Is it true that $$ H^k(K(\mathbb{Z},n);\mathbb{Q})\cong H^k(K(\mathbb{Q},n);\mathbb{Q}) $$ for all $k$?
Asked
Active
Viewed 227 times
2
-
1Yes, more precisely the map $K(\mathbb Z, n) \to K(\mathbb Q, n)$ induces a rational equivalence, i.e., an isomorphism on cohomology. – Justin Young Oct 28 '13 at 09:34