment added Torsten Ekedahl Sorry, I came on a little bit too strong I rather meant something like "should be close to the additive pushout only in very simple cases". What do you mean by the small diagonal?
Oct 4, 2010 at 13:16 comment added Charlie Frohman I get it, the check is to indicate that I have the option of saying its the answer. Sorry about that. I would conjecture that the example I gave above is not the only one, but that there are a couple of universal constructions that should produce families of examples where its true. Maybe, suppose $X$ has no torsion in its homology and $P$ and $Q$ are components of the small diagonal of $X^n$ then its true?
Oct 4, 2010 at 10:23 comment added Charlie Frohman Does the check mark mean that you determined that you answered my question? The remark at the end of the first paragraph " it is close to the additive pushout only in very simple cases" Seems to be headed towards an answer, but mostly the answer is magnifying the statement in your remark, which is that additive and multiplicative pushouts are in general different. It certainly works when $X$ is the cartesian product of $CP(1)$'s and $P$ and $Q$ are half dimensional components of the small diagonal. Is that the only time it works, or is are there families of examples where it works?
Oct 4, 2010 at 5:15 history answered Torsten Ekedahl CC BY-SA 2.5