Comparison theorem — Let be first quadrant spectral sequences of flat modules over a commutative ring and a morphism between them. Then any two of the following statements implies the third:
is an isomorphism for every p.
is an isomorphism for every q.
is an isomorphism for every p, q.
Illustrative example
As an illustration, we sketch the proof of Borel's theorem, which says the cohomology ring of a classifying space is a polynomial ring.[citation needed]
First of all, with G as a Lie group and with as coefficient ring, we have the Serre spectral sequence for the fibration . We have: since EG is contractible. We also have a theorem of Hopf stating that , an exterior algebra generated by finitely many homogeneous elements.
Next, we let be the spectral sequence whose second page is and whose nontrivial differentials on the r-th page are given by and the graded Leibniz rule. Let . Since the cohomology commutes with tensor products as we are working over a field, is again a spectral sequence such that . Then we let
Note, by definition, f gives the isomorphism A crucial point is that f is a "ring homomorphism"; this rests on the technical conditions that are "transgressive" (cf. Hatcher for detailed discussion on this matter.) After this technical point is taken care, we conclude: as ring by the comparison theorem; that is,